Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Failure of L’Hospital’s rule to $\lim_{x\rightarrow\infty} x\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right)$? I rewrote the limit
$$\lim_{x\rightarrow\infty} x\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right) =\lim_{x\rightarrow\infty} \frac{\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\rig... | Alternatively, with the trigonometric identity,
$$ \tan^{-1} (a) - \tan^{-1}( b) = \tan^{-1} \left( \dfrac{ a - b}{1 + ab} \right) $$
Set $a = \frac{x+1}{x+2} $ and $ b = 1 , $
$$ \tan^{-1} \left( \dfrac{x+1}{x+2} \right) - \dfrac \pi4 = \tan^{-1} \left( \dfrac1{-2x-3} \right) $$
So the limit is $$ L :=\lim_{x\to\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$ in two ways gives different answers
Evaluate
$$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$$
I started by rationalising followed by dividing numerator and denominator by $x$.
$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$
=$\lim_{x\rightarrow-\infty}\frac{{(\... | Hint note that if $x < 0$ then $-2x > 0$, but $-2 < 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let ABCD be a square, E be midpoint of AB. Let F be on BC and G be on CD and EF is parallel to AG. Prove FG is tangent to the incircle of the square.
I tried using OC and finding the slope so that maybe OC would be perpendicular to FG. I know that EF and AC will have the same slope. I also know that I have to use the ... | Line $EF$ has the parametric form $\pmatrix {-1 \\ 0}+\mu\pmatrix {1+x \\ 1}$.
Line $AG$ has the parametric form $\pmatrix {-1 \\ -1}+\lambda\pmatrix {1+x \\ 1}=\pmatrix {1 \\ y}$.
$\therefore\lambda=\dfrac{2}{1+x}$
$\therefore y=-1+\dfrac{2}{1+x}=\dfrac{1-x}{1+x}$
Line $FG$ has the parametric form $\pmatrix {x \\ 1}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4289089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now... | A better approach would be to treat the given expression as a quadratic in $x$ and, so we can re-write the expression as $ 4x^2 -2x(y+2) +3(y-1)$. Set it equal to zero, and solve for $x$ using the quadratic formula.
$x = \frac{ 2(y+2) \pm \sqrt{4(y+2)^2-48(y-1)}} {8}$
This gives us the solutions $x=\frac{3}{2} $ and $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Unsure how to solve this quadratic formula problem and how to expand and substitute into the formula Find all values of a and b that will make $(ax+5)^2 = 16x^2 + bx + 25$ true for all $x$.
I believe that this is an equation that needs to be solved using the quadratic formula but am not sure how to expand and substitut... | The quadratic formula is not needed here
\begin{align*}
(ax+5)^2 &= 16x^2 + bx + 25\\
\implies
b &= a^2 x- 16 x + 10 a \\
\implies \mathbf{(ax+5)^2} &= 16x^2 + \bigg( a^2 x- 16 x + 10 a \bigg)x + 25\\
&=\mathbf{(ax+5)^2}
\end{align*}
We can see from
$\quad (ax+5)^2=(ax+5)^2\quad $
how all values of $\space x\space$ r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show this sequence always is rational number let $\{a_{n}\}$ such $a_{1}=-8$,and such
$$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
show that
$$a_{n}\in Q,\forall n\in N^{+}$$
I try let $a_{2}=x$,and for $n=1$, then we have
$$-8+5\sqrt[3]{x}=3\sqrt[3]{-49(x+1)}\Longrightarrow x=-1/8$$
and fo... | This isn't an answer, really more of a musing on the problem, a full expression for $a_{n+1}$ in terms of $a_n$, and a way forward. Assuming that this problem is well posed (that is, $a_n$ can always be calculated from $a_{n-1}$ without ambiguity) then to get this expression, define $a_n=b_n^3$. Then the formula relati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4298049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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proving $\left[\left(\cos(x)+1\right)+i\sin(x)\right]^{n} = 2^n\cos^{n}(\frac{x}{2})\left(\cos(\frac{nx}{2})+i\sin(\frac{nx}{2})\right)$ How can I prove the following where i is the imaginary unit.
$$\left[\left(\cos(x)+1\right)+i\sin(x)\right]^{n} = 2^n\cos^{n}(\frac{x}{2})\left(\cos(\frac{nx}{2})+i\sin(\frac{nx}{2})\... | $$\left[\left(\cos(x)+1\right)+i\sin(x)\right]^{n} \implies\left(2\cos^{2}(\frac{x}{2})+2i\sin(\frac{x}{2})\cos(\frac{x}{2})\right)^{n}\\ \implies2^n\cos^n(\frac{x}{2})\left(\cos(\frac{x}{2}) + i\sin(\frac{x}{2})\right)^{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Maximum likelihood estimator for beta distribution I'm trying to derive MLE for $\theta $ parameter in beta distribution $B(\theta, 2)$:
$$f(x, \theta) = \theta(\theta+1)x^{\theta-1}(1-x) {1}_{\{(0, 1)\}}(x)$$
My strategy will be a standard one - calculate derivative and look for zero of a function.
$$L(X, \theta) = \t... | For $a_n < 0$, $$\frac{-a_n - 2 \pm \sqrt{a_n^2 + 4}}{2a_n} = -\frac{1}{2} - \frac{1}{a_n} \mp \frac{1}{2}\sqrt{1 + \frac{4}{a_n^2}}.$$ Notice the reversal in sign due to the fact that $a_n < 0$ implies $a_n = -\sqrt{a_n^2}$. Therefore, as $a_n \to -\infty$, we have $\theta_1 \to -\frac{1}{2} + \frac{1}{2} = 0$, but ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to evaluate $232^2-62^2\times14$ by factoring or using identities or...? The expression $232^2-62^2\times14$ can be calculated directly ($53824-3844
\times14=8$). But is it possible to evaluate it for example by factoring or using identities?
