Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ Can some one help me out on where to go?
If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ ?
| Your problem is as follows:
\begin{align*}
&\max \,2a+b\\
\\
&\text{s.t. } a^2 + b^2=1
\end{align*}
Using substitution, let $f(b) = 2a + b = 2\sqrt{1 - b^2} +b$. Now, we wish to find the maximum of this function with respect to $b$.
Using simple derivatives, we get
$$
\frac{-2b}{\sqrt{1-b^2}} + 1 = 0\implies 2b=\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $(a+1/a)^2+(b+1/b)^2\ge 25/2$
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$
My work:
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\b... | By QM-AM, $\displaystyle a^2+b^2 \geq \frac{1}{2}(a+b)^2 = \frac{1}{2}$. QM-AM again gives $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)^2$. By AM-HM, $\displaystyle \frac{2}{\frac{1}{a}+\frac{1}{b}} \leq \frac{a+b}{2} = \frac{1}{2}$, whence $\displaystyle\frac{1}{a}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding sum of sequence $\sum\limits_{k=2}^n \frac{1}{k^2-1}$ Here is a problem I need to solve:
$$ \sum_{k=2}^n \frac{1}{k^2-1} $$
It came with another one alreay done in the same task:
$$ \sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{(k+1)-k}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+1} =... | Hint
$$\frac{1}{k^2-1}=\frac 1 2\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\frac 1 2\left(\frac{1}{k-1}-\frac 1 k+\frac 1 k-\frac{1}{k+1}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$ How to prove that the minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$
| If $\displaystyle2x+2y+2z=n\pi,$
$$F=\cos2x+\cos2y+\cos2z=2\cos(x+y)\cos(x-y)+2\cos^2z-1$$
As $\displaystyle x+y=n\pi-y, \cos(x+y)=(-1)^n\cos z$
$$\implies2\cos^2z\pm2\cos z\cos(x-y)-(1+F)=0$$ which is a Quadratic Equation in $\cos z$
So, the discriminant must be $\ge0$
$$\implies (2\cos(x-y))^2\ge 4.\cdot2(-1-F)\iff ... | {
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"url": "https://math.stackexchange.com/questions/639890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all solutions of $z^5+a^5=0$ The task is as follows:
Find all solutions of $z^5+a^5=0$, where $a$ is a positive real number.
My initial attempt
(which leads nowhere)
My guess is that i'll have to find the 5 5th roots of $-z^5$:
$w_1 = |-z^5|^5(cos(\frac{\theta}{5})+isin(\frac{\theta}{5})) \\
w_1 = |z|(cos(\frac{... | I figured it out (delayed rubber duck?)
I define $z=(r,\theta)$. Then $z^5=(r^5,5\theta)$.
Since $z^5$ have to be a real negative number,
$$ 5\theta=\pi+2\pi n \\ \theta=\frac{\pi+2\pi n}{5}$$
And since $$r^5=a^5 \\ r=a$$
Then $z=(a, \frac{\pi+2\pi n}{5})$.
Or, with a more straight forward notation:
$z=a(cos(\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/640965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Sum of consecutive square roots inside a square root $$\large\sqrt{1+\sqrt{1+2+\sqrt{1+2+3+\sqrt{1+2+3+4+\cdots}}}}$$
I saw this somewhere in the internet but, the website didn't provide me any further information. What is the sum of the equation above? What is it called?
|
What is the sum of the equation above? What is it called?
It's neither an equation nor a sum, but is a continued radical defined as the limit (as $n\to\infty$) of the sequence of the nested radicals
$$
u_n = \sqrt{a_1+\sqrt{a_2 + \cdots + \sqrt{a_n}}}
$$
where $a_n = \sum_{i=1}^n i = n(n+1)/2$ (the $n^{th}$ triang... | {
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"url": "https://math.stackexchange.com/questions/642210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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If $A,B,C$ are the angle of a triangle, then show that $\sin A+\sin B-\cos C\le \dfrac3 2$ If $A,B,C$ are the angle of a triangle, then show that $\sin A+\sin B-\cos C\le \dfrac32$
I tried substituting $C=180^\circ-(A+B)$ and got stuck. I also tried using the formula $\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}... | So, we have $\displaystyle \sin\frac{A+B}2=\sin\left(\frac\pi2-\frac C2\right)=\cos\frac C2$
Again, $\displaystyle\cos C=2\cos^2\frac C2-1$
Let $\displaystyle\sin A+\sin B-\cos C=y$
$\displaystyle\implies2\cos\frac C2\cos\frac{A-B}2-\left(2\cos^2\frac C2-1\right)=y$
$\displaystyle\implies2\cos^2\frac C2-2\cos\frac C2\c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the system using Gaussian elimination with back-substitution or Gauss-Jordan elimination Here is the system:
$$
\left\{
\begin{aligned}
x_1-3x_3&=-2 \\
3x_1+x_2-2x_3&=5 \\
2x_1+2x_2+x_3&=4
\end{aligned}
\right.
$$
This is my very first problem actually using a matrix so here is my attempt
First I setup the augmen... | Your mistake is at row: $(-1)R_2 + R_3 \to R_3$.
| {
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"question_score": "2",
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Is this polynomial solvable by radicals? The polynomial
$p(x) = x^6-9x^4-4x^3+27x^2-36x-23$.
has at least one (real, irrational) root that is expressible by radicals (can you find it?).
Are all the roots of $p$ expressible by radicals and if so, how can one find the expressions?
| Maple says
$$
16(x^6-9x^4-4x^3+27x^2-36x-23) =
\left( i\sqrt {3}\sqrt [3]{2}-2\,\sqrt {3}-\sqrt [3]{2}-2\,x
\right) \\
\left( i\sqrt {3}\sqrt [3]{2}+2\,x+2\,\sqrt {3}+\sqrt [3]{2}
\right) \left( i\sqrt {3}\sqrt [3]{2}-2\,\sqrt {3}+\sqrt [3]{2}+2\,x
\right) \\
\left( i\sqrt {3}\sqrt [3]{2}+2\,\sqrt {3}-\sqrt [3]{2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $A+B+C+D=\pi$, what is $\min(\cos{A}+\cos{B}+\cos{C}+\cos{D})$ Let $A,B,C,D \in R ~|~ A+B+C+D=\pi$, what is the minimum of the following function
$$f(A,B,C,D)=\cos{A}+\cos{B}+\cos{C}+\cos{D}$$
I found a post about a similar problem, butI think an even number of variables is harder than an odd number of variables, be... | So we want to minimize $f(A,B,C) = \cos A + \cos B + \cos C + \cos(\pi-A-B-C)$. We have
\begin{align*}\partial_A f(A,B,C) &= \sin(\pi-A-B-C) - \sin A\\
\partial_B f(A,B,C) &= \sin(\pi-A-B-C) - \sin B\\
\partial_C f(A,B,C) &= \sin(\pi-A-B-C) - \sin C
\end{align*}
In a mininum, we have $f'(A,B,C) = 0$, this gives
$$ \si... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\lef... | Old thread, but this question came up again and was marked a duplicate, so I have to post my physics lecture here.
You know the symmetries of the problem that the electric field of a line charge must be purely radial and that its magnitude has no axial or azimuthal dependence $\vec E=E(r)\hat e_r$. Thus we may apply ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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"answer_id": 1
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Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?
Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coeff... | We have $\displaystyle x+10x^3+100x^5=x\frac{1000x^6-1}{10x^2-1}$. (A geometric progression)
Hence $\displaystyle -2(1000x^6-1)=x \frac{1000x^6-1}{10x^2-1}$
Hence either $1000x^6-1=0$ or $x=-2(10x^2-1)$.
Therefore $20x^2+x-2=0$ for second equation.
Solving we get
$$x=\frac{-1\pm \sqrt{161}}{40}$$.
Comparing $m=-1, n=1... | {
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"url": "https://math.stackexchange.com/questions/651024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Incorrect Solution for Problem 7 of Pinter's Book of Abstract Algebra, Chapter 2? I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:
$$ x * y = \frac{xy}{x+y+1}$$
The book solution:
$$ (x * ... | As far as I can tell it's associative. You've proved it twice yourself already, but my nitty-gritty arithmetic is:
\begin{align*}
(x * y) * z &= \left(\frac{xy}{x+y+1}\right) * z \\
&= \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} \\
&= \frac{\left(\frac{xyz}{x+y+1}\right)}{\frac{xy+(z+1)(x+y+1)}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/651396",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Induction: show that $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$ The question:
Induction: show that:
$$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$$
for $n \geq 1$
My attempt at a solution:
First w... | You have proved it; you just need to write the 6 inequalities in the reverse order, and connect them by implication $\implies$ or keep the ordering and show that each is equivalent $\iff$ with the next.
| {
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How to find all polynomials P(x) such that $P(x^2-2)=P(x)^2 -2$? I am trying the fallowing exercise :
Solve $P(X^2 -2)=P(X)^2 -2$ with P a monic polynomial (non-constant)
My attempt :
Let P satisfying $P(X^2-2) = (P(X))^2-2$
Then $Q(X)=P(X^2-2) = (P(X))^2-2$
Therefore, $$Q(X^2-2) = (P(X^2-2))^2-2 = (P(X)^2-2)^2-2 = Q... | Lemma : If $P(x)^2$ is a polynomial in $x^2$, then so is either P(x) or P(x)/x.
