Integration by partial fractions; how and why does it work? Could someone take me through the steps of decomposing
$$\frac{2x^2+11x}{x^2+11x+30}$$
into partial fractions?

More generally, how does one use partial fractions to compute integrals
$$\int\frac{P(x)}{Q(x)}\,dx$$
of rational functions ($P(x)$ and $Q(x)$ are polynomials) ?

This question is asked in an effort to cut down on duplicates. See Coping with *abstract* duplicate questions.

Also see List of Generalizations of Common Questions.

What you are trying to do is a partial fraction decomposition.
Idea. Imagine your calculator is broken and a bunch of hoodlums stop you on the street, and demand at knifepoint that you compute $\frac{191}{105}$ as a decimal (part of their initiation into the Mathies Gang, you see), or else they will slit your throat. Unfortunately, since your calculator is broken, you can really only do divisions if the divisor is a single digit number (so you can use your fingers to do the operations; you're very good with those, because you know the  multiplication tables of single digit numbers...). Luckily, you do notice that $105 = 3\times 5 \times 7$. Is there some way you can save your neck using this observation?
Well, you do know that $191 = 105 + 86$, so at least you know that $\frac{191}{105} = 1 + \frac{86}{105}$, so that takes care of the integer part of the fraction. What about $\frac{86}{105}$? Aha! Here's a clever idea: maybe $\frac{86}{105}$ is really the result of a sum of fractions! If you had a sum of fractions of the form
$$\frac{A}{3} + \frac{B}{5} + \frac{C}{7}$$
then to write it as a single fraction you would find the common denominator, $105$, and then do a bunch of operations, and end up with a fraction $\frac{\mathrm{something}}{105}$. If you can find an $A$, $B$, and $C$ so that the something is $86$, then instead of computing $\frac{86}{105}$ you can do $\frac{A}{3}$, $\frac{B}{5}$, and $\frac{C}{7}$ (which you can do, since the denominators are single digit numbers), and then add those decimals to get the answer. Can we? We do a bit of algebra:
$$\frac{A}{3} + \frac{B}{5} + \frac{C}{7} = \frac{35A + 21B + 15C}{105}$$
so you want $35A + 21B + 15C = 86$. As luck would have it, $A=B=1$ and $C=2$ works, so
$$\frac{86}{105} = \frac{1}{3} + \frac{1}{5} + \frac{2}{7}.$$
And now, all is well:
\begin{align*}
\frac{191}{105} &= 1 + \frac{86}{105}\\\
&= 1 + \frac{1}{3} + \frac{1}{5} + \frac{2}{7}\\\
&= 1 + (0.3333\ldots) + (0.2) + (0.285714285714\overline{285714}\ldots)
\end{align*}
and you can give those dangerous hoodlums their answer, and live to derive another day.

**Your problem.** You want to do something similar with the polynomial quotient, with denominators that "easy"; in this case, degree $1$. The first task is to make the fraction "less than $1$", by making sure the denominator has degree greater than the numerator. You do this with polynomial long division (see also this recent answer). Doing the long division mentally, we have: to get $2x^2$ from $x^2+11x+30$ we multiply by $2$:
$$2x^2 + 11x = (x^2+11x+30)(2+\cdots)$$
that produces unwanted $11x + 60$, (well, you get $22x + 60$, but you *do* want $11x$ of those, so you only have a leftover of $11x+60$); nothing to do about them, except cancel them out after the product is done. So you have
$$2x^2 + 11x = (x^2+11x+30)(2) - (11x+60).$$
So you can write
$$\frac{2x^2+11x}{x^2+11x+30} = 2 + \frac{-(11x+60)}{x^2+11x+30}.$$
Now we got the "integer part", and we work on the "fraction part". The denominator factors as $(x+5)(x+6)$, so we want to think of that fraction as the end result of doing a sum of the form
$$\frac{A}{x+5} + \frac{B}{x+6}.$$
Because the sum is "smaller than $1$" (numerator of degree smaller than the denominator) each of these fractions should also be "smaller than one". So both the $A$ and $B$ will be constants.
So we have:
\begin{align*}
\frac{-11x - 60}{x^2+11x+30} &= \frac{A}{x+5} + \frac{B}{x+6} \\\
&= \frac{A(x+6)}{(x+5)(x+6)} + \frac{B(x+5)}{(x+6)(x+5)}\\\
&= \frac{A(x+6) + B(x+5)}{(x+5)(x+6)}\\\
&= \frac{Ax + 6A + Bx + 5B}{x^2+11x+30} = \frac{(A+B)x + (6A+5B)}{x^2+11x+30}.
\end{align*}
For this to work out, you need $(A+B)x + (6A+5B) = -11x-60$. That means we need $A+B=-11$ (so the $x$s agree) and $6A+5B = -60$ (so the constant terms agree).
That means $A=-11-B$ (from the first equation). Plugging into the second equation, we get
$$-60 = 6A+5B = 6(-11-B)+5B = -66 -6B + 5B = -66-B.$$
So that means that $B=60-66 = -6$. And since $A+B=-11$, then $A=-5$.
(An alternative method for finding the values of $A$ and $B$ is the Heaviside cover-up method; from the fact that
$$-11x - 60 = A(x+6)+B(x+5)$$
we know that the two sides must take the same value for every value of $x$; if we plug in $x=-6$, this will "cover up" the $A$ on the right and we simply get $B(-6+5) = -B$; this must equal $-11(-6)-60 = 6$; so $6 = -B$, hence $B=-6$. Then plugging in $x=-5$ "covers up" the $B$ to give us $-11(-5)-60 = A(-5+6)$, or $-5=A$. So we obtain $A=-5$, $B=-6$, same as before.)
That is,
$$\frac{-11x-60}{x^2+11x+30} = \frac{-5}{x+5} + \frac{-6}{x+6}.$$
Putting it all together, we have:
$$\frac{2x^2+11x}{x^2+11x+30} = 2 + \frac{-11x-60}{x^2+11x+30} = 2 - \frac{5}{x+5} - \frac{6}{x+6}.$$
And <triumphant trumpet cue>ta da!</triumphant trumpet cue> Your task is done. You've decomposed the quotient into partial fractions.
Caveat. You need to be careful if the denominator has repeated factors, or has factors that are of degree $2$ and cannot be further factored (e.g., $x^2+1$); I talk about that below in the general discussion.
Now, let's hope the Mathies Gang doesn't ask for a proof of Goldbach's Conjecture from its next batch of pledges...

General Idea.
So let's discuss this idea for the general problem of integrating a rational function; that is, a function of the form
$$\frac{P(x)}{Q(x)}$$
where $P(x)=a_nx^n+\cdots +a_1x+a_0$ and $Q(x)=b_mx^m+\cdots + b_1x+b_0$ are polynomials.
Integrating polynomials is easy, so the first task is to do long division in order to write the fraction as a polynomial plus a rational function in which the numerator has degree smaller than the denominator. So we will consider only the case where $n\lt m$ going forward.
First, let us make sure we can actually do those fractions with "easy denominators." To me, "easy denominator" means (i) linear, $ax+b$; (ii) power of a linear polynomial, $(ax+b)^n$; (iii) irreducible quadratic; (iv) power of an irreducible quadratic. So let's talk about how to integrate those:

When $Q(x)$ has degree $1$ and $P(x)$ is constant. For example, something like
$$\int \frac{3}{2x-5}\,dx.$$
These integrals are very easy to do: we do a change of variable $u=2x-5$, so $du=2dx$. We simply get
$$\int\frac{3}{2x-5}\,dx = 3\int\frac{dx}{2x-5} = 3\int\frac{\frac{1}{2}du}{u} = \frac{3}{2}\ln|u|+C = \frac{3}{2}\ln|2x-5|+C.$$

In fact, the idea above works whenever the denominator is a power of a degree $1$ polynomial and the numerator is a constant. If we had something like
$$\int\frac{3}{(2x-5)^6}\,dx$$
then we can let $u=2x-5$, $du=2dx$ and we get
$$\begin{align*}
\int\frac{3}{(2x-5)^6}\,dx &= 3\int\frac{\frac{1}{2}du}{u^6} = \frac{3}{2}\int u^{-6}\,dx \\&= \frac{3}{2}\left(\frac{1}{-5}u^{-5}\right)+C\\ &= -\frac{3}{10}(2x-5)^{-5} + C.\end{align*}$$

What if the denominator is an irreducible quadratic? Things get a little more complicated. The simplest example of an irreducible quadratic is $x^2+1$, and the easiest numerator is $1$. That integral can be done directly:
$$\int\frac{1}{x^2+1}\,dx = \arctan(x)+C.$$
If we have an irreducible quadratic of the form $x^2+a$, with $a\gt 0$, then we can always write it as $x^2+b^2$ with $b=\sqrt{a}$; then we can do the following: factor out $b^2$ from the denominator,
$$\int\frac{dx}{x^2+b^2} = \int\frac{dx}{b^2((\frac{x}{b})^2 + 1)};$$
and now setting $u=\frac{x}{b}$, so $du = \frac{1}{b}\,dx$, we get:
$$\int\frac{dx}{x^2+b^2} = \frac{1}{b}\int\frac{1}{(\frac{x}{b})^2+1}\left(\frac{1}{b}\right)\,dx = \frac{1}{b}\int\frac{1}{u^2+1}\,du,$$
and now we can do the integral easily as before.
What if we have a more general irreducible quadratic denominator? Something like $x^2+x+1$, or something else with an $x$ term?

The magic phrase here is *Completing the square". We can write $x^2+x+1$ as
$$x^2 + x + 1 = \left(x^2 + x + \frac{1}{4}\right) + \frac{3}{4} = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}.$$
Then setting $w=x+\frac{1}{2}$, we end up with an integral that looks just like the previous case! For instance,
$$\int\frac{dx}{x^2+x+1} = \int\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}} = \int\frac{dw}{w^2+\frac{3}{4}},$$
and we know how to deal with these.
So: if the denominator is an irreducible quadratic, and the numerator is a constant, we can do the integral.

What if the denominator is an irreducible quadratic, but the numerator is not constant? Since we can always do the long division, then we can take the numerator to be of degree $1$. If we are lucky, it's possible we can do it with a simple substitution; for example, to do
$$\int\frac{2x+3}{x^2+3x+4}\,dx$$
(note the denominator is irreducible quadratic), we can just let $u=x^2+3x+4$, since $du = (2x+3)\,dx$, exactly what we have in the numerator, so
$$\int\frac{2x+3}{x^2+3x+4}\,dx = \int\frac{du}{u} = \ln|u|+C = \ln|x^2+3x+4|+C = \ln(x^2+3x+4) + C.$$
If we are not lucky? Well, we can always make our own luck. For instace, if we had
$$\int\frac{3x}{x^2+3x+4}\,dx$$
then we can't just make the substitution $u=x^2+3x+4$; but if we wanted to do that anyway, we would need $2x+3$ in the numerator; so we play a little algebra game:
$$\begin{align*}
\frac{3x}{x^2+3x+4} &= 3\left(\frac{x}{x^2+3x+4}\right)\\
&= 3\left(\frac{\frac{1}{2}(2x)}{x^2+3x+4}\right)\\
&=\frac{3}{2}\left(\frac{2x}{x^2+3x+4}\right) &\text{(still not quite right)}\\
&= \frac{3}{2}\left(\frac{2x+3-3}{x^2+3x+4}\right)\\
&= \frac{3}{2}\left(\frac{2x+3}{x^2+3x+4} - \frac{3}{x^2+3x+4}\right).
\end{align*}$$
What have we accomplished? The first summand, $\frac{2x+3}{x^2+3x+4}$, is an integral we can do with a simple substitution; and the second summand, $\frac{3}{x^2+3x+4}$, is an integral that we just saw how to do! So in this way, we can solve this kind of integral. We can always rewrite the integral as a sum of an integral that we can do with a substitution, and an integral that is as in case 3 above.

What if the denominator is a power of an irreducible quadratic? Something like $(x^2+3x+5)^4$? If the numerator is of degree at most $1$, then we can play the same game as we just did to end up with a sum of two fractions; the first one will have numerator which is exactly the derivative of $x^2+3x+5$, and the second will have a numerator that is constant. So we just need to figure out how to do an integral like
$$\int\frac{2dx}{(x^2+3x+5)^5}$$
with constant numerator and a power of an irreducible quadratic in the denominator.

By completing the square as we did in 3, we can rewrite it so that it looks like
$$\int\frac{dw}{(w^2+b^2)^5}.$$
Turns out that if you do integration by parts, then you get a Reduction formula that says:
$$\int\frac{dw}{(w^2+b^2)^n} = \frac{1}{2b^2(n-1)}\left(\frac{w}{(w^2+b^2)^{n-1}} + (2n-3)\int\frac{dw}{(w^2+b^2)^{n-1}}\right)$$
By using this formula repeatedly, we will eventually end up in an integral where the denominator is just $w^2+b^2$... and we already know how to do those. So all is well with the world (with a lot of work, at least).
Okay. Do we now need to go and discuss what to do when the denominator is a cubic polynomial, then a fourth degree polynomial, then a fifth degree polynomial, etc.?
No! We can play the same game we did with fractions above, and take an arbitrary rational function and rewrite it as a sum of fractions, with each fraction a power of a degree 1 or an irreducible quadratic polynomial. The key to this is the Fundamental Theorem of Algebra, which says that every polynomial with real coefficients can be written as a product of linear and irreducible quadratic polynomials. In fact, one of the major reasons why people wanted to prove the Fundamental Theorem of Algebra was to make sure that we could do integrals of rational functions in the manner we are discussing.
So here is a method for integrating a rational function:

To compute
$$\int\frac{P(x)}{Q(x)}\,dx$$
where $P(x)$ and $Q(x)$ are polynomials:

If $\deg(P)$ is equal to or greater than $\deg(Q)$, perform long division and rewrite the fraction as a polynomial plus a proper fraction (with numerator of degree strictly smaller than the denominator). Integrate the polynomial (easy).

