Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Rational bounds for $\tan^{-1} x$ I want help with this question.
Show that for all $x>0$, $$ \frac{x}{1+x^2}<\tan^{-1}x<x.$$
Thank you.
| $$ \frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2} < \frac{d}{dx} x =1$$
and $\tan^{-1}(0)=0$ so that the last inequality is proved.
$$ \frac{d}{dx} \frac{x}{x^2+1} = \frac{(x^2+1) - x(2x) }{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2} < \frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2} $$
and $\frac{x}{1+x^2}(0)=0=\tan^{-1}(0)$ so th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/588930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Difference of squares Determine if $5^{36} - 1$ is divisible by $13$ using the difference of squares. I tried splitting it up by difference of squares a couple of times but can't seem to get to a point where I can determine divisibility.
| We have $5^{36}-1=(5^{18}+1)(5^{18}-1)$. Also, $5^{18}+1=(5^6)^3+1=(5^6+1)((5^6)^2-5^6+1)$, and $5^6+1=(5^2)^3+1=(5^2+1)((5^2)^2-5^2+1)$. We are done, because $5^2+1=26=2\cdot13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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eigenvalues of integral matrices Is it possible that a $3$-by-$3$ matrix with integer values and determinant 1 has a real eigenvalue with algebraic multiplicity 2, that is not equal to $\pm 1$?
Doing some elementary computations one can rephrase the question as follows. Do there exist integers $k$ and $m$ such that $a... | As in Ewan's proof, write
$$x = \frac{2 a^3+1}{a^2} \quad y = \frac{a^3+2}{a}.$$
Notice that
$$a = \frac{xy-9}{2 (x^2-3y)}. \quad (\ast)$$
So, if $x$ and $y$ are integers then either $a$ is rational or else $xy=9$ and $x^2=3y$. The latter case implies $x (x^2/3) = 9$, so $x=y=3$ and $a=1$, which we already ruled out.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Logarithm / exponential equation, not sure what to make of this, (simple) Solve for $a:(2 \log_a x)(3 \log_{x^2} 4) = 3$
No idea how to approach this problem other than moving the 2 and the 3 into an exponent..
| This answer is full of seemingly useless derivations, but would help you in the future problems.
There are three interesting results of the base-change property:
First property: $\log _a b = \dfrac{1}{\log _b a }$
Proof. $\log_a b = \dfrac{\log b}{\log a}$ and $\dfrac{1}{\log_b a} = (\log _b a)^{-1} = \left(\dfrac{\lo... | {
"language": "en",
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calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\c... | Given $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$$
First we will simplify $$\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$$
$$\displaystyle \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
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Proving $4(a^3 + b^3) \ge (a + b)^3$ and $9(a^3 + b^3 + c^3) \ge (a + b + c)^3$
Let $a$, $b$ and $c$ be positive real numbers.
$(\mathrm{i})$ Prove that $4(a^3 + b^3) \ge (a + b)^3$.
$(\mathrm{ii})$Prove that $9(a^3 + b^3 + c^3) \ge (a + b + c)^3.$
For the first one I tried expanding to get $a^3 + b^3 \ge a^2b+ab... | Actually both inequalities follow directly form Holder's Inequality. Obviously $q=\frac 32$ and $p = 3$ satisfy the condition $\frac 1p + \frac 1q = 1$. Then we have:
$$(1^q + 1^q)^{\frac 1q}(a^p + b^p)^{\frac 1p} \ge (a+b)$$
Substitute and we have:
$$(1^{\frac 32} + 1^{\frac 32})^{\frac 23}(a^3 + b^3)^{\frac 13} \ge (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomials with rational zeros Find all polynomials $F(x)={a_n}{x^n}+\cdots+{a_1}x+a_0$ satisfying
*
*$a_n \neq0$;
*$(a_0, a_1, a_2, \ldots ,a_n)$ is a permutation of $(0, 1, 2 ... n)$;
*all zeros of $F(x)$ are rational.
| The complete list of such polynomials is
$$x, \quad 2x^2+x, \quad x^2+2x, \quad 2x^3+3x^2+x, \quad x^3+3x^2+2x.$$
We now prove this. The first observation to make is that for a polynomial satisfying your hypotheses, all rational roots are non-positive, and that $0$ occurs as a root at most once and only if the constant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596589",
"timestamp": "2023-03-29T00:00:00",
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How can I find this integral? I want to find the integral $$\int{\dfrac{\cos^2 x}{\sin^2 x + 4\sin x \cos x}} \mathrm{d} x.$$ I tried put $t=\tan \frac{x}{2}$, I got
$$\int{-\dfrac{1}{4}\dfrac{(t^2-1)^2}{t(t^2 + 1)(2t^2-t-2)} \mathrm{d} t.}$$
But, it's too difficult for me to calculate this integral.
| HINT:
Diving the numerator & the denominator by $\sec^4x,$
$$I=\int{\dfrac{\cos^2 x}{\sin^2 x + 4\sin x \cos x}} dx=\int\frac{\sec^2xdx}{(\tan^2x+4\tan x)(\tan^2x+1)}$$
Setting $\tan x=u,$
$$I=\int\frac{du}{(u^2+4u)(u^2+1)}$$
Using Partial Fraction Decomposition
$$\frac1{u(u+4)(u^2+1)}=\frac Au+\frac B{u+4}+\frac{Cu+D}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/599906",
"timestamp": "2023-03-29T00:00:00",
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The limit $\lim_{x\to \infty}\frac{x-\frac{1}{2}\sin x}{x+\frac{1}{2}\sin x}$ How to find the value of the limit:
$$ \lim_{x\to\infty}\frac{x-\frac{1}{2}\sin x}{x+\frac{1}{2}\sin x} $$
(l'Hopital not working here, right?)
| Numerator and denominator are both $\sim x$, because $\lim_{x \to \infty} \frac{x \pm \frac 1 2 \sin x}{x}=1$, thus your limit is also 1.
Two functions $f$ and $g$ are equivalent at some point $a$ (possibly $\infty$), and you denote $f \sim g$, if: $f(x) = (1+\varepsilon(x))g(x)$ with $\varepsilon(x) \to 0$ for $x\to a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/601115",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find all solutions for a system of linear equations over a given field My Problem is: to find all Solutions for the following given System of linear equations over the Field $K = \mathbb{Z}_{/7}$
The System is given with:
$$\begin{equation}
\begin{split}
x_1\quad\quad\quad + 3x_3 &= 5 \\
5x_1 + 3x... | \begin{align}
(A\ \ b)=
\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
5 & 3 & 6 & 3 \\
6 & 2 & 5 & 6
\end{array}\right)
&\to
\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
0 & 3 & 5 & 6 \\
0 & 2 & 1 & 4
\end{array}\right)\quad R_2-5R1, R_3-6R_1
\\
&\to
\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
0 & 1 & 4 & 2 \\
0 & 2 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/601912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of inequality with ratio of odds to evens. A professor gave this to the class as a challenge. It may or may not have to do with calculus, but I'm running out of ideas.
Prove that for any $n > 1$
$$ \frac{1}{2\sqrt{n}} < \frac{1}{2} \cdot \frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2n -1}{2n} < \frac{1}{\sqrt{2n}} $$... | Let $\displaystyle \frac1A = \frac{1}{2} \cdot \frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2n -1}{2n}$. Then we have
$$\displaystyle \frac{1}{2\sqrt{n}} < \frac1A < \frac{1}{\sqrt{2n}} \iff 2\sqrt{n} > A > \sqrt{2n} \iff 4n > A^2 > 2n$$
For the lower bound, we have to show
$$\frac{2^2}{1} \cdot \frac{4^2}{3^2}\cdot \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604431",
"timestamp": "2023-03-29T00:00:00",
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Prove $1 + \tan^2\theta = \sec^2\theta$ Prove the following trigonometric identity: $$1 + \tan^2\theta = \sec^2\theta$$
I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant.
| Assuming the First Pythagorean Trigonometric Identity,
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing by $\cos^2\theta$,
$$\Rightarrow \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$
$$\Rightarrow \left(\frac{\sin\theta}{\cos\theta}\right)^2 + \left(\frac{\cos\theta}{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Divisibility by large powers Show that $2^{111} + 1$ divides $2^{555} + 1$ but does not divide $2^{444} + 1$.
My try:
$(2^{111}+1)^5 = 2^{555}+5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111}+1$
$2^{555} + 1$ = $(2^{111}+1)^5$ - $(5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111})$
= $(2^{111}+1)^5$ - ($5*2^{333}(2^{111}+1)+5*2^{222}... | Note that $2^{111}\equiv -1\pmod{2^{111}+1}$. Thus $2^{555}\equiv (-1)^5=-1\pmod{2^{111}+1}$.
