Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
showing $1 + z + z^2 + \dots $ uniformly converges to $\frac{1}{1-z}$ for $|z| < 1$ What test can I use to show that $1 + z + z^2 + \dots $ uniformly converges to $\frac{1}{1-z}$ for $|z| < 1$.
I know $\displaystyle 1 + z + z^2 + \dots +z^n = \frac{1-z^{n+1}}{1-z}$ and as $n \to \infty$, $1 + z + z^2 + \dots = \frac{1}... | Let denote
$$S_n(z)=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$$
the partial sum of the series so we have
$$\left|R_n(z)\right|=\left|S_n(z)-\sum_{k=0}^\infty z^k\right|=\frac{|z|^{n+1}}{|1-z|}$$
so it's clear that
$$\sup_{|z|<1}|R_n(z)|=+\infty$$
so $(S_n(z))$ does not converge uniformly to $\frac{1}{1-z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/374463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Condition for fourth degree polynomial to have all real roots For what range of values of $a$ will the following fourth degree polynomial have all real roots:
$$x^4 - 2ax^2 + x + a^2 -a = 0$$
| First transform this equation into the equation with unknown $a$:
$a^{2}-(2x^{2} + 1)a + x^{4} + x = 0$
Solving the equation with unknown obtain two second degree equations:
$x^{2} + x - a = 0 $ and $ x^{2} - x + 1 - a = 0$
For $a < - \frac{1}{4}$ equation has no real roots
For $a = - \frac{1}{4}$ equation has two equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculus Reduction Formula
For any integer $k > 0$, show the reduction formula
$$\int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx = C_k \int^{2}_{-2} x^{2k-2} \sqrt{4-x^2} \, dx$$
for some constant $C_{k}$.
(original image)
I thought this would be fairly straightforward but im a little confused. Do I start out by doing a ... | Let $I_k = \displaystyle \int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx$.
Let $u = x^{2k-1}$ and $dv = x\sqrt{4-x^2} \, dx$.
We then have $du = (2k-1)x^{2k-2} \, dx$ and
$\begin{align*}
v &= \int x\sqrt{4-x^2} \, dx\\
&=\displaystyle -\frac{1}{3} (4-x^2)^{3/2}\\
&=-\frac{1}{3}(4-x^2)\sqrt{4-x^2}\end{align*}$
Applying integra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2... | Take $f (x) = \frac {5} {2} \log x + \log 5 - x$. Obviously, $f$ is defined on the interval $]0, \infty[$. Since $f ' (x) = \frac {5} {2x} - 1$, it is easy to see that $f$ increases on $]0, \frac {5} {2}]$, take its maximum at $x = \frac {5} {2}$ and decreases on $[\frac {5} {2}, \infty[$. Also, since $f_{\max} = f (x_... | {
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"timestamp": "2023-03-29T00:00:00",
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How do you orthogonally diagonalize the matrix? How do you orthogonally diagonalize the matrix A?
Matrix A =
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{bmatrix}
$$
| Since the matrix $A$ is symmetric, we know that it can be orthogonally diagonalized. We first find its eigenvalues by solving the characteristic equation:
$$0=\det(A-\lambda I)=\begin{vmatrix}
1-\lambda & 1 & 1 \\
1 & 1-\lambda & 1 \\
1 & 1 & 1-\lambda
\end{vmatrix}=-(\lambda-3)\lambda^2 \implies \left\{\begin{array}{l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-... | If you consider $f(x)=2^x+2^{-x}$ and notice how it is symmetrical around $x=0$ and has its minimum there as for example $f(1)=2.5$ and $f(2)=4.25$, which the left hand side has a maximum value of 2, thus there is only one possible root though I'll leave that for you to see.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric Formula I am stuck with the simple expression
$$
\frac{\cos^2(\theta + \alpha)}{1 - \cos^2(\theta - \alpha)} = \text{const.}
$$
where $\theta$ is a variable and $\alpha$ is the number satisfying
$$
\alpha = \tan^{-1} (\frac{4}{3})\,.
$$
I cannot see it to be immediate, somehow I am missing a particular tr... | Note that $1 - \cos^2(\theta - \alpha) = \sin^2(\theta - \alpha)$
That gives you:
$$\frac{\cos^2(\theta + \alpha)}{1 - \cos^2(\theta - \alpha)} = \text{const.} = \frac{\cos^2(\theta + \alpha)}{\sin^2(\theta - \alpha)}$$
For the numerator: $\cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha.\tag{1}$
For the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/381520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$. Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$.
Some solutions I found are $f\equiv0,f\equiv1$, $f(x)=0$ if $x\neq0$ and $f(x)=1$ if $x=0$.
| Perform the substitution $x=\frac{a+b}{2}, y=\frac{a-b}{2}$ to get $f(\frac{a^2+b^2}{2})=f(a)f(b)$. We shall denote this statement with $P(a, b)$.
$P(a, a)$: $f(a^2)=f(a)^2$. In particular, when $a=0$, we have $f(0)=f(0)^2$ so $f(0)=0$ or $1$.
If $f(0)=0$, then $P(a, 0)$: $f(\frac{a^2}{2})=0$, so $f(x)=0 \, \forall x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/381724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})... | We show that $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle = \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$
Since, $$\frac{5+\sqrt{17}}{2}=2+\frac{1+\sqrt{17}}{2},$$ we have $$\frac{5+\sqrt{17}}{2}\in\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle,$$ thus $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/382188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$ Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$
I have tried to fiddle with it as follows:
$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividi... | Let us put $2^x=a,3^x=b$ to remove the indices to improve clarity
So, $8^x=(2^3)^x=(2^x)^3=a^3$ and similarly, $27^x=b^3$
$12^x=(2^2\cdot3)^x=(2^x)^2\cdot3^x=a^2b$ and similarly, $18^x=ab^2$
So, the problem reduces to $$\frac{a^3+b^3}{ab(a+b)}=\frac76$$
$\displaystyle\implies 6(a^2-ab+b^2)=7ab$ as $a+b\ne0$
$\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/384090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
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Why can I divide a fraction like this? Suppose I have a fraction: $$\frac{2^n}{2^{2n}+1}$$
I can simplify it to become: $$\frac{1}{2^{n}+\frac{1}{2^n}}$$
Now obviously, this is just dividing both the numerator and the denominator of the fraction by $2^n.$ My question is why I can do this. Can anyone explain the algebra... | One way of looking at it, is that you are multiplying the numerator and the denominator by $\frac{1}{2^n}$.
In the numerator you get $\frac{2^n}{2^n}$ which is $1$.
In the denominator you get:
$$\frac{1}{2^n}\cdot (2^{2n} + 1) = \frac{2^{2n}}{2^n} + \frac{1}{2^n} = 2^{2n-n} + \frac{1}{2^n} = 2^n + \frac{1}{2^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Finding $\lim_{x \to 0}\frac{\tan x-x}{x^3}$ Feeling like i did this wrong
$\displaystyle \lim_{x \to 0}\frac{\tan x-x}{x^3}$ $\to$ $\displaystyle \lim_{x \to 0}\frac{\sec^2x-1}{3x^2}$
$\displaystyle \lim_{x \to 0}\frac{2\tan x\sec^2x}{4x}$ $\to$ $\displaystyle \lim_{x \to 0}\frac{2\sec^2x\sec^2x+2\tan x(2\tan x\sec^... | The Pythagorean Identity is of help here: starting from your first application of l'Hopital's Rule, you could then write
$$\lim_{x \rightarrow 0} \ \frac{\sec^2 x \ - \ 1}{3 x^2} \ ^{*} \ = \ \lim_{x \rightarrow 0} \ \frac{(\tan^2 x \ + \ 1 ) \ - \ 1}{3 x^2} \ = \ \lim_{x \rightarrow 0} \ \frac{\tan^2 x }{3 x^2} $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (s... | We could use Carleman's Inequality which states that,
$$
\int_0^\infty f(x)dx \ge 1/e\int_0^\infty exp(1/x*\int_0^xln(f(x))dx)dx
$$
Then, by substituting our function as f(x), we can prove the statement by proving that
$$
1/e\int_0^\infty exp(1/x*\int_0^xln(f(x))dx)dx \gt 1
$$
This is done by noting that for all x >0
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Derivative of Trig Functions (Intuition Help?) Looking for some intuition help here.
I have the following exercise and these are the steps I take:
$$
y = \sin\left(\frac{1}{x}\right)
$$
$$
u=\frac{1}{x}
$$
$$
y = \sin u,\;\;\frac{dy}{du} = \cos u= \cos\left(\frac{1}{x}\right)
$$
$$
u=x^{-1};\;\frac{du}{dx} =-x^{-2}=-\f... | Here is where parentheses come in handy:
You found, correctly, $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$.
But the scope of $\cos$ is restricted to its argument: $\left(\dfrac 1x\right)$ ONLY:
The FUNCTION $\dfrac{dy}{du} = \cos\left(\dfrac 1x\right)$ is multiplied by the function $\dfrac{du}{dx} = -\dfrac 1{x^2}$. That is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute the remainder when $67!$ is divided by $71$. This is how far I've been able to get.
