Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Which is the easiest and the fastest way to find the remainder when $17^{17}$ is divided by $64$? Which is the easiest and the fastest way to find the remainder when $17^{17}$ is divided by $64$?
| $17^4=(16+1)^4=16^4+4\cdot 16^3+6\cdot 16^2+4\cdot 16+1 \equiv 1 \pmod {64}$
(Just notice that all numbers $16^4$, $16^3$, $16^2$ and $4\cdot 16$ are multiples of $64$.)
$17^{17} = (17^4)^4\cdot 17 \equiv 1\cdot 17 \pmod {64}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the limit of $\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$ $$\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$$
Can't find a means to resolve. The answer is $10$ by graphing.
| Completing the squares:
$x^2+5x+2 = (x + 2.5)^2 + O(1)$
$x^2+0.5x+1 = (x + 0.25)^2 + O(1)$
So:
$\sqrt{x^2+5x+2} = x + 2.5 + o(1)$
$\sqrt{x^2+0.5x+1} = x + 0.25 + o(1)$
Now the answer is immediate.
Updated to address didier's concerns:
We show using elementary methods that $\sqrt{u^2+O(1)}=u+o(1)$, which is the ... | {
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Help with a geometry problem The problem says:
A triangle has its lengths in an arithmetic progression,
with difference d. The area of the triangle is t. Find the dimensions.
the solution says:
the notation can be even better if we make it more symmetrical, by making the
side lengths $b − d, b,$ and $b + d$ .
by Heron... | We have $4d^4=(2d^2)^2$, and therefore $4d^4+\frac{3}{16}t^2>(2d^2)^2$ and therefore
$$\sqrt{4d^4+\frac{3}{16}t^2}>2d^2.$$
It follows that $2d^2-\sqrt{4d^4+\frac{3}{16}t^2}<0$, so $2d^2-\sqrt{4d^4+\frac{3}{16}t^2}$ does not have a (real) square root.
| {
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Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Possible Duplicate:
Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction
How do I prove the following?
$$\sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{... | For any $r \in [1, n]$, we have
$$
\frac{1}{\sqrt{r}} \geqslant \frac{1}{\sqrt{n}}.
$$
with strict inequality for $1 \leqslant r \lt n$. Adding all these $n$ inequalities,
$$
\sum_{r=1}^{n} \frac{1}{\sqrt{r}} \gt n \cdot \frac{1}{\sqrt{n}} = \sqrt{n}.
$$
Proof using induction. The OP requested a proof using induct... | {
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$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
How can you compute the limit of
$\sum \limits_{n=1}^{\infty} n(2/3)^n$
Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?
| Just a curioursity.
Using Fourier series one comes to:
$$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx = 6\ \frac{13 \cos x-12}{(13-12\cos x)^2}$$
hence:
$$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n = \sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx \Bigg|_{x=0} =6\ \frac{13 \cos x-12}{(13-12\cos x... | {
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Simplifying rational fractions I can't get this one either for whatever reason, spent about 20 minutes on it and I have made no progress at all.
$$\frac{x^2}{(x^2-4)} - \frac{x+1}{x+2}.$$
I know that I can simplify this into one fraction so I make it $$\frac{x^2}{x+2}-\frac{(x+1)(x^2-4)}{(x^2-4)(x+2)}$$
I then can simp... | As first step we may use the common denominator $(x^{2}-4)=(x-2)(x+2)$
because the $\text{lcm}\left( (x-2)(x+2),(x+2)\right) =(x-2)(x+2)$
$$
\begin{eqnarray*}
\frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2} &=&\frac{x^{2}}{(x-2)(x+2)}-\frac{
\left( x+1\right) (x-2)}{\left( x+2\right) (x-2)} \\
&=&\frac{x^{2}-\left( x+1\ri... | {
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solve complex equation $x^8 = \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt[8]{\frac{2}{\sqrt{2}}}(\cos \frac{\pi}{4} + i \sin{\frac{\pi}{4}})}{2 \cos \frac{\pi}{6} + i \sin \frac{3\pi}{2}}$
What's the way to solve this kind of equation? I think there must be 8 solutions.
I tried to solve the following two equations
$a^6 = 1+... | $(1+i)/\sqrt(2) = e^{i \pi/4}$
$(\sqrt(3)-i)/2= e^{-i \pi/6}$
You should obtain something like $z^8= a e^{i b}$ that should be solvable!
| {
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How to get sine / cosine value out of tangens I know that: $\tan(\alpha) = 1/2$.
How can I get clean values for sine / cosine without the calculator?
Is there a relationship?
I know that $\sin(\arctan(1/2))$ is a way ... But I hope you get the point.
Thank you!
| Here is an algebraic solution.
Recall that $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$. Now your equation states that
$$\sin \alpha = \frac{1}{2}\cos \alpha.$$
Squaring both sides gives
$$\sin^2 \alpha = \frac{1}{4} \cos^2 \alpha = \frac{1}{4}(1 - \sin^2 \alpha).$$
Or
$$\sin^2 \alpha = \frac{1}{5}.$$
Similarly one ... | {
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Integration by partial fraction mistake I am trying to understand my mistake using the partial fraction method to solve $ \displaystyle\int{\frac{2x-5}{x^2-4x+4}} $.
Here's what I have so far:
$$ \frac{2x+5}{x^3-4x^2+4x} = \frac{2x+5}{x(x-2)^2} = \frac{A}{x} + \frac{B}{(x-2)^2}$$
And its easy to see that $A=\displayst... | If the denominator has no repeated factor, then it's easy to express it into partial fraction. For example
$$\frac{-2x+3}{x^3-x} =\frac{-2x+3}{x(x-1)(x+1)}= \frac{A}{x} + \frac{B}{(x-1)} + \frac{C}{(x+1)},$$
$$\frac{7}{x^4-3x^2+2} =\frac{-2x+3}{(x-2)(x+2)(x-1)(x+1)}= \frac{A}{x-2} + \frac{B}{x+2} + \frac{C}{x-1}+\frac{... | {
"language": "en",
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How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers? This was Problem 3 (first day) of the 1990 IMO. A full solution can be found here.
How many rationals of the form $\large \frac{2^n+1}{n^2},$ $(n \in \mathbb{N} )$ are integers?
The possible values of $n$ that i am able to find is $n=1$ and $n=3... | Andre's modification of a wrong answer :)
If $n=3^k$, then
$$2^n+1=2^{3^k}+1=2^{3 \cdot 3^{k-1}}+1= (2^{3^{k-1}}+1)( 2^{2 \cdot 3^{k-1}}-2^{ \cdot 3^{k-1}}+1) $$
The second bracket is never divisible by $9$, thus by induction one can prove that $3^{2k-1}$ doesn't divide $2^n+1$.
Note: Since Geoff's answer was wrong, a... | {
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How could we solve $x$, in $|x+1|-|1-x|=2$? How could we solve $x$, in $|x+1|-|1-x|=2$?
Please suggest a analytical way that I could use in other problems too like this $ |x+1|+|1-x|=2$ and of this genre.
Thank you,
| (This isn’t really an answer, but it’s too big for a comment. It’s just continuing from where Nana stopped in their answer, revealing that you can’t find the solution using that approach.)
We have just squared both sides to try to get rid of the square roots.
$$ \left( \sqrt{(x+1)^2} \right)^2 = \left( 2 + \sqrt{(1-x)^... | {
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How can I solve a complicated induction exercise? (formula for sum of fourth powers) I have an induction exercise:
It is for $n \in \mathbb{N_0}$. Show:
$$ \sum\limits_{k=1}^{n} k^4 = \frac 1 {30} n(n+1)(2n+1)(3n^2+3n-1) $$
As far as I understand it, you have to put in $(n+1)^4$ at the end and you have to resolve it to... | By induction we show that:
$$1+2^4+3^4+....+n^4 =\frac 1 {5}n^5+\frac 1 {2}n^4+\frac 1 {3}n^3-\frac 1 {30}n$$
For $n=1$:
$$1=\frac 1 {30}\cdot2\cdot3\cdot5=1$$
Now suppose that is true for $n$, and show that and show that it is also true for $n+1$, that is:
$$1+2^4+3^4+....+n^4+(n+1)^4 =\frac 1 {5}(n+1)^5+\frac 1 {2}(n... | {
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Solve $\frac{4}{x+1}+\frac{2}{x-2} \leq 3$ Solve
$$\frac{4}{x+1}+\frac{2}{x-2} \leq 3$$
I must be making a very stupid mistake somewhere ... been stuck on this for 1hr+ or even more ...
| One solution is given below. Another one, very close in spirit to your attempt, is given in a comment at the end.