Similarly, $2^{444}\equiv (-1)^4=1\pmod{2^{111}+1}$, and therefore $2^{111}+1$ divides $2^{444}-1$. It follows that $2^{111}+1$ cannot divide $2^{444}+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/610148",
"timestamp": "2023-03-29T00:00:00",
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Parity through Series expansion By Maclaurin series; $\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-....$, and we know that period of $\sin{x}$ is $2π$ because $\sin{(x+2π)} = \sin{x}$. But if we consider only the RHS of the above series then how we can tell that this expression (series) is also of period$=2π$?
| Expanding on the comment I have given I first give the proof for $\sin (x + y)$ formula. First we define $$\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots $$ and by differentiation of power series $(\sin x)'= \cos x, (\cos x)' = -\sin x$.
Let us define $$f(z) = 1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate $\int{ dx \over \sqrt{1 + x^2}}$ How to integrate $dx \over \sqrt{1 + x^2}$?
Answer should be $\log ( x + \sqrt{1 + x^2})$
Please help as possible...
Thank you
| Let $\displaystyle 1+x^2=y^2\Rightarrow \sqrt{1+x^2} = y,$ and $2xdx = 2ydy$
$$\displaystyle xdx = ydy\Rightarrow \frac{dx}{y} = \frac{dy}{x} = \frac{d(x+y)}{(x+y)}$$ (Using Ratio and Proportion).
Now $$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \int \frac{dx}{y} = \int \frac{d(x+y)}{(x+y)} = \ln \left|x+y\right|+C$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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max and min $x y - \ln(x^2 + y^2)$ Find max and min
$$x y - \ln(x^2 + y^2) , 1/4 \le x^2 + y^2 \le 4$$
Problem: With this hairy expression as my partial derivatives, I do not know what to do.
Attempt:
| \begin{align}
x-\frac{2y}{x^2+y^2}=0\\
y-\frac{2x}{x^2+y^2}=0
\end{align}
can be combined to, as you have done, to
\begin{align}
(x+y)\left(1-\frac2{x^2+y^2}\right)=0\\
(x-y)\left(1+\frac2{x^2+y^2}\right)=0
\end{align}
and also, since $(x,y)\ne(0,0)$, to
\begin{align}
2xy-2=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/613864",
"timestamp": "2023-03-29T00:00:00",
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Need Sine form of Cotangent equation $$(b^2 - c^2)\cot A + (c^2 - a^2)\cot B + (a^2 -b^2)\cot C=0$$
I want this equation to be in the Sine form.
Please help me with steps.
Thanks a lot
| Assuming that $a,b,c$ are the sides of a Triangle
Using Law of sines
$$T=(b^2-c^2)\cot A=4R^2(\sin^2B-\sin^2C)\frac{\cos A}{\sin A}$$
Again using
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ or Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$
$\displaystyle\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$
But as $A+B+C=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The second degree polynomial $f(x)$ satisfying $f(0) = 0$, $f(1) = 1$ and $f'(x)>0$ for all $x\in[0,1]$ The second degree polynomial $f(x)$ satisfying $f(0) = 0$, $f(1) = 1$ and $f'(x)>0$ for all $x\in[0,1]$
Then which of the following is right?
Options
(a) $ax+(1-a)x^2$, $a\in \mathbb{R}$
(b) $ax+(1-a)x^2$, $a\in (0,2... | So you found out that $f(x) = (1 - b)x^2 + bx$
Then $f'(x) = (2 - 2b)x + b$. This we know is a linear equation.
So, if this line has a positive gradient, then $2 - 2b > 0 \Rightarrow 1 > b$ We also want the points $f'(0)$ and $f'(1)$ to both be greater than 0.
$f'(0) > 0 \Rightarrow b > 0, f'(1) > 0 \Rightarrow 2 - b >... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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show that $\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$ show that
$$I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$$
I find this is Nice equalition!
My try: let
$$\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}$$
so
$$dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt$$
so
$$I=\frac{3}{7}\int_0^1 \frac{t^... | You can write the two parts of the integrals as follows:
$$
\int_0^1 \sqrt[3]{1-x^7} \, dx = \int_0^1 \int_0^\sqrt[3]{1-x^7} \, dy\,dx
$$
and
$$
\int_0^1 \sqrt[7]{1-y^3}) \, dy = \int_0^1 \int_0^\sqrt[7]{1-y^3} \, dx\,dy.
$$
Now you can show that the regions of both these integral are the same. Namely the sets $\{(x, y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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how to find out the region between two surfaces Describe the region cut out of the ball $x^2+y^2+z^2\le4$ by the elliptic cylinder $2x^2+z^2=1$ i.e the region inside the cylinder and ball
I equated $4-x^2-y^2=1-2x^2$ getting $y^2-x^2=3$. I guessed the region has to be
$\sqrt{1-2x^2}\le z \le \sqrt{4-x^2-y^2}$, $-\sqrt... | If we want to describe the region in $1/4$ of all space, i.e; $$x\ge0, z\ge0$$ then it becomes:
$$\sqrt{1-2x^2}\le z \le \sqrt{4-x^2-y^2},~~ 0\leq x\leq\frac{\sqrt{2}}{2},~~-\sqrt{x^2+3}\le y \le\sqrt{x^2+3}\\ 0\le z \le \sqrt{4-x^2-y^2},~~ \frac{\sqrt{2}}{2}\leq x\leq 2,~~-\sqrt{x^2+3}\le y \le\sqrt{x^2+3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with derivative of $y=x^2\sin^5x+x\cos^{-5}x$
Find $y^{\prime}$ of $y=x^2\sin^5x+x\cos^{-5}x$
My try:
$\dfrac{d}{dx}(x^2\sin^5x)=x^2(-5\sin^4x)+(2x\sin^5x)$
$\dfrac{d}{dx}(x\cos^{-5}x)=x(-5\cos^{-6}x)+1(\cos^{-5}x)$
This doesn't seem right. Can you please show how to do it correctly?
The answer is $y^{\prim... | $\dfrac{d}{dx}(x^2\sin^5x)=x^25\sin^4x \cos x+(2x\sin^5x)$
$\dfrac{d}{dx}(x\cos^{-5}x)=5x\cos^{-6}x \sin {x}+1(\cos^{-5}x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$3$ never divides $n^2+1$ Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.
Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.
If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.
And that is where I am stuck. I try ... | $n= 0 \pmod3 \implies n^2 + 1 = 1\pmod3$,
$n = 1\pmod3 \implies n^2 + 1 = 2\pmod3$,
$n = 2\pmod3 \implies n^2 + 1 = 2\pmod3$.
So $n^2 + 1$ is not a multiple of $3$ for any $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 4
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Set of integers with a particular additive property I have a set of integers $S = \{a_1,a_2,\ldots,a_n\}$. Let $K = a_1+a_2+\ldots+a_n$.
Consider the space of all $n$-tuples whose values are taken from the set $S$.
For example $(a_1,a_2,\ldots,a_n)$ is one $n$-tuple and its sum is $K$, $(a_1,a_1,a_2,a_2,\ldots,a_k)$ i... | The set of squares will not work in general. Any notrivial solution to $a^2+b^2=c^2+d^2$ allows us to drop $a^2$ and $b^2$ and instead have $c^2$ and $d^2$ twice. And such solutions are easy to find since the equation is equivalent to $(a-d)(a+d)=a^2-d^2=c^2-b^2=(c-b)(c+b)$. For example $1\cdot 15 = 3\cdot 5$, which le... | {
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"url": "https://math.stackexchange.com/questions/621756",
"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$ let $a,b,c,d,e$ are positive real numbers,show that
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+e}+\dfrac{d-e}{e+a}+\dfrac{e-a}{a+b}\ge 0$$
My try:
I have solved follow Four-variable inequality:
let $a... | For $n$ variables $x_i > 0$, (with the convention $x_{n+k}=x_k$) we need to show that
$$\sum_i \dfrac{x_i - x_{i+1}}{x_{i+1} + x_{i+2}} \geqslant 0 \iff \sum_i \left( \dfrac{x_i - x_{i+1}}{x_{i+1} + x_{i+2}} +\frac12 \right) \geqslant \frac{n}2 \iff \sum_i \dfrac{2x_i - x_{i+1}+x_{i+2}}{x_{i+1} + x_{i+2}} \geqslant n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding difference of angles in triangle . If $\sin A+\sin B+\sin C=0$, $\cos A+\cos B+\cos C=0$, then prove that $$A-B=B-C=C-A=\dfrac{2\pi}3$$
| Adding a new answer due to the volume of the other answer and these methods are markedly different from the other.
