A conjecture on bounded complex partial sums A friend of mine has made the following conjecture, but we don't know how to prove it.

Let $(a_n)_{n\in \mathbb {N}}$ be a strictly increasing sequence of natural numbers.
Suppose that for every complex $z\neq 1 $ with $|z|=1$, there exists a constant $C_z>0 $ verifying
$$\forall N\geq 0, \qquad \left|\sum_{i=0}^N z^{a_i}\right|  <C_z$$
Then the sequence $(a_n) $ contains all but finitely many natural numbers.

For exemple, if $a_n=2n$, the property is not verified with $z=-1$.

Can you find a proof or a counterexample?

I have failed to prove this conjecture and am currently trying to find a counterexample.
Any linear transformations (eg $$a_n = k*n + r$$)  won't work as a counterexample since unit roots will be a problem so I've tried looking at sequences like $$a_n = n^2$$
I have not proven that the series converges (it doesn't seem easy since usual tests won't work) but according to wolfram alpha it does converge even if |z| = 1. Which would mean that the conjecture is false.

http://m.wolframalpha.com/input/?i=sum+of+z%5E%28n%5E2%29+from+0+to+infinity

I don't really know what the strange looking function is though (when I look at the latex code it's written "curly theta") and it might be above my level to show that the series converge if |z| =1.


How to integrate $\int^{\pi/2}_0\sin^4xdx$ Sorry if the question is lame but here it

The following was given in my textbook $$\int^{\pi/2}_0\sin^4xdx$$

so i integrated it this way

$$\implies\int^{\pi/2}_0\sin^4xdx = \int^{\pi/2}_0\frac{\sin^5x}{5}(-cosx)$$$

and then substituted the values but i am getting a wrong answer which 0 and probably my approach is wrong because in my textbook a totally different approach is there. So what and were i am wrong please help me.

Thanks

Akash

$\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#00f}{\large\sin^{4}\pars{x}} &= \bracks{1 - \cos\pars{2x} \over 2}^{2}
={1 \over 4}\bracks{1 - 2\cos\pars{2x} + \cos^{2}\pars{2x}}
={1 \over 4} - \half\,\cos\pars{2x} + {1 \over 4}\,{1 + \cos\pars{4x} \over 2}
\\[3mm]&=\color{#00f}{\large{3 \over 8} -\half\,\cos\pars{2x} + {1 \over 8}\,\cos\pars{4x}}
\end{align}
Now, integrate each term in the right hand side.

Hint: use integration by parts with $u= \sin^3 x $


How do I find the critical points of this function? My online homework question reads:
Let $f(x) = (x - 3)^3 + (x - 3)^{-3}$.

I set up to find the critical points by first finding the derivative of the equation. I used the power rule at first and got $3(x-3)^2-\frac{3}{(x-3)^4}$. Then I was attempting to use the power rule, however, I think I am going about the problem all wrong.
Any assistance on how to find the critical points for the equation would be helpful! Thank you

You're on the right track.  Set the first derivative equal to zero:

$$0 = 3(x-2)^2 - \frac{3}{(x-3)^4}$$
$$(x-3)^6 = 1 \to x = 4, 2$$

Then continue from there.


Compute this complex limit I am having difficulties in proving that the below limit exists:
$$\LARGE{ \lim_{n \to \infty} \frac{\sum_{k=1}^n \frac{1}{(k+x)^{x + i\cdot t}}}{\sum_{k=1}^n \frac{1}{k^{x + i\cdot t}}}}$$ where $x \in (0,1)$ and $t \in \mathbb{R}_+$
I would like to have an upper bound on the modulus of this thing, for instance as a function of $t,x$.
My observations:
for $x = 0$ the above is $1$, for $t = 0$ one can use Stolz-Cesaro and prove that the limit is $ \lim_{n \to \infty} \left( \frac{n}{n+x} \right)^{x} = 1$. For $x = 1$ one obtains something like $\lim_{n \to \infty} \frac{S_n - 1}{S_n}$ where $S_n$ is the denominator of the above limit. Again, if $|S_n| \to \infty$ then the limit apporoaches $1$ in modulus. How can I prove this for the general case? If the function is holomorphic, then can we use the identity theorem for holomorphic functions to say that the limit is $1$ for all $t$ since it is $1$ for all $\{x + i \cdot t | t = 0, x \in (0,1) \}$ (which has an accumulation point)


Prove that gcd exists and it is unique How can I show using division algorithm on $\mathbb{N}$ that there is a gcd for every pair of number $a,b \in \mathbb{N}$ and this gcd is unique?

There are multiple methods, here are 3 :

1) you can show that the euclidian algorithm is a finite sequence (see http://en.wikipedia.org/wiki/Euclidean_algorithm) and thus when it stops you get the pgcd.

2) you can also using the division of $\mathbb{N}$ show that the ring $\mathbb{Z}$ is principal, hence just take a positive generator of the sum of the ideals generated by $a$ and $b$ in $\mathbb{Z}$. This is less elementary but not less illuminating.

3) assuming the existence and unicity of decompositions of $a$ and $b$ has a product of power of prime, you can define the gcd by taking for any prime $v_p(gcd(a,b))=\min (v_p(a),v_p(b))$.

If $d$ is the greates common divisor of $a$ and $b$, then by def.  $d|a$ and $d|b$ and if $c$ is any other common divisor then $c|d$. Now suppose on the contrary. Suppose $a$ and $b$ have another greatest common divisor $d'$ then $d'|d$ and $d|d'$ i.e $d=d'h_1$ and $d'=dh_2$  where $h1, h2 \in \mathbb{N}$
$d'=d'h_1h_2$ which means
$h_1h_2=1$. This is possible only when $h_1=h_2=1$.
So $d=d'$.
Hence proved that the gcd of $a$ and $b$ is unique.


Product of units modulo m Help!!! I can find units in a given $Z_m$ manually by testing out every single value which is relative to m,,, but in general I do not know a way how to find such units... hence I cant answer this question: Prove that the product of units mod m is congruent to $\pm 1$ mod m.

Any idea?

More generally: the product of all elements in any finite abelian group has order $2$ or $1$, and the order is $2$ iff there is exactly one element of order $2$ in the group.

The first claim is easy to show: For each element that does not have order $2$, its inverse is also in the product, so they pair up and disappear from the final result. Thus the product of all elements is the same as the product of all elements of order $2$, which necessarily has order $2$ or $1$.

From here on, I will just disregard all the other elements and concentrate on the subgroup consisting of elements of order $2$ (plus the identity), since all the other elements cancel anyways and don't contribute to the final product. If this subgroup is trivial, then we are done, since there is no element of order $2$, and the product must therefore have order $1$.

Otherwise, take any non-identity element $g$ in the group, and divide all the elements of the group up in pairs so that $h$ is paired with $gh$ for all elements $h$. The product of the two elements in a pair is $g$, so it remains to count the number of pairs.

In fact, the number of elements in the group (and therefore the number of pairs) is a power of $2$: If you want to count the elements, start with the identity, that's one element. Now take a $g_1\neq e$. We now have two elements. Take a new $g_2\neq e, g_1$. This $g_2$ gives two new elements, because neither $g_2$ nor $g_1g_2$ have been counted already. Next, we pick a new element $g_3$, and get four new elements: $g_3, g_1g_3, g_2g_3, g_1g_2g_3$. We keep going, for each new element doubling the number of elements we can reach. Thus when we are done we must have ended up on a power of $2$.

So we see that if there is exactly one element of order $2$, then we just get one pair, so the final product is that $g$. Otherwise, the number of pairs must be even, and so the final product becomes $e$.

If the class of of $[p]$ is invertible mod $m$, $a'p=am+1$ or $am-1$, you have
$(am+1)(bm+1)=m(mab+a+b)+1, (am+1)(bm-1)=m(abm-a+b)-1$

$ (am-1)(bm-1)=m(abm-a-b)+1$. This shows the product of two invertible numbers mod $m$ is $1$ mod $m$ or $-1$ mod $m$.


Continuously Compounded Interest What exactly does it mean? By continuously compounded it makes me think it is almost like multiplied as time goes on. Could someone also explain what the constant e is and how it originated? Also how do they get the compounded interest formula from the constant e?

So lets look at the what we mean by compound interest. Let use $r$ as a rate for a year. If we want to define a semiannual compound rate we do the following
$$
\left(1+\frac{r}{2}\right)\left(1+\frac{r}{2}\right) = \left(1+\frac{r}{2}\right)^{2}
$$
for quarterly we have
$$
\left(1+\frac{r}{4}\right)^4
$$
since at the end of the year we have $4$ growths where we have invested the increased value.
in general we have for $n$-periods we have
$$
\left(1+\frac{r}{n}\right)^n
$$
now if we have a continuous growth then we are effectively saying that period becomes infinitesimally small i.e. $n\to\infty$
Mathematically speaking
$$
\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^n\to e^r
$$
where $\mathrm{e}$ is the exponential constant. A really special function in mathematics is $\mathrm{e}^x$ where the gradient at any point of the graph $\mathrm{e}^x$ is proportional to itself, so ideal for growth formula.

Now we can extend this from one year to many years by using
$$
\left(1+\frac{r}{n}\right)^{nt}
$$
where we can adjust the index by $nt=k$ we get
$$
\left(1+\frac{rt}{k}\right)^{k}
$$
if we have $r,t<\infty$ then we have
$$
\lim_{k\to\infty}\left(1+\frac{rt}{k}\right)^{k} \to \mathrm{e}^{rt}
$$

Let denote the interest per annum (year) as $i$. Therefore the compound interest is $1+i$. Now you can compound twice a year. But you take only the half of the interest rate, because the period between the compoundings are only half as long as before.

$\left(1+\frac{i}{2} \right) ^2$

Then you compound 4 times a year (every three months). Thus you divide i by 4 and exponentiate it by 4.

$\left(1+\frac{i}{4} \right) ^4$

Now imagine you do compound infinitely times in a year. The expression becomes

$$\lim_{n \to \infty}\left(1+\frac{i}{n} \right)^n=e^i $$

This means that the interest per year is effectively $(e^i-1)\cdot 100\% \neq i$

$\color{blue}{\text{numerical example}: i=0.05}$

$(e^{0.05}-1)\cdot 100\% =5.127\%\neq 5\%$

You see, that this tansformation from yearly compounding to continuous compounding is not equivalent.


Laplace functional of sum of independent uniformly distributed random variables I'm doing some of the exercises in Cinlar's "Probability and Stochastics" to better understand the material. This exercise (VI.1.17) is taken from page 247:

  Fix an integer $n \geq 1$. Let $X_1,\ldots,X_n$ be independent and uniformly distributed over $E = [0,1]$. Define M by $$Mf(\omega) = \sum_i f \circ X_i(\omega), \qquad \omega \in \Omega, $$ where $f$ is a Borel function on E. Note that this makes M a random measure on $(E,\mathcal{E})$. Compute the Laplace functional of M.

The Laplace functional is defined as a mapping $$f \mapsto \mathbb{E}[e^{-Mf}].$$
I'm having a little trouble calculating it, here's my current approach:

$$\begin{align}\mathbb{E}[e^{-Mf}] &= \mathbb{E}[e^{\sum_{i=1}^n -f(X_i)}]\\
&= \mathbb{E}[\prod_{i=1}^n e^{-f(X_i)}]\end{align}$$
I think the next step would be to use independence to put the product outside, and then "insert" the uniform distribution - but in the book it's not really mentioned what the uniform distribution "means", so I don't know what exactly to insert.

Some good hints or the next steps of the calculation would be appreciated.

I'm assuming:

$f$ is not a Borel set but a Borel-measurable function $\mathbb R \to \mathbb R$.
$\mathcal{E} = \mathscr{M}(E) \ or \ \mathscr{B}(E)$

$$\mathbb{E}[\prod_{i=1}^n e^{-f(X_i)}]$$

$$= \prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$

Now

$$\mathbb{E}[e^{-f(X_i)}] = \int_{\Omega} e^{-f(X_i)} d\mathbb P$$

$$= \int_{E} e^{-f(x_i)} d \mathcal{L}_{X_i}(t)$$

$$= \int_{E} e^{-f(x_i)} f_{X_i}(x_i) dx_i$$

Now insert the uniform distribution $f_{X_i}(x_i) = 1_{E}(x_i)$

$$= \int_{E} e^{-f(x_i)} 1 dx_i$$

$$= \int_{E} e^{-f(x_i)} dx_i$$

Hence,

$$\prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$

$$= \prod_{i=1}^n [\int_{E} e^{-f(x_i)} dx_i]$$


Is Lebesgue integral area under the curve? We know that Riemann integral is defined to be the area between the graph of a function $y=f(x)$ and $x$-axis. Actually, many people state the definition works in both ways (area under the curve is defined to be equal to Riemann integral).

If a Riemann integral of $f$ exists, then Lebesgue integral also exists and is equal to Riemann integral. However, there are Lebesgue-integrable functions that are not Riemann-integrable.