Completely factor the denominator $Q(x)$ into linear terms and irreducible quadratics. This can be very hard to do in practice. In fact, this step is the only obstacle to really being able to do these integrals always, easily. Factoring a polynomial completely and exactly can be very hard. The Fundamental Theorem of Algebra says that there is a way of writing $Q(x)$ that way, but it doesn't tell us how to find it.

Rewrite the fraction as a sum of fractions, each of which has a denominator which is either a power of linear polynomial or of an irreducible quadratic. (More about this below.)

Do the integral of each of the fractions as discussed above.

How do we rewrite as a sum?
Write $Q(x)$ as a product of powers of distinct polynomials, each linear or irreducible quadratic; e.g., $Q(x) = x(2x-1)(x+2)^3(x^2+1)(x^2+2)^2$.
For each power $(ax+b)^n$, use $n$ fractions of the form:
$$\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \cdots+\frac{A_n}{(ax+b)^n},$$
where $A_1,\ldots,A_n$ are constants-to-be-determined-later.
For each power of an irreducible quadratic, $(ax^2+bx+c)^m$, use $m$ fractions of the form:
$$\frac{C_1x+D_1}{ax^2+bx+c} + \frac{C_2x+D_2}{(ax^2+bx+c)^2} + \cdots + \frac{C_mx+D_m}{(ax^2+bx+c)^m},$$
where $C_1,D_1,\ldots,C_m,D_m$ are constants-to-be-determined-later.
So in the example above, we $Q(x) = x(2x-1)(x+2)^3(x^2+1)(x^2+2)^2$, we would get:
$$\begin{align*}
&\quad+\frac{A}{x} &\text{(corresponding to the factor }x\text{)}\\
&\quad+\frac{B}{2x+1} &\text{(corresponding to the factor }2x-1\text{)}\\
&\quad+\frac{C}{x+2} + \frac{D}{(x+2)^2}+\frac{E}{(x+2)^3} &\text{(corresponding to the factor }(x+2)^3\text{)}\\
&\quad+\frac{Gx+H}{x^2+1} &\text{(corresponding to the factor }x^2+1\text{)}\\
&\quad+\frac{Jx+K}{x^2+2} + \frac{Lx+M}{(x^2+2)^2}&\text{(corresponding to the factor }(x^2+2)^2\text{)}
\end{align*}$$
And now, the final step: how do we figure out what all those constants-to-be-determined-later are? We do the operation and compare it to the original! Let's say we are trying to calculate
$$\int\frac{3x-2}{x(x+2)^2(x^2+1)}\,dx$$
(not the same as above, but I want to do something small enough that we can do it).
We set it up as above; then we do the algebra. We have:
$$\begin{align*}
\frac{3x-2}{x(x+2)^2(x^2+1)} &= \frac{A}{x} + \frac{B}{x+2} + \frac{C}{(x+2)^2} + \frac{Dx+E}{x^2+1}\\
&= \frac{\small A(x+2)^2(x^2+1) + Bx(x+2)(x^2+1) + Cx(x^2+1) + (Dx+E)x(x+2)^2}{x(x+2)^2(x^2+1)}.
\end{align*}$$
Now we have two options: we can do the algebra in the numerator and write it as a polynomial. Then it has to be identical to $3x-2$. For example, the coefficient of $x^4$ in the numerator would be $A+B+D$, so we would need $A+B+D=0$; the constant term would be $4A$, so $4A=-2$; and so on.
The other method is the Heaviside cover-up method. Since the two expressions have to be the same, the numerator has to be the same as $3x-2$ when evaluated at every value of $x$. If we pick $x=0$ and plug it into the left numerator, we get $-2$; if we plug it into the right hand side, we get $A(2)^2(1) = 4A$, so $4A=-2$, which tells us what $A$ is. If we plug in $x=-2$, th left h and side is $3(-2)-2 = -8$, the right hand side is $C(-2)((-2)^2+1) = -10C$, so $-10C = -8$, which tells us what $C$ is; we can then simplify and continue doing this (you'll note I selected points where a lot of the summands on the right simply evaluate to $0$) until we get the value of all the coefficients.
And once we've found all the coefficients, we just break up the integral into a sum of integrals, each of which we already know how to do. And so, after a fair amount of work (but mainly just busy-work), we are done.

Here are some formal aspects of the delightful answer ( by Arturo Magidin ) above.
Let $ f(x), g(x) $ be polynomials ( $ g \neq 0 $ ). Our goal now is to integrate $ \frac{f}{g} $.
Dividing $ f $ by $ g $ gives that $ f = qg + r $ with $ \deg(r) < \deg(g) $, so $ \frac{f}{g} $ is a polynomial $ q $ plus a proper fraction $ \frac{r}{g} $ [ We'll call a  fraction $ \frac{p}{q} $ of polynomials  "proper" if $ \deg{p} < \deg{q} $ ]. Since the $ q, r $ here are unique, we'll call $ q $ the "polynomial part" and $ \frac{r}{g} $ the "proper fraction part" of $ \frac{f}{g} $.
Now we need only figure out how to integrate proper fractions, so let's take $ \deg(f) < \deg(g) $ to begin with.
$ g $ can be broken into a product consisting of linear factors, and quadratic factors with negative discriminant [ Proved here, for instance ]. Also, any proper fraction $ \displaystyle \frac{f}{g_1 \ldots g_k} $ where $ g_i $s are mutually coprime can be written as a sum of proper fractions with denominators $ g_1, \ldots, g_k $ [ Let's show it for $ k = 2 $, the general case follows by induction. Since $ g_1, g_2 $ are coprime, by Euclidean algorithm $ g_1 u_1 + g_2 u_2 = 1 $ for some polynomials $ u_1, u_2 $. Now $ \displaystyle \frac{f u_1}{g_2} + \frac{f u_2}{g_1} = \frac{f}{g_1 g_2} $, and since right hand side is proper the polynomial parts of $ \displaystyle \frac{f u_1}{g_2}, \frac{f u_2}{g_1} $ sum up to $ 0 $. So ( proper fraction part of $ f u_1 / g_2 $ ) + ( proper fraction part of $ f u_2 / g_1 $ ) = $ f / g_1 g_2 $ , done. ]
So $ \displaystyle \frac{f}{g} $ breaks into a sum of proper fractions of the form $ \displaystyle \frac{s(x)}{t(x) ^l} $ and $ \displaystyle \frac{p(x)}{h(x) ^k} $, where the $ t(x) $s are linear polynomials and $ h(x) $s are negative discriminant quadratics .
The terms here can be broken down further. Any proper fraction of the form $ \displaystyle \frac{f}{g^k} $ is a sum of proper fractions, with denominators $ g, \ldots, g^k $ and numerators having lower degree than $ g $ ( The $ f, g $ here are separate from the original problem. Notation repeated to try keep things more readable ). This is because : Division gives $ f = qg + r $ with $ \deg(r) < \deg(g) $, so $ \displaystyle \frac{f}{g^k} = \frac{q}{g^{k-1}} + \frac{r}{g^k} $. All the terms here are proper and $ \deg(r) < \deg(g) $. Now continuing onto $ \displaystyle \frac{q}{g^{k-1}} $, we're done.
So our original proper fraction $ \displaystyle \frac{f}{g} $ is a sum of proper fractions, of the form $ \displaystyle \frac{m}{(x+p)^l} $, and of the form $ \displaystyle \frac{ax+b}{(x^2 + cx + d)^k} $ where $ x^2 + cx + d $ has negative discriminant.
Proper fractions of the form $ \displaystyle \frac{m}{(x+p)^l} $ are easily integrated, for instance by substituting $ u = x + p $.
So finally, the problem is reduced to integrating fractions of the form $ \displaystyle \frac{ax+b}{(x^2 + cx + d)^k} $ where $ x^2 + cx + d $ has negative discriminant. Writing $ x^2 + cx + d $ as $ \displaystyle (x + \frac{c}{2})^2 + (d - \frac{c^2}{4}) $ and substituting $ u = x + \frac{c}{2} $ makes the integral take the form $ \displaystyle \int \frac{Au + B}{(u^2 + C^2)^k} du $. We can integrate the $ \displaystyle \frac{Au}{(u^2 + C^2)^k} $ part by substituting $ v = u^2 + C^2 $. And the $ \displaystyle \frac{B}{(u^2 + C^2)^k} $ part, as mentioned in the above answer, can be handled using integration by parts.


Possible values of $b$ for $x^2 + bx + 6 = 0,$ for integer solutions The question is essentially in the title. The actual question is here(and also my working out) if there's any confusion in the title http://m.imgur.com/a/lgNzt

My method was to find the discriminant = 0 and then use trial and error since this question was multiple choice.

Is there a way of doing this properly, as in algebraically without trial and error?

Continuing your's work:
$\sqrt{b^2-24}=s\ \text{, s is integer}\implies b^2-s^2=24\implies (b-s)(b+s)=24$

Now, factor $24\begin{cases}24 &=1\cdot 24\\ 24 &=2\cdot 12\\ 24 &=3\cdot 8\\ 24 &=4\cdot 6\\ &\vdots\end{cases}$

Of which two products adds to an odd integer will not work for e.g. $(b-s)(b+s)=24=1\cdot 24$ here two product adds to $24+1=25$, odd,  hence kick it out. Similarly $24=8\cdot 3$ will be kicked out.  Leaving rest to verify it yourself.

Let A and B be the roots. So A and B are integers.
(X-A)×(X-B)=X^2+bX+6

X^2+bX+6=X^2-(A+B)X+AB

So AB=6 . And they are integers. So possible value are

1×6 , -1×-6 , 2×3 , -2×-3.

b=-(A+B)

Putting above values , b=1+6 , -1+-6 , 2+3 ,-2,+-3
                                            7 , -7 , 5 , -5


Explain why a line can never intersect a plane in exactly two points. Why can a line never intersect a plane in exactly two points?

I know this seems like a really simple question, but I'm having a hard time figuring out how to answer it. I also tried googling the question but I couldn't find an answer for exactly what I'm looking for.

If you pick two points on a plane and connect them with a straight line then every point on the line will be on the plane.

Given two points there is only one line passing those points.

Thus if two points of a line intersect a plane then all points of the line are on the plane.

First, some definitions:

A line is straight and infinitely long.
A plane is flat and infinitely long and wide.
An intersect is a point shared by the line and the plane.

This leads to three relations between a line and a plane:

It is parallel and not part of the plane. This means it has 0 intersects.
It is parallel and part of the plane. This means it has infinite intersects.
It is not parallel. This means at some point it intersects exactly once.

To make a line intersect the same plane twice it can't be parallel, and either the line or the plane must be curved.


How prove this number $(a,b,n)=(1,1,4k)$ Let $a,b,n\in \mathbb{N}$, $\gcd(a,b)=1$, $\left(\sqrt{a}+i\sqrt{b}\right)^n\in \mathbb{R}$

Show that
$(a,b,n)=(1,1,4k)$ or $(1,\sqrt{3},3k)$ or $(\sqrt{3},1,6k)$ where $k\in \mathbb{N}$, $i=\sqrt{-1}$

Let $x=\sin^{-1}\frac{\sqrt{b}}{\sqrt{a+b}}$, then $\cos 2x=\frac{a-b}{a+b} \in \mathbb{Q}$. Since $nx=k\pi$, using the well known theorem (see for example http://yaroslavvb.com/papers/olmsted-rational.pdf), $\cos 2x=0, \pm 1, \pm \frac{1}{2}$.


Probability of Random Selection Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random, one at a time with replacement. What is the probability that the largest number appearing on a selected coupon be 9?

My attempt: I tried to break down the sum into 7 parts and find the probability in each case. I do not know how to find the probability of choosing n particular choices( all less than or equal to 9) out of m choices remembering the fact that 9 should appear in atleast one of the 7 draws

For a with replacement selection each draw is independent. Consider the $k^{th }$ draw. Any ticket numbered less than $9$ can be chosen, the probability of chosing such a ticket will be $\frac9{15}$, as any ticket has the same probability, $\frac1{15}$, of being chosen. Hence the probability that in all the $7$ draws the tickets chosen are $\le9$, will be $$\left(\frac9{15}\right)^7$$
Similarly, the probability that in all the $7$ draws the tickets chosen are $\le8$, will be $$\left(\frac8{15}\right)^7$$
But see that in the former case the maximum of the numbers drawn will he $\le9$, while in the latter case the maximum of the numbers drawn will be $\le8$. So the probability that the maximum of the numbers drawn is exactly $9$ will be  $$\left(\frac9{15}\right)^7- \left(\frac8{15}\right)^7$$


Cauchy–Schwarz Inequality for Negative Numbers If $a_i, b_i, p_i \ge 0$, we can show by Cauchy–Schwarz inequality that $$\left(\sum_i {a_i p_i}\right)\left(\sum_i {b_i p_i}\right) \ge \left(\sum_i {p_i\sqrt{a_i b_i}} \right)^2.$$
What happens when $p_i$ can take negative values, but $\sum_i {a_i p_i} \ge 0$, and $\sum_i {b_i p_i} \ge 0$. Can we still say that $$\left(\sum_i {a_i p_i}\right)\left(\sum_i {b_i p_i}\right) \ge \left(\sum_i {p_i\sqrt{a_i b_i}} \right)^2?$$
Can any other restrictions on $p_i$ ensure that the above result holds?

