How can I improve my explanation that the ratio $\theta=\frac{s}{r}$ that defines the radian measure holds for all circles? I'm trying to demonstrate why the ratio $\theta=\frac{s}{r}$ (where $s$ is an arc measuring some $s$-units in length and $r$ is the radius of the circle) which defines the radian measure holds for all circles.  I'm attempting to show this by using two circles, one of radius $1$ and another radius $r$.
I would like to know if there is anything pertinent that I can add, or that is wrong with this explanation, or perhaps you could put forward a better explanation of your own. I don't think it's rigorous, so go easy. This is what I have so far:

Consider the Unit Circle centered at the origin $O$. Let $P$ be the point on the terminal side of the central angle $\theta$ that lies on the Unit Circle. Let the point $A(1,0)$ be the point on the Unit Circle that lies on the positive $x$-axis.
Consider another circle with radius $r$ (with $r>1$) that is also centered at the origin $O$. Let $Q$ be the point on the terminal side of the central angle $\theta$ that lies on the circle with radius $r$. Let the point $B(r,0)$ be the point on the circle with radius $r$ that lies on the positive $x$-axis.
Here's a diagram:

Given that arcs $QB$ and $PA$ subtend the same central angle $\theta$ they are similar, and as the circle segments are congruent this must mean the radiuses are also similar, so we can write:
$\frac{\text{arc PA}}{\text{arc QB}}=\frac{OP}{OQ}$
As $OP=1$ and $OQ=r$ we have:
$\frac{\text{arc PA}}{\text{arc QB}}=\frac{1}{r}$
Both arc $PA$ and $QB$ are segments of the circumference, and can be written as, $PA=2\pi r k$ where $0<k<1$ and represents the fraction of the circumference that the arc constitutes. As $r=1$ in the case of arc $PA$ we have $PA=2\pi (1) k=2\pi k $. Similarly arc $QB=2\pi r k$, so we now have:
$\frac{2\pi k}{2\pi r k}=\frac{1}{r}$
$\frac{2\pi k}{1}=\frac{2\pi r k}{r}$
This equals the ratio which we have defined to be the radian measure: $\theta=\frac{s}{r}=\frac{2\pi r k}{r}$. Therefore we can conclude.

I lost my way towards the end as I don't know how to conclude this; help would be appreciated. Thanks.

My additions in bold.

Let $\theta$ be any angle. Define the radian measure $\rho(\theta,r)$ to be the ratio $s/r$, where $s$ is the length of the arc which belongs to a circle of radius $r$ and which is bounded by the $x$-axis and a ray through the origin which makes an angle $\theta$ with the $x$-axis (measured in the usual counterclockwise way). We wish to prove that $\rho$ is in fact a constant function of $r$ for any fixed $\theta$ and has therefore the same value for all circles.

Consider the unit circle centered at the origin $O$. Let $P$ be the point on the terminal side of the central angle $\theta$ that lies on the unit circle. Let the point $A(1,0)$ be the point on the unit circle that lies on the positive $x$-axis.

Consider another circle with radius $r$ (with $r\mathbf{\neq} 1$) [your assumption $r>1$ is unnecessary] that is also centered at the origin O. Let $Q$ be the point on the terminal side of the central angle $\theta$ that lies on the circle with radius $r$. Let the point $B(r,0)$ be the point on the circle with radius r that lies on the positive x-axis.

[The rest of your argument.]

This equals the ratio which we have defined to be the radian measure: $\rho(\theta,r)=s/r=2πrk/r=2\pi k=\rho(\theta,1)$. Since $r$ was chosen arbitrarily and $\rho(\theta,1)$ is independent of $r$, we conclude that $\rho(\theta,r)$ is the same for all circles and that we may write $\rho(\theta)$.

Towards the "anything pertinent that I can add" is the use of $\tau = 2\pi$. This reduces the unnecessary repeats of $2\pi$ which makes everything cleaner.


When does equality in Markov's inequality occur? Markov's inequality states that given any nonnegative random variable and $a>0$ then we have:
$$P(X \geq a) \leq \frac{E(X)}{a}$$
At which $a$ is equality supposed to hold?

First assume $a$ is finite, otherwise the result is trivial. Define the random variable $Y_{a}=X-a1_{\{X \geq a\}}$, where $$1_{\{X \geq a\}}=
\begin{cases}
1, \:\text{if}\; X\geqslant a\\
0, \:\text{if}\; X< a
\end{cases}.$$

Observe that $Y_a$ is non-negative. Taking the expectation yields $$E(Y_a)=E(X)-aP(X \geq a).$$

Hence Markov's inequality holds with equality if and only if $E(Y_a)=0$. Since $Y_a$ is non-negative, this is equivalent to $P(Y_{a}=0)=1$. Note that $Y_{a}=0$ if and only if $X=0$ or $X=a$.

Therefore, Markov's inequality holds with equality if and only if $P(X\in\{0,a\})=1$.

You get equality when $X$ is a random variable that takes on value $a$ with probability $1$. For any non-constant non-negative random variable, you will always have $P(X \geq a) < E(X)/a$ because the quantity $a P(X \geq a)$ is less than or equal to a strict partial contribution to the expectation $E(X)$.


How to solve $5000 n \log(n) \leq 2^{n/2}$ I'm trying to solve the following problem:

What is the smallest value of n so that an algorithm with a runtime of $5000 n \log(n)$ runs faster than an algorithm with a runtime of $2^{2/n}$ on the same computer?

So I figured it was just a matter of solving the equation $5000 n \log(n) \leq 2^{n/2}$ but I have no idea how I should do it, I've tried for a while using algebra but that's leading me nowhere.

Take log of both sides (I'm assuming that "log" here means $\log2$) to get
$$
\log(5000) + \log(n) + \log \log n \le \frac{n}{2}
$$
Since $\log n$ and $\log \log n$ are both small compared to $n$, you can say that $n$ is somewhere around $2\log(5000) \approx 25$. So start with $n = 20$ and $n = 52$, say; for one of these the left side should be smaller than the right; for the other, it should be greater. Apply bisection at most 6 times to get your answer.

some modification you have: $ ln(5000)=8.52<n/2*(0.7)-ln(ln(n)*n)$
Then:
$ 8.52<0.35*n+ ln(n*ln(n))$
For n= 30 we have 8.52>7.some
For n = 40 we have 8.52<9.some
We conclude that number is bwthween 30 and 40.
Metohd od half use 35: we have:  8.52>7.5
=> number is in [35, 40]

Use n=38 => 8.52>8.37 n is in [39, 40]
Us n=39 ->8.52<8.68 => n=39 is final soultuon(min n with that property).
Maybe it would be some mistakes, but this method you may use.


Show that $3\mid xy$ in $x^2 + y^2 = z^2$ Part 1:

If $x$, $y$, and $z$ are positive integers for which $gcd(x, y, z) = 1$ and $x^2 + y^2 = z^2$, show that $3\mid xy$

Part 2:

Now again if $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) = 1$ and $x^2 + y^2 = z^2$, can a similar result to the one in (part 1) be said modulo 5? Not that $5|xy$, but one of $x,y,z$?

Note that $3$ is a prime, so $3 \mid xy \iff 3\mid x$ or $3\mid y$. Suppose to the contrary that $3 \nmid x$, and $3\nmid y$, then:$x = 1,2$ mod $3$ and $y = 1,2$ mod $3$. Then:

$x^2+y^2 = 2$ mod $3$, and $z^2 = 0,1$ mod $3$. Contradiction, proving the claim.

Primitive Pythagorean triplets can be characterized precisely.

  If $x^2+y^2=z^2$, where $x,y,z \in \mathbb{Z}$, and $\gcd(x,y,z) =1$, then $x=m^2-n^2$, $y=2mn$ and $z=m^2+n^2$, where $\gcd(m,n) = 1$ and are of opposite parity.

From the above, we can immediately conclude that $3$ divides $xy$. Similarly, we can conclude that $4$ divides $xy$ and $5$ divides $xyz$. Look here for more properties on Pythagorean triplets.


D - Part in PID doesn't impact steady state error? Why does the D-part in a PID controller not do any impact on the steady state error. I mean if tries to resist changes, should it not then make it stay at wanted configuration?

A PID controller generates a control input

$$u(t) = u_\text{P}(t)+u_\text{I}(t)+u_\text{D}(t).$$

The contributions of each of the terms can be written explicitly by using the error $e$.

$$=k_\text{P}e(t)+k_\text{I}\int_{\tau=t_0}^{t}e(\tau)~d\tau + k_\text{D}\dfrac{d}{dt}e(t).$$

Now, we assume that our system is in a state with constant error $e(t) = \text{const}$. What will happen to the different components? It is clear that the derivative term $k_\text{D}\dot{e}(t)$ will vanish. Only the proportional and the integral terms will provide a non-zero term for the total control input. The proportional term will not increase its action even if we have a remaining constant error. The integral term will accumulate errors and increase its action. Hence, only the integral term will increase its action for a steady state error. Note, that the integral term might lead to a phenomenon called integrator windup if the control input is saturated.

If the steady state error is nonzero but settles down then the derivative is zero hence no correction. That's why there is always, though small, integral action added in the controller. Such that when it builds up, the system is forced to return to the designated reference.


What's the fuss about performing operations from left to right? When I learned about BODMAS / PEMDAS rules in mathematics. I was told to follow certain rules.

For example if I have this expression: $3 − 2 + 4 − 1$ and to solve it I was told to always go from left to right because addition and subtraction in this case have the same rank or weight.

Or, some rule that goes like "collect positive terms first and then negative term ..."

But I found that it doesn't really matter in which order I perform or solve it as I would get a same answer.

$3 + 4 = 7 - 2 = 5 - 1 = 4$
$3 - 1 = 2 - 2 = 0 + 4 = 4$
$3 - 2 = 1 - 1 = 0 + 4 = 4$
$-2 + 4 = 2 + 3 = 5 - 1 = 4$
$-1 + 4 = 3 + 3 = 6 - 2 = 4$
$4 - 2 = 2 + 3 = 5 - 1 = 4$

Am I right or wrong?

You may rearrange the terms how you wish before actually adding and subtracting without changing the value (assuming you do it the right way). That's not what "go from left to right" is about.

Subtraction is not an associative operation, and it doesn't associate with addition either. That means that given an expression with subtraction (and possibly addition), inserting parentheses in different ways may change the outcome. For instance, in your example, we can do
$$
((3-2)+4)-1=4
$$
Or we could do
$$
3-(2+(4-1))=-1
$$
or any of a number of different orders.

Going left-to-right means that the top one is the correct one. You evaluate the leftmost binary operation (which is $3-2=1$) first, then the next one (which becomes $1+4=5$) and finally the rightmost one ($5-1=4$).

So left-to-right isn't about the terms, it's about the operations. You even follow this yourself since you in your calculations always put the next operation to the right of the previous one (abusing the $=$ sign in ways I don't really approve of, but that's another story).

You seem to have internalised the fact that in an expression like $3-2+4$, the $-$ in some sense "belongs" to the $2$, and only the $2$. That's not a bad thing in general, but in this specific case I think it made you confused about what the whole left-to-right thing is about. I personally prefer the "belongs to" interpretation over the "go from left to right" imperative, but that's not usually what's taught in schools, so it's a realisation most of us have to come to on our own.

You are correct. For any numbers $a,b$, we have $a+b=b+a$ (this works whether or not $a$ or $b$ is positive or negative), so you can permute the terms of a sum any way you want and you will still get the same result.


Abelian group is not cyclic Suppose that $n\ge3$ where n is odd and $D_n$ is a dihedral group of order $2n$. I need to prove or disprove that if $H$ is an abelian finite subgroup of $D_n$, then $H$ is cyclic. I cannot understand what does $H$ being a finite subgroup of dihedral group means. Like, if $D_3=[e,r,r^2,s,sr,sr^2]$, then how can I construct an abelian subgroup of this so that it is not cyclic? Can someone elaborate on this please?

Thanks in advance!

Let us recall some definitions:

One says that $H$ is a subgroup of $D_n$ (or any group) when $H$ is a subset of $D_n$ this is itself a group with respect to the inherited operations. More explicitly it is a subset that contains the identity element and is closed under products and taking inverses. That it is finite just means that the set $H$ is finite; in  this case this is automatic as every subset of $D_n$ is finite.
A group (or subgroup) is called abelian when the group law satisfies the claw of commutativity.

Thus what you are supposed to decide is: if $H$ is a (finite) subset of $D_n$ that contains the identity and is closed under multiplication and taking inverses (that is, a subgroup) and such that $gh = hg$ for all $h,g \in  H$ (so that it is an abelain subgroup), is it then true that $H$ is cyclic, that is it is can be generated by a single element.

It is true that to disprove this it would suffice to exhibit an example of an abelian subgroup of $D_n$ that is not cyclic.


What is the probability that if five hats are distributed among five boxes that box $B_1$ has hat $H_1$ or hat $H_2$ but not both? There are $5$ different boxes $B_1, B_2, B_3, B_4, B_5$ and $5$ different hats $H_1, H_2, H_3 H_4, H_5$. The hats are to be distributed among the different boxes. Each box can accommodate all the hats.

A) What is the probability that $B_1$ has either $H_1$ or $H_2$ but not both?

B) If $B_1$ and $B_2$ have the hats $H_1$ and $H_2$ among themselves, in how many ways can you arrange the hats among the $5$ boxes?

Answer of A part is $2 \cdot 5^4-5^3/5^5=9/25$

Answer of B part is $4 \cdot 5^3$

Can anyone please explain this to me?

Each of the different ways of distributing the hats can be described as a 5-tuple, $(x_1,x_2,x_3,x_4,x_5)$, where $x_i$ represents which box hat $i$ was placed in.