By the lemma, there is a polynomial Q such that $P(x)=Q(x^2-2)$ or $P(x)=xQ(x^2-2)$.
Then $Q((x^2-2)^2-2)=Q(x^2-2)^2−2$ or $(x^2-2)Q((x^2-2)^2-2)=x^2Q(x^2-2)^2-2$
Substituting $x^2-2=y$ yields $Q(y^2-2)=Q(y)^2-2$ and $yQ(y^2-2)=(y+2)Q(y)^2+... | {
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"url": "https://math.stackexchange.com/questions/654461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Solve the difference quotient
So I tried a variety of answers and they all came up incorrect. I guess I don't understand difference quotient because I feel like it's pointless and a slow way to do the problem. That being said... can't get the answer
| Evaluate $f(4)$:
$$f(4) = \frac{4+6}{4+4}=\frac{10}{8}$$
Plug in what you know into the difference quotient:
$$\frac{\frac{x+6}{x+4}-\frac{10}{8}}{x-4}$$
Simplifying the numerator:
$$\frac{x+6}{x+4}-\frac{10}{8}=\frac{8(x+6)-10(x+4)}{8(x+4)}=\frac{8-2x}{8(x+4)}=\frac{-2(x-4)}{8(x+4)}=-\frac{(x-4)}{4(x+4)}$$
$$\frac{-\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding maximum value for a function I was working on this question to find the following function's maximum value.Let $$y=f(x)={{(\sqrt{-3+4x-x^2}+4)}}^2 + (x-5)^2$$ where $$1 \le x \le 3$$.I have to find it's maximum value.
I tried by simple method of taking $$\frac{dy}{dx}=0$$ and got the answer.But this path was pr... | You have simply$$y=(\sqrt{-3+4x-x^2}+4)^2 + (x-5)^2 \\
y=(\sqrt{1-(x-2)^2}+4)^2 + (x-5)^2 \\
y'=2(\sqrt{1-(x-2)^2}+4).\frac{-2(x-2)}{2\sqrt{1-(x-2)^2}} + 2(x-5)\\
y'=(\sqrt{1-(x-2)^2}+4).\frac{-2(x-2)}{\sqrt{1-(x-2)^2}} + 2(x-5)\\
y'=\frac{-2(x-2)\sqrt{1-(x-2)^2}-8x+16+2(x-5)\sqrt{1-(x-2)^2}}{\sqrt{1-(x-2)^2}}\\
y'=\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit $\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$ The limit is
$$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$
The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal O$-notation.
| I assume you know about the geometric series and Taylor's theorem to integrate them, but you might not. I apologize if you don't, as this will not be of much use.
We have
$$
\frac 1{\log(x+1)} - \frac 1x = \frac{x-\log(x+1)}{x \log(x+1)} = \frac{1 - \frac{\log(x+1)}{x}}{\log(x+1)}.
$$
Now
$$
\frac 1{1+x} = \frac 1{1-(-... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin... | All your work is okay.
Equations last but one, last but two and last but three can be deleted and you can straight away jump to last result because you already know $ \sec,\csc $ are inverse trig functions of $\cos, \sin$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
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For a positive integer $n$ both $5n+1$ and $7n+1$ are perfect squares. Show that $n$ is divisible by 24. My try:
$5n + 1 = k^2$
$7n +1 = \frac{7k^2-2}5$
Just don't know how to proceed after this. Please help.
| If $x\equiv1(mod\ 3)$, we have $7n+1\equiv2(mod\ 3)$.
If $x\equiv2(mod\ 3)$, we have $5n+1\equiv2(mod\ 3)$.
But a perfect square cannot be congruent 2 mod 3, so x must be divisible by 3.
The remainders modulo 8 of the numbers are :
? for(n=0,7,print(n," ",Mod(7*n+1,8)," ",Mod(5*n+1,8)))
0 Mod(1, 8) Mod(1, 8)
1 Mod(0,... | {
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"timestamp": "2023-03-29T00:00:00",
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Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
| First note that if $a=0$ or $b=0$, then the question is easy. So assume that both are non-zero. Consider the value $ab$. We can show that we must have $|ab|\le1$.
Assume by contrary that we have: $|ab| > 1$. This means $$\frac{1}{|b|}<|a|$$
Expanding: $$0\le(b^2-1)^2$$
we get:$$2\le \frac{1}{b^2}+b^2,$$
Thus, using t... | {
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"url": "https://math.stackexchange.com/questions/666217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Pseudo-pythagorean theorem Pythagoras' theorem is a special case of the Cosine theorem for a angle of $90°$. But also for an angle of 60° and 120°, "aesthetical" special cases derive:
$c^2=a^2+b^2\pm ab$
First question:
Are there further angle $x°$ with a rational number $x$, so that $\cos x$ is rational as well, thus ... | To find all integer triples $a,b,c$ for which $c^2=a^2+b^2\pm ab$, it's enough to consider only $c^2=a^2+b^2+ab$ because changing the sign of $a$ or $b$ gives a solution to the other equation and vice versa.
We'll only look for primitive solutions, i.e, satisfying $\gcd(a,b,c)=1$. Then at least one of $a$ or $b$ is odd... | {
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"url": "https://math.stackexchange.com/questions/668422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
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Help needed for Partial Fraction Decomposition The problem I need help with is this:
$$\frac{x^4} {(x-1)^3}$$
I've already put it into the form: $$ \frac{x^4} {(x-1)^3}=\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$$
Then, I multiplied the whole equation by the LCD ($(x-1)^3$) then got: $$x^4=A(x-1)^2+B(x-1)+C... | You can't use partial fraction decomposition since the degree of the numerator is greater than the degree of the denominator. Consider using polynomial division first before applying partial fraction decomposition. Given the equation
$$\dfrac{x^4}{(x - 1)^3}$$
Expanding the denominator expression, we have
$$\dfrac{x^... | {
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"timestamp": "2023-03-29T00:00:00",
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Help with a general solution for $\int \tan^a \theta$ Working out the integrals of $\tan^6 \theta$ I saw a pattern. I would like to put it in series representation. The pattern is as follows:
$$\int \tan^a d\theta= \int\tan^{a-2}\theta(\sec^2\theta-1)\ d\theta= $$ $$ \int \tan ^{a-2}\theta\sec^2\theta\ d\theta-\int\ta... | Simply you have
$$I_a=\int \tan^a d\theta= \int\tan^{a-2}(\sec^2-1)d\theta
\\=\frac{1}{a-1}\tan^{a-1}\theta-\int \tan^{a-2} d\theta
\\ I_a=\frac{1}{a-1}\tan^{a-1}\theta-I_{a-2}$$from this last one you can start by applying it many times to get your required formula. If $n$ is even we get
$$I_a=\frac{1}{a-1}\tan^{a-1}\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Derivative of $\sqrt{\frac{9+x}{x}}$ using first principle I am trying to find the derivative of $\sqrt{\dfrac{9+x}{x}}$ using first principle to differentiate.
For sake of simplifying the formulas, I will omit the /h part to the first principles formula, unless someone is able to show me how to integrate it with thes... | For $f(x)=\sqrt {\dfrac{9+x}{x}}$,
\begin{align}
&\dfrac{f(x+h)-f(x)}{h}
\\=&\dfrac{\sqrt{\dfrac{9+x+h}{x+h}}-\sqrt{\dfrac{9+x}{x}}}{h}
\\\\\\=&\dfrac{{\dfrac{9+x+h}{x+h}}-{\dfrac{9+x}{x}}}{h\left(\sqrt{\dfrac{9+x+h}{x+h}}+\sqrt{\dfrac{9+x}{x}}\right)}\tag{Multiplying by the conjugate}
\end{align}
The numerator simplif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
| Hint:
$$\left( x + \frac{1}{x}\right)^5 = x^5 + \frac{1}{x^5} + 5\left(x^3 + \frac{1}{x^3}\right)
+ 10\left(x + \frac{1}{x} \right)$$
$$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x} \right)\left(x^2 + \frac{1}{x^2} - 1 \right)$$
$$\left( x + \frac{1}{x}\right)^2 - 2 = x^2 + \frac{1}{x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/678650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 1
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Proof of Divisibility of $n(n^2+20)$ by 48. This is a question from Bangladesh National Math Olympiad 2013 - Junior Category that still haunts me a lot. I want to find an answer to this question. Please prove this.