First of all, as coffeemath has observed $A,B,C$ can not be angles of a triangle as $\displaystyle0<A,B,C<\pi\implies$ all sine ratios will positive $\implies\sin A+\sin B+\sin C>0$
Method $1:$
From the o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Minimum value of: $x^7(yz-1)+y^7(zx-1)+z^7(xy-1)$
$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of:
$$x^7(yz-1)+y^7(zx-1)+z^7(xy-1)$$
I put it in the form $x^6y +x^6z+y^6x+y^6z+z^6x +z^6y$. I tried AM-GM but it's not helping.
| $$x^6y +x^6z+y^6x+y^6z+z^6x +z^6y \ge 6 (xyz)^{7/3}$$
Also $xyz = x+y+z \ge 3 \sqrt[3]{xyz} \implies xyz \ge 3\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If the coefficients of the quadratic equation $ax^2+bx +c$ are u.i.i.d ran variates in $(0,1)$ what is the probability of roots being real? If the coefficients a,b,c(taken in order ,c being the constant term) of a quadratic equation are randomly and independenly chosen in the open interval(0,1) what is the probability ... | $a,b,c$ are limited to the interval $(0,1)$, so that
$$
P(b^2 < 4ac) = \int_{a=0}^1\int_{c=0}^1\int_{b=0}^{\min(2\sqrt{ac},1)}da\,db\,dc =
\int_{a=0}^1\int_{c=0}^1 \min(2\sqrt{ac},1)\,da\,dc,
$$
so that
$$
P(b^2 < 4ac) = \int_{a=0}^1\int_{c=0}^{\min(1/(4a),1)}2\sqrt{ac}\,da\,dc
+ \int_{a=0}^1\int_{c=\min(1/(4a),1)}^1\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
... | As $(x-4)-(x-5)=1$
$\displaystyle(\sqrt{x-4}-\sqrt{x-5})(\sqrt{x-4}+\sqrt{x-5})=1$
As $\displaystyle\sqrt{x-4}-\sqrt{x-5}=-1, \sqrt{x-4}+\sqrt{x-5}=-1$
Adding we get $\displaystyle\sqrt{x-4}=-1$ which is impossible as $\sqrt{x-1}\ge0$ for real $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Compute $\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$ Compute the value of the following expression
$$\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\left ( \frac{1}{2}+\cdots + \frac{1}{n}\right )^2+\cdots+\left (\frac{1}{n-1}... | We can solve this by recursion relations. Denoting the sum as $S_n$, we have,
$$ S_{n+1} = \left (1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1} \right )^2+\left ( \frac{1}{2}+\cdots + \frac{1}{n} +\frac{1}{n+1}\right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} +\frac{1}{n+1} \right )^2+\left (\frac{1}{n} +\frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
prove $ b^6c^8-b^5c^7+b^8c^6-b^3c^6+b^2c^6-b^7c^5-b^6c^3+b^6c^2-bc^2+c^2-b^2c+b^2 \ge 0$ $b>0,c>0, $ prove $g(b,c)=b^6c^8-b^5c^7+b^8c^6-b^3c^6+b^2c^6-b^7c^5-b^6c^3+b^6c^2-bc^2+c^2-b^2c+b^2 \ge 0$
this is from a middle process of a inequality. (I am sure it is correct because the inequality is proved.)
edit: the inequa... | It seems the following.
AM-GM inequality implies the following inequalities:
$$\frac 12 b^8c^6+\frac 16 b^6c^8+\frac 13 b^6c^2\ge b^7c^5,$$
$$\frac 12 b^6c^8+\frac 16 b^8c^6+\frac 13 b^2c^6\ge b^5c^7,$$
$$\frac 13 b^8c^6+\frac 12 b^6c^2+\frac 16 b^2\ge b^6c^3,$$
$$\frac 13 b^6c^8+\frac 12 b^2c^6+\frac 16 c^2\ge b^3c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/626526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to calculate Taylor expansion of $\cos(\sin x)$ I know that Taylor expansion of $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^6)$
and that of $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - O(x^7)$.
But how do I calculte the Taylor Expansion of $\cos(\sin x)$.
| We know that for any $x$ (close to $0$):
$$
\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - O(x^6)\\
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - O(x^7)
$$
We can find $\cos\sin x$ by substituting $x\to \sin x$ in the expansion of $\cos x$:
$$
\cos x = 1 - \frac{(x - \frac{x^3}{3!} + \frac{x^5}{5!} - O(x^7))^2}{2!} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Calculation of $f(x) = \int_{0}^{\frac{\pi}{4}}\ln \left(1+x\cdot \tan z\right)dz$ If $\displaystyle f(x) = \int_{0}^{\frac{\pi}{4}}\ln \left(1+x\cdot \tan z\right)dz,$ where $x>-1$. Then value of $\displaystyle f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right) = $
$\bf{My\; Try::}$ Given $\displaystyle f(x) = \int_{... | There is a somewhat simpler way to procede: once you get to
$$f^\prime(x) = \int_0^{\pi/4} \frac{\tan z}{1+x \tan z} dz,$$ make the substitution $u = 2 z,$ so the integral becomes
$$\frac12\int_0^{\pi/2} \frac{\tan u/2}{1+x \tan u/2} du.$$ Now make the substitution $\tan u/2 = t,$ to get
$$\int_0^1 \frac{t}{(1+ x t)(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| Try looking at the terms in binary:
2^10 10000000000
2^9 1000000000
2^8 100000000
2^7 10000000
2^6 1000000
2^5 100000
2^4 10000
2^3 1000
2^2 100
+ 2^1 10
---------------------
= 11111111110
+ 2 10
= 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
Range Of Quartic Polynomial Of Two Variables $a$,$b$ are real numbers such that $~3\leq a^{2}+ab+b^{2}\leq 6$.
I would like to find the range of $~a^{4}+b^{4}$.
Is it possible to find it with (well-known) AM-GM, CS, etc...?
| $ab \le \dfrac{a^2+b^2}{2} \implies \dfrac{3(a^2+b^2)}{2} \ge 3 \implies a^2+b^2 \ge2 \implies a^4+b^4 \ge 2(\dfrac{a^2+b^2}{2})^2 \ge 2 $ when $a=b=1$ hold "="
$a^4+b^4=(a^2+b^2+\sqrt{2}ab)(a^2+b^2-\sqrt{2}ab)=(u+(\sqrt{2}-1)ab)(u-(\sqrt{2}+1)ab)=u^2-2abu-a^2b^2, u=a^2+ab+b^2 \le 6$
it hints that when $ab<0$, the max... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)... | Setting $\frac1n=h$
$$F=\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)$$
$$=\lim_{h\to0}\frac{\sqrt{1+h^2+20h^3+7h^4}-\sqrt{1+h^2+h^4}}{h^2}$$
Rationalizing the numerator,
$$F=\lim_{h\to0}\frac{(1+h^2+20h^3+7h^4)-(1+h^2+h^4)}{h^2(\sqrt{1+h^2+20h^3+7h^4}+\sqrt{1+h^2+h^4})}$$
As $h\to0,h\ne0,$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Maximum number of edges that a bipartite graph with $n,m$ vertices can have when it doesn't contain $4$-cycle Let $A_{n,m}$ be the maximum number of edges that a bipartite graph with $n,m$ vertices can have when it doesn't contain $4$-cycle. I have calculated some values:
$A_{2,2}=3$, $A_{3,3}=6$, $A_{4,4}=9$, $A_{4,5}... | I will give simple proofs of the evaluations $A_{4,6}=12,\ A_{5,6}=14,\ A_{6,6}=16,\ A_{6,7}=18,\ $ and $A_{7,7}=21$.
1. Since the ratio $\dfrac{A_{m,n}}{mn}$ is a nonincreasing function of $m$ and $n$, we have the recursive upper bound
$$A_{m,n+1}\le\lfloor\frac{n+1}nA_{m,n}\rfloor.$$
2. It's easy to see that $A_{1,n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
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 $2a + b$ ?
| We are given that $ a^2 + b^2 = 1$, and $ a, b \geq 0$.
Applying AM-GM or Cauchy-Schwarz, we get that
$$ 20 = 5(a^2 + a^2 + a^2 + a^2 + 4b^2) \geq (a+a+a+a+2b)^2 = (4a+2b)^2$$
Hence $ 2a + b \leq \frac{20}{16} = \frac{5}{4} $, with equality when $ a = 2b, b = \sqrt{\frac{1}{5} }$.
| {
"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": 0
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How to find nth derivative of $1/(1+x+x^2+x^3)$ I was trying to solve a differentiation question but unable to understand .