For example, $(1,1,2,1,4)$ corresponds to box1 containing hats 1 2 and 4, box2 containing hat3, and box4 containing hat5.

From rule of product we know that there are $5^5$ possible such 5-tuples and therefore $5^5$ different ways to distribute the hats.

The problem statement does not explicitly state how we randomly distribute the hats, but a common assumption for problems like these is that each hat is distributed uniformly and independently at random amongst the boxes, implying that each of those $5^5$ possible ways of distributing the hats are equally likely to occur.  This is an important observation because it allows us to use counting techniques to continue the problem.

For the first question, finding the probability that the first box has hat1 or hat2 but not both (while also possibly having additional hats beyond those), we count how many 5-tuples correspond to such an arrangement.  Again, approach via rule of product breaking it down into steps as follows:

Pick whether it is hat1 or hat2 that is in box1  (2 options)
Whichever wasn't picked in the first step, pick which box it does go to remembering that it cannot be box1 (4 options)
Pick which box hat3 is in (5 options)
Pick which box hat4 is in (5 options)
Pick which box hat5 is in (5 options)

Applying the rule of product, we have then $2\cdot 4\cdot 5\cdot 5\cdot 5 = 8\cdot 5^3$ such arrangements of hats satisfying the phrase "exactly one of (but not both of) the first two hats are in box1."

Using what we know about finite equiprobable sample spaces then, the probability of this happening is the ratio of the number of favorable outcomes divided by the total number of outcomes.  I.e. a probability of $\frac{2\cdot 4\cdot 5^3}{5^5}=\frac{8}{25}$.

This contradicts your answer in the original post which might have caused some of your confusion.  An answer of $\frac{9}{25}$ actually corresponds to the probability of at least one of the first two hats being in the first box, including the possibility of both.  To have calculated this, we could approach similarly to above, having first broken into cases based on whether exactly one or both of the first two hats were in the first box, giving a total number of valid sequences being $2\cdot 4\cdot 5^3+1\cdot 1\cdot 5^3$ and the probability then as $\frac{2\cdot 4\cdot 5^3+1\cdot 1\cdot 5^3}{5^5}=\frac{2\cdot 4+1}{5^2}=\frac{9}{25}$.  Alternatively, you could have calculated this using inclusion exclusion arriving at the same answer.

The second part of the problem is very similar to the first.  The steps for rule of product being:

Pick location of hat 1
Pick location of hat 2
Pick location of hat 3
Pick location of hat 4
Pick location of hat 5

   With 2 options for each of the first two steps and 5 options for each of the remaining steps, that gives the total number of valid arrangements as $2\cdot 2\cdot 5\cdot 5\cdot 5=4\cdot 5^3$ valid sequences.  Taking the ratio then, this gives the probability as $\frac{4\cdot 5^3}{5^5}=\frac{4}{25}$

My reasoning seems too simplistic and is probably flawed but does give the solution to the first problem. The chance of hat1 not being in box1 is $\frac 4 5$ and likewise for hat2. $\frac 4 5 \times \frac 4 5 = \frac {16} {25}$. Thus the chance of one or both being in box1 is $\frac {25} {25} - \frac {16} {25} = \frac 9 {25}$. The chance of both being in box1 is $ \frac 1 5 \times \frac 1 5 = \frac 1 {25}$. So for only one of them we have $\frac 9 {25} - \frac 1 {25} = \frac 8 {25}$.


Riemann-integrability of $f(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}$ on $[0,1]$ Determine whether  $\displaystyle f(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}$
is Riemann-integrable on $\displaystyle [0,1]$

Attempt: I can clearly see that $\displaystyle f$ is derivative of $\displaystyle g(x)=x^2\sin\frac{1}{x}$

$\displaystyle g(1)-\lim_{x \to 0} g(x)=\sin(1)$

Is it correct? Given function $f(x)$ is discontinuous only at $x=0$ in $[0,1]$. So, it should be Riemann-integrable.

clearly $f(x)$ is bounded in $[0, 1]$ and continuous also except for $x = 0$. Hence it is Riemann integrable in $[0, 1]$. and you don't need to worry about the anti-derivative of $f(x)$.

How I think about this: suppose I divide $[0, 1]$ into $N$ equal partitions indexed from $0$ to $N - 1$.  Then partition $i$ goes from $i \over N$ to $i + 1 \over N$.

For reasonable resolution on $\cos({1\over x})$ we need the fractional change in $x$ to be less than some threshold $\delta$, so $(i + 1) < i (1 + \delta)$ or $i > {1 \over \delta}$.  So assume the portion of the integral for $i > {1 \over \delta}$ is good and the portion $i < {1 \over \delta}$ is "bad".

$i$ goes from $0$ to $N - 1$ so the portion for which $i < {1 \over \delta}$ is approximately ${1 \over \delta N}$ .  $\delta$ is finitely small, $N$ goes to the limit of $+\infty$, so the fraction of the integral which is "bad" goes to $0^+$.  Since the value is bounded the contribution of the "bad" fraction to the integral thus has limit $0$.  That means as the number of samples goes to $\infty$ the integral will converge on the correct value.


Find the circle about the origin containing all roots of $X^7+X^2+1=0$ We need to  find the circle about the origin containing  all roots of $X^7+X^2+1=0$,
I am not getting any hint where to start from. It is a problem of Munkres General topology section $56$, chapter Fundamental Group.

Since the question is tagged "complex analysis" I thing a good approach could be to use Rouche's Theorem, so

Hints:

(1) Put $\,f(z):=z^2+1\;,\;\;g(z):=z^7\,$

(2) Check that on $\,C:=\left\{z\in\Bbb C\;;\;|z|=\frac{5}{4}\,\right\}\,$ , for example,  we have $\,|f(z)|<|g(z)|\,$

(3) How many roots does $\,g(z)\,$ have within $\,C\,$ ?...

Hint: From the equation, you know that $X$ must be negative and less than negative one. You need to determine the relationship between $X^7$ and $X^2$ that allows their summation ($X^7$ will be negative) to be equal to $-1$.


Most accurate finite difference result for first derivative I got a problem in my assignment: obtain the most accurate finite difference results possible for the first derivative of f (x) = exp(cos(x)) at x=1, h = 0.5, 0.25, 0.125,...2^{16}. I have to do this in MATLAB. Can any one explain how to find such accurate result, so that I can implement it in MATLAB

You can compute the exact derivative to compare against.

The central difference formula
$$
f'(x)=\frac{f(x+h)-f(x-h)}{2h}+O(h^2)
$$
has two error contributions, the discretization error $\sim f''(x) h^2$ and the evaluation error $\sim |f(x)|\mu/h$ where $\mu$ is the floating point machine constant.

The overall error is minimal where both contributions are about equal, which happens at $h\sim\sqrt[3]\mu\approx 2^{-17}\approx 8·10^{-6}$, the last for double precision floats, with an error of about $2^{-34}\approx 6.4·10^{-11}$

If you use then Richardson extrapolation or other methods to get a higher order formula, the next best result would be $O(h^4)$. By the same estimates, the best result is obtained for $h\sim\sqrt[5´]\mu\approx 2^{-10}\approx 10^{-3}$ with error in the region of $2^{-40}\approx 10^{-12}$.

This trend continues, the higher the order, the larger the intersection between the error contributions. However, the gains in using ever higher order formulas quickly reduce to a trickle.

See here and here. You can get arbitrarily accurate results by using arbitrarily high order approximations, so your question is somewhat underspecified.


Can we find positive integers $a$ and $k \geq 2$ with $2^n - 1 = a^k$? I would appreciate if somebody could help me with the following problem:

  For a given positive integer $n$, can we find positive integers $a$ and $k$ ($k\geq 2$) such that $2^n-1=a^k$?

The answer would be no, I  guess.

Please give me some ideas!

No. This is a special case of Mihăilescu's theorem / Catalan's Conjecture, which says that the only consecutive powers are 8 and 9.

Proof by counter example.

Let's pick n=2.
Then $2^2-1=3$.
I'm afraid that no integer $k \ge 2$ will fit.


Find the functions satisfying $ \frac { f ( a b ) } { a b } = \frac { f ( a + b ) } { a + b } \cdot \frac { f ( a - b ) } { a - b } $ Find all the functions $ f : \mathbb R \to\mathbb R $ such that
$$ \frac { f ( a b ) } { a b } = \frac { f ( a + b ) } { a + b } \cdot \frac { f ( a - b ) } { a - b } \text . $$

I think $ f ( x ) = x $, but I don't know how to get it.

First of all, note that you have $ f : \mathbb R \to \mathbb R $, and in particular $ 0 $ is in the domain of $ f $. But the functional equation
$$ \frac { f ( x y ) } { x y } = \frac { f ( x + y ) } { x + y } \cdot \frac { f ( x - y ) } { x - y } \tag 0 \label 0 $$
can only hold when $ x , y \ne 0 $ and $ x \ne \pm y $. That means \eqref{0} doesn't tell you anything about $ f ( 0 ) $, and given any solution $ f $ of \eqref{0}, you can change the value of $ f ( 0 ) $, and still get a solution; i.e. any function $ \tilde f : \mathbb R \to \mathbb R $ with $ \tilde f ( x ) = f ( x ) $ for all $ x \ne 0 $ is a solution. Thus the better question is to ask for all $ f : \mathbb R \setminus \{ 0 \} \to \mathbb R $ satisfying \eqref{0} for all $ x , y \in \mathbb R \setminus \{ 0 \} $ with $ x \ne \pm y $. Then, defining $ g : \mathbb R \setminus \{ 0 \} \to \mathbb R $ with $ g ( x ) = \frac { f ( x ) } x $, \eqref{0} gives
$$ g ( x y ) = g ( x + y ) g ( x - y ) \tag 1 \label 1 $$
for all nonzero $ x $ and $ y $ with $ x \ne \pm y $. Conversely, any $ g : \mathbb R \setminus \{ 0 \} \to \mathbb R $ satisfying \eqref{1} gives rise to an $ f : \mathbb R \setminus \{ 0 \} \to \mathbb R $ defined with $ f ( x ) = x g ( x ) $, which satisfies \eqref{0}. So we will try to solve this equivalent problem. It's straightforward to check that the constant functions $ g ( x ) = 0 $ and $ g ( x ) = 1 $ satisfy \eqref{1} (equivalently, the constant zero function and the identity function both satisfy \eqref{0}). We will show that these are the only solutions.
Setting $ y = 1 $ and then $ y = - 1 $ in \eqref{1} we get
$$ g ( x ) = g ( x + 1 ) g ( x - 1 ) = g ( x - 1 ) g ( x + 1 ) = g ( - x ) $$
for all nonzero $ x $ with $ x \ne \pm 1 $. Similarly, substituting $ \frac x 2 $ for $ x $ in \eqref{1} and once setting $ y = 2 $ and once $ y = - 2 $, we get
$$ g ( x ) = g \left( \frac x 2 + 2 \right) g \left( \frac x 2 - 2 \right) = g \left( \frac x 2 - 2 \right) g \left( \frac x 2 + 2 \right) = g ( - x ) $$
for all nonzero $ x $ with $ x \ne \pm 4 $. Thus we have
$$ g ( - x ) = g ( x ) \tag 2 \label 2 $$
for all nonzero $ x $.
Now, suppose there is a nonzero $ a $ with $ g ( a ) = 0 $. Without loss of generality, we can assume that $ a > 0 $, because if $ a < 0 $, by \eqref{2} we have $ g ( - a ) = 0 $ while at the same time $ - a > 0 $. Letting $ x = y + a $ in \eqref{1} we get $ g \big( y ( y + a ) \big) = 0 $ for all positive $ y $ (note that since $ a > 0 $, thus $ 0 \ne y + a \ne \pm y $). This means that we have $ g ( x ) = 0 $ for all positive $ x $, as when $ x > 0 $ there is a positive $ y $, namely $ y = \frac { - a + \sqrt { a ^ 2 + 4 x } } 2 $, such that $ y ( y + a ) = x $. Using \eqref{2}, this can be generalized for negative $ x $ as well, which means that $ g $ is the constant zero function, one of the mentioned solutions.
Finally, consider the other case where for all nonzero $ x $, $ g ( x ) $ is not equal to $ 0 $. Suppose that $ y $ is negative. Then we have $ 0 \ne \frac y { y - 1 } \ne \pm y $, and we can let $ x = \frac y { y - 1 } $ in \eqref{1}. In that case, we have $ x y = x + y = \frac { y ^ 2 } { y - 1 } $, and by the fact $ g \left( \frac { y ^ 2 } { y - 1 } \right) $ is not equal to $ 0 $, we find out that $ g ( x - y ) = g \left( \frac { y ^ 2 - 2 y } { y - 1 } \right) = 1 $. This means that we have $ g ( x ) = 1 $ for all negative $ x $, since for $ x < 0 $ there is a negative $ y $, namely $ y = \frac { x + 2 - \sqrt { x ^ 2 + 4 } } 2 $, such that $ \frac { y ^ 2 - 2 y } { y - 1 } = x $. By \eqref{2}, that is also true for positive $ x $, and $ g $ is the constant one function, the other mentioned solution.

just hint

For $b=1,$

put
$$g (n)=\ln (\frac {f (n)}{n}) $$

then

$$g (n+1)-g (n)+g (n-1)=0$$

the characteristic equation is
$$x^2-x+1=0$$
and

$$g (n)=A\cos (n\frac {\pi}{6})+B\sin (n\frac {\pi}{6})$$


$\forall n\in\mathbb{N}$, $a_n\not=0$, $\lim_{n\to\infty} a_n=0$. Find $\lim_{n\to\infty} \frac{\sqrt{3+a_n}-\sqrt{3}}{a_n}$ I already multiplied by the conjugate so I get
$$\lim_{n\to\infty} \frac{1}{\sqrt{3+a_n}+\sqrt{3}}$$
but I dont know how to prove that $\lim_{n\to\infty} \sqrt{3+a_n}=\sqrt{3}$.
Also I already tried with the definition with the limit $\frac{1}{2\sqrt{3}}$, but is more hard. Can anyone help me? Also important to say that $a_n$ is a sequence.