It turns out that the numbers fit together perfectly! Rewrite our inequality as
$$\frac{4}{x+1}-4 +\frac{2}{x-2}-1 \le 0.$$
This simplifies to
$$-\frac{4x}{x+1}+\frac{x}{x-2} \le 0.$$
The problem now bre... | {
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Simple Partial Fractions Question For practice, I am integrating,
$$\int \frac{x}{3x^2 + 8x -3}dx$$
So, I can then factor it as,
$$\int \frac{x}{(3x-1)(x+3)}dx$$
By partial fractions, I decompose
$$\frac{x}{(3x-1)(x+3)}= \frac{A}{3x-1} + \frac{B}{x+3}$$
For finding $A$, I multiply both sides by $3x-1$, which gives
$$\f... | Wolfram forgot the absolute value signs. Further both answers are the same.
$$\frac{9}{30}=\frac{3}{10}$$
| {
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How to solve the recurrence $T(n) = \frac{n}{2}T(\frac{n}{2}) + \log n$ I am trying to solve the recurrence below but I find myself stuck.
$T(n) = \frac{n}{2}T(\frac{n}{2}) + \log n$
I have tried coming up with a guess by drawing a recurrence tree. What I have found is
number of nodes at a level: $\frac{n}{2^{i}}$
ru... | Consider the equality
$$
1+x+x^2+...+x^n=\frac{x^{n + 1} -1}{x-1}
$$
Differentiate it by $x$, then multiply by $x$:
$$
x+2x^2+3x^3+...+n x^n=\frac{nx^{n+2} - (n + 1)x^{n+1} + x}{(x-1)^2}
$$
Now we can substitute $x=\frac{1}{2}$ and obtain
$$
\sum\limits_{i=0}^n\frac{i}{2^i}=n\left(\frac{1}{2}\right)^{n} - (n + 1)\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to set up the existence condition of an absolute value $$
\frac{\sqrt{4 + \arccos\left|\frac{2-x}{x+3}\right|}}{\sqrt{x^2 - 4x + 5} - 3}
$$
I'm trying to find the natural domain of the function above. I set up this conditions:
$$
\begin{cases}\sqrt{x^2 - 4x + 5} - 3\neq0&(denominator)\\x^2 - 4x + 5\ge0&(root)\\4... | We first do it in a slow and tedious way, and then a quick way. For the $\arccos$ to be defined, we need $\frac{|2-x|}{|3+x|} \le 1$ (it is naturally non-negative).
Thus we need the inequality $|2-x|\le |x+3|$. First suppose that $x \ge 2$. Then $|2-x|=x-2$, so we want $x-2 \le x+3$, which is true.
For $-3 <x<2$, we h... | {
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differential equation nondevelopable I try to solve this differential equation whose solution seems not to be constructable in power series
$y''+(x+a/x^2+b)y=0$, where $a$ and $b$ are some positive real numbers.
If one can help me please?
| Hint:
$y''+\left(x+\dfrac{a}{x^2}+b\right)y=0$
$x^2y''+(x^3+bx^2+a)y=0$
Let $y=x^ku$ ,
Then $y'=x^ku'+kx^{k-1}u$
$y''=x^ku''+kx^{k-1}u'+kx^{k-1}u'+k(k-1)x^{k-2}u=x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u$
$\therefore x^2(x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u)+(x^3+bx^2+a)x^ku=0$
$x^{k+2}u''+2kx^{k+1}u'+(x^3+bx^2+k(k-1)+a)x^ku=0$
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding Limit $\lim_{x \to \infty} (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$ Finding Limit
$$\lim_{x \to \infty} (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$$
So I let
$$y = (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$$
$\ln$ both sides:
$$\ln{y} = \frac{1}{x} \ln {(2^x + 3^x + 5^x + 7 ... | From the idea in the question here, for $x>0$:
$$
13^x<2^x+3^x+5^x+7^x+11^x+13^x <6\cdot 13^x;
$$
whence
$$
13 <(2^x+3^x+5^x+7^x+11^x+13^x )^{1/x}<6^{1/x}\cdot13.
$$
Now use the squeeze theorem.
| {
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Integration of rational functions.. Some rational function is giving me some trouble...
\begin{aligned}
\ \int \frac {x^2-9x+16}{(x-1)(x^2+6x-7)} dx
\end{aligned}
I simplified it like so:
\begin{aligned}
\ \int \frac {x^2-9x+16}{(x-1)^2(x+7)} dx
\end{aligned}
If I get it right I can transform it into this:
\begin{a... | You have:
$$
{x^2-9x+16\over(x-1)^2(x+7)}={A\over x-1\vphantom{(^2}}+{B\over (x-1)^2}+{C\over x+7\vphantom{(^2}}.
$$
Clearing denominators in the above equality gives:
$$\tag{1}
x^2-9x+16=A(x-1)(x+7)+B(x+7)+C(x-1)^2.
$$
This holds for all $x$. We can be clever and take advantage of this. Set $x=1$ in $(1)$ to obt... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$ Evaluating
$$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$$
Its $\frac{\infty}{\infty}$ so I should use L Hospital rule, but the terms are exponential and differentiation won't do much good? I am thinking maybe I somehow use $\ln$ both sides? But how? Or perhaps I ... | You can rewrite it as
$$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}=\sum_{n=1}^\infty \frac{\frac{1}{2}\cdot2^{2n}}{5\cdot 5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{2^{2n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{4^{n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n$$
which is a geometric series with rat... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Try to prove $\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$ $$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$
I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continu... | Let's cheat and use one of Euler's many results:
$$\sum_{i=1}^{n} \frac{1}{i} = \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$
Note that:
$$A_n + \sum_{i=1}^{n} \frac{1}{i} = \sum_{i=1}^{2n} \frac{1}{i}$$
Substituting Euler's result for both summations, we get:
$$A_n + \ln n + \gamma + \frac{1}{2n} + O\l... | {
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Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem:
Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$
I attempted to solve it by induction, but when proving for $n=... | Joe, since your original question was to prove that it is an integer:
An extension of Sivaram Ambikasaran's answer:
First claim that
$\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30} = 1^4 + 2^4 +\cdots + n^4$
And show your claim by induction. (This should be straightforward)
Once your claim is proven, the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$ As we know, for the $$f(x)f(y)=f(x+y)$$ $f(x)=\mathrm e^{\alpha x}$ is a solution.
What about
$f(x)f(y)=f(\sqrt{x^2+y^2})$?
Does anybody know about the solution of the function equation?
I tried to find $f(x)$.
See my attempts below to find $f(x)$.
$$f(x)=a_0+a_1x+\... | If you set $$g(x) := f(\sqrt{x})$$ for $x \in [0, \infty)$ then you get $$g(x)g(y) = f(\sqrt{x})f(\sqrt{y}) = f(\sqrt{x+y}) = g(x+y)$$
You see that $g(x) \geq 0$, and if $g(x) = 0$ for some $x > 0$ then $g \equiv 0$. Thus, you can look at $$h(x) := \log(g(x))$$ It satisfies $$h(x) + h(y) = \log(g(x)g(y)) = \log(g(x+y))... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 1
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how to find out remainder of $3^{256}$ divided by $13$ here is a question that about finding the remainder when dividing $3^{256}$ divided by $13$. Can anyone suggest how to find the solution
| In general when the modulus $m$ is small, you can check the successive powers $x^r$ modulo $m$, the remainders will cycle quickly.
For instance let's examine $4^r\pmod 7$
$\begin{cases}
4^0\equiv 1\pmod 7\\
4^1\equiv 4\pmod 7\\
4^2\equiv 16\equiv 2\pmod 7\\
4^3\equiv 4^2\times 4\equiv 2\times 4\equiv 8\equiv 1\pmod 7\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/116103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
Help to evaluate determinant I want to evaluate the determinant of the $n \times n$ matrix
$\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\
0 & 0 & \ldots & 0 & -a\\
0 & 0 & \ldots & -a & 0\\
&&&\vdots \\
0 & -a & 0 &\dots & 0
\end{array}\right|.$
So I try to say that it is $(-1)^{ n + (n-1) + \ldots n-(n-2)}(... | In fact, your answer is correct and agrees with the given answer. Here is the reason:$$\frac{(n-1)(n+2)}{2} + n-1=\frac{n(n-1)+2(n-1)}{2}+n-1=\frac{n(n-1)}{2}+2(n-1),$$
which implies that
$$(-1)^{\frac{(n-1)(n+2)}{2} + n-1}=(-1)^{\frac{n(n-1)}{2}+2(n-1)}=(-1)^{\frac{n(n-1)}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/116240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the value of equation? Suppose
\begin{align*}
a+b+c &= 20\\
\frac 1a + \frac 1b + \frac 1c &= 30
\end{align*}
Then find the value of
$$
\frac ab + \frac ba + \frac ac + \frac ca + \frac bc + \frac cb
$$
How i can apply A.M. $\ge$ G.M $\ge$ H.M. in this?