Here is what I have tried,
$$(58\times4)^2-62^2\times14=58^2\t... | The numbers involved are easy to factor, so take out the common factors to get
$$232^2-62^2\times14=(2^3\times29)^2-(2\times31)^2\times(2\times7)
=2^3\times(2^3\times29^2-7\times31^2).$$
The last factor looks similar to a difference of two squares; note that
\begin{eqnarray*}
2^3\times29^2-2^3\times31^2&=&2^3\times(29^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Can two circles intersect each other at right angles, such that one circle passes through the center of the other circle? More precisely, do there exist two intersecting circles such that the tangent lines at the intersection points make an angle of 90 degrees, and one of the circles passes through the center of the ot... | Let
\begin{align}
\vec{r}_1(\alpha)&=R\left(\cos \alpha, \sin \alpha\right)\\
\vec{r}_2(\beta)&=\lambda R\left(1+\cos \beta, \sin \beta\right)
\end{align}
be equations of the given circles where $\lambda$ and $R$ are both positive. The parameters $\alpha$ and $\beta$ span from $0$ to $2\pi$.
Assume their intersections ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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In how many distinguishably different ways can a pair of indistinguishable dice come up? The answer is $21$, but why doesn't $(6\times 6)/2$ work? It accounts for the overlap where $2-4$ and $4-2$ are the same cases.
| Suppose we have a pair of indistinguishable white dice. There are $\binom{6}{2}$ outcomes in which they show different numbers and $\binom{6}{1}$ outcomes in which they show the same number. Hence, there are a total of
$$\binom{6}{2} + \binom{6}{1} = 21$$
outcomes when two indistinguishable dice are thrown.
Suppose w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To determine the integration of $ \int_{0}^{+\infty} \exp\!\Big(-\Big(\frac{ax^2+bx+c}{gx+h}\Big)\Big) dx$. What is the integration of the following function:
$$ \int_\nolimits{0}^{+\infty} \exp\!\bigg(-\bigg(\frac{ax^2+bx+c}{gx+h}\bigg) \bigg)dx.$$
What I have done is as follows:
Here, $\kappa=c-\Big(\frac{bg-ah}{g^... | Defining the incomplete Bessel function as
$$
K_\nu(x,y) = \int_1^{\infty} t^{-\left(\nu+1 \right)}e^{-\left(xt + \frac{y}{t} \right)}\, \mathrm{d}t
$$
We get
\begin{align}
\int_{0}^{\infty} e^{-\frac{ax^2+bx+c}{gx+h}} \, \mathrm{d}x & \overset{\color{purple}{x = \frac{h}{g}(t - 1)}}{=}\frac{h}{g}e^{\frac{2ah}{g^2}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Write $x^k - y^k$ as the product of two factors. Write $x^k - y^k$ as the product of two factors.
$x^3 - y^3 = (x-y)(x^2+xy+y^2)$
$x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3)$
$x^5 - y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$
$x^k - y^k = (x-y)(x^{k-1}+\underbrace{x^{k-2}y + \c... | Here is a variation without using induction. We want to use the sigma symbol and start with an example for small $n=4$ to better see what's going on.
We have
\begin{align*}
x^4-y^4&=(x-y)(x^3+x^2y+xy^2+y^3)\\
&=(x-y)(x^3y^0+x^2y^1+x^1y^2+x^0y^3)\tag{1}\\
&=(x-y)\sum_{k=0}^3 x^{3-k}y^k
\end{align*}
Since we can write $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
volume of the solid which is the intersection of the solid sphere $x^2+y^2+z^2\leqq1 \; , 2x^2+y^2-2x=0$ i'm trying to find the volume of the solid which is the intersection of the solid sphere $x^2+y^2+z^2\leqq1 \; , 2x^2+y^2-2x=0$
My attempt:
I tried with cylindrical coordinates $$x=rcos(\phi)\;,y=rsin(\phi)\;,z=z$$
... | Well, first of all, the second equation describes a shifted elliptical cylinder. From $2x^2+y^2-2x=0$, divide by $2$ and complete the square, yielding
$$
(x-1/2)^2+y^2/2=1/4,
$$
which writing in standard form yields
$$
\frac{(x-1/2)^2}{1/4}+\frac{y^2}{1/2}=1.