We know that for positive values, $\sqrt{ \cdot }$ is a continuous function, meaning that the limit of the function is the function of the limit.  So $\lim_{n \rightarrow \infty} \sqrt{3+a_n}=\sqrt{\lim_{n \rightarrow \infty}(3+a_n)}$.  Find $\lim_{n \rightarrow \infty} 3+a_n$ and you should be good!

$$\begin {align} \lim_{n\to \infty}\sqrt {3+a_n}=\sqrt {\lim_{n\to \infty}\left(3+a_n\right)}=\sqrt {\lim_{n\to \infty }3+\lim_{n\to \infty} a_n}=\sqrt {3+0}=\sqrt 3\end{align} $$


Determinant of a particular $3\times 3$ block matrix Let $m\geq 0,n\geq 1$ be fixed integers. I want to compute the determinant of the $(2n+m)\times (2n+m)$ matrix given by
$$a_{ij}=\left\{\begin{array}{cl}n+m & \text{if }i=j\leq 2n,\\
2n& \text{if }i=j> 2n,\\
0 & \text{if $i\neq j$ and $i,j\leq n$, $n< i, j\leq2n$ or $2n<i,j$,}\\
-1 &\text{otherwise.}\end{array}\right.$$
So, it's the sum of a matrix with diagonal blocks $0$ and every other block $-1$, and a diagonal matrix with the first $2n$ elements being $n+m$ and the rest $2n$.

Or just:
$$\begin{bmatrix}D_n({n+m}) & -1 & -1\\
-1 & D_{n}(n+m) & -1\\
-1 & -1 & D_m({2n})
\end{bmatrix}$$

where $D_r(s)$ denotes the diagonal $r\times r$ matrix with diagonal $s$.

I'd like to use something like this, but here the blocks have different sizes and there are some rectangular ones, too.

For the context, I was trying to compute the number of spanning trees of the tripartite graph $K_{n,n,m}$ using the Matrix-Tree Theorem and ended up with that.

One detail is that in the problem I have to remove a row and a column corresponding to one vertex, which I choose to be the last one, but the resulting matrix is still with the same form. Also, for the case of $m=0$ ($m=1$ in the problem) this can be easily solved by this (or any) method since there are only 4 blocks, all of which have the same size, commute and are very simple.

If you're using the matrix tree theorem, then you've forgotten to delete one row and one column; the determinant of your current matrix is $0$, because the all-ones vector is in the null space.

But in general, we can compute the determinant of an $(m+n+p)\times (m+n+p)$ matrix of the form
$$
  \begin{bmatrix}aI_m & -1 & -1 \\ -1 & bI_n & -1 \\ -1 & -1 & cI_p\end{bmatrix}
$$
by looking at the eigenvalues.

First of all, $a$, $b$, and $c$ are each eigenvalues with multiplicity $m-1,n-1,p-1$ respectively. (Or, if $a=b$ as in this case, we add the multiplicities.) To see this, just subtract the appropriate multiple of the identity matrix and look at the rank.

The remaining three eigenvalues come from the block structure. The block matrix acts on a vector of the form $(x,x,\dots,x, y,y,\dots, y,z,z,\dots,z)$ in exactly the same way that the $3\times 3$ matrix
$$
   \begin{bmatrix}
       a & -n & - p \\ -m & b & -p \\ -m & -n & c
   \end{bmatrix}
$$
acts on the vector $(x,y,z)$. So just compute the determinant of this $3\times 3$ matrix, multiply by $a^{m-1}b^{n-1}c^{p-1}$, and you get the determinant of the original block matrix.

Your matrix is Laplacian. In particular, all row/column sums of your $A$ are zero. Hence $\det(A)=0$.


Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$ Compute

$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$

Put $x = \tan\theta$, then your integral transforms to $$I= \int_{0}^{\pi/4} \log(1+\tan\theta) \ d\theta. \tag{1}$$

Now using the property that $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx,$$ we have $$I = \int_{0}^{\pi/4} \log\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} \log\biggl(\frac{2}{1+\tan\theta} \biggr) \ d\theta.\tag{2}$$

Adding $(1)$ and $(2)$ we get $$2I = \int_{0}^{\pi/4} \log(2) \ d\theta\Rightarrow I= \log(2) \cdot \frac{\pi}{8}.$$

You can use the residue theory to calculate
$$J=\int_{-\infty}^\infty \frac{\ln|1+x|}{(1+x^2)} \, dx $$
by putting
$$f(z)=\frac{\ln(1+z)}{(1+z^2)}$$ and we obtain
$$J = 2\pi i\frac{\ln(1+i)}{(i+i)}=\pi\ln\left(\sqrt2e^{\frac14\pi i}\right)=\pi\left(\frac12\ln2+\frac14\pi i\right)=\frac{\pi}{2}\ln 2+\frac{\pi^2i}4$$

$$J=\int_{-\infty}^{-1}\frac{\ln(1+x)}{1+x^2} \, dx
+\int_{-1}^0 \frac{\ln(1+x)}{1+x^2} \, dx
+\int_0^1\frac{\ln(1+x)}{1+x^2} \, dx
+\int_1^\infty \frac{\ln(1+x)}{1+x^2} \, dx$$

Then by putting $(t-1)e^{\pi i}=1+x$ in the first integral we obtain:
$$\int_{-\infty}^{-1}\frac{\ln(1+x)}{1+x^2}dx=\int_1^{\infty}\frac{\pi i+\ln(t-1)}{t^2+1}dt=\frac{\pi^2i}4+\int_1^{\infty}\frac{\ln(t-1)}{t^2+1}dt$$

Then by putting $t=-x$ in the second integral we obtain:
$$\int_0^1\frac{\ln(1-x^2)}{1+x^2}dx+\int_1^{\infty}\frac{\ln(x^2-1)}{1+x^2}dx=\frac{\pi}2\ln2$$

then by putting $t=1/x$ in the second integral we obtain:
$$\int_0^1\frac{\ln(1-x^2)}{1+x^2}dx-\int_0^1\frac{\ln x}{1+x^2}dx=\frac{\pi}4\ln2$$

and we put $x=(1-t)/(1+t)$ in the first integral
we obtain after calculus :
$$\int_0^1 \frac{\ln \, (1+x)}{1+x^2}= \frac{\pi}{8} \ln 2$$


Can't argue with success? Looking for "bad math" that "gets away with it" I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare").

One example would be "cancelling" the 6's in

$$\frac{64}{16}.$$

Another one would be something like

$$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;.$$

Yet another one would be

$$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;.$$

Note that I am specifically not interested in mathematical fallacies (aka spurious proofs).  Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for all cases where one arrives at valid ends by (shockingly) wrong means.

Edit: fixed typo in last example.

I was quite amused when a student produced the following when cancelling a fraction:

$$\frac{x^2-y^2}{x-y}$$

He began by "cancelling" the $x$ and the $y$ on top and bottom, to get:

$$\frac{x-y}{-}$$

and then concluded that "two negatives make a positive", so the final answer has to be $x+y$.

Well, using $\frac{dy}{dx}$ as a ratio is also one such example. It carries no meaning at all to use  $\frac{dy}{dx}$ as ratio but it works each time to get you to the right result instead of appreciating the limit process behind it.
I think it is as much abuse of notation as is eliminating 6's from a fraction, it is just that in this case the notation is designed such that each time you abuse it, you will get the right answer but you won't know actually why if you just rely on the notation. For example:
$\frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}$ due to chain rule and not due to treating $\frac{dy}{dx}$ as a fraction just like $\frac{x^2-y^2}{x-y}=x+y$ by algebra and not by

$$\frac{x^2-y^2}{x-y}$$

"cancelling" the $x$ and the $y$ on top and bottom, to get:

$$\frac{x-y}{-}$$

and then conclude that "two negatives make a positive", so the final answer is be $x+y$.

Now if you see a student prove $\frac{5^2-3^2}{5-3}=8$ by the wrong method, will he be correct as this method of proof will hold for all $x,y \in R$ until $x \neq y$.

Also in three dimensions you have stuff like $ \frac{\partial x}{\partial y}\cdot\frac{\partial y}{\partial z}\cdot\frac{\partial z}{\partial x}=-1. $

Also, $dy$ isn't even defined , let alone be manipulated like a real number.


Writing a polynomial as a product of irreducible polynomials over another field One root of the polynomial $f(x) = 2x^5 + x^4 + 44x^3 + 22x^2 - 150x - 75$ over $\Bbb{C}$ is
$ -5i$.

How can I write this as a product of irreducible polynomials over $\Bbb{R}$ and $\Bbb{C}$?

I have no clue how to go about this. I used grouping to factor out products to write it as a product of irreducible polynomials over $\Bbb{Q}$ however I don't know if that is correct either.

Let $f(x) = 2x^5 + x^4 + 44x^3 + 22x^2 - 150x - 75$.

Since the coefficients of $f$ are real and $-5i$ is a root of $f$, it follows that $5i$ is also a root of $f$.

Since $5i$ and $-5i$ are roots of $f$, it follows that $x-5i$ and $x+5i$ are factors of $f$, hence $x^2 + 25$ is a factor of $f$.

By long division, the other factor comes out as
$$2x^3+x^2-6x-3$$
The rational root test yields the root $x = -\frac{1}{2}$, which yields the factor $2x+1$.

By long division, the remaining factor is $x^2-3$.

Hence:

$\;\;\;{\small{\bullet}}\;$Over $\mathbb{Q}$, $f$ can be factored as
$$(x^2+25)(2x+1)(x^2-3)$$
$\;\;\;{\small{\bullet}}\;$Over $\mathbb{R}$, $f$ can be factored as
$$(x^2+25)(2x+1)(x+\sqrt{3})(x-\sqrt{3})$$
$\;\;\;{\small{\bullet}}\;$Over $\mathbb{C}$, $f$ can be factored as
$$(x-5i)(x+5i)(2x+1)(x+\sqrt{3})(x-\sqrt{3})$$

by conjugate root theorem, 5i is also a root since 5i is conjugate of -5i.

Then we can have (x-5i) and (x+5i) as factors of the polynomial.

(x-5i)(x+5i)is (x^2+25)

Finally you could do long division using this.

answer should be (x-5i)(x+5i)(2x+1)(x^2-3)

Hope this helps ;)


Addition composition problem? I'd like to know how you go about solving the problem below. Does a trick exist other than evaluating substituted values until a solution is found?

Furthermore, is there a generic name for such problems? I'll happily change the title of the question once I know if a name exists for such a class of problems.

  A car dealership sells used cars for \$7,000 and new cars for \$16,000. If a total of 17 cars were sold for \$191,000, how many of the cars sold were used?

You're right, such problems do have a specific name: simultaneous equations, or more formally, these ones are a system of linear equations.

Let's think about the maths of it with a little bit of algebra. If the number of new cars the dealership sold can be labelled $n$, and the number of used cars $u$, then the statement of the problem can be written as a pair of equations:

$$
(1)
$$
$$
n + u = 17
$$

and

$$
16000n + 7000u = 191000
$$

or

$$
(2)
$$
$$
16n + 7u = 191
$$

by simply dividing both sides by $1000$.

The fact that these equations are both true and both involve the same variables ($n$ and $u$) is what gives the pair the name simultaneous.

Now, for how to go about solving them:

Solving Simultaneous Equations

The key is to realise that both equations are true. Therefore, anything that can be said using one can be assumed to be true in the other (after all, if we have some number of used cars sold, knowing an extra fact about it doesn't change that number!). We use this to our advantage to cast the problem in terms of a single variable.

Equation $(1)$ tells us:

$$
n + u = 17
$$

so, by re-arranging it:

$$
u = 17 - n
$$

Now, we know that this is true in both equations, so we can go ahead and substitute $u$ for $17 - x$ wherever it appears in equation $(2)$! From there, it's easy to reduce things down:

$$
16n + 7u = 191
$$

$$
16n + 7(17 - n) = 191
$$

$$
16n + 119 - 7n = 191
$$

$$
9n = 72
$$

$$
n = 8
$$

Et voila! $n = 8$. We take this and put it back into equation $(1)$ (or our re-arranged form) to get a value for $u$:

$$
u = 17 - n = 9
$$

And there we have it; the dealership sold $8$ new cars and $9$ used cars.

Summary

The overall approach we've got here holds for any such problem - we've got two variables and two known equations in those variables. The key is to re-arrange one to get a representation of one of the variables in terms of the other, and then plug that into the other equation.

In general, you might like to know that any set of $N$ equations involving $N$ variables can be solved by doing the exact same thing over and over again. But even one less equation to give you information, and you'll never manage it!

This kind of problem is called a system of linear equations in two variables.  In particular it's also a word problem; many similar problems exist, for example here.

The method for solving them is to make variables for the unknown quantities.  Let $U$ be the number of used cars sold, and $N$ be the number of new cars sold.  Then you get
$$
U + N = 17
$$

$$
7000U + 16000N = 191000
$$

and solving the problem amounts to solving the system of equations.


Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?

Can someone point me to a proof, or explain if it's a simple answer?

What I'm looking for is the point where it becomes understood that trigonometric functions and pi can be expressed as series.  A lot of the information I find when looking for that seems to point back to arctan.