If $a_i,b_i>0$ and $p_i<0~\forall i$ then the following inequality will obviously hold: $$\left(\sum_ia_i|p_i|\right)\left(\sum_ib_i|p_i|\right)\geq\left(\sum_i|p_i|\sqrt{a_ib_i}\right)^2$$
$$\Rightarrow\left(\sum_ia_i(-p_i)\right)\left(\sum_ib_i(-p_i)\right)\geq\left(\sum_i(-p_i)\sqrt{a_ib_i}\right)^2~~~~~~~~~~~~~~~[since ~|p_i|=-p_i]$$
$$or,~\left(-\sum_ia_ip_i\right)\left(-\sum_ib_ip_i\right)\geq\left(-\sum_ip_i\sqrt{a_ib_i}\right)^2$$
$$or,~\left(\sum_ia_ip_i\right)\left(\sum_ib_ip_i\right)\geq\left(\sum_ip_i\sqrt{a_ib_i}\right)^2$$

The inequality would hold whenever $p_ia_i>0,p_ib_i>0~\forall i$. Because then take $\tilde{x}=(\sqrt{a_1p_1},\sqrt{a_2p_2},\cdots,\sqrt{a_np_n})'$ and $$\tilde{y}=(\sqrt{b_1p_1},\sqrt{b_2p_2},\cdots,\sqrt{b_np_n})'$$, then we know $$\tilde{x}.\tilde{y}=||\tilde{x}||.||\tilde{y}||\cos{\theta}\leq||\tilde{x}||.||\tilde{y}||$$
where $\theta$ is the angle between $\tilde{x}$ and $\tilde{y}$


how to solve this question using set theory? Consider a set U of 23 different compounds in a chemistry lab. There is a subset S of U of 9 compounds, each of which reacts with exactly 3 compounds of U.

Consider the following statements:

Each compound in U \ S reacts with an odd number of compounds.
At least one compound in U \ S reacts with an odd number of compounds.
Each compound in U \ S reacts with an even number of compounds.

Which one of the above statements is ALWAYS TRUE?

A.  Only I

B.  Only II

C.  Only III

D.  None

==============================================================

Answer given in book is- The sum of the degrees of all the vertices in a graph is equal to twice the number of edges.

Here, Set S has 9 elements and each element has a degree of 3 as each element in S  reacts to exactly 3 elements. So 9*3 + Sum. of degrees of vertices in U-S = 2* e

Since 2*e is even no, Sum of degrees of vertices in U-S is  odd .

Thus,some vertex in U-S must have odd degree. .At least one compound in U-S reacts with an odd number of compounds. Option B is the correct answer.

But How to solve it using set theory? Also, is there any other way solve this question!

Here is a restatement of the book solution which does not use graph theory. For shorthand, let's call a compound "even" if it reacts with an even number of compounds, and "odd" if reacts with an odd number of compounds.

Since each compound of $S$ reacts with exactly $3$ compounds of $U$, there are a total of $27$ reactions between $U$ and $S$. Distributing those $27$ reactions between the $13$ compounds in $U \setminus S$ can be done in many ways, but we know that it's impossible for all $13$ compounds in $U \setminus S$ to be even, because then the total number of reactions would be even, but $27$ is odd.

In fact, no matter how we distribute the reactions, the number of odd compounds in $U \setminus S$ is odd, because the total must be odd. (If we add up $k$ odd numbers, and $13-k$ even numbers, then the total is odd when $k$ is odd, and even when $k$ is even.)

So far, it seems that statement II is true. But we haven't yet considered reactions between two compounds in $U \setminus S$.

It turns out that this doesn't make a difference. There are three cases:

If we add a reaction between an even compound and an odd compound, then the even compound becomes odd and the odd compound becomes even, but the number of odd compounds doesn't change.
If we add a reaction between two even compounds, they both become odd. So there is still an odd number of odd compounds in $U \setminus S$ (their number has gone up by $2$, so it stays odd).
If we add a reaction between two odd compounds, they both become even. So there is still an odd number of odd compounds in $U \setminus S$ (their number has gone down by $2$, so it stays odd).

As a result, no matter how many reactions between compounds in $U \setminus S$ there are, one compound in $U \setminus S$ will still be odd, proving statement II.

Consider all compounds as numbers like {1,2,,.....,23} and let’s consider subset S={1,2,.,8,9}.

Now if we assign 3 compound of U\S to each element of set S  in such a way that.

1: 10,11,12

2: 13,14,15

3: 16,17,18

4: 19,20,21

5: 22,23,10  (here you can see we have to start again from 10 as we have only 10-23
              compounds with us)

6: 11,12,13
7: 14,15,16

8: 17,18,19

9: 20,21,22

Now you can see from above assignments that 10-22 reacts with even no. of compounds but 23 reacts with odd number of compounds.

So option (B) is the right one.

Note: There is a subset S  of U of 9 compounds, each of which reacts with exactly 3 compounds of U. If you consider that all S elements reacts with itself too..then in that case 3 elements in U\S react with 1 (odd) element of S .

please let me know if anything wrong with this method.


Show $P(A \subset B) = (\frac{3}{4})^n$ Let $S = \{1, 2, \dots, n \}$ and suppose that $A$ and $B$ are,
  independently, equally likely to be any of the $2^n$ subsets of $S$.

  Show that $P(A \subset B) = (\frac{3}{4})^n$ and show that $P(A \cap B = \emptyset) = (\frac{3}{4})^n$.

Applying the law of total probability ,

$$P(A \subset B)  =\sum_{k=0}^n P(A \subset B \Large \mid \normalsize |B| = k)P(|B| = k) $$

If $k = 0$ then $B = \varnothing $. Therefore  $P(A \subset B \Large \mid \normalsize |B| = k) = 0$ because there do not exist any sets $A$ such that $A \subset B$.

If $k = 1$ and $A \subset B$, then there exists precisely one set $A$ such that $A \subset B$, and this $A = \varnothing$. Therefore

$$\begin{align}
&P(A \subset B \mid \normalsize |B| = k) =  \frac{1}{2^n} = \frac{2^1 - 1}{2^n} = \frac{2^k - 1}{2^n} \\
&P(|B| = k) = \frac{\binom{n}{k}}{2^n}
\end{align}$$

If  $A \subset B$, then $A \in  \mathcal{P}(B) - B$. Thus, there are $|\mathcal{P}(B) - B| = 2^{k} - 1$ possibilities for $A \subset B$ out of a possible $2^{|S|} = 2^n$ sample space.

Therefore I have

$$\begin{align}
&P(A \subset B)  = 0 + \sum_{k=1}^n P(A \subset B \Large \mid \normalsize |B| = k)P(|B| = k) = \sum_{k=1}^n \frac{2^k - 1}{2^n}\frac{\binom{n}{k}}{2^n}  \\
&\mathsf{but} \ \ \  \sum_{k=1}^n \frac{2^k - 1}{2^n}\frac{\binom{n}{k}}{2^n} \ne (\frac{3}{4})^n
\end{align}$$

Where is my mistake? Hints only please.

It is easy to see that $P(A\subseteq B)=(\frac34)^n$ so the first part of what you're trying to show can't be true.

Namely, suppose we choose for each of the $n$ possible elements first whether it is in $A$ and then whether it is in $B$. All of the $2n$ choices are independent and uniformly distributed between yes and no.

$A\subseteq B$ means that each pair of answers is one of (yes, yes), (no, yes) or (no, no). For each pair the probability of that happening is $\frac 34$, so the probabilty of that happening for all $n$ pairs is $(\frac{3}{4})^n$.

Similarly $A\cap B=\varnothing$ means that each pair of answers is one of (yes,no), (no, yes), or (no,no). Again, this has probability $\frac 34$ and must happen $n$ times, so the probability of disjoint $A$ and $B$ is $(\frac 34)^n$.

Trying to count possibilities, grouped by the size of the sets is definitely the long way around to an answer here.

If you want to calculate strictly about the case of $A\subset B$, from my persepective there is nothing wrong about your calculation. But if you want to consider about the case of $A\subseteq B$， the above answer is direct and simple. I  would like to give a solution using the law of total probability.
$$
P(A \subseteq B)=\sum\limits_{i=0}^{n} P(A\subseteq B \mid |B|=i)\cdot P(|B|=i)=\sum\limits_{i=0}^{n} \frac{2^i}{2^n} \cdot \frac{\left(\begin{array}{c}
n\\i\end{array}\right)}{2^n}=\frac{\sum\limits_{i=0}^{n}\left(\begin{array}{c}
n\\i\end{array}\right)2^i1^{n-i}}{4^n}=\frac{3^n}{4^n}
$$
The second question is similar.


Find all possible multiplicand who results in given number I have some random figure let's say 400, I need equation to find all possible combinations (of integer) whose multiplication will results in 400. Condition is number of factors (multiplicand) will be dynamic.

For example, for 400 with two factors I need output as

1 * 400
2 * 200
4 * 100
5 * 80
etc...

means all a and b where a * b = 400.

Same with 3 factors,

1 * 1 * 400
1 * 2 * 200
2 * 2 * 100
etc...

mean all a, b, and c where a * b * c = 400. We can do this by iterating loop for each possibility, but this approach takes longer time. I need solution that can give answer in minimum possible time. Thanks.

It looks like you're looking only at positive integers, so we'll stick with that.  Also, you're looking for an algorithm more than you're looking for an equation.

First, find the prime factorization of the number.  Let's say it's composed of the product of $N$ prime factors (which may be repeated, of course).

All of the terms in any of your products will need to be these factors, or products of these factors, or $1$, or the number itself.

To find all of the possible products with $M$ terms, you'll need to find all of the possible products with $1, 2, \dots, min(M,N)$ terms, but for each case, you'll only need to find them once.  (The function $min(N,M)$ means the minimum of $M$ and $N$.)  Here's why.

You show three cases for three-term solutions for the product of $400$:  $$1 \times \ 1 \times 400 = 400, \\ 1 \times \ 2 \times 200 = 400, \\ 2 \times \ 2 \times 100 = 400.$$

The first one is really a trivial extension of a one-term answer:  $400.$  It's just padded with $1$'s.

The second one is really a trivial extension of a two-term answer:  $2 \times 200.$  It's just padded with a $1$.

The third one is a non-trivial three-term answer.

So the set of all possible products with $M$ terms ($M \leq N$) is the set of non-trivial products with $M$ terms, plus all of the trivial ones created by padding products of $1, 2, \dots, M-1$ terms with the appropriate number of $1$'s.

If $M > N$, then there are no non-trivial products left, so you'll simply pad all of the products (trivial and non-trivial) with $M-N$ $1$'s, and those will be the products in addition to the ones you found for $M \leq N$.


Divide a number by $0$ I don't know much about maths, I just started thinking about this the other day and would like to ask the experts.

I believe in maths it is said that $\frac 10$ is 'undefined'.

This may be because mathematicians look at it like this:

$$y = \lim_{x\to0}\frac 1x $$ is a very big number, because you can subtract a very small fraction from $1$ many many times before you get to zero.

I think division is actually subtraction.

$$1 - 0 = 1$$

Now, maybe mathematicians say, "How many times can you subtract $0$ from $1$? We can't say."

Teacher: Johnny, if you have $1$ apple and I take $1$ apple from you, how many apples do you have?
Johnny: None.
Teacher: Correct Johnny. Now, if you have $1$ apple and I take $0$ apples from you, how many apples do you have?
Johnny: Then you didn't do anything and I still have $1$ apple. Stupid question if you ask me.

If you 'subtract $0$' you have in fact done nothing. If you do nothing to $1$, $1$ does not change.

$$\frac 10 = 1$$

You seem to think of division as repeated subtraction.  That's fine, it turns out there's a way to think of it like that.

When you divide $a$ by $b$ you're essentially asking "how many times can I subtract $b$ from $a$ until I get $0$?"  For instance $10/5 = 2$ tells us that we can subtract $5$ twice from $10$ to get $0$.  $-4/2 = -2$ tells us that we can "subtract" $2$ negative two times to get $0$: but it's easy to see that that just means add $2$ twice.  It's a little harder to see what something like $4/5$ would mean: you'd have to subtract $5$ from $4$ a total of $0.8$ times to get $0$.  But somehow that sort of makes sense because you couldn't possibly subtract all of one $5$ from $4$ or you'd overshoot.

Now we turn our attention to $1/0$.  How many times do we have to subtract $0$ from $1$ until we get $0$?

Only 0/0 is 'undefined', because any number multiplied by 0 leads to 0. Any number divided by 0 tends to infinity, because the lower the divisor, the higher the result (if the dividend is different form 0, of course).


Show that $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n\in R[x]$ is nilpotent iff $a_0,a_1,a_2,\ldots,a_n$ are nilpotent Let $R$ be a commutative ring. Show that $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n\in R[x]$ is nilpotent if and only if $a_0,a_1,a_2,\ldots,a_n$ are nilpotent.