How i can achieve this?
| $$\frac ab + \frac ba + \frac ac + \frac ca + \frac bc + \frac cb $$ can be simplified as
$$ \frac{a^2c+b^2c+a^2b+bc^2+ab^2+ac^2}{abc} $$
which is
$$ \frac{\left[ac(a+c)+bc(b+c)+ab(a+b)\right]}{abc}$$
and that is
$$ \frac{\left[ac(20-b)+bc(20-a)+ab(20-c)\right]}{abc}$$
$$ = \frac{\left[20(ab+bc+ca)-3abc\right]}{abc} $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, ... | I think it would be easier for you to recall
$$\left(1+x+x^2+\cdots+x^{n-1}\right)(x-1) = x^n-1$$
and put $x=\dfrac{b}{a}$
$$\eqalign{
& \left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right)\left( {\frac{b}{a} - 1} \right) = \frac{{{b^n}}}{{{a^n}}} - 1 \cr
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 9,
"answer_id": 5
} |
Limit of a Sequence involving $\frac {n^2} {\sqrt{n^{6}+k}}$ I am stuck on this problem:
Compute the limit of the sequence $(a_{n})_{n=1}^{\infty}$ defined by
$$a_{n}:=\frac {n^2} {\sqrt{n^{6}+1}}+\frac {n^2} {\sqrt{n^{6}+2}}+\cdot \cdot \cdot + \frac {n^2} {\sqrt{n^{6}+n}}=\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}... | The conclusion is correct, but the step $$\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}=n$$ is clearly wrong, a fact that the other answers unfortunately failed to mention: each of the $n$ denominators is greater than $1$, so each term of the sum is less than $1$, and the sum itself is less than $n$.
The largest term in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Check my solution to $x^2 + x + 1 > 0$ I spent an hour or so yesterday trying to solve the inequality $x^2 + x + 1 > 0$. Since I'd spent so long on a problem didn't seem like it should be that difficult, I decided I'd call it a day and try it again later.
I just had another look at it and this solution became immediate... | First I will show that with small modifications, your approach can be made to work. (It is, however, simpler to use one of the methods in other answers.) At the end, I suggest an additional simple approach.
It is clear that $x^2+x+1 \ge 1$ if $x \ge 0$. There is potential trouble only when $x<$, so suppose that $x<0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
For complex $z$, $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0$
If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$
I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:
$$\frac{1 - x}{1 + x}$$
I tried a number of mani... | Yet another approach, is to note that $Re(\overline z_1z_2)=z_1 \cdot z_2$, where $\cdot$ denotes the euclidean dot product in $R^2$.
Let $z=x+iy$ for $x,y$ in $R$, then we have $Re(\frac{1-z}{1+z})=\overline{1-z}{\frac{1}{1+z}}=(1-x,y)\cdot (\frac{1+x}{1+2x+x^2+y^2},\frac{-y}{1+2x+x^2+y^2})=\frac{1-(x^2+y^2)}{2+2x}=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 9,
"answer_id": 4
} |
Sums of Double series where $a_{m,n}=\frac {m-n} {2^{m+n}}\frac {\left( m+n-1\right) !} {m!n!}$ I am trying to show if $$a_{m,n}=\dfrac {m-n} {2^{m+n}}\dfrac {\left( m+n-1\right) !} {m!n!}$$ such that $$\left( m,n>0\right) $$ that $\sum _{m=0}^{\infty }\left( \sum _{n=0}^{\infty }a_{m,n}\right) =-1$, $\sum _{n=0}^{\inf... | It's not clear how to define $a_{0,0}$, since $(-1)!$ is undefined. But if you take $a_{0,0}$ to be $0$ you should have $\sum_{n=1}^\infty a_{0,n} = -1$ while $\sum_{n=0}^\infty a_{m,n} = 0$ otherwise. Note that $$a_{m,n} = 2^{-m-n} \left({m+n-1 \choose n} - {m+n-1 \choose m} \right)$$
and
$$ { m+n-1\choose n} = (-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how many zeroes does 2012! have at the end?
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
How many zeroes does $2012!$ end with?
My idea is:
402 zeroes come from $2\times 5$, 80 from $2\times 25$, 16 from $2\times 125$ and 3 from $2\times 625... | The correct answer is 501.
In order to find the number of zeros is same as finding the number of factors of powers of $5$. There are more factors of powers of $2$ than the factors of powers of $5$.
For instance $10! = 3628800 = \hspace{3pt}2^8 \hspace{3pt} 3^4 \hspace{3pt}5^2\hspace{3pt} 7$
$$\left \lfloor \frac{n}{p} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Please help me prove by induction that $n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)$ Please help me prove by induction that
$$
n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)
$$
| This is proof without induction.
$$
1\cdot 3\cdot 5\cdot\ldots\cdot(2n-1)=
\frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}=
\frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}{(2\cdot 1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot\ldots\cdot (2\cdot n)}=
$$
$$
\frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
How to find curve equation from data? How do I find the formula when I only know some data points ?
Usually I just use the Trendline option for diagrams in Excel, but this one eludes me.
I expect it to be something like : Ax^2 + or - Bx + or - C.
Sample data:
X Y
1 4
2 8
3 13
4 18
5 24
6 30
7 37
8 4... | From plotting you data (Wolfram|Alpha link), it does not look linear. So it better be fit by a polynomial. I assume you want to fit the data:
X Y
1 4
2 8
3 13
4 18
5 24
..
using a quadratic polynomial $y = ax^2 + bx + c.$ If so, then put your data in a matrix form (note that $x^0, x^1, x^2, y$ below are no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$ is a $C^1$-function I would like to show that the function:
$$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$$
is a $C^1$-function.
$$ \frac{\partial f}{\partial x}(x,y)=\frac{\sin(y)-y\cos(x)}{x^2+y^2}+\frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}$$
$$ \frac{\partial f... | Because of the Taylor expansion $\sin(x) = x - {x^3 /6} + ...$, if $x$ and $y$ are small enough you can write $\sin(x) = x + E(x)$ and $\sin(y) = y + E(y)$, where $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$. So you have
$${2x(y\sin(x) - x\sin(y)) \over (x^2 + y^2)^2} = {2x(yx + yE(x) - xy - xE(y)) \over (x^2 + y^2)^2}$$
$$= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $n$ such that $\displaystyle{2^{n-2005}} | n!$ Let $n$ such that $\displaystyle{2^{n-2005}} | n!$
Prove that this number has at most
$2005$ non-zero digits when written in base $2$.
| there are $\sum_{k\ge 1} \lfloor \frac{n}{2^k}\rfloor$ 2s in $n!$. So we must have $n - 2005 \le \sum_{k\ge 1} \lfloor \frac{n}{2^k}\rfloor$. Now write $n = \sum_{i} 2^{\ell_i}$ in base $2$ with $\ell_1 < \ldots < \ell_m$. We have
\begin{align*}
\left \lfloor \frac{n}{2^k}\right\rfloor &=
\sum_{i:\ell_i \ge k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either
$(1) 0 \leq a,b \leq 1$
OR
$(2) ab \geq 3$
Since this question was under ... | Nice approach, you are very very close but this is how you finish it off: let $u^2=\cos x$ and $t^2=\cos y$ . Substituting, you have $u^2+t^2 \ge 2ut$, or $u^2-2ut+t^2 \ge 0$ , and this can be factored into $(u-t)^2 \ge 0$ , which is always true of course.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Plot $|z - i| + |z + i| = 16$ on the complex plane
Plot $|z - i| + |z + i| = 16$ on the complex plane
Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellip... | Don't we all love algebraic solutions...
The ellipse has an equation of the form
$$(\frac {x} {a})^2 + (\frac {y} {b})^2 = 1$$
The focals are aligned on the y axis, therefore
*
*a is the side of a rectangle triangle, the other site being 1 and the hypothenuse 16/2:
$$a = \sqrt{8^2-1}$$
*
*b is 16/2.
The equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Solving $\lim\limits_{(x,y)\to(0,0)}\;\frac{x^5 + \,y^5}{x^3+\,y^3}$ How do I solve the limit $$\lim_{(x,y)\to(0,0)}\;\frac{x^5+y^5}{x^3+y^3}\quad ?$$ I have tried using polar coordinates, but I don't think an answer would be valid because theta is not fixed. What else can I do?
| Hint
$$ \frac{x^5+y^5}{x^3+y^3} =\frac{x^5+x^2y^3}{x^3+y^3}+ \frac{x^3y^2+y^5}{x^3+y^3} - \frac{x^2y^2(x+y)}{x^3+y^3} $$
And
$$x^2-xy+y^2 \geq |xy| \,.$$
OK, to make it more clear. If you combine the two hints, you get:
$$\left|\frac{x^5+y^5}{x^3+y^3} \right| \leq x^2+y^2+|xy| \,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
how to find x in the following formula For a software project, I need to calculate some things. on of the formulas looks like:
26280 = 2 * SQRT((149,598,000,000 - x) / 1.4) + x / 10,217,588,910,387,300,000
My colegue says you can't solve the above equation (you won't be able to find x) I quite convinced you should be ... | First, it's useful to write the expression in the standard quadratic form. For ease of reading/writing, I'm going to set $A = 20,435,177,820,744,600,00$ and $B = 149,598,000,000$.
$$
\begin{align*}
\left(\frac{26280 - x}{A}\right)^2 &= \frac{B - x}{1.4}\\
\frac{(26280 - x)^2}{A^2} &= \frac{B}{1.4} - \frac{x}{1.4}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/128346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Differentiate respect to $x$ $(x^2+2x+1)^3$
let u=$x^2+2x+1$ $\frac{du}{dx} = 2x$ or $2x+2$ $\frac{dy}{dx}=3u^2 $
if $\frac{du}{dx} = 2x$ then
$3(x^2+2x+1)^2 (2x)$
answer is $6x(x^2+2x+1) $
Or if $\frac{du}{dx} = 2x+2$ then
$3(x^2+2x+1)^2 (2x+2)$
$6x(x^2+2x+1)+2$
however the right answer is $6(x+1)^5$
can ple... | Hint
$$\begin{align}\dfrac{du}{dx}&=2x+2=2(x+1)\\x^2+2x+1 &=(x+1)^2\\(x^a)^b&=x^{ab}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$ Finding $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$$
I suppose I need integration by parts and trigo substitution
Let $u=\frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2}$, $\frac{1}{x^2} = \tan{\theta}$. Is my substitution OK?
S... | $$I = \int \frac{\sqrt{x^4+1}}{x^3} \mathrm{d}x$$
Substitute $u=\sqrt{x^4+1}, \mathrm{d}u=\frac{2x^3}{\sqrt{x^4+1}}\mathrm{d}x$
$$\int \frac{u^2}{2(u^2-1)^{\frac{3}{2}}} \mathrm{d}u$$
Now substitute $u=\sec \theta$ $\mathrm{du} = \sec \theta \tan \theta \mathrm{d}\theta$
Then $(u^2-1)^{\frac{3}{2}} = \tan^3 \theta$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
| The imaginary parts of $\left(\cos\left(\frac{k\pi}5\right)+i\sin\left(\frac{k\pi}5\right)\right)^5=(-1)^k$ give
$$
\sin^5\left(\frac{k\pi}5\right)-10\sin^3\left(\frac{k\pi}5\right)\cos^2\left(\frac{k\pi}5\right)+5\sin\left(\frac{k\pi}5\right)\cos^4\left(\frac{k\pi}5\right)=0
$$
Dividing by $\sin\left(\frac{k\pi}5\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 18,
"answer_id": 15
} |
Proving $\sqrt 3$ is irrational. There is a very simple proof by means of divisibility that $\sqrt 2$ is irrational. I have to prove that $\sqrt 3$ is irrational too, as a homework. I have done it as follows, ad absurdum:
Suppose
$$\sqrt 3= \frac p q$$
with $p/q$ irreducible, then
$$\begin{align}
& 3q^2=p^2 \\
& 2q^2=p... | Here are some proofs I've found (link at bottom):
If $\sqrt 3 = m/n$:
$$ \frac{m}{n} = \sqrt 3 \frac{\sqrt 3 - 1}{\sqrt 3 - 1} = \frac{3-\sqrt 3}{\sqrt 3 - 1}
= \frac{3-m/n}{m/n-1} = \frac{3 n - m}{m-n}$$
and the right side has a smaller denominator, since $m < 2n$ (i.e., $\sqrt 3 < 2$).
$x = \sqrt{3} - 1$ is a root o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 10,
"answer_id": 7
} |
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but ... | $n^5 -n = n^5 + 6n -n = n(n^(4) - 1) + 6n
=n(n+1)(n^2 + 1)(n-1) + 6n
6(n^2 +1) m + 6n
6[{n^2+1+n}]$
by fermat 's theorem** $n^5$ is congruent to $n\mod 5$
$5|n^5 -n$
hence,
$30|n^5-n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 21,
"answer_id": 20
} |
How do I solve this equation? $\displaystyle \large \cos 2x + 1 - \sin 2x=\frac{2 \cos 2x \cos x}{\cos x + \sin x}$
I've been trying for a long time but I can't get it.
| Using the identities
$$
{\cos 2x=\cos^2 x -\sin^2 x},\quad \cos 2x=2\cos^2 x-1,\quad \sin 2x=2\sin x\cos x
$$
We have, if $\cos x+\sin x\ne0$
$$\eqalign{
{2 {\cos 2x} \cos x\over \cos x+\sin x}
&={2( {\cos^2 x-\sin^2 x}) \cos x\over \cos x+\sin x}\cr
&={2(\cos x+\sin x)(\cos x-\sin x) \cos x\over \cos x+\sin x}\cr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/133217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to calculate the improper integral $\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx$ How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$
| Here I give another way to solve the problem. We rewrite the integral as
\begin{eqnarray}
I&=&\int_0^\infty\log(x+\frac{1}{x})\frac{1}{1+x^2}dx\\
&=&\int_0^\infty\frac{\log(1+x^2)}{1+x^2}dx-\int_0^\infty\frac{\log x}{1+x^2}dx
\end{eqnarray}
Note
$$ \int_0^\infty\frac{\log(1+x^2)}{1+x^2}dx=\pi\log 2$$
from Evaluating $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions I have been trying to solve the recurrence:
\begin{align*}
a_{n+1}=\frac{2(n+1)a_n+5((n+1)!)}{3},
\end{align*}
where $a_0=5$, via generating functions with little success. My progress until now is this:
Let $A(x)=\sum_{n=0} ^{\infty} a_nx^n$. ... | A first order linear non-homogeneous recurrence:
$$
a_{n + 1} - c_n a_n = f_n
$$
can be reduced to a telescoping sum by dividing by the summing factor $s_n = \prod_{0 \le k \le n} c_n$:
$$
\begin{align*}
\frac{a_{n + 1}}{s_n} - \frac{a_n}{s_{n - 1}}
&= \frac{f_n}{s_n} \\
\sum_{0 \le n \le m - 1} \frac{a_{n + 1}}{s_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
Finding $\int e^{2x} \sin{4x} \, dx$ Finding $$\int e^{2x} \sin 4x \, dx$$
I think I should be doing integration by parts...
If I let $u=e^{2x} \Rightarrow du = 2e^{2x}$,
$dv = \sin{4x} \Rightarrow v = -\frac{1}{4} \cos{4x}$
$\int{ e^{2x} \sin{4x}} dx = e^{2x}(-\frac{1}{4}\cos 4x) - \color{red}{\int (-\frac{1}{4} \cos ... | To find $\int e^{2x}\sin4x\,dx,$ we can let $u=e^{2x}$ and $dv=\sin4x\,dx,$ which implies $du=2e^{2x}\,dx$ and $v=-\frac{1}{4}\cos4x.$ Therefore, $$\begin{align}\int e^{2x}\sin4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(-\frac{1}{4}\cos4x\right)-\int-\frac{1}{4}\cos4x\left(2e^{2x}\right)\,dx\\ &= -\frac{1}{4}e^{2x}\cos4x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/136595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Showing $\int_{0}^{\infty} \frac{1}{(x^2+1)^2(x^2+4)}=\frac{\pi}{18}$ via contour integration I want to show that:
$$\int_{0}^{\infty} \frac{1}{(x^2+1)^2(x^2+4)}=\frac{\pi}{18}$$
so considering:
$$\int_{\gamma} \frac{1}{(z^2+1)^2(z^2+4)}$$ where gamma is the curve going from $0$ to $-R$ along the real axis, from $-R$ ... | Consider the contour $C$ that spans along $-R$ to $R$ and around the arc $Re^{i\theta}$ for $0\le\theta\le \pi$.
Letting
$$f(z):=\frac{1}{(z^2+1)^2(z^2+4)}=\frac{1}{(z+i)^2(z-i)^2(z+2i)(z-2i)}$$
and we see the poles are located at $\pm i$ and $\pm 2i$. Letting $R \to \infty$, it is very clear that the denominator ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/137167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How do I show that the sum $(a+\frac12)^n+(b+\frac12)^n$ is an integer for only finitely many $n$?
Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$.
I tried hard but nothing seems to work. :(
| First of all, rewrite the equation as others have, to yield $(2a+1)^n+(2b+1)^n = m\cdot 2^n$, or in other words $(2a+1)^n+(2b+1)^n \equiv 0\pmod {2^n}$. Now, if $n=2k$ is even then for the equation to hold $\bmod 2^{2k}$ it must certainly hold $\bmod 4$; i.e., $\bigl((2a+1)^k\bigr)^2 + \bigl((2b+1)^k\bigr)^2\equiv 0\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ I need to show absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ is $$\frac{a^ab^bc^c}{(a+b+c)^{a+b+c}}$$
So I I do have $$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$
then I went on to equating each of the... | You were doing fine, the argument from the textbook that you are quoting is more symmetrical, that's all.
You obtained $x+\frac{b}{a}x+\frac{c}{a}x=1$. That almost finishes things! Multiply through by $a$. You get $(a+b+c)x=a$ and therefore
$$x=\frac{a}{a+b+c}.$$
From your $y=\frac{b}{a}x$ you can then get $y=\dfrac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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How to prove that $\int_0^1\left(\sum\limits_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx$ American Mathematical Monthly problem 11611 essentially asks you to show that
$$\lim_n\ n \int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx=2\log(2).\tag1$$
This wou... | Let $\mathcal{R}_n$ denote the integral on the right-hand-side of eq. (2):
$$
\mathcal{R}_n = \int_0^1 2 x^{n-1} \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x
$$
Consider
$$
\begin{eqnarray}
(n+1) \mathcal{R}_{n+1} - n \mathcal{R}_n &=& \int_0^1 2 \left( (n+1) x^{n} - n x^{n-1} \right) \log\left(1+\frac{1}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
} |
A trigonometric identity: $(\sin x)^{-2}+(\cos x)^{-2}=(\tan x+\cot x)^2$ I've been trying to prove it for a while, but can't seem to get anywhere.
$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$
Could someone please provide a valid proof?
I am not allowed to work on both sides of t... | Look at the largest triangle.
(There's a reason my avatar --the logo of my software company-- is a stylized version of this figure.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Solutions to $z^3 - (b+6) z^2 + 8 b^2 z - 7+b^2 = 0, b\in \mathbb R, z \in \mathbb C$ $z_1 = 1+i$ is a given solution.
I guess what I have to find is $z_2$ and $z_3$ in
$(z - (1 + i))(z - z_2)(z-z_3) = z^3 - (b+6) z^2 + 8 b^2 z - 7+b^2$.
I tried to divide the polynomial by $(z - (1 + i))$, but that didn’t seem to work ... | Consider the cubic polynomial
$$\begin{equation*}
P(z)=z^{3}+Az^{2}+Bz+C
\end{equation*}\tag{1},$$
where the coefficients $A,B$ and $C$ are real numbers. If we denote its
roots by $z_{1},z_{2}$ and $z_{3}$, then it factors as
$$\begin{eqnarray*}
P(z) &=&\left( z-z_{1}\right) \left( z-z_{2}\right) (z-z_{3}) \\
&=&z^{3}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/144512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using contour integration, or other means, is there a way to find a general form for $\int_{0}^{\infty}\frac{\sin^{n}(x)}{x^{n}} \, dx$? While studying some CA, I ran across methods of evaluating $$\int_0^\infty \frac{\sin x}{x} \, dx, \;\ \int_0^\infty \frac{\sin^2 x}{x^2} \, dx, \;\ \text{and} \ \int_0^\infty \frac{\... | I have a generalized elementary method for this problem，If f (x) is an even function, and the period is $\pi$,we have:
$$\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx=\int_{0}^\frac{\pi}{2}f(x)g_n(x)\sin^nxdx \qquad (1)$$
Where the $g_n(x)$ in (1) is as follows
$$g_n(x)=\begin{cases}\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_... | A "functorial approach" using the exterior product: If $\phi: V \rightarrow V$ is an endomorphism of a vector space, you may calculate the determinant of the endomorphism $\phi$ as the induced map
$$\wedge^n (\phi): \wedge^n V \rightarrow \wedge^n V$$
where $n:=\dim(V)$. Since $\wedge^n$ is a functor you get a canonica... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 1... | You've already solved the case $n=1$, so I'll not repeat that there.
Assuming as the induction hypothesis that $n$ has the property that $3|n(n^2+8)$, we can rewrite $(n+1)((n+1)^2+8)$ to obtain
$$
\begin{split}
(n+1)((n+1)^2+8) & = (n+1)(n^2+2n+1+8)\\
& =n(n^2+8)+n(2n+1)+((n+1)^2+8)\\
& = n(n^2+8)+3n^2+3n+9
\end{split... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/150425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 0
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Is $n = k \cdot p^2 + 1$ necessarily prime if $2^k \not\equiv 1 \pmod{n}$ and $2^{n-1} \equiv 1 \pmod{n}$? $p$ is an odd prime and $k$ is a positive integer.
Let $n=k \cdot p^2+1$.
If $2^k \not\equiv 1 \pmod n$ and $2^{n-1} \equiv 1 \pmod n$, is $n$ prime? If it is, why?
| Not necessarily. Let $p = 3$ and $k = 154$. Then $n = 3^2 * 154 + 1 = 1387$, $2^{154} \equiv 1024 \not\equiv 1 \pmod{1387}$, and $2^{1386} \equiv 1 \pmod{1387}$. But $n = 1387 = 19 * 73$ is not prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral of $\int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$ I am trying to find this by using integration by parts but I am not sure how to do it.
$$\int_0^{\pi/2} (\sin x)^7 (\cos x)^5 \mathrm{d} x$$
I tried rewriting as
$$\int_0^{\pi/2} \sin x\cdot\ (\sin x)^6\cdot\ (\cos x)^5 \mathrm{d} x = \int_0^{\pi/2}\s... | Beta version:
$$\begin{aligned}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{7}\left(\cos x\right)^{5}dx & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{6}\left(\cos x\right)^{4}2\sin x\cos xdx\\
& =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\left(\sin x\right)^{2}\right)^{3}\left(1-\left(\sin x\right)^{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/151516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
} |
Proving $\sin x + \sin x \cdot \cot^2 x = \csc x $ The exercise is to prove the trig identity by rewriting each side of the equation into the same form. However only the following identities can be used in the process:
$$\begin{align*}
\tan \theta &= \frac{\sin \theta}{\cos \theta}\\
(\sin \theta)^2 + (\cos \theta)^2 &... | Taking the left hand side, try cancelling the $\sin^2{x}$ with the $\sin{x}$ in the numerator
(i.e. $\sin{x} .(\frac{\cos^2{x}}{\sin^2{x}}) = \frac{\cos^2{x}}{\sin{x}}$)
Then add the remaining two terms ($\sin{x} + \frac{\cos^2{x}}{\sin{x}}$) and use the identity $\sin^2{x}+\cos^2{x}=1$. That should do the trick :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/152583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
The calculation of a series Calculate the series
\begin{equation}
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}.
\end{equation}
| We let
$$f(z)=\dfrac{1}{(4z+1)(4z+3)}$$
There are two poles of $f(z)$. They are $4z_0+3=0 \implies z_0=-\frac{3}{4}$ and $4z_1+1=0 \implies z_1=-\frac{1}{4}$.
Residue calculus tells us that
$$\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(\text{sum of residues of }\pi\cot(\pi z)f(z))$$
We calculate the residues ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/152636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Using the roots of polynomial finding the value of sum. If $a,b$ and $c$ are the roots of $x^{3}+px^{2}+qx+r$, then how can we find the value of $\displaystyle \sum \frac{b^{2}+c^{2}}{bc}$.
| I think you are asking for
$$\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca},\tag{$1$}$$
or perhaps twice this quantity.