The derivative of the arc tangent is
$$\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}.$$

From the formula for geometric series (see for example this answer for a proof) shows that
$$1+y+y^2+y^3+\cdots = \frac{1}{1-y}\qquad\text{if }|y|\lt 1.$$
Plugging in $-x^2$ for $y$, we get that
$$\begin{align*}
\frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\
&= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\
&= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots
\end{align*}$$
provided that $|-x^2| \lt 1$; that is, provided $|x|\lt 1$. All the computations below are done under this hypothesis (see comments at the end).

So we have that:
$$\frac{d}{dx}\arctan(x) = 1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\qquad\text{if }|x|\lt 1$$
Because this is a Taylor series, it can be integrated term by term. That is, up to a constant, we have:
$$\begin{align*}
\arctan(x) &= \int\left(\frac{d}{dx}\arctan (x)\right)\,dx \\
&= \int\left(1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\right)\,dx\\
&= \int\left(\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\right)\,dx\\
&= \sum_{n=0}^{\infty}\left(\int (-1)^{n}x^{2n}\,dx\right)\\
&= \sum_{n=0}^{\infty}\left((-1)^{n}\int x^{2n}\,dx\right)\\
&= \sum_{n=0}^{\infty}\left((-1)^{n}\frac{x^{2n+1}}{2n+1}\right) + C\\
&= C + \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} +\cdots\right).
\end{align*}$$
Evaluating at $x=0$ gives $0 = \arctan(0) = C$, so we get
$$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} + \cdots,\qquad\text{if }|x|\lt 1.$$
the equality you ask about.

Note however that this does not hold for all $x$: it certainly works if $|x|\lt 1$, by the general properties of Taylor series. But the arc tangent is defined for all real numbers. The series we have here is
$$\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}.$$
Using the
Ratio Test, we have that
$$\begin{align*}
\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} &= \lim_{n\to\infty}\frac{\quad\frac{|x|^{2n+3}}{2n+3}\quad}{\frac{|x|^{2n+1}}{2n+1}}\\
&= \lim_{n\to\infty}\frac{(2n+1)|x|^{2n+3}}{(2n+3)|x|^{2n+1}}\\
&= \lim_{n\to\infty}\frac{|x|^2(2n+1)}{2n+3}\\
&= |x|^2\lim_{n\to\infty}\frac{2n+1}{2n+3}\\
&= |x|^2.
\end{align*}$$
By the Ratio Test, the series converges absolutely if $|x|^2\lt 1$ (that is, if $|x|\lt 1$) and diverges if $|x|\gt 1$. At $x=1$ and $x=-1$, the series is known to converge. So the radius of convergence is $1$, and the equality is valid for $x\in [-1,1]$ only (that is, if $|x|\leq 1$; we gained two points in the process).

However, the arc tangent has a nice property, namely that
$$\arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} - \arctan(x),$$
So, given a value of $x$ with $|x|\gt 1$, you can use this identity to compute $\arctan(x)$ by computing $\arctan(\frac{1}{x})$ instead, and for this argument the series is valid.

You can show that this series converges for $r = 1$ using the alternating series test.  The sequence of the absolute values of the terms of this series is monotonically decreasing and approach zero. Hence the series converges for $r = 1$.

Also roughly speaking you need ten times as many terms in the partial sums for the approximation of pi to be one decimal place more accurate.  i.e. If if the first 100 terms have about two decimal places of accuracy, then the first 1000 will give you about three decimal places of accuracy.  Thus, the current record of 12 trillion digits would require a summation of about $10^{12000000000000}$  terms with this series.


Eigenvalues of rank 2 perturbation of the identity Let $I$ be the identity matrix, and $u, v\in\mathbb{R}^n$. What can we say about the eigenvalues of the following matrix?

$$2I+vu^{\mathsf{T}}+uv^{\mathsf{T}}$$

I'm particularly interested in the minimum eigenvalue of this matrix in terms of $u$ and $v$.

Suppose $u, v$ are linearly independent. Then
$$x = \|v\|u + \|u\|v$$
is non-zero, and
\begin{align*}
Ax &= (uv^\top + vu^\top)(\|v\|u + \|u\|v) \\
&= \|v\|uv^\top u + \|u\|uv^\top v + \|v\|vu^\top u + \|u\|vu^\top v \\
&= \|v\|(u \cdot v) u + \|u\| \|v\|^2 u + \|v\|\|u\|^2 v + \|u\|(u \cdot v) v \\
&= (\|u\| \|v\| + u \cdot v)x.
\end{align*}
Hence $\|u\| \|v\| + u \cdot v$ is an eigenvalue for $A$. Note that, since $u, v$ are linearly independent, this eigenvalue is strictly positive, making it the maximum eigenvalue.

Similarly, if $y = \|v\|u - \|u\|v \neq 0$, then $Ay = (u \cdot v - \|u\| \|v\|)y$, hence we also obtain a strictly negative eigenvalue $u \cdot v - \|u\| \|v\|$.

Since $A$ is the sum of two rank $1$ matrices, it is of rank at most $2$. Hence, the only other eigenvalue is $0$, making $u \cdot v - \|u\| \|v\|$ the minimum eigenvalue.

If $u = 0$ or $v = 0$, then $A = 0$, and the eigenvalues are all $0$. Note that this agrees with the given formula for the minimum eigenvalue.

If $v = ku \neq 0$ for some $k$, then $A$ is rank $1$ with eigenvector $u$ and corresponding non-zero eigenvalue $2k\|u\|^2$. If $k < 0$, then this is a negative number, and thus the minimum eigenvalue. Note that $2k\|u\|^2 = u \cdot v - \|u\|\|v\|$, which is to say that the formula continues to hold. If $k > 0$, then the minimum eigenvalue is $0$, which is again equal to $u \cdot v - \|u\| \|v\|$.

Therefore, in any case, the minimum eigenvalue for $A$ is $u \cdot v - \|u\| \|v\|$, and indeed the eigenvalues are, up to multiplicity, in ascending order,
$$u \cdot v - \|u\| \|v\|, \underbrace{0, \ldots, 0}_{n - 2 \text{ times}}, u \cdot v + \|u\| \|v\|.$$
The eiegenvalues for $A + 2I$ are the same as $A$, but shifted by $2$.

Rank of $vu^T$ is $1$ same is the case for $uv^T$.

Both $n\times n$, so in total $n$ eigenvalues for both.

Now out of those $n$, $n-1$ will be $0$ for both

The last eigenvalue will be $<u,v>$, that is the inner product of $ u$ and $v$. Can you see this?

Finally minimum eigenvalues will be $$ 2+ min\{0, <u,v>\}$$


$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$ Prove that

$$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$$

Is true for $x$, $y$ and $z$ being positive integers, with $x$ and $y$ being co-prime and $3z<x<y$.

There are 2 cases in this question:

Case 1 - $x$ is odd and $y$ is even (or $x$ is even and $y$ is odd), in this case $z$ is odd.

Case 2 - Both $x$ and $y$ are odd, in this case $z$ is even.

I tried to show some contradiction, we know that for the statement on the RHS to be true, the prime factors must be present in multiples of 3's, however I struggled going any further than that.

Any help would be much appreciated.

It is often a proven method to look for coprime factors.

Suppose $q \neq 3$ is a prime dividing $3z-x$. Then $q$ divides $(3z)^3$, hence $z$. This implies that $q$ divides also $x$. Given the symmetry in $x$ and $y$, $q$ cannot divide $3z-y$ as $x$ and $y$ are coprime. Also $x+y$ is not divisible by $q$ as $y$ isn't.

If $3$ divides $3z-x$ then $3 \mid x$. Hence $3 \nmid y$ as $x$ and $y$ are coprime, and therefor $3 \nmid x+y$ and $3 \nmid 3z-y$.

We conclude that $x+y$, $3z-x$ and $3z-y$ are mutually coprime.

As $(3z)^3=27z^3$ is a third power, there exist positive coprime integers $u,v,w$ such that $z=uvw$ and either:

A:
    $x+y=9u^3$
    $3z-x=v^3$
    $3z-y=w^3$

or

B:
    $x+y=u^3$
    $3z-x=9v^3$
    $3z-y=w^3$

Note that we can skip the case $3z-y=9w^3$ as the Diophantine equation is symmetric in $x$ and $y$.

A: We have $6z-x-y=6uvw-9u^3=v^3+w^3$. Hence, $6uwv=9u^3+v^3+w^3$. As in general $\sqrt[3]{x_1x_2x_3} \le \frac{x_1+x_2+x_3}{3}$, we have $6uvw<3\sqrt[3]{9}.uvw \le 9u^3+v^3+w^3$, so there is no solution.

B: Likewise we have $6uvw<3\sqrt[3]{9}.uvw \le u^3+9v^3+w^3$, so there is also no solution.

We conclude that there exists no positive integers $x,y,z$ such that $x$ and $y$ are coprime and fulfill the Diophantine equation $(3z)^3=3(x+y)(3z-x)(3z-y)$. QED

A final remark: $z$ is even. The case that $x$ is even and $y$ is uneven implies that $(3z-x)(3z-y)$ is an even number, hence $(3z)^3$ is even.

Above equation shown below:

$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$

Assume:

$\left(3z\right)^3= 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$ ---(1)

For the special case take,

$(x,y,z)=[3m,3n,3(m+n)]$

substituting value of, $(x,y,z)$ in equation (1) we get:

$m^2-mn+n^2=0$  ----(2)

It is well known that equation (2)

has no integer solution's. Hence a contradiction.

Hence our assumption is wrong & there is no

solution for "OP" question for the special case.


Between which two integers does $\sqrt{2017}$ fall? Between which two integers does $\sqrt{2017}$ fall?

Since $2017$ is a prime, there's not much I can do with it. However, $2016$ (the number before it) and $2018$ (the one after) are not, so I tried to factorise them. But that didn't work so well either, because they too are not perfect squares, so if I multiply them by a number to make them perfect squares, they're no longer close to $2017.$ How can I solve this problem?

Update:
Okay, since $40^2 = 1600$ and $50^2 = 2500$, I just tried $45$ and $44$ and they happened to be the answer - but I want to be more mathematical than that...


How many ways are there to travel from the top left corner to the bottom right corner? Came up with this question when studying combinatorics.

  How many ways are there to travel from the top left hand corner to the bottom left hand corner of a $m\times n$ block if: (1) in every moves you can only go right a block or go down a block; (2) two consecutive moves cannot be the same (which means, you can't go right twice or going down twice).

Thanks!

HINT: Clearly you must make $m+n$ steps altogether, $m$ down and $n$ to the right. Think of a route as a string of $m+n$ symbols, each of which is either $R$ (go right) or $D$ (go down).

How many such strings are there?
Once you know where the $D$’s are, you also know where the $R$’s are. How many ways are there to choose $m$ places in the string for the $D$’s?

If two consecutive moves cannot be the same, the only possible strings are alternating strings, $DRDR\ldots$ or $RDRD\ldots\;$. This puts some very stringent limitations on the possible combinations of $m$ and $n$; what are they? Once you know what combinations of $m$ and $n$ are possible, it’s not hard to count the possible paths.

Expanding on possible combinations of m and n for given conditions from previous answer, if m+n is even number, then you are bound to have m=n i.e a square block, in which case you have two alternate paths RDRD.. or DRDR.. If m+n is oddnumber, then you are bound to have m and n as two consecutive integers. In that case if m>n, then you have only one path i.e DRDR.. otherwise you have only RDRD..

Correct me if I am wrong.


Is a matrix multiplied with its transpose something special? In my math lectures, we talked about the Gram-Determinant where a matrix times its transpose are multiplied together.

Is $A A^\mathrm T$ something special for any matrix $A$?

The main thing is presumably that $AA^T$ is symmetric. Indeed $(AA^T)^T=(A^T)^TA^T=AA^T$. For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real.

Moreover if $A$ is invertible, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$

Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors.

Last but not least if one is interested in how much the linear map represented by $A$ changes the norm of a vector one can compute

$$\sqrt{\left<Ax,Ax\right>}=\sqrt{\left<A^TAx,x\right>}$$

which simplifies for eigenvectors $x$ to the eigenvalue $\lambda$ to

$$\sqrt{\left<Ax,Ax\right>}=\sqrt \lambda\sqrt{\left<x,x\right>},$$

The determinant is just the product of these eigenvalues.

The matrix $A$ represents a rotation if $A^TA=I$ and Det A=1 (where $A^T$ means $A$ transpose)


Probability of their's constituting a triangle? We have a line segments with length $l$ then you choose two random points and cut it from these points so that we
have three piece of line segments.

What is the probability that these piece constitute a triangle ?

We need to use triangle inequality but I could not manage it. Thanks.

I'm sure there is a nicer solution:

label the points $0$ to $l$. And call the points $p_1$ and $p_2$ with $p_1<p_2$

notice $p_1<.5$ and $p_2>.5$

now, suppose $p_1$ has already been decided, then the fraction of points that would be suitable $p_2$'s is $p_1$ since we need $|p_1-p_2|<.5$.

So the answer is $\int_0^.5p_1 d p_1$ which is $\frac{1}{4}$


Why does the discriminant in the Quadratic Formula reveal the number of real solutions? Why does the discriminant in the quadratic formula reveal the number of real solutions to a quadratic equation?

That is, we have one real solution if
$$b^2 -4ac = 0,$$
we have two real solutions if
$$b^2 -4ac > 0,$$
and we have no real solutions if
$$b^2 -4ac < 0.$$

Think about it geometrically $-$ then compute.

Everyone knows $x^2$ describes a parabola with its apex at $(0,0)$. By adding a parameter $\alpha$, we can move the parabola up and down: $x^2+\alpha$ has its apex at $(0,\alpha)$. Looking at the graph as it moves up and down you immediately see how the number of zeros depends on $\alpha$:

for $\alpha>0$ we have no zeros.
for $\alpha=0$ we have a single zero.
for $\alpha<0$ we have two zeros.