Now since $f(x)$ is nilpotent then $f^n=0$ for some $n$. Now $f^2=c_0+c_1 x+\cdots+c_n x^n$ where $c_i=\sum _{k=0}^i  a_k a_{i-k}$.

But how to write it for $f^n$? How to take it from here?

If $n_i$ be the least positive integer such that $a_i^{n_i}=0$ then  what will the positive integer such that $a_i^n=0$ for all $i$??

I think it's easier to do this with induction on $n=\text{deg}(f)$. The case $n=0$ is obvious.

For the induction step, if $g(x)=a_0+a_1x+a_2x^2+\cdots+a_{n+1}x^{n+1}$ is nilpotent, then $a^{n+1}$ is nilpotent by looking at the coefficient of the highest term. That means that $a_{n+1}x^{n+1}$ is nilpotent, so $g(x)-a_{n+1}x^{n+1}=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ is nilpotent. (Sum of nilpotents is nilpotent) By induction hypothesis $a_0,\cdots,a_n$ are all nilpotent.

Conversely if $a_0,\cdots,a_{n+1}$ are all nilpotent, then in particular $a_0,\cdots,a_{n}$ are nilpotent, so $(a_0+\cdots+a_nx^n)^k=0$ for some $k$. Also, $(a_{n+1}x^{n+1})^l=0$ for some $l$. Now expand $((a_0+\cdots+a_nx^n)+(a_{n+1}x^{n+1}))^{k+l}$ using the binomial theorem to see that it's $0$.

Only if part is trivial, for the if part show that if $a_i ^{N_i}=0$ then $(\sum_{i=1}^{n}a_ix^i)^{\sum_{i=1}^{n}N_i+1}=0$ by showing that any general term is 0.


Is the outer measure of $A\cup B$ equal to the sum of their outer measures if $A\cap B=\varnothing$? I understand that Lebesgue outer measure on $\mathbb R$ is not countably additive. But if there are two disjoint sets, does the outer measure of their union equal the sum of their outer measure? Can someone give me a counterexample?

Use transfinite induction to construct $2$ (or $2^{\aleph_0}$) disjoint subsets $A_i$ of the unit interval $[0,1]$, such that each $A_i$ has nonempty intersection with every uncountable closed subset of $[0,1]$. (These are called Bernstein sets.) Then each $A_i$ has outer measure $1$, as does the union of all the $A_i$.

Countably additive holds only If the sets $\{ E_n \} $ are measurable and pairwise disjoint: That is. If $\{ E_n \}_{n\geq 1} $ are measurable, then

$$ \mu^{*} ( \bigcup E_n ) = \sum \mu^*(E_n)  $$


Proof on disjoint union of sets $A$ and $B$ Here is my attempt at a solution for a proof about disjoint unions of sets $A$ and $B$. Can you please point out the mistakes? Thank you all.

Let $A$ and $B$ be any set. Prove:

$A$ is the disjoint union of $A\setminus B$ and $A\cap B$
$A\cup B$ is the disjoint union of $A\setminus B$, $A\cap B$, and $B\setminus A$.

Part 1

Suppose $A$ and $B$ are not disjoint. Then,

$$
\{x | x\in A\setminus B\text{ and }x\in A\cap B\}
$$

Since $A\subseteq A\setminus B$ and $A\subseteq A\cap B$, $A\setminus B = A\cap B$. So, $A\setminus B\nsubseteq B$, or $A\setminus B\subseteq B^{C}$, and $A\cap B\subseteq B$.

Then, $B\cup B^{C} = \emptyset$, and $A$ and $B$ are disjoint.

Part 2

Suppose $A\cup B$ is not disjoint. Then,

$$
\{x | x\in A\setminus B\text{ and }x\in A\cap B\text{ and }x\in B\setminus A\}
$$

$A\setminus B\subseteq B^{C}$, $A\cap B\subseteq A$, $A\cap B\subseteq B$, and $B\setminus A\subseteq A^{C}$. But since $B\cup B^{C} = \emptyset$ and $A\cap A^{C} = \emptyset$, $A\cup B$ is disjoint.

In the first one, $A$ is not a subset of $A\setminus B$, but rather the other way around, that is $A\setminus B\subseteq A$ (Consider $A=\{0,1,2\}$ and $B=\{1\}$ as a counterexample to your statement).

Also $B\cap B^c=\varnothing$, and rather $B\cup B^c$ is everything.

You need to argue, however, $x\in A\setminus B$ then $x\notin B$, therefore $x\notin A\cap B$; and vice versa (that is $x\in A\cap B$ then $x\notin A\setminus B$). Then you need to show that $x\in A$ then either $x\in A\cap B$ or $x\in A\setminus B$ (which really boils down to the fact that either $x\in B$ or $x\notin B$).

In the second one the argument is completely unclear to me. Using the first part you can write $A=(A\setminus B)\cup (A\cap B)$ as a disjoint union, as well to apply the same argument on $B = (B\setminus A)\cup (B\cap A)$.

Now use the fact that $A\setminus B$ and $B\setminus A$ are disjoint to prove that the decomposition of $A\cup B=(A\setminus B)\cup(B\setminus A)\cup (A\cap B)$ is a disjoint union.

Lastly (after the $\LaTeX$ was fixed by cardinal) note that:

$$A\cup B=\{x\mid x\in A\ \mathbf{or}\  x\in B\}$$

While you wrote that this is "$x\in A\setminus B$ and $x\in A\cap B$ and $x\in B\setminus A$" which would be the intersection, which you can prove is empty.

Your statements 1. and 2. are so obvious that any "formal proof" actually weakens their credibility.

Ad 1. Any element $x\in A$ either does not belong to $B$ or is  an element of $B$ as well.

Ad 2. For any element of $x\in A\cup B$ exactly one of the following is true: $x$ belongs solely either to $A$ or to  $B$, or to both of them.


A polynomial problem, Source: AOPS Let $f(x)$ be a polynomial of real coefficients such that if $n\equiv 5 \pmod{10}$ or $n\equiv 8\pmod{10}$, then $f(n)$ is an integer ($n$ is an integer). Is it true that $f(0)$ is an integer?

I am sure the answer is yes. But why? Please help!

Suppose $f \in \mathbb{C}[x]$ is such that $f(10u+5)$ and $f(10u+8)$ are integers, for all integers $u$.

Claim:$\;f(0)$ is an integer.

Proof:

Since $f(v)$ is an integer for infinitely many integers $v$, it follows, by Lagrange interpolation, that $f$ has rational coefficients.

Let $g\in\mathbb{Q}[x]$ be such that $f(x)=xg(x) + f(0)$.

Let $G\in \mathbb{Z}[x]$ be such that $g(x) = {\large{\frac{G(x)}{D}}}$, where $D$ is a positive integer.

Write $D=(2^a)(5^b)D'$, where $a,b$ are nonnegative integers, and $D'$ is a positive integer, with $\gcd(D',10)=1$.

Choose $m,n\in\mathbb{Z}$ such that

$D'{\,\mid\,}(2m+1)$, and $5^b{\,\mid\;}5(2m+1)$$\\[4pt]$
$D'{\,\mid\,}(5n+4)$, and $2^a{\,\mid\;}2(5n+4)$

Then
\begin{align*}
&f(10m+5)\in \mathbb{Z}\\[4pt]
\implies\;&(10m+5)g(10m+5)+f(0)\in \mathbb{Z}\\[4pt]
\implies\;&\left(5(2m+1)\frac{G(10m+5)}{D}\right)+f(0)\in \mathbb{Z}\\[4pt]
\implies\;&2^af(0)\in \mathbb{Z}
\end{align*}
and
\begin{align*}
&f(10n+8)\in \mathbb{Z}\\[4pt]
\implies\;&(10n+8)g(10n+8)+f(0)\in \mathbb{Z}\\[4pt]
\implies\;&\left(2(5n+4)\frac{G(10n+8)}{D}\right)+f(0)\in \mathbb{Z}\\[4pt]
\implies\;&5^bf(0)\in \mathbb{Z}
\end{align*}
hence, when reduced to lowest terms, the denominator of $f(0)$ is both a power of $2$ and a power of $5$.

It follows that $f(0)$ is an integer.

Suppose $f$ is linear.
Then
$f(n) = an+b$.

$f(10n+5)
=a(10n+5)+b
=50n+5a+b
$
and
$f(10m+8)
=a(10m+8)+b
=50m+8a+b
$.

Since these are integers,
as are $n$ and $m$,
so are
$5a+b=u$
and
$8a+b=v$.

Therefore
$8u-5v
=3b$
is an integer.

Similarly,
$5u-3v
=a+2b
$
is an integer.

Therefore
$3a = 3(a+2b)-2(3b)$
is an integer.
(Also from
$v-u = 3a$)

If
$a=b=\frac13$,
then
$5a+b = 2$
and
$8a+b = 3$,
so
$\frac13 n+\frac13$
works
without integer coefficients.


Find the multiplicative order of $15$ mod $ 257$ could someone please help me with this problem?  Thanks in advance!

Find the multiplicative order of $15$ modulo $257$.

Elementary theory in modular arithmetic says that $x^p-x\equiv0\mod p$ for any prime $p$. In this case, we have $15^{257}-15\equiv0\mod257$ so $15^{256}\equiv1\mod257$. The order of $15$ then, is a divisor of $256=2^8$. So it must be one of the numbers $1,2,4,8,16,32,64,128,256$.


How do you calculate the rate of change of the volume of a cone at a given height and radius when the radius is decreasing and the height is increasing?

i have to calculate the partial derivatives, right? but then do i add the values?

like V = partial derivative height + partial derivative radius??

It would help if you gave example numbers of some sort, but here is the general solution.
The radius $r$ is changing at the rate of $r'$, and the height $h$ is changing at the rate of $h'$. The volume $V$ has a rate of change of $V'$.

Now we know that $V = (\frac13\pi) r^2 h$. If you take the derivative of that, then you get (using product rule):

$V' = \frac13\pi\frac{d}{dt}(r^2h) = (\frac13\pi)(2rr'h + r^2h')$
All you have to do is plug in your current $r$ and $h$ values, and the rate of changes $r'$ and $h'$.

Just to make sure it is clear, $x'$ is the derivative of $x$.

The Radius of a cylinder is decreasing at a constant rate of 1 in per minute. The volume remains constant 415 cubic inches.  at the instant when the radius of the cylinder is 4 inches, what is the rate of change of the height?


How to find $N(\varepsilon)$ such that $f(n)N(\varepsilon)$ Where $f(n)=\dfrac{2}{8n^2-5}$ and $n\in \mathbb{N}$, $\varepsilon\in\ \mathbb{R}^{+}$, $N:\mathbb{R}^{+}\rightarrow \mathbb{N}$.
I realise that this is related to epsilon-delta limit proofs but I can't figure out how to get the answer.
I have attempted this by just solving for $f(n)=\varepsilon$, however I bieleve that I need to solve it using epsilon delta proof which I have no idea how to do.
I've researched the epsilon-delta limit proofs as an attempt to understand the problem but all the example problems give you the limit and make you prove, whereas this seems very different.
Sorry for my bad English.

Let $n \ge 3$.

$f(n)= \dfrac{2}{8n^2-5} \lt$

$\dfrac{2}{8n^2-n^2}\lt \dfrac{2}{2n^2} \lt 1/n.$

Let $\epsilon \gt 0$ be given.

Archimedean principle:

There is a $n_0$, positive integer, such that

$n_0> 1/\epsilon$.

For $n \ge n_0$ :

$f(n) \lt 1/n \le 1/n_0 \lt \epsilon$.

We have $0 <f(n) \le \frac{1}{n^2}\le \frac{1}{n}$ for all $n$.

Can you proceed ?


Proving the triangle inequality for the $l_2$ norm $\|x\|_2 = \sqrt{x_1^2+x_2^2+\cdots+x_n^2}$ I want to prove the triangle inequality for the l2-norm $\|x\|_2$:
$$\|x\|_2 = \sqrt{x_1^2+x_2^2+\cdots+x_n^2}$$

$$\begin{align}
  \sqrt {\sum\limits_{i = 1}^n (x_i + y_i)^2 }  &\leqslant \sqrt {\sum\limits_{i = 1}^n x_i^2 }  + \sqrt {\sum\limits_{i = 1}^n y_i^2} \cr
  \sum\limits_{i = 1}^n {(x_i + y_i)}^2  &\leqslant \sum\limits_{i = 1}^n x_i^2  + 2\sqrt {\sum\limits_{i = 1}^n x_i^2} \sqrt {\sum\limits_{i = 1}^n y_i^2 }  + \sum\limits_{i = 1}^n y_i^2 \cr
  \sum\limits_{i = 1}^n (x_i^2 + 2x_i y_i + y_i^2)  &\leqslant \sum\limits_{i = 1}^n x_i^2 + 2\sqrt {\sum\limits_{i = 1}^n x_i^2 } \sqrt {\sum\limits_{i = 1}^n y_i^2 }  + \sum\limits_{i = 1}^n y_i^2 \cr
  \sum\limits_{i = 1}^n x_i^2 + \sum\limits_{i = 1}^n y_i^2 + 2\sum\limits_{i = 1}^n x_i y_i  &\leqslant \sum\limits_{i = 1}^n x_i^2 + \sum\limits_{i = 1}^n y_i^2 + 2\sqrt {\sum\limits_{i = 1}^n x_i^2 } \sqrt {\sum\limits_{i = 1}^n {y_i^2} }  \cr
  \sum\limits_{i = 1}^n x_i y_i  &\leqslant \sqrt {\sum\limits_{i = 1}^n x_i^2 } \sqrt {\sum\limits_{i = 1}^n y_i^2 }  \end{align}$$
After this I squared both sides and multiplied out. It gets really messy, and then I got stuck. I think you have to realize some factorization and then conclude $0\leq \text{some terms}$

1.How do I finish this proof?  All help is greatly appreciated!