Does Lebesgue integral also give the area under the curve of such functions? Can it be proven formally? Actually this is an important question, because it depends on whether the area under a curve is defined as the value of integral or not.

I would say the answer to your question is yes, because if the function is non-negative, you can write the Lebesgue integral as a Riemann integral:
$$ \int_X f(x) \, d\mu(x) = \int_{0}^{\infty} \mu{\{x \in X: f(x) \geq t \}} \, dt $$

I was searching for an answer to that same question (which led me to this page, albeit 6 years too late). Since I didn't find anything completely satisfying I made my own attempt at an answer with an example where to me the concept of area isn't clear. Consider $f: X \rightarrow Y$
with domain $X = \{a,b,c\}$ and $Y=\{ 0,1\}$ (this happens to be an indicator function).
And $f = \{(a,0),(b,0),(c,1) \} \subset X \times Y$.
If we use counting measure (cardinality)
the Lebesgue integral
$$ \int_X f d \mu =  0 \cdot \mu(f^{-1}(\{0\}) + 1 \cdot \mu(f^{-1}(\{1\})
= 0 \cdot |\{a,b\}| + 1 \cdot |\{c\}| = 1$$
Here, $f$ is a discrete function so is there really a curve or an area under it? It seems to me the answer by Chappers produces "area under the measure". Does that translate to area under $f$?
It seems to me since there is nothing connecting the discrete points of $f$ this wouldn't be an area but this is still a question for me too so if this logic is incorrect, I would like to know why (maybe leave a comment so I can update it)


Limits of $f(x)=x!\frac{e^{x}+1}{x^{x}}$? I have to say that I've no idea, but I just want to calculate $\lim\limits_{x\to\infty}f(x)$
$\lim\limits_{x\to\infty}\bigl(x!\dfrac{e^{x}+1}{x^{x}}\bigr)$

Use the Stirling approximation: $n! \approx \sqrt{2\pi n} \frac{n^n}{e^n}$. Thus, $x! \frac{e^x+1}{x^x} \approx \sqrt{2\pi x} \frac{e^x+1}{e^x} \approx \sqrt{2\pi x}$ for large $x$. If you meant $e^{x+1}$ instead, you get $x! \frac{e^{x+1}}{x^x} \approx \sqrt{2\pi x} e$.

The limit of $f(x)=x!\frac{e^{x}+1}{x^{x}}$ when $x\to\infty$ is $0$.


Multiplicative property of exponents doesn't apply to all bases, why? Let's say we have 2 = 2^(2/2), then using the multiplicative property of exponents to turn it into a square root: ((2)^2)^(1/2) = 2, it just works.

But in the case of -2 = (-2)^(2/2), if we use the property: ((-2)^2)^(1/2) = 2, because the ^1/2 notation means the principal root, we must pick the positive result, which is 2.

Basically (-2)^(2/2) != ((-2)^2)^(1/2), which proves that the multiplicative property of exponents doesn't apply to all bases, what is bothering me is where this property doesn't work? For example, -2 = (-2)^(3/3) = ((-2)^3)^(1/3) = -2, it works. I have a feeling that it is when the denominator is even and the base is negative, but i can't prove it or make a logical reasoning to why and where it doesn't holds.

In general, the rule $a^{bc}=(a^b)^c$ only applies for non-negative values of $a$ (and always remember that $0^0$ is undetermined). This is because general exponentiation is usually defined via

$$a^b:= \exp(b\cdot\log(a))$$

where $\exp$ and $\log$ can be defined by e.g. power series or differential equations. The natural logarithm $\log(a)$ has a canonical value only for positive $a$, and we can extend the defnition to $a=0,b\not=0$ by setting $0^b=0$.

Writing e.g. $(-2)^{1/2}$ requires more explanation and some arbitrary choices in order to uniquely determine the value. It is dangerous to write it this way because of this ambiguity. One preferably talks about "the two solutions of $x^2=-2$" instead.

One other situation in which the rule $a^{bc}=(a^b)^c$ can be applied to all $a$, is when $b$ and $c$ are integers (still $0^0$ is excluded). That this works out well with the definition of $a^b$ given above requires some knowledge about complex numbers (I can add the reasoning if requested).

Actually the multiplicative property of exponents is valid regardless of the base but I presume  that you would require prior knowledge about complex number analysis. What you say is correct if we are dealing with numbers that are strictly real and therefore can't have imaginary or complex solutions. Hope this helps...


Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series) Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series).

I have , somehow, to find the sum of $\sum_{n=1}^\infty \frac{1}{n^4}$ using Parseval's theorem.

I tried some things that didn't work so I won't post them.

Can you please explain me how do I find the sum of this series using Parseval's identity?

Thanks

Let $f(x)=x^2$ for $x\in(-\pi,\pi)$. Computing the Fourier coefficients gives

$$a_n=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 e^{i n x} dx=\frac{2 \cos(\pi n)}{n^2}=2\frac{(-1)^n}{n^2}$$

for $n\in\mathbb{Z}$, $n\not=0$, and $a_0=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 dx=\frac{\pi^2}{3}$.

Therefore $|a_n|^2=\frac{4}{n^4}$ for $n\in\mathbb{Z}$, $n\not=0$ and $|a_0|^2=\frac{\pi^4}{9}$.

By Plancherel/Parseval's theorem,

$$\frac{\pi^4}{9}+8\sum_{n=1}^\infty \frac{1}{n^4}=\sum_{n=-\infty}^\infty |a_n|^2=\frac{1}{2\pi}\int_{-\pi}^\pi x^4 dx=\frac{\pi^4}{5}$$

Simplifying, this gives

$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{8}\left(\frac{1}{5}-\frac{1}{9}\right)=\frac{\pi^4}{90}$$

The series $\sum_{n\ge 1}\frac{1}{n^4}$ also appears in the following funny way. Consider the operator ${\rm d}^2/{\rm d}x^2$ on $[0,\pi]$ s.t. Dirichlet boundary conditions. Find Green's function $G(x,s)$. Find the normalized in $L^2[0.\pi]$ eigenfunctions $u_n(x)$ of the operator; the eigenvalues are $\lambda_n=n^2$.
Evaluate the integral $\int_0^\pi\int_0^\pi G^2(x,s){\rm d}x~{\rm d}s$ in two ways, a) using the explicit formula for $G(x,s)$ and b) using the representation $G(x,s)=\sum_{n\ge 1}\frac{u_n(x)u_n(s)}{\lambda_n}$.
After simplifications, you will find $\sum_{n\ge 1}\frac{1}{n^4}=\frac{\pi^4}{90}$.


Why gradient vector is perpendicular to the plane I know what gradient vector or $\nabla F$ is and I know how to prove that it is orthogonal to the surface (using calculation - not intuitive).
In a particular case, in which we have a three variable function, I want to know why the gradient vector is perpendicular as mentioned. I mean, not in theoretical terms, but intuitive.

Let $F:\mathbb{R}^3 \to \mathbb{R}$ be a function of three variables.  Say we are looking at the surface defined by $F(x,y,z) = 0$.  By definition, the total derivative of $F$ at $p \in \mathbb{R}^3$ is the best linear approximation to $F$ near $p = (a,b,c)$, in other words

$$dF\big|_p (\Delta x,\Delta y,\Delta z)= \dfrac{\partial F}{\partial x} \Delta x + \dfrac{\partial F}{\partial y} \Delta y +\dfrac{\partial F}{\partial z} \Delta z$$

and

$$F(a+\Delta x,b +\Delta y,c + \Delta z) \approx F(p)+dF\big|_p (\Delta x,\Delta y,\Delta z)$$

This is really what the multivariable derivative is all about.  Now we can notice that

$$dF\big|_p (\Delta x,\Delta y,\Delta z)= \langle\dfrac{\partial F}{\partial x}, \dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z}\rangle \cdot \langle \Delta x,\Delta y,\Delta z\rangle$$

Now defining $\nabla F = \langle\dfrac{\partial F}{\partial x}, \dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z}\rangle$, we see that

$$dF\big|_p (\textbf{v})= \nabla F\cdot \textbf{v}$$

This one formula packages a lot of mathematics.

Hopefully this gives you some more intuition about what the gradient does.  In words, to see how much $F(\textbf{p})$ changes when you move away a little bit to $\textbf{p}+\textbf{v}$, just dot product $\nabla F\big|_p$ with $\textbf{v}$.

Armed with this intuitive understanding of the gradient, we can see why it must be perpendicular to the level curves of $F$ quite intuitively.

If $p$ is a point of the surface $F(x,y,z) = 0$, then the tangent vectors $\textbf{v}$ to the surface must satisfy $dF\big|_p(v) = 0$, because moving in the direction of the surface should not change the value of $F$ much since the value of $F$ is constant on the surface.  Translating this into a gradient statement we see that $\nabla F\big|_p \cdot \textbf{v} = 0$ for each tangent vector to the surface.

This just says that $\nabla F\big|_p$ is perpendicular to the tangent plane to the surface at $p$!

Simply put: The gradient at a point p(a,b) is the greatest rate of change of z(a,b). The level curve has constant value z. Therefor for these two "lines" to satisfy there definition, they must be perpendicular to one another.


What is the intuitive explanation for the CDF of any random variable to follow uniform distribution (0,1)? If $X$ is a continuously distributed random variable and $F$ is the c.d.f. of its distribution, then $F(X)$ is uniformly distributed in the interval $(0,1).$

While i'm clear with the mathematical proof, I'm looking for an intuitive answer to this.

Assuming we are dealing with continuous random variables

the CDF $F(X)$ is a continuous increasing function of $X$ where the probability that $F(X) \le p$  for $p \in (0,1)$ is the probability that $X \le q$ when $F(q)=p$, which is $p$
so $F(X)$ has a uniform distribution on the unit interval, since the probability it is less than or equal to a given $p \in (0,1)$ is $p$


Why does $\lvert \Bbb{Q} [\sqrt[n]{2}] : \Bbb{Q} \rvert = n$ if $ \Bbb{Q} [\sqrt[n]{2}]$ is formed by adding only one element? I know that the degree of an extension the dimension of, in this case, $\Bbb{Q} [\sqrt[n]{2}]$ seen as a vector space over  $\Bbb{Q} $. But I thought that the degree of an extension field formed by adding just one element to a given field was one. So, I find it confusing that

$\lvert \Bbb{Q} [\sqrt[n]{2}] : \Bbb{Q} \rvert = n$

Given that $ \Bbb{Q} [\sqrt[n]{2}]$ is formed by adding only one element, namely $\sqrt[n]{2}$, to the field of rational numbers. Any help would be very much appreciated. Thank you.

You are confusing the vector space obtained by adding one element with the field obtained by adding only one element.

Since $\mathbb Q[\sqrt[n]{2}]$ is a field, when we add $\sqrt[n]{2}$ we are also adding  $\sqrt[n]{2}\cdot  \sqrt[n]{2},...,\sqrt[n]{2}\cdot...\cdot\sqrt[n]{2}$ where in the last term there are $n-1$ products. It is enough to stop here since the product of n terms
$$\sqrt[n]{2}\cdot...\cdot\sqrt[n]{2}=2 \in \mathbb Q$$
is already here.

$\mathbb{Q}[\sqrt[n]{2}]$ is a vector space over the field $\mathbb{Q}$. The equation
$$[\mathbb{Q}[\sqrt[n]{2}]:\mathbb{Q}]=n$$
says that the dimesnion of this vector space over $\mathbb{Q}$ is $n$.

In general, if $K$ and $L$ are fields such that $K\subseteq L$, then $L$ can be considered as vector space over $K$. Then $[L:K]$ denotes the dimension of this vector space.


Can you prove this product equal to sum identity involving exponentiation of the sum? Can you prove this identity?

$$\exp \left(\sum _{n=1}^{\infty} \frac{1}{n \left(1-c^n\right)}\right)=\prod _{n=1}^{\infty} \left(1-\frac{1}{c^n}\right)$$

where $|c|>2$.

I found this conjecture by entering the expression similar to the expression in the left hand side:
$$\sum _{n=1}^{\infty} \frac{1}{n \left(1-2^n\right)}$$ into Excel and looking up the number in the advanced version of the inverse symbolic calculator which told me that:

$$\exp \left(\sum _{n=1}^{\infty} \frac{1}{n \left(1-2^n\right)}\right)=\prod _{n=1}^{\infty} \left(1-\frac{1}{2^n}\right)$$

After that I generalized by guessing that $2$ could be replaced by $c$.