Now we can introduct a second paramter $\beta$ to move the parabola left and right: $(x-\beta)^2+\alpha$ has its apex at $(\beta,\alpha)$.

Note: we used the fact that given a function $f(x)$, the graph of the function $f(x-\beta)$ looks exactly like the one of $f$ but shifted to the right by an amount $\beta$.

But of course, shifting a function left and right does not alter the amount of zeros. So it still only depends on $\alpha$. We expand the term a bit:

$$(x-\beta)^2+\alpha=x^2-2\beta x+\beta^2+\alpha.$$

Would your quadratic equation be given in this form, you would immediately see the amount of zeros as describes above. Unfortunately it is mostly given as

$$ x^2+\color{red}px+\color{blue}q=0$$

So instead, you have to look at what parts of the $\alpha$-$\beta$-form above corresponds to these new parameters $p$ and $q$:

$$ x^2\color{red}{-2\beta} x+\color{blue}{\beta^2+\alpha} = 0.$$

So we have $p=-2\beta$ and $q=\beta^2+\alpha$. If we only could extract $\alpha$ from these new parameters, we would immediately see the amount of zeros. But wait! We can!

$$\alpha=q-\beta^2=q-\left(\frac p2\right)^2.$$

This is exactly what you know as (the negative of) the discriminant.

I used the form $x^2+px+q=0$ and you used $ax^2+bx+c=0$. I hope this is not confusing you. Just divide by $a$ (if $a$ is non-zero):

$$x^2+ \color{red}{\frac ba}x+\color{blue}{\frac ca}=0$$

If you set $p=b/a$ and $q=c/a$ and plug this into my discriminant from above you obtain the one you know:

$$\left( \frac {\color{red}p}2 \right )^2-\color{blue}q = \frac{(\color{red}{b/a})^2}4-\color{blue}{\frac ca}=\frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}.$$

Because $4a^2$ is always positive it suffices to look at $b^2-4ac$ as you did in your question.

Say your quadratic equation is
$$
ax^2 + bx +c = 0
$$
and suppose that $a>0$ , so that we can imagine $p(x) = ax^2 + bx +c$ to be a parabola which is open at the top. Let's call $m$ the minimum value of $p(x)$ . Graphically it is just the lowest point of the parabola.

If you now draw a few examples for parabolas, you will notice that, depending on the sign of $m$, the parabola will cut, touch or never touch the horizontal line ($x$-axis). Depending on the situation, you can see how many roots there are:
\begin{align*}
&\text{if} ~~m>0~~ \text{we don't touch the line $\longrightarrow$ no roots} \\
&\text{if} ~~m = 0~~ \text{we touch the line in one point $\longrightarrow$ one root}
\\
&\text{if} ~~m < 0~~ \text{we cross the line $\longrightarrow$ two roots}
\end{align*}
To find the $x$ coordinate of our minimum $m$, we can use the derivative, since it has to be $0$ there.
$$
0 = p'(x) = 2ax + b  ~~~\Longleftrightarrow ~~~ x = -\frac{b}{2a}
$$
so we get
$$
m
= p \left( -\frac{b}{2a} \right)
= a \frac{b^2}{4a^2} - b\frac{b}{2a} +c
= \frac{-b^2 + 4ac}{4a}
$$
Since we are only interested in the sign of $m$, and since $a>0$ , the sign of $m$ will be the same as the sign of $-b^2 +4ac$ . If we combine this with earlier observation, we have
\begin{align*}
&\text{if} ~~~4ac - b^2 >0~~~ \text{we don't touch the line $\longrightarrow$ no roots} \\
&\text{if} ~~~4ac - b^2 = 0~~~ \text{we touch the line in one point $\longrightarrow$ one root}
\\
&\text{if} ~~~4ac - b^2 < 0~~~ \text{we cross the line $\longrightarrow$ two roots}
\end{align*}


Find $\lim _{n\to \infty}\frac{n-1}{n-2}$ I'm taking a university real analysis course and I have been tasked with proving that the sequence $x_n = \frac{n-1}{n-2}$ converges using first principles.

First fix $\epsilon >0$.

Using the algebra of limits we find the limit in question.

$x_n = \frac{n-1}{n-2} = \frac{1-\frac{1}{n}}{1-\frac{2}{n}} \rightarrow x_n = 1 $ as $n\rightarrow\infty$

We observe a limit if for any $\epsilon > 0$ there is an $N \in \mathbb{N}$ such that

$n \geq N \implies \mid \frac{n-1}{n-2}- 1 \mid < \epsilon$

Provided $n \geq N$

For all $n>3$, we can remove absolute value.

$\frac{n-1}{n-2} - \frac{n-2}{n-2} = \frac{1}{n-2} \leq \frac{3}{n} <\epsilon$ $\iff \frac{3}{\epsilon}<n$

Thus any integer $N$ greater than $max(3, \frac{3}{\epsilon})$ has the required property

Any feedback would be much appreciated!

Your proof is correct, but I think you should explicitly specify a choice of $N$ as a function of $\epsilon$. Something like this: For any $\epsilon$ and for all $n' \geq N = \max\{\frac{4}{\epsilon}, 3\}$, then
$$
\left\lvert \frac{n'-1}{n'-2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{3}{N} = \frac{3}{4} \epsilon < \epsilon.
$$
Thus the sequence $(x_n) \rightarrow 1$.

Here's another bound that works: For any $\epsilon$ and for all $n' \geq N = \max\{3, \lceil 2 + \frac{2}{\epsilon} \rceil\}$, then
$$
\left\lvert \frac{n' - 1}{n' - 2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{1}{ 2 - 2 + \frac{2}{\epsilon}} = \frac{1}{2}\epsilon < \epsilon
$$
Thus the sequence $(x_n) \rightarrow 1$.


rewrite $\sum_{n=0}^\infty \frac{(-1)^nx^{n-1}}{2^{n+1}}$ as $-\sum_{n=0}^\infty (x+1)^{2n}$ Rewrite $\sum_{n=0}^\infty \frac{(-1)^n x^{n-1}}{2^{n+1}}$ as $-\sum_{n=0}^\infty (x+1)^{2n}$.

These two series are equal because they can both be derived from $\frac{1}{x^2+2x}$. To get the first one, you split it into $\frac{1}{x}\cdot \frac{1}{x+2}$ and then you use the power series. To get the second series, you complete the square with $\frac{1}{x^2+2x+1-1}$.

The point is those two are obviously equal but I can't figure out how to manipulate them while leaving them in series form to make them both equal each other (without using $\frac{1}{x^2+2x}$ as an in between equals). This was a problem that came up while my teacher was lecturing and he couldn't figure it out.

One tricky part is that you have the singularity at $x=0$ built into the left side, but not the right.

So you'll need to use that $\frac{1}{x}=\frac{-1}{1-(1+x)}=-\sum (1+x)^k$.

These two are equal only on the interval $(-2,0)$ since the left side converges on $(-2,2)\setminus\{0\}$ and the right side converges on $(-2,0)$.

Writing $$f(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{n-1}}{2^{n+1}}$$

We rewrite it as: $$\begin{align}f(x)
&=\frac{1}{2x}+\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2^{n+1}}\\
&=\frac{1}{2x}+\sum_{m=0}^{\infty}\frac{(-1)^{m+1}x^m}{2^{m+2}}\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^mx^m\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^m\left(1+x-1\right)^m\\
&=\frac1{2x}-\frac14\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^m\sum_{k=0}^{m}\binom{m}{k}(1+x)^k(-1)^{m-k}\\
&=\frac{1}{2x}-\frac14\sum_{k=0}^{\infty}\sum_{m=k}^{\infty}\binom{m}{k}(1+x)^k(-1)^k\left(\frac 12\right)^m\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{k=0}^{\infty}(-1-x)^k\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}\\
\end{align}$$

I'll prove below: $$\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}=2.$$

So you get:

$$f(x)=\frac{1}{2x}-\frac{1}{2}\sum_{k=0}^{\infty}(-1)^k(1+x)^k$$

Also, $$\frac{1}{2x}=\frac{-1}2\frac1{1-(1+x)}=\frac{-1}{2}\sum_{k=0}^{\infty}(1+x)^k$$

Now $$(1+x)^k+(-1)^k(1+x)^k=\begin{cases}2(1+x)^k&k\text{ even}\\0&k\text{odd}\end{cases}$$

So you get: $$f(x)=-\sum_{k=0}^{\infty}(1+x)^{2k}$$

Lemma: For integer $k\geq0$, we have $$\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}=2.$$

Proof: We'll use a probability proof.

Flip a fair coin repeatedly, and stop when you have exactly $k+1$ heads. Let $X$ be the random variable equal to the number of tosses you needed.

Then $P(X=m+1)=\binom{m}{k}\frac{1}{2^{k+1}}$, because you must get exactly $k$ heads in the first $m$ tosses, and then the next toss must be a heads.

But then $\sum_{m=0}^{\infty}P(X=m+1)=1$. Multiplying by $2$ gives us our lemma.

[To deduce that last step, you technically also have to prove that $P(X<\infty)=1$ - that is, that is, there is zero probability that you never get $k+1$ heads.]

If you instead want to prove that:

$$\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^{n+1}}=-x\sum_{n=0}^{\infty}(1+x)^{2n}$$ when both are defined, you can do so fairly easily by noting that:

$$\begin{align}-x\sum_{n=0}^\infty(1+x)^{2n}&=(1-(1+x))\sum_{n=0}^{\infty}(1+x)^{2n}\\
&=\sum_{n=0}^{\infty}(1+x)^{2n}-\sum_{n=0}^{\infty}(1+x)^{2n+1}\\
&=\sum_{m=0}^{\infty}(-1)^m(1+x)^{m}
\end{align}$$

Then a variant of the proof above proof converts $xf(x)$ to this same series:

$$\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^{n+1}}&=
\sum_{n=0}^{\infty}\frac{(-1)^n(1+x-1)^n}{2^{n+1}}\\
&=\sum_{k=0}^{\infty}(1+x)^k(-1)^k\sum_{n=k}^{\infty}\binom{n}{k}\frac{1}{2^{n+1}}\\
&=\sum_{k=0}^{\infty}(-1)^k(1+x)^k
\end{align}$$

$\sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{2^{n+1}}
=\sum_{n=0}^\infty (-1)^n(x/2)^{n+1}
$
converges for
$-2 \lt x \lt 2$
and
$-\sum_{n=0}^\infty (x+1)^{2n}
$
converges for
$-1 < x+1 < 1$
or
$-2 < x < 0$.

They are representations of
$\frac{1}{x^2+2x}
$
in different regions.

You can get another representation by
$\frac{1}{x^2+2x}
=\frac1{x^2}\frac{1}{1+2/x}
=\frac1{x^2}\sum_{n=0}^{\infty} (-1)^n (2/x)^n
$
which converges for
$|x| > 2$.

Look up analytic continuation
for fun and profit.


Is there such a thing as a countable set with an uncountable subset? Is there such a thing as a countable set with an uncountable subset?

Actually I know the answer. Well, I believe I know the answer, which is NO.
Unfortunately, the professor in a Theory of Computation class said that yes, there is such a subset.

This is to settle a discussion with fellow students. A discussion that is going nowhere so we go to the internets for a verdict.

Thanks in advance for weighing in on this question.

Of course it is easy to see that any subset of a countable set is countable. While enumerating the larger set, just skip over any elements that are not in the subset, and you have an enumeration of the subset.

Nevertheless, since you mentioned you were in a computability theory course, there are a few computability-like senses in which something like a positive answer to the question is possible.

Every infinite computably enumerable set has numerous subsets that are not computably enumerable. This is simply because every infinite set has uncountably many subsets, and most of these will not be computably enumerable, since there are only countably many c.e. sets. But from a computability perspective, computably enumerable is often the right analogue of countable, since those are the sets with a computable enumeration function, and so this is a very reasonable sense in which the claim is nearly true.
There can be c.e. sets whose complement is infinite, but contains no infinite c.e. subset. Thus, you can enumerate the set $S$, and infinitely many numbers are not in $S$, but there is no computable way to enumerate an infinite set of numbers not in $S$. These are known as the simple sets, and were studied from the time of Post.
Without AC, it is consistent that you can have a set $A$ with an equivalence relation $\sim$ on it, such that the number of equivalence classes of $\sim$ on $A$ is strictly larger than the cardinality of $A$. For example, we can make an equivalence relation $\sim$ having exactly $\mathbb{R}+\omega_1$ many equivalence classes, by saying that two reals are equivalent iff they are equal, or else they both code relations on $\omega$ that are well-orders of the same length. But under the Axiom of Determinacy, there is no $\omega_1$-sequence of distinct real numbers, and so this cardinality is strictly larger than $\mathbb{R}$.

Consider a set which contains a countable collection of uncountable sets.

One example would be the set whose $m^{th}$ member (even though they are not ordered) is the set of real numbers between $m$ and $m+1$.

Then I can say that such a set has a countable subset, and that each of these subsets contains an uncountable number of elements.

However, I could not say that such a set contains an uncountable subset of a countable subset of real numbers, merely that it contains a countable subset of (uncountable) sets.


Why isn’t ‘because’ a logical connective in propositional logic? In simple terms, could someone explain why there is not a logical connective for ‘because’ in propositional logic like there is for ‘and’ and ‘or’?

Is this because the equivalent of ‘because’ is the argument of the form ‘if p, then q’, or am I missing something?

Please illustrate your answer with example(s) if possible.

It is because 'because' is not truth-functional.

That is, knowing the truth-values of $P$ and $Q$ does not tell you the truth-value of '$P$ because of $Q$'

For example, the two statements 'Grass is green' and 'Snow is white' are both true, but 'Grass is green because snow is white' is an invalid argument, and hence, as a statement as to the validity of that argument, a false statement.