If $n$ is an even integer, prove that $48$ divides $n(n^2+20)$.
| $48=2^4\cdot 3$, so we will work modulos $16$ and $3$.
We have $n(n^2+20)=n^3+20n$. Now in modulo 3, one notices that $n^3\equiv n\pmod 3$ for any n, so that $n^3+20n\equiv n+20n\equiv 21n\equiv 0\pmod 3$. So we're done with divisibility by 3.
We now need to show divisibility by 16. We have $n(n^2+20)\equiv n(n^2+4)\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/679509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 2
} |
Recurrence relations (Big-O notation) Say I'm given a recursive function such as:
function(n) {
if (n <= 1)
return;
int i;
for(i = 0; i < n; i++) {
function(0.8n)
}
}
how would I go about applying recurrence relations to the find the Big-O run time
(as a function of n)?
| I solved the recurrence relation as the following: $$\text{With: }0.8 = \frac{4}{5}$$
$$
\\
T(n) = \sum_{i = 0}^{n - 1}(T(\frac{4}{5}n) + c), \text{ with } T(n) = 1 \text{ when } n \leq 1
\\\\
T(n) = nT(\frac{4}{5}n) + nc
\\\\
T(n) = (\frac{4}{5})n^2T([\frac{4}{5}]^2n) + (\frac{4}{5})n^2c + nc
\\\\
T(n) = (\frac{4}{5})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$
for all $n\gt 0$. Show that the sequence is upper bounded.
My idea: since
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+... | We have
$$\frac{a_{n+1}}{a_n} = 1 + \frac{a_n}{n^2}$$
Therefore, if the product
$$\prod_{n=1}^{\infty} ( 1+ \frac{a_n}{n^2})$$
is bounded, then so is $a_n$. Since $1+x \le e^x$, it is enough to show that $\sum_{n\ge 0} \frac{a_n}{n^2} < \infty$. Now, this would follow from $\frac{a_n}{n} < n^{1-\epsilon}$ for some $\ep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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"answer_id": 2
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How to go about solving this recurrence relation? In my discrete math class we were given the problem of finding an explicit formula for this recurrence relation and proving its correctness via induction.
$$a_n=2a_{n-1}+a_{n-2}+1, a_1 = 1, a_2=1.$$
I feel confident that I could prove the correctness of the formula thro... | Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write your recurrence as:
$$
a_{n + 2} = 2 a_{n + 1} + a_n + 1
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = 2 \frac{A(z) - a_0}{z} + A(z) + \frac{1}{1 - z}
$$
Solve for $A(z)$:
$$
A(z)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Visual explanation of $\pi$ series definition Can you visually explain why the following is true:
$$
\frac{\pi}{4} = \sum\limits_{k=0}^\infty \frac{(-1)^k}{2k + 1} = \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\ldots\approx 78.5\%
$$
By visually I mean use a circle and square to help me understand rather... | You should know that:
$$\dfrac{\pi}{4}=\tan^{-1}(1)$$
This means that:
$$\dfrac{\pi}{4}=\int_0^1 \dfrac{1}{1+x^2} \ dx$$
We will use the rule that $\dfrac{1}{1+x^2}=1-x^2+x^4-x^6+x^8\dots \ (-1\le x \le 1)$.
$$\dfrac{\pi}{4}=\int_0^1 1 -x^2+x^4-x^6+x^8\dots \ dx$$
We will solve the indefinite integral $\int 1 -x^2+x^4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/682532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{\cos{A}\cos{B}}{\cos{C}}}\ge\frac{3\sqrt{2}}{2}$ let $x,y,z>0$,and such $$x^2y^2+y^2z^2+x^2z^2+2x^2y^2z^2=1$$
show that
$$x+y+z\ge\dfrac{3\sqrt{2}}{2}$$
My idea: in $\Delta ABC$,we have
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
so let
$$xy=\cos{A... | The claim is false: If we let $x = 0$, $y=1$ and $z=1$, then we have
$x+y+z = 2 < 3 \sqrt{2} / 2$. (This conclusion holds by continuity for some $x,y,z > 0$, too.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/683612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Vector fields on $\Bbb S^2$ with zero construct a vector field on ${S} ^ 2$ with only one zero?
$$\mathbb{S}^2= \left\{(x_1,x_2,x_3) \in R^3: x_1^2+x_2^2+x_3^2=1\right\}$$
| Write down the stereographic projection from the plane $R^2$ to the sphere, with projection from the north pole. Call that F.
For each $(x, y)$ in $R^2$, take
$$
DF(x, y) \begin{bmatrix} 1/(1 + x^2 + y^2) \\ 0 \end{bmatrix}
$$
That will be a vector at $F(x, y) \in S^2$. It'll be nonzero everywhere except the north pol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Showing that $\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}}$ without using complex variables The identity $$\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}} \ , \ |a| <1$$
can be derived by using the fact that $ \displaystyle \sum_{k=0}^{\infty} a^{k} \cos(kx)... | Using the identity,
$$\cos{\left(nx\right)}=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n}{2k}\sin^{2k}{\left(x\right)}\cos^{n-2k}{\left(x\right)},$$
the infinite series in question may be rewritten as a double infinite series over a triangle. Changing the order of summation (if you're like me and the transforma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
} |
Sum this series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\ldots$ upto $n$ terms
Sum this series: $$\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+\ldots$$ upto $n$ terms.
My approach:
$$(1-n^6)=(1-n^2)(1+n^2+n^4)\implies \dfrac{n}{1+n^2+n^4}=\dfrac{n(1-n^2)}{1-n^6}$$
So, the above series can be written as $$\sum\limit... | HINT:
As $\displaystyle1+n^2+n^4=(1+n^2)^2-n^2=(1+n+n^2)(1-n+n^2)$
$$\frac{2n}{1+n^2+n^4}=\frac{(1+n+n^2)-(1-n+n^2)}{(1+n+n^2)(1-n+n^2)}=\cdots$$
Again if $\displaystyle f(n)=\frac1{n^2-n+1}, f(n+1)=?$
So, we are dealing with a Telescoping series
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/691109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proof Using the Monotone Convergence Theorem for the sequence $a_{n+1} = \sqrt{4 + a_n}$ Consider the recursively defined sequence $a_0 = 1$
$a_{n+1} = \sqrt{4 + a_n}$
How to prove that the sequence converges using the Monotone Convergence Theorem, and find the limit?
| Step 1. $\{a_n\}$ is strictly increasing, i.e., $a_{n+1}> a_n$.
This is shown inductively. For $n=1$, We have that $$a_2=\sqrt{4+a_1}=\sqrt{4+1}=\sqrt{5}>1=a_1.$$
Assume now that it is true for $n=k$, i.e., $a_{k+1}>a_{k}$. Then $a_{k+1}+4>a_{k}+4$, and hence
$$
a_{k+2}=\sqrt{4+a_{k+1}}>\sqrt{4+a_{k}}=a_{k+1},
$$
and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/692906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$? What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$?
How can we calculate this expression ?
I've applied the binomial theorem formula and got $91$ terms but I am not sure if it is right or wrong.
| You could use the binomial theorem twice. Let $[x^{k}]$ denote the coefficient of $x^k$ from the polynomial $P(x)=\sum_{j=0}^{n}a_jx^j$, i.e. $[x^{k}]P(x)=a_{k}$.
Now, $$\begin{eqnarray}[x^4y^5](x+y+2)^{12}&=&[x^4y^5](x+(y+2))^{12}\\&=&[x^4y^5]\sum_{j=0}^{12}\binom{12}{j}x^j(y+2)^{12-j}\\&=&[y^5]\binom{12}{4}(y+2)^{12-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do i simplify this $\cos{\theta}=\frac{1-\sin^2{\theta}}{1+\cos{\theta}}$ I was wondering if someone could help me solve this equation: $$\cos{\theta}=\frac{1-\sin^2{\theta}}{1+\cos{\theta}}$$
| $\cos{\theta}=\dfrac{1-\sin^2{\theta}}{1+\cos{\theta}}$:
1.) Multiply through by $1 + \cos \theta$:
$\cos \theta + \cos^2 \theta = 1 - \sin^2 \theta$;
2.) Add $\sin^2 \theta$ to each side:
$\cos \theta + \cos^2 \theta + \sin^2 \theta = 1$;
3.) Use $\cos^2 \theta + \sin^2 \theta = 1$ on the left:
$\cos \theta + 1 = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Help me prove the result of this limit: $\lim_{x\to \infty} {\log_{x^3+1} (x^2+1) \over \log_{2e^x+1} (x^2+5x)}$ The limit is
$$\lim_{x\to \infty} {\log_{x^3+1} (x^2+1) \over \log_{2e^x+1} (x^2+5x)}$$
I know that it should be equal to $\infty$ but i have yet to prove it. Please help me do so.
| Hint 1: $\displaystyle \log_a b = \frac{\ln b}{\ln a}$. So $$\frac{\log_{x^3+1} (x^2+1)}{\log_{2e^x+1} (x^2+5x)} = \frac{\frac{\ln (x^2+1)}{\ln (x^3+1)}}{\frac{\ln (x^2+5x)}{\ln (2e^x+1)}} = \frac{\ln (x^2+1) \ln (2e^x+1)}{\ln (x^3+1) \ln (x^2+5x)}$$
Hint 2: For very large $x$, $x^2+1 \approx x^2$ and $x^3 + 1 \approx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reducing the System of linear equations \begin{align*}
x+2y-3z&=4 \\
3x-y+5z&=2 \\
4x+y+(k^2-14)z&=k+2
\end{align*}
I started doing the matrix of the system:
$$ \begin{pmatrix} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & k^2-14 & k+2 \end{pmatrix} $$
But I still don't know how can I reduce this, could you explain me w... | This is not very different from other answers. And I think that the approach using Gaussian elimination is more systematic. I wanted to show how this can be done with a little of guesswork.