If you bring Expression $(1)$ to a common denominator, you will get
$$\frac{a^2c+b^2c+b^2 a+c^2 a+c^2b+a^2b}{abc}.$$
Note that
$$(a+b+c)(ab+bc+ca)=a^2c+b^2c+b^2 a+c^2 a+c^2b+a^2b+3abc.$$
Th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating $\int \frac{dx}{1+3\sin x+\cos x}$ Can I get a detailed answer on integrating this? I am currently stumped.
$$
\int \frac{dx}{1+3\sin x+\cos x}
$$
Thanks.
| Let $y=x+\arctan(1/3)$, then because $\sin(x+\arctan(1/3))=\frac{3}{\sqrt{10}}\sin(x)+\frac{1}{\sqrt{10}}\cos(x)$
$$
\int\frac{\mathrm{d}x}{1+3\sin(x)+\cos(x)}=\int\frac{\mathrm{d}y}{1+\sqrt{10}\sin(y)}\tag{1}
$$
The standard substitution $z=\tan(y/2)$ yields $\sin(y)=\dfrac{2z}{1+z^2}$ and $\mathrm{d}y=\dfrac{2\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Integrate a complex function Suppose $\gamma$ is the circle $|z|=1$ traversed counterclockwise. Evaluate
$$
\int_\gamma\frac{\cot(z)}{z^2}dz
$$
I was able to show $0$ is a pole of order $3$ for the integrand, and I think I have to use residues, but I am stuck. Is this right so far?
| Yes, you are correct. The only pole inside the contour for the integrand is at $0$ with a multiplicity of $3$. You can make use of the Laurent series for $\cot(z)$ at $z=0$. $$\cot(z) = \dfrac1z - \dfrac{z}{3} + \mathcal{O}(z^3)$$
Hence, $$\dfrac{\cot(z)}{z^2} = \dfrac1{z^3} - \dfrac1{3z} + \mathcal{O}(z)$$
Hence, $$\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate these summations? How to find the values of these kind of summations:
$$\large\sum_{i=0}^6(6-i)\;\ast\;\sum_{j=1}^6(7-j)\;\ast\;\sum_{k=2}^7(8-k)\;\ast\;\sum_{\ell=3}^8(9-\ell)$$
| Use that $$\begin{align}\sum_{t=a}^b(c-t)&=\left(\sum_{t=a}^bc\right)-\left(\sum_{t=a}^b t\right)\\\\&=\left(\sum_{t=a}^bc\right)-\left(\sum_{s=0}^{b-a} (s+a)\right)\\\\ &=\left(\sum_{t=a}^bc\right)-\left(\sum_{s=0}^{b-a} s\right)-\left(\sum_{s=0}^{b-a}a\right)\\\\ &=(b-a+1)c-(b-a+1)a-\left(\sum_{s=0}^{b-a} s\right)\\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Reducing a fraction in algebra? How would I reduce the following fraction?
$$\frac{4A^2-B^2}{4A^2-4AB+B^2}$$
I am not sure how I would reduce it.
| The numerator is a difference of squares, so it factors:
$$4A^2 - B^2 = (2A)^2 - B^2= (2A-B)(2A+B).$$
The denominator is a perfect square:
$$4A^2 - 4AB + B^2 = (2A)^2 - 2(2A)B + B^2 = (2A-B)^2.$$
Then you can cancel one factor:
$$\frac{4A^2 - B^2}{4A^2 - 4AB+B^2} = \frac{(2A-B)(2A+B)}{(2A-B)^2} = \frac{2A+B}{2A-B}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving a biquadratic $x^{4}-2x^3 + x^2 - 2x +1 =0$ How do I find the roots of $$x^{4}-2x^3 + x^2 - 2x +1 =0$$
I am not able to find any roots by trial and error.
| *
*Divide throughout by $x^2$.
*Then you have $x^{2}-2x + 1 - \frac{2}{x} + \frac{1}{x^2} = 0$
*You can re-write as $x^{2}+\frac{1}{x^2} - 2(x+\frac{1}{x}) + 1 =0$
*Then use $x^{2}+\frac{1}{x^2} = (x+\frac{1}{x})^{2} -2$ and then reduce it to a quadratic by putting $y=x+\frac{1}{x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Multiplication in the field $F = \mathbb{Z}_2[x]/f(x)$ Let $f(x) = x^6 + x + 1$ and define the field $F = \mathbb{Z}_2[x]/f(x)$
Compute the following in this field:
1. $(x^5 + x + 1)(x^3 + x^2 +1)$
I start by multiplying (in $\mathbb{Z}_2[x]$):
$(x^5 + x + 1)(x^3 + x^2 +1)$ = $(x^8 + x^7 + x^5 +x^4 + x^2 + x + 1)$
T... | Your first idea is right, and I think you need to read the extended Euclidean algorithm over again (because $x^6+x+1$ being irreducible should not deter you.)
Dividing $\frac{x^6+x+1}{x+1}$ you'll find $(x+1)(x^5+x^4+x^3+x^2+x)+1=x^6+x+1$, and so $(x+1)(x^5+x^4+x^3+x^2+x)=1$ in the quotient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/156718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Simplifying a radical? How would I simplify the following two radicals.
$$\sqrt{\frac{X^3}{50}}$$
For my answer I got $\frac{X^2}{50X}$ but I am not sure if this is correct.
My second question is how would I simplify $\sqrt{\frac{1}{12}}$
| No, your simplification is incorrect.
The first radical only makes sense if $x\geq 0$. Assuming this is the case, remember that:
*
*If $a$ and $b$ are both nonnegative, then $\sqrt{ab}=\sqrt{a}\sqrt{b}$, and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$.
*$(\sqrt{a})^2 = a$ for any $a\geq 0$.
So:
$$\sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution to a system of quadratics I am learning about a Bell State, and am trying to show that they are entangled. I believe that the required proof is to show that the system
$$\alpha_0^2+\alpha_1^2=1$$
$$\beta_0^2+\beta_1^2=1$$
$$\alpha_0\beta_0=1/\sqrt{2}$$
$$\alpha_1\beta_1=1/\sqrt{2}$$
has no solutions. I have tr... | From (3) and (4), we have:
\begin{align*}
\beta_0^2 &= \frac{1}{2\alpha_0^2} \\
\beta_1^2 &= \frac{1}{2\alpha_1^2}
\end{align*}
Plug into (2):
\begin{align*}
\frac{1}{2\alpha_0^2} + \frac{1}{2\alpha_1^2} &= 1 \\
\frac{\alpha_1^2}{\alpha_0^2\alpha_1^2} + \frac{\alpha_0^2}{\alpha_0^2\alpha_1^2} &= 2 \\
\frac{\alpha_0^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac... | The eigenvalues of an $n$ by $n$ matrix consisting entirely of 1's are $n$ and $0.$ An eigenvector for $n$ can be all entries 1. $0$ has multiplicity $n-1,$ for $1 \leq i \leq n-1$ take an eigenvector to have mostly 0's, but $-1$ at position $i$ and 1 at position $n.$ Adding the identity matrix makes the eigenvalues $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Assume: $a,b,c >0$ prove that :
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
| Using the AM-GM inequality we obtain:
$$\frac{b+c}{a^{2}}+\frac{c+a}{b^{2}}+\frac{a+b}{c^{2}}=\frac{b^{3}c^{2}+b^{2}c^{3}+a^{2}c^{3}+a^{3}c^{2}+a^{3}b^{2}+a^{2}b^{3}}{a^{2}b^{2}c^{2}}\ge2\frac{a^{3}bc+b^{3}ac+c^{3}ab}{a^{2}b^{2}c^{2}}=2\frac{a^{2}+b^{2}+c^{2}}{abc}$$
Now a bit of juggling around proves an even stronger... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Finding all integer solutions for $x^2 - 2y^2 =2 $ I'd love your help with finding all the integer solutions to the following equation:
$x^2 - 2y^2 =2 $. I want to use Pell's theorem so I changed the equation to $-\frac{1}{2}x^2+ y^2 =-1$, Can I use Pell's Theorem now? I got a private solution for $-\frac{1}{2}x^2+ y^2... | Let's say $\alpha_n$ and $\beta_n$ the $n$-solution of the equation $x^2 - 2y^2 = 2$. We have: $$\left\{\begin{matrix}
\alpha_0 = 2
\\\beta_0 = 1
\end{matrix}\right.
\land \left\{\begin{matrix}
\alpha_1 = 10
\\\beta_1 = 7
\end{matrix}\right.
\land \left\{\begin{matrix}
\alpha_2 = 58
\\\beta_2 = 41
\end{matrix}\right.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integral of determinant Good evening. I need help with this task
$$
\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi{\det}^2\begin{Vmatrix}\sin \alpha x&\sin \alpha y&\sin \alpha z\\\sin \beta x&\sin \beta y&\sin \beta z\\\sin \gamma x&\sin \gamma y&\sin \gamma z\end{Vmatrix} \text{d}x\,\text{d}y\,\te... | This is a bit too long to comment. A change in notation $(\alpha, \beta, \gamma) \to (a,b,c)$.
If $a=0$ or $b=0$ or $c=0$ or $a^2 = b^2$ or $b^2=c^2$ or $c^2 = a^2$, the integral is zero.