By using Wilson's Theorem:
$$\begin{align}
70! &\equiv -1 \pmod{71} \\
67!(68)(69)(70) &\equiv -1 \pmod{71} \\
67!(68)(69)(-1) &\equiv -1 \pmod{71} \\
67!(68)(69) &\equiv 1 \pmod{71} \\
\end{align}$$
EDIT: Here is how I proceede... | Recall that if
$$xy \equiv a \pmod{n}$$
then
$$x \equiv ay^{-1} \pmod{n}$$ if $y$ is invertible in $\pmod{n}$. In your case, $n$ is $71$ a prime, which guarantees that any $y \not \equiv 0 \pmod{71}$ is invertible in $\pmod{71}$. Hence,, we have
$$67! \times 68 \times 69 \equiv 1 \pmod{71} \implies 67! \equiv (69)^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$ I'm trying to solve this recurrence relation:
$$
a_n = \begin{cases}
0 & \mbox{for } n = 0 \\
5 & \mbox{for } n = 1 \\
6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
\end{cases}
$$
I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - ... | I did not check your work, so I’ll outline what you need to do to finish. You have something of the form
$$A(x)=\frac{a}{1-x}+\frac{b}{5-x}+\frac{c}{(1-x)^2}=\frac{a}{1-x}+\frac{b/5}{1-\frac{x}5}+\frac{c}{(1-x)^2}\;.$$
Expand these three terms into power series:
$$\begin{align*}
A(x)&=a\sum_{n\ge 0}x^n+\frac{b}5\sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390644",
"timestamp": "2023-03-29T00:00:00",
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A closed-form expression for the integral $\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx$ Is there a closed-form expression for this integral:
$$\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx,$$
where $\text{Ci}(x)=-\int_x^\infty\frac{\cos z}{z}\mathrm dz$ is the cosine integral?
$\text{Ci}(x)$ and $\text{Ci}^{2}(x)$ have ... | Expanding on Start wearing purple's answer, the following is an evaluation of $$\int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, dx .$$
First notice that by making the substitution $ u = \frac{t}{x}$, we get
$$ \operatorname{Ci}(x) = - \int_{x}^{\infty} \frac{\cos (t)}{t} \, dt = - \int_{1}^{\infty} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
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Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$ How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?
Note: Init... | I will be solving the first integral. If I can figure out the second integral I'll post my solution as another answer.
$$I=\int\bigg(\frac{\cos x}{1+\sin^2x}\bigg)^2\mathrm{d}x$$
$$I=\int\frac{\cos^2x}{(1+\sin^2x)^2}\mathrm{d}x$$
$$I=\int\frac{\sec^2 x\ \mathrm{d}x}{(2\tan^2x+1)^2}$$
$t=\tan x\Rightarrow \mathrm{d}t=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Solving two algebraic equations I'm trying to prove an identity from physics. I have the following two equations ($M$ is a constant):
$$e^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$ and $$r = \frac{l^2}{2M}\left[1 - \sqrt{1 - 12\left(\frac{M}{l}\right)^2}\right]$$ and I need to show that $$\frac{l}... | let $x=\dfrac{l^2}{M^2},a=\dfrac{2r}{M} \to a=x-\sqrt{x^2-12x} \to x^2-12x=(x-a)^2 \to x=\dfrac{a^2}{2a-12}$
$\sqrt{1-\dfrac{12}{x}}=\sqrt{\dfrac{a^2-12*2a+12*12}{a^2}}=\dfrac{|a-12|}{a}$
now you can simplify more easy.
ya,if 12 is wrong, you give a right number but the middle result is same.
But it seems the last one ... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the general equation of a cubic polynomial? I had this question:
"Find the cubic equation whose roots are the the squares of that of $x^3 + 2x + 1 = 0$"
and I kind of solved it. In that my answer was $x^3 - 4x^2 + 4x + 1$, but it was actually $x^3 + 4x^2 + 4x - 1 = 0$.
I took the general equation of a cubic equ... | Let $a$ be a root of $x^3+2x+1=0$ and $b=a^2$ be a root of the required equation
So, $a^3+2a+1=0\implies a\cdot b+2a+1=0\implies a=-\frac1{b+2}$
As $a$ be a root of $x^3+2x+1=0$, put this value of $a$ in $x^3+2x+1=0$
On simplification, I get $(b+2)^3-2(b+2)^2-1=0\iff b^3+4b^2+4b-1=0$
So, the required equation will b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393312",
"timestamp": "2023-03-29T00:00:00",
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Using the hypothesis $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ to prove something else Assuming that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$
Is it possible to use this fact to prove something like:
$$\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}+b^{2013}+c^{2013}... | A simpler way to see that one of $a = -b$ or $b = -c$ or $c = -a$ is true.
wlog, we can assume $abc = 1$ (why?).
So we can assume $a,b,c$ are roots of $x^3 - px^2 + qx -1 = 0$.
Your relation gives us that $q = \frac{1}{p}$ (Vieta's formulas)
Thus we get $a,b,c$ are roots of
$$x^2(x - p) + \frac{1}{p}(x - p) = 0$$
Thus ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How can I find all the solutions of $\sin^5x+\cos^3x=1$
Find all the solutions of $$\sin^5x+\cos^3x=1$$
Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.
| $\sin^5x\le \sin^2x, \cos^3x\le \cos^2x$, and $\sin^2x+\cos^2x=1$, so $\sin^5x+\cos^3x$
can be equal to 1 if and only if $$\sin^5x=\sin^2x$$ and $$\cos^3x=\cos^2x$$, i.e.
$$\sin^2x(1-\sin^3x)=0$$ and $$\cos^2x(1-\cos x)=0$$
The first equation has solution: $x=k\pi$ or $x=\pi/2+2m\pi, k,m $ integers.
The second equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Finding the number of non-neg integer solutions? How would I find the number of non negative integer solutions to this problem?
$$x_1 + x_2 + x_3 + x_4 = 12$$ if $0 \leq x_1 \leq 2$.
| $$x_1 + x_2 + x_3 + x_4 = 12, 0 \leq x_1 \leq 2$$
According to http://oeis.org/wiki/User:Adi_Dani_/Restricted_compositions_of_natural_numbers
there is derived formula
$${\binom{m}{k}}_{s}=\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}\binom{m+k-si-1}{m-1}\,$$
for number of solutions of $$x_0+x_1 + x_2 +...+ x_{m-1} =k,0\leq x_i\l... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd
If $a,b \in \mathbb{Z}$ and odd, show $8 \mid (a^2-b^2)$.
Let $a=2k+1$ and $b=2j+1$. I tried to get $8\mid (a^2-b^2)$ in to some equivalent form involving congruences and I started with
$$a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}$$
$$\Righta... | The given condition says that $a,b$ both are odd. So let, $a=2k+1$ and $b=2l+1$ where $k,l$ are some integers.
Now, $a^2-b^2=4k^2+4k+1-(4l^2+4l^2+1)=4(k^2-l^2+k-l)$
$$=4((k^2-k)(l^2-l))=4(k(k-1)+l(l-1)$$.
Notice that $k(k-1),l(l-1)$ are products of two consecutive integers and hence divisible by $2$. So, you can writ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/397830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$
If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.
If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinkin... | Continue as you started: As $3 \nmid a$, we have $a \equiv \pm 1 \pmod 3$, giving
$$ a^4 - 1 \equiv 1 - 1 = 0 \pmod 3 $$
which means $3 \mid a^4 + 1$.
For $5$, we have either $a \equiv \pm 1 \pmod 5$, giving
$$ a^4 - 1 \equiv 1 - 1 \equiv 0 \pmod 5 $$
or $a \equiv \pm 2 \pmod 5$,
$$ a^4 - 1 \equiv 16 - 1 = 15 \equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of volumes of revolution: bisector surface
$y = (16-x^2)^{0.5}$ is rotated around the $x$-axis to give a sphere of radius $4$ units. Find the equation of the straight line that passes through $(-4,0)$, such that when also rotated around the $x$-axis, will create a surface that bisects the volume of the sph... | The volume of your sphere is $\,\displaystyle{\frac43\pi 4^3=\frac{256\pi}{3}}\,$ . You need a line $\,y=kx+4k\,$ , which intersects $\,y=\pm\sqrt{16-x^2}\,$ at
$$16-x^2=k^2x^2+8k^2x+16k^2\implies(k^2+1)x^2+8k^2x+16(k^2-1)=0\implies$$
$$\Delta=64k^4-64(k^4-1)=8^2\implies$$
$$x_{1,2}=\frac{-8k^2\pm8}{2(k^2+1)}=\begin{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the general solution for $2\cos 2x−5\cos x+2=0$ Determine the general solution for $2\cos 2x−5\cos x +2=0$
my answer I got was : $1.05+n\pi, 4.19+n\pi$
| First, notice that $\cos(2x) \equiv \cos^2x-\sin^2x$ and so
$$2\cos(2x)-5\cos x + 2 \equiv 2\cos^2x - 2\sin^2x - 5\cos x + 2$$
Then we know that $\sin^2x = 1-\cos^2x$, meaning that
$$2\cos(2x)-5\cos x + 2 \equiv 4\cos^2x - 5\cos x$$
We can factorise to give $(4\cos x - 5)\cos x$. The solutions are then just $\cos x = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Help finding eigenvectors? The given matrix is:
$$
\begin{pmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{pmatrix}\qquad
$$
I got the characteristic polynomial of $$x^3 - x^2 - 5x - 3 = 0$$
which factors down to $$(x+1)^2 * (x-3) = 0$$
I see that it has eigenvalues of -1 and 3.