You want to prove that $$\lVert x+y\rVert \leq \lVert x\rVert+\lVert y\rVert$$
We use $\langle x,y\rangle$ to denote the inner (dot) product $$x\cdot y=\sum_{k=1}^n x_ky_k$$
Note that $$\lVert x+y\rVert ^2=\lVert x\rVert^2+2\langle x,y\rangle+\lVert y\rVert ^2$$
Continuing, use the Cauchy-Schwarz inequality, you get  $$\lVert x\rVert^2+2\langle x,y\rangle+\lVert y\rVert ^2\leq \lVert x\rVert^2+2 \lVert x\rVert \cdot \lVert y\rVert +\lVert y\rVert ^2=\left(\lVert x\rVert+\lVert y\rVert\right)^2 $$ and since all is positive, we can obtain your inequality by taking square roots.
The last equation you got is almost $$|\langle x,y\rangle|\leq \lVert x\rVert \cdot \lVert y\rVert $$
which is known as the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality is true for any inner product space, and here you have the canonical proof. Your inequality is a particular case of the Minkowski inequality for $L_p$ norms.
ADD To clear out the proof in Wikiepdia
Recall that the nature of roots of the polynomial $aX^2+bX+c$ are intimately related to its discriminant: $\Delta=b^2-4ac$ which appears in the formula $$X=\frac{-b\pm\sqrt{\Delta}}{2a}$$
If a quadratic polynomial is proven to satisfy $p(X)\geq 0$ then $\Delta \leq 0$: either there is one real root $\Delta=0$ or there are no real roots ($\Delta<0$) because the square root of the negative discriminant will give rise to complex numbers.

I did exactly what you did above. That was actually very good proof--better than using u and v since it is more abstract hence more generalized than the latter one. You should proceed with you work, since you are almost there.

To continue with what you wrote above, you should notice that the left side of the very last line of equation you got is the inner product of vector x and vector y: =x^Ty=||x||||y||*cos(theta) where theta is the angle between x and y. Now cos(theta) is always less or equal to 1. Hence you have the inequality. Hope this helps.


Function that is non-zero only at one point. I am searching for, if there exists, a continuous function $f(x)$ such that $f(x) = 0$ for all values of $x$, with the exception of one point (say $\tilde x$) where $f(\tilde x)\neq0$.

This is possible if and only if $\tilde x$ is an isolated point of the domain.

Example 1: For example, the domain could be $$(-\infty,\tilde x - \epsilon)\cup \{\tilde x\}\cup (\tilde x +\epsilon,\infty)$$ if it is a subset of $\mathbb R$ (which, incidentally, is not specified) and the function rule could be
$$f(x) = \begin{cases}
0 & \textrm{ if }x <\tilde x - \epsilon\\
c & \textrm{ if }x = \tilde x \\
0 & \textrm{ if }x > \tilde x +\epsilon
\end{cases}
$$

where $\epsilon$ is a positive constant and $c$ is a nonzero constant.

For example, you could have $$f:(-\infty,-1)\cup \{0\}\cup (1,\infty) \rightarrow \mathbb R$$

with

$$f(x) = \begin{cases}
0 & \textrm{ if }x <-1\\
1 & \textrm{ if }x = 0 \\
0 & \textrm{ if }x >1
\end{cases}
$$

Example 2: An even simpler example of such a function would be
$$f:\{0,1\} \rightarrow \mathbb R$$
with
$$f(x) = \begin{cases}
1 & \textrm{ if }x = 0 \\
0 & \textrm{ if }x =1
\end{cases}
$$

Everyone seems to be mistaken about what continuity of a function actually means. The Dirac Delta function answers the question (as others have already suggested). The Dirac delta is a continuous function; as defined by its integrability i.e. the n-dimensional integral of any n-dimensional Dirac delta function is unity (it would be naturally non-integrable if it was discontinuous).

Looking at it from a slightly more intuitive angle; it can be recognised that the Dirac delta is derived via the integration of a complex valued exponential function (which are naturally continuous).


What conditions can we implement for a function to determine it to be one-one using differentiation? My teacher told me that for a function to be one-one it's derivative needs to be greater than or equal to zero or less than or equal to zero. While it's true for the greater than or less than part I have some confusion regarding the equal to zero condition.

It's true for the function $x^3$, but it can never be true for $x^2$ or $x^4$.  If anyone can help me out it'd be greatly appreciated.

The function is on one if for every different x in domain  you get different y from codomain.  Even in $x^2$ or $ x^4$ the points at x=0 .
Let's consider y= $x^2$ for all $x\le 0$ function is decreasing and one one including point x=0. Also for $x\ge 0 $ function is increasing including point x=0. But for all $x\in R $ the function is many one.


If $a,b,c\in(0;+\infty)$, prove that $\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$. If $a,b,c\in(0;+\infty)$ and $$\frac{c}{1+a+b}+\frac{a}{1+b+c}+\frac{b}{1+c+a}\ge\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ca}{1+c+a}$$Prove that $$\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$

I know that $$a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}\ge ab+bc+ac$$

So we could prove that $$a+b+c+3\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$

I.e. $$2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le a+b+c+3$$

By using AM-GM we see that we could prove that: $$2(a+b+c)\le a+b+c+3\Rightarrow a+b+c\le3$$

So this didn't work.

The given condition is equivalent to:
$$\sum_{cyc}\frac{c-ab}{1+a+b} \ge 0$$
$$\implies \sum_{cyc}\frac{c(1+a+b)-ab}{1+a+b} \ge  \sum_{cyc}\frac{c(a+b)}{1+a+b} \qquad \text{adding to both sides}$$
$$\implies \sum_{cyc}c \ge  \sum_{cyc}\frac{c(a+b)+ab}{1+a+b} = (ab+bc+ca)\sum_{cyc}\frac{1}{1+a+b} \tag{1}$$

By Cauchy-Schwarz, we also have
$$\left(\sum_{cyc}\frac{1}{1+a+b}\right)\left(\sum_{cyc}(c+ca+bc)\right)\ge \left(\sum_{cyc} \sqrt{c} \right)^2 \tag{2} $$

Using this in $(1)$, we have:
$$a+b+c \ge (ab+bc+ca)\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{a+b+c+2(ab+bc+ca)}$$

Cross multiplying and expanding we get,
$$(a+b+c)^2 + 2(ab+bc+ca)(a+b+c)\ge (ab+bc+ca)(a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca})$$

Expanding the first term, cancelling part of the second term with RHS and dividing throughout by $ab+bc+ca$, we get the result.

Here are some thought I have, but no proof (yet):
Both the condition and the inequality to be proven have equality when $(a,b,c)=(x,x,x)$ and when $(a,b,c)=(1,1,4)$ (or permutations). We know that the inequalities of means only have equality when the term we apply them to are equal. Thus, we can never use AM-GM on $a$, $b$ and $c$, because $(1,1,4)$ wouldn't give equality and thus the inequality is to lossy.
If we apply AM-GM on $f(a,b,c)$, $g(a,b,c)$ and $h(a,b,c)$, we want $f(1,1,4)=g(1,1,4)=h(1,1,4)$, $f(1,4,1)=g(1,4,1)=h(1,4,1)$ and the same for $(1,1,4)$. I think it is not easy to find functions that satisfy this (apart from $f=g=h$, but then, you won't get any information from the inequality).
We can thus only use the given inequality and use rewriting of terms. Maybe we can also use Cauchy-Schwarz, because it has not-so-trivial cases for equality.


Why $\lim_{h\to 0}\frac{|f(a+h)-f(a)-\nabla f(a)\cdot h|}{\|h\|}=0$ only if $f$ is $\mathcal C^1$? Let $f:\mathbb R^n\longrightarrow \mathbb R$ and $a\in\mathbb R^n$ s.t. $\nabla f(a)$ exist.

I know that for a function $g:\mathbb R\longrightarrow \mathbb R$, it suffice that $g'(a)$ exist to have that $g$ is differentiable in $a$. Indeed,
$$\lim_{h\to 0}\frac{g(a+h)-g(a)-g'(a)h}{h}=\lim_{h\to 0}\left(\frac{g(a+h)-g(a)}{h}-g'(a)\right)=0$$
since $$\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}=g'(a).$$
Now I was wondering why in $\mathbb R^n$ the continuity of the gradient is requiert. So I try to compute :

$$\lim_{h\to 0}\frac{f(a+h)-f(a)-\nabla f(a)\cdot h}{\|h\|},$$
but I don't see in what the fact that $\nabla f$ is continuous at $a$ is necessary to conclude that the limit above is $0$. I mean, when we do the calculation, when do we need the fact that $\nabla f$ is continuous at $a$ ?

I think you have some confusion about what the theorems say. I don't have the proofs or the counter examples available to me right now. But here is at least something for you.

The definition of the derivative of a function $f:R^2 \to R$ is as follows. At $a$, a derivative of $f$ exists at $a$ if there is a linear function $L: R^2 \to R$ such that
$$\lim_{h \to 0} \frac{f(a+h)-f(a)-L(h)}{||h||} = 0.$$
Remember that $h$ is a vector. We call $L$ the derivative.

This is a stronger condition than the partial derivatives of $f$ existing. That is, for $f(x,y)$, both
$$\lim_{\delta x \to 0} \frac{f(y,x+\delta x)-f(x,y)}{\delta x}$$
and
$$\lim_{\delta y \to 0} \frac{f(y+\delta y,x)-f(x,y)}{\delta y}$$
existing does not necessarily imply that there is a linear map $L$ that satisfies the above. If you want there to be just a linear map $L$, then you need additional conditions. Namely, that both $f_x$ and $f_y$ are continuous. I don't have the proof available to me so I forget the precise reason why. There is a proof of this in Rudin's Principles of Analysis.

The continuity of $\nabla f$ at $a $ is not necessary !

If  all partial derivatives of $f$ at $a$ exist, then we have:

f is differentiable at $a$ $\iff$

$\lim_{h\to 0}\frac{f(a+h)-f(a)-\nabla f(a)\cdot h}{\|h\|}=0$


Calculating run times of programs with asymptotic notation When calculating the run time of programs using asymptotic notation, I know how to set up the sums for things like for loops, but I'm getting stuck on summing them up.

Sorry if this is a dumb question but say we have something like

for i=4 to n^2
   for j = 5 to 3i log (i)
      s = s + ij

I would set this up with outer loop = sum from i=4 to n^2, inner loop sum from j = 5 to 3i log i, and finally a constant.  I'm getting hung up on how to sum everything up.  Can someone please explain this to me?

A simple way to determine runtime is to use summations.  Each loop becomes a sum, where the lower index is the starting value of the loop and the upper index is the ending value of the loop.  Because you're performing a constant time operation in the innermost loop, we are simply summing $1$:
\begin{align}\sum_{i=4}^{n^2} \sum_{j=5}^{3i\lg i} 1 &= \sum_{i=4}^{n^2}3i\lg i - 5 + 1\\
&\approx\sum_{i=1}^{n^2}i\ln i\end{align}
(I'm using $\approx$ to slide between expressions that are asymptotically equivalent throughout the problem.)

The last summation above is kind of difficult (its solution involves the hyperfactorial function).  So, we switch to an approach that is accessible to more students: a summation may be approximated by an integral.  So, we now compute:
\begin{align}
\sum_{i=1}^{n^2}i\ln i&\approx \int_1^{n^2}x\ln x\,dx \\
&\approx x^2\ln x + \mathcal O(x^2)\Bigg |_1^{n^2}\\
&\approx n^4\ln n
\end{align}

So, your program executes with $\mathcal{O}(n^4\ln n)$ time complexity.

Consider two loops $A$ and $B$ that are not nested. The body of A executes $n$ times and the body of B executes $k$ times. If we execute $B$ in the body of $A$, then $A$ still executes $n$ times, but $B$ executes whenever the body of $A$ executes. The body of $B$ executes $k$ times whenever the body of $A$ executes, and the body of $A$ executes $n$ times, so the body of $B$ executes $n \cdot k$ times.

When calculating the running time of an algorithm, you want to compare the number of steps in the algorithm to the input size. Say the input size for the algorithm $B$ nested inside of $A$ is $n$. You first want everything in terms of the input size, so say we know that $k = n/2$. Then the number of steps taken is $n \cdot k = 1/2 \cdot n^2 = c \cdot n^2 = O(n^2)$.

I hope this helps.


Rules for Product and Summation Notation When we deal with summation notation, there are some useful computational shortcuts, e.g.:
$$\sum\limits_{i=1}^{n} (2 + 3i) = \sum\limits_{i=1}^{n} 2 + \sum\limits_{i=1}^{n} 3i = 2n + \sum\limits_{i=1}^{n}3i$$

However, I don't think I know all the useful shortcuts here. Are there other computational tricks one should be aware of? What's a good way for thinking about this?

More importantly, consider product notation:
$$\prod\limits_{i=1}^{n} (\sqrt{2} - \sqrt[n]{2})$$

I don't know what the shortcuts here are. What are some of the more effective ways of attacking such computations?