Update 25.1.2017:

Let
$$s=0 \text{ or } s=1$$
$$\sum _{n=1}^{\infty} \frac{1}{\left(1-c^n\right) n^s}=\sum _{n=1}^{\infty} \lim_{z\to s} \, (z-2) \zeta (z) \left(1-\frac{1}{\left(\frac{c^n}{c^n-1}\right)^{z-1}}\right)$$

Associated Mathematica code of these two small attempts to generalize as above:

(*Mathematica start*)
Clear[s, z, n]
s = 0;
c = N[2, 50];
Sum[1/(n^s*(1 - c^n)), {n, 1, 200}]
Sum[Limit[(z - 2)*Zeta[z]*(1 - 1/(c^n/(c^n - 1))^(z - 1)),
  z -> s], {n, 1, 100}]

s = 1;
c = N[2, 50];
Sum[1/(n^s*(1 - c^n)), {n, 1, 200}]
Sum[Limit[(z - 2)*Zeta[z]*(1 - 1/(c^n/(c^n - 1))^(z - 1)),
  z -> s], {n, 1, 100}]
(*end*)

-1.6066951524152917637833015231909245804805796715058

-1.606695152415291763783301523190135719575358660

-1.2420620948124149457978454818946296689734039782504

-1.242062094812414945797845481893840808068182966

Possible hint.

$$e^{\sum_{n=1}^{\infty}\frac{1}{n \left(1-c^n\right)}} = e^{\frac{1}{1\cdot \left(1-c^1\right)} + \frac{1}{2\cdot \left(1-c^2\right)} + \frac{1}{3\cdot \left(1-c^3\right)} + \ldots} = e^{\frac{1}{1\cdot \left(1-c^1\right)}} e^{\frac{1}{2\cdot \left(1-c^2\right)}} e^{\frac{1}{3\cdot \left(1-c^3\right)}} e^{\ldots} = \prod_n e^{\frac{1}{n\cdot \left(1-c^n\right)}}$$


Conditional Variance: Is it a random variable? Let's X, Y are random variables (i.e. each one maps elements of a sample space to real numbers). In particular, let's  X is such that  $X:\Omega \rightarrow \mathcal{R}$, where $\Omega=\{\omega_1, ..., \omega_n \}$  and $X(\omega_i)=i$

Then the conditional variance $Var(Y|X(\omega_i))$ is a random variable because it maps an element of a sample space of $\Omega$ to $\mathcal{R}$.

Now consider $Var(Y|X(\omega_i)=1 \bigcup X(\omega_i)=2)$ This expression maps a set $\{\omega_1, \omega_2 \}$ to  $\mathcal{R}$. So, it seems no more to satisfy the definition of a random variable. Then what is it? A measure? Perhaps no.
Thank you for answering.

By definition,
$$
\textrm{Var}(Y\mid X):=\mathbb E\bigl[\bigl(Y-\mathbb E(Y\mid X)\bigr)^2\mid X\Bigr]=\mathbb E(Y^2\mid X)-\mathbb E(Y\mid X)^2.
$$
Thus, the conditional variance is a random variable, in the same way that the conditional expectation $\mathbb E(Y\mid X)$ is. Conceptually, the variance is the "same type of object" as the expectation, in this regard.

Now, one may also consider an event $A\subseteq \Omega$ (the sample space) and ask what is $\textrm{Var}(Y\mid A)$. And it follows the exact same behavior as the conditional expectation, namely that we define
$$
\textrm{Var}(Y\mid A):=\mathbb E\bigl[\bigl(Y-\mathbb E(Y\mid A)\bigr)^2\mid A\Bigr]=\mathbb E(Y^2\mid A)-\mathbb E(Y\mid A)^2.
$$

By definition, $$\mathbb E(Y\mid A):=\frac{\mathbb E(Y\cdot 1_A)}{\mathbb E(1_A)},$$
where $1_A$ denotes the indicator of the set $A$. It is a random variable taking the value $1$ on $A$ and $0$ off $A$. Note also that $\mathbb E(1_A)=\mathbb P(A)$, I just wrote it that way in the denominator of the formula for consistency with the numerator.

Per the discussion below, there was an even more basic question that I should clarify. A random variable is a function from the sample space $\Omega$ to the real numbers. This means it assigns a real number to each element $\omega\in \Omega$. On the other hand, when we condition on an event we obtain a set function on $\Omega$, or in other words, a function that assigns values to subsets of $\Omega$ and not to individual elements of $\Omega$. In this case, being even more precise, we have a partially defined set function which means that not every subset is assigned a value - it is only those subsets which are measurable and are assigned a positive measure for which the conditional variance is defined.

To compare and contrast the two types of mathematical objects, conditional variance with respect to a random variable is a function from $\Omega$ to $\mathbb R$, whereas conditional variance with respect to an event is a partially defined function from $P(\Omega)$ to $\mathbb R$ (the power set of $\Omega)$.

Actually as $\{ \omega_1, \omega_2 \} \in \Omega \oplus \Omega$ and that the latter is an equally valid sample space, it would no less satisfy the definition of a random variable.


Expected Value of 3 numbers I don’t understand expected value, can somebody tell me and help me do this problem?

Generalization: what is the expected value of the biggest number if we choose $k$ numbers from $1,2,...n$.

$$
\begin{aligned}
E&=\frac{1}{\binom{n}{k}}\sum_{i=k}^{n}{i\binom{i-1}{k-1}}\\
&=\frac{1}{\binom{n}{k}}\sum_{i=k}^{n}{k\binom{i}{k}}\\
&=\frac{k}{\binom{n}{k}}\sum_{i=k}^{n}{\binom{i}{k}}\\
&=\frac{k}{\binom{n}{k}}\binom{n+1}{k+1}\\
&=\frac{k}{\binom{n}{k}}\frac{n+1}{k+1}\binom{n}{k}\\
&=\frac{k}{k+1}(n+1)
\end{aligned}
$$

in Your case, substitute $n=10$ and $k=3$

The expected value of a variable is the average of possible values weighted by their probabilities. It will not necessarily be an whole number even if only whole numbers are possible values.

The largest of the three numbers picked cannot be 1 or 2 because there are insufficient smaller numbers available.

If the largest number is 3 then there is only 1 possible combination for the other numbers (1 2).

If the largest number is 4 then there are 1+2=3 possible combinations for the other numbers (1 2) & (1 3)(2 3).

If the largest number is 5 then there are 1+2+3=6 possible combinations for the other numbers (1 2) & (1 3)(2 3) & (1 4)(2 4)(3 4).

Following the same pattern, if the largest number is 6 then there are 1+2+3+4=10 possible combinations for the other numbers. This pattern is the triangular numbers. The next are 15, 21, 28, 36.

The expected value of the largest number picked is therefore
$$E(X)=\frac{3*1+4*3+5*6+6*10+7*15+8*21+9*28+10*36}{1+3+6+10+15+21+28+36}=\frac{990}{120}=8.25$$


Coin tosses probability to calculate I'm stuck at solving the following problem: launch 3 fair coins independently. Let A the event: "you get at least a head" and B "you get exactly one tail".
Then what is the probability of the event $A \cup B$?

The probability that you want is everything except if you get all $3$ tails, since everything else is in the union of $A$ and $B$. So the answer = $1 - \frac{1}{8} = \boxed{\frac{7}{8}}$.

Enumerate all 8 possible outcomes (e.g., HHH, HHT, etc.)
Count how many outcomes satisfy A or B.
Divide by 8.


Evaluate: $\int x\sqrt{\frac{a-x}{a+x}}\hspace{1mm}dx$ Evaluate: $$\int x\sqrt{\dfrac{a-x}{a+x}}\hspace{1mm}\mathrm{d}x$$

I don't know where to start. Hints/suggestions will be appreciated

Hint: Let $x=a\sin t$ and $dx=a\cos t dt$

$\displaystyle\int x\sqrt{\dfrac{a-x}{a+x}}dx=\int a\sin t\sqrt{\dfrac{a-a\sin t}{a+a\sin t}} a\cos t dt=a^2\int \sin t\sqrt{\dfrac{1-\sin t}{1+\sin t}} \cos t dt$

Now, multiply the inside of the square root by $\frac{1-\sin t}{1-\sin t}$

$\displaystyle a^2\int \sin t\sqrt{\dfrac{(1-\sin t)^2}{1-\sin^2 t}} \cos t dt= a^2\int \sin t\dfrac{(1-\sin t)}{\cos t}\cos t  dt$

Can you proceed?

This is not the exact problem, but I am sure this will help: http://www.slader.com/textbook/9780618149186-larson-calculus-7th-edition/418/exercises/40b/

Main Idea is to substitute $x = a\sin \theta$


If $g\circ f$ is injective and $f$ is surjective then $g$ is injective Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions, prove that if $g\circ f$ is injective and $f$ is surjective then $g$ is injective.

I need advise or correction if something is incorrect with my proof. Thank you beforehand.

We must show that $g$ is injective, i.e for $x,y\in B, g(x)=g(y)\implies x=y$

Let $x,y\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \in A$ such that $f(a)=x$ and $f(b)=y$

$\implies g(f(a))=g(f(b))$

$\implies g\circ f(a)=g\circ f(b)$

$\implies a=b$ (by injectivity of $g\circ f$)

$\implies f(a)=f(b)$

$\implies x=y$

Would appreciate any correction in proof writing also!

Your proof is correct. I myself would prove it exactly the same. But, I think it's useful to know more than one way, so here is an alternative solution. It's not profoundly different, but I think it's still worth mentioning.

I'm assuming that $A$ is nonempty (and, since there is a map from $A$ to $B$, $B$ is also nonempty). When $A$ is empty there's not much to prove.

The solution uses left and right inverses. A function with non-empty domain is injective iff it has a left inverse, and a function is surjective iff it has a right inverse.

So, we know that $g\circ f$ has a left inverse $h:C \to A$ and $f$ has a right inverse $k: B \to A$. We want to show that $g$ has a left inverse. Just observe that
$$
(f \circ h) \circ g = (f \circ h) \circ g \circ (f \circ k) = f \circ (h \circ g \circ f) \circ k = f \circ \mathrm{id}_A \circ k = f \circ k = \mathrm{id}_B,
$$
so $(f \circ h)$ is a left inverse for $g$. It follows that $g$ is an injection.

PS: this solution is actually worse than your original one, because this one relies on the axiom of choice (it is used when we say that surjectivity is equivalent to having a right inverse). But it is good in the sense that we don't look at particular elements and manipulate maps as "opaque" objects.

I've proved it on my own like this:

Pick two arbitrary elements of $B$, $y_1$ and $y_2$, with $g(y_1)=g(y_2)$. Since $f$ is surjective, $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1,x_2 \in A$. Then $g(f(x_1))=g(f(x_2))$. Since $f \circ g$ is injective, $x_1=x_2$, and so $f(x_1)=f(x_2)$, or $y_1=y_2$. Finally, $g$ is injective.


Intermediate value theorem with continuous function Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous with $f(0)=f(1)$. If $h \in (0,\frac{1}{2})$ is not of the form $\frac{1}{n}$, there does not necessarily exist $|x-y|=h$ satisfying $f(x)=f(y)$. Provide an example that illustrates this using $h=\frac{2}{5}$

I'm thinking I need to use some kind of modification to a sin function to get this to work. Not sure how to come up with an explicit formula, though. I could just draw a picture, but I want to find some explicit formula so I can show that it's true.

If you accept a piecewise function as an explicit formula, then the function in the figure below works.

A modified $\sin$ function could work, but I think it would be easier to reason using a piecewise linear function. Try defining a function like the one below. The $x$-intercepts are specifically chosen to be $0, \frac16, \frac13,\frac12,\frac23,\frac56,1$. What has to be true about $|x-y|$ if $f(x)=f(y)$? (If you really want to use $\sin$, then modify your $\sin$ function so that it has $x$-intercepts at the same places as this function. That should also work.)


How to find a bounded linear functional which is nonzero at given point without using Hahn-Banach Theorem Let $X$ be a normed vector space. For any $0\neq x \in X$, there exists
$l \in X^*$ such that $l(x)\neq 0$. This statement can be easily proved based on the Hahn-Banach Theorem. My question is how to prove this without using the theorem.

Here $X^{*}$ means the dual space of $X$. The dual space consists of all bounded linear functionals from $X$ to $\mathbb{R}$ or $\mathbb{C}$.

Take $0 \neq x \in X$ and complement $x$ by a family of vectors $(e_j)_{j \in J}$ to get a basis $(x,(e_j)_{j \in J})$ of $X$. This uses Axiom of choice.

Then define $l(x)=1$ and $l(e_j)=0$ for $j \in J$. $l$ satisfies the requested properties.


Why is the support of Dirac distribution $\{0\}$? Distributions are of two types: those that are obtained from locally integrable functions, and those that aren't. For the first type, the support of distribution is simply the support of the function. For the other kind of distribution, for example, the Dirac Delta 'function', we can't find the support this way.

The support of the Dirac Delta distribution is given to be the set $\{0\}$. Can someone help me in understanding why?

Well the definition of the support is defined as the part of the domain where the distribution is non-zero. There are a few definitions of the Dirac delta but all agree that $x\neq 0 \implies \delta(x) = 0$.


Probability: 6 Dice are rolled. Which is more likely, that you get exactly one 6, or that you get 6 different numbers? Here's the question:

6 Dice are rolled. Which is more likely, that you get exactly one 6, or that you get 6 different numbers?

Here's what I've done:

The number of possible outcomes is $6^6 = 46656$.