$$
Hence, the natural $(x,y)$ parameterization is
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the quadratic equation with real coefficients and solutions $x_1,x_2$ if you know that $\Delta =b^2-4ac=-36$ and $x_1+x_2=6m$ What I've done so far is:
We know that $\Delta \lt 0$, which means that $x_1$ and $x_2$ are two conjugate complex numbers with the form: $x_1=r+n\cdot i \space\text{ and }\space x_2=r-n\cdo... | You already concluded from the negative discriminant that the zeroes of the quadratic polynomial must be a "conjugate pair" $ \ \alpha \ + \ \beta·i \ \ . $ The difference of the zeroes is
$$ x_1 \ - \ x_2 \ \ = \ \ \frac{\sqrt{\Delta}}{a} \ \ \Rightarrow \ \ 2·\beta·i \ \ = \ \ \frac{\sqrt{-36}}{a} \ \ = \ \ \frac{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What's the measure of the radius of the biggest circle below? For reference: In figure A is the tangency point::
LE=2(TE),
$\overset{\LARGE{\frown}}{AN}=60^o$
$\frac{(TE)^2}{R-r}= 10$
Calculate R.(Answer:80)
My progress
$\triangle AON: \angle AON = 60^o, AO=r, ON=r \implies\\
AN =r \therefore \triangle AON(equilatera... | First note that $A, O $ and $O_1$ are collinear and as $A, O$ and $T$ are collinear, all four points $A, O, T$ and $O_1$ must be collinear. As $\angle O_1AL = 60^\circ$ and $O_1A = O_1L = R$, $\triangle O_1AL$ is equilateral with side length $R$.
Now there are multiple ways to get to the answer. One approach would be,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of solution of ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ Number of solution of the equation ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ in the interv... | Starting with: $ 2abc + ab\left( {a + b} \right) + bc\left( {b + c} \right) + ac\left( {a + c} \right) = 0$
$$ ab(c + a + b) + bc(a+b+c) + ac\left( {a + c} \right) = 0$$
$$ b(c + a + b)(a+c) + ac\left( {a + c} \right) = 0$$
$$ (a+c)\bigl(b(c + a + b)+ac\bigr) = 0$$
$$ (a+c)(bc + ab + b^2+ac) = 0$$
$$ (a+c)\bigl(c(a+b) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\varepsilon$-$\delta$ for multivariables I have a pracitce question that I have been troubling proving: $\lim_{(x, y) \to (2, 1)} \dfrac{x^2 - 2xy}{x^2 - 4y^2} = \dfrac{1}{2}$ using the $\varepsilon$-$\delta$ definition. But here is my attempt of the solution.
If $\displaystyle \lim_{(x, y) \to (2, 1)} \dfrac{... |
$\lim_{(x, y) \to (2, 1)} \dfrac{x^2 - 2xy}{x^2 - 4y^2} = \dfrac{1}{2}$
Actually, the above assertion is false, for the following reason. In any (small) neighborhood of radius $\delta$ around $(2,1)$, you will find $(x,y) \neq (2,1)$ such that $x - 2y = 0$. If you examine the original fraction that is being interro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find inverse of element in a binary field The question states:
Let us consider the field $GF(2^4)$ with multiplication modulo $x^4+ x^3+1$
Find all y such that $1010(y + 0011) = 1111$, in other words find y that satisfies $(x^3+x)(y +x+1) = x^3+x^2+x+1$
I tried to find the inverse for element $1010$ to multiply both si... | You can use the Extended Euclidean algorithm to find the inverse of $x^3+x$ modulo $x^4+ x^3+1$ in $GF(2)[x]$
Given $a$ and $b$ you will find $u$ and $v$ such that
$$au+bv=\gcd(a,b)$$
in our example $\gcd(a,b)=1$ because $a$ is an irreducible polynomial and $\deg(a)>\deg(b)$. So we have
$$bv\equiv 1\pmod(a)$$
and $v$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Use Principle of Mathematical Induction to show that for all $n\:\in \mathbb{N}$, $a_n=2^{n+2}\cdot 5^{2n+1}+3^{n+2}\cdot 2^{2n+1}$ is divisible by 19 Here's what I did:
Base Case: $n = 1$, $2^3*5^3+3^3*2^3$ = $1216$ = $19(64)$
Induction Hypothesis (IH): $a_k=2^{k+2}\cdot 5^{2k+1}+3^{k+2}\cdot 2^{2k+1}$ for some $k\:\i... | You're almost there! Hint for the finish:
$$
2^{k+2}\cdot 5^{2k+1}\cdot 50+3^{k+2}\cdot 2^{2k+1}\cdot 12\\
= 12\left(2^{k+2}\cdot 5^{2k+1}+3^{k+2}\cdot 2^{2k+1}\right) + 38\left(2^{k+2}\cdot 5^{2k+1}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the sum of geometric progression Evaluate the sum:
$$\sum_{x=0}^\infty x(x-1) {2+x \choose x}(0.008)(0.8)^x $$
I was able to make this into:
$$0.004\sum_{x=0}^\infty x(x-1) (x+1)(x+2)(0.8)^x $$
Let $x=n-2$ then $n=x+2$:
$$0.004\sum_{n=2}^\infty n(n-1)(n-2)(n-3)(0.8)^{n-2} $$
$$0.004\sum_{n=4}^\infty (n)_3~(0.8)... | $$(x-1) x \binom{x+2}{x}=\frac{1}{2} (x-1) x (x+1) (x+2)$$ Rewriting
$$\frac{1}{2}(x-1) x (x+1) (x+2)= \frac{1}{2}x (x - 1) (x - 2) (x - 3) + 4 x (x - 1) (x - 2) + 6 x (x - 1) $$
$$S=\sum_{x=0}^\infty x(x-1) {2+x \choose x}p^x=\frac{1}{2}p^4\sum_{x=0}^\infty x (x - 1) (x - 2) (x - 3)p^{x-4}+$$ $$4p^3\sum_{x=0}^\infty x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving $1/(a+b) + 1/(b+c) + 1/(c+a) > 3/(a+b+c)$ for positive $a, b, c\,$? I have to prove that:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} > \frac{3}{a+b+c},$$
where $a, b , c$ are positive real numbers.