Let's assume $a,b,c \in \mathbb{Z} \backslash \{0\}$ and are distinct. $$
D(x,y,z,a,b,c) = {\det}^2\begin{Vmatrix}\sin a x&\sin a y&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/163574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
} |
Finding $3$ distinct prime numbers $a| (bc+b+c)$, $b|(ac+a+c)$, $c|(ab+a+b)$ How to find $3$ prime numbers $a,b$ and $c$ such that:
$$a| (bc+b+c)$$
$$b|(ac+a+c)$$
$$c|(ab+a+b)$$
| There are no such primes:
Without loss of generality assume $a<b<c$. Clearly 2 cannot be one of the primes, so $a\geq 3$ , $b\geq 5$ and $c\geq 7$.
Now $abc+ab+bc+ca+a+b+c$ is divisible by $a$, by $b$ and by $c$ so it is also divisible by $abc$,
but this is impossible since
$1<\frac{abc+ab+bc+ca+a+b+c}{abc}<\frac{a+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/165038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $6m \mid (2m+3)^n + 1$ if and only if $4m \mid 3^n + 1$ Let $m$ and $n$ be positive integers. How to prove that $$6m \mid (2m+3)^n + 1$$ if and only if $$4m \mid 3^n + 1$$
| $\newcommand{\jaco}[2]{{\left(\frac{#1}{#2}\right)}}$I have tried to solve this - I hope I did not make too many mistakes there. Maybe there is much easier solution than this one.
Let us start by a few easy observations.
Since $2m+3 \equiv 3 \pmod m$, we see that
$$m\mid (2m+3)^n+1 \Leftrightarrow m\mid 3^n+1. \tag{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/165237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Value of $n$ for which an improper integral is convergent. A question from the Calculus book that I'm self-studying is asking me to determine the value of $n$ for which the improper integral below is convergent:
$$\int_1^{+\infty}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$$
My attempt is below:
Using the de... | Your method is correct and it is easy to formalize your neglecting of minor terms.
We can see that $$\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right) - \ln\left( \frac{b^n}{(2b^2)^{3/4}}\right)= \ln \left( (1+1/b)^n (1+n/(2b^2))^{-3/4} \right) \to \ln 1=0$$
as $b\to \infty.$ So the limit of the first term equals the li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Inequality $\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\cdots+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$ Let:$a,b,c,d>0$ be real numbers ,how to prove that :
$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\frac{b^2+c^2+d^2}{b^5+c^5+d^5}+\frac{c^2+d^2+a^2}{c^5+d^5+a^5}+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$$.
Ed... | This question can be solved only using the AM-GM inequality. Building on an idea that Pan Yang suggests in his comment, it suffices to show the following.
$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}\le\frac{d}{abcd}=\frac{1}{abc}$$
This is equivalent to showing that
$$a^3 bc+b^3ca+c^3ab\le a^5+b^5+c^5 $$
However, this is true b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Analytic expression for the primitive of square root of a quadratic Can an analytic expression be given for
$$\int \sqrt{ax^2 + bx +c} \, dx$$
I think substitution doesn't work in this case (I need to compute the integral $\int_0^t \ldots$).
| When you see
$$
ax^2 + \underbrace{{}\quad bx\quad{}}_\text{1st-degree term} + c,
$$
it may help to remember that there is a standard technique in algebra for reducing problems involving quadratic polynomials with a first-degree term to problems involving quadratic polynomials with no first-degree term. It's called "c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
inequality with sum of powers How to prove the following inequality:
$$\forall n\geqslant 4:\dfrac {3^{n}+4^{n}+\cdots +\left( n+2\right) ^{n}} {\left( n+3\right) ^{n}} < 1$$
| Induction base: verify for $n=4$:
$$
\underbrace{3^4 + 4^4 + 5^4 + 6^4}_{2258} < 7^4 = 2401
$$
Induction step: Assuming inequality true for $n$, prove it true for $n+1$:
$$ \begin{eqnarray}
\sum _{k=3}^{n+3} k^{n+1} &=& \sum _{k=3}^{n+2} k^{n+1}+(n+3)^{n+1} \leqslant (n+2)\sum _{k=3}^{n+2} k^{n}+(n+3)^{n+1} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Inequality for cosines Is the following inequality in a triangle known?
$$4(\cos A + \cos B + \cos C) \le 3 + \cos \left(\frac{B-C}{2}\right) + \cos \left(\frac{C-A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$$
It looks correct to me but I would appreciate if someone confirm it.
| WLOG let $C$ be the largest angle, let
$x=\frac{A-B}{2}, y=\frac{A+B}{2}$ then $-\frac{\pi}{4}<x<\frac{\pi}{4}, 0<y<\frac{\pi}{2}$.
$$
A = x+y,~~ B=y-x,~~ C = \pi-2y \\
\cos\left(\frac{B-C}{2}\right) = \sin\left(\frac{3y-x}{2}\right), ~~
\cos\left(\frac{C-A}{2}\right) = \sin\left(\frac{3y+x}{2}\right) \\
\sin\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
Finding a pair of elements to satisfy an inequation Let $F$ be a field of characteristic 2 with more than 2 elements. Show that there are elements $a$ and $b$ in $F$ such that $(a+b)^3 \not= a^3 + b^3$.
$F$ couldn't possibly have less than 2 elements, and if it had exactly 2 — that is, $F = \mathbb Z_2$ —, $(a+b)^3$ w... | It might help to know the addition and multiplication tables of $F_4$, the smallest extension of the simplest Galois field of characteristic 2.
Suppose $F_4 = \{0, 1, \alpha, \alpha+1\}$, then addition works as you might expect, and multiplication is such that $\alpha^2 = \alpha + 1$. This lets you work out the full ad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determining and (dis)proving if $ \sum_{n = 1}^{\infty} (-1)^{n + 1} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) $ converges I am trying to determine if $ \sum_{n = 1}^{\infty} (-1)^{n + 1} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) $ converges using an alternating series test. The test in qu... | Notice that $$\lim_{n\to\infty} n \ln\left( \frac{n + 1}{n} \right)=\lim_{n\to\infty} \ln\left( 1+ \frac{1}{n} \right)^n=\ln e=1.$$
So $\lim_{n \to \infty} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) = 0$ as required and we are left with monotonicity:
$$
1- (n + 1) \log \left( \frac{n + 2}{n + 1} \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ How would I verify the following double angle identity.
$$
\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B
$$
So far I have done this.
$$
(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)
$$But I am not sure how to proceed.
| I will prove the result by starting with the right hand side of the identity:
$$\begin{align}\sin^2A-\sin^2B&=(\sin A+\sin B )(\sin A-\sin B)\\
&=(2\sin\frac{A+B}{2}\cos\frac{A-B}{2})(2\sin\frac{A-B}{2}\cos\frac{A+B}{2})\\
&=(2\sin\frac{A+B}{2}\cos\frac{A+B}{2})(2\sin\frac{A-B}{2}\cos\frac{A-B}{2})\\
&=\sin(A+B)\sin(A-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 6
} |
Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\... | Just use the sum formula twice:
$$
\begin{align*}
\tan(A+B+Y)=\tan((A+B)+Y) &= \frac{\tan(A+B)+\tan Y}{1-\tan(A+B)\tan Y}\\
&=\frac{\left(\frac{\tan A +\tan B}{1-\tan A\tan B}\right)+\tan Y}{1-\left(\frac{\tan A +\tan B}{1-\tan A\tan B}\right)\tan Y}\\
&= \frac{(\tan A + \tan B)+ \tan Y(1-\tan A\tan B)}{(1-\tan A\tan B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Cauchy principal value integral Doing numerical integration I encountered the following calculation of the Cauchy principal value of the following integral:
$$\int_0^{\pi/2}d\varphi\frac{1}{\cos(\varphi)-c},$$
with $0<c<1$.