I know I'm almost there, I plugged in ... | it is enough you find:
$$\left(\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}- 3I\right)\begin{bmatrix}x
_1 \\ x_2 \\ x_3\end{bmatrix}=0\qquad
\text{and} \qquad
\left(\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}+ I\right)\begin{bmatrix}x
_1 \\ x_2 \\ x_3\end{bmatrix}=0 \ .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Compute $\int_{0}^{1}\left[\frac{2}{x}\right]-2\left[\frac{1}{x}\right]dx$ The question is to find
$$\int_{0}^{1}\left(\left[\dfrac{2}{x}\right]-2\left[\dfrac{1}{x}\right]\right)dx,$$
where $[x]$ is the largest integer no greater than $x$, such as $[2.1]=2, \;[2.7]=2,\; [-0.1]=-1.$
Is there any nice method to solve thi... | You just have to figure out where the integrand is nonzero. In this case, the integrand really takes the value
$$\left\lfloor \frac{2}{x} \right\rfloor - 2\left \lfloor \frac{1}{x} \right\rfloor = \begin{cases} \\0 & x \in \left [ \frac{2}{2 n+1},\frac{2}{2 n}\right )\\ 1 & x \in \left [ \frac{2}{2 n},\frac{2}{2 n-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 0
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Determine if vectors are linearly independent Determine if the following set of vectors is linearly independent:
$$\left[\begin{array}{r}2\\2\\0\end{array}\right],\left[\begin{array}{r}1\\-1\\1\end{array}\right],\left[\begin{array}{r}4\\2\\-2\end{array}\right]$$
I've done the following system of equations, and I think ... | You just stopped too early:
Since you have 3 varibles with 3 equations, you can simply obtain $a,b,c$ by substituting $c = 0$ back into the two equations:
*
*From equation $(3)$, $c = 0 \implies b = 0$.
*With $b = 0, c = 0$ substituted into equation $(1)$ or $(2)$, $b = c = 0 \implies a = 0$.
So in the end, sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
"answer_count": 4,
"answer_id": 1
} |
Solve the following in non-negative integers: $3^x-y^3=1$. Solve the following in non-negative integers: $$3^x-y^3=1$$
Of course $(x,y)=(0,0)$ is a trivial solution. After seeing that I proceeded like this:
$$3^x-y^3=1$$$$\implies3^x-1=y^3$$$$\implies2(3^{x-1}+3^{x-2}+ \cdots +3^1+1)=y^3$$$$\therefore2|y$$
So let $y=2k... | HINT:
If $d$ divides $y+1,y^2-y+1$ it will divide $y^2-y+1-y(y+1)=1-2y$
As $d$ divides $y+1,1-2y$ it will divide $2(y+1)+1-2y=3$
If $d=1,$ either $y+1=1$ or $y^2-y+1=1$ or both $=1$
If $d=3,$ either $y+1=3$ or $y^2-y+1=3$ or both $=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412681",
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"source": "stackexchange",
"question_score": "4",
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Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac... | We can attack the other integral
$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2}$$
in a manner similar to what @O.L. outlined in his answer for the other case, but with a different contour. To wit, consider
$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{(1+z^2)^2}$$
where $C$ is the following contour
This is a ke... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 6,
"answer_id": 4
} |
Limit $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}+a_n\right)^n=e$ if $\lim\limits_{n\to\infty}na_n=0$
Let $\{a_n\}$ be any sequence of real numbers such that $\lim_{n\rightarrow\infty}na_n=0$. Prove that $$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}+a_n\right)^n=e$$
I thought about using binomial theorem. So $\left... | Here is a technique
$$\left(1+\frac{1}{n}+a_n\right)^n= e^{n\ln\left(1+\frac{1}{n}+a_n\right)}= e^{n\left ( (\frac{1}{n}+a_n) -\frac{1}{2}(\frac{1}{n}+a_n)^2+\dots. \right) }=e^{\left( (1+na_n) -\frac{n}{2}(\frac{1}{n}+a_n)^2+\dots.\right)} $$
$$ = e^{\left( (1+na_n) -\frac{n}{2}\frac{(1+na_n)^2}{n^2}+\dots.\right)} .$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't h... | Notice that $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}} = \frac{\frac{-x}{1-2x}}{\frac{-1}{1-2x}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Prove that $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ converges when $n \to \infty$ I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.
By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.
It seem... | $
\text {Assumption: } \big(x_n\big)_{n\ge2} \text { is strictly monotonically decreasing} \tag{1} \\
$
$
\text {Assumption: } \sqrt2 \le \big(x_n\big)_{n\ge2} \tag{2}
$
Strictly monotonic & bounded $\Longrightarrow$ convergent.
$$\begin{align*}
\text {ad (1):} && x_{n+1} &< x_n \\
&& \frac{x_n}2+\frac 1{x_n} &< x_n \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/416274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Local maxima of Legendre polynomials When I plotted the (normalized) Legendre polynoials, I couldn't help noticing that all the local maxima lay on a really nice curve:
What is the equation of the curve (and how can we arrive to that equation)?
| I claim that that curve is
$$y=\pm \frac{\sqrt{2/\pi}}{ \sqrt[4]{1-x^2}}.$$
This argument will not be rigorous, and will cite a source I haven't fully understood.
Take a look at Whittaker and Watson, A course in Modern Analysis, p. 316. They write:
$$P_n(\cos \theta) = \frac{4}{\pi} \frac{2 \cdot 4 \cdots (2n)}{3\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
"answer_id": 1
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Determining a basis for solution set
For a given equation system (over field $R$), determine the basis for
the space $\operatorname{Sol}(U)$ (where $\operatorname{Sol}$ is a set of solutions):
\begin{cases} 2x + 7y + 3z+t=0 \\ 3x + 5y + 2z + 2t=0 \\ 9x + 4y + z+7t=0 \end{cases}
It's a new type of problem for me and... | Put the system into an augmented matrix:
$$\left[\begin{array}{rrrr|r}
2 & 7 & 3 & 1 & 0 \\
3 & 5 & 2 & 2 & 0 \\
9 & 4 & 1 & 7 & 0
\end{array}\right]$$
Use Gauss-Jordan elimination to get it into reduced row echelon form.
$$\left[\begin{array}{rrrr|r}
1 & 0 & -\frac{1}{11} & \frac{9}{11} & 0 \\
0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
Find $x,y,z \in \mathbb Q$ such that: $$x + \frac 1y, y + \frac 1z, z+
\frac 1x \in \mathbb Z$$
Here is my thinking:
$$x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z\\ \implies \left ( x + \frac 1y\right ) \left ( y +... | It is most interesting to consider $x>0,y>0,z>0$, because there is a lots of solutions in other case.
A.
If $(x,y,z)$ is a solution, then one can construct other solutions:
$$\begin{array}{ccc}
(x,y,z) & (y,z,x) & (z,x,y) \\
(-x,-y,-z) & (-y,-z-,x) & (-z,-x,-y) \\
(\frac{1}{z},\frac{1}{y},\frac{1}{x}) & (\frac{1}{x},\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
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How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
Let
$$\begin{align*}x &= \frac{1}{1} + \fr... | Your proof is not right since $x=\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/420047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Show that $\frac 1 2 <\frac{ab+bc+ca}{a^2+b^2+c^2} \le 1$
If $a,b,c$ are sides of a triangle, then show that $$\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2} \le 1$$
Trial: $$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0\\\implies a^2+b^2+c^2 \ge ab+bc+ca $$ But how I prove $\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2}$ . Please help.
| We know from triangle inequality
In a triangle the sum of any two sides is greater than the third side, from which it follows that
$(a+b-c)(a+c-b)+(b+c-a)(b+a-c)+(c+a-b)(c+b-a)>0$
Just expand this to get the inequality
$\Rightarrow a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ac+c^2-a^2-b^2+2ab>0$
$\Rightarrow -a^2-b^2-c^2+2(ab+bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/424367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Given that $ x + \frac 1 x = r $ what is the value of: $ x^3 + \frac 1 {x^2}$ in terms of $r$? Given that $$ x + \cfrac 1 x = r $$
what is the value of: $$ x^3 + \cfrac 1 {x^2}$$ in terms of $r$?