On the other hand,'Grass is green because grass is green' is a true statement as to the validity of this as an argument, but yet again it involves two true statements.

This shows that with $P$ and $Q$ both being true, the statement '$P$ because of $Q$' can either be true or false, and hence it is not truth-functional.

You can define things however you want. (Be careful; you may accidentally be inconsistent.)

Either:

"because" is logically equivalent to a binary operator
or it's not

If it is, it's probably the same as "only if" (Or take your pick of the other 15 operators). Adding a "because" overload would create an additional word to remember: unneeded complexity. We like simplicity.

If it isn't, you can define a binary function because(a, b) however you want.


What is difference between undefined & indeterminant form? Then whats about 0^0? I really cant underatand what is exactly difference. By their it can be verified that undefined has no value at all but inderminant form we can able to evaluate its limit by various way(such as L'Hospital's rule) whether it exists.

Indeterminate forms generally refer to combining limits.  If you have two functions $f(x)$ and $g(x)$ such that $\lim_{x \to \infty} f(x)=a, \lim_{x \to \infty} g(x)=b$ you can say $\lim_{x \to \infty} f(x)+g(x)=a+b$.  Other operations can be done similarly.  It is also true that if $\lim_{x \to \infty} f(x)=+\infty, \lim_{x \to \infty} g(x)=b$ you can say $\lim_{x \to \infty} f(x)+g(x)=+\infty$ if you define infinite limits in the usual way.  Indeterminate forms come about when you have something like $\lim_{x \to \infty} f(x)=+\infty, \lim_{x \to \infty} g(x)=-\infty$ where you would like to say $\lim_{x \to \infty} f(x)+g(x)=+\infty-\infty$.  This is indeterminate because you can get any value.  The simple combination is not sufficient and you need to look how fast $f$ and $g$ go off to $\infty$ to come up with the result.

Undefined means exactly that.  Division by zero is undefined.  $0^0$ is undefined.  This is not in reference to limits at all.

Undefined is something that is not and will never be defined. e.g dividing by zero. Is just an expression that has no sense and no possible value. On the other side, indeterminate is an expression that you can't know its value at simple sight, but it can be anything (even undefined). Some indeterminate forms are $0/0$ or $\infty/\infty$. These kind of expressions are usually solvable using limits, and they can have any value.

In summary, undefined expressions are hopeless, indeterminate ones may have an answer if you work them.

The case $0^0$ is just another example of indeterminate form, and it can be worked in various ways usually taking natural log will bring the exponent down. For example:

$$\lim_{x\to0} x^x=e^{\lim_{x\to0}ln(x^x)}=e^{\lim_{x\to0}xln(x)}=e^{\lim_{x\to0}\frac{ln(x)}{1/x}}$$

So now you can use LH:

$$e^{\lim_{x\to0}\frac{1/x}{-1/x^2}} = e^0=1$$


Proof involving gamma function, infinite product and Gauss How can I rigorously and directly prove that $$\Gamma (z)=\lim_{n\rightarrow \infty }\frac{n!n^{z}}{z(z+1)\cdots(z+n)}$$

For simplicity define $z=k$ ,so we prove

$$\lim_{n\to \infty}\frac{n! \,\, n^k}{k(k+1)\cdots (k+n)}=(k-1)!$$

Note that

$$\frac{(k+n)!}{(k-1)!} = k(k+1)\cdots (k+n)$$

so we have

$$ (k-1)! \lim_{n\to \infty}\frac{ \, n! \,\, n^k}{(k+n)!}=(k-1)!\lim_{n\to \infty}\frac{n^k}{(k+n)(k+n-1)\cdots (n+1)}$$

Now this can be written as

$$(k-1)! \lim_{n\to \infty} \frac{n\cdot n\cdot \cdots n}{(n+k)(n+k-1)\cdots (n+1)}=(k-1)!$$

The gamma function is an extension of the factorial by the relation

$$\Gamma(k)=(k-1)!$$

We define the function

$$\psi_n(t):=\begin{cases}
\left(1-\dfrac{t}{n}\right)^n t^{x-1} & \text{if}\;0\le x\le n\\
0 & \text{if}\;x>n\\\end{cases}.$$

First we prove: for any $\varepsilon>0$ and any $t_0>0$ there exists a $N_0\in\mathbb N$ such that

$$\left\vert e^{-t} t^{x-1}-\psi_n(t)\right\vert<\varepsilon$$

for $n\ge N_0$ and $t\in[0,t_0]$. With

$$\chi_n(k):=\begin{cases}
1 & \text{if}\;0\le k\le n\\
0 & \text{if}\;k>n\\\end{cases}$$

we establish

$${\left\vert e^{-t} t^{x-1}-\psi_n(t)\right\vert}=
{\left\vert t^{x-1}\left\{ e^{-t} -\left(1-\dfrac{t}{n}\right)^n\right\}\right\vert}=
{\left\vert t^{x-1}\left\{ e^{-t} -\sum_{k=0}^{n}(-1)^k\binom{n}{k}\left(\dfrac{t}{n}\right)^k\right\}\right\vert}=
{\left\vert t^{x-1}\left\{ e^{-t} -\sum_{k=0}^{n}\dfrac{(-1)^k}{k!}t^k\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\right\}\right\vert}=
{\left\vert t^{x-1}\left\{ \sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!}t^k -\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!}t^k\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\chi_n(k)\right\}\right\vert}=
{\left\vert t^{x}\sum_{k=1}^{\infty}\dfrac{(-1)^k}{k!}t^{k-1}\left\{1-\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\chi_n(k)\right\}\right\vert}\le
{t^{x}\sum_{k=1}^{\infty}\dfrac{t^{k-1}}{k!}\left\{1-\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\chi_n(k)\right\}}.$$

With

$$b_k:=\dfrac{1-\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\chi_n(k)}{k!}$$

we have $0\le b_k\le \dfrac{1}{k!}$.  Therefore the series has the majorant $\sum_k\dfrac{t^{k-1}}{k!}$. The radius of convergence of this  series is $R=\infty$ with the quotient criterion

$$\dfrac{\left(\dfrac{t^k}{(k+1)!}\right)}{\left(\dfrac{t^{k-1}}{k!}\right)}=\dfrac{t}{k+1}.$$

Hence there exists a $K_1\in\mathbb N$ such that

$$\tag{1}0\le{t^{x}\sum_{k=K_1+1}^{\infty}\dfrac{t^{k-1}}{k!}\left\{1-\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\chi_n(k)\right\}}<\dfrac{\varepsilon}{2}$$

for $t\in[0,t_0]$. With $\zeta=\dfrac{1}{n}$ and $k\le n$ we write  for the coefficient $b_k$ above

$$b_k=b_k(\zeta)=\zeta\cdot p_k(\zeta)$$

with a polynompial $p_k(\zeta)$. We obtain for $n\ge K_1$

$$t^{x}\sum_{k=1}^{K_1}b_kt^{k-1}=\zeta\cdot t^xp(t,\zeta)$$

with another polynomial $p(t,\zeta)$. On the closed set $[0,t_0]\times[0,1]$ the function $\vert t^xp(t,\zeta)\vert$ has a maximum so that we get

$$\left\vert t^xp(t,\zeta)\right\vert\le B$$

for some $B>0$. Now we only have to choose a $N_0\in\mathbb N$, $N_0\ge K_1$, such that

$$\tag{2}0\le t^{x}\sum_{k=1}^{K_1}b_kt^{k-1}=\dfrac{1}{n}\cdot t^xp\left(t,\dfrac{1}{n}\right)\le\dfrac{B}{n}\le\dfrac{\varepsilon}{2}$$

for $n\ge N_0$ and $t\in[0,t_0]$. The inequalities (1) and (2) prove now

$$\tag{3}{\left\vert e^{-t} t^{x-1}-\psi_n(t)\right\vert}<\varepsilon$$

for $n\ge N_0$ and $t\in[0,t_0]$.

For $t>0$ we get from

$${\lim_{n\to\infty}\left\vert e^{-t} t^{x-1}-\psi_n(t)\right\vert}\to 0$$

the well-known limit

$$\tag{4}\lim_{n\to\infty}\left(1-\dfrac{t}{n}\right)^n=e^{-t}$$

that holds for $t=0$ too. Next we prove for $t\in[0,n)$

$$\tag{5}0\le\left(1-\dfrac{t}{n}\right)^n\le\left(1-\dfrac{t}{n+1}\right)^{n+1}.$$

The first inequality is obvious. The derivitave of the function $h(z):=\left(1-\dfrac{t}{z}\right)^z$ is

$$h'(z)={\dfrac{d}{dz}e^{z \ln\left(1-\dfrac{t}{z}\right)}}=
\left\{\ln\left(1-\dfrac{t}{z}\right)+\dfrac{\dfrac{t}{z}}{1-\dfrac{t}{z}}\right\}\left(1-\dfrac{t}{z}\right)^z.$$

With

$${\ln\left(1-\dfrac{t}{z}\right)+\dfrac{\dfrac{t}{z}}{1-\dfrac{t}{z}}}=
-\sum_{k=1}^\infty\dfrac{1}{k}\left(\dfrac{t}{z}\right)^k+\sum_{k=1}^\infty\left(\dfrac{t}{z}\right)^k$$

we see that $h'(z)\ge0$ for $0\le t<z$ and the inequality (5) is proven. Because the function $\psi_n(t)$ is not negative, grows with n and converges to $e^{-t}t^{x-1}$ we obtain

$$\tag{6}0\le{\int_0^\infty\left\{e^{-t}t^{x-1}-\psi_n(t)\right\}dt}\le
{\int_0^\infty e^{-t}t^{x-1}dt}=:\Gamma(x).$$

Because the integral on the right side converges, for any $\epsilon>0$ there exists a $T_0>0$ such that

$$0\le{\int_{T_0}^\infty\left\{e^{-t}t^{x-1}-\psi_n(t)\right\}dt}\le
{\int_{T_0}^\infty e^{-t}t^{x-1}dt}<\dfrac{\varepsilon}{2}.$$

We choose a $N_1\in\mathbb N$ such that

$${\left\vert e^{-t} t^{x-1}-\psi_n(t)\right\vert}\le\dfrac{\varepsilon}{2T_0}$$

for $n\ge N_1$ in the inequality (3). The last $2$ inequalities render for $n\ge N_1$

$$0\le{\int_0^\infty\left\{e^{-t}t^{x-1}-\psi_n(t)\right\}dt}=
{\int_0^{T_0}\left\{e^{-t}t^{x-1}-\psi_n(t)\right\}dt+\int_{T_0}^\infty\left\{e^{-t}t^{x-1}-\psi_n(t)\right\}dt}\le
\int_0^{T_0}\dfrac{\varepsilon}{2T_0}dt+\dfrac{\varepsilon}{2}\le\varepsilon$$

or

$$\Gamma(x)={\lim_{n\to\infty}\int_0^\infty\psi_n(t)dt}=
{\lim_{n\to\infty}\int_0^n\left(1-\dfrac{t}{n}\right)^n t^{x-1}dt}.$$

You get your formula by integration by parts:

$${\int_\alpha^n\left(1-\dfrac{t}{n}\right)^n t^{x-1}dt}=
{\left[\left(1-\dfrac{t}{n}\right)^n \dfrac{t^x}{x}\right]_\alpha^n}+{\dfrac{1}{x}\int_\alpha^n\left(1-\dfrac{t}{n}\right)^{n-1} t^{x}dt}.$$

For $\alpha\to 0+$ we obtain

$${\int_0^n\left(1-\dfrac{t}{n}\right)^n t^{x-1}dt}=
{\dfrac{1}{x}\int_0^n\left(1-\dfrac{t}{n}\right)^{n-1} t^{x}dt}.$$

Consecutively integrating by parts renders

$${\int_0^n\left(1-\dfrac{t}{n}\right)^n t^{x-1}dt}=
{\dfrac{1}{x}\int_0^n\left(1-\dfrac{t}{n}\right)^{n-1} t^{x}dt}=
{\dfrac{n-1}{n}\cdot\dfrac{1}{x(x+1)}\int_0^n\left(1-\dfrac{t}{n}\right)^{n-2} t^{x+1}dt}=\cdots=
{\dfrac{(n-1)(n-2)\cdots 1}{n^{n-1}}\cdot\dfrac{1}{x(x+1)\cdots(x+n-1)}\int_0^nt^{x+n-1}dt}={\dfrac{n-1!}{n^{n-1}}\cdot\dfrac{1}{x(x+1)\cdots(x+n-1)}\cdot\dfrac{n^{x+n}}{x+n}}=
{\dfrac{n!n^x}{x(x+1)\cdots(x+n)}}.$$

In an alternative proof we could also have taken the well-known limit (4) in the case $x\ge1$. For any $\varepsilon>$ and $t\in[0,t_0]$ we would have determined a $N_0(t_0)\in
\mathbb N$ such that

$$\left\vert \left(1-\dfrac{t_0}{N_0(t_0)}\right)^{N_0(t_0)} t_0^{x-1}-e^{-t_0}t_0^{x-1} \right\vert\le\dfrac{\varepsilon}{2}.$$

Because the functions $\left(1-\dfrac{t}{N_0(t_0)}\right)^{N_0(t_0)} t^{x-1}$ and $e^{-t}t^{x-1}$ are continuous there exists an open interval $\mathbb M(t_0)$ around the point $t_0$ such that

$$\tag{7}\left\vert\left(1-\dfrac{t}{N_0(t_0)}\right)^{N_0(t_0)} t^{x-1}-e^{-t}t^{x-1}\right\vert\le\varepsilon$$

for $t\in\mathbb M(t_0)$. Because the function $\left(1-\dfrac{t}{n}\right)^n t^{x-1}$ grows with n, the inequality (7) is true then for any $n\ge N_0(t_0)$. For an interval $[0,t_0]$ there exists a finite number of open intervals $M(t_k)$ such that $[0,T_0]\subset\bigcup_k\mathbb M(t_k)$ with the theorem of Heine-Borel. We determine for such a finite choice

$$N_0=\max_k N_0(t_0)$$

and obtain the inequality once again. However for $0<x<1$ we have

$$\lim_{t\to 0+}e^{-t}t^{x-1}\to +\infty$$

and the inequality is not immediate for $t$ about $0$. Hence in this case it is not enough that the function $\psi_n(t)$ is growing and converging to the function $e^{-t}t^{x-1}$. However we can find a $\delta>0$ such that

$$0\le\int_0^\delta\left\{e^{-t}t^{x-1}-\psi_n(t)\right\}dt\le
\int_0^\delta e^{-t}t^{x-1}dt\le\varepsilon,$$

and then we restrict our attention to the interval $[\delta,T_0]$.