If you notice that the sum of the first two equations is $4x+y+2z=6$, then you see that every solution of your system also fulfil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/697288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Determine the convergence of a sequence given by $a_n= \frac{a_{n-1} + a_{n-2}}2$ let $a_0$ and $a_1$ be any two real numbers, and define
$a_n= \dfrac{a_{n-1} + a_{n-2}}{2}$
Determine the convergence of a sequence.
Alright, what I have so far. I have two cases $a_0$ > $a_1$ or $a_0$ < $a_1$
for $a_0 > a_1$
The sequen... | $a_n = \dfrac{a_{n-1}+a_{n-2}}{2}$. One overkill way of finding the limit, is getting the $a_n$ in terms of $n$ via a generating function.
Write $F(x) = \sum_{i=0}^{\infty}a_i x^i$. Thus,
$$
\frac{F(x)-a_0}{x}=\sum_{i=0}^{\infty}a_{i+1}x^{i}
$$
$$
\frac{F(x)-a_0-a_1}{x^2} = \sum_{i=0}^{\infty}a_{i+2}x^{i}
$$
So,
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/699640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Contour integration question help I am trying to evaluate the following integral using complex analysis:
$$\int^{2\pi}_0\frac{\cos(2x)}{5-4\cos(x)}dx$$
I have written it in the following form:
$$\frac{-i}{2}\int_{|z|<1}\frac{z^4+1}{z^2(z-2)(1-2z)}dz$$
However, the residue for $z=1/2$ is $-17/6$ whereas the residue for ... | \begin{align}
\operatorname{Res}\limits_{z=1/2} \frac{z^4+1}{z^2(z-2)(1-2z)} &=
\lim_{z\to 1/2} (z-\frac12) \cdot \frac{z^4+1}{z^2(z-2)(1-2z)} \\
&= \lim_{z\to 1/2} \frac{z^4+1}{z^2(z-2)\cdot(-2)} = \frac{(1/2)^4+1}{(1/2)^2\cdot(-3/2)\cdot(-2)} = \frac{17}{12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/701423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $a,b,c>0$ and $a+b+c=1$ prove inequality: $\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$ If $a,b,c>0$ and $a+b+c=1$ prove inequality: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$
| By C-S
$$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^2}{ab+ac^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a^2b)}.$$
Thus, it remains to prove that
$$4(a+b+c)^2\geq9(ab+ac+bc)+9(a^2b+b^2c+c^2a)$$ or
$$4(a+b+c)^3\geq9(ab+ac+bc)(a+b+c)+9(a^2b+b^2c+c^2a)$$ or
$$\sum_{cyc}(4a^3-6a^2b+3a^2c-abc)\geq0$$
and since by AM-GM $\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Proof this curious trigonometric identity Proof that $$\cos^2{10^\circ} + \cos^2{50^\circ} - \sin{40^\circ}\sin{80^\circ} = \frac{3}{4}$$
I notice that $10^\circ + 80^\circ = 90^\circ$, and $50^\circ +40^\circ = 90^\circ$. I tried doing some manipulation but my efforts were futile. Any hints?
| Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$\displaystyle\cos^210^\circ+\cos^250^\circ=\cos^210^\circ-\sin^250^\circ+1=\cos60^\circ\cos40^\circ+1=\frac{\cos40^\circ}2+1$
Use Werner Formulas to find $\displaystyle2\sin40^\circ\sin80^\circ=\cos40^\circ-\cos120^\circ$
Now, $\displaystyle\cos12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate a recursive equation in terms of theta I am struggling with the following equation for one week!
Please help me solve it.
$$T(n)=T(\frac{n}{2})+\frac{n}{logn}$$
So far, I have come to the equation $T(n)=\Sigma \frac{2^x}{x}$
| Suppose you have $T(0)=0$ and $T(1) = 1$ and your recurrence for $n>1$ is
$$T(n) = T(\lfloor n/2 \rfloor) + \frac{n}{\lfloor\log_2 n \rfloor}.$$
By unrolling the recursion we find the exact formula for $n\ge 2:$
$$T(n) = 1 +
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} \frac{1}{\lfloor\log_2 n \rfloor - k}
\sum_{j=k}^{\lf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/704393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$3^{2x} - 34(15^{x-1}) + 5^{2x} = 0$ I've never seen anything like this. Do somebody have a way to solve it?
I've tried the basci exponential functions techniques but it does not work. Even substitution does not work...
I'm really intersted in learning, this is not a homework.
| Another way to look at this is to consider that $ \ a^{2x} = (a^x)^2 \ , $ from the properties of exponents. Suppose you had a binomial square $ \ (3^x + 5^x)^2 \ $ then: its expansion would be
$$ \ (3^x)^2 \ + \ 2 \cdot 3^x \cdot 5^x \ + \ (5^x)^2 \ = \ 3^{2x} \ + \ 2 \cdot (3 \cdot 5)^x \ + \ 5^{2x} \ = \ 3^{2x} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/704664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How do I prove this trigonmetric identity? I need to prove that the following identity is true:
$$
\frac{\cos^2x-\sin^2x}{1-\tan^2x}=\cos^2x
$$
This isn't homework; just a practice exercise. But I keep getting stuck! Thanks much.
| This is sort of the "reverse" of MPW's suggestion, using the difference of two squares:
$$ \frac{\cos^2 x \ - \ \sin^2 x}{1-\tan^2x} \ = \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{(1 - \tan x) \ (1 + \tan x)} $$
$$ = \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{(1 - \tan x) \ (1 + \tan x)} \ \cdot \ \frac{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/704732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
Solve $7\sin^2(x) - 9\cos(2x) = 0$ I need to solve for x in the polynomial
$$7\sin^2(x) - 9\cos(2x) = 0$$
I have tried approaching the problem in multiple ways. I am only looking for some hints, not the actual answer. Thanks :D
| There are two ways to solve this all leading to the same answer:
1) 7sin^2(x)−9cos(2x)=0
using $cos(2x) = $cos^2x - sin^2x$
=> $7sin^2(x)−9(cos^2x - sin^2x)=0$ {as cos(2x) = $cos^2x - sin^2x$}
=> $7sin^2(x)−9(cos^2x - sin^2x)=0$
=> $7sin^2(x)−9(1 - sin^2x - sin^2x)=0$ {as $cos^2x = 1 - sin^2x$}
=> $7sin^2(x)−9(1 - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/706479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.
Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
| When you have these kinds of limits it's always a good idea to multiply by conjugate. In your case you have:
$$
\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)} - x\right) = \lim_{x\to+\infty}\frac{\left(\sqrt{(x+a)(x+b)} - x\right)\left(\sqrt{(x+a)(x+b)} + x\right)}{\sqrt{(x+a)(x+b)} + x} = \\ = \lim_{x\to+\infty}\frac{(x+a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Can $\frac {100-100}{100-100}=2$? \begin{align*}
\frac{0}{0} &= \frac{100-100}{100-100} \\
&= \frac{10^2-10^2}{10(10-10)} \\
&= \frac{(10+10)(10-10)}{10(10-10)} \\
&= \frac{10+10}{10} \\
&= \frac{20}{10} \\
&= 2
\end{align*}
I know that $0/0$ isn't equal to $2$, the what is wrong in this proof? I wasn't satisfied after... | I think the best way to see the folly here is to work backwards. The problem occurs when we encounter the "equality"
$$
\frac{10+10}{10}=\frac{(10+10)(10-10)}{10(10-10)}
$$
But to obtain this equality, we must multiply the rational number $\displaystyle\frac{10+10}{10}$ by the undefined expression $\displaystyle\frac{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/710551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 0
} |
Probability [Distinct balls in distinct boxes] 6 different balls are put in 3 different boxes, no box being empty. The probability of putting balls in the boxes in equal numbers is ? Could anyone pleases give the answer and explain it.
| Note that the balls and boxes are both distinct.