My question is :
find the $n^{th}$ derivative of $1/(1+x+x^2+x^3)$
I know that if we divide the numerator by denominator then the expression would be :
$1- x(1+ x^2 + x )/(1+x+x^2+x^3)$
But now h... | Hint: Note that $(1-x)(1+x+x^2+x^3)=1-x^4$, so $$\frac1{1+x+x^2+x^3}=\frac{1-x}{1-x^4}=\frac{1-x}{(1-x^2)(1+x^2)}=\frac{1}{(1+x)(1+x^2)}.$$
(After the fact, this factorization is easy to see directly.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/636718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$ Let $a,b,c\in R$ and satisfying $a^2+b^2+c^2=1$
Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$
| $3u=a+b+c,3v=ab+bc+ac \implies 9u^2-6v=1$
edit:
$P \ge \sqrt{(a+b+c)^2+(3-(3b+bc+ac))^2}=\sqrt{(9u^2+(3-3v)^2}=\sqrt{1+6v+9-18v+9v^2}=\sqrt{(2-3v)^2+6} $
$3v=ab+bc+ac \le a^2+b^2+c^2 =1 $
$ 2-3v \ge 1 \implies \sqrt{(2-3v)^2+6} \ge \sqrt{7}$
last "=" will hold when $a=b=c=\pm \dfrac{1}{\sqrt{3}}$
another "=" will hold... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solving recurrence relation with generating functions - Nearly got the answer I'm trying to solve the following recurrence relation (Find closed formula) using generating functions:
$f(n)=10f(n-1)-25f(n-2)$, $f(0)=0$, $f(1)=1$
I'm having a small difficulty at the end and can use a nudge in the right direction.
My solu... | An orderly way of attacking such problems is as follows: Define $F(z) = \sum_{n \ge 0} f(n) z^n$, write the recurrence so there are no substractions in indices:
$$
f(n + 2) = 10 f(n + 1) - 25 f(n)
$$
Multiply by $z^n$, sum over $n \ge 0$, recognize:
\begin{align}
\sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/638354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Complex numbers - Exponential numbers - Proof Let $z$ be a complex number, and let $n$ be a positive integer such that $z^n = (z + 1)^n = 1$. Prove that $n$ is divisible by 6.
For this problem I am stumped...how should I begin?
Also there's a hint for it:
From $z^n = 1$, prove that $|z| = 1$. What does the equation $(z... | We can say: $$z^n=1 \implies |z|^n=1 \implies |z^n|=1 \implies |z|=1$$
Using similar logic, we can also get: $$(z+1)^n=1 \implies |z+1|^n=1 \implies |(z+1)^n|=1 \implies |z+1|=1$$
Let's set $z=a+bi$: $$|z|=|z+1| \implies a^2 + b^2 = (a+1)^2 + b^2 \implies a^2 + b^2 = a^2 + b^2 + 2a + 1$$ $$\implies 2a + 1 = 0 \implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/643024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
limit problem - can't get rid of $0$. I am trying to evaluate limit:
$$\lim_{x\rightarrow 0}\frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}$$
I tried to use known limit in denominator to get:
$$\lim_{x\rightarrow 0}\frac{\frac{\arcsin}{x} - \frac{\arctan x}{x}}{x( \frac{e^x-1}{x}\cdot \frac{1}{x}+\frac{1-\cos x}{x^2}... | Using the Taylor series
$$f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f^{(3)}(0)+o(x^3)$$
for $f$ is $\arcsin$ and $\arctan$ we find
$$\arcsin x=x+\frac{x^3}{6}+o(x^3)$$
and
$$\arctan x=x-\frac{x^3}{3}+o(x^3)$$
and we have
$$e^x-\cos x-x^2-x=1+x+\frac{x^2}{2}+\frac{x^3}{6}-1+\frac{x^2}{2}-x^2-x+o(x^3)=\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/643787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Infinite Geometric Series Issue i have came across a series, i am trying to find its sum knowing the fact that, if it converges and its common ratio ex. r is: -1 < r < 1, then i can use the specified formula $\frac{a}{1-r}$ , which specifically means first term of series over 1 minus common ratio
here is the series
$\... | Since you know it converges, let $S=\sum_{n=1}^\infty\frac{2n-1}{2^n}$. Then we have
$$\begin{align}
S &=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\cdots&\text{so}\\
\frac{1}{2}S&=\phantom{\frac{1}{2}}+\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\cdots&\text{subtracting, we get}\\
S-\frac{1}{2}S&=\frac{1}{2}+\frac{2}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/645560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to get the man's age. I am very much in need of solution for this problem. I can't figure out the answer for this. does anybody know about this problem below?
thanks,,
Problem:
A man's boyhood lasted for 1/6 of his life, he played soccer for the next 1/12 of his life, and he married after 1/7 more of his life. a d... | We can set up this problem by adding all the fractions of his life so they equal $1$ (his whole life). Also, let's say that he lived to $x$ years old.
Now,
$\frac{1}{6}$ of his life was boyhood.
$\frac{1}{12}$ of his life was soccer.
$\frac{1}{7}$ of his life was end of soccer to marriage.
After $5$ years, his daughter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/647035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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How many zero elements are there in the inverse of the $n\times n$ matrix How many zero elements are there in the inverse of the $n\times n$ matrix
$A=\begin{bmatrix}
1&1&1&1&\cdots&1\\
1&2&2&2&\cdots&2\\
1&2&1&1&\cdots&1\\
1&2&1&2&\cdots&2\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
1&2&1&2&\cdots&\cdots
\end{bmatri... | I am excavating this post because this post needs an affirmative answer.
After some try I feel that this can be (ironically) done fastest by brute force. Let $A^{-1} = (a_{ij})$ and we just solve column-wisely $AA^{-1} = I$.
For the first column, by subtracting equations derived from neighboring row of $A$, you will ge... | {
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"url": "https://math.stackexchange.com/questions/648340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Am I understanding induction correctly? Here is an induction proof that I have written for my homework and I want to know if I am understanding this correctly:
Prove that for:
$ \sum\limits_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6}$
My proof:
Check the base-case, let mine be n=1 (or do I have to choose 0..? The lowest one)
... | If it is to be shown by induction that $$\sum_{i=0}^{n}f\left(i\right)=g\left(n\right)$$
then $f\left(0\right)=g\left(0\right)$ must be proved (the base case) and
secondly it must be shown that: $$\sum_{i=0}^{n+1}f\left(i\right)=g\left(n+1\right)$$
on base of: $$\sum_{i=0}^{n}f\left(i\right)=g\left(n\right)$$
That amou... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\frac{x}{1+x}<\log(1+x)0$ using the mean value theorem I want to show that $$\frac{x}{1+x}<\log(1+x)<x$$ for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.
$$\frac{x}{1+x}<\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) <0$$
Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$... | By Definition of log(which already a kind of mean value theorem ) we have
For any $x>0$ We have
$$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1} \le \int_{0}^x\frac{dt}{1} = x $$
Thus,
$$\frac{x}{x+1} \le \ln(x+1 ) \le x $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/652581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Find value of unending continued fraction I am trying to find the limit if it exists for the following unending continued fraction: $$1+{1\over{2+{1\over2+{1\over{2+...}}}}}$$
I have discovered this is the continued fraction for $\sqrt2$, but I am unsure how to show this.
Also I am being asked to assume that the even-o... | Let $x = 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}}$. Then \begin{align*}
x - 2 &= \frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}} \\
\frac{1}{x-2} &= 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}} \\
\frac{1}{x-2} &= x
\end{align*}
We find two roots to this quadratic equation, $1+\sqrt{2}$ and $1-\sqrt{2}$. Since $x = 2+\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to solve $\mathrm{diag}(x) \; A \; x = \mathbf{1}$ for $x\in\mathbb{R}^n$ with $A\in\mathbb{R}^{n \times n}$? I would like to solve the following equation for $x\in\mathbb{R}^{n}$
$$\mathrm{diag}(x) \; A \; x = \mathbf{1}, \quad \text{with $A\in\mathbb{R}^{n\times n}$},$$
where $\mathrm{diag}(x)$ is a diagonal matr... | Do these least squares solutions help?
$n=2$
$$
\begin{align}
%
\mathbf{D} \mathbf{A} x &= \mathbf{1} \\
%
\left[
\begin{array}{cc}
a_{\{1,1\}} & a_{\{1,2\}} \\
a_{\{2,1\}} & a_{\{2,2\}} \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
x_{\{1\}} & 0 \\
0 & x_{\{2\}} \\
\end{array}
\right]
%
\left[
\begin{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proving a property of about the Fermat numbers
Show that the last digit in the decimal expansion of $F_n=2^{2^n}+1$ is $7$ for $n \geq 2$.
For our base step we let $n=2$. Now we have $2^{2^2}=16$. So the assertion holds for our base case. Then we assume it holds for $n$. For the $n+1$ case, is there a way to de... | Observe that $\displaystyle2^1\equiv2,2^2\equiv4,2^3\equiv8,2^4=16\equiv6,2^5=32\equiv2\pmod{10}$
So, $\displaystyle2^m\equiv6\pmod{10}\iff 4|m$
$\displaystyle\implies2^{2^n}\equiv6\pmod{10}\iff 4|2^n\iff n\ge2$
Alternatively, as $\displaystyle6^{m+1}-6=6(6^m-1)\equiv0\pmod{30}$ as $6^m-1$ is divisible by $6-1=5$
$\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$8$ cards are drawn from a deck of cards without replacement I draw $8$ cards randomly from a shuffled deck without replacement. What is the expected value of the sum of the largest $3$ cards? The ace is given a value of 1 and the jack, queen and king are all given a value of $10$. I can generate a simulation answer bu... | For each of the possible values for the sum, $S$, of the three highest cards -- integers on $[6,30]$ -- figure out the number of combinations of eight-card hands that would give the three highest cards that sum.