What is the strategy to solve it?
| The indefinite integral is elementary, can be found using half-angle substitution:
$$
\int \frac{\mathrm{d} \varphi}{\cos(\varphi)-c} = \frac{2}{\sqrt{1-c^2}} \operatorname{arctanh}\left(\frac{1+c}{\sqrt{1-c^2}} \tan\frac{\varphi}{2} \right)+C =: F(\varphi)
$$
The Cauchy principal value is found as a symmetric limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\sqrt{x} +y = 4$, $x+ \sqrt{y}= 6$, find the solution $(x,y)$ $\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, find the solution (x,y). $NOTE$ : $\sqrt{4}+1= 4-1$, $\sqrt{1} +4 =1+4$
| Let, $\sqrt{x}=s$, $\sqrt{y}=t$
we have, $s^4 -12s^2+s+32=0$, which is a 'biquadratic' equation of the form,
$$(s^2+ks+l)(s^2-ks+m)=0$$
i.e. $$s^4 -12s^2+s+32=(s^2+ks+l)(s^2-ks+m)$$
now by equating coefficients, we have
$$l+m-k^2 = -12, k(m-l) = 1, lm = 32$$
from the first two of these equations, we obtain
$$2m=k^2-12+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int_0^\infty (u\log(a^2+u^2))/(e^u-1)\, du$ Does the following definite integral have a known "closed form" value?
$$\int_0^\infty\frac{u\log(a^2+u^2)}{e^u-1}~du,$$
or can anyone see a way to integrate it?
| I was only able to find the following formula: for $a = 2\pi\alpha$,
$$\begin{align*}
\int_{0}^{\infty} \frac{u \log(4\pi^2\alpha^2+u^2)}{e^u - 1} \; du
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\
&\quad + \pi^2 \alpha \left(2\alpha \log\alpha - \alpha + 2 - 4 \log\Gamma(\alpha+1)\right) \\
&\quad + 4\pi^2 \int_{0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/181448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| We have:
$$\sqrt{\frac{a^2+b^2+c^2}{3}}\underset{Q\geq G}{\geq} \sqrt[3]{abc} = 1$$
But $\sqrt{x} \geq 1 \iff x \geq 1 \iff x\geq \sqrt{x}$ so:
$$\frac{a^2+b^2+c^2}{3} \geq\sqrt{\frac{a^2+b^2+c^2}{3}}\underset{Q\geq A}{\geq}\frac{a+b+c}{3}$$
$$a^2+b^2+c^2 \geq a+b+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 6
} |
Solve / Simplify for x $$\frac{1}{\sqrt{a^{2}-x^{2}}}+\frac{1}{\sqrt{b^{2}-x^{2}}}-\frac{1}{c}=0$$
Hello
I'm wondering whether anyone can help me rearrange this to solve for $x$, where $a$, $b$, $c$ are constants. I initially thought about a trig substitution $x=b\cdot \cos(z)$ but that just lead to another brick wall... | Put $1/c$ on one side of the equation, square both sides, separate out the square root and do again so that
$x$ is a root of $$\begin{multline}x^8+(-2b^2-2a^2+2c^2+2a^2c^2)x^6\\+(-4b^2a^2c^2+a^4-2a^4c^2-4a^2c^2-2b^2c^2+a^4c^4\\+b^4+4b^2a^2-2a^2c^4+c^4)x^4\\+(4b^2a^2c^2+2b^2a^2c^4-2a^4c^4b^2+4a^4c^2b^2+2b^4a^2c^2-2b^4a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/182788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Using the sum of squares formula to solve more complex sums. I'm studying integration and trying to figure out how to use the sum of squares formula to solve more complicated sums.
For example: knowing that $$\sum_{i=1}^n i^2 = \frac{n (n+1) (2 n+1)}{6}$$
how can we simplify
$$\sum_{i=1}^n (i/n)^2$$
| The key here is what anon wrote in his comment. It's helpful to expand out the terms of the sum:
$$ \begin{eqnarray}
\sum_{i=1}^n (i/n)^2 & = & \sum_{i=1}^n \frac{i^2}{n^2} \\
&=& \frac{1^2}{n^2} + \frac{2^2}{n^2} + \dots + \frac{n^2}{n^2} \\
&=& \Big( \frac{{1^2} + {2^2} + \dots + {n^2}} {n^2} \Big) \\
& = & \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that: $\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $ If $a\geq0$, $b\geq 0$ then the following inequality holds:
$$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $$
There are at least three things to try here:
a). Use AM-GM for the denominator of the second fraction, $(a+b)\le \frac{(a+b+1)^2}{4}... | To prove:$$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1$$
Prove:$$\frac{b}{(a+b+1)^2}+\frac{a+b+1}{(a+b+1)^2}\le\frac{1}{a+1}$$
$$\frac{a+2b+1}{(a+b+1)^2}\le\frac{1}{a+1}$$
$$\color{green}{a^2}+\color{blue}{2ab}+\color{red}{a+a+2b+1}\le\color{green}{a^2}+b^2+\color{blue}{2ab}+\color{red}{2a+2b+1}$$
$$0\le b^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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How to construct minimal polynomial? This is an exam question from last semester.
We have the finite field
$$ \mathbb F_{81} = \mathbb Z_3 [x]/(x^4+x^2+x+1)$$
(a) Prove that the polynomial $$ x^4+x^2+x+1 $$
is irreducible
(b) Construct the minimal polynomial of the element $$ x^3+x^2+x+1 \space\epsilon\space Z_3 [x]/(x... | I will answer only (b).
By abuse of notation, we write $x$ for the image of $x$ in $\mathbb{Z}_3[x]/(x^4 + x^2 + x + 1)$.
Let $y = x^3 + x^2 + x + 1$.
Since $y = (x^2 + 1)(x + 1)$, $y^2 = (x^2 + 1)^2(x + 1)^2$.
Hence $y^2 = (x^4 + 2x^2 + 1)^2(x + 1)^2 = (-x + x^2)(x + 1)^2 = x(x - 1)(x + 1)^2 = x(x^2 - 1)(x + 1) = x(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the... | The claim is false, for example
$$n=2\Longrightarrow n^4+4n^2+11=16+16+11=43$$
which is not a multiple of 16. Check your expression.
Now, if $\,n=2k+1\,$ is odd, then the claim is true, since then
$$n^4+4n^2+11=8k(k+1)(2k^2+2k+3)+16$$
and since $\,8k(k+1)=0\pmod {16}\,$ no matter what parity $\,k\,$ has, we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
How to compute this limit related to series? How to compute $\lim_{N\rightarrow+\infty}\frac{\ln^2N}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^{2\ln N-2}$? thank you.
| Because $f(x) =x^{2 \ln(n) - 2}$ is increasing function of $x$ for $n \geqslant 3$, we have
$$
\int_0^n \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \frac{\mathrm{d} k}{n} =
\sum_{m=0}^{n-1} \int_{m}^{m+1} \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \frac{\mathrm{d} k}{n} \lt \frac{1}{n} \sum_{m=0}^{n-1} \left( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/187292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$ let $ABC$ be an acute triangle with all angles greater than $45^o$
Prove that
$$\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$$
I let $\tan A=a$, $\tan B=b$, $\tan C=c$ with $a+b+c=abc$ then the inequali... | Let's use Lagrange multiplier method to maximize
$$
f(a,b,c) = \frac{2}{1+a} + \frac{2}{1+b} + \frac{2}{1+c}
$$
subject to the constraint $0 = g(a,b,c) = a+b+c-a b c$. Derivatives of $f - \lambda g$ read:
$$ \begin{eqnarray}
\frac{\partial(f- \lambda g)}{\partial a} &=& -\frac{2}{(1+a)^2} + \lambda (b c -1) = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/187938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, ... | Here is a closed form solution to your recurrence relation obtained by Maple,
$$ s(n)={n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}}s \left( 1
\right) +\frac{{n}^{3}}{3}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}}
\left( \left( -1 \right)^{{\frac {\ln \left( n \right) }{\ln
\left( 2 \right) }}} \right)^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Limit of equation as x tends to -1 I was given the following expression and had to find the limit as: $$ x \rightarrow 1, x \rightarrow - 1, x \rightarrow \infty $$
$$ \lim_{x \to -1} \frac{x^2 +3x +2}{x^2 -1} = \lim_{x \to -1} \frac{\frac{x^2}{x^2} + \frac{3x}{x^2} + \frac{2}{x^2}}{\frac{x^2}{x^2} - \frac{1}{x^2}} = \... | $$ \lim_{x \to -1} \frac{x^2 +3x +2}{x^2 -1}$$
$$=\lim_{x \to -1} \frac{(x+1)(x+2)}{(x+1)(x-1)} $$
$$=\lim_{x \to -1} \frac{(x+2)}{(x-1)} $$ $$\text{as}:x \to -1, x≠-1$$
$$=\frac{-1+2}{-1-1}=-\frac{1}{2}$$
Now $ \lim_{x \to 1} \frac{x^2 +3x +2}{x^2 -1}$
$=\lim_{x \to 1} \frac{(x+2)}{(x-1)} $ as $\lim_{x \to 1}, x≠-1$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving an equality involving compositions of an integer Let's consider various representations of a natural number $n \geq 4$ as a sum of positive integers, in which the order of summands is important (i.e. compositions). The task is to prove the number $3$ appears altogether $n2^{n-5}$ times in all of them.
I know t... | It appears that the generating function approach is quite simple here. We have by inspection that the bivariate generating function of compositions with the number three marked is
$$M(z, u) = \sum_{q\ge 1}\left(\frac{z}{1-z} - z^3 + uz^3\right)^q.$$
Therefore the generating function of the total number of ocurrences is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
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