Select 2 balls in each of 3 boxes, count the permutations of the six balls. Ways to arrange 6 balls, 2 in each box: $N(2,2,2)=\frac{6!}{2!2!2!}$ That's permutations of all six balls, divided by permutations of balls in each box (since the order in the box does not matte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Inequality $\sum_{cyc}\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}\ge{12}$ when $ab+bc+ca=6$ Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$.
Prove that:
$$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}... | We'll prove a stronger inequality: $\sum\limits_{cyc}\frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}}\geq24$.
Indeed, let $a=\sqrt2x$, $b=\sqrt2y$, $c=\sqrt2z$, $x+y+z=3u$
and $xy+xz+yz=3v^2$, where $v\geq0$.
Hence, $v=1$, $u\geq v$ and we need to prove that $\sum\limits_{cyc}\frac{(x+y)^3}{\sqrt[3]{2(x+y)(x^2+y^2)}}\geq12.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/719523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Getting incorrect result on limit of square root I have a limit problem:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x$
According to Wolfram|Alpha the answer is $\frac{1}{2}$
However, my calculation gives $1$. Please help me understand what I'm doing wrong. The process is:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x = \lim_{x \t... | You've lost your 2 in the derivation!
$\frac{d}{dt}\sqrt{1+t} = \frac{1}{2}\frac{1}{\sqrt{1+t}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/721187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find Max value of this expression: $P=(x-2yz)(y-2zx)(z-2xy)$ Let $x$, $y$ and $z$ be positives and satisfying $x^2+y^2+z^2=2xyz+1$ .
Find a maximum of this expression:
$$P=(x-2yz)(y-2zx)(z-2xy).$$
| Let $x=1$ and $y=z=\frac{1}{2}.$
Hence, $P=\frac{1}{8}$.
We'll prove that it's a maximal value.
Indeed, let $x\geq y\geq z$.
If $x\leq2yz$ then we obtain:
$$2xy\geq2xz\geq2yz\geq x\geq y\geq z,$$
which says that $P\leq0$.
Thus, we can assume that $x>2yz$.
*
*$x=1$.
In this case the condition gives $y=z$ and
$$P=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Recurrence relation for number sequence Let $a_n$ be the number of sequences of $n$ numbers, consisting of $0's, 1's$ and $2's$, such that a number $1$ on the $j$-th place isn't followed by a $1$ or $2$ on the $j+1$-th place for $1\leq j\leq n-1$.
I'm asked to prove that the correct recurrence relation for this sequenc... | Got the recurrence, now solve it. You have:
$$
a_{n + 2} = 2 a_{n + 1} + a_n
$$
There is 1 zero-length sequence, and 3 one-length sequences; this checks $a_2 = 7$.
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$. Then recognize:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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determine all polynomials $P(x)$ such that $(x+1)P(x-1)-(x-1)P(x)$ is a constant polynomial Determine all polynomials $P(x)$ with real coefficients such that $(x+1)P(x-1)-(x-1)P(x)$ is a constant polynomial.
*
*clearly we have to show $(x+1)P(x-1)-(x-1)P(x)=c$ for all values of $x$ ($c$ is a real constant)
*i have ... | Let $(x+1)P(x−1)−(x−1)P(x)=C=const$. Set $x=2$ in order to get $$P(2)=3P(1)-C.$$ Set $x=3$ to obtain $$P(3)=6P(1)-\frac{5}{2}C.$$
This motivates us to try to prove by induction that
$$P(n)=\frac{n(n+1)}{2}P(1)-\frac{(n-1)(n+2)}{4}C$$ for every integer $n\geq 2.$ Indeed, if we assume the last equality holds for some $n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/727398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Calculation of a Residue Does anyone know of a good way to calculate the residue at zero of the following function? I was able to calculate it with the higher order pole formula for residues and then used Mathematica to find the limit. Really just looking for a nice trick without having to get too dirty.
$$ f(z)=\frac... | $$\begin{eqnarray}
\frac{1}{z^2(e^{-z}-1)}
&=& \frac{1}{z^2\left(1-z+\frac{z^2}{2}-\frac{z^3}{6} +\ldots - 1\right)} \\
&=& -\frac{1}{z^3\left(1-\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right)\right)} \\
&=& -\frac{1}{z^3}\left(1+\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right)+\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $ and $ \ x+y+z \ = \ xyz \ $ Consider the system of equations in real numbers $ \ x,y,z \ $ satisfying
$$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $$
and $ \ x+y+z \ = \... | Starting with $\frac{sint}{4} = \frac{sinu}{5} = \frac{sinw}{6}$, then use the law of sines to get: $\frac{a}{4} = \frac{b}{5} = \frac{c}{6}$. So $a = \frac{2c}{3}$, $b = \frac{5c}{6}$.
So apply the law of cosines to have: $a^2 = b^2 + c^2 - 2bccosA$. Substituting these values of $a$ and $b$ into the above equation we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find closed formula $f(n)$ from generating function I'm asked to find a closed formula for
$f(n)=6f(n-1)-9f(n-2)$ for $n>1$ with $f(0)=-1. f(1)=0$,
using the ordinary generating function $F(X)$.
I found $F(X)=-1/(1-3x)^2$ but from there I don't manage to get a satisfactory formula for $f(n)$. Can anyone give me a hint... | The reason you're not getting the closed form from your generating function is that the generating function is wrong. Let $F(x) = \sum_{n \ge 0} f(n) x^n$, then
$$\begin{align} F(x)
&= f(0) + f(1)x + \sum_{n \ge 2} f(n) x^n \\
&= f(0) + f(1)x + \sum_{n \ge 2} (6f(n-1) - 9f(n-2))x^n \\
&= f(0) + f(1)x + 6x\sum_{n \ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/729434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Solving a Linear Recurrence Relation I made quick progress on this, and then of course got stumped, so here's the problem:
$$a_0 = -1, a_1 = -2, a_n = 4a_{n-1} - 3a_{n-2}$$
So, following the way I was taught to solve this type of problem, I do the following:
Convert it to $$x^n = 4x^{n-1} - 3x^{n-2}$$
Divide by $x^{n-2... | A solution using generating functions: Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence as:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n \qquad a_0 = -1, a_1 = -2
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$, recognize:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/734331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Need help to solve taylor series of $e^{\sin x}$ How to derive the taylor series of $e^{\sin x}$, up to $x^5$?
i just don't know how to get the answer
$$f(x) = 1 + x + \frac{x^2}{2} - \frac{x^4}{8} -\frac{x^5}{15}$$
really need some help. Thanks
| $$\sin x \sim x - \frac{x^3}{6} + \frac{x^5}{120}$$
$$e^x \sim 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120}$$
so $$e^{\sin x} \sim 1 + \sin x + \frac{\sin ^2 x}{2} + \frac{\sin ^3 x}{6} + \frac{\sin ^ 4 x}{24} + \frac{\sin ^5x}{120} $$
Now substitute the expansion of $\sin x$, and you shou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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What is being represented by this 2 images? image 1
image 2
It's possible that image 1 is showing some kind of methods for building polygons out of trigonometric functions ?
It's also possible that image 2 is a quadratic bezier curve ?
| Here's how we can produce the figures in image 1. The graphs are showing the $ \ y-$ coordinate of the point being "swept along" on each curve.
The circle is simple because familiar trigonometric functions can be defined in terms of a circle. For a circle of radius $ \ a \ , $ the coordinates of the point are just $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)... | Use the fact that
$$a^2+a^2+a^2+b+c \geq5\sqrt[5]{a^2.a^2.a^2.b.c}=5\sqrt[5]{a^6bc}=5a$$ since $abc=1$
Similarly, we get two more results, and adding them we get:
$$3(a^2+b^2+c^2)+2(a+b+c) \geq 5(a+b+c) $$ which gives us our result.
$$a^2+b^2+c^2 \geq a+b+c$$
$$Q.E.D.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/740518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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2x2 Fibonacci matrix singular value decomposition $A = \left[\begin{array}[c]{rr}1 & 1\\1 & 0\end{array}\right]$
I am supposed to find all the eigenvalues and vectors for this matrix so that $Av=σu$ and then form a singular value decomposition $UΣVᵀ$ and show that $A=UΣVᵀ$.