For $S=30$, we can have anywhere from three to eight $10$s, so we consider those separately:
$$P(S=30, 10, 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How do I solve $y''+4y=0$? This problem is in Penney's Elementary Differential Equations, listed as a reducible, 2nd-order DE. The chapter has taught two techniques to be used, which are for when either $x$ or $y(x)$ is missing. It didn't show how to solve when $y'(x)$ is missing.
I checked with WolframAlpha, and it ... | $y''+4y=0$ is a second order differential equation.
First, change the equation to
$r^2+4 = 0$
This equation will will have complex conjugate roots, so the final answer would be in the form of $e^{\alpha x}(c_1\sin(\beta x) + c_2 \cos (\beta x))$
where $\alpha$ equals the real part of the complex roots and $\beta$ equal... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Rationalize the denominator of the surd, giving your answer in the simplest form. Rationalize the denominator of the surd, giving your answer in the simplest form.
$\frac {3}{\sqrt2+5} $
Please help me...
It must be like this right?
$\frac {3}{\sqrt2+5} * \frac{\sqrt2-5}{\sqrt2-5}$
| Yes. in general we do not leave the denominator irrational. So we multiply and divide the fraction by the conjugate of denominator.
In this case the conjugate of $\sqrt{2}+5$ is $\sqrt{2}-5$.
$$\frac{3(\sqrt{2}-5)}{(\sqrt{2}+5)(\sqrt{2}-5)}$$
$$\frac{3\sqrt 2 - 15}{2-25} = \frac{3\sqrt 2 - 15}{-23}$$
$$\frac{15 - 3\sqr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\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... | Your steps are correct, but just keep in mind that robotically converting everying into $\sin$s and $\cos$s isn't the only option available to you.
Note that
$$\cot\theta = \frac{\cos\theta}{\sin\theta}=\frac{\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}}=\frac{\csc\theta}{\sec\theta}$$
that
$$\cot\theta\tan\theta=\frac{1... | {
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"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
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How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$ How prove this $$\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$$
This equation How prove it? Thank y... | By way of enrichment here is another algebraic proof using basic
complex variables, quite similar to the accepted answer.
Note that the second binomial coefficient in both sums controls the
range of the sum, so we can write our claim like this:
$$\sum_{k=0}^{n+1} {n+1\choose k} (-1)^k {2n-3k\choose n-3k}
= \s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
| Expand both sides and we have:
$$\frac{x^2 + y^2}{2} \ge \frac{x^2 + 2xy + y^2}{4}$$
$$\iff x^2 + y^2 \ge {2xy}$$
$$\iff (x-y)^2 \ge 0$$
Which is well-know fact.
Also there are lot of other ways to prove it, here's "fancy" one using Cauchy-Scwarz inequality:
$$(1 + 1)(x^2 + y^2) \ge (x + y)^2$$
Multiply both sides by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
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
| Yet another method, inspired by looking at the problem geometrically (try drawing the region $a^2+b^2\leq2$ and the line $a+b=2$): let $s=a+b$, $t=a-b$. Then $a=\frac12(s+t)$ and $b=\frac12(s-t)$, so $a^2+b^2=\frac14\bigl((s^2+2st+t^2)+(s^2-2st+t^2)\bigr) = \frac12(s^2+t^2)$ and $a+b=s$, so the problem becomes:
if $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/666217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Generating functions - difficulty with question
Find the number of non-negative integers solutions to the equation $$x_1+x_2+x_3+x_4=12$$
when $x_1=2x_2+2$ and $x_3 \le x_4$.
My try:
Iv'e substituted $x_1$, thus the equation is $3x_2+x_3+x_4=10$. Second, we may define $x_4=x_3+y$ where $y \ge 0$ and the final equat... | One way to simplify the computation would be to take out the $\frac{1}{1-x^3}$, so that you only have to do partial fractions on $\frac{1}{(1-x^2)(1-x)}$. Then you get
\begin{align*}
\frac{1}{(1-x^2)(1-x)}
&= \frac{1}{4(1+x)} + \frac{1}{4(1-x)} + \frac{1}{2(1-x)^2} \\
&= \frac{1}{4} \sum_{i=0}^\infty (-x)^i
+ \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/666558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Does $b^2 = 4a + 2$ have integer solutions? I got the following question on an exam I had today: $$b^2 = 4a+2$$
I said that it didn't since $b^2 -2$ was not a multiple of $4a$ or in other words, was not divisible by $4$. Is this correct?
| We assume, on the contrary, that $b^2 = 4a + 2$ does indeed have integer solutions. Then, $b^2 \equiv 2 \mod 4 \implies 2$ is a quadratic residue modulo $4$.
We now aim to obtain a contradiction by proving (as demanded by the rigorous standards of mathematics) that $2$ cannot be a quadratic residue modulo $4$, i.e. $b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/669042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Optimization with a few Variables (AMC 12 question) In the 2013 AMC 12B, question 17 says:
Let $a$,$b$, and $c$ be real numbers such that
$a+b+c=2$, and $a^2+b^2+c^2=12$
What is the difference between the maximum and minimum possible values of $c$?
I was wondering if there is a quick and easy solution using multivariab... | Both extremes occur when $a=b.$
This is a geometric observation, the plane $a=b$ is a plane of symmetry for both the sphere and the (orthogonal) plane $a+b+c=2.$ Put another way, switching $(a,b)$ does not change anything, you still get a point in the intersection.
So, there is an argument that makes smaller demands on... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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For given positive integers $n,k$ prove that there always exists some positive integer $x$ for which $2^n\mid \dfrac{x(x+1)}{2}-k$ For given positive integers $n,k$ prove that there always exists some $x$ for which $2^n \mid \dfrac{x(x+1)}{2}-k.$
My work:
$\dfrac{x(x+1)}{2}$ is the sum of all positive integers upto $... | $$P_{n,k}(x) \equiv 2^n | \frac{x^2 + x}2 - k$$
By mechanical inspection we can see that the $x$ which satisfy $P_{n,k}$ this are all of the form $x_z = z\cdot 2^{n+1} + B$. So it suggests that we should consider this equation this equation in modulo $2^{n+1}$.
Consider possible solutions of the form $x = B \pmod {2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/670893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$ let $x,y,z$ are postive numbers,show that
$$\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}+\dfrac{y+z}{\sqrt{y^2+yz+z^2+zx}}+\dfrac{z+x}{\sqrt{z^2+zx+x^2+xy}}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$
My try: Wit... | We can use here the Ji Chen's lemma:
https://artofproblemsolving.com/community/c6h194103
for $p=\frac{1}{2}.$
For which it's enough to prove that:
1.$$\sum_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq2+\frac{xy+xz+yz}{x^2+y^2+z^2},$$
2.$$\sum_{cyc}\frac{(x+y)^2(y+z)^2}{(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}\geq1+\frac{2(xy+xz+yz)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
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Solve exponential-polynomial equation Solve the equation in $\mathbb{R}$
$$10^{-3}x^{\log_{10}x} + x(\log_{10}^2x - 2\log_{10} x) = x^2 + 3x$$
To be fair I wasn't able to make any progress. I tried using substitution for the logarithms, but it didn't help at all.
This is a contest problem, so there should be a nice s... | Clearly $x > 0$. Let $x = 10^y$. Then we have $10^{-3}x^y + x(y^2 - 2y) = x^2 + 3x$.
$$\iff 10^{-3}x^{y-1} + (y-3)(y+1) = x$$
$$\iff 10^{y^2-y-3} + (y-3)(y+1) = 10^y$$
$$\iff \underbrace{10^y(10^{(y-3)(y+1)}-1)} + \underbrace{(y-3)(y+1)} = 0$$
Now note that if $(y-3)(y+1) > 0$, both terms on the left are positive, he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Real number multiplicative inverses expressed in another form I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have:
$$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer... | We have $a+b\sqrt c$, where $a,b,c$ are known, and $c$ is not a perfect square, and we want to find $x$ and $y$ to make $$(x +y\sqrt c)(a + b\sqrt c) = 1.$$
Multiplying out, we get $$(ax + bcy) + (ay+bx)\sqrt c = 1.$$
The first term, $ax+bcy$, is an integer and its sum with $(ay+bx)\sqrt c$ is $1$, another integer, so... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Summation - relatively simple? I have a question which might be too simple for this site but I really tried many ideas without coming to a solution. This is assignment from elementary school in which I am trying to help and the solution should be relatively simple but somehow I cannot figure out the correct approach.