NOTE: it is $\cfrac 1 {x^2}$ and not $ \cfrac 1 {x^3} $
Where I reached so far:
$$ \Big(x^3 + \cfrac 1 {x^2}\Big) + \cfrac 1... | If $g$ is a function such that $g(x)=f(x+\frac{1}{x})$ then $g(1/x)=g(x)$.
Your $g(x)=x^3+x^{-2}$ therefore cannot be written as $f(x+1/x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to derive compositions of trigonometric and inverse trigonometric functions? To prove:
$$\begin{align}
\sin({\arccos{x}})&=\sqrt{1-x^2}\\
\cos{\arcsin{x}}&=\sqrt{1-x^2}\\
\sin{\arctan{x}}&=\frac{x}{\sqrt{1+x^2}}\\
\cos{\arctan{x}}&=\frac{1}{\sqrt{1+x^2}}\\
\tan{\arcsin{x}}&=\frac{x}{\sqrt{1-x^2}}\\
\tan{\arccos{x}}... | Here is astart
$$ \sin({\arccos{x}})= \sqrt{1-\cos(\arccos(x))^2}=\sqrt{1-x^2}, $$
since $\cos(\arccos(x))=x.$ You need to use some trig. identities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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CLT for random variables with varying distributions The following is not a homework problem. Here it is:
Suppose $\{X_k\}_{k=1}^{\infty}$ is a sequence of independent random variables such that
$$
P(X_k=1)=P(X_k=-1)=\frac{1}{2}-\frac{1}{2k^2}
$$
$$
P(X_k=k)=P(X_k=-k)=\frac{1}{2k^2}.
$$
Find the limiting distribution of... | Suppose $\{X_k\}_{k=1}^{\infty}$ is a sequence of independent random
variables such that $$ P(X_k=1)=P(X_k=-1)=\frac{1}{2}-\frac{1}{2k^2}
$$
$$ P(X_k=k)=P(X_k=-k)=\frac{1}{2k^2}. $$ Find the limiting distribution of $\frac{1}{\sqrt{n}} \sum\limits_{k=1}^n X_k$.
The characteristic function of $X_k$ is
$$\phi_k(\xi)=(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/429627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How does one [easily] calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$? How does one [easily] calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$, where $\mathrm{pop}(n)$ counts the number of bits '1' in the binary representation of $n$?
Is there any trick to calculate the sum? From wha... | There is one more way to calculate this expression. Consider the following recurrence: $a_0 = 0, a_1 = 1, a_2 = 1$ and $$a_{2n} = a_n, \ a_{2n+1} = a_n + 1, \ n \ge 1.$$
It is easy to see that $a_n$ counts the number of ones in the binary expansion of $n$. We wish to calculate
\begin{align*}
\sum_{n = 1}^{\infty} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 1
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Non-homogenous recurrence relation I need to solve the recurrence relation $A(n)=2A(n-2)+ 2^{n-2}$. I tried writing out equations up to the $A(2)$ and multiplying by powers of two and adding all the equations together then all the terms cancelled but after that I couldn't find the sum of the powers of two. I forgot t... | It never hurts in such problems to start by gathering some numerical data:
$$\begin{array}{rcc}
n:&2&4&6&8&10&12\\
A(n):&2&8&32&128&512&2048\\
A(n):&2^1&2^3&2^5&2^7&2^9&2^{11}
\end{array}$$
There’s a very obvious pattern here, that leads to the conjecture that $A(n)=2^{\text{what function of }n?}$ if $n$ is even. Compl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/433964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Trig substitution $\int x^3 \sqrt{1-x^2} dx$ $$\int x^3 \sqrt{1-x^2} dx$$
$x = \sin \theta $
$dx = \cos \theta d \theta$
$$\int \sin^3 \theta d \theta$$
$$\int (1 - \cos^2 \theta) \sin \theta d \theta$$
$u = \cos \theta$
$du = -\sin\theta d \theta$
$$-\int u^2 du$$
$$\frac{-u^3}{3} $$
$$\frac{\cos^3 \theta}{3}$$
Wi... | You can avoid all the trig by making a much simpler substitution:$$u^2=1-x^2$$so: $$x^2=1-u^2$$ $$ u= \sqrt{1-x^2}$$ $$2u du=-2xdx$$
Rewriting the integral, factoring out one $x$:$$\int x^3 \sqrt{1-x^2} dx=\int x^2 \sqrt{1-x^2} xdx=-\int (1-u^2) u^2 du$$Multiply out the integrand, integrate with the power formula term ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Find the derivative of $y = f(x^2 - 2x + 7)$ where $f'(10) = 2$ Determine the derivative if $y = f(x^2 - 2x + 7)$ and $f'(10) = 2$
Ok so honestly, I dont know how to solve this, or even know where to start. All i know is that we are given a point $(10, 2)$. But what is $f(x^2 - 2x +7)$ supposed to mean?
| Suppose $y=f(z)$ = well $f$ can be anything. Then we put $z=x^2-2x+7$ and get a new function $g(x)=f(x^2-2x+7)$. Some examples:
$$f(z)=z; f(x^2-2x+7)=x^2-2x+7=g(x)$$
$$f(z)=z^2+3; f(x^2-2x+7)=(x^2-2x+7)^2+3=g(x)$$
$$f(z)=\sin (z); f(x^2-2x+7)=\sin (x^2-2x+7)=g(x)$$
Now we want to see what we know about the derivative o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ Find all real numbers such that
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier metho... | Hints:
$$x=\sqrt{x-\frac1x}+\sqrt{1-\frac1x}\implies x\sqrt x=\sqrt{x^2-1}+\sqrt{x-1}\implies$$
$$ x^3=x^2+x-2+2\sqrt{x^3-x^2-x+1}\implies(x^3-x^2-x+2)^2=4(x^3-x^2-x+1)$$
Well, now you can put $\,w:=x^3-x^2-x+1\;$ and solve
$$(w+1)^2=4w\iff (w-1)^2=0\iff w=1\;\ldots\;\text{and etc.}\ldots$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Express $f(x) = x^2 \cos(2x) + \sin^2(x)$ as a power series
Express $f(x) = x^2 \cos(2x) + \sin^2(x)$ as a power series
What I know:
I know that $$x^2\cos(2x) = x^2 \cdot \sum_{n=0}^{\infty} {(-1)^n \cdot \frac{(2x)^{2n}}{(2n)!}} = \sum_{n=0}^{\infty} {(-1)^n \cdot \frac{2^{2n} \cdot x^{2n+2}}{(2n)!}}$$
But how do ... | HINT
$$ \sin^2 {x} = 1 - \cos^2 {x} = \dfrac{1 - \cos(2x)}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this $\frac{1}{\cos{A}}-\frac{\sin{\frac{A}{2}}}{\sin{\frac{B}{2}}\sin{\frac{C}{2}}}=4$ in $\Delta ABC $ not An equilateral triangle, let
$$\begin{vmatrix}
2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\
4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{B}&4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\si... | I'm getting the result without using the formula.
Note that, since $\Delta ABC$ is a triangle, $$2\sin A\sin C-\cos B=\cos (A-C)-\cos(A+C)-\cos C=\cos(A-C)$$
Similarly, $$2\sin A\sin B-\cos C=\cos (A-B)$$
Now, observe that if $B=C$, the determinant condition is automatically satisfied and in that case $$\large \cos A=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$ Problem :
How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$
I am unable to find out how to proceed in this problem.. this is a problem of arithmetic progress... | HINT:
As $1+r^2+r^4=(1+r^2)^2-(r)^2=(1+r+r^2)(1-r+r^2)$
and $(1+r+r^2)-(1-r+r^2)=2r,$
$$\frac r{1+r^2+r^4}$$
$$=\frac12\cdot\frac{2r}{(1+r+r^2)(1-r+r^2)}$$
$$=\frac12\cdot\frac{(1+r+r^2)-(1-r+r^2)}{(1+r+r^2)(1-r+r^2)}$$
$$=\frac12\left(\frac1{1-r+r^2}-\frac1{1+r+r^2}\right)$$
Put the values of $r=1,2,\cdots.. n-1,n$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The pr... | In the expression $\frac{3}{\frac{\sqrt{5}}{5}}$, to find the reciprocal of that expression's denominator, $\frac{\sqrt{5}}{5}$, simply swap its numerator and denominator.