I am thinking about using arithmetical and geometrical averages:
$$A_{3} = \frac{a_{1}+a_{2}+a_{3}}{3},$$
... | Since $a > 0$, $b > 0$, and $c > 0$, we have
$$ 0 < a +b < a+b + c, $$
and therefore upon dividing both sides of this inequality by $(a+b)(a+b+c) > 0$, we get
$$
0 < \frac{1}{a+b+c} < \frac{1}{a+b},
$$
which implies
$$
\frac{1}{a+b} > \frac{1}{a+b+c}. \tag{1}
$$
Similarly, we have the inequalities
$$
\frac{1}{b+c} > \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the area of this function. I want to calculate the region bounded by $x^2-2xy+y^2+x+y=0$ and $x+y+2=0$.
I have drawn a sketch of this region in Geogebra:
I noticed I can use $\iint1dD$ to solve for the integral. I can apply the mapping: $(x^2-2xy+y^2)=(x-y)^2=u$, and $x+y=v$, so $x^2-2xy+y^2+x+y=u+v=0$ and $... | The answer is: $\frac{8\sqrt{2}}{3}$.
Explanation:
We want to calculate the area of the region $D$ bounded by $$x^{2}-2xy+y^{2}+x+y=0\quad \text{and}\quad x+y+2=0.$$
The area of the region $D$ is given by $$\color{red}{\boxed{A(D)=\iint_{D}{\rm d}A}}.$$
Setting the change of variable,
$$u\longmapsto u(x,y)=(x-y)^{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4331437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solution of $\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$ Find the solution of ${\left( {{x^2} - 7x + 11} \right)^{\left( {{x^2} - 13x + 42} \right)}} = 1$
My approach is as follow $\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$
Case 1: $f\left( x \right) = 1;x = 2,5$
Case 2: $f\le... | $$\left ( x^2-7x+11 \right )^{\left ( x^2-13x+42 \right )}=1\Leftrightarrow \left ( x^2-7x+11 \right )^{\left ( x^2-13x+42 \right )} =\left ( x^2-7x+11 \right )^0$$
$$\left ( x^2-7x+11-1 \right )\left ( x^2-13x+42 \right )=0\Leftrightarrow \left ( x^2-7x+10 \right )\left ( x^2-13x+42 \right )=0$$
$$x^2-7x+10=0\Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Contour integral $\oint_{|z|=1}\frac{z^2\sin(1/z)}{z-2}dz$ How do I compute this integral?
$$\oint_{|z|=1}\frac{z^2\sin(1/z)}{z-2} \, dz$$
I tried substituting $1/z$ with $z$ and ended up with
$$\oint_{|z|=1}\frac{\sin(z)}{z^3(1-2z)} \, dz$$
At this point I thought of using the residue theorem and got $2i\pi(2-4\sin(\f... | Where are you getting $\pi i/6$? I am also getting a residue of
$$2-4\sin\left(\frac{1}{2}\right)$$
We can see this directly using power series centered at zero:
$$\frac{z^2\sin(1/z)}{z-2}=$$
$$z^2\cdot\left(\frac{1}{1!z}-\frac{1}{3!z^3}+\cdots\right) \cdot\left(\frac{-1/2}{1-z/2}\right)=$$
$$z^2\cdot\left(\frac{1}{1!z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $1If $a$, $b$, $c$ are three positive real numbers such that:
$a+b>c$
$b+c>a$
$c+a>b$
$a+b+c=2$
$a$, $b$ and $c$ might be or might not be equal in value
Show that:
$$1<ab+bc+ca-abc<\frac{28}{27}$$
From the first half of the question, I realised that $a$, $b$ and $c$ are the sides of a triangle whose perimeter is ... | Proof:
Let
$$x = \frac{a + b - c}{2} > 0, \quad y = \frac{b + c - a}{2} > 0, \quad z = \frac{c + a - b}{2} > 0.$$
Correspondingly, $a = z + x, b = x + y, c = y + z$ (the so-called Ravi's substitution).
We have $a + b + c = 2(x + y + z)$. Thus, $x + y + z = 1$.