T... | I think there isn't a nice closed form for this sum. But if we let $H_n$ denote $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$, then
$\frac{1}{1 + 2} + \frac{1}{2 + 3} + ... + \frac{1}{99 + 100} = \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{199} = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{200} - 1 - (\frac{1}{2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/679430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Eliminate the parameter from the parametric equations $$x=\frac{3t}{1+t^3} , y=\frac{3t^2}{1+t^3} , t \neq -1,$$
and hence find an ordinary equation in x and y for this curve, The parameter t can be interpreted as the slope of the line joining the general point $(x,y)$ to the origin. Sketch the curve and show that th... | Note that
$$
t=\frac yx
$$
Then
$$
x=\frac{3\frac yx}{1+\frac{y^3}{x^3}}=\frac{3x^2y}{x^3+y^3}\implies x^3+y^3-3xy=0
$$
As $x$ gets big, note that the major terms are $x^3$ and $y^3$, so if we scale $x$ and $y$ by $\lambda$, we get
$$
x^3+y^3-\frac3\lambda xy=0\to0=x^3+y^3=(x+y)(x^2-xy+y^2)
$$
Since there is no solutio... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integer solutions of $ x^3+y^3+z^3=(x+y+z)^3 $ Consider the equation
$$ x^3+y^3+z^3=(x+y+z)^3 $$
for triples of integers $(x, y, z) $.
I noticed that this has infinitely many solutions: $ x, y $ arbitrary and $ z=-y $.
Are there more solutions?
| For such equations :
$$(x+y)^3+x^3+y^3+z^3=(x+y+z)^3$$
You can write the formula.
$$x=3p^2+18ps-s^2$$
$$y=15p^2-6ps-5s^2$$
$$z=3p^2-6ps+7s^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/682666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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How find this integral $I=\int_{-\infty}^{+\infty}\frac{x^3\sin{x}}{x^4+x^2+1}dx$ Find this integral
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea:
$$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because
$$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$
so
$$I=\int... | Note that
$$I(a) = \int_{-\infty}^\infty \frac{t\sin at} {t^2+1}dt
= \pi e^{-a}
$$
(derived from $I’’(a)=I(a)$, $I(0)= -I’(0) =\pi$). Then
\begin{align}
\int_{-\infty}^{\infty}\frac{x^3\sin x}{x^4+x^2+1}\> dx
= &\frac12\int_{-\infty}^{\infty}
\overset{x=\frac{\sqrt3t -1}2}{\frac{(x+1)\sin x}{x^2+x+1}} +\overset{x= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=1$? I'm working on an assignment where part of it is showing that $S_k=0$ for even $k$ and $S_k=1$ for odd $k$, where
$$S_k:=\sum_{j=0}^{n}\cos(k\pi x_j)= \frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}}+e^{-ik\pi x_{j}}) $$
Here $x_j=j/(n+1)$.
So, workin... | Using trigonometric identities
\begin{align}
\cos (k \pi x_j) &= \frac{\sin (k \pi x_{j+1}) - \sin (k \pi x_{j-1})}{2 \sin (\frac{k \pi}{n+1})} \\
\sum_{j=0}^{n} \cos (k \pi x_j) &= 1+ \frac{1}{2 \sin (\frac{k \pi}{n+1})} \sum_{j=1}^{n} {\sin (k \pi x_{j+1}) - \sin (k \pi x_{j-1})} \\
\end{align}
In the summation on th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to find the sum of this power series $\sum\limits_{n=0}^\infty \frac {x^{5n}} {(5n)!}$ How to prove that
$$
\sum\limits_{n=0}^\infty \frac {x^{5n}} {(5n)!}=
\frac{2}{5} e^{-\cos \left( 1/5\,\pi \right) x}\cos \left( \sin
\left( 1/5\,\pi \right) x \right) +\frac{2}{5}\, e^{\cos \left( 2/5\,
\pi \right) x}\cos \... | Hint. Clearly
$$
\frac{1}{5}\sum_{j=1}^5\mathrm{e}^{\omega^j x}=\sum_{n=0}^\infty\frac{x^{5n}}{(5n)!}
$$
where $\omega=\mathrm{e}^{2\pi i/5}$, since
$$
\sum_{j=1}^5 \omega^{jn}=\left\{\begin{array}{ccc} 5&\text{if}& 5\mid n, \\ 0&\text{if} &5\not\mid n. \end{array}\right.
$$
But
$$
\omega=\cos (2\pi/5)+i\sin (2\pi/5),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that a pair of irrational numbers is the solution to a quadratic polynomial. Suppose a, b are two irrational numbers such that ab is rational and a+b is rational. Then a, b are the solution to a quadratic polynomial with integer coeffecients.
| Suppose $a$, $b$ are two irrational numbers such that $ab$ is rational and $a+b$ is rational. Then $a$, $b$ are the solution to a quadratic polynomial with integer coefficients.
Proof:
Because $a+b$ and $ab$ are rational, from the definition
of a rational number we can write
$$a+b = \frac{m}{n}, \quad\mbox{and}\quad a... | {
"language": "en",
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"source": "stackexchange",
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Factoring a difference of 2 cubes I am trying to factorize the expression $(a - 2)^3 - (a + 1)^3$ and obviously I would want to put it in the form of $(a - b)(a^2 + ab + b^2)$
So I start off with the first $(a - b)$ and I get $(a - 2) - (a + 1)$ which I simplify from $(a^2 + a -2a -2)$ to $(a^2 -3a -2)$
Now I'm up to $... | Brute force algebra gives us
$$
(a-2)^3 - (a+1)^3
= (a^3 -6a^2 + 12a - 8) - (a^3 + 3a^2 + 3a +1)
= -9(a^2 - a + 1)
$$
But $a^2 - a + 1$ doesn't have any factors (not real ones, anyway).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/689671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all triples $(p; q; r)$ of primes such that $pq = r+ 1$ and $2(p ^ 2+q ^ 2) =r ^ 2 + 1$. We have to find all triples $(p; q; r)$ of primes such that $pq = r+ 1$ and $2(p ^ 2+q ^ 2) =r ^ 2 + 1$. This question was asked in the 2013 mumbai region RMO but i could not find a solution to it. Can you please help me out w... | First, we square the first equation. We have: $p^2q^2=r^2+2r+1$. If we plug $r^2+1$ we would have: $p^2q^2=2(p^2+q^2+r)$. Since $p$ and $q$ are prime numbers and the RHS is even, at least one of them should be $2$. Without loss of generality we take $p=2$. We have: $4q^2=2(4+q^2+r)$ so $q^2=r+4$. If we plug this into t... | {
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"url": "https://math.stackexchange.com/questions/689884",
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"source": "stackexchange",
"question_score": "5",
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How to solve simple trigonometry equation. So we are learning trigonometry in school and I would like to ask for a little help with these. I would really appreciate if somebody can explain me how I can solve such equations :)
*
*$\sin 3x \cdot \cos 3x = \sin 2x$
*$2( 1 + \sin^6 x + \cos^6 x ) - 3(\sin^4 x + \cos^4 ... | For the second:
$$
2\left(1+\sin^6 x+\cos^6 x\right)-3\left( \sin^4 x + \cos^4 x\right) - \cos x= 0 \\
2\left(1+\left( \sin^2 x + \cos^2 x \right) \left(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x\right)\right)-3\left( \sin^4 x + \cos^4 x\right) - \cos x = 0 \\
2\left(1+\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x\right)-3\le... | {
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"source": "stackexchange",
"question_score": "2",
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Please explain where the $2x$ came from in this cubic identity: The question is to factorize the difference of cube identity $(x + 1)^3 - y^3$
I obviously want to put it in the form of $(a - b)(a^2 + ab + b^2)$
My working out:
$(a - b) = (x + 1) - (y) = (x + 1 - y)$
$(a^2 + ab + b^2) = (x^2 + 1) + (x + 1)(y) + (y^2) = ... | Note that
$$a^2+ab+b^2={\mathbf{\color{red}{(x+1)^2}}}+(x+1)y+y^2$$
and $(x+1)^2\neq x^2+1$.
| {
"language": "en",
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Find the coefficients such that all four roots of $(x^2-px+q)(x^2-qx+p)$ are natural numbers
Find all ordered pairs $(p,q)$ of natural numbers such that all $4$ the roots of $$f(x)=(x^2-px+q)(x^2-qx+p)$$ are natural numbers.
I got a solution of the problem (see below) but I want some alternatives.