Thus the reciprocal of $\frac{\sqrt{5}}{5}$ is $\frac{5}{\sqrt{5}} $, and $$ \frac{3}{\frac{\sqrt{5}}{5}} = 3 \cdot \frac{5}{\sqrt{5}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to evaluate the integral $\oint_{C}\frac{z + 2}{z^2(z^2-1)}dz$ I've been trying to solve the following complex-analysis problem:
Evaluate
$\oint_{C}\frac{z + 2}{z^2(z^2-1)}dz$
Where $C = \{z : |z + 1| = \frac{3}{2}\}$
What I've tried to do was the following: I detected that there are singularities at $z = -1$ and ... | You should expand $$\frac{z+2}{z^{2}(z^{2}-1)}$$ as $$\frac{1}{z(z^{2}-1)}+2(\frac{1}{z^{2}-1}-\frac{1}{z^{2}})=\frac{1}{2z}(\frac{1}{z-1}-\frac{1}{z+1})+(\frac{1}{z-1}-\frac{1}{z+1})-\frac{2}{z^{2}}$$where the first part can be further expanded as $$\frac{1}{2}(\frac{1}{z-1}-\frac{1}{z}-(\frac{1}{z}-\frac{1}{z+1}))=\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many squares are there modulo a Mersenne prime? Mersenne primes are primes of the form $M_n = 2^n - 1$. I'm wondering how many distinct natural numbers result from squaring the naturals modulo $M_n$.
As an example, $M_3 = 7$. If we take the naturals less than seven, we get:
$$1^2 \equiv 1 \bmod 7$$
$$2^2 \equiv 4... | For any odd prime $p$, there are $\frac{p-1}{2}+1=\frac{p+1}{2}$ squares modulo $p$.
To show this, note first that $0$ is a square modulo $p$. It is, modulo $p$, $0^2$, or, if you prefer, it is congruent to $p^2$. (But it is not called a *quadratic residue of $p$.)
Now consider the squares of numbers in the interval $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Deriving the generating function of a divide and conquer type recurrence relation I am working through Analysis of Algorithms by Sedgewick/Flajolet
On problem 3.44 I am given the recurrence, and I need to come up with a generating function. I have tried the various methods in the section (it isn't linear, I don't see ... | I think the first equation in the problem statement should be
$f_{2n}=f_{2n-1}+f_n $ with $n \ge 1$ and $f_0=0$.
Otherwise, as noted by RGB, there is no way to find $f_2$.
Let $F(x) = \sum_{n=0}^{\infty}f_{n} x^{n}$.
Note that we have
$\sum_{n=0}^{\infty}f_{2n} x^{2n}= \sum_{n=0}^{\infty} f_{2n-1} x^{2n} + \sum_{n=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sqrt 2 + \sqrt 3$ is irrational I have proved in earlier exercises of this book that $\sqrt 2$ and $\sqrt 3$ are irrational. Then, the sum of two irrational numbers is an irrational number. Thus, $\sqrt 2 + \sqrt 3$ is irrational. My first question is, is this reasoning correct?
Secondly, the book wants me... | Note that $\sqrt{2}+\sqrt{3}$ is a solution to the equation: $$x^4-10x^2+1=0$$Does this polynomial have any rational roots?
Edit: To find this polynomial, note that if $x=\sqrt{2}+\sqrt{3}$, then: $$x^2=5+2\sqrt{6}$$and: $$x^4=49+20\sqrt{6}.$$You need $-10x^2$'s to get rid of the $20\sqrt{6}$ in $x^4$, and $x^4-10x^2=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 9,
"answer_id": 6
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Hypergeometric value Is their a closed form for the following
$${}_2F_1 \left(a,b;c;\frac{1}{2} \right)$$
I would use the following
$${}_2F_1 \left(a,b;c;x \right)= \frac{\Gamma(c)}{\Gamma(c-b)\Gamma(b)} \int^1_0 t^{b-1}(1-t)^{c-b-1} (1-xt)^{-a} \, dt $$
But it wasn't a success !
Edit: Corrected integral representat... | There are closed forms for the 2F1 but their exact form depend on the values and the relations between the parameters.
*
*The trivial case is if $a$ or $b$ is a negative integer: the function becomes a finite sum of rational expressions in the parameters. if you choose $a=-4$ and if you name $d$ the value you have f... | {
"language": "en",
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Partial Fractions I here would like to clear my doubt on the question below:
$$\frac{1}{x(x-1)(x-2)}\;,$$
that is, we want to bring it into the form:
$$\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}\;,$$
in which the unknown parameters are $A,B$, and $C$. Multiplying these formulas by $x(x − 1)(x − 2)$ turns both into polynom... | Examine the form $$A(x−1)(x−2)+Bx(x−2)+Cx(x−1)=1$$
This is supposed to be true for every value of $x$. If we try $x=0$ we get $2A=1$, $x=1$ gives $=B=1$, and $x=2$ gives $2C=1$.
This short-cut method (known as the "cover-up" rule) can be used to find the numerators for partial fractions where all the denominators are l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The values of $N$ for which $N(N-101)$ is a perfect square
For how many values of $N$ (integer), $N(N-101)$ is a perfect square number?
I started in this way.
Let $N(N-101)=a^2$
or $N^2-101N-a^2=0.$
Now if the discriminant of this equation is a perfect square then we can solve $N$.
But I can't progress further.... | We want $N(N-101)$ to be a perfect square. Note that the gcd of $N$ and $N-101$ is $1$ or $101$.
Suppose it is $101$. Let $N=101M$. We get that $101M(101M-101)$ is a perfect square. This is the case iff $M(M-1)$ is a perfect square. That happens iff $4M^2-4M$ is a perfect square.
But $4M^2-4M$ is a perfect square iff $... | {
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"source": "stackexchange",
"question_score": "5",
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Initial value problem differential equation $y' = (x-1)(y-2)$ $$y' = (x-1)(y-2)$$
$y(2)= 4$
$$\frac{1}{y-2}dy = (x-1)dx$$
$$\int \frac{1}{y-2}dy =\int (x-1)dx$$
$$\ln(y-2) = \frac{x^2}{2} - x + c$$
$$y - 2 = e^{\frac{x^2}{2} - x + c} $$
$$y = e^{\frac{x^2}{2} - x + c} + 2$$
plug in the inital value
$$y = e^{\frac{2^2}{... | For $x=2$ we have $y=4$ so $e^c=4-2=2$ and finally
$$y = 2e^{\frac{x^2}{2} - x } + 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/460608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of Solutions to Diophantine Equation $(a)$ Let $c < 2\pi$ be a positive real number. Show that
there are infinitely many integers $n$ such that the equation
$x^2 + y^2 + z^2 = n$
has at least $c\sqrt n$ integer solutions.
$(b)$ Find a constant $C > 0$ such that there are infinitely many $n$ for
which the equation
... | HINT:Consider $$F(t):=\sum_{n\le t} \#\{x^a+y^b+z^c=n\} $$ and find its asymptotic formula.
Take your problem (a) as an example.
$$\sum_{n\le t} \#\{x^2+y^2+z^2=n\}=\sum_{x,y,z\in\mathbb{Z}}\mathbb{1}_{\{x^2+y^2+z^2\le t\}}=Area\ of\{x^2+y^2+z^2\le t\}+o(Area\ of\{x^2+y^2+z^2\le t\})=\frac{4}{3}\pi t^\frac{3}{2}+o(t^\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differential of normal distribution Let
$$f(x)=\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}$$
(Normal distribution curve) Where $\sigma$ is constant. Is my derivative correct and can it be simplified further?
$$\begin{align} f'(x)
&=\frac d{dx}\left(\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\... | If you want to use the method that you started, then you should note that we can only "pull out the negative sign" from $$\left(-\frac{x^2}{2\sigma^2}\right)^n$$ when $n$ is odd. It would be better to keep a $(-1)^n$ term around, because then, $$\begin{align}\frac d{dx}\exp\left(-\frac{x^2}{2\sigma^2}\right) &= \sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A tricky integral (flux of a point charge through a disk) The integrals:
$$
\oint \frac{r\,dr\,d\phi}{\left(L^2+r^2+h^2+2Lr\cos\phi\right)^{3/2}}\\
\oint \frac{dx\,dy}{\left((L+x)^2+y^2+h^2\right)^{3/2}}
$$
If we have a point charge at the origin and we want to find the flux through a disk of radius $R$ which is locate... | The solid angle can be expressed in terms of complete elliptic integrals of first and third kind.
I believe this is first derived by F. Paxton in 1959.
THE REVIEW OF SCIENTIFIC INSTRUMENTS. VOLUME 30, NUMBER 4 APRIL, 1959.
Solid Angle Calculation for a Circular Disk. F. PAXTON
A copy of the article can be found
he... | {
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If $n$ is an even perfect number $ n> 6$ show that the sum of its digits is $\equiv 1 (\bmod 9)$ If $n$ is an even perfect number $ n> 6$ show that the sum of its digits is $\equiv 1 \mod 9$.
I know perfect numbers are of the form $(2^{p-1})(2^{p}-1)$. I have a few trials that I have done and they check out but as far ... | Note that if $\mathfrak{d}(n)$ is digit sum of $n$, then $$\mathfrak{d}(n) \equiv n \pmod{9}.$$ Hence we wish to show that the perfect number $n$ is congruent to $1$ modulo $9$. Now note that $$2^0 \equiv 1 \pmod{9}$$$$2^1 \equiv 2 \pmod{9}$$ $$2^2 \equiv 4 \pmod{9}$$ $$2^3 \equiv 8 \pmod{9}$$ $$2^4 \equiv 7 \pmod{9}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Need help with logarithmic differentitation I have the expression
$$y = \sqrt{x^2(x+1)(x+2)}.$$
I have tried looking at videos but I still cannot arrive at the correct answer and don't know how to get there.