The probability of rolling exactly one 6 = $\frac{1}{6}\times(\frac{5}{6})^5 = \frac{3125}{46656}$

The probability of getting six different numbers is:
$C(6,1)\times C(5,1)\times C(4,1)\times C(3,1)\times C(2,1)\times C(1,1) = \frac{720}{46656}$

Therefore if everything I've said above is true, then it is more likely that you will roll exactly one six. However I'm really not sure about the last part. Is this correct way to solve this type of problem and can the combinations part be simplified?

EDIT: Since I'm also looking for a better idea of how to solve this type of problem, rather than just this specific case, so could you please include how to solve this problem for rolling 5 dice, as well as/or instead of 6 dice in your answer, so that I can see the pattern of what is happening? Many thanks.

The question is very easy to answer without computing probabilities. Every combination with six different numbers contains exactly one six. There are then additional combinations which contain exactly one six - e.g. $111116$. So the probability of exactly one six is greater.

The trick is to compare the sets of possible outcomes for each condition. Whichever condition's result set has a higher cardinality -in other words, whichever set has more possible outcomes- is more likely to occur, and if the two sets are equal in size, then they are equally likely.

When the number of dice is the same as the number of sides per die -for example, 6d6- you can take a shortcut by showing that one condition is a proper subset of the other. This is how @MarkBennet's answer works. Because there are as many dice as sides, every outcome with all different numbers must show exactly one of each number. So as long as it's possible for your chosen number to come up at all (your example uses a 6 on a six-sided die), then every outcome with all different numbers must also have exactly one of the chose number. You can also show outcomes with exactly one six which don't have all different numbers, like (2, 2, 3, 4, 5, 6), but you can't show outcomes with all different numbers but not exactly one six. Therefore, the set of outcomes with exactly one six is larger, and therefore it's more likely to roll exactly one six.

When there are more dice than sides, you have a different shortcut: the pigeonhole principle. Every die must show some number, so since there are more dice than sides, there must always be at least one number that comes up twice. This means that the set of outcomes with all different numbers is zero, so as long as it's possible for your chosen number to come up at all, then the set of outcomes with exactly one of that number has a greater cardinality. This makes exactly one six more likely than all different numbers, because there are no possible outcomes with all different numbers.

But when there are more sides than dice, there are no shortcuts. You can still show outcomes which satisfy both conditions at the same time, like (2, 3, 4, 5, 6) on 5d6: all different numbers, and exactly one six. And you can still show outcomes with exactly one six but all different numbers: continuing our example, (3, 3, 4, 5, 6) on 5d6 is one of them. But now, you can also show outcomes containing all different numbers but not exactly one six: e.g. (1, 2, 3, 4, 5).

When there's no shortcut, you have to compare the cardinalities of both sets, which is what you did in the question (and @MattS did in his answer). Whichever set has a higher cardinality is more likely to occur.


Positive semidefinite but non diagonalizable real matrix - proof real parts of eigenvalues are non-negative I have a question about positive semidefinite matrices that are non diagonalizable.
Example:
\begin{equation}
A= \left(\begin{array}{cc}
2 & 1\\
0 & 2\\
\end{array}\right)
\end{equation}
Clearly the (real part of the) eigenvalues of $A$ are non-negative.

But how do I prove in general that the real part of the Eigenvalues of a positive semi-definite real matrix are non-negative?
(I have seen the proof where they use diagonalization of the matrix ($B=T^{-1}DT$) but this is not possible for all positive semi-definite real matrices.)

Of course, we assume that $A$ is a real matrix. The matrix $A$ may have non-real eigenvalues as this one $A=\begin{pmatrix}1&1\\-1&1\end{pmatrix}$. The correct result is:

If $A$ is real and for every $x\in\mathbb{R}^n$, $x^TAx\geq 0$, then the eigenvalues of $A$ have a non-negative real part.

cf. Does non-symmetric positive definite matrix have positive eigenvalues?

Now, I think that to call such a matrix $A$, a positive semi-definite matrix, is a very bad idea. Note that this condition is equivalent to  $A+A^T$ is $\geq 0$ in the usual sense. Moreover we encounter questions about this subject on MSE and often the OP does not even report that the studied matrix is not assumed to be symmetric!!

I guess PSD matrices are symmetric and a symmetric matrix is orthogonally diagonalizable.We allways consider symmetry because otherwise eigen values can be complex and then it loses the essence of psd.


Is composition of analytic functions itself analytic? Is composition of analytic functions itself analytic?

Is there a proof that, say,

$$f(x)=e^{\frac{x^2+1}{x^2-1}}$$

analytic?

The composition of analytic functions indeed is analytic. The fastest proof surely relies on complex analysis: every analytic function of one real variable is the restriction of a holomorphic function of one complex variable, so the statement is a consequence of the (complex variable) chain rule.

Your example is a composition/combination of several functions
$$
u(x) = e^x \\
v(x) = x^2 + 1\\
h(x) = x^2  - 1 \\
m(x, y) = x / y \\
f(x) = u( m( v(x), h(x) ) )
$$
The function $m$ is not very nice when $y = 0$, but $y = 0$ happens to be in the image of $h$.

What are you assuming is the domain of $f$? And are you asking "Is it analytic on that domain?"?


Parabola: x-coordinate is negative when it should be positive. I'm teaching myself math. I was working on $y = -x^2 + 8x$ and I used the quadratic equation to determine the x-coordinates. I ended up with $(0,0)$ and $(-8,0)$. The vertex is $(4,16)$ and the y-intercept is $(0,0)$. The thing is, x coordinate of root should be a positive number$=+8,$  otherwise I cannot have a parabola.

Where is my error?

You have that $-x^2 + 8x = -x(x-8)$ so the intersections of the parabola with the $x$-axis are $(0,0)$ and $(8,0)$ as you remarked. You probably made a sign mistake somewhere along your computations.

The thing is, 8 should be a positive number, otherwise I cannot have a parabola.

This makes no sense.  Also, all your calculated points are correct.  It may relieve your confusion for you to know that the parabola is concave (i.e opens downward) and the vertex is above the curve and on the positive side of the y-axis.


Prove that the union of three subspaces of $V$ is a subspace iff one of the subspaces contains the other two. Prove that the union of three subspaces of V is a subspace iff one of the subspaces contains the other two.

I can do this problem when I am working in only two subspaces of $V$ but I don't know how to do it with three.

What I tried is:
If one of the subspaces contains the other two, Then their union is obviously a subspace because the subspace that contains them is a subspace. (Is this sufficient??).

If the union of three subspaces is a subspace.....
How do I prove that one of the subspaces must contain the other two from here?

*When proving this for two I said that there is an element in one of the subspaces that is not the other and proved by contradiction that one of the subspaces must be contained in the other. How would I do this for three?

The statement is false. Consider the following counterexample:

Consider the vector space $V=(\mathbb{Z}/2\mathbb{Z})^{2}$ where $F=\mathbb{Z}/2\mathbb{Z}$. Let $V_{1}$ be spanned by
$(1,0)$. Let $V_{2}$ be spanned by $(0,1)$. Let $V_{3}$ be spanned by $(1,1)$. Then we have $V=V_{1}\cup V_{2}\cup V_{3}$, but none of the $V_{1},V_{2},V_{3}$ are subspace of another.

You can usually count on field of characteristic $2$ to give you counterexample. There are many similar counterexample, too. In finite dimension, I think all counterexamples can be constructed this way. My intuition tells me that there are infinite dimensional counterexamples of other form, but have not checked clearly.

EDIT. Here is a proof of the statement with the restriction $F\not=\mathbb{Z}/2\mathbb{Z}$:

Without loss of generality, we can assume the whole space $V$ is in fact $V_{1}+V_{2}+V_{3}$. Easily seen that in fact we must also have $V=V_{1}\cup V_{2}\cup V_{3}$.

There exist $a,b\in F$ such that $a,b\not=0$ and $a-b=1$ (take $a$ to be anything except $0,1$, and take $b=a-1$).

Assume $V_{1}$ and $V_{2}$ neither contains another (otherwise this reduce to the 2-subspace case). For any $u\in V_{1}\setminus(V_{1}\cap V_{2})$ we take an arbitrary $w\in V_{2}\setminus(V_{1}\cap V_{2})$ (it exists due to the fact that neither $V_{1}$ nor $V_{2}$ contains another). Then $au+w$ is in neither $V_{1}$ nor $V_{2}$ (if in $V_{1}$ then since $au\in V_{1}$ we must have $w\in V_{1}$ so $w\in V_{1}\cap V_{2}$ contradiction; same for the other case but now using the fact that $a\not=0$), so $u+aw\in V_{3}$. Same argument apply to show $bu+w\in V_{3}$. Hence $u=(bu+w)-(au+w)\in V_{3}$. Hence $V_{1}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Same argument apply to show $V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Now for any $v\in V_{1}\cap V_{2}$ we pick a $w\in V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Then $w+v\notin V_{1}\cap V_{2}$ (otherwise $w\in V_{1}\cap V_{2}$). But $w+v\in V_{2}$. Hence $w+v\in V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Thus $v=(w+v)-w$ so $v\in V_{3}$. Hence $V_{1}\cap V_{2}\subset V_{3}$. Therefore $V_{1},V_{2}\subset V_{3}$.

Suppose $W=V_1\cup V_2\cup V_3$ is a subspace. Since $W$ contains $V_1,V_2,V_3$ and is closed under linear combinations, we have $V_1+V_2+V_3\subset W$ as well. The reverse inclusion also holds, since each $V_1,V_2,V_3$ lie in the sum. Thus $$W=V_1+V_2+V_3.$$

Observe the inclusions
$$
W=V_1\cup V_2\cup V_3\subset (V_1+V_2)\cup V_3\subset V_1+V_2+V_3=W.
$$
Thus $(V_1+V_2)\cup V_3=W$, so by the result for two subspaces, either $V_3\subset V_1+V_2$ or $V_1+V_2\subset V_3$. In the former case, $W=V_1+V_2$ so repeating the two-subspace argument yields $W=V_1$ or $W=V_2$. In the latter case, $W=V_3$. The result follows.

Note: The same proof extends by induction to any finite union of subspaces.


Show that if $a > 0$, then $ax^2 + bx + c ≥ 0$ for all values of $x$ if and only if $b^2 − 4ac ≤ 0$. So I'm fairly new to proofs and was wondering how to solve this problem. This is my train of thought.

I listed the terms as so: $ax^2 + bx + c ≥ 0 ≥ b^2 − 4ac$.

From there I tried to complete the square and ended up with this:

$$x ≥ \sqrt{-c/a} ≥ \sqrt{(b^2/a) - 4ac - c/a}.$$

I'm not quite exactly sure how to show that if $a > 0$, then $ax^2 + bx + c ≥ 0$ for all values of $x$ if and only if $b^2 − 4ac ≤ 0$ and would greatly enjoy any help.

Thanks. Forgive my formatting.

Note that $$ax^2+bx+c=a\cdot\left(x+\frac b{2a}\right)^2+c-\frac{b^2}{4a}=a\cdot\left(x+\frac b{2a}\right)^2-\frac{b^2-4ac}{4a}.$$

1.If the $b^2-4ac < 0$ then equation has no real root so graph of the $ax^2+bx+c$ will always be above $X-axis$.

  2.If $b^2-4ac=0$ then equation will have only one real root.

  3.If $ax^2+bx+c$ is Non negative then it has at most one real root so $b^2-4ac$ is nonpositive

so $ax^2+bx+c$ is always non negative iff $b^2-4ac$ is non positive


Can we say that $\text {tr}\ (A) = 0\ $? Let $A$ be an $n \times n$ real matrix with $A^3 + A = 0.$ Can we say that $\text {tr}\ (A) = 0\ $?
I think it's true but can't prove it. Any help will be highly appreciated.
Thanks in advance.

The minimal polynomial of $A$ must divide $x^3+x$.
Then the real Jordan form of $A$ can have, consequently, two kinds of blocks:

0 blocks of dimension 1
2-dimensional blocks associated to rotations of  $\pi/2$

In both cases the trace is equal to 0.
Recalling that the trace of $A$ is invariant for conjugation you have done.
Observe that this is not true over the complex numbers: $A=i I$ satisfies $A^3+A=0$ but $trA \neq 0$.

the trace of the matrix is the sum of its eigenvalues, Here  $i,-i,0$ are  the eigenvalues of the given equation $A^3+A=0$ and the sum of its eigenvalues is zero,
So the trace of the matrix is $0$.
A matrix with real entries often has non-real eigenvalues.


Why not include as a requirement that all functions must be continuous to be differentiable? Theorem: Suppose that  $f : A \to \mathbb{R}$ where $A \subseteq \mathbb{R}$. If $f$ is differentiable at $x \in A$, then $f$ is continuous at $x$.