For calculating the gamma function you better go to Mathmatics:Algorithm to compute Gamma function. The product formula of the function $\Gamma(x)$ is mostly used in order to prove other formulas like

$$\dfrac{\pi}{\sin\pi x}=\Gamma(x)\Gamma(1-x)$$

or

$$\dfrac{\sqrt{\pi}}{2^{2x-1}}\Gamma(2x)=\Gamma(x)\Gamma(x+\dfrac{1}{2}).$$


Show that the following matrix has $4$ as an eigen value of multiplicity $3$ Show that the following matrix has $4$ as an eigen value of multiplicity $3$

$\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&0&1\\1&1&1&0&3&0\\1&1&1&1&0&4\end{bmatrix}$.

How should I try this?Isn't this equivalent to finding the entire spectrum of the above matrix?

Or there are some tricks involved here?Please help.

Call the matrix $A$. It is real symmetric, hence diagonalisable. Therefore, it suffices to show that the rank deficiency of $A-4I$ is $3$.


Piecewise Function vs Regular Function I'm into graphing functions and I'm currently working on some project of mine. I'm a little confused, what's the main distinction of a Piecewise Function with just a Regular / $f(x)$ Function? I mean, most Piecewise functions posses the same format of equations with an $f(x)$ Function.

Generally a piece-wise function has at least one jump discontinuity in it. For example $F(x) =1$ is a "regular" function as you refer to it; whereas
$G(x)= \begin{cases}
      -1 & x< 0 \\
      0 & 0\leq x\leq 100 \\
      1 & x > 100
   \end{cases}$ is piece-wise (it has jump discontinuities at $x=0$ and $x=100$). For more info see the Wikipedia article on piecewise functions.


What is the slope of the adjacent/opposite in a right triangle? Suppose I have the lengths of the adjacent, opposite, and hypotenuse of a right triangle (in this instance $a, b$ and $c$ respectively), and coordinates of point $\mathrm{B}$ and $\mathrm{C}$ ($\mathrm{A}$ is unknown).  My question is, how do I calculate the slope of the opppsite? (the slope made with points $\mathrm{A}$ and $\mathrm{C}$).

2 options:

to simply get the slope you can take the slope of $CB$, convert it to angle using $\tan^{-1}$ (you'd also have to add $\pi$ if $B$ is to the left of $C$), get the angle $ACB$ using $\tan^{-1}$, then add those angle together and take $\tan$ of sum. This can be simplified by using formula for tangent of sum:
$$\tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$
Long story short, if $s_1$ is the slope of $CB$ and $s_2 = \frac{a}{b}$, then the slope of $CA$ is simply
$$\frac{s_1+s_2}{1-s_1s_2}.$$
If you want to find the coordinates of the point $A$, you may want to use Rational parametrization of a circle. Given $M = \frac{1}{2}(B+C)$, $v = B -M$; $cos = \frac{b^2-a^2}{c^2}$, $sin = \frac{2ab}{c^2}$ and finally $$R = \begin{pmatrix}cos & sin\\ -sin & cos\end{pmatrix};$$
you get $A = M + Rv$.

Well, if you do not know how to work with matrices and vectors, you may want to ignore the 2nd option.

Assuming your triangle is oriented so that the adjacent side is on the $x$-axis we have that slope is $\frac{rise}{run} = \frac{opposite}{adjacent}$


Graphs Theory Hint about k-connected Show That G is a k-edge-connected graph if and only if there is no non-empty proper subset W to V (G), such that the number of edges joining W to V (G) - W is less k.
Can anyone help me or give me a hint? please

Lets prove it by the method of contradiction.

Consider W such that the number of edges joining W to V (G) - W is greater than k. These are two disconnected components that are connected by greater than k number of edges in V (G), which contradicts the statement that the graph is k-connected.


Solving Recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n\geq 1, a_0=2$? How can I solve recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n \geq1 $ and $a_0 = 2$?

general sequence is $a_n= (3^n)a_0 +n \times 5\times 3^n$
substitute $a(n-1)$ for $a(n-2)$ and check the pattern you can easily figure it out.


How to sketch the level curves of $f(x,y) = x^2 - y^2$ I've been practising functions of several variables for college and I've been working with circles all the time $(x^2 + y^2)$, however, I still can't figure out how to solve non circular shapes, as far as I know by research $(x^2 - y^2) $ represent two queues that pass through the origin, and when you plot the function on 3D using the Google's plot you get this:

Google search for function's plot

The thing I don't understand is why does it look like a wave?, how do you determinate the Z/Height values for each queue and why X queue starts from below 0 and gains Z while Y starts from 0 and loses Z?

I would like that someone demonstrate to me (also including level curves) how to solve this exercise, sorry if it's an obvious question but I can't figure it out, really, plus, teachers don't help.. Thank you.

If you freeze $y$ (hence work in the vertical plane $y=Y$), you get curves $$z=x^2-Y^2$$ which are upward parabolas with their vertices at $(0,Y,-Y^2)$. The latter curve is obviously a downward parabola with its vertex at the origin.

Well, in order to determine the shape of the graph, you need to understand that the          [(x^2/a) -(y^2)/b = z/c] will result in a saddle shape or a hyperbolic paraboloid. Basically, the crest, or bottom, of the level curve will be at (0,0,0) and there will a parabolic curve on the z-x plane with a decline on the 3D space:


Find the values of the positive constants $k$ and $c$ such that $-37\le k(3\sin\theta + 4\cos\theta) +c\le 43$ for all values of $\theta$ Hi how do i go about solving this?

Find the values of the positive constants $k$ and $c$ such that $$-37\le k(3\sin\theta + 4\cos\theta) +c\le 43 $$for all values of $\theta$ $$\rightarrow-37\le k(5(\sin\theta + 53.1)) +c\le 43 $$
Then what?

Cheers

By Cauchy-Schwarz

$$(3\sin(x)+4\cos(x))^2 \leq 25 (\sin^2(x)+\cos^2(x))=25$$

and equality is possible.

Then

$$-5 \leq 3\sin(x)+4\cos(x) \leq 5 $$

this shows that

$$-5k+c \leq - k(3\sin\theta + 4\cos\theta) +c\le 5k+c \,.$$

and the lower/upper bounds can be atatined. You can finish it easely.

Hint: $|\sin \alpha| \leq 1$.

Hence, this implies that $ c = \frac {43+(-37)} {2} $.


What does the dot product of two vectors represent? I know how to calculate the dot product of two vectors alright. However, it is not clear to me what, exactly, does the dot product represent.

The product of two numbers, $2$ and $3$, we say that it is $2$ added to itself $3$ times or something like that.

But when it comes to vectors $\vec{a} \cdot \vec{b}$, I'm not sure what to say. "It is $\vec{a}$ added to itself $\vec{b}$ times" which doesn't make much sense to me.

The dot product tells you what amount of one vector goes in the direction of another. For instance, if you pulled a box 10 meters at an inclined angle, there is a horizontal component and a vertical component to your force vector. So the dot product in this case would give you the amount of force going in the direction of the displacement, or in the direction that the box moved. This is important because work is defined to be force multiplied by displacement, but the force here is defined to be the force in the direction of the displacement.

http://youtu.be/KDHuWxy53uM

When directions are considered, we essentially bring a new dimension to the perception of the entity. (Speed vs Velocity: 5km/h vs 5km/h towards east).
Bringing the sense of direction, the question arises, how the entities interact?

In dot product, diagrammatically, what we find is, essentially, the area that is affected by the two entities taken together.

Consider Tetris. You have built a foundation already. Now, a new part is falling and you have the arrow keys to move it around. Two competing vectors, your movement and the falling of the brick/part, will determine how the new part is arranged. The area covered by the falling part would be determined by the dot product of the said vectors.


Find out the subsequential limits of a sequence Let $$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$ and find the upper and lower limits of the sequence $\{x_n\}_{n=1}^\infty$.

We put $n=1, 2, 3 \dots $ then $x_1=-9, x_2 = \frac{15}{2}, \dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right?
Please explain.

Notice

$\underset{n \to \infty}{\lim \sup}\, x_n = \inf \{x_{2n} : n \in \mathbb{N}\} $,

and

$\underset{n \to \infty}{\lim \inf}\, x_n = \sup \{x_{2n-1} : n \in \mathbb{N}\} $.

Moreover, $2$ is a lower bound for the set $\left\{2+\frac{3^{2n}}{(2n)!}+\frac{1}{n^2} : n \in \mathbb{N}\right\}$, and $-2$ is an upper bound for the set $\left\{-2-\frac{3^{2n-1}}{(2n-1)!}-\frac{4}{(2n-1)^2}  : n \in \mathbb{N}\right\} $.

Also observe that we have $\frac{3^n}{n!} < 2^{6-n}$ whenever $n>6$.

Let $\varepsilon>0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that
\begin{aligned}&N_1  > 7 - \frac{\log\varepsilon}{\log 2}, \text{ and} \\&
N_2\varepsilon > 8 .
\end{aligned}
Set $N=\max\{7, N_1, N_2\}$. So if $n \geq N$, then we have
\begin{equation}\frac{3^n}{n!}+\frac{4}{n^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .
\end{equation}
So there is $N \in \mathbb{N}$ such that
\begin{aligned} &x_{2N}<2+\varepsilon, \text{ and} \\& x_{2N-1}>-2-\varepsilon.
\end{aligned}
Therefore, $\underset{n \to \infty}{\lim \sup}\, x_n =2$ and $\underset{n \to \infty}{\lim \inf}\, x_n =-2$.

Note: Since $N \geq 7$, we know that $2N-1>N$ and $2N>N$.

Here is a less rigorous but quicker way to look at the problem.

We have
$$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$

Let's look at
$$\lvert x_n\rvert=2+\frac{3^n}{n!}+\frac{4}{n^2}$$

We know that
$\dfrac4{n^2}$ is decreasing and $2$ is constant.

The first few terms of $\dfrac{3^n}{n!}$ are
$$\dfrac{3}{1},\,
\dfrac{3\cdot3}{2\cdot1},\,
\dfrac{3\cdot3\cdot3}{3\cdot2\cdot1},\,
\dfrac{3\cdot3\cdot3\cdot3}{4\cdot3\cdot2\cdot1},\,
\dfrac{3\cdot3\cdot3\cdot3\cdot3}{5\cdot4\cdot3\cdot2\cdot1},\cdots$$

which has 2nd and 3rd terms equal and 4th terms onwards decreasing.

All in all, $\lvert x_n\rvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.


Number of integral solutions for $|x | + | y | + | z | = 10$ How can I  find the number of integral solution to the
equation

$|x | + | y | + | z | = 10.$

I am using the formula,

Number of integral solutions for $|x| +|y| +|z| = p$ is $(4P^2) +2 $, So the answer is 402.

But, I want to know, How we can find it without using formula.

any suggestion !!!

Please help

The number $n_1(k)$ of solutions of
$$|x|=k$$
is quite obviously given by
$$n_1(k)=\begin{cases}2&\text{if }k>0\\1&\text{if }k=0\\0&\text{if }k<0.\end {cases}$$
Then the number of solutions $n_2(k)$ of
$$|x|+|y|=k$$
can by obtained as
$$n_2(k)=\sum_{i\in\mathbb Z}n_1(i)\cdot n_1(k-i)=\begin{cases}2+(k-1)\cdot 4+2=4k& \text{if }k>0\\1&\text{if }k=0\\0&\text{if }k<0.\end{cases}$$
Finally, the number $n_3(k)$ of solutions of
$$|x|+|y|+|z|=k$$
is (using $\sum_{i=1}^n i=\frac{n(n+1)}2$)
$$n_3(k)=\sum_{i\in\mathbb Z}n_2(i)\cdot n_1(k-i)\\=\begin{cases}2+2\sum_{i=1}^{k-1}4i + 4k=2+4k(k-1)+4k=4k^2+2& \text{if }k>0\\1&\text{if }k=0\\0&\text{if }k<0.\end{cases}$$

In general, the number of solutions of
$$|x_1|+\cdots +|x_m|=k$$
is given by
$$n_m(k)=\begin{cases}P_m(k)&\text{if }k>0\\1&\text{if }k=0\\0&\text{if }k<0,\end{cases}$$
where $P_m$ is some polynomial of degree $m-1$. The $P_m$ can be obtained recursively, e.g. via the relations
$$P_m(X+1)-P_m(X)= P_{m-1}(X)+P_{m-1}(X+1)$$
$$P_m(1)=2m$$

I think it must 8 times the number of intergral solution to equation:
$$x+y+z=10$$
So the result is :
$$8C_{12}^2=528$$


If $a<b$, then $a^2<\frac{1}{3}(a^2+ab+b^2)<b^2$ I am trying to prove $a^2<\frac{1}{3}(a^2+ab+b^2)<b^2$ if $a<b$. I am having trouble with the left inequality:
$a^2<\frac{1}{3}(a^2+ab+b^2) \Rightarrow 2a^2<ab+b^2$.
If $a>0$, then all is well and adding $a^2<b^2$ to $a^2<ab$ will yield the result I want, but I do not know how to deal with the case $a<0$, or if there is any simpler ways of proving the inequality.