Also, we have
\begin{align*}
&ab + bc + ca - abc \\
=\,& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how this simplification was done in the integral? I'm stuck at simplification
Here is the Integral ,
$$I=\int\frac{x^2+x+1}{\sqrt{x^2+2x+3}} \ dx$$
To solve this ,
First let's Substitute, we got
$$x^2+2x+3=t^2$$
$$\implies {x}=\sqrt{(t^2-2)}-1$$
$$\implies dx=\frac{t}{\sqrt{t^2-2}}dt$$
Putting this back into the Integr... | If you let $x=\sqrt{t^2-2}-1,$ then $\mathrm{d}x=\frac{t}{\sqrt{t^2-2}}\,\mathrm{d}t,$ $x^2+2x+3=t^2,$ and $x^2+x+1=x^2+2x+3-x-2=t^2-\left(\sqrt{t^2-2}-1\right)-2=t^2-1-\sqrt{t^2-2}.$ Therefore, $$\int\frac{x^2+x+1}{\sqrt{x^2+2x+3}}\,\mathrm{d}x=\int\frac{t^2-1-\sqrt{t^2-2}}{\sqrt{t^2}}\frac{t}{\sqrt{t^2-2}}\,\mathrm{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4355890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Identity for $\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d}$ I am looking for an identity for the following derivative:
$$\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d},$$
where $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$ are column vectors.
My approach to expand ... | I think I arrive at the same conclusion but in slightly different way.
Using differentials
$$
d\mathbf{f} = (d\mathbf{a})^T\mathbf{b} \mathbf{c}+
\mathbf{a}^T d\mathbf{b} \mathbf{c}+
\mathbf{a}^T\mathbf{b} d\mathbf{c}
$$
Now using
$d\mathbf{a}=\mathbf{J}_a d\mathbf{x}$... and rearranging terms,
you will find that the J... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdo... | you have that:
$$(x-x_1)(x-x_2)=0$$
since these are the roots, so:
$$x^2-(x_1+x_2)x+x_1x_2=0$$
so:
$$x_1+x_2=a\qquad x_1x_2=a-1$$
now we can say:
$$(x_1+x_2)^2=x_1^2+2x_1x_2+x_2^2=(x_1^2+x_2^2)+2(a-1)=a^2$$
which finally gives us:
$$y=x_1^2+x_2^2=a^2-2a+2$$
now to find the minimum differentiate:
$$\frac{dy}{da}=2a-2$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove $\left(\sqrt{x}+\sqrt{y}\right)^2\ge 2\sqrt{2\left(x+y\right)\sqrt{xy}}$, with $x,\:y\in \mathbb{R}$ and $x\ne 0$ My thinking:
Starting with the basic fact $\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0$ which is trivial.
$\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0\:$
$\rightarrow \:x^2-4\sqrt{yx}\sqrt{x}+6xy-4\sqrt{xy}\sqrt{... | The given inequality is equivalent to the $t\ge0$ case of
$$(t+1)^4\ge8t(t^2+1)\tag1$$
(Let $t=\sqrt{y/x}$.) Taking the difference in (1),
$$(t+1)^4-8t^3-8t=(t-1)^4,$$
which proves (1) for arbitrary $t.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve the Diophantine equation $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$? How can one solve the following Diophantine equation in $x, y \in \mathbb{Z}$?
$$x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$$
| Since this is a Diophantine equation, we only seek for integer solutions. Notice that $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$ implies $(y-x^2-x+3)(y+x^2+x-1)=-9$, we only have $6$ cases: $((y-x^2-x+3),(y+x^2+x-1))$ must be one of $(1,-9),(9,-1),(3,-3),(-1,9),(-9,1),(-3,3)$. However, after checking all these cases there are no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rolling three dice and getting exactly two 4s - am I doing this right?
Toss $3$ dice. What's the probability that a "4" will come up exactly
twice?
Since we need two of the dice to come up 4, and the third can be any number except 4 we can have the following options:
[4, 4, $\neq$ 4], [4, $\neq$ 4, 4], or [$\neq$ 4, ... | You are very close, but in order to solve it you need to use the binomial distribution: $$\binom{\textrm{No. of attempts}}{\textrm{number of successes}} \cdot (\textrm{success chance})^{\textrm{No. of successes}} \cdot (1 - \textrm{success chance})^{\textrm{No. of failures}}.$$ In your case, it is
$$\binom{3}{2} \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$ Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sq... | You can do pairwise comparisons. For instance, to compare $\sqrt[4]4$ with $\sqrt[5]5$, you only need to compute $4^5$ and $5^4$. (And you can rule out $\sqrt[6]6$ easily by comparing it with $\sqrt[3]3$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Using De Moivre's theorem to solve $(z−3+2i)^4 = z^4$ What are all the solutions to: $(z−3+2i)^4 = z^4$?
I know I have to use De Moivre's theorem which states:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$
| Specifically addressing your edit, $\sqrt{20 + 48i} = 2 \sqrt{5 + 12i}$.
Now if $\sqrt{5 + 12i} = a + bi$, then squaring both sides, $5 + 12i = (a^2 - b^2) + 2abi$, or that $5 = a^2 - b^2, 12 = 2ab$. Since $b = \frac{6}{a}$, you can form a quadratic with this information, which gives $a + bi = ±(3 + 2i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\frac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\frac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}$ Prove that $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\dfrac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}, a>0, x>0,x\ne a.$$ My try $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}... | Just to make things easier to type let $c =\sqrt[4]a$ and $y=\sqrt[4]x$ so we are asked to prove
$\frac {c^4 - y^4}{c^2-y^2} -(\frac {c^4 + cy^3}{c^2 + cy} - cy)= 2cy$.