This theorem is equivalent (by the contrapositive) to the result that if $f$ is not continuous at $x \in A$ then $f$ is not differentiable at $x$.

Why then do authors in almost every analysis book, not take continuity of $f$ as a requirement in the definition of the derivative of $f$ when we (seemingly) end up with equivalent results?

For example I don't see why this wouldn't be a good definition of the derivative of a function

  Definition: Let $A \subseteq \mathbb{R}$ and let $f : A \to \mathbb{R}$ be a function continuous at $a$. Let $a \in \operatorname{Int}(A)$. We define the derivative of $f$ at $a$ to be $$f'(a) = \lim_{t \to 0}\frac{f(a+t)-f(a)}{t}$$
  provided the limit exists.

I know this is probably a pedagogical issue, but why not take this instead as the definition of the derivative of a function?

Definitions tend to be minimalistic, in the sense that they don't include unnecessary/redundant information that can be derived as a consequence.

Same reason why, for example, an equilateral triangle is defined as having all sides equal, rather than having all sides and all angles equal.

The "obvious" (in Cauchy's or Riemann's view) properties of differentiation or continuity lead to the current definitions. The epsilon-delta definition of continuity seems reasonable; it's a bit difficult to find something else. This definition equates to "can draw without lifting the pencil."

The idea of differentiable has to slopes with slopes. One checks (f(x)-f(x-h))/h for existence and other properties. Then the result follows.


First order quasilinear Partial Differential Equation methods What methods, not including the one of characteristic curves and surfaces, can I use to solve first-order quasilinear equations like this one?
$$u_x+u_y=u^2$$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dy}{dt}=1$ , letting $y(0)=y_0$ , we have $y=t+y_0=x+y_0$

$\dfrac{du}{dt}=u^2$ , we have $u(x,y)=\dfrac{1}{f(y_0)-t}=\dfrac{1}{f(y-x)-x}$


Matrix rows notation I'm working with a set of $M$ vectors $ \{\mathbf{w}_i \in \mathbb{R}^N, \, i = 1, \ldots, M \}$. Since single vectors are usually considered as column vectors, I'm defining a matrix

$$
\mathbf{W} = [\mathbf{w}_1, \ldots, \mathbf{w}_M] \in \mathbb{R}^{N \times M}
$$

by placing the vectors as matrix columns.

However, for some descriptions, I need to refer to the matrix rows.

Is there an elegant notation to refer to this matrix rows (preferably with less notation overhead)?

You can represent the rows of $\mathbf W$ by the  $M$-(column-)vectors $\mathbf w'_i :=\mathrm{col}_i\mathbf W^\mathsf T\;(i=1,\dots,N).$ There is no standard notation for this; I chose the prime notation for convenience. You would also need to state your chosen notation explicitly.

If you write
$$\mathbf{W} = [\mathbf{w}_1, \ldots, \mathbf{w}_M]^t \in \mathbb{R}^{M \times N}$$
then the rows of $\mathbf W$ are $\mathbf{w}_1^t, \ldots, \mathbf{w}_M^t$.


Math identities, Poisson summation formula I have derived this following formulas using Poisson summation formula. Can anyone kindly refer if this identity is derived or quoted somewhere? I wanted to be sure.

\begin{equation}
    \sum_{n=0}^{\infty}e^{-\pi\left(n+\frac{1}{2}\right)^2 a }=\frac{1}{2\sqrt{a}}+\sum_{n=1}^{\infty}(-1)^n e^{-\pi n^2/a}, \quad a>0
\end{equation} and
\begin{equation}
    \sum_{n=-\infty}^{\infty}e^{-\pi\left(n+\frac{1}{2}\right)^2 a }=\frac{1}{\sqrt{a}}+2\sum_{n=1}^{\infty}(-1)^n e^{-\pi n^2/a}, \quad a>0
\end{equation}

Here is the proof:

The Poisson summation formula can be stated as
\begin{equation}
    \sum_{n\in\mathbb{Z}} f(n) = \sum_{n\in\mathbb{Z}} \hat{f}(\nu),
\end{equation}
where $\hat{f}(\nu)$ is the Fourier transform of $f(n)$ and
\begin{equation}
    \hat{f}(\nu) = \int_{-\infty}^\infty dn\, f(n) e^{i 2\pi n\nu}.
\end{equation}
Then,
\begin{align}
\sum_{n=-\infty}^{\infty} e^{-\pi a (n+1/2)^2} & =  \sum_{\nu=-\infty}^\infty \int_{-\infty}^\infty dn\, e^{-\pi a (n+1/2)^2 + i 2\pi n\nu} \nonumber\\
& =  \sum_{\nu=-\infty}^\infty \frac{1}{\sqrt{a}} e^{-i \nu\pi -\pi \nu^2/a} \nonumber\\
&= \sum_{n=-\infty}^\infty \frac{1}{\sqrt{a}} (-1)^n e^{-\pi n^2/a} \nonumber\\
&= \frac{1}{\sqrt{a}} + 2\sum_{n=1}^\infty \frac{1}{\sqrt{a}} (-1)^n e^{-\pi n^2/a}
\end{align}
Now the left hand side of above eq.
can be re written as,

\begin{eqnarray}
\sum_{n=-\infty}^{\infty} e^{-\pi a (n+1/2)^2} &=&
    \underbrace{e^{-\pi a/4} }_\text{$n=0$} +
    \underbrace{e^{-\pi a(3/2)^2} }_\text{$n=1$} +      \underbrace{e^{-\pi a(5/2)^2 } }_\text{$n=2$} +     \underbrace{e^{-\pi a(7/2)^2} }_\text{$n=3$} +     \underbrace{e^{-\pi a(9/2)^2} }_\text{$n=4$} +   \nonumber......\\
   &&      \underbrace{e^{-\pi a/4} }_\text{$n=-1$} +
    \underbrace{e^{-\pi a(3/2)^2} }_\text{$n=-2$} +     \underbrace{e^{-\pi a(5/2)^2 } }_\text{$n=-3$} +     \underbrace{e^{-\pi a(7/2)^2} }_\text{$n=-4$} +     \underbrace{e^{-\pi a(9/2)^2} }_\text{$n=-5$} +    ......
\end{eqnarray}
Therefore one can see the $n=0$ term matches with $n=-1$,
$n=1$ term matches with $n=-2$,
$n=2$ term matches with $n=-3$,
$n=3$ term matches with $n=4$ and so on.
As a result the eq. above  can be written as
\begin{eqnarray}
\sum_{n=-\infty}^{\infty} e^{-\pi a (n+1/2)^2} &=&
    2(\underbrace{e^{-\pi a/4} }_\text{$n=0$} +
    \underbrace{e^{-\pi a(3/2)^2} }_\text{$n=1$} +      \underbrace{e^{-\pi a(5/2)^2 } }_\text{$n=2$} +     \underbrace{e^{-\pi a(7/2)^2} }_\text{$n=3$} +     \underbrace{e^{-\pi a(9/2)^2} }_\text{$n=4$} +   \nonumber......)\nonumber\\
     &=&
     2\sum_{n=0}^{\infty}e^{-\pi\left(n+\frac{1}{2}\right)^2 a }.
     \end{eqnarray}
     Combining equations above
     we obtain,
     \begin{eqnarray}
     \sum_{n=0}^{\infty}e^{-\pi\left(n+\frac{1}{2}\right)^2 a }=\frac{1}{2\sqrt{a}}+\sum_{n=1}^{\infty}(-1)^n e^{-\pi n^2/a}\nonumber
 \end{eqnarray}

If anyone thinks any error in derivation, kindly let me know. If anyone knows any reference where this formulas are quoted let me know

I think this are wrong formulas. Consider the first one at $a \gg 1$. The left hand side is exponentially small positive. Whereas the sum at the right hand side tends to the limiting value $-1/2$. Thus this formula can't be correct.


Prove that the union of two normal subgroups is also a normal subgroup I know that the statement "the union of two normal subgroups is also a normal subgroup" is false.

Is there a counterexample to show this?

I can prove that the intersection is normal, but I can't disprove this.

$3\mathbb{Z}$ and $2\mathbb{Z}$ are subgroups of $(\mathbb{Z}, +)$ but the union has elements $2$ and $3$. But $2 + 3 = 5 \notin 3\mathbb{Z} \cup 2\mathbb{Z}$. $(\mathbb{Z}, +)$ is cyclic, and hence abelian, and hence all subgroups of $(\mathbb{Z}, +)$ are normal. But the union of these 2 subgroups is not even a subgroup.

Consider $(\mathbb{Z}, +)$, the group of integers under addition. We can prove that $(2\mathbb{Z}, +)$ and $(3\mathbb{Z},+)$ are normal subgroups of $(\mathbb{Z}, +)$. Consider $2\mathbb{Z}\cup 3\mathbb{Z}= \{x\in \mathbb{Z}\mid  x\in 2\mathbb{Z}\ or \ x\in3\mathbb{Z}\}$. We know that $2,3\in 2\mathbb{Z}\cup 3\mathbb{Z}$, but $2+3=5 \notin 2\mathbb{Z}\cup 3\mathbb{Z}$ because $5 \notin 2\mathbb{Z}$ and $5\notin 3\mathbb{Z}$. Thus, $2\mathbb{Z}\cup 3\mathbb{Z}$ is not a subgroup of $\mathbb{Z}$. Consequently, $2\mathbb{Z}\cup 3\mathbb{Z}$ is not normal subgroup of $\mathbb{Z}$. Therefore, the union of any two normal subgroups need not be a normal subgroup of a group.


Evaluating $\int 7^{2x+3} dx$ I have this integral to evaluate: $\int 7^{2x+3} dx$

u substitution should work and you are left with $\frac{1}{2}\int7^udx$

And the final answer should be: $$\frac{7^{2x+3}}{2\ln7}$$

I wasn't too sure about this... so did I do this correctly?

You do not show detail, but you presumably found $\int 7^u\,du$ by looking up a formula for $\int a^u\,du$. That is perfectly correct. But I  would use a slightly different approach, which is a bit more complicated but does not rely on remembering $\int a^u\,du$.

Note that
$$7^{2x+3}=(e^{\ln 7})^{2x+3}=e^{(\ln7)(2x+3)}. \qquad\qquad (\ast)$$
Let $v=(\ln 7)(2x+3)$. Then $dv=(\ln 7)(2) \,dx$. Substituting, we find that
$$\int e^{(\ln7)(2x+3)}\,dx=\int \frac{1}{2\ln 7}e^v\,dv=\frac{1}{2\ln 7}e^v+C.$$
Finally, by $(\ast)$, $e^v=7^{2x+3}$ so our integral is
$$\frac{1}{2\ln 7}7^{2x+3}+C.$$

Yes it is correct.

except for the + Constant


Are there any calculus/complex numbers/etc proofs of the pythagorean theorem? I have been looking for proofs for the pythagorean theorem that don't use area calculation but calculus, complex numbers or any other interesting ways to proof it.

I would love to see any interesting proof,
Shay

Explanation in terms of linear algebra. From  this blog post by Terence Tao

  The statement $a^{2}+b^{2}=c^{2}$ is equivalent to the assertion that the
  matrices $%
\begin{pmatrix}
a & b \\
-b & a%
\end{pmatrix}%
$ and $%
\begin{pmatrix}
c & 0 \\
0 & c%
\end{pmatrix}%
$ have the same determinant. But it is easy to see geometrically that the
  linear transformations associated to these matrices differ by a rotation,
  and the claim follows.

An elegant proof.  Consider the vector $\mathbf{c}$ in the complex plane: $\mathbf{c}= a + \mathbf{i}b$, where $a$ and $b$ are scalar values, and $a,b > 0$.  Now consider the reflection of $\mathbf{c}$ in the line at an angle of 45 degrees (pi/4 radians) passing through the origin, call that $\mathbf{d}$. $\mathbf{d} = b + \mathbf{i}a$, and $\mathbf{d}$ clearly has the same length as $\mathbf{c}$.

Now take the vector product of c and d.  That has the same length as the square of the length of c, and in terms of polar coordinates, it is at an angle of pi radians (by virtue of the multiplicative properties of complex numbers expressed as polar coordinates).

so $\mathbf{c\cdot d} = i(c^2) = (a+ib)\cdot (b+ia) = ab-ba+i(a^2 + b^2) = i(a^2 + b^2)$.