My solution
Let $... | Here's my take at the problem. Let $a\leq b$ be the roots of the first equation and $c\leq d$ of the second one. Vieta's formulas then tell us that
$$\begin{array}{ccccc}
a+b & = & p & = & cd \\
c+d & = & q & = & ab \\
\end{array}$$
Adding the two and moving everything to one side yields
$$ab + cd - (a+b) - (c+d) = 0$$... | {
"language": "en",
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"source": "stackexchange",
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Prove that any palindrome with an even number of digits is divisible by 11 Confusing myself here, need some clarification..
First, we consider the palindrome $abccba$. We can see this can be written as
$$a(10^5 + 10^0) + b(10^4 + 10^1) + c(10^3 + 10^2) = a(10^5 + 1) + 10b(10^3 + 1) + 100c(10 + 1)$$ So essentially we se... | Also, notice that in your inductive proof:
$10^{2k+3} + 1$ = $10^{(2k+2)+1}+1$ = $10^{2\overline{k} +1}+1$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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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... | Let our augmented matrix be $$A=\begin{pmatrix} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & k^2-14 & k+2 \end{pmatrix}$$ as you pointed out. So, if we do $$-3R_1+R_2\to R_2~(I),~~~ -4R_1+R_3\to R_3~(II),~~~R_2-R_3\to R_3~(III)$$ then we get $$A\to B=\begin{pmatrix} 1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & 0 & 16-k^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/697288",
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"source": "stackexchange",
"question_score": "2",
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What will be the solution of this equation? What will be the solution of the equation.
$(x^2+m^2)\frac{\partial^2y}{\partial(x^2+m^2)}+(x+m)\frac{\partial y}{\partial (x+m)}+(x^2+m^2-n^2)=0$ where $m$ may be a constant
| Hint:
$(x^2+m^2)\dfrac{\partial^2y}{\partial^2(x^2+m^2)}+(x+m)\dfrac{\partial y}{\partial(x+m)}+x^2+m^2-n^2=0$
$(x^2+m^2)\dfrac{\partial}{\partial(x^2+m^2)}\left(\dfrac{\partial y}{\partial(x^2+m^2)}\right)+(x+m)\dfrac{\partial y}{\partial x}+x^2+m^2-n^2=0$
$\dfrac{x^2+m^2}{2x}\dfrac{\partial}{\partial x}\left(\dfrac{1... | {
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EXERCISE VERIFICATION: Find where $f(x):=|x|+|x+1|$ is differentiable and calculate its derivative Could someone verify my exercise?
a) $f(x):=|x|+|x+1|$
First, analyse the roots of each absolute value, where they go to zero:
$$|x|:=\left\{\begin{matrix}
& x& x>0 \\
& x- & x<0\\
& 0 & x=0
\end{matrix}\right.$$
$... | Since $|x|=\sqrt{x^2}$, we have $|x|'=\dfrac{x}{|x|}$ for $x\neq0$, hence $f'(x)=\dfrac{x}{|x|}+\dfrac{x+1}{|x+1|}$ for $x\notin\{-1,0\}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Having problems using Hensel Lifting for Prime Power Moduli We have $P(x) = x^2 + 3x + 9 ≡ 0 \mod(3^i)$, for $i=1,2,3,4$
for $\mod(3): x^2 + 3x + 9 ≡ x^2$, so $x ≡ 0 \mod(3)$ is a solution.
From Hensel lifting, we have $s = x + a\times 7,\ a = [-P(0)\times (1/3)] \times (P'(0))^{-1}$.
Computing for $a$, we get $a = -3 ... | We go through the lifting process. The case $i=1$ has been dealt with. We now deal with $i=2$. Let $p=3$. Instead of $i$ we will use $k$.
We have $P'(x)=2x+3$. This is congruent to $0$ modulo $p$. In that case, the Hensel lifting goes as follows:
1) If we have a root $a$ modulo $p^k$, and $p^{k+1}$ divides $P(a)$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/700640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cousin of the Vandermonde binomial identity The Vandermonde binomial identity can be expressed as
\begin{align*}
\sum_{i+j=r} \binom{m}{i} \binom{n}{j} = \binom{m+n}{r} && r \leq m +n.
\end{align*}
While working on an algebra problem, I stumbled on a formally similar, but distinct identity:
\begin{align*}
\sum_{i+j=r} ... | It can be adjusted to be a convolution by shifting the upper and lower indices and then there must be a formal power series $f(x)$ for which the second identity
equates the $x^t$ terms in $f(x)^p f(x)^q = f(x)^{p+q}$,
just as the first identity compares the $x^r$ coefficients of $(1+x)^m (1+x)^n = (1+x)^{m+n}$.
T... | {
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"url": "https://math.stackexchange.com/questions/706190",
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"source": "stackexchange",
"question_score": "9",
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Prove that if $(u,v)$ is chosen randomly from $S$, there is at least a $50\%$ chance that $u\ne\pm v \bmod N$ I need serious help with this problem.
Suppose $N$ is an odd composite number and $S=\{(x,y) \in \mathbb Z^2 : x^2 \equiv y^2 \mod N\}$
*
*Prove that if $(u,v)$ is chosen randomly from $S$, there is at leas... | Let $n=15$, and fix $y$. We find the probability that if $0\le x\lt 15$, and $x^2\equiv y^2\pmod{15}$, then $x\equiv \pm y\pmod{15}$.
It is not hard to verify that if $y$ is relatively prime to $15$, then there are precisely $4$ values of $x$ such that $x^2\equiv y^2\pmod{15}$. Thus for such $y$ the probability is exac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show $\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$?
Show that $\,\displaystyle\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$.
I'm thinking right now (though not getting anywhere with it) that I want to expand out the summation portion to $i!/2!(i-2)!$ and simplify from there? Not sure if that will help, not to men... | Well, if $n-1 =2k$ $$\displaystyle\sum_{i=1}^{n} \binom{i}{2} = \displaystyle\sum_{i=1}^{n}\frac{i\cdot (i-1)}{2}=\frac{1\cdot 0}{2}+\frac{2\cdot 1}{2}+\frac{3\cdot 2}{2}+\frac{4\cdot 3}{2} + \ldots +\frac{n\cdot (n-1)}{2} = \frac{2\cdot 1+3\cdot 2+4\cdot 3+ \ldots +n\cdot (n-1)}{2} = \frac{2^2\cdot2+4^2\cdot2+6^2\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/707256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ according as $n$ is odd or even.
Since for any $n\ge3$,$2^{2^n}\equiv2^{2^{n-2}}\pmod9$, if $n=2k+3$
then
$$2^{2^n}=2^{2^{2k... | Note that $2^6 \equiv 1 \pmod 9$.
Also $2^n \equiv 2,4 \pmod 6$ depending on the parity of $n$. If $n$ is odd then $2^n \equiv 2 \pmod 6$, otherwise $2^n \equiv 4 \pmod 6$. This means that $2^n = 6k + 2$ or $2^n = 6k + 4$. Now just substitute and use the first conclusion.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Coin probability example I have a bag of 10 coins and 1 of them has heads on both sides you choose 1 of the coins from the bag at random and then flip it 100 times observing heads every time and then tell me what the probability you chose a fair coin is.
| TL;DR The answer is $\frac{9}{2^{100}+9}$
Explanation
The a priori probability that you got a fair coin is $\frac{9}{10}$. The probability of 100 heads in a row with a fair coin is $\frac{1}{2^{100}}$. So, the total probability that you got a fair coin and flipped 100 heads in a row is $\frac{9}{10\cdot2^{100}}$.
The a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$
where $C$ is Catalan's constant.
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\r... | Integrating by parts twice,
$$ \begin{align} \int_{0}^{1} (\arctan x)^{2} \ dx &= x (\arctan x)^{2} \Big|^{1}_{0} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \arctan(x) \ln(1+x^{2}) \Big|^{1}_{0} + \int_{0}^{1} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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How to divide polynomials? The title says it all : How to divide polynomials?
I don't understand the method below taught in my school.
Would any of you mind explaining it or even better, suggest an alternative way to solve this?
| Here is another method that you asked for. Its main advantage is that you can easily do this in a text editor, which helps immensely in accurately tracking your work. It is completely done with algebraic manipulation.
So, with $A$ as the highest term in numerator and $B$ as the sum of all remaining terms. And $C$ a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the integral : $\int\frac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$ Find the integral : $\int\dfrac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$
Please guide which substitution fits in this I am not getting any clue on this .. thanks..
| If $x=m^6$, then $dx=6m^5 dm$ and $x^{\frac{1}{2}}=(m^6)^{\frac{1}{2}}=m^3$ and $x^{\frac{1}{3}}=(m^6)^{\frac{1}{3}}=m^2$, then
$$\int \frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\int \frac{6m^5dn}{m^3+m^2}=6\int\frac{m^5dm}{m^2(m+1)}=6\int \frac{m^3dm}{m+1}. $$
Now, $$\frac{m^3}{m+1}=m^2-m+\frac{m}{m+1}, $$ then
$$\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/717902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\fr... | $$
\begin{aligned}
I&=\frac{1}{2} \int \cos 2 x \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)^{2} d x \\
&=\frac{1}{2} \int \cos 2 x \ln \left(\frac{1+\sin 2 x}{1-\sin 2 x}\right) d x \\
& =\frac{1}{4} \int \ln \left(\frac{1+y}{1-y}\right) d y, \text { where } y=\sin 2 x\\
&\stackrel{y=\sin 2x}{=} \frac{1}{4}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718719",
"timestamp": "2023-03-29T00:00:00",
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Need help with Proof by Strong Induction question So, here is the question:
For any position integer $n$, let $T(n)$ be the number 1 if $n<4$ and the number $T(n-1) + T(n-2) + T(n-3)$ if $n \geq 4$.