By the way, the correct answer is
$$y' = \frac{4x^2+9x+4}{2\sqrt{(x+1)(x+2)}}.$$
Please, help.
| Alternatively, write $$y^2=x^2(x+1)(x+2)$$
Then $$2yy'=2x(x+1)(x+2)+x^2(x+1)+x^2(x+2)$$ whence
$$y'=\frac{2x(x+1)(x+2)+x^2(x+1)+x^2(x+2)}{2x\sqrt{(x+1)(x+2)}}$$
$$y'=\frac{2(x+1)(x+2)+x(x+1)+x(x+2)}{2\sqrt{(x+1)(x+2)}}\\=\frac{4x^2+9x+4}{2\sqrt{(x+1)(x+2)}}$$
assuming you wanted $x\sqrt{(x+1)(x+2)}$ since $\sqrt{x^2}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me ... | Here's a nice way to get the expansion the OP used:
The expression
$$P(a,b,c)=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$
is clearly a homogeneous polynomial of degree $4$, symmetric in its three variables. It's also clear that the coefficient of $a^4$ (hence also $b^4$ and $c^4$) is $-1$. Moreover,
$$P(-a,b,c)=P(a,-b,c)=P(a,b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/465103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Factorise: $2a^4 + a^2b^2 + ab^3 + b^4$ Factorize : $$2a^4 + a^2b^2 + ab^3 + b^4$$
Here is what I did:
$$a^4+b^4+2a^2b^2+a^4-a^2b^2+ab^3+b^4$$
$$(a^2+b^2)^2+a^2(a^2-b^2)+b^3(a+b)$$
$$(a^2+b^2)^2+a^2(a+b)(a-b)+b^3(a+b)$$
$$(a^2+b^2)^2+(a+b)((a^2(a-b)) +b^3)$$
$$(a^2+b^2)^2+(a+b)(a^3-a^2b+b^3)$$
At this point I don't... | After minar:
$2x^4+x^2+x+1= 2x^4+2x+x^2-x+1= 2x(x+1)(x^2-x+1)+x^2-x+1=(2x^2+2x+1)(x^2-x+1) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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An irreducible polynomial of degree $4$ in $\mathbb{Z}_5[x]$ Q: Find an irreducible polynomial of degree four in $\mathbb{Z}_5[x]$.
My Answer: $x^{4} - 2$
Is my answer correct?
| This is correct, but you need to show it's irreducible. Here's one way to do it:
We immediately verify that $x^4-2$ has no roots in $\mathbb Z _5$. Thus, if it is reducible, it must split into two monic quadratic terms. Say $x^4-2=(x^2+ax+b)(x^2+cx+d)=x^4+x^3(a+c)+x^2(b+d+ca)+x(ad+bc)+bd$. Now, matching coefficients, w... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What is? $ \lim_{x \to \infty} \left(\frac{e}{\left( 1 + \frac {1}{x} \right)^x} \right)^x $
Find $$ \lim_{x \to \infty} \left( \dfrac{e}{\left( \left( 1 + \frac {1}{x} \right)^x \right)} \right)^x. $$
$ \lim_{x \to \infty} \left( \dfrac{e}{\left( \left( 1 + \frac {1}{x} \right)^x \right)} \right)^x = \lim_{x \to \i... | Take the logarithm,
$$\begin{align}
\log \left(\frac{e}{\left(1+\frac1x\right)^x}\right)^x &= x\log \frac{e}{\left(1+\frac1x\right)^x}\\
&= x\left(1 - x \log \left(1+\frac1x\right)\right)\\
&= x\left(1 - x\left(\frac1x - \frac{1}{2x^2} + O(x^{-3})\right) \right)\\
&= x\left(1 - 1 + \frac{1}{2x} + O(x^{-2})\right)\\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Cyclic polynomial proof If$$ x+y+z = 0 $$
Then prove,
$$ (x^2+xy+y^2)^3+(y^2+yz+z^2)^3+(z^2+zx+x^2)^3$$
$$=3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)$$
| HINT:
Observe that if we put $x^2+xy+y^2=a$ etc., we need to prove $a^3+b^3+c^3=3abc$
From this, the above proposition will be true
either if $a+b+c=0$
or if $a=b=c$ for real $a,b,c$
Now, $x^2+xy+y^2-(y^2+yz+z^2)=x^2-z^2+xy-yz=(x-z)(x+z+y)=0 $
Alternatively eliminating $x,$
$x^2+xy+y^2=x(x+y)+z^2=\{-(y+z)\}(-z)+y^2$... | {
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} |
Finding the transition matrix for the rational canonical form Let $A$ be the $3\times3$ matrix
$$\begin{bmatrix} 3 & 4 & 0 \\-1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix}$$
The characterisitc and minimal polynomials are both $(x-1)^2(x+1)$
The eigenspace for $\lambda=1$ is
$$\left\{ \begin{bmatrix} 2 \\-1 \\ 1 \end{bmatrix... | Hints:
We are given:
$$\tag 1 A = \begin{bmatrix} 3 & 4 & 0 \\-1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix}$$
Jordan Form
*
*Correct on eigenvalues
*You need three independent eigenvectors, and one of those ends up being a generalized one.
*You would have, $A = P \cdot J\cdot P^{-1}$, where $J$ represents the Jordan No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
a divisibility problem if $a$ is an integer , such that it is not divisible by 2 or 3, prove that $ 24 $ divides $ a^2+ 23$ .
I took cases ,case $1$ : when $a$ is divisible by only $2$ and not $3$ then we can write $a = 6k+2, 6k+4 $ and case 2: when $a= 6k+3$ and here $a$ is divisible by only $3$ and not $2$.
and th... | Method $1:$
HINT:
Using Carmichael Function, $\lambda(24)=2$
If $(a,2)=(a,3)=1, (a,2^m\cdot3^n)=1$ for any integer $m\ge0,n\ge0$
So, $a^2\equiv1\pmod{24}$ if $(a,24)=1$
Now, $a^2+23\equiv a^2-1\pmod{24}$
Method $2:$
If $(a,3)=1, a\equiv\pm1\pmod 3$
$\implies a^2\equiv1\pmod 3\implies a^2+23\equiv24\pmod{3}\equiv0$
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). ... | HINT:
If $A+Bw+Cw^2=0$ where $w$ is one of the three cube roots of unity
$\implies -A=Bw+Cw^2$
Cubing we get, $(-A)^3=(Bw+Cw^2)^3$
$\implies -A^3=B^3w^3+C^3w^6+3\cdot Bw\cdot Cw^2(Bw+Cw^2)=B^3+C^3+3BC(-A)$
$\implies A+Bw+Cw^2$ is a factor of $A^3+B^3+C^3-3ABC$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/475354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 6,
"answer_id": 0
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How to calculate the intersection of two planes? How to calculate the intersection of two planes ?
These are the planes and the result is gonna be a line in $\Bbb R^3$:
$x + 2y + z - 1 = 0$
$2x + 3y - 2z + 2 = 0$
| Let $x$ be in the intersection of the planes
\begin{equation*}
\langle m, x-b \rangle = 0,\\
\langle n, x-c \rangle = 0
\end{equation*}
and let $p = m \times n$.
Then for some $\lambda$,
\begin{equation*}
\begin{pmatrix}
m_0 & m_1 & m_2 \\
n_0 & n_2 & n_3 \\
p_0 & p_1 & p_2 \\
\end{pmatrix}
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 7,
"answer_id": 3
} |
How many zeros are there in $1 \times 2^2 \times 3^3 \times \cdots \times 100^{100}$? How many zeros are there at the end of $1 \times 2^2 \times 3^3 \times \cdots \times 100^{100}$ ?
I tried it by grouping all the $2$'s and $5$'s and $5$'s and $6$'s but cant get my answer...
| To get a zero you need to multiply a factor $5$ by a factor $2$. There are clearly more factors of $2$ than of $5$ in the product, so you need to find a systematic way of counting the factors of 5.
Well these come from the terms $5^5\cdot 10^{10} \cdot 15^{15} \dots 100^{100}$
All of these terms provide a factor $5$ so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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On seventh powers $x_1^7+x_2^7+\dots+x_n^7 = 2$? We have,
$$(-6m^3 + 1)^3 + (6m^3 + 1)^3 + (-6m^2)^3 = 2$$
$$(-8m^5 + 1)^5 + (8m^5 + 1)^5 + (-8m^6 + 2m)^5 + (-8m^6 - 2m)^5 + 2(8m^6)^5=2$$
The first identity has been long known, while the second is by Ajai Choudhry. Anybody knows if there is anything similar for 7th pow... | The best I can do is in terms of the radical $\sqrt{3}$:
$$2=(9 m^7 + 1)^7 + (-9 m^7 + 1)^7 + (\sqrt{3} m - 9 m^8)^7 + (-\sqrt{3} m -
9 m^8)^7 + 2 (9 m^8)^7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/478769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Direction and Magnitude of a Dog Running Problem: A dog in an open field runs 12.0m east and then 30.0m in a direction 54 degrees west of north.