The inequality doesn't generally hold if $a<0$ so you don't need to worry about it. e.g. consider $a=-1$ and $b=1$.


Difficult Cauchy Problem Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$
I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.
I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.
I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.

$$\sum{a\over bc}=3$$
$$\sum{a\over b^2c^2}\times(a+b+c)\geq(\sum{a\over bc})^2$$ by Cauchy.

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[1]: https://i.stack.imgur.com/VF8FZ.jpg            hi this is my answer is it true?


What is the derivative of a summation with respect to its upper limit? For the moment, consider the corresponding problem involving integration. Let $s(x)$ be the explicit solution to the following integral.

$
\displaystyle s(x)=\int_a^x f(t) \, dt
$

The function $s'(x)$ is equivalent to the derivative of the integral with respect to it's upper limit and may be expressed in integral form.

$
\displaystyle s'(x)=\partial _x\left(\int_a^x f(t) \, dt\right)=f(a)+\int_a^x f'(t) \, dt
$

Now let $s(x)$ be the explicit solution to the following summation.

$
\displaystyle s(x)=\sum _{t=a}^x f(t)
$

The function $s'(x)$ is equivalent to the derivative of the summation with respect to it's upper limit. What is the derivative of $s(x)$ expressed in summation form?

$
\displaystyle s'(x)=\partial _x\left(\sum _{t=a}^x f(t)\right)=\ ?
$

In order to express continuous summation, define an auxiliary function $s(x,f)$ equal to the $RHS$ of the Euler-Maclaurin formula with zero remainder, such that, the index variable $n$ is replaced with the real variable $x$.
$$
\displaystyle s(x,f)\text{:=}\int_a^x f(t) \, dt+\frac{f(x)}{2}+\frac{f(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} }{(2 k)!} \left(f^{(2 k-1)}(x)-f^{(2 k-1)}(a)\right)
$$

Clearly, if $x$ is an integer $n$, then $s(n,f)$ is equal to the summation of $f(t)$.

$
\displaystyle s(n,f)=\sum _{t=a}^n f(t)
$

We can now derive a differential equation for $s$. Differentiate $s$ with respect to $x$ and simplify.

$
\displaystyle \frac{\partial s(x,f)}{\partial x}=f(x)+\frac{f'(x)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} f^{(2 k)}(x)}{(2 k)!}
$

Substitute $f'(t)$ for $f(t)$.

$
\displaystyle s\left(x,f'\right)=f(x)-f(a)+\frac{f'(x)}{2}+\frac{f'(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} }{(2 k)!} \left(f^{(2 k)}(x)-f^{(2 k)}(a)\right)
$

Calculate the difference and simplify.

$
\displaystyle \frac{\partial s(x,f)}{\partial x}-s\left(x,f'\right)=\sum _{k=0}^{\infty } \frac{B_k}{k!} f^{(k)}(a)
$

Observe that the term $\frac{f^{(k)}(a)}{k!}$ is the coefficient for the Taylor expansion of $f(t)$, hence.

$
\displaystyle f(t)=\sum _{k=0}^{\infty } \frac{f^{(k)}(a)}{k!} (t-a)^k=\sum _{k=0}^{\infty } c(k) (t-a)^k
$

$
\displaystyle \frac{\partial s(x,f)}{\partial x}-s\left(x,f'\right)=\sum _{k=0}^{\infty } c(k) B_k
$

Consider what happens if we restrict $x$ to be an integer $n$.

$
\displaystyle \left.\frac{\partial s(x,f)}{\partial x}\right|_{x=n}=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t)
$

Now, let's get creative and allow the following symbolic equivalence.

$
\displaystyle \left.\frac{\partial s(x,f(t))}{\partial x}\right|_{x=n}\equiv \partial _n\sum _{t=a}^n f(t)
$

The derivative of a summation with respect to it's upper limit may then be expressed as,

$$
\displaystyle \partial _n\sum _{t=a}^n f(t)=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t)
$$

Example

As an example consider the following summation identity with integer $m\geq 1$.

$
\displaystyle \sum _{t=0}^n t^m=\frac{B_{m+1}(n+1)-B_{m+1}}{m+1}
$

Define the functions $f$ and $s$.

$
\displaystyle f(t)=t^m
$

$
\displaystyle s(x,f)=\frac{B_{m+1}(x+1)-B_{m+1}}{m+1}
$

Verify the differential equation.

$
\displaystyle \partial _n\sum _{t=a}^n f(t)=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t)
$

$
\displaystyle \left.\partial _x\left(\frac{B_{m+1}(x+1)-B_{m+1}}{m+1}\right)\right|_{x=n}=B_m+\sum _{t=0}^n m t^{m-1}
$

$
\displaystyle B_m(n+1)=B_m+m\left(\frac{(B_m(n+1)-B_m)}{m}\right)
$

$
\displaystyle B_m(n+1)=B_m(n+1)
$

Application

As an application consider the following summation identity that uses the incomplete gamma function, $\Gamma (a,z)$.

$
\displaystyle \sum _{t=0}^n \frac{1}{\Gamma (t+1)}=\frac{e \Gamma (n+1,1)}{n!}
$

Define the functions $f$ and $s$.

$
\displaystyle f(t)=\frac{1}{\Gamma (t+1)}
$

$
\displaystyle s(x,f)=\frac{e \Gamma (x+1,1)}{\Gamma (x+1)}
$

Let the constant $C$ represent the Bernoulli sum.

$
\displaystyle C=\sum _{k=0}^{\infty } c(k) B_k
$

Because $C$ is a constant, we may choose any valid $n$ to solve for $C$, choose $n=0$.

$
\displaystyle C=-e \text{Ei}(-1)
$

In conclusion, assemble the differential equation and simplify. The variable $t$ is replaced with the standard index variable $k$. The result is a summation identity that makes use of these additional functions: (1)
$\, _2\tilde{F}_2$, (2) $\text{Ei}$, and (3) $\psi ^{(0)}$.

$$
\displaystyle \frac{-1}{e}\sum _{k=0}^n \frac{1}{k!}\psi ^{(0)}(k+1)=\text{Ei}(-1)+\psi ^{(0)}(n+1) \left(1-\frac{1}{n!}\Gamma (n+1,1)\right)\\ +n! \, _2\tilde{F}_2(n+1,n+1;n+2,n+2;-1)
$$

Peace

Consider the following integral equation with constant $C$.

$
\displaystyle \partial _x\left(\int_a^x f(t) \, dt\right)-\int_a^x f'(t) \, dt=C
$

Is there a corresponding summation equation?

$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)-\sum _{t=a}^x f'(t)=C
$

The proof of this result is based on the Euler-Maclaurin formula assuming zero remainder.

$
\displaystyle \sum _{t=a}^x f(t)=\int_a^x f(t) \, dt+\frac{f(x)}{2}+\frac{f(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} }{(2 k)!} \left(f^{(2 k-1)}(x)-f^{(2 k-1)}(a)\right)
$

Differentiate with respect to x.

$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=f(x)+\frac{f'(x)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k}}{(2 k)!} f^{(2 k)}(x)
$

Apply the formula to $f'(t)$.

$
\displaystyle \sum _{t=a}^x f'(t)=f(x)-f(a)+\frac{f'(x)}{2}+\frac{f'(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k}}{(2 k)!} \left(f^{(2 k)}(x)-f^{(2 k)}(a)\right)
$

Calculate the difference and simplify.

$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)-\sum _{t=a}^x f'(t)=\sum _{k=0}^{\infty } \frac{B_k}{k!} f^{(k)}(a)\text{  }\ \square
$

Observe that the term $\frac{f^{(k)}(a)}{k!}$ is the coefficient for the Taylor expansion of $f(t)$, hence.

$
\displaystyle f(t)=\sum _{k=0}^{\infty } \frac{f^{(k)}(a)}{k!} (t-a)^k=\sum _{k=0}^{\infty } c(k) (t-a)^k
$

The derivative of a summation with respect to it's upper limit is therefore,
$$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=\sum _{k=0}^{\infty } c(k)B_k+\sum _{t=a}^x f'(t)
$$

Example

As an example let $f(t)=e^{r t}$ and $a=0$. The summation of $e^{r t}$ is just the geometric series in $e^{r}$, hence.

$
\displaystyle \sum _{t=0}^x f(t)=\sum _{t=0}^x e^{r t}=\sum _{t=0}^x \left(e^r\right)^t=\frac{\left(e^r\right)^{x+1}-1}{e^r-1}=\frac{e^{r (x+1)}-1}{e^r-1}
$

$
\displaystyle \partial _x\left(\sum _{t=0}^x f(t)\right)=\partial _x\left(\sum _{t=0}^x e^{r t}\right)=\partial _x\left(\frac{e^{r (x+1)}-1}{e^r-1}\right)=\frac{r e^{r (x+1)}}{e^r-1}
$

$
\displaystyle \sum _{t=0}^x f'(t)=\sum _{t=0}^x r e^{r t}=\frac{r\left(e^{r (x+1)}-1\right)}{e^r-1}
$

The Bernoulli sum in terms of the coefficient function $c(k)$.

$
\displaystyle e^{r t}=\sum _{k=0}^{\infty } \frac{r^k}{k!} t^k
$

$
\displaystyle c(k)=\frac{r^k}{k!}
$

$
\displaystyle \sum _{k=0}^{\infty } c(k) B_k=\sum _{k=0}^{\infty } \frac{r^k}{k!} B_k
$

Notice that this is the generating form for the Bernoulli numbers.

$
\displaystyle \frac{r}{e^r-1}=\sum _{k=0}^{\infty } \frac{B_k}{k!} r^k
$

Now assemble the equation and simplify.

$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=\sum _{k=0}^{\infty } c(k)B_k+\sum _{t=a}^x f'(t)
$

$
\displaystyle \frac{r e^{r (x+1)}}{e^r-1}=\frac{r}{e^r-1}+\frac{r\left(e^{r (x+1)}-1\right)}{e^r-1}
$

$
\displaystyle r=r
$

Peace


Analysis question on Integration bounds I have to find all functions f(t) such that

$\int_x^{x^2} f(t) d t=\int_1^x f(t) dt$

I think the solutions are all the functions of the form a/(x+b) because the logarithm would divide the two powers and ln(1) is 0 so that would go away..

Not sure if there are any others though..

We are going to assume that the function is well defined on the interval $(1,x)$ and $(x,x^2)$ since it is integrable there apiori.

$$\int_x^{x^2} f dt = \int_x^1 f dt + \int_1^{x^2} f dt  = \int_1^x f dt$$

The only thing that we need to worry about is if the integral is well defined on $(1, x^2)$. If $x>1$ then it is the case, because $(1,x^2)$ is the disjoint union of the other two well defined integrals.  If $x=0$, we only get the trivial result (so we always ignore this).  If $x<0$ then $(x^2,1) \subset (x,1)$, so it is still well-defined.  This yields the following result

$$\int_1^{x^2} f dt = 2 \int_1^x f dt $$

From here, I recommend you look up the definition of the weak derivative, this should yield a nice answer (if you want to assume for example that your function is not even continuous)

We must make the assumption that f(t) is defined at least on the interval (1, $x^2$) and integrable there.  Then $\int_x^{x^2}f(t)dt = \int_1^{x^2}f(t)dt - \int_1^xf(t)dt$.

By hypothesis  $\int_x^{x^2}f(t)dt =  \int_1^xf(t)dt$.   So we have
 $\int_1^{x^2}f(t)dt = 2\int_1^xf(t)dt \hspace{50px}$ (1)

Assuming f is continuous we can differentiate both sides of (1) to get f($x^2$) = 2f(x).  As per TimG's thought the answer here is f(x) = log(x). $\hspace{100px}$ (2)

This log can be to any base, so there is a family of functions saisfying the condition.

This leaves two questions:  what to do if f is not continuous; and even if f is continous are there any other solutions to (2).

There are two kinds of "not continuous" -- discontinous at a finite or countably many number of  discrete points; or discontinuous at uncountably many points.  For the first case, we can still go with the solution f(x) = log(x) (piecewise), because those discontinuities do not affect the integral.

In the event of uncountably many discontinuities, the function is not Riemann integrable.  It may be Lebesgue integrable, and if anyone wants to address this question they are welcome, but I don't know.

Now about other solutions.  If there is a g(x) which is another solution it must satisfy $\int_1^x g(t)dt = \int_1^x log(t)dt$.  Now if g is continuous we can differentiate both sides and conclude g(x) = log x.  But we can construct a discontinuous g (other than a log x with holes) which satisfies the condition.  Picking a log to a specific base we set

$\begin{pmatrix}
g(x) = logx -\epsilon & t < (x-1)/2\\
g(x) = logx +\epsilon& t \ge (x-1)/2\\
\end{pmatrix}$

The integral of g will be the same as that of f. This opens up endless possibilities.


A question about Riemann integral Let $h $ be Riemann integrable on $[c,d]$.Let $a\in [c,d]$ and $g(x)=\int_a^x h(t)dt $. Suppose $f $ is Riemann integrable on $g([c,d])$
How do you show the integral $\int_c^df(g(x))h(x)dx$ exists and is equal to  $\int_{g(c)}^{g(d)}f(x)dx  $ ?

With a substitution. Let $y=g(x)$. Then $dy=g'(x) dx= h(x) dx$ where we have used the fundamental theorem of calculus. Thus

$\int_c^df(g(x))h(x)dx =\int_{g(c)}^{g(d)}f(y)dy$

and you are done.