$c^2 = a^2 + b^2$


Solve a differential equation using variable separation I am trying to solve the following differential equation:

$y'=xy^{2/3}$ with the initial condition: $y(0) = 0$

Here is my progress so far:

We separate the fariables: $\int y'y^{-2/3}dy = \int x dx$

Therefore, $\int y'y^{-2/3}dy = yy^{-2/3} + \dfrac{3}{2}\int yy^{-5/3} = y^{1/3} +\dfrac{3}{2}\int y^{-2/3} = y^{1/3} - 3\dfrac{3}{2}y^{1/3} = y^{1/3} = -\dfrac{9}{2}y^{1/3}$

I know the answer to the problem, and I am supposed to find $3y^{1/3} = 3x^2+C$ which is obviously different from my result

It is $y'=\frac{dy}{dx}$

Therefore the equation is

$y'=xy^{2/3}\Rightarrow \frac{dy}{dx} =xy^{2/3}$

Multiplying both sides by $dx$

$dy =xy^{2/3} \, dx$

Now you can go on with seperating the variables.

$$y'=xy^{2/3}\implies \frac{dy}{dx} =xy^{2/3}\implies \frac{dy}{y^{2/3}}=x\,dx $$ So $$\int \frac{dy}{y^{2/3}}=\int x\,dx \implies 3 \sqrt[3]{y}=\frac 12 x^2+C$$ Then $$\sqrt[3]{y}=\frac 16 x^2+C\implies y=\left(\frac 16 x^2+C\right)^3$$


Can decimal numbers be considered "even" or "odd"? Is the concept of even/odd numbers applicable to decimal numbers? For e.g. - 4.222 is a even number?

I am not saying that this is a way to consider "even" and "odd" numbers, but I think it's quite cool that it looks somehow as a generalization of this notion.

It doesn't make sense to look at the digits of an irrational number to determine its "parity", hence let's just stick with rational numbers.

When we were looking at integers, we said that an integer was even when it was divisible by $2$ and odd when it is not. Instead of using words, we could give much better information, which is to define a map
$$
\begin{align}
\nu_2 : \mathbb N & \to \mathbb N \cup \{ 0 \} \\
n=2^k m & \mapsto k
\end{align}
$$
where in the latter, $m$ is odd. The map $\nu$ then measures how even the integer $n$ is, in the sense that when there are more powers of 2 that divide $n$, it is considered "more divisible by two".

This concept can be generalized to the rational numbers : write $x = a/b$ with $a$ and $b \neq 0$ integers. Write $a = 2^{k_1} a'$ and $b=2^{k_2} b'$, so that $x = 2^{k_1 - k_2} (a'/b')$ and now both $a'$ and $b'$ are odd. Now define $\nu$ again in a similar manner :
$$
\begin{align}
\nu_2 : \mathbb Q & \to \mathbb Z \\
x = \frac{2^{k_1}a'}{2^{k_2}b'} & \mapsto k_1 - k_2.
\end{align}
$$
When $x$ is an integer, we recover the map we had before, so this can be considered as an extension of this map to the rational numbers.

I don't know if you have done some abstract algebra, but I am going to work things out a little more so that readers that might be interested find good information here.

If we replace $\nu_2$ by $\nu_p$, where $p$ is a prime number, all that we did works just fine and we get a map from $\mathbb Q$ to $\mathbb Z$ which indicates how much can a rational be divisible by $p$ (by allowing "negative divisibility" in some sense, because it also measures how many times $p$ divides the denominator). We are getting close to the concept of a discrete valuation, which is defined as the following : given a field $K$, a discrete valuation $\nu$ is a function from $K^{\times}$, the group of units, to $\mathbb Z$, which satisfies the following properties :
$$
\begin{align}
\nu(ab) & = \nu(a) + \nu(b) \\
\nu(a+b) & \ge \min \{ \nu(a),\nu(b) \}
\end{align}
$$
and must also be surjective. This function has many interesting algebraic properties. Recalling informations we had before such as "the set of all even integers form a ring", and just by considering this map and its properties, we can show that
$$
\{ x \in K^{\times} \, | \, \nu(x) \ge 0 \} \cup \{ 0 \}
$$
is a subring of $K$ which contains the identity of $K$. For some examples of how this is relevant, consider this : if we take the map $\nu_2$ we had before (over $\mathbb Q$), proving this fact means that the set of all rational numbers with odd denominator form a subring of $\mathbb Q$ (and so does the set of all rational numbers with denominators coprime to $p$). Valuations are extensively studied, and you can probably look it up on Wikipedia or ask for a reference if you are more interested in this.

Hope that helps!

There are several mistakes in the answers given here. First, rational numbers may have a repeating decimal, like 1/6 = 0.16666... etcetera, so they do not necessarily have a finite expression when written in decimal form. The definition of 'even' and 'odd' makes most sense in the context of exponentiation. If we have an even exponent, then we have two roots (real or imaginary). If we have an uneven exponent, then we have only one.

Now, if r is a rational number r = m/n, make sure you first express it as an irreducible fraction first, so the numerator m and denominator n have no other common divisors than 1 (or –1 when considering negative numbers). But let's look at positive numbers first. If we write r as an irreducible fraction m/n, then m and n cannot both be even. Why not? Because m and n can then both be divided by 2 and m/n is not an irreducible fraction in that case. Let's assume m is even. Hence, n must be odd in that case. We can then write a^(2k/n) as [a^(k/n)]^2. This number will always be positive, because we are squaring something. So it doesn't matter if the term between the brackets, i.e. a^(k/n) has one or two roots: we'll square them and so the result will always be positive.

Now let's assume that m is odd (the second possibility). We can then write a^(m/n) as [a^(1/n)]^m. So now it will depend on whether or not n is even. If n is even, we have two real roots, if n is uneven, then we have only one. You can work a few examples yourself.

So we have two roots if m is odd and n is even, and only one root in all other cases. However, we said that m and n cannot both be even, hence, if n is even, m must be odd. In short, we can say that a rational exponent m/n is even (i.e. there will be two roots) if n is even.

So where are these even numbers on the real line? They are everywhere: we can start from 1/2 and then change the numerator: 3/2, 5/2, etcetera. It's all fine, as long as we use an odd number. However, we can also go down and change the denominator: 1/4, 1/6, 1/8 etcetera. And then we can, of course, take odd multiples of these fractions once again, such as 1025/1024 = 1.0009765625, for example, or on the other side, 1023/1024 = 0.9990234375. So we have two even numbers here right next to the odd number 1. We may increase the precision: we could take 3587/3588 and 3589/3588 for example.

Of course, you may have noticed something here. The first thing, of course, is that we've defined these two even numbers 1.0009765625 and 0.9990234375 with a precision of 10 digits behind the decimal point, i.e. 1/1024 = 1/(2^10) = 0.0009765625. The second point to note is that the last digit of these two rational coefficients, when expressed as a decimal, was 5. Now, you may think that should always be the case because of that 1/2 factor. But it's not true: as mentioned above, 1/6 is an example of a rational number that, written in decimal form, will yield 0.166666... This is an expression with a recurring decimal. And 1/10, of course, just yields 0.1. So there's no easy rule here. You need to look at the fraction itself, and rational numbers are either as a finite decimal or an infinite repeating decimal. Of course, there are rules for that, but you can Google more stuff on that if you're interested in this.


Are all fractional deriviatives/integrals of $e^x$ equal to $e^x$? I have learned through calculus that the derivatives and the indefinite integrals of the exponential function are the same (at integer arguments) but was wondering if this holds true for fractional derivatives/integrals, such as $\frac{1}{2}$, or do they turn into something monstrous?

I would test this in Wolfram Alpha but I cannot seem to get the entry correct for a fractional derivative/integral.

Not monstruous, but not very simple. The fractional integrals/dérivatives of the exponential function involves the Incomplete Gamma function :
Page 10, section 6, in the paper "The fractionnal derivation"
http://www.scribd.com/JJacquelin/documents

This depends on the definition of fractional derivatives but the definition based on Newton series interpolation over positive derivatives gives all fractional derivatives to be the same, $e^x$:

$$f^{(s)}(x)=\sum_{m=0}^{\infty} \binom {s}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$

The same goes for the definition based on Fourier transform:

$$f^{(s)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\omega^s \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

The both definitions coincide, and I consider them natural unlike other definitions that may produce a different result.


Is there a norm on ${\Bbb R}^{\Bbb N}$ Let $E={\Bbb R}^{\Bbb N}$ be the real vector space of real sequences.

1) Is there a norm on $E$?

2) Is there a norm $N$ on $E$ such that the restriction of $N$ to $\ell^2$ is finer than the standard norm $N_2$ of $\ell^2$ ?

Pick a basis, and define the norm of a vector to be the sum of the absolute values of the coefficients of that vector in the basis.

For your extra requirement: Find a subspace $U$ of $\mathbb R^\mathbb N$ such that $U\oplus\ell^2=\mathbb R^\mathbb N$. Put on $\mathbb R^\mathbb N$ the norm which is the sup of the norm of $\ell^2$ and any norm (constructed as in my first paragraph) on $U$. This restricts to the norm of $\ell^2$, of course.

There is no norm on the entire $E=\Bbb R^{\Bbb N}$ (with the property that $||x||=0 \implies x=0$).

Suppose such norm exists. Let's denote the elements of $E$ as $a=(a_1,...,a_n,...)$ and suppose $||a_i||=\alpha_i>0$. Then form a vector $b=(b_1,...,b_n,...)$ such that $||b||=\infty$.

Here's how you do that: start with $b_1=2^1\frac{a_1}{||a_1||}$, then choose between $b_2=\pm 2^2\frac{a_2}{||a_2||}$, depending for which $b_2$ the norm is greater, etc. This would make the norm of $b$ unbounded.


Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$. Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

I found a solution, but it involves a lot of case work. Can someone help me find a solution which doesn't involve a lot of case work?

Let $x$, $y$ and $z$ be natural numbers satisfying
$$x+2y+2z=xyz.$$
If $xyz=0$ then $x=y=z=0$, so suppose $x,y,z>0$. A bit of algebra shows that
$$(xy-2)(xz-2)=x^2+4.\tag{1}$$
Of course without loss of generality $y\leq z$. If $x=1$ then the above simplifies to
$$(y-2)(z-2)=5,$$
and so $y=3$ and $z=7$. Otherwise $x\geq2$ and then from $y\leq z$ it follows that
$$(x+1)^2>x^2+4=(xy-2)(xz-2)\geq(xy-2)^2,$$
and hence that $x\geq xy-2$, or equivalently $2\geq x(y-1)$. This means that either $x=y=2$, in which case $z=3$, or $y=1$, in which case $(1)$ simplifies to
$$(x-2)(xz-2)=x^2+4.\tag{2}$$
Clearly $x\neq2$, and so from the identity
$$xz-2=\frac{x^2+4}{x-2}=x+2+\frac{8}{x-2},$$
we see that $x-2$ divides $8$, so $x-2\in\{1,2,4,8\}$. Plugging these values into $(2)$ yields
\begin{eqnarray*}
1\cdot(3z-2)&=&13,\\
2\cdot(4z-2)&=&20,\\
4\cdot(6z-2)&=&40,\\
8\cdot(10z-2)&=&104.
\end{eqnarray*}
The first three have the solutions $z=5$, $z=3$ and $z=2$, respectively, and the last has no integral solution.
In summary, up to permutation of $y$ and $z$, the solutions $(x,y,z)$ are
$$(0,0,0),\quad(1,3,7),\quad(2,2,3),\quad(3,1,5),\quad(4,1,3),\quad(6,1,2).$$


Proving that $\lim_{n\to \infty}\frac{(-1)^n(n+1)!}{n^{n+1}}=0$ Prove that $$\lim_{n\to \infty}\frac{(-1)^n(n+1)!}{n^{n+1}}=0$$

I noticed that it can be expressed in the form of $n+1$ terms product but I have no idea with the upper bound.

I have trouble with how to operate $(-1)^n$

Let us ignore the $(-1)^n$ for the time being

$$0<\dfrac{(n+1)!}{n^{n+1}}=\dfrac{n+1}n\cdot \prod_{r=2}^n\dfrac rn\cdot\dfrac1n<\dfrac{n+1}n\cdot\dfrac1n$$


Show that the angles satisfy $x+y=z$ How can I show that $x+y=z$ in the figure without using trigonometry? I have tried to solve it with analytic geometry, but it doesn't work out for me.

Imagine that all the the squares are 1 by 1 and so the rectangle has base 3 and a height of $1$.

There are three right-angled triangles in the diagram. The one with angle $x$ has base 3, and height 1. The triangle with angle $y$ has base 2 and height 1. The triangle with angle $z$ has base 1 and height 1.

Using the standard trig' ratio $\tan \theta = \frac{\mathrm{opp}}{\mathrm{adj}}$, we get $\tan x = \frac{1}{3}$, $\tan y = \frac{1}{2}$ and $\tan z= \frac{1}{1}=1$.

There is a well-know formula for angle addition:

$$\tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$

Applying this formula to the case of $\alpha =x$ and $\beta = y$ gives:
$$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\cdot\frac{1}{2}}=1$$
It follows that $\tan(x+y)=\tan z$. Since $0^{\circ} < x<y<z < 90^{\circ}$ it follows that
$$\tan(x+y) = \tan z \iff x+y = z$$


Do we need _strict_ monotonicity for the fixed point lemma for normal functions? I was reading about the fixed-point lemma for normal functions and I was wondering if we could weaken strict monotonicity to non-decreasing. I see why it's necessary for the proof, but is there a continuous non-decreasing ordinal operation without a fixed point? I'm probably missing something obvious but I can't think of any.