Part A: In what direction must the dog then run to end up 12.0m south of her original starting point?
My Answer: Let the first path the dog runs A, the second path B, the thi... | Almost. Note that the angle of path $B$ is $54^\circ$ West of North (not North of West). Hence, you mixed up your $\sin$ and $\cos$. You should obtain:
\begin{align*}
C_x &= 30\sin(54^\circ)-12 \\
C_y &= -30\cos(54^\circ)-12 \\
\theta &= \arctan(|C_x/C_y|) = \arctan \left(\frac{30\sin(54^\circ)-12}{30\cos(54^\circ)+12}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \inft... | As the denominator contains $x^2+1$ we put $x=\tan y\implies dx=\sec^2ydy$
For $x^2-1$ with $x\ge1$ we need to put $x=\sec y$
and for $1-x^2$ with $x\le1$ we need to put $x=\sin y$
$$\implies \int\frac{x^2dx}{(x^2+1)^2}=\int\frac{\tan^2y\sec^2ydy}{\sec^4y}=\int\sin^2ydy=\frac12\int(1-\cos2y)dy=\frac{y}2-\frac{\sin2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
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Show that this relation is an implicit solution of the following differential equation Differential equation:
$$\frac{dy}{dx}=\frac{xy}{x^2+y^2}$$
Relation:
$$2y^2 \ln{y} - x^2 = 0$$
From this, I end up getting:
$$\frac{dy}{dx} = \frac{x}{2y\ln{y} + y} $$
The missing step would be to put in the form given at the top.... | Multiply by $y/y$ first. Note from our relation $2y^2\log y-x^2=0$ that adding $x^2$ to both sides yields $2y^2\log y=x^2$. Substitute for $2y^2\log y$ and you are done.
$$\begin{align*}\frac{dy}{dx}&=\frac{x}{2y\log y+y}\\&=\frac{xy}{2y^2\log y+y^2}\\&=\frac{xy}{x^2+y^2}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/484461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Integrating $\cos^3 (x) \, dx$ I am wondering whether I integrated the following correctly.
*
*$\int \cos^3 x \, dx$
I did
\begin{align}
\int \cos^3 x \, dx &= \int \cos(x)(1-\sin^{2}x) \, dx \\
&= \int \cos(x)-\sin^{2}x \cos x \, dx \\
&= \sin(x)-\frac{u^{3}}{3} + c, \quad(u=\sin(x)) \\
&= \sin(x)-\dfrac{\sin^{3}x... | Your first integral is correct.
The second has two sign errors: $$u = \cos x \implies du = -\sin x\,dx$$
So evaluating the second integral should yield
$$\begin{align} \int \Big(\cos^2x\sin x-\cos^4x\sin x\Big)\,dx & = -\int \cos^2x\Big(-\sin x\,dx\Big) - (-)\int \cos^4x\Big(-\sin x\,dx\Big) \\ \\& = -\int u^2 du + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$ let $a,b$ are positive numbers,and such $ab=8$ find this minum
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$
My try:
and I find when $a=4,b=2$,then
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$ is minum $5\sqrt{5}$
it maybe use Cauchy-Schwarz inequality
Thank you
| by the Cauchy-Schwarz inequality,we have
$$(a^2+64)(1+4)\ge(a+16)^2$$
$$(b^2+1)(4+1)\ge (2b+1)^2$$
then
$$\sqrt{a^2+64}+\sqrt{b^2+1}\ge\dfrac{1}{\sqrt{5}}(a+2b+17)$$
and By AM-GM,we have
$$a+2b\ge 2\sqrt{2ab}=8$$
so
$$\sqrt{a^2+64}+\sqrt{b^2+1}\ge 5\sqrt{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequalities question I've been having trouble solving these kind of inequalities ;
$\sqrt { -7x+1 } - \sqrt {x+10} \gt \ {7}$
Attempt at a Solution; We first find the official boundaries of the inequality; x is limited to the section: $\ -10 \le x \le \frac{1}{7} $
Then, squaring both sides we get:
${-7x+1} -2 \... | From $\sqrt{-7x+1} - \sqrt{x+10} \gt 7$, your assessment of the interval $x \in [-10, \frac{1}{7}]$is clearly right.
Let us make sure we have positive quantities on both sides first, so rewrite as
$\sqrt{-7x+1} \gt 7 + \sqrt{x+10} $
Squaring, $- 7x + 1 > 49 + x + 10 + 14\sqrt{x+10} $
or $-4x -29 > 7\sqrt{x+10} $
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Is this correct method to prove that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, when $a,b,c \geq 0$? Can I prove it like this: Let's say that $a=b=c$ so we get "If $a \geq 0$ then $3a^2 ≥ 3a^2$" Now I take the negation of that statement and get "If $a \geq 0$ then $3a^2 < 3a^2$" The anti-thesis is obviously wrong which makes th... | \begin{align}
a^2 + b^2 + c^2 - ab - bc - ca
&= (1/2)(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca)\\
&= (1/2)(a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2)\\
&= (1/2)[(a-b)^2 + (b-c)^2 + (c-a)^2]
\end{align}
Now we know that, the square of a number is always greater than or equal to zero.
Hence, $(1/2)[(a-b)^2 + (b-c)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How to get all solutions to equations with square roots I would like to find all solutions to
$$b-a\sqrt{1+a^2+b^2}=a^2(ab-\sqrt{1+a^2+b^2})$$
$$a-b\sqrt{1+a^2+b^2}=b^2(ab-\sqrt{1+a^2+b^2})$$
I found some solutions. For example, $a = 1, b = \pm i$ and $b = 1 , a = \pm i$. How can I find all solutions?
| One technique with this kind of problem is to isolate the square root, and then square the equation (which brings in possible extra roots from the negative value of the square root).
[Now corrected for error pointed out by Anush in copying over the equations]
Beginning that process here we find that the first equation ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $(a,b)$ if $x^2-bx+a = 0, x^2-ax+b = 0$ both have distinct positive integers roots
If $x^2-bx+a = 0$ and $x^2-ax+b = 0$ both have distinct positive integers roots, then what is $(a,b)$?
My Try:
$$\displaystyle x^2-ax+b = 0\Rightarrow x = \frac{a\pm \sqrt{a^2-4b}}{2}$$
So here $a^2-4b$ is a perfect square.
Simila... | Let the roots of $x^2 - ax +b$ be $r$ and $s$, and the roots of $x^2 - bx + a$ be $u$ and $v$. Then
$$\begin{align}
r+s &= a\\
rs &= b\\
u+v &= b\\
uv &= a.
\end{align}$$
Let, without loss of generality, $a \leqslant b$. Thus
$$\begin{align}
uv &\leqslant u+v\\
\iff uv - u - v + 1 & \leqslant 1\\
\iff (u-1)(v-1) & \leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/494679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Finding vertex and focus of parabola given an equation I am defeated to complete square on the following parabolic equation. Please help.
Find the vertex and focal width for the parabola: $$ x^2+6x+8y+1=0 $$
I am hoping to get an equation in this form
$$(x−h)^2=4p(y−k)$$
| $$x^2 + 6x \color{blue}{+ 9} + 8y + 1 \color{blue}{- 9} = 0$$
$$(x + 3)^2 + 8(y - 1) = 0$$
$$(x + 3)^2 = -8(y - 1) = 4(-2)(y-1)$$
Now, you should be able to "read off" the vertex of the parabola. From there, see if you can find.
With respect to completing the square: you have
$$(x + 3)^2 + 8 y + 1 = 9$$
Subtract $9$ fr... | {
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"url": "https://math.stackexchange.com/questions/494835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral (square root function multiplied by exponential function) did i do it right? I'm trying to determine $\int x^3\sqrt{x^2 +1}\, dx$
I said that $u(x) = x^2 + 1$
and then that $dx = 2x\,dx$
so I rewrote the integral as
$$\int x^3\sqrt{x^2 +1}\,2x\,dx$$
which is also
$$\int2x^4\sqrt{x^2 +1} \,dx$$
and then it is... | \begin{align}
&\int x^{3}\,\sqrt{x^{2} + 1\,}\,{\rm d}x
=
\int x^{2}\,{\rm d}\left[{1 \over 3}\,\left(x^{2} + 1\right)^{3/2}\right]
\\[3mm]&=
x^{2}\,{1 \over 3}\left(x^{2} + 1\right)^{3/2}
-
\int{1 \over 3}\left(x^{2} + 1\right)^{3/2}
\,{\rm d}\left(x^{2} + 1\right)
\\[3mm]&=
{1 \over 3}\,x^{2}\left(x^{2} + 1\right)^{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Can an integer of the form $4n+3$ written as a sum of two squares? Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact?
| Let's assume x^2+y^2 = 4n+3, then either x or y has to be even. Let's assume x = 2z and write
(2z)^2+y^2 = 4n+3. This can also be written as follows:
(2z+y)^2-4zy = 4n+3 by rearrangement we can write
(2z+y)^2-1^2 =4n+2+4zy
(2z+y)^2-1^2=2(2n+2zy+1) and further
(2z+y-1)(2z+y+1)=2(2n+2zy+1)
The left side is product of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
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Finding sequence that's defined by a recurrence relation This is the problem that I ran into when doing practice in my textbook.