We have $T(1)=1, T(2)=2, T(3)=3$ $$T(4)=T(3)+T(2)+T(1) = 1+1+1+1 = 3$$
$$T(5) = T(4)+T(3)+T(2) = 3+1+1 = 5$$
Prove that:... | Suppose that for all $m\lt n$ we have $T(m)\lt 2^m$. We want to show T(n)\lt 2^n$.
This is certainly true up to $n=3$. Past that, we have $T(n)=T(n-1)+T(n-2)+T(n-3)$. So by the induction assumption we have
$$T(n-1)\lt 2^{n-1},\qquad T(n-2)\lt 2^{n-2},\qquad\text{and}\qquad T(n-3)\lt 2^{n-3}.$$
It follows that
$$T(n)\l... | {
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"timestamp": "2023-03-29T00:00:00",
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Question about the number of primes greater than $3$ in a sequence of consecutive integers. I recently noticed that for any $x > 16$, it follows that there are at least $2$ integers in the any sequence of 3 consecutive integers that are divisible by a prime greater than $3$.
For example, for $10,11,12$, we have $5|10$ ... | I believe that I've worked out the answer for a sequence of $n$ consecutive integers: $x+1, x+2, \cdots, x+n$
We can assume at least one integer $x+c$ where $c \le n$ is not divisible by a prime greater than $3$ so that $x+c = 2^m3^n$ where $m,n \ge 0$
The question is under what circumstances will $2^m3^n \pm d = 2^u3^... | {
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"timestamp": "2023-03-29T00:00:00",
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Does $7$ divide $2 x^2 - 4y^2$ for all $x,y$? Does $7$ divide $2 x^2 - 4y^2$ for all $x,y \in \mathbb{Z}$?
| ...suppose the result holds true, then,
$2x^2-4y^2=0 \pmod7$
since $2$ does not divide $7$, we have $x^2-2y^2=0 \pmod7$, but $2$ is congruent to $9$ modulo $7$, so, $x^2-9y^2=0 \pmod7$ or
$x=3y \pmod7$, so the result holds true $iff$ $x=3y+7k$, for some integer $k$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Is this proof by induction that $13 \mid 4^{2n+1}+3^{n+2}$ correct/sufficient? To show: $13\, |\, 4^{2n+1}+3^{n+2}$
I used induction beginning successfully with $n=0$ $($or $n=1),$
then making the step to $n+1:$
An $x$ exists so that $13x = 4^{2n+3}+3^{n+3}$
$$\begin{align*}13x &= 16\cdot4^{2n+1}+3\cdot3^{n+2}\\
\frac{... | Yes, it seems to be correct. However, the person who would read the proof might be confused a little. Instead, you can re-write it differently, like
$$
\frac{4^{2n+3} + 3^{n+3}}{13} = \frac{13 \cdot 4^{2n+1} + 3\left(4^{2n+1} + 3^{n+1} \right)}{13} = \frac{13\cdot 4^{2n+1} + 3 \cdot 13m}{13} \in \mathbb{Z}
$$
which is ... | {
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Combinatorics for integer solutions How many integer solutions are there to the following equation?
$x_1 + x_2 + x_3 = 17$
a) if $x_1 > 1, x_2 > 2, x_3 > 3$
b) if $x_1 < 6, x_3 > 5$ and $x_2$ can be any integer.
c) if $x_1 < 4, x_2 < 3, x_3 < 5$
What is a systematic way to approach and solve these types of problems?
| Use generating functions. I.e., the value of $x_1$ is represented by $z^2 + z^3 + \cdots = z^2 (1 - z)^{-1}$, and similarly the others. For all three variables it is in case (a):
$$
[z^{17}] z^2 \cdot z^3 \cdot z^4 \cdot (1 - z)^{-3} = (-1)^{17} \binom{-3}{17}
= [z^8] (1 - z)^{-3}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Irreducibility of $X^6+X^3+1$ in $\mathbb{Q}[X]$
Could anyone advise me on how to prove $X^6+X^3+1$ is irreducible in $\mathbb{Q}[X] \ ?$
I'm thinking of substituting $X=Y+1$ into the equation, do some tedious computations to simplify and use Eisenstein's criterion. May I know if that is the correct approach?
Thank y... | If you don't want to do the multiplication, write $X^9-1=(X^6+X^3+1)\cdot(X^3-1)$. Now use your $X=Y+1$ trick and reduce mod 3:
$$\begin{align}
(Y+1)^9-1&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot((Y+1)^3-1)\\
Y^9+1^9-1&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot(Y^3+1^3-1)\\
Y^9&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot Y^3\\
Y^6&=(Y+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $5\sin^2(x) + \sin(2x) - \cos^2(x) = 1$ I tried
$$5\sin^2(x) + 2\sin(x)\cos(x)- (1-\sin^2(x)) = 1.$$
Simplifying,
$$6\sin^2(x) + 2\sin(x)\cos(x) -2 = 0$$
Then I'm stuck!
| Using the twice-angle formulas, this becomes
$$ -3 \cos(2x) + \sin(2x) = -1 $$
Let $\alpha = \arctan(1/3)$ so $\cos(\alpha) = 3/\sqrt{10}$ and
$\sin(\alpha) = 1/\sqrt{10}$. The equation can be written as
$$ - \cos(\alpha) \cos(2x) + \sin(\alpha) \sin(2x) = - 1/\sqrt{10}$$
or $$\cos(\alpha + 2 x) = 1/\sqrt{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/727276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proof of convergence of $s_n = \frac{1}{2\cdot 1} + \frac{1}{3\cdot 2} + \cdots + \frac{1}{n(n+1)}$ I am really bad at this, but I am trying my best to understand.
If I am given a sequence of partial sums such that
$$s_n = \frac{1}{2\cdot1} + \frac{1}{3\cdot2} + \cdots + \frac{1}{n(n+1)}$$
Prove that $s_n \to 1$.
I a... | Can you show by induction that
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}. $$
Indeed, for $n=1$ is true because
$$\frac{1}{1\cdot 2}=\frac{1}{2}=\frac{1}{1+1}. $$
Now, suppose that the equality is true for $n=k$, i.e.,
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{k(k+1)}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/729724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How prove this sum $\sum_{n=1}^{\infty}\binom{2n}{n}\frac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}$ show that
$$\sum_{n=1}^{\infty}\binom{2n}{n}\dfrac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}=5+4\sqrt{2}\left(\log{\dfrac{2\sqrt{2}}{1+\sqrt{2}}}-1\right)$$
where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$
My try: we let
$$s(x)=\sum... | There's a transformation formula (might have been derived from generalized Euler series transformation):
$$
f(x)=\sum_{n=0}^\infty \, \binom{2n}{n}\, a_n \, x^n = \dfrac{1}{\sqrt{1+4\, x}}\, g\left(\dfrac{x}{1+4\, x}\right)
$$
where
$$a_n=\sum_{k=0}^n\, \binom{n}{k}\, (-1)^{n-k}\, b_k$$
and
$$g(x)=\sum_{n=0}^\infty \, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/730885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 1
} |
How find this $\frac{3x^3+125y^3}{x-y}$ minimum value
let $x>y>0$,and such $xy=1$, find follow minimum of the value
$$\dfrac{3x^3+125y^3}{x-y}$$
My idea: let $x=y+t,t>0$
then
$$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3(y+t)^3+125y^3}{t}=3t^2+3yt+3y^2+\dfrac{128y^3}{t}$$
and $$(y+t)y=1$$
I think this can use AM-GM inequalit... | $\dfrac{3x^3+125y^3}{x-y}=\dfrac{3x^6+125x^3y^3}{x^4-x^3y}=\dfrac{3x^6+125}{x^2(x^2-1)}=\dfrac{3(p+1)^3+125}{p(p+1)},p=x^2-1>0$
$\dfrac{3(p+1)^3+125}{p(p+1)}=3p+6+\dfrac{3}{p}+\dfrac{125}{p(p+1)}$
$3p$ is mono increasing function, $\dfrac{3}{p}+\dfrac{125}{p(p+1)}$ is mono decreasing function,so there must be only a mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/733770",
"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.