I have attempted this so many times and just... | We are given:
$$A = \begin{bmatrix}1 & 1\\1 & 0 \end{bmatrix}$$
We have:
$$W = A^T A = \begin{bmatrix}2 & 1\\1 & 1 \end{bmatrix}$$
The characteristic polynomial and eigenvalues of W are:
$$\lambda^2 +3 \lambda -1 = 0 \implies \lambda_1 = \frac{1}{2} \left(3-\sqrt{5}\right), ~ \lambda_2 = \frac{1}{2} \left(3+\sqrt{5}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that $ a^2 + b^2 + c^2 +2abc < 2$ let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that
$ a^2 + b^2 + c^2 +2abc < 2$
I used the fact that $ a+b-c>0 $ and multiplied the 3 cyclic equations but coudn't reach the final equation.
| Let's homogenize the equation by multiplying with $ a+ b + c = 2$.
We need to show that
$$ 2 ( a + b + c) (a^2 + b^2 + c^2) + 8abc < (a+b+c)^3.$$
Expanding this, it is equivalent to showing that
$$ (a+b-c) ( b+c-a) ( c+a-b) > 0. $$
Hence we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/743099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Area of part of parametric function I need to get area of function: $x= 2\sqrt{2}\cos ^3 t$ and $y= 4\sqrt{2}{\sin ^3 t}$, but only the part when $x\geq1$. How can I do that? I know that area of full function would be
$$S= \int_a^b y(t)x'(t) \, dt $$
$a \leq t \leq b$
| UPDATE. The area from $x=1$ to $x(0)=a=2\sqrt{2}$ enclosed by the parametric curve
\begin{equation*}
\left( x(t),y(t)\right) =\left( 2\sqrt{2}\cos ^{3}t,4\sqrt{2}\sin
^{3}t\right)
\end{equation*}
is
\begin{equation*}
S=2\int_{1}^{a}y\,dx.
\end{equation*}
Since $x^{\prime }(t)=-6\sqrt{2}\cos ^{2}t\sin t$ and
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/743772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integral $ \int \frac{dx}{\cos^3 x+2\sin(2x)-5\cos x}$ $$
I\equiv \int \frac{dx}{\cos^3 x+2\sin(2x)-5\cos x}.
$$
This integral does have a closed form. I am not sure where to start. We can factorize the denominator as
$$
\cos^3 x+2\sin(2x)-5\cos x=(\cos^2 x+4\sin x-5)\cos x,
$$
but I am not sure where to go from ther... | The present integrand is a rational function (i.e. a ratio of two polynomials) of $\cos x$ and $\sin x$, which is considered a common question. So I take it partially from the answer by Arturo Magidin to the question Evaluating $\int P(\sin x, \cos x) \text{d}x$ by Aryabhata, which is an entry of the List of Generaliz... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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Help with this combinatorial proof $\sum\limits_{k=1}^nk^2(k-1){n\choose k}^2 = n^2(n-1){2n-3\choose n-2}$ considering $n\ge2$
$\displaystyle\sum\limits_{k=1}^nk^2(k-1){n\choose k}^2 = n^2(n-1)
{2n-3\choose n-2}$ considering $n\ge2$
Can somebody help with this combinatorial proof?
I'm struggling a lot.
Thanks.
EDIT: ... | Hint: Note that because choosing $k$ elements from a set of $n$ is the same as choosing the complement of the $k$ elements, we have
$$
\binom{n}{k}=\binom{n}{n-k}\tag{1}
$$
and since choosing a team of $k$ people and then a leader from those chosen is the same as choosing a leader and then choosing the remaining $k-1$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proof that $ p $ | ${ p^n \choose k } $ for any prime $p$ and $ k < p^n$ I know how to prove the fact that $ p$ | ${ p \choose k } $ (when writing it as a fraction, $p$ cannot be divided by any of the $1\times2\times...\times k$ or $1\times2\times...\times(p-k)$ because $p$ is prime).
When I try to apply the same ratio... | The highest prime power $p$ dividing $n!$ is expressed as $$\sum_{x=1}^\infty \lfloor \frac{n}{p^x} \rfloor.$$
Since $\binom{p^n}{k}=\frac{(p^n)!}{k!(p^n-k)!},$ we compare the highest prime power $p$ which divide numerator and denominator.
$$v_p((p^n)!)=\sum_{x=1}^n \lfloor \frac{p^n}{p^x} \rfloor,$$
$$v_p(k!(p^n-k)!)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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How to find centre,vertics,foci,focal radii,letus rectum... when exists of a general quadratic equation in x and y Is there a generalized way( a particular conic section of any shape,for instance an ellipse without determining its major/minor axis) to find the centre,vertices,foci,focal radii,letus rectum,eccentricity,... | One can use a list of equations to determine the property you require. Note, however, that in many cases it is easier to use derived value(s) as opposed to using equations that rely solely on the coefficients of the equation in general quadratic form. Below are examples of equations one can use.
Properties of an elli... | {
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"url": "https://math.stackexchange.com/questions/756220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$ I have the following question
$$\int x \sqrt{1 - x^4} \,\mathrm{d}x$$
I know we have to use trig. substitution for this and therefore, I did the following by letting $x = \sin \theta$ and $dx = \cos \theta \,\mathrm{d}\theta$
\begin{align}
&\int x \sqrt{1-x^4} \,\mathrm{d}... | HINT :
Let $u^2=1-x^4\;\Rightarrow\; x^2=\sqrt{1-u^2}\;\Rightarrow\; x\ dx=-\dfrac{u\ du}{2\sqrt{1-u^2}}$, then rewrite
$$
\begin{align}
\int x\sqrt{1-x^4}\ dx&=-\frac12\int\dfrac{u^2}{\sqrt{1-u^2}}\ du\\
&=\frac12\int\dfrac{1-u^2-1}{\sqrt{1-u^2}}\ du\\
&=\frac12\left(\int\dfrac{1-u^2}{\sqrt{1-u^2}}\ du-\int\dfrac{1}{\... | {
"language": "en",
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} |
Geometric intuition: Seeing the regions in double integrals Context: solving double integrals.
I had the formula $$x^2+y^2=1-x-y$$ yet I could not see what shape it had. This is even more true with 3D pictures like $$2x^2+2y^2 \le 1+z^2.$$ Is there a summary somewhere of shapes to learn, so that I can get this.
| Note
\begin{align}
x^2+y^2&=1-x-y \\
\Leftrightarrow x^2+y^2+x+y & =1 \\
\Leftrightarrow x^2+x+\frac{1}{4}+y^2+y+\frac{1}{4}&=1+\frac{1}{4}+\frac{1}{4} \\
\Leftrightarrow \left(x+\frac{1}{2}\right)^2+ \left(y+\frac{1}{2}\right)^2 & =\frac{3}{2}
\end{align}
Last equation describes a circular cilinder with axis parallel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/757988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimal polynomial: is $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$? I was wondering about the minimal polynomial of real number
$$u=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$$
over field $\mathbb{Q}$.
As you can see here, I worked out that $u$ is a root of monic rational polynomial $x^3+3x-4$. This is not irreducible... | A calculation shows that $(1+\sqrt{5})^3=16+8\sqrt{5}$, so the real cube root of $2+\sqrt{5}$ is $\frac{1}{2}(1+\sqrt{5})$.
Similarly, or by using conjugation, the real cube root of $2-\sqrt{5}$ is $\frac{1}{2}(1-\sqrt{5})$.
Add.
| {
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"url": "https://math.stackexchange.com/questions/759725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Range of f(x) = $\frac{\sqrt3\,\sin x}{2 + \cos x}$ Can you give any idea about the range of the following function?
$$f(x) = \frac{\sqrt{3}\,\sin x}{2 + \cos x}$$
| Yes:
*
*$-1 \leq \sin x \leq 1 \implies -\sqrt{3} \leq \sqrt{3}\sin x \leq \sqrt{3}$
*$-1 \leq \cos x \leq 1 \implies 1 \leq 2+\cos x \leq 3$
Hence: $$-\sqrt{3} \leq \frac{\sqrt{3}\sin x}{2+\cos x} \leq \sqrt{3}$$
Of course, you can do better than this using $f'(x)$ in order to calculate min/max points of $f(x)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/762243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove this identity: $ \tan(2x)-\sec(2x) =\tan(x-\pi/4)$ I've been having a time with this problem. I tried to start with the left side but I hit a dead end quick... I then tried the right side and had a little more luck but I've hit a block. I first used the tan difference identity then converted everything into sins ... | You're almost there. Dividing the numerator and denominator by $\cos x$ gives $$\frac{\sin x - \cos x}{\cos x + \sin x} = \frac{\tan x - 1}{1 + \tan x},$$ and then observe that $\tan \frac{\pi}{4} = 1$. Then recall the tangent addition identity $$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/762487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculating absolute value and argument of a complex number I want to calculate the absolute value and argument of the complex number $a = \left(\sqrt{3} - i\right)^{-2}$.