so lets do that
$\frac {c^4 - y^4}{c^2-y^2} -(\frac {c^4 + cy^3}{c^2 + cy} - cy)=$
$\frac {(c^2-y^2)(c^2+y^2)}{c^2-y^2} -\frac {c(c^3+ y^3)}{c(c+y)} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is ... | The two given equations imply that
\begin{align}
&(1-\tan x)^2=\frac{\sin^2y}{1-\sin^2y},\tag{1}\\
&\cos x=\frac{1}{\sqrt{2}\sin y}.\tag{2}
\end{align}
Suppose both $x$ lies inside the first quadrant. From $(2)$ we see that $y$ must lie inside the first or the second quadrant. Also,
\begin{align}
\tan x&=\frac{\sqrt{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that for all $n$ there exist positive integers $x_1, ..., x_n$ such that $\frac{1}{x_1} + \frac{1}{2x_2} + ... + \frac{1}{nx_n} = 1$ Prove that for all $n$ there exist positive integers $x_1, ..., x_n$ such that $\frac{1}{x_1} + \frac{1}{2x_2} + ... + \frac{1}{nx_n} = 1$
I actually managed to write a short proof,... | I think the idea is correct, but you do not need induction.
You suggested that $\frac{1}{2*1}+\frac{1}{3*2}+...+ \frac{1}{n(n-1)}+\frac{1}{n}=1$. This is true: the sum is $\frac11-\frac12+\frac12-\frac13+....+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve Summation of Catalan Convolution? Can someone help me solve this?
$$
\sum_{k=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{2 k+4} 4\left(\begin{array}{c}
2 k+3 \\
k
\end{array}\right)}{k+4}
$$
converges.
Wolfram alpha says it converges to $1$.
But actually, I have no idea, how to solve it.
| Using a version of the binomial power series
$$\sum_{k\geq 0}{2k+\alpha \choose k}z^k=\frac 1 {\sqrt{1-4z}}\left(\frac {1-\sqrt{1-4z}}{2z}\right)^\alpha$$
With $\alpha=3$:
$$\sum_{k\geq 0}{2k+3\choose k}z^k=\frac 1 {\sqrt{1-4z}}\left(\frac {1-\sqrt{1-4z}}{2z}\right)^3$$
Consequently
$$\sum_{k\geq 0}{2k+3\choose k}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Can $(x^2+a^2)^{-\frac{3}{2}}$ be integrated without using trigonometric substitutions? I know that
$$
\int\frac{1}{(x^2+a^2)^{\frac{3}{2}}}\,dx
$$
can be solved using a trigonometric substitution, but is there a trick to calculate this integral without using trigonometric substitutions? I think something in the same s... | A very simple approach would be:-
$$\int\frac{1}{(x^2+a^2)^{\frac{3}{2}}}\,dx=\int\frac{1}{(x^3)(1+\frac{a^2}{x^2})^{\frac{3}{2}}}\,dx$$
Substituting $(1+\frac{a^2}{x^2})$ as $t^2$,we get :
$$\int\frac{-1}{a^2t^2}dt=\frac{1}{a^2t} +c=\frac{x}{(a^2){(x^2+a^2)^{\frac{1}{2}}}} +c $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Is there an elementary proof for $\int_{0}^{\infty} \frac{d x}{x^{n}+1}= \frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) $ When I first encounter the integral
$$\int_{0}^{\infty} \frac{d x}{x^{n}+1},$$
I am trying to resolve the integrand into partial fractions. Then I found it is very tedious and complicated and look fo... | Here is an elementary proof using partial fractions, avoiding complex analysis, infinite series, or special functions commonly employed to evaluate this integral
\begin{align}
& \int_{0}^{\infty} \frac{1}{1+x^{n}} dx\\
=&\int_{0}^{1} \frac{1}{1+x^{n}}dx+ \int_{1}^{\infty} \frac{1}{1+x^{n}}\overset{x\to 1/x}{dx }
= \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why is $ 2 \cos(\frac{2π}{5}) + \left(2 \cos(\frac{2π}{5})\right)^2 = 1$? I was able to prove $(a)$, but I have trouble with $(b).$ Can somebody give me a hint?
Let $ \alpha := e^{
\frac{2πi}{5}} \in \mathbb{C}$, $K:=\mathbb{Q}(\alpha) $ and $f(X) := X^4 + X^3 + X^2 + X + 1$
(a) Prove that $f(\alpha) = 0$.
(b) Let $\... | Note that $1,\alpha,\alpha^2,\alpha^3,\alpha^4$ are roots of $x^5-1=0$
Now we have $\alpha\bar{\alpha}=1\implies\alpha\bar{\alpha}=\alpha^5\implies\bar{\alpha}=\alpha^4$
Therefore $\beta=\alpha+\alpha^4$ and $\beta+\beta^2=\alpha+\alpha^4+\alpha^2+\alpha^8+2\alpha^5=2+\alpha+\alpha^2+\alpha^3+\alpha^4=2-1=\boxed{1}$
He... | {
"language": "en",
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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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In a triangle ABC, if certain areas are equal then P is its centroid Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at
$D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centr... | This is not a trivial geometry problem... Let $x=AF/FB, y=BD/DC, z=CE/EA$, then we have $xyz=1$. Furthermore, we can calculate that the area of $AFB$ is $\frac{x}{(1+z+yz)(1+x)}$. Similarly we can calculate two others, so, we have
$$\frac{x}{(1+z+xz)(1+x)}=\frac{y}{(1+x+yx)(1+y)}=\frac{z}{(1+y+zy)(1+z)}$$
Notice that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to compute $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}}$ Trying to prove $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}} = 2e^{-1/2}$
I tried using ratio-root criteria with $a_n= \frac{1}{n^n} \l... | This result for the logarithm that you've got can be calculated further. We have
\begin{align} & -\log(n) + \sum_{k=1}^n \frac{k}{n^2} \log(k^2+n^2) = \\
&= -\log(n) + \sum_{k=1}^n \frac{k}{n^2} \Big(2\log n + \log(1+\frac{k^2}{n^2})\Big) = \\
&= -\log(n) + \frac{2\log n}{n^2}\frac{n(n+1)}{2} + \frac{1}{n} \sum_{k=1}^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving that $\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}$ when $n$ is very large This is an example from Mathematical Methods in the Physical Sciences, 3e, by Mary L. Boas.