I suppose $\prod\limits_{i=1}^{n}(x)$ means multiplying from 1 to n, which is n factorial. So maybe it will help. Also $(\sqrt{2} - \sqrt[n]{2})$  does not contain any i.


Prove Borel sigma-algebra translation invariant Can anyone explain: Let $B$ be a Borel set and $B + a = \{ x + a : x \in B\}$. Why is $B + a$ a Borel set?

I think I have to use some good set principle but not sure how to complete the proof.

Translation ($T_a(x) = x+a$) is continuous, hence Borel measurable. Hence $B+a = T_{-a}^{-1} B$ is (Borel) measurable.
Alternative proof:
Let ${\cal B}$ be the Borel sets, and let ${\cal B'} = \{ A \in {\cal B} | A+ \{x\} \in {\cal B} \}$. Note that ${\cal B'} \subset {\cal B}$ and a little bit of work shows that ${\cal B'}$ is a $\sigma$ field.
Since the open intervals are in ${\cal B'}$ we have ${\cal B} \subset {\cal B'}$ and hence they are equal.

Borel measurable functions mean the preimage of Borel set is Borel. Therefore we use subtraction instead of addition.
    Take $B\subset \mathbb R$, and $t\in\mathbb R$, then take $f(b) = b-t$ $\forall b\in B$. This is continuous, thus Borel measurable. Therefore, $f^{-1}(B)$ is Borel measurable. However, $f^{-1}(B)$ is exactly $t+B$, therefore, $t+B$ is Borel measurable.


Pigeonhole principle for a triangle Consider a equilateral triangle of total area 1. Suppose 7 points are chosen inside. Show that some 3 points form a triangle of area $\leq\frac 14$.

Choose one point $p$ and draw a line from $p$ to each of the other six points.

We can order the six points $a,b,c,d,e,f$ going clockwise around point $p$ and draw lines from $a$ to $b$, from $b$ to $c$, from $c$ to $d$, from $d$ to $e$ and from $e$ to $f$.

This gives five disjoint triangles which fit inside the triangle of area $1$.

Therefore the total area of the five triangles is less than or equal to one and at least one triangle has area less than or equal to $\frac 15$.

I have seen answers with 9 points (in comment link above) and 5 points (above).

However, not with 7 points. Please see below:

Drawlines from midpoints of all 3 sides to each other

Now we have 6 points (3 vertices and 3 mid points)

we have 4 triangles of equal area 1/4. (equilateral triangle)

To add the 7th point will create a triangle of area less than 1/4.

QED


Percentages question John brought a flat at $\\540,000$ and he sold it to Tanny at a loss. Tanny then sold it at a profit to Kimberly at $\\535,000$ .

If the loss percentage on the cost of John is equal to the percentage profit of Tanny on his cost , find the price that Tanny paid for the flat

I'm stuck by just reading the question... Can I get a hint on how to solve this ?

Thanks in advance ..

Since you don't know how much Tanny paid, assign a name to that quantity.
The letter $x$ is a good choice. Then $x$ is the unknown value you want to
find out.

Take the way you would have calculated John's loss and Tanny's profit
if you had known the price Tanny paid, but use the symbol $x$ instead
of a known price.
Write out the calculations of how much John lost (as a percentage of
his cost, which was $540,000$) and how much
Tanny gained (as a percentage of his cost, which was $x$).
You will not be able to get a simple numerical percentage out of these
because of the unknown value $x$, but you will have two expressions
involving some known numbers, the unknown $x$, and some operations,
each representing a percentage.

The "is equal to" part of the problem tells you these two
percentages are equal. So you can put them on each side of an $=$ sign.
Now you have an equation, and the only unknown quantity in the equation
is $x$. You have to solve for $x$.

Hint: If the first step goes OK, on one side you'll have a fraction that is kind of ugly because it has $x$ in both the numerator and the denominator. You may want to recall rules such as cross-multiplying so you don't have an $x$ in a denominator.

Second hint: you'll probably end up wanting to solve a quadratic equation in order to finish.

It will be harder to follow these hints than if I had written out all
the formulas explicitly. What I'm trying to do here is to explain a little
of what would go through my head when looking at a problem like this,
and how I would get past the "what does this question even mean?" phase
and get something I could actually use some math to solve.

$$540000(1-\frac{x}{100})(1+\frac{x}{100})=535000$$
on solving, $x = 9.62\%$
Cost paid by tanny = $488038$

Reference for formulas used: www.prep4paper.com


Monotone Convergence Theorem One simple case of the monotone convergence theorem for integration is:

Let $E \subset \mathbb{R}^n$ and suppose that $f_k : E \rightarrow \mathbb{R}$ is a sequence of non-negative measurable functions which increases monotonically to a limit $f$. Then $f : E \rightarrow \mathbb{R}$ is measurable and
\begin{equation*} \lim_k \int_E f_k = \int_E f \end{equation*}
where here we mean the Lesbegue integral on $\mathbb{R}^n$.

I know one proof of this theorem which is very measure-theoretic, and I was wondering is there a non-measure theoretic proof for the theorem if we only assume that the $f_k$ are Riemann integrable?

There are proofs of the monotone and bounded convergence theorems for Riemann integrable functions that do not use measure theory, going back to Arzelà in 1885, at least for the case where $E=[a,b]\subset\mathbb R$.  For the reason t.b. indicated in a comment, you have to assume that the limit function is Riemann integrable.  A reference is W.A.J. Luxemburg's "Arzelà's Dominated Convergence Theorem for the Riemann Integral," accessible through JSTOR.  If you don't have access to JSTOR, the same proofs are given in Kaczor and Nowak's Problems in mathematical analysis (which cites Luxemburg's article as the source).

In the spirit of a comment by Dylan Moreland, I'll mention that I found the article by Googling "monotone convergence" "riemann integrable", which brings up many other apparently helpful sources.

It is precisely uniform convergence that gives us the Riemann integrability of $f$.  Monotonicity is not sufficient, as t.b.'s comment shows.  Assuming $f$ is Riemann integrable to begin with is sort of cheating, although it may be useful in specific examples.


are non-degenerate critical points always isolated? I have a question regarding the isolation of critical points of a function:

Suppose $f : \mathbb{R}^n \to \mathbb{R}$ is a $C^\infty$ function such that $f$ has a non - degenerate critical point at $0 \in \mathbb{R}^n$. That is, we have $\triangledown f (0) = 0$, and the Hessian $\left((\partial^2 f/ \partial x_i \partial x_j ) (0) \right)$ is invertible.

Can I deduce from this that the critical point at $0$ is an isolated critical point ? My guess is to say yes, because the fact that the Hessian is non-degenerate forces $f$ to change its value in the vicinity, and in all directions. But I am unsure, in particular with regards to the last statements ("in all directions") - though that should be encoded by the fact all eigenvalues of the Hessian are non - zero.

Is this the right way to think about the question of whether the critical point is isolated ?
Thanks for your thoughts !

Yes, this is a right way of thinking about the question.  You can see this in several ways.

1: The Morse lemma:  The Morse lemma declares that about any nondegenerate critical point $p$ of a smooth function $f$ there is a coordinate neighborhood $x_i$ so that in those coordinates, $f(x_i) = x_1^2 + \cdots + x_i^2 - x_{i+1}^2 - \cdots - x_n^2 + f(p)$.  It's an easy exercise to compute the gradient in these coordinates, and you can immediately see that the only zero of the gradient in the neighborhood is at $p$.  (In this case, $p = 0$ and the claim is that you can precompose $f$ with a diffeomorphism $\varphi$ of $\mathbb{R}^n$ so that $f\circ\varphi = \sum x_i^2 - \sum x_j^2$.)

2: Directly:  Take a curve (easiest, a straight line) $\gamma$ through $0$, and for simplicity's sake, parametrize the curve so that $\gamma(0)=0$.  Consider $f|_\gamma$.  Then $f'(t) = (\nabla f\cdot \gamma')(t)$ and $f''(t) = (\gamma'^THess_f\gamma')(t)$.  Leaving out some details you should verify as an exercise, the Hessian is invertible, so in a neighborhood of $0$, $f$ has a local extremum, hence the zero of $f'$ is isolated.  This is true for any curve, thus, every direction, so $0$ is an isolated critical point.  This formalizes your intuition about "every direction."

In Neal answer ,there is a flaw in the proof. Hessian(x0) being non-degenerate does not imply f is P.D or N.D hence x(0) not necessarily a local extremum. e.g saddle point.
Here is my version: for x in N(0), the Hessian(x) is non-degenerate since eigenvalues of matrix continously depends on entry hence x.
in N(0) any x,y we have
grad(x)-grad(y)=H(x-y)  +o(|x-y|^2)
H is non-degenrate ,then the map from x to grad(x) is one-to one linear map in N(0).therefore grad(x) implies x=0


Can someone help me solve this limits question? $$\begin{align}\lim x → ∞\end{align}$$

$$\begin{align}
f(x) = {\frac{2^{x+1}+{3^{x+1}}}{2^x + 3^x}} \\
\end{align}$$

I tried using L Hopitable but that gives the same expression. Also tried using substitution but I didn't get anywhere. Help would be appreciated.

Here is another idea. Try writing,
\begin{align*}
\frac{2^{x+1}+3^{x+1}}{2^x+3^x} &= \frac{2 \cdot 2^x+3 \cdot 3^x}{2^x+3^x} \\
&=\frac{3\cdot 2^x - 2^x+3 \cdot 3^x}{2^x+3^x} \\
&=\frac{3\left( 2^x+3^x \right)-2^x}{2^x+3^x} \\
&=3-\frac{2^x}{2^x+3^x} \\
&=3-\frac{\frac{2^x}{2^x}}{\frac{2^x}{2^x}+\frac{3^x}{2^x}} \\
&=3-\frac{1}{1+\left( \frac{3}{2} \right)^x}.
\end{align*}
Now most people would just look at this and say
$$\lim_{x \rightarrow \infty} \left( 3-\frac{1}{1+\left( \frac{3}{2} \right)^x} \right)=3.$$

To begin with
$$f(x)=\frac{2\cdot(\frac23)^x+3}{(\frac23)^x+1}$$
it follows that
$$\lim_{x\to+\infty}f(x)=3.$$
However
$$f(x)=\frac{2+3\cdot(\frac32)^x}{1+(\frac32)^x}$$
in such a case, we get that
$$\lim_{x\to-\infty}f(x)=2.$$


Trace-Determinant Plane I'm a bit confused about why a spiral sink or spiral source in a linearization guarantees that the original, nonlinear system also has a spiral sink or spiral source, respectively.  Let's state the rational for why a center in a linearization is inconclusive on the type of the critical point in the original, nonlinear system.

  If the center in the linearization is the point in the
  trace-determinant plane ($0$, $D_L$), then the associated point in the
  nonlinear system may not be ($0$, $D_L$), since the linearization is
  only an approximation; it may be a nearby point ($T_N$, $D_N$)
  instead, where $T_N \neq 0$.

My question is, why isn't it possible for there to be a different system of ODEs where ($T_N$, $D_N$) is the linearized point and ($0$, $D_L$) is the nonlinearized point, in which case the spiral sink or spiral source in the linearization would also be inconclusive?

First of all, it's misleading (and maybe actually the cause of your confusion?) to say that the nonlinear system corresponds to another point in the trace-determinant plane, since “trace” and “determinant” refers specifically to properties of linear systems (only). The nonlinear system corresponds to the same point in the trace-determinant plane as its linearization; it's just that you have additional nonlinear terms which are not taken into account by looking only at the trace and the determinant.

But anyway, the intuitive explanation goes as follows:

If the trace of the linearized system is zero, then the linear system has closed orbits, which is a very delicate property; the slightest nonlinear perturbation might cause the orbit to not return exactly where it started, and instead go into a spiral.

But if the linearized system is a spiral sink, then its orbits are spirals going inwards (the orbits return a bit inwards from where they started, after one lap), and if the perturbation is sufficiently small (which it is if you start close enough to the equilibrium point), then the orbits of the nonlinear system will do the same (maybe they will return a bit further out, but still inwards from where they started, so they too go into a spiral).


How to compute matrices raised to large powers? Compute the matrix $$ Q = \begin{pmatrix} \frac{3}{4} &   4-\frac{\sqrt 3}{4}\\ 4+\frac{\sqrt 3}{4} &  -\frac{1}{4}\end{pmatrix}$$ raised to power $2005$.

Is there any algebraic property that can be used always as a formula? I don't have a very good intuition for algebra so help me if I am missing this because of poor basics. Thank you :)

I am sure this has been answered before, but I am unable to find the original. (Edit: found it!) In general, suppose $\lambda_1,\lambda_2$ are the two eigenvalues of a $2\times2$ matrix $Q$ (note: $Q$ doesn't need to be diagonalisable). Then Cayley-Hamilton theorem says that $(Q-\lambda_1I)(Q-\lambda_2I)=0$. Now suppose we want to find $Q^m$ for some integer $m\ge0$. By long division, we may write
$$
x^m=(x-\lambda_1)(x-\lambda_2)p(x)+ax+b\tag{1}
$$
for some polynomial $p$ and some numbers $a$ and $b$. It follows that $Q^m=aQ+bI$ and all we need is to determine $a$ and $b$. In case $\lambda_1\ne\lambda_2$, by substituting $x=\lambda_1,\lambda_2$ into $(1)$, we get $\lambda_1^m=a\lambda_1+b$ and $\lambda_2^m=a\lambda_2+b$. Solving them, we get $a=\dfrac{\lambda_1^m-\lambda_2^m}{\lambda_1-\lambda_2},\ b=\dfrac{\lambda_1\lambda_2^m-\lambda_2\lambda_1^m}{\lambda_1-\lambda_2}$.