In order to calculate these two values I tried to reform the number into the form $z = x + y \cdot i$:
$$a =\left(\sqrt{3} - i\right)^{-2} = \frac{... | Rewrite:
\begin{align}
(\sqrt{3}-i)^{-2}&=\frac{1}{(\sqrt{3}-i)^{2}}\\
&=\frac{1}{2-2\sqrt{3}\ i}\\
&=\frac{1}{2-2\sqrt{3}\ i}\cdot\frac{2+2\sqrt{3}\ i}{2+2\sqrt{3}\ i}\\
&=\frac{2+2\sqrt{3}\ i}{2^2-(2\sqrt{3}\ i)^2}\\
&=\frac{2+2\sqrt{3}\ i}{16}\\
&=\frac18+\frac18\sqrt{3}\ i
\end{align}
It should be easy from this to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Three-Digit numbers divisbile by 3 How many three digit numbers are divisible by 3 and have an additional property that the sum of of their digits is 4 times the middle digit?
My approach: let the number be $abc$ so $$abc \equiv 0\pmod{3}$$ and $$a + b+ c= 4b$$
I'm stuck now. Any help?
| First, $b$ should be dividable by $3$, since $a+c=3b$, we have $0<3b<18$. So $b$ can be $3$ and $6$.
For $b=3$, we have $a+c=9$, there are 9 choices;
for $b=6$, we have $a+c=18$, then the only choice is $a=c=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the limit $\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$ I am trying to evaluate the following limit
$$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$$
If we use the binomial expansion of the numerator term, the answer is $\frac{1}{4}$. The same answer is obtained i... | For example, use
\begin{align}
&\;\;\;\frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8}}{x}=\\&=\frac{1}{x}\cdot\frac{(1+x)^\frac{1}{4}-(1-x)^\frac{1}{4}}{(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}}=\\
&=\frac{1}{x}\cdot\frac{(1+x)^\frac{1}{2}-(1-x)^\frac{1}{2}}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding an integer $c$ for which $a+c^2 \equiv 0 \pmod {2b-2c}$, $a$ & $b$ constant Looking for a solution for such a challenge, I have a decision problem that is solved if there is a positive integer $c$, which for given integer constants $a$, $b$ satisfies the equation
$a+c^2 \equiv 0 \pmod {2b-2c}$
or simply to say ... | There are many choices for $c$, depending on the given integers $a$ and $b$.
Given
$a+c^2 \equiv 0 \pmod {2b-2c}$, subtract $b^2$ from both sides;$a+c^2-b^2 \equiv -b^2\pmod {2b-2c}$
multiplying through by $2$;
$2a+(2c-2b)(c+b) \equiv -2b^2 \pmod {2b-2c}$
or
$2(a+b^2) \equiv 0 \pmod {2b-2c}$. Or $(a+b^2) \equiv 0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show the limit of the following is $\dfrac{1}{12}$
Show that $$\lim_{n\to \infty}n^2 \log \left(\dfrac{r_{n+1}}{r_{n}} \right)=\dfrac{1}{12}$$ where $r_{n}$ is defined as;
$$r_{n}=\dfrac{\sqrt{n}}{n!} \left(\dfrac{n}{e} \right)^n$$.
Now I simplified $\dfrac{r_{n+1}}{r_{n}}$ to $$\dfrac{r_{n+1}}{r_{n}}= \left(\dfr... | Here is the remaining part -
$$\lim_{n\to \infty} n^{2}((n + 0.5)\log(1+\frac{1}{n})-1)\\\\\,\\
\lim_{n\to \infty} n^{2}((\frac{n+0.5}{n}-\frac{n+0.5}{2n^2}+\frac{n+0.5}{3n^3} - ...) -1)\\\\\,\\
\lim_{n\to \infty} n^{2}(\frac{0.5}{n}-\frac{n+0.5}{2n^2}+\frac{n+0.5}{3n^3} - ...)\\\\\,\\
\lim_{n\to \infty} (0.5n-0.5n - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/769805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How should I simplify this $\tan^{-1}$ expression? Integral question I have to integrate this function
$$I = \int_0^4\frac{20x-5x^2}{x^2+9} \mathrm{d}x$$
obtaining
$$20\ln(5/3) + 15\tan^{-1}(4/3) -20.$$
However, my calculator, even after somehow simplifying it a bit, gives this:
$$\frac{40\ln(5/3) -30\tan^{-1}(3/4) ... | By first simplifying the fraction, we get that $$\frac{40\ln(5/3) -30\tan^{-1}(3/4) +15\pi -40}{2} = 20\ln(5/3) -15\tan^{-1}(3/4) +\frac{15}{2}\pi -20$$
Subtracting $20\ln(5/3) + 15\tan^{-1}(4/3) -20$ from it, we get $$(20\ln(5/3) -15\tan^{-1}(3/4) +\frac{15}{2}\pi -20) - (20\ln(5/3) + 15\tan^{-1}(4/3) -20)$$
$$ = \fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove $n^2(n^4-1)$ is divisible by 60 using Mathematical Induction. Base step:
p(2)=4 * 15= 60
Inductive Hypothesis:
Assuming p(k) = $k^2(k^4-1)$ = 60q
Induction:
p(k+1)= $(k+1)^2[(k+1)^4-1]$
= $(k+1)^2[(k+1)^2 + 1][(k+1)^2 - 1]$
= $(k+1)^2(k^2+2k+2)(k^2+2k+1- 1)$
= $(k+1)^2(k^2+2k+2) k (k+2)$
=... | Does it have to be by induction?
Write $n^2(n^4-1)= n^2 (n-1) (n+1) (n^2+1) = n(n-1) n (n+1) (n^2+1)$.
$2$ divides $n(n-1)$ and $n(n+1)$ and so $4$ divides $n^2(n^4-1)$.
$3$ divides $(n-1)n(n+1)$ and so $3$ divides $n^2(n^4-1)$.
Since $60=4\cdot3\cdot5$, it remains to prove that $5$ divides $n^2(n^4-1)=n(n^5-n)$. But t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$ Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$.
I have no clue how to proceed and tried to prove that the whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I think... | $$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Let $\displaystyle z^7=-1=\cos\pi+i\sin\pi$
Using this, $\displaystyle z=\cos\frac{(2n+1)\pi}7+i\sin\frac{(2n+1)\pi}7$ where $n\equiv-3,-2,-1,0,1,2,3\pmod7$
So, the equation whose roots are $\displaystyle z=\cos\frac{(2n+1)\pi}7+i\sin\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Closed form of $ \int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx$ Hello I am trying to solve an incredible integral given by
$$
\int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx=\pi \ln\bigg[\frac{1}{2}\left(\cos^2\alpha +\sqrt{\cos^4 \alpha +\cos^2\frac{\beta}{2} \si... | Here is a sketchy calculation. It matches numerically for all values of $a,b$ that I tried.
Abbreviate $a = \cos^2 \alpha$ and $b = \sin^2 \beta$. Then $0 \leq a,b \leq 1$.
We calculate
$$
I(a,b)= \int _{0} ^{\pi/2} \ln(1-\cos^2 x(1-a-b \sin^2 x)) \,dx
$$
We have
$$
\partial_aI(a,b)= \int _{0} ^{\pi/2} \frac {\cos^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Expectation with square root I don't know how to calculate the expectation when there is some square root in the expression. My problem is this: we have three real random variables $X,Y,Z$, independent and with standard normal distribution $N(0,1)$ and we want to calculate
$$E\left(\frac{1}{2} \left(X + Z + \sqrt{X^2 +... | I am assuming that $X$, $Y$, and $Z$ are mutually independent, standard Gaussian.
The expression of your expectation can be simplified.
Let us denote your expectation as $E$.
First, notice that $X$ and $Y$ are zero-mean, so
$$
E = \frac{1}{2}\mathsf{E}\left[\sqrt{X^2 + 4Y^2 - 2XZ + Z^2}\right]
$$
Grouping terms, we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/777573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Show $ \frac {a^3}{a^2+b^2+c^2-bc}+\frac {b^3}{a^2+b^2+c^2-ca} +\frac {c^3}{a^2+b^2+c^2-ab} \le \frac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $ If $a,b,c$ are positive real numbers then how do we prove that $ \dfrac {a^3}{a^2+b^2+c^2-bc}+\dfrac {b^3}{a^2+b^2+c^2-ca} +\dfrac {c^3}{a^2+b^2+c^2-ab} \le \dfrac {(a+b+c)^2... | $(a+b+c)^2\geq3(ab+ac+bc)$.