My question is,
\begin{equation}
\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}
\end{equation}
can also be written as,
\begin... | You can also apply Lagrange's theorem to $f(x)=\frac{1}{x^2}$ in the interval $[n, n+1]$:
$$
f(n+1)- f(n) = f'(\xi_n)(n+1-n), \quad \xi_n \in (n, n+1),
$$
i.e.,
$$
\frac{1}{(n+1)^2} - \frac{1}{n^2} = -\frac{2}{\xi_n^3}, \quad \xi_n \in (n,n+1)
$$
or
$$
\frac{2}{(n+1)^3} \leq \frac{1}{n^2}-\frac{1}{(n+1)^2}\leq \frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
From homogeneous to non-homogeneous linear recurrence relation I'm trying to do the following exercise:
Find a non-homogeneous recurrence relation for the sequence whose general term is
$$a_n = \frac{1}{2}3^n - \frac{2}{5} 7^n$$
From this expression we can obtain the roots of the characteristic polynomial $P(x)$, which... | Just start with your favorite term containing $a_n$ and $a_{n-1}$, say
$a_n - a_{n-1}$, calculate the difference, here
$$ a_n - a_{n-1} = \frac 12 3^n - \frac 25 7^n - \frac 12 3^{n-1} + \frac 25 7^{n-1} $$
giving you the inhomogeneous recurrence
$$ a_n = a_{n-1} + \frac 12 3^n - \frac 25 7^n - \frac 12 3^{n-1} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluat... | If you like hypergeometric functions, there is an antiderivative
$$J_n=\int\frac{\log(x)}{\left(x^{2}+a\right)^{n}}\,d x=\frac x {a^n} \left( \,
_2F_1\left(\frac{1}{2},n;\frac{3}{2};-\frac{x^2}{a}\right) \log(x)-\,
_3F_2\left(\frac{1}{2},\frac{1}{2},n;\frac{3}{2},\frac{3}{2};-\frac{x^2}{a}\right)
\right)$$ So
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
$\text { Show that } \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x \text { if }-1 \leq x \leq-\frac{1}{\sqrt{2}} $
Show that
$$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x$$ if $-1 \leq x \leq-\frac{1}{\sqrt{2}}.$
I tried solving this question a lot but I’m unable to. My answer co... | hint
Let
$$f(x)=\arcsin(2x\sqrt{1-x^2})-2\arccos(x)$$
assuming $ f $ is differentiable at $ [-1,-\frac{1}{\sqrt{2}}] $,
prove that
$$f'(x)=0$$
So
$$f(x)=Cte=f(-1)=-2\pi$$
Other approach
Put $$x=\cos(\frac{\theta}{2})$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can we determine the sum of a series where each term is a product of two integers using method of differences? I have been trying to find out the sum of a series up to the $n^{th}$ term but failed.$$S_n=1 \cdot 3+2 \cdot 4+3 \cdot 5+4 \cdot 6+ \ldots + n(n+2)$$
my work:
\begin{align*}
S_n & =\frac{n(n+2)(n+4)-(n-2)... | $S_n$ will be a cubic function of $n$, because there are $n$ terms of $2nd$ degree (quadratic). Then, we can write the formula for $S_n$ as $S_n = An^3 + Bn^2 + Cn + D$.
When $n=0$, $S_n = 0\cdot2 = 0 \Rightarrow D = 0$
When we plug in $n=1,2,3$, we get this system of equations:
\begin{align}
\tag{1}
A+B+C = 3
\newline... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the sum of natural numbers that have $5$ digits or less, and all of the digits are distinct? $1+2+3+\dots+7+8+9+10+12+13+\dots+96+97+98+102+103+104+\dots+985+986+987+1023+1024+1025+\dots+9874+9875+9876+10234+10235+10236+\dots+98763+98764+98765=$
The only thing I can do is to evaluate a (bad) upper bound by eval... | This is the approach of Empy2 (which I believe didn’t deserve any downvotes) spelled out.
Consider the numbers for each length separately.