You don't need that $f$ is strictly increasing to have at least one fixed point. Consider the sequence $\alpha_n = f^n(0)$. By induction it follows that this is non-decreasing. If $\alpha_n=\alpha_{n+1} $ for some $n$ then we are done as this means that $\alpha_n = f(\alpha_n)$. Else consider the sup of this sequence and follow by continuity that this is a fixed point.

Of course you cannot get arbitrarily large fixed points as witnessed by the constant $0$ function.

With $\ell(\alpha)=\min\{\,\lambda\mid \lambda\text{ is limit ordinal}, \lambda>\alpha\,\}$, let
$$ f(\alpha)=\begin{cases}\ell(\alpha)&\text{if }\alpha\text{ is limit ordinal}&\\\ell(\ell(\alpha))&\text{otherwise}\end{cases}$$
Then $f$ is continuous, non-decreasing, and has no fixed point.


How to approach proving the limit of $[\cos(x) - 1]$ doesn't exist? I'm wondering what the most graceful/accurate way to state the limit of $\cos(x) - 1$ as $x$ approaches $-\infty$ doesn't exist. Intuitively, this limit can't exist; but I'm unsure if using the fact that all subsequences of a sequence must converge for the limit to exist would work here (My original approach) as this seems more like the limit of a function rather than a sequence (Even when you define a sequence as having elements equivalent to this).
Could I get some opinions/headings?

Hint:

Check the sequences $${x_n}=-\frac{2n+1}{2}\pi$$ and $${y_n}=-4n\pi$$, converging to negative infinity. Then put them in the function and find it violates seqential definition of limit of function.

A periodic function can't have a limit at infinity.
Because a periodic function is not constant and in a period you will find a value that differs from the would-be limit by a finite amount. This difference repeats in every period.

Formally:
$$\forall\epsilon>0:\exists M:\forall x>M\implies |p(x)-L|<\epsilon.$$
Let us take any $x_0$ such that $|p(x_0)-L|=\eta>0$; such an $x_0$ necessarily exists as the function is non-constant. Then by periodicity,
$$\forall\epsilon>0:\exists M: x_0+kT>M\implies |p(x_0+kT)-L|=\eta<\epsilon$$ is false for $\epsilon<\eta$.


Finding the remainder of $823^{823}$ in the division by 11 I wish to find the remainder $823^{823}$ in the division by $11$.

I used the  Euler-Fermat theorem  that tells that: $a^{\phi(n)} \equiv 1$ $(\mod n)$, whenever $(a,n) =1$
First, Euler- Fermat tells us that $a^{\phi(11)} \equiv 1 (\mod 11)$ whenever $a$ is coprime with to 11. Since 823 is prime, we have, $(823,11) = 1$.
Now $823 \equiv 9 (\mod 11)$

But I don't know how to continue.. Because both $823$ and $11$ are prime ; do I have to use Fermat's little theorem?

Any help appreciated

Note that you also have $823\equiv -2\mod11$, which will make calculating powers by hand easier. Now by lil' Fermat,
$$823^{823}\equiv (-2)^{823\bmod 10}=-8\equiv 3\mod 11.$$

Note that $$823^5\equiv 1 \mod 11$$ and then you can square the equation.


How many lines are needed to create 6 triangles on W? Basically, the question started with a little argument I had with my friend. My friend said he thinks it's possible to draw only 2 lines on the letter "W" and make 6 triangles, and I played around with it, but I couldn't really do it, so I told him I don't think it's possible, and we need at least 3 lines.

We kept arguing, but I really couldn't prove that we need more than 2 lines on letter "W" to create 6 triangles except that I couldn't find a way of doing so.
Is there any proof for this kind of problem?

PS. Sorry. I didn't clarify. You CANNOT combine one or more triangles to form a triangle and count that separately. Triangles must be INTERNALLY disjoint.
For instance, if triangles ABC, BCD can be combined to form triangle ADC, then we don't count ADC as a triangle.

No one said the W couldn't be "cursive"...

Perhaps not what the OP was looking for (and I'm aware this post was brought back from YEARS AGO, but I just thought I'd share this) but this solution does contain $6$ triangles with only $2$ lines

Three parallel lines one at the top, one middle, one bottom, and, if tri inside tri is allowed, you have $6$. Of these, $3$ are true and $3$ more are built around their respective parent.


Eigenvalues of the transpose I have tried this problem and I'm unsuccessful.

Let $T: M(n,n)\rightarrow M(n,n)$ be the mapping defined by $ T(A) = A^T$. Show that the only eigenvalues of $T$ are $\pm 1$.

@JohannesHahn has the best answer but here's something a bit more explicit.

Your map is $T:M_{n}\to M_n$ defined by $T(A)=A^\top$. An eigenvector of $T$ is a nonzero $n\times n$ matrix $A$ such that $A^\top=\lambda\cdot A$. Since $A$ is nonzero, it has a nonzero entry $a_{ij}$. But the equation $A^\top=\lambda\cdot A$ implies
$$
a_{ji}=\lambda a_{ij}\tag{1}
$$
and
$$
a_{ij}=\lambda\cdot a_{ji}\tag{2}
$$
Now, plugging (1) into (2) gives
$$
a_{ij}=\lambda^2 a_{ij}
$$
so $\lambda^2=1$.

The map $T$ that sends a matrix to its transpose is a linear map between vector spaces. This means that once you choose a basis you can write $T$ as a linear map and then find its eigenvalues.


Find the number of integral solutions of $(x,y)$ Given this equation: $4x^3+5=y^2$

Find the ordered pairs of $(x,y)$ where $x,y\in Z$

This is an elliptic curve, and it would appear that it has infinitely-many rational points (generated by (1,3)).  It is also an example of "Mordell's Equation" - curves of the form $y^2 = x^3 + D$ (in your case D = 80).  Many things are known about its integral solutions.  You might find this article by Keith Conrad to be interesting.  The Wikipedia article on the subject links to a large source of data, as well.

$4x^3+5=y^2$

(1)$4(x^3+1)=y^2-1$ Therefore $x=-1,y=∓1$ are solutions

(2) $4(x^3-1)=y^2-9$ Therefore $x=+1,y=∓3$ are solutions

You need integer solutions, No?. Then there is  no any other integer solutions. If you can’t prove this using High school mathematics  let me know.

P.Ranawaka


What are "complementary pair-wise comparable functions"? I got this term while studying periodic functions; my book writes:

  If $f_1(x)$ & $f_2(x)$ are periodic functions with periods $T_1$ & $T_2$ respectively, then we have $h(x) = f_1(x) + f_2(x)$ has period, as $\dfrac{1}{2} \text{L.C.M. of }\; \{T_1 ,T_2\}$, if $f_1(x)$ & $f_2(x)$ are complementary pair-wise comparable functions.

Don't know how the author deduced the rule. But my main question is what do this "complementary, pair-wise, comparable function" mean?  I've googled this but it was in vain.

I think it means if you put $(π/2-x)$ in place of $x$ in one of the functions and get the second function, then they are called complementary. Like $sin(x)$ and $cos(x)$.

Consider a set of functions $F$.

The set is pairwise comparable if, for any two functions $f(x)$ and $g(x)$ within the set $F$, we know that either:

$f(x) \ge g(x)$ for any value of $x$, or
$f(x) \le g(x)$ for any value of $x$.


period of product markov chain Consider $Z_n := (X_n,Y_n)$ where $(X_n)_{n\in \mathbb{N}}$ and $(Y_n)_{n\in \mathbb{N}}$ are irreducible markov chains with periods $\lambda$ and $\mu$.
We know that $(Z_n)_{n\in \mathbb{N}}$ is a Markov Chain then. Work on the following problem:

$\mu$ and $\lambda$ have no common divisor $\iff$ $(Z_n)_{n\in \mathbb{N}}$ is irreducible; further $(Z_n)_{n\in \mathbb{N}}$ has period $\lambda \mu$ then.

I do not manage. See below:

We need to show that $\forall i,j,i',j' \exists n \in \mathbb{N}: P(Z_n = (i',j') | Z_0 = (i,j)) > 0$ $\iff$ $\operatorname{gcd}(\mu,\lambda) = 1$.

$\Rightarrow$ direction:

Consider irreducibilty, then the probabilities are always greater than zero. For an $n \in \mathbb{N}$ especially we have $P(Z_n = (i,j) | Z_0 = (i,j)) > 0$. So we have $n|\mu$ and $n|\lambda$.

But how does this impy that $\lambda$ and $\mu$ have no common divisor?

$\Leftarrow$ direction:

I do not know. It seems that if you have no common divisor you will manage to reach every tuple $(i',j')$ starting from any $(i,j)$ since you can reach every $i'$ from $i$ since X is irreducible and the same goes for the $j$'s. But I do not manage to get this formally written down..

Hints? Is this the right way to go?

[Drawing out a simple case, e.g. two circles of length 2 might help visualising the problem (their product becomes a "cross").]

So as OP suggests, for irreducible $Z$ we need $\forall i, i', j, j'$ $\exists n : P_{(i,j)(i',j')}^{(n)}>0$, i.e. there is some $n$ for which the $n$-step transition probability from $(i,j)$ to $(i',j')$ is nonzero (for all pair of vertex tuples). Since $X$ and $Y$ are both irreducible, the real question here is whether it is possible to go $i \rightarrow i'$ (in $X$) and $j \rightarrow j'$ (in $Y$) in the same number of steps.

Let the directed distances  $i \rightarrow i'$ be a mod $\bf\lambda$
and  $j \rightarrow j'$ be b mod $\bf\mu$. This is unique (well-defined) for any pair as both chains are irreducible with periods $\lambda$ and $\mu$.* Then we want $\forall a, b$ $\exists$ $l,m,n$ so that $n=l\lambda +a$ and $n=m\mu+b$, which means that we can get $i\rightarrow i'$ (in $X$) in $a$ steps then we wait there for $l$ full periods while a similar thing happens in $Y$; all these during $n$ steps (for some $n$). This allows to get $(i,j)\rightarrow(i',j')$ in exactly $n$ steps.

WE then turn to number theory. Eliminating $n$ and rearranging, we get $l\lambda-m\mu=b-a$. Having a solution $l,m$ for any $a,b$ corresponds to an irreducible $Z$. This is a linear Diophantine equation in variables $l$ and $m$, and has a solution iff gcd$(\lambda,\mu)=1$ (otherwise there is some $a-b$ which is not divisible by gcd$(\lambda,\mu)$, whereas the left hand side is). In fact, if there is one solution then there are infinitely many.

To sum up: ($Z$ is irreducible) $\Leftrightarrow$ (equation has a solution) $\Leftrightarrow$ gcd$(\lambda,\mu)=1$. The above also shows that in case of a larger gcd there will be multiple irreducible classes which are not communicating with each other.

*If there were two paths $i\rightarrow i'$ with different lengths (mod $\lambda$), then the difference would be smaller than $\lambda$, so (if we go back $i'\rightarrow i$ on the same path) there are also two circles $i\rightarrow i$ the lengths of which (mod $\lambda$) differ by less than $\lambda$. But this contradicts the definition of period, which requires $\lambda$ to be a divisor of all circle lengths.


Joint convexity of the following function Let $x,x_0 \in \mathbb{R}^n$ and $Y \in \mathbb{R}^{n \times n}$ Is the following function convex as a function of $(x,Y)$?

$f(x,Y) = ||(x-x_0)-Y(x-x_0) ||^2_2$

I know that this function is convex w.r.t. each argument alone.

First of all, if $E$ is a (Banach) space and $g\,\colon\,E\to\mathbb{R}$ is a linear function, then $g^2$ is convex since

$$g^2\left(tx + (1-t)y\right) = g^2(tx) + 2g(tx)g\left((1-t)y\right) + g^2\left((1-t)y\right) = t^2g^2(x) + (1-t)^2g^2(y) + 2t(1-t)g(x)g(y) = tg^2(x) + (1-t)g^2(y) + t(1-t)\left(g(x) + g(y)\right)^2.$$

Then, $\|Yx\|_2^2$ is a sum of squares of linear functions hence a sum of convex functions hence convex. Finally,

$$f(x, Y) = \left\|(Y-I)(x-x_0)\right\|_2^2$$

is the same function shifted by $(x_0, I)$ hence also convex (here $I$ stands for the identity matrix).


How does one attack a divisibility problem like $(a+b)^2 \mid (2a^3+6a^2b+1)$? In my current line of investigation, I am running into [many] divisibility questions like the one in the title, i.e.
$$
  (a+b)^2 \mid (2a^3+6a^2b+1),  \qquad(\star)
$$
where $a > b \ge 1$ are integers. For example, maxima calculations suggest that $(\star)$ implies $(a,b)=(4,1)$.

I really haven't the first clue how to effectively attack this problem. Any pointers and references would be greatly appreciated.

EDIT: I've now cross-posted a partial solution to this.