Thus, it remains to prove that
$$\sum_{cyc}\frac{a^3}{a^2+b^2+c^2-bc}\leq\frac{a^4+b^4+c^4}{2abc}$$ or
$$\sum_{cyc}\left(\frac{a^3}{2bc}-\frac{a^3}{a^2+b^2+c^2-bc}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^3(a^2+b^2+c^2-3bc)}{bc(a^2+b^2+c^2-bc)}\geq0$$ or
$$\sum_{cyc}\frac{a^4(a^2+b^2+c^2-3bc)}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Show $a^2+b^2+c^2=1$ $\implies$ $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $ If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that
$ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $
( don't apply Schur's in... | By C-S
$$\sum_{cyc}\frac{b^2c^2}{1+a^2}=\sum_{cyc}\frac{b^2c^2}{a^2+b^2+a^2+c^2}\leq$$
$$\leq\sum_{cyc}\frac{b^2c^2}{4}\left(\frac{1}{a^2+b^2}+\frac{1}{a^2+c^2}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b^2c^2}{a^2+b^2}+\frac{b^2c^2}{a^2+c^2}\right)=$$
$$=\frac{1}{4}\sum_{cyc}\left(\frac{b^2c^2}{a^2+b^2}+\frac{a^2c^2}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/779330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Show that an integer of the form $8k + 7$ cannot be written as the sum of three squares. I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed.
My attempt at the program was as follows:
A number of the form, $8k +... | A variation:
Let $n = 8k + 7 = x^2 + y^2 +z^2$.
Then $n \equiv 3 \pmod 4$. Every square is congruent to $0$ or $1$ mod $4$, so $x^2, y^2, z^2 \equiv 1 \pmod 4$ for these numbers to add up to $ 3 \bmod 4$. Then $x, y, z$ are odd.
One checks immediately that $(\pm 3)^2 \equiv 1 \pmod 8, (\pm 1)^2 \equiv 1 \pmod 8$, which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/779784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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How to expand out negative powers for complex numbers I have the following expansions but I don't know how my teacher gets them. Apparently there is a formula for it (though the guy who told me didn't know it well), but I cannot find it in my notes.
For example, let $z$ be a complex number. Then,
$$\frac{1}{z^2+1}=\fra... | Start from the original equation
$$
\frac{1}{z^2+1} = \frac{i/2}{z+i} - \frac{i/2}{z-i}
$$
And just square it.
$$
\frac{1}{(z^2+1)^2} = \left(\frac{i/2}{z+i}\right)^2 + \left(\frac{i/2}{z-i}\right)^2 - 2\frac{i/2}{z+i}\frac{i/2}{z-i}
= -\frac{1/4}{(z+i)^2} - \frac{1/4}{(z-i)^2} + \frac{1/2}{z^2+1}
$$
Reuse the first eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/780063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove that the following function is convex? I want to prove convexity of the following function:
$$f(x) = log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)$$
for any fixed $a, b \in (0, 1)$ and:
*
*$x\in(0,1)$
*$x\in(1, \infty)$
I'm trying to solve it for a very long time, tried to investigate the sign ... | You can write
$$
f(x) = \log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)
= \frac{\ln \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)}{\ln x},
$$
and prove that $f''(x) > 0$ where needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let f be a continuous function defined on [-2009,2009] such that f(x) is irrational for each $x \in [-2009,2009]$ ... Problem : Let f be a continuous function defined on [-2009,2009] such that f(x) is irrational for each $x \in [-2009,2009]$ and $f(0) =2+\sqrt{3}+\sqrt{5}$ Prove that the equation $f(2009)x^2 +2f(0)x +... | A continuous function that takes only irrational values on an interval must be constant. Hence $f(2009)=f(0)\ne0$ and the quadratic is equivalent to $x^2+2x+1=0$ (with double root $x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it sp... | Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form:
$$
f(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} - x
$$
where
$$
n,k \in \Bbb N \\
a_k \in \Bbb R
$$
Consider the follo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 1
} |
Solve $X_n=\lfloor \sqrt{X_n} \rfloor+X_{n-1}$ How do I solve $X_n=\lfloor \sqrt{X_n} \rfloor+X_{n-1}$? The initial terms are $1,2,3,5,7,10,13,17,21,26,31$. A search on oeis.org/ gave $\lfloor n/2 \rfloor\cdot\lceil n/2 \rceil$ + 1 which should be proven by induction. Is there a different approach?
| The difference equation
\begin{align}
x_{n} = \lfloor \sqrt{x_{n}} \rfloor + x_{n-1},
\end{align}
where $x_{0}=1 $ and $x_{1}=2$ can, upon writing out several terms, ie $x_{2}$, $x_{3}$, $x_{4}$, and so on can the given set of values stated. It is without much difficulty to show that the difference equation for $x_{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/784302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove Fibonacci sequence with matrices? How do you prove that:
$$
\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}^n
=
\begin{pmatrix}
F_{n+1} & F_n\\
F_{n} & F_{n-1}
\end{pmatrix}$$
| $$\begin{align}
F(n+1) &= 1\,F(n) + 1\,F(n-1)\\
F(n) &= 1\,F(n) + 0\,F(n-1)\\
\\
\begin{bmatrix} F(n+1) \\ F(n) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} F(n) \\ F(n - 1) \end{bmatrix} \\
\begin{bmatrix} F(n+1) \\ F(n) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/784710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
Computing the sum $\sum_{n=2}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n-1)^2}}$ Find the value of $$\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \sqrt{1 + \frac{1}{3^2} + \frac{1}{4^2}} +...+ \sqrt{1 + \frac{1}{2010^2} + \frac{1}{2011^2}}$$
I have not been able to simplify the ter... | Let us first note that
$$n^2+(n-1)^2+n^2(n-1)^2=(n^2-n+1)^2,$$
and therefore the sum can be rewritten as
$$\sum_{n=2}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n-1)^2}}=\sum_{n=2}^{2011}\frac{n^2-n+1}{n(n-1)}=\sum_{n=2}^{2011}\left(1+\frac{1}{n-1}-\frac1n\right)=2011-\frac{1}{2011}.$$
The sum of $1$'s at the last step giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Understanding substitution to compute primitive To compute the primitive
$$\int x^2 \sqrt[3]{x^3+3}\ dx$$
I am trying this:
$t = x^3+3$
$dt = 3x^2dx$
but
$\int x^2 \sqrt[3]{x^3+3}\ dx=\ ?$
How to continue from here?
| Take the differential form ($dx$) out from underneath the squareroot sign. It reads $\int x^2\cdot (x^3+3)^{1/3} dx = 1/3 \int (3x^2 dx)\cdot (x^3+3)^{1/3} = 1/3\cdot (x^3+3)^{4/3} / (4/3) + c = \frac{(x^3+3)^{4/3}}{4} + c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Integral $\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c}$ Hi I am trying to prove this result $$
I:=\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c},\quad c>1.
$$
Thanks. Since $x\in[0,1]
$ we can write$$
I=\s... | Using the identity
\begin{align}
\psi(1-x) - \psi(x) = \pi \cot(\pi x)
\end{align}
then the derivative with respect to $x$ yields
\begin{align}
\psi_{1}(1-x) + \psi_{1}(x) = \pi^{2} \csc^{2}(\pi x).
\end{align}
Now the integral
\begin{align}
I &= \int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/791870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
$(X, \circ)$ is a group, define $a\circ b$ Let $X$ be $\{1,2,3\}$ and $\circ$ a binary operation over $X$ such that $(X,\circ)$ forms a group, and $3\circ 3 = 3$.
Define $a\circ b$ for all $ a,b \in X$.
I think that the identity element is 3. But I am not sure how to proceed.
| Knowing only $3\circ 3=3$, we start with the table: $$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & & & \\ 2 & & & \\ 3 & & & 3\end{array}$$
By definition, a group is closed, associative, has a unique identity element, and every element has an unique inverse.
Closed: $\forall a,b\in X \implies a\circ b \in X$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
no. of ways of forming a garland using $6$ red roses and $4$ white roses. The no. of ways that the garland can be made out of $6$ red and $4$ white roses so that all white roses not come together.
$\bf{My\; Solution::}$ First we will form a garlad using $6$ red roses, This can be done in $\displaystyle (6-1)!$ ways.
No... | This problem can be solved with the Polya Enumeration Theorem
(PET). We assume that garlands that can be transformed into each other
by rotations or reflections are considered the same.
Furthermore we do not treat the problem where the forbidden garlands are
precisely those that contain four white roses a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
To solve $2\cdot(5^y)-7^x=1$ How do we find all non-negative integers $x,y$ such that $2\cdot(5^y)-7^x=1$ ?
Does there exist any solution with $x>2, y>2$ ?
| If $y \geq 3$, then taking $\pmod{125}$, $7^x \equiv -1 \pmod{125}$ so $x \equiv 10 \pmod{20}$. Write $x=10a$ where $a$ is odd. Then $2(5^y)=7^x+1 \equiv 7^{10a}+1 \equiv (-1)^a+1 \equiv 0 \pmod{7^{10}+1}$. However $281 \mid 7^{10}+1$, so we get a contradiction.
Thus there are no solutions with $y \geq 3$. We may check... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.