You can for example do matrix logarithm + exponential followed by usage of power laws:

$$\log(X^k) = k \log(X) \Rightarrow \\X^k = \exp(k\log(X))$$

which means if the matrix logarithm converges, you can first calculate it, then multiply with the scalar $k$ and then exponentiate the result.

One simple (but not extremely efficient) way to calculate exp and log is by their usual Taylor series from calculus in one variable.


Show the Beta function $B(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$ is defined for $x,y > 0$ without actually integrating An old exam question: Show that
$$B(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$$ exists for all $x,y>0$.

I'm sure because of time allotment that it's not in the scope of the question to actually integrate.

Is there an elegant way to show that the integral exists without calculating it explicitely?

Hint:

It is easy to just integrate $t^{x-1}$ for $x>0$, and since $(1-t)^{y-1}\approx 1$ in a neighborhood of $t=0$, the integral converges near $t=0$.

Likewise, the same holds for $t=1$ and $(1-t)^{y-1}$.

$\frac{\Gamma{(\alpha+\beta)}}{\Gamma{(\alpha)}\Gamma{(\beta)}}t^{x-1}(1-t)^{y-1}$ is the Beta density and the area under the curve is 1

Then $\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma{(\alpha)}\Gamma{(\beta)}}{\Gamma{(\alpha+\beta)}}$


Name for geometry that differs by a translation I have a simplistic question:  If I have two triangles, and there exists a translation that makes them equivalent (all their vertices would be the same after the translation), then is there a special term in geometry that I would use to describe the relationship between the two triangles?

We say that one triangle is the translate of the other. You could also say that they are translationally equivalent.

Yes, there is.

We say that the triangles (or shapes in general) are congruent. The relationship is known as congruency.


An asymptotic from below to number of solutions to $xyz + x + y = n$ Let $n$ be a positive integer. One can show (not that easy, but still via elementary methods) that the number of triples $(x,y,z)$ of positive integers satisfying $xyz + x + y = n$ is $O(n^{\frac{1}{3}+\varepsilon})$ for any $\varepsilon > 0$. (Use that $xz+1$ divides $n-x$, take without loss of generality $x< n^{\frac{1}{3}}$ or $z<n^{\frac{1}{3}}$, etc.)

Hence I was wondering - is it also true that this number is at least $Cn^{\frac{1}{3}}$ for some constant $C>0$?

No. I'll show there are arbitrarily large $n$ with at most $c(\log n)^2$ solutions.

Let $s(n)$ be the number of solutions $(x, y, z)$ of $xyz + x + y = n$, and $t(n)$ the number of solutions of $xyz \leq n$, so we have $t(n) \geq \sum_{k=1}^n s(k)$, since $xyz + x + y \leq n$ implies $xyz \leq n$. Note that for given $x, y$, the number of $z$ with $xyz \leq n$ is $\lfloor n/xy \rfloor$, hence
$$t(n) = \sum_{1 \leq x, y \leq n} \left \lfloor \frac{n}{xy} \right\rfloor \leq \sum_{1 \leq x, y \leq n} \frac{n}{xy} = nH_n^2 \leq 2n(\log n)^2$$
holds for sufficiently large $n$. If we also had $s(n) > 16(\log n)^2$ for large $n$, this would mean
$$t(n) \geq \sum_{k=\lceil n/2 \rceil}^n s(k) > \sum_{k=\lceil n/2 \rceil}^n 16(\log k)^2 \geq (n/2) 16(\log (n/2))^2 \geq 2n(\log n)^2$$
for sufficiently large $n$, a contradiction, so there are arbitrarily large $n$ with $s(n) \leq 16(\log n)^2$.

xyz+x+y=n.  ----- (1)

Solution is,

(x,y,z,n)=((p-38),(p-11),(1),(p^2-47p+369))

for  p>38

example, p=40, (x,y,z,n)=(2,29,1,89)

since 'p' can take many value's, equation (1) has infinite number of solutions.


Question about 2D PDE Suppose I have this PDE
$$
\frac{\partial A(r,\theta)}{\partial r} + \frac{\partial B(r,\theta)}{\partial \theta} = 0
$$
Is it ok to conclude (without losing any generality) that
\begin{align}
A(r,\theta) &= \frac{\partial \psi(r,\theta)}{\partial \theta} \\
B(r,\theta) &= -\frac{\partial \psi(r,\theta)}{\partial r}
\end{align}
for some unknown function $\psi$? This would allow me to write the PDE in terms of one unknown instead of 2. Of course I am assuming all functions are smooth, differentiable, etc.

It's OK locally, that is, as long as $A(r, \theta)$ and $B(r, \theta)$ are considered to be defined in some open ball $B(p, \epsilon) = \{ q \in \Bbb R^2 \mid \vert q - p \vert \ < \epsilon \}$ about some point $p = (r_0, \theta_0) \in \Bbb R^2$.

I will use subscript notation for partial derivatives throughout; thus

$A_r = \dfrac{\partial \psi}{\partial r}, \; B_\theta = \dfrac{\partial B}{\partial \theta}, \tag 0$

and so forth.

I will exploit the language of differential forms.

Consider the one form defined on $B(p, \epsilon)$,

$\omega(r, \theta) = -B(r, \theta) \; dr + A(r, \theta) \; d\theta; \tag 1$

we have

$d\omega= -dB \wedge dr + dA \wedge d \theta$
$= -(B_r \; dr \wedge dr + B_\theta \; d \theta \wedge dr) + A_r \; dr \wedge d\theta + A_\theta \; d\theta \wedge d\theta; \tag 2$

since

$dr \wedge dr = d\theta \wedge d\theta = 0, \tag 3$

(2) becomes

$d\omega = -B_\theta \; d\theta \wedge dr + A_r \; dr \wedge d\theta$
$= B_\theta \; dr \wedge d\theta + A_r \; dr \wedge d\theta = (B_\theta + A_r) \; dr \wedge d\theta = 0; \tag 4$

thus $\omega$ is a closed one-form; as such it is also exact, by the Poincare Lemma; thus there is a zero-form, or function, $\psi$ on $B(p, \epsilon)$ such that

$-Bd\theta + A dr = \omega = d\psi = \psi_r \; dr + \psi_\theta \; d\theta; \tag 5$

then

$A = \psi_r \tag 6$

and

$-B = \psi_\theta, \tag 7$

or

$B = -\psi_\theta. \tag 8$

Of course, all of this may be stated in a more old-school style by using the notation of vector calculus; the reader may easily check that the condition

$A_r + B_\theta = 0 \tag{9}$

is sufficient to conclude that

$\nabla \times (-B, A) = (\partial/\partial r, \partial/\partial \theta) \times (-B, A) = 0; \tag{10}$

thus $(-B, A)$ must be a gradient, that is,

$(-B, A) = (\psi_\theta, \psi_r), \tag{11}$

which is the requisite result.

That would be nice, but it's not ok, I'm afraid. The problem is that writing $A$ and $B$ that way forces a relationship between $A$ and $B$ that the original PDE does not assume.

Now, if you have additional information, like you're supposed to have harmonic functions, that would be another matter.


Evaluate limit without L'Hopital's rule How can this limit be evaluated without the use of L'Hopital's rule. I already understand how to evaluate it with the use of it.

\begin{equation}
\lim_{x\to 0} (\frac{e^{3x}-1}{x})
\end{equation}

I am wondering if there is some sort of substitution I can use to prove it can evaluate to 3. Assume knowledge of first-principals.

This fact was used in the answer for this question.
Evaluating Limit Without L'Hopital

I do not follow the solution to this question Evaluate $\lim_{x\to 0} \frac{a^x -1}{x}$ without applying L'Hopital's Rule. If someone can expand further on that that would also be great!

I am not looking for similar answers to these: Show $\lim\limits_{h\to 0} \frac{(a^h-1)}{h}$ exists without l'Hôpital or even referencing $e$ or natural log

Recall that by standard limit as $x\to 0$

$$\frac{e^x-1}{x}\to 1$$

then by $y=3x\to 0$

$$\frac{e^{3x}-1}{x}=3\frac{e^{3x}-1}{3x}=3\frac{e^y-1}{y}\to 3$$

Put $t=e^{3x} -1$ and find $x$ using log transformation.
You will be left with limit as $\frac{3t}{\ln(1+t)}$ then use log expansion you will get required answer.


Is there a way of solving integrals where the numerator is an integral of the denominator? Is there a way of solving integrals where the numerator is an integral of the denominator?

I was evaluating the integral $$\int \frac{x-\sin x}{1-\cos x}\mathrm{d}x$$. I separated the numerator into two fractions and solved it.

However, the numerator seems to be the integral of the denominator. So, is there an alter method to do this integral by taking advantage of this fact?

$\int \frac{x-\sin x}{1-\cos x}\mathrm dx$
$\int\frac x{1-\cos x}dx-\int\frac{\sin x}{1-\cos x} dx$
Integrate by parts for the first one and then substitute $t=\tan(\frac{x}2)$
For the second one simple put $\cos x=u$
As far as I know, for $\int\frac{f(x)}{f'(x)}dx$, I don't think there is a unique method.


About the maximal condition on subgroups and normality Let $G$ be a group which satisfies the maximal condition on subgroups. Now, let $H$ be a subgroup containing a non trivial conjugate (i.e. there exists an element $x\in G,\,\notin H $ such that $H^x\leq H$). Is it true that, in this case, $H$ is normal in $G$? How can we prove that? Is solvability of any help?

Note

A group $G$ is said to satisfy the maximal condition on subgroups iff it is impossibile to find an infinite strictly ascendant chain of subgroups.

Observe that any finite nilpotent group $\;G\;$ both fulfills the maximal condition for subgroups and also for any proper subgroup $\;H\;$ we have that $\;H\lneqq N_G(H)\;$ which means there exists $\;x\in G\setminus H\;\;s.t.\;\;H^x\le H\;$ ...and this doesn't make the subgroup $\;H\;$ normal.

I may be missing something, but the maximal condition simply tells you that if $H^x \leq H$ then $H^x = H.$ In other words, $H$ has normalizer strictly containing $H.$ It does not seem to imply that $H$ is normal.


How solve for x in an Infinite exponent How would one solve for x in the following equation:

$x^{x^{x^{x^{\cdots}}}} = 4$

The exponent continues forever... So what is the value of x?

Thank you for helping

An interesting dynamical system is hidden behind the curtain. In the socalled power tower previously mentioned one usually looks at $0<x<1$ for which the map $f_x(y)=x^y$ is decreasing and there is a well-defined meaning to the tower-limit which is simply the unique fixed point of the map. There is a catch to the problem, however, when $x>1$. In that case there are two fixed points for $1<x<1.444...$ In the case mentioned, starting with $y=x$, and iterating $y\mapsto x^y$ you will end up at the fixed point $y=2$ and not 4 as expected! On the other hand $y=4$ is indeed a fixed point, though unstable.  Try on a pocket calculator or similar (not maple, though) to iterate $y\mapsto (\sqrt{2})^y$ starting with $y=4$. Takes about 50 iterations before you see that the fixed point is indeed unstable.

The difficulty is perhaps: how do you define $x^{x^{x\ldots}}$?

We could define it as the limit of successive towers, defining the sequence $y_{n}=x^{y_{n-1}}$ and $y_{1}=x$. Then, assuming this sequence has a limit (!), this limit satisfies $x^{y}=y$, by continuity of exponentiation.

If we insist, say, that $x$ is positive, then the equation $y=4$ is equivalent to the equation $x^{y}=x^{4}$, i.e. $x^{4}=4$. This has the positive solution $x=\sqrt{2}$.


$n^{th}$ derivative of $\log(x)/x$ The problem
Given: $$x\, f(x) = \log(x) \qquad \forall\; x > 0$$
we need to prove that the $n^{th}$ derivative of $\,f(x)\,$ at $\,x = 1\,$ is: $$f^{(n)}(1) = (-1)^{n+1}\, n! \, \left( 1 + \frac{1}{2} + \ldots + \frac{1}{n} \right)$$
What I tried
Proved for  $\,n=1, 2\,$  and tried applying mathematical induction, wherefrom I am getting  $\,f^{(n)}(x)\,$  from  $\,f^{(n-1)}(x)$,$\:$  but  $\,f^{(n)}(1)\,$ seems underivable from $\,f^{(n-1)}(1)$.

The suggestive form in which the function is written can make things a bit neater, rather than just taking a bunch of derivatives of $\ln(x)/x$ (which could be messy).
So I'll take one derivative of both sides of your equation, using the product rule on the left side, and get
$$xf'(x)+f(x)=\frac{1}{x}.$$
That tells you what $f'$ is. Take another derivative to get
$$xf''(x)+f'(x)+f'(x) = xf''(x)+2f'(x) = -\frac{1}{x^2},$$
which allows you to solve for $f''$ in terms of $f'$. Let's try once more to try to see a pattern. We find
$$xf^{(3)}(x)+f''(x)+2f''(x)=xf^{(3)}(x)+3f^{(2)}(x)=\frac{2}{x^3}.$$
Maybe one more derivative gives
$$xf^{(4)}(x)+4f^{(3)}(x) = -\frac{2 \cdot 3}{x^4}.$$
At this point, I'd be comfortable positing that the following is true for all $n \geq 1$:
$$xf^{(n)}(x)+nf^{(n-1)}(x)=(-1)^{n-1}\frac{(n-1)!}{x^n}$$
You should be able to prove this easily by induction. Plugging in $x=1$, you can then use this to prove your claim by induction.