$(a+b)^2|2a^3 +6a^2b+1$ iff $k(a^2+2ab+b^2) = 2a^3 +6a^2b+1$ or $-2a^3+a^2(k+6b)+a(2kb)+(kb^2+1)=0$. This is a a cubic equation so one could find $(a,b)$ by solving for $a$, but then one would still have to iterate through the $k$ and $b's$ to find an integer value of $a$ and $b$. This is also an elliptic curve, specifically $y^2 = -2a^3+a^2(k+6b)+a(2kb)+(kb^2+1)$. One could reduce this into Weiertrauss form using the cubic reduction formula to obtain something like $y^2 = a^3 + pa + q$. Anyway, elliptic curves have either finitely many or infinitely many solutions. We are searching for integer values of $a,b$ so we can apply what we know about solving for rational points on elliptic curves. If we are looking for rational points, then the rational points $E(\mathbb{Q})$ form an abelian group and is finitely generated, i.e. there is a finite number of rational points that one can use to find the rest of the points.

Solving your problem is the same as finding rational points on an Elliptic Curve and there are lots of tricks to do that, many of which can be found online. As far as I know, there is no formula for solving elliptic curves, just methods to find solutions. However, your question is even more complicated than that, because two of values in your elliptic curve are undetermined.

This link provides some examples of how to solve Elliptic Curves if your $k,b$ were determined. However, even with them determined, one would still use an algorithm to find the points. http://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf

This is not even close to a complete solution, it will just help to reduce the number of calculations that you have to do on Maxima.

Consider $2a^3 + 6a^2b + 1 \equiv 1 \pmod{2}$ and $(a+b)^2 \equiv 1 \pmod 2$. We also have that $2a^3 + 6a^2b + 1 \equiv 0 \pmod{3}$ and $(a+b)^2 \equiv 1 \pmod{3}$. Combining these we have that $a+b = 6k+1$ for some $k\in\mathbb{Z}$.


Using Strong Induction The statement is: If $n$ is an integer with $n > 1$, then $n$ has a prime factor. I know you have to use $P(k+1)$ but I'm not sure how to implement that into the proof. Is this even going in the right direction?

All $n > 1$ are a factor of themselves, therefore all primes have a prime factor which is themselves.  If we assumed there was a number X that was the smallest number to have no prime factors, then it would not be able to prime and it would have to have at least one factor p, smaller than itself (otherwise it would be prime).  If p itself has a prime factor r, then r would also be a factor of X - since it can be seen that if X = qp, and q = rs, then  r divides into X sp times

so p can have no prime factors - but we said that X was the smallest number with no prime factors, and p is smaller than X, so by contradiction, X can not exist.

This is a perfect example of strong vs. weak induction.  We don't care in the slightest bit that $n+1$ is the number immediately after $n$.  We care only that $n+1$ is larger than $1....n$.

Base case: $n =2$.  $n = 2 $ has a prime factor because $2$ itself is prime.

Induction step:  Assume that every number from $2....n$ have prime factors.

$n+1$ has factors.  Because all numbers have factors.  $n+1$ has $1$ and $n+1$ as factors.  Does it have any others?

If not, then $n+1$ is prime and that is a prime factor.

Otherwise if $n+1$ has a factor $k$, $k \ne 1; k \ne n+1$ where $k|n+1$....

Well, here are some questions for you:

Is  $k > n+1$?

If not then ... what is it compared to $n+1$?
..... So what else can we assume about $k$?  ... because we are doing strong induction so we are making an assumption of what...?

So what does that tell us about $n+1$?

====

The trick is on weak induction we would assume the property is true for $n$ to get that it is true for $n+1$.

In this strong induction we are not doing that.  We are assuming it is true on some $k$ that is not $n$ and using that it is true on $k$, and therefore it is true on $n+1$.  We are using i) $k|n+1$ and $k < n+1$.  The number $n$... is completely off the table and has nothing to do with anything.


About the solvability of certain equations How one can see if this equation has real solutions:

$$x^{2^{k}}-x-a=0$$ where $x$ is the unknown, $k$ is a positive integer and $a$ is a real constant.

Use Descartes' Rule of Sign: Looking at the polynomial

$$ 1\cdot x^{2^k} +0\cdot x^{2^k-1} + 0\cdot x^2+1\cdot x -a =0$$

you can count the number of sign changes in the sequence of coefficients. The number of sign changes is an upper bound on the number of positive roots: If there are $c$ sign changes, then the polynomial has $c-2k$ positive roots, where $k$ is integer. Hence, if $c=1$, the polynomial has exactly one positive root, if $c=2$, the polynomial has either no or two positive roots, etc.

In our case the sequence of coefficients is $(1,0,0,\dots,0,1,-a)$. There is exactly one sign change in the polynomial, hence there is exactly one positive root. If you replace $x$ by $-x$, you get

$$ x^{2^k} +x -a =0$$

which has again only one sign change. Hence there is exactly one negative root.


For the line $y = mx$, let $m = tan(\theta)$. Write $f(x, mx)$ as a function of $\theta$..? I have a problem, and I am not sure how to solve it.
This is the problem from my book:

  let $f(x, y)$ be given by the function: $$ f(x, y) =  \begin{cases}
       \frac{2xy}{x^2 + y^2}, & (x, y) \ne (0, 0) \\
       \
      0, & (x, y) = (0, 0)      \end{cases} $$ for the line $y = mx$, let $m = tan(\theta)$. Write $f(x, mx)$ as a function of $\theta$.

  Use the
  answer to the above to show how the value of $f(x, y)$ varies with the
  line’s angle of

  inclination.

Hint: Start by trying to write $f$ as a function of $r$ and $\theta$ (as in polar coordinates).


Show that any cyclic group of even order has exactly one element of order $2$ Show that any cyclic group of even order has exactly one element of order $2$

Attempt:

Let $G$ be a finite group of even order.

Assuming a set $A=\lbrace g \in G \vert g \neq g^{-1} \rbrace$, I can show that  $A\cup\{e\}$ has odd number of elements as $A$ has even number of elements. So $G$ has at least one element $a$ such that $a$ does not in $A$, ie $a=a^{-1}$ ie $a^2=e$. That is $G$ has atleast one element of order $2$. "More specifically, $G$ has odd number of elements of order $2$ --- ???"

But how to show that exactly one such element if $G$ be cyclic.

  NB: Please don't use isomorphic properties.

If $G$ is a finite cyclic group of order $n$, then $G$ has a unique subgroup of order $m$ for each $m$ dividing $n$. Now, if $n$ is even, then $G$ has a unique subgroup of order $2$, hence one element of order $2$ (as if $G$ had two distinct elements of order $2$, then we'd have two distinct subgroups of order 2).

Let $g$ be a generator of $G$ and $G$ has order $n$.  Suppose, $(g^m)^k=e$. Then $km=nr$ for some integer $r$. Let $d=(m,n)$ and $l=mn/(m,n)$.  Note that $l$ is the least common multiple of $n$ and $m,n$ and $km=nr=lt$ for some $0\le t<d$. Now, $k=n/d$ and $o(g^k)=d$. Letting $m=2$, we get that $o(g^{m/2})$ is the only element of order $2$.


trigonometry solution $$\frac{\sin A - \cos A}{\sin A+\cos A}=\frac{X}{Y}$$
Then prove that $X^2 + Y^2 = 2$.
Answer :
$\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{x}{y}$
$\implies x=k(\sin A-\cos A)$ and
         $y=k(\sin A+\cos A)$.

Squaring on both sides and adding

$x^2+y^2=k^2(\sin A-\cos A)^2+k^2(\sin A+\cos A)^2$

$\implies x^2+y^2=k^2[\sin^2A+\cos^2A-2\sin A\cos A+\sin^2A+\cos^2A+2\sin A\cos A]$

$\implies x^2+y^2=2 k^2$
But how can eliminate k.means by what logic we can say k=+1 or -1.
Please suggest. Thanks.

It is impossible to prove that result, mainly because it is not correct. Take $A = 0$ then $$\frac{\sin A - \cos A}{\sin A + \cos A} = -1$$ now take $X = k$ for some $k \neq \pm1$ and take $Y = -k$ so that $$\frac{X}{Y} = -1$$ but since $k \neq \pm1$ then $$X^2 + Y^2 = 2k^2 \neq 2.$$

You can simply substitute $X$ with $(\sin A - \cos A)$ and $Y$ with $(\sin A + \cos A)$. If we do this we see that the first equality is obviously satisfied. The $k$ term that you multiply this by is unnecessary. You can see how I proved this here.


What’s the difference between analytical and numerical approaches to problems? I don't have much (good) math education beyond some basic university-level calculus.

What do "analytical" and "numerical" mean? How are they different?

Analytical approach example:

Find the root of $f(x)=x-5$.

Analytical solution: $f(x)=x-5=0$, add $+5$ to both sides to get the answer $x=5$

Numerical solution:

let's guess $x=1$: $f(1)=1-5=-4$. A negative number. Let's guess $x=6$: $f(6)=6-5=1$. A positive number.

The answer must be between them. Let's try $x=\frac{6+1}{2}$: $f(\frac{7}{2})<0$

So it must be between $\frac{7}{2}$ and $6$...etc.

This is called bisection method.

Numerical solutions are extremely abundant. The main reason is that sometimes we either don't have an analytical approach (try to solve $x^6-4x^5+\sin (x)-e^x+7-\frac{1}{x} =0$) or that the analytical solution is too slow and instead of computing for 15 hours and getting an exact solution, we rather compute for 15 seconds and get a good approximation.

The easiest way to understand analytical and numerical approaches is given below:
pi=22/7 is the approximate value which is numerical
1/2=0.5 is the exact value means analytic.


finding $P(X_1>X_2\mid X_1>X_3)$ in Exponential distribution suppose $X_1,X_2,X_3$ are independent random variables with exponential distribution and means $\frac{1}{\lambda1},\frac{1}{\lambda2},\frac{1}{\lambda3}$ , how can find $P(X_1>X_2\mid X_1>X_3)$

$$\begin{eqnarray}
  \Pr\left(X_1>X_2 \mid X_1>X_3\right) &=& \frac{\Pr(X_1 > X_2, X_1>X_3)}{\Pr(X_1>X_3)} \\ &=& \frac{\mathbb{E}\left( \Pr(X_1 > X_2, X_1>X_3 \mid X_1) \right)}{\Pr(X_1>X_3)} \\
&=& \frac{\mathbb{E}\left( \Pr(X_2<X_1\mid X_1)\Pr( X_3<X_1 \mid X_1) \right)}{\Pr(X_1>X_3)}
\\
&=& \frac{\mathbb{E}\left( F_{X_2}(X_1)F_{X_3}(X_1) \right)}{\mathbb{E}\left( F_{X_3}(X_1) \right)} \\
 &=& \frac{\int_0^\infty \left(1-\mathrm{e}^{-\lambda_2 x_1}\right)\left(1-\mathrm{e}^{-\lambda_3 x_1}\right) \lambda_1 \mathrm{e}^{-\lambda_1 x_1} \mathrm{d}x_1}{\int_0^\infty \left(1-\mathrm{e}^{-\lambda_3 x_1}\right) \lambda_1 \mathrm{e}^{-\lambda_1 x_1} \mathrm{d}x_1 } \\ &=& \frac{\lambda_2 \left(2 \lambda_1+\lambda_2+\lambda_3\right)}{\left(\lambda_1+\lambda_2\right) \left(\lambda_1+\lambda_2+\lambda_3\right)}
\end{eqnarray}
$$

Hint: Memoryless property implies $$P(X_1 > X_2\mid X_1 > X_3) = P(X_1 > X_2-X_3).$$


How to show that $ALL_{DFA}$ is in P How can I show that $ALL_{DFA}$ is in P ?

$ALL_{DFA}  = \{ \langle A \rangle \mid A \text{ is a DFA and } L(A) = \Sigma^* \}$

Note that a DFA accepts $\Sigma^*$ if and only if all reachable states from the start state, $q_0$, are accepting. This can easily be decided in polynomial-time by performing a breadth- or depth-first search on the DFA from $q_0$.  If at any time a non-accepting state is visited, reject, otherwise, if only accepting states are found, accept.

Interestingly, this problem is much harder for NFAs; $\{ \langle A \rangle \mid A \text{ is an NFA and } L(A) = \Sigma^* \}$ is NP-hard.

My answer to this problem on a recent homework was originally similar to the other answer on this question: Perform a breadth first search on the input, If a non-accept state is visited reject, Otherwise accept.  However, this solution is wrong. This decider will accept a DFA that does not accept all inputs if, for example, there is no transition for one of the characters in Σ. The following DFA (with an alphabet of 0 and 1) is an example of this. It will not accept inputs with a 1, but there is no reject state that will cause this DFA to be rejected by the decider.

The answer is actually to construct the complement to the input and test whether the language of the complement is an empty set.  You can do this by doing a breadth or depth first search on the complement and if an accept state is found, reject the input and otherwise, accept.