How do I find the sequence (call it $a_n$) that's defined by recurrence relation whose generating function is $\frac 4 {-x^2-2x+3}$?
Help appreciated!
| We use partial fractions. So we want to find integers $A$ and $B$ such that
$$\frac{4}{-x^2 -2x+3}=\frac{A}{3+x}+\frac{B}{1-x}.$$
There are general procedures, but we can see that $A=1$, $B=1$ work.
Now
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots=\sum_0^\infty x^n.$$
Expanding $\frac{1}{3-x}$ takes more work. Rewrite as
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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The sum of an infinite series with integral $1+\dfrac{1}{9}+\dfrac{1}{45}+\dfrac{1}{189}+\dfrac{1}{729}+\dots=\sum\limits_{n=1}^\infty \dfrac{1}{(2n-1)\cdot 3^{n-1}}$
I got:
$\sum\limits_{n=1}^\infty \dfrac{1}{(2n-1)\cdot 3^{n-1}}=\sum\limits_{n=1}^\infty \dfrac{\int\limits_0^1 x^{2n-2}\,dx}{3^{n-1}}=\dots$
And no ide... | After writing $\frac{1}{2n-1}$ as an integral, you have the series
$$\sum_{n=1}^\infty \int_0^1 \left(\frac{x^2}{3}\right)^{n-1}\,dx.$$
Since the geometric series
$$\sum_{k=0}^\infty \left(\frac{x^2}{3}\right)^k$$
converges uniformly on the interval $[0,1]$, we can interchange summation and integration, and obtain
$$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$ I found the following two relational expressions in a book without any additional information:
$$\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac13(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25})$$
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\f... |
For the identity in the title, use the identity $x^3+1=(x+1)(x^2-x+1)$.
Let $x=\sqrt[3]{2}$, then $(x^2-x+1)(x+1)=x^3+1=3$ hence the RHS is $\sqrt[3]{3}/(x+1)$ and the LHS divided by the RHS is the cube root of one third of $(x-1)(x+1)^3=(x^3-2)(x+2)+3$. Since $x^3=2$, this is $3$ and the result follows.
For the oth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Constant-Coefficient Systems. Find a real general solution of the following system. Find a real general solution of the following system. (Show the
details.)
$y'_1 = 9 y_1 + 13.5 y_2 \;\;\;\;\;\;\; y′_2 = 1.5 y_1 + 9 y_2$
$y' = \begin{pmatrix} 9& 13.5\\1.5& 9\end{pmatrix}$y
$\det$ (A - $\lambda$I$)= \begin{pmatrix} 9-\... | We are given:
$$A = \begin{bmatrix}9 & 27/2\\3/2 & 9\end{bmatrix}$$
The characteristic equation and eigenvalues are found by solving $[A-\lambda I] = 0$, which gives:
$$1/4 (2 \lambda-27) (2 \lambda-9) = 0 \rightarrow \lambda_1 = \dfrac{27}{2}~,\lambda_2 = \dfrac{9}{2}$$
We now find the eigenvectors for each distinct e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Convergence of $x_{n}=\frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\cdots+\frac{n^2}{\sqrt{n^{6}+n}}$ Let $$x_{n}=\frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\frac{n^2}{\sqrt{n^{6}+3}}+...+\frac{n^2}{\sqrt{n^{6}+n}}.$$ Then $(x_{n})$ converges to
(A)$1$
(B)$0$
(C)$\frac{1}{2}$
(D)$\frac{3}{2}$
My Tr... | You can use the following inequality:
$$
\frac{n.n^2}{\sqrt{n^{6}+n}}\leq \frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\frac{n^2}{\sqrt{n^{6}+3}}+...+\frac{n^2}{\sqrt{n^{6}+n}}\leq \frac{n.n^2}{\sqrt{n^{6}+1}}
$$
Then take the limit of both sides and you can see the limit is 1.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all natural numbers n such that $7n^3 < 5^n$ Find all natural numbers $n$ such that $7n^3 < 5^n$.
I drew a graph which showed that $n \geq 4$
wolfram
How can I prove that? I guess I need to use induction with the base case $n=4$?
But I am stuck because the induction hypothesis uses a $\geq$ sign so I can not subst... | Base Case: For $n=4$, we have $7(4)^3 = 448 < 625 = 5^4$, which works.
Induction Hypothesis: Assume that the claim holds true for all $n \in \{4,\ldots,k\}$, where $k>3$.
It remains to prove the inequality true for $n=k+1$. Indeed, observe that:
\begin{align*}
7(k+1)^3 &= 7(k^3 + 3k^2 + 3k + 1) \\
&< 7(k^3 + 3k^2 + 9k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Geometric/Visual Solution - Shortest Vector for which Dot Product = x + 2y = 5. (Strang P21 1.2.26) Much as I referenced this same exercise, I'm questing after an exclusively geometric solution.
Question: If $\color{#0070FF}{\vec{v} = (1,2)}$ draw all vectors $\vec{w} = (x,y)$ in the plane $x,y$ with $\color{#0070FF}... | To answer your first question:
$$
\mathbf{v \cdot w} = 5 \Longrightarrow
\mathbf{v} \cdot (k\mathbf{v}) = 5 \Longrightarrow
k \|\mathbf{v}\|^2 = 5 \Longrightarrow k=1
$$
Regarding the second question ...
By definition, we have $\mathbf{v \cdot w} = \|\mathbf{v}\| \cdot \|\mathbf{w}\| \cos\theta$, where $\theta$ is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I get the integral of $\frac{1}{(x^2 - x -2)}$ I'm working with this problem $$ \int \frac{1}{x^2 - x - 2}$$
I'm thinking I break up the bottom so that it looks like this $$\int \frac{1}{(x-2)(x+1)} $$
Then I do $$x^2 - x -2 = \frac{A}{x-2} + \frac{B}{x+1} $$
Multiple both sides by the common denominator and co... | Your error is in the step where you write:
$$x^2 - x -2 = \frac{A}{x-2} + \frac{B}{x+1}$$
What you have, actually, is $$\dfrac 1{x^2 - x - 2} = \dfrac A{x-2} + \frac B{x+1}$$
So $A(x+1) + B(x - 2) = 1$.
If $x = -1$, $$A(-1 + 1) + B(-1 - 2) = 1 \iff -3B = 1 \iff B = -\frac 13$$
If $x = 2$, $$A(2 + 1) + B(2 - 2) = 1 \if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Is $X^2-Y^2/ \sqrt{X^2+Y^2}$ normal where $X,Y\sim N(0,1)$ Can I say that $(X^2-Y^2)/ \sqrt{X^2+Y^2}$ is normal since $X,Y\sim N(0,1)$ and $X,Y$ are normal and independent?
I am trying to do those problem using polar coordinates with $X=r \cos \theta$ and $Y=r \sin \theta$, but I got stuck when I was trying to simplif... | Let $(R,\Theta)$ be the polar coordinates corresponding to the Cartesian coordinates $(X,Y)$, with, say, $0 \le \Theta < 2 \pi$. Thus $X = R \cos(\Theta)$, $Y = R \sin(\Theta)$,
$R = \sqrt{X^2 + Y^2}$. Then $R$ and $\Theta$ are independent, and $\Theta$ is
uniform on $[0,2\pi)$.
Now $(X^2 - Y^2)/\sqrt{X^2 + Y^2} = (R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Limit $ \sqrt{2\sqrt{2\sqrt{2 \cdots}}}$
Find the limit of the sequence $$\left\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots\right\}$$
Another way to write this sequence is $$\left\{2^{\frac{1}{2}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}... | Hint: take a logarithm of every element in your sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result. Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result.
I've attempted to apply the Squeeze Theorem as such:
$\frac{-1}{x^2 + y^4} \leq \frac{\sin{(x^3 + y^5)}}{x^2 + y^4} \leq \frac{1}{x^2 + y^4}$. Cl... | Here's another hint/warm-up problem:
$$\lim_{(x,y)\to (0,0)}\frac{x^6+y^5}{x^4+y^4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability of two consecutive head or tail or any one of them in a row? Fair coins are tossed and when either four consecutive heads and tails appear the process will be stopped. What is the probability of two consecutive head or tail or any one of them in a row?
| Because heads and tails play symmetric roles in the stopping criterion (and presumably have equal chances in each Bernoulli trial), it suffices to find the probability of getting two consecutive heads before the process stops.
If the process stops with four consecutive heads, obviously that means also we got two consec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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