Instead we could find the $n$th derivative of
$$g(x) = f(x+1) = \frac{\log(1+x)}{1+x}$$
at $x=0$. We have that
$$xg(x) + g(x) = g(0)+\sum_{n=1}^{\infty} \left[\frac{g^{(n)}(0)+ng^{(n-1)}(0)}{n!}\right]x^n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n$$
which gives us the recurrence relation
$$\begin{cases}g^{(n)}(0)+ng^{(n-1)}(0) = (-1)^{n+1}(n-1)! \\ g^{(1)}(0) = 1 \\ \end{cases}$$
The recurrence relation can be solved by making your ansatz exactly the proposed solution.


Induction proof for a function being in $C^{\infty}(\mathbb{R})$ Let $$f(t) = \begin{cases}
     0 & \text{: } x \le 0\\
     e^{-\frac{1}{t^2}}& \text{: } x > 0
\end{cases}$$
Prove that $f(t)$ is in $C^{\infty}(\mathbb{R})$.

I know that for a function to be in $C^{\infty}(\mathbb{R})$ it has to be infinitely times differentiable and each derivative must be continuous. I'm a bit stuck on how to prove this. I'm assuming the answer will need to be proved inductively. Ie:
Suppose the $f^{(n)}(t)$ is continuous. I need to then prove that $f^{n+1}(t)$ exists and is continuous. I'm not quite sure how to do this, I'm new to proofs by induction. I can also image proving continuity at $t=0$ may be a challenge.

Hint.
A standard example of a smooth non-analytic function is
$$
g(x)=\left\{\begin{array}{ll}e^{-\frac{1}{x}} & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{array}\right.
$$
A detailed proof for smoothness is given in this Wikipedia article:
https://en.wikipedia.org/wiki/Non-analytic_smooth_function#The_function_is_smooth
You can adapt the techniques for your example. A key step is to note that for any positive integer $m$,
$$
\lim_{x\downarrow 0}\frac{e^{-\frac1{x}}}{x^{m}}=0
$$
Alternatively, observe that
$$
f(x)=g(h(x))
$$
where $h(x)=x^2$ for $x\ge 0$ and $h(x)=0$ for $x<0$. You can show that $h$ is smooth. The result then follows from the chain rule.

Find all derivatives of $f(x)$ everywhere on $\mathbb{R}$ except for $t = 0$. Then prove that all of them in fact are continuous at zero.


How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$ let $x,y,z>0$, find the minimum of the value
  $$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$

I think we can use AM-GM inequality to find it.

$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4]{xy^3}$$
$$3x+z=x+x+x+z\ge 4\sqrt[4]{x^3z}$$
but this is not true,because not all four equalities can hold at once.

This problem is from china middle school test,so I think have without Lagrange methods,so I think this inequality have AM-GM inequality

Here is another way, using Holder's inequality:
$$\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz} \geqslant \frac{(\sqrt[4]{5y \cdot 5\cdot x\cdot z}+\sqrt[4]{2\cdot2z\cdot3y\cdot3x})^4}{xyz} \\ = (\sqrt5+\sqrt6)^4=241+44\sqrt{30}$$

with equality iff $\dfrac{5y}2=\dfrac5{2z}=\dfrac{x}{3y}=\dfrac{z}{3x} \iff (x, y, z) = \left(1, \sqrt\frac2{15}, \sqrt\frac{15}2 \right)$.


Number of different possible game scores on a quiz game with n questions On a quiz game, players get 5 points for a correct answer, 2 points for not answering, and 0 points for a wrong answer. If there are n questions, find a way to determine how many possible scores there are.

I'm thinking this isn't as hard as I think it is. All even numbers are a multiple of 2, and odd numbers ≥5 are given by 5 + multiples of 2. Would it be all integers between the maximum possible score (5n) and 0, discard scores 1 and 3?
Even in this way, though, don't know how the 0 points for a wrong answer would be incorporated.

Edit: 1 and 3 don't work. I think 5n -1 , 5n - 2, 5n - 4 and 5n-7 don't work for a total of 6 solutions that don't work? So 5n - 6

On a quiz with $n$ questions, the score will be between 0 and $5n$ so there are at most $5n+1$ possibilities.

As observed above, scores of 1 and 3 are not possible at the ''bad'' end of the score range. If we instead consider the ''good'' end of possible high scores, we can observe the following. For any question not answered we lose 3 points, and for every incorrect answer we lose 5 points. Thus the scores
$$5n - 1,\, 5n-2,\, 5n-4,  \quad\text{and} \quad5n-7$$
are not possible, since the numbers $1, 2, 4,$ and $7$ cannot be written as a sum of $3$'s and $5$'s.

In all we have found 6 scores that are not possible: 2 in the ''bad'' range, and 4 in the ''good'' range. If we additionally assume that $5n - 7 > 3$, i.e. $n > 2$, then these are in fact 6 distinct scores that are not possible so there are at most $5n-5$ scores which are possible.

I'll make the claim that all remaining $5n-5$ scores are possible, and leave this proof as an exercise.

Thus the answer is $5n-5$ when $n\geq 3$, and  the remaining cases can be computed separately: 6 possibilities with 2 questions and 3 possibilities with 1 question.

If you are looking for the number of possible scores as stated in your first paragraph rather than all the literal point values you could use combinations:
for the first 4 questions this formula translates simply to the combinations formula without repetition:
$$
\frac{(r+n-1)!}{r!(n-1)!)}
$$
This formula is chosen rather than a permutation formula because the order in which you get different scoring possibilities does not matter, and you are able to get repetitions of the same score. note: that n in this example is not the same as "n" questions, I'll use "r" for questions.

To test if it worked later and to make it easier- turn the scoring n possibilities into variables:

5 points=A
2 ponts = B
0 points = C
n=3 possibilities of scores, or n=3
there are 3 possibilities for every r questions. lets use two examples

EXAMPLE 1: 2 questions
n=3
r=2
$$
\frac{(2+3-1)!}{2!(3-1)!)}=6
$$
you can test this by checking the combinations of points by using the assigned variables from above:
1: AA =10 points
2: AB, BA= 7 points
3: AC, CA =5 points
4: BB =4 points
5: BC, CB =2 points
6: CC =0 points
Note: even if you swap around these variables you still get the same points.. AB=BA=7 points

EXAMPLE 2: 4 questions
n=3
r=4
$$
\frac{(4+3-1)!}{4!(3-1)!)}=15
$$

There are special cases:

for 5, 6, and 7 questions the possible scores are:
5 questions =20
6 questions = 25
7 questions = 30

because after 4 questions, the different combinations can end up resulting in the same number of points.

for every n>=4, the formula is:
5(n-1)


Prove that $\lim_{n \to \infty} \frac{n!}{n^n} = 0$ Prove that

  $\lim_{n \to \infty} \frac{n!}{n^n} = 0$

I've already considered using l'Hoptials rules but I cannot take the derivative of a factorial (as it is a discrete function).

Thanks

The easiest way is to get an explicit upper bound for the fraction:

$0 < \dfrac{n!}{n^n} = \dfrac{1\cdot 2\cdot ...\cdot n}{n\cdot n\cdot n\cdot...\cdot n} < \dfrac{1}{n}$, and $\dfrac{1}{n} \to 0$. So the answer follows by squeeze theorem.

Another approach: Stirling's approximation tells you $n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$. That tells you the thing you're taking the limit of is approximately $\frac{2 \pi n}{ e^n}$. This goes to 0 as $n \to \infty$.

Alternatively, you can use the bound from Stirling's approximation is $n! \leq e \sqrt{n} n^n e^{-n}$. Then, $0 \leq \frac{n!}{n^n} \leq e \sqrt{n} e^{-n} \to 0$.


Sum of positive definite matrices still positive definite? I have two matrices, which are square, symmetric, and positive definite.
I would like to prove that the sum of the two matrices still have the same properties,
that is square, symmetric, and positive definite.
The first two properties are obvious, what about the positive definite property.
Any clue to the proof?
Thank you.

A real matrix $M$ is positive-definite if and only if it is symmetric and $u^TMu > 0$ for all nonzero vectors $u$.

Now if $A$ and $B$ are positive-definite, $u^T(A+B)u = \ldots?$

A+B, or B+A, is positive definite if both A and B are positive definite. Suppose A is a m1*n1 matrix and B is a m2*n2 matrix. Because you can sum them up, m1=m2, n1=n2. Since then, as you add up these two matrices, the properties of leading principal minors will not change from the old. Since then, A+B is positive definite. Good luck!


When is the sum of a quasi-convex function and a convex function quasi-convex? I know that the sum of a convex function and a quasi-convex function is not necessarily convex, it is easy to construct counter-examples like $f(x)=-x$ and $g(x)=x-1/2|x|$.
However, I have a setting (dynamic programming with minimization function involved, so I cannot simplify it easily) where I strongly assume that my sum of convex function and quasi-convex function is a quasi-convex function.
Within my search for this, I am a bit stuck on where to start. Are there some well-known properties on one or both of my functions that would make the sum a quasi-convex function?

This is not always true. In fact, constructing a counterexample is an exercise in Bauschke & Combettes' book (volume 2), Exercise 10.15. Without seeing your particular functions, it is hard to say whether or not their sum will be quasiconvex.


Is there a bijection between uncountable sets? I know that there is a bijection between naturals and rationals. I also know that there is no bijection between naturals and reals (diagonal argument).

But, I have never heard of the existence of a bijection between uncountable sets (ex aleph-one). Is there a way to create a (computable ?) function that takes an element from an uncountable set and outputs (in infinite time ?) an element from another uncountable set ?

(I do not have a strong mathematical background, so please keep it simple or use terms of computer science)

[EDIT]
It seems that my question was very trivial. An answer would be y = f(R) where f is just one-to-one. I was hoping for something more sophosticated :( . Sorry for the inconvenience.

[EDIT2]
How we would construct a bijection between these sets ?
A = reals
B = reals without naturals
C = reals without primes

Example: from $\Bbb R$ to $(-\pi/2,\,\pi/2)$ with $\arctan x$.

Use $\sim$ to indicate the existence of a bijection between two sets.

Thanks to a theorem of Cantor, there is no set $A$ for which
$$A\sim \mathcal P(A)$$
so not every uncountable set can be made into bijection.
Moreover thanks to Cantor-Bernstein theorem, two infinite sets $A,B$ can be made in a bijection if and only if
$$\exists A_1 \subset A: A_1\sim B$$
$$\exists B_1 \subset B: B_1\sim A$$
lastly, the question about computability does not hold since Turing machines only deal with a countable imput


Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.

Suppose it is not true that  $1\ge\frac{3}{x(x-2)}$, so  $1\lt\frac{3}{x(x-2)}$.  Then  $\frac{3}{x(x-2)}-1\gt0$.  Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.

$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$

${3-(x(x-2)\gt0}$

${3-x^2-2x\gt0}$

${-x^2-2x+3\gt0}$

${-1(x^2+2x-3)\gt0}$

$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$

${(x-1)(x+3)\lt0}$

At this point I really do not know what to do after this point or if I really even went about it the right way.  Thank you for the help.

$x > 3
\implies x-2 > 1
\implies x(x-2) > 3
\implies 1 > \dfrac{3}{x(x-2)}
$

Your method is correct. You missed the sign in this step:
$$3-(x(x-2))>0 \Rightarrow -x^2\color{red}{+}2x+3>0 \Rightarrow (x+1)(x-3)<0.$$
The solution is $x\in (-1,3)$, which contradicts the condition $x>3$. So the original claim is true.

@fleablood's example: If $-1<x<3$, prove $1\le \frac{3}{x(x-2)}$.

Assume for $-1<x<3$, it is true: $1>\frac{3}{x(x-2)}$.

Consider two cases:
$$\text{Case 1:} \ \ x\in (-1,1)\cup (2,3): \ \ \ 1>\frac{3}{x(x-2)} \Rightarrow x(x-2)>3 \Rightarrow x\in (-\infty,-1)\cup (3,+\infty) \Rightarrow \text{contradiction}; \\
\text{Case 2:} \ \ x\in (1,2): \ \ \ 1>\frac{3}{x(x-2)} \Rightarrow x(x-2)<3 \Rightarrow x\in (-1,3) \Rightarrow \text{not contradiction}.$$
Since Case 2 is not contradicting, the original claim is not true.


Integration using maple I am trying to evaluate the following integral using maple but it returns the unsolved integral. Can anybody help me in using maple to solve this integral?

$$
\int {2\pi\lambda R e^{-\lambda \pi R^2}} D^{a\pi R^2} (a \pi R^2)^C dR
$$

Regrouping and ignoring non-essential constants, the integral can be rewritten
$$\int R^{2c+1}\alpha^{R^2}dR.$$
And by the change of variable $t=R^2$:
$$\int t^c\alpha^tdt.$$
Maple should recognize the exponential integral function.

Maple finds it, when putting D (which is reserved for the differentiation operator) to be a local variable and the other parameters to be real:
$$local D; int(2*Pi*R*exp(-lambda*Pi*R^2)*D^{a*Pi*R^2}*(a*Pi*R^2)^C, R) $$  $$\, assuming \, real$$ outputs that,
 where the $\Gamma$ function is described here.