First allow a leading zero. Then there are $\frac{10!}{(10-k)!}$ different numbers with $k$ digits, all distinct, and they sum to $10^k-1$ in pairs, so their sum is $\frac{10^k-1}2\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What am I doing wrong to finding the perpendicular distance between a point and a plane? From this post How to prove that a plane is a tangent plane to a sphere?, I went and attempted to show that $4x+3z+29=0$ is tangent to the sphere $(x+1)^2+(y-3)^2+z^2=25$. However I keep getting a different distance from the plane ... | The $\sqrt{1^2+3^2+7^2}$ should be $\sqrt{4^2+3^2}$.
The length of the projection of $\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}$ onto $\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}$ is given by
$$\left|\frac{\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}\cdot\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}}{\color{red}{\sqrt{4^2+3^2}}}\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4396973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using diagonalisation of a symmetric on quadratic form Let $f (x, y) = 7x_2 + 4y_2 + 4xy$. Use the diagonalisation of a symmetric matrix to write this quadratic form in the form $λ_1u_2^2 +λ_2v_2^2$, with u and v linear combinations of x and y.
Here's what I have tried:
step 1. get the symmetric matrix
$$f (x, y) = 7x_... | This is a longer comment rather than an full answer:
If you want to "diagonalize" a quadratic form, completing the square is simpler than your approach via the eigendecomposition. Note that
$$ f (x, y) = 7x^2 + 4y^2 + 4xy = 6 x^2 + (2 y +x)^2 = 6 u^2 + v^2$$
with
$$u = x, \qquad v = 2 y +x$$
is a diagonalization in ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality with real exponents I had the following problem on a past homework asignment:
For $x,p \in \mathbb{R}^+$
$$\frac{p}{(1+x)^{p+1}}<\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
So, I think one way to show this for $1\leq p$ would be with rules for real exponents for real exponents (since $x^{-1} >-1$):
$$(1+\frac{1}{x})... | Let $f(x)=\dfrac{1}{x^p}$, then $f'(x)=-\dfrac{p}{x^{p+1}}$. Your inequality becomes $$f'(x+1) \ge f(x+1)-f(x) = \dfrac{f(x+1)-f(x)}{(x+1)-x}$$
This follows easily from that $f$ is convex when $p>0$.
Edit: Since $f$ is convex, then for any $0<t<1$, $$f(x+t)=f((1-t)x+t(x+1))<(1-t)f(x)+tf(x+1) \implies \dfrac{f(x+1)-f(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $\displaystyle \int \limits _{0}^{\infty} \dfrac{\mathrm dx}{\sqrt{x^n+a} + \sqrt{x^n+b}}$ I have a question which asks to find the value of:
$\displaystyle \tag*{} \int \limits _{0}^{\infty} \dfrac{\mathrm dx}{\sqrt{x^n+a} + \sqrt{x^n+b}}$
Where, $a,b >0$ and $n >2$
I tried to rationalize the denomin... | We'll evaluate the following auxiliary integral:
$$\int_0^\infty \sqrt{x^n + \alpha} - x^{\frac{n}{2}} \, \mathrm{d}x = \frac{\alpha^{\frac1n + \frac12} }{(n+2)\sqrt{\pi}}\Gamma\left(\frac{1}{2} - \frac{1}{n} \right)\Gamma\left( \frac{1}{n} \right)\qquad \text{for} \quad \alpha >0, \, n>2$$
Proof: Taking the substitu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the minimum of $a^4+2b^4+3c^4$ Suppose that $a, b,$ and $c$
are positive real numbers satisfying
$a+b+c=3$. Find the minimum of $$a^4+2b^4+3c^4$$
We know $f(x)=x^4$ is convex on the positive reals so by Jensen's Inequality, we have $$a^4+b^4+c^4\ge \frac{1}{27}$$
Hence $$a^4+2b^4+3c^4\ge \frac{1}{27}+b^4+2c^4$$
W... | In fact if you use Lagrange multipliers you get:
$\vec{\nabla}(a^4+2b^4+3c^4)=(4a^3,8b^3,12c^3)\ \ \propto\ \ \vec{\nabla}(a+b+c-3)=(1,1,1)$ which gives
$$a^3=2b^3=3c^3$$
same as Macavity $a^3:b^3:c^3=1:\frac 12:\frac 13$ ratio presentation.
Since $x\mapsto x^3\ $ bijective this is also $\ a=\sqrt[3]{2}\,b=\sqrt[3]{3}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to integrate $\int \frac{dx}{x\sqrt{x^4-1}}$? $$\int \dfrac{dx}{x\sqrt{x^4-1}}$$
I need to solve this integration.
I solved and got $\dfrac12\tan^{-1}(\sqrt{x^4-1}) + C$, however the answer given in my textbook is $\dfrac12\sec^{-1}(x^2) + C$
How can I prove that both quantities are equal? Is there something wrong ... | I tell my students that inverse trig functions are angles. So if you write
$$\tan^{-1}\sqrt{x^4-1} = \theta,$$
then
$$\tan\theta = \sqrt{x^4-1}.$$
A right triangle that tells this story has $\theta$ as one angle, $\sqrt{x^4-1}$ as the opposite side and $1$ as the adjacent side. Using Pythagorean theorem we can work o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove $8abc(a+b+c)^3\leq27(a^2+bc)(b^2+ca)(c^2+ab)$ Let $a\geq0, b\geq0,c\geq0$. Prove that:
$$8abc(a+b+c)^3\leq27(a^2+bc)(b^2+ca)(c^2+ab).$$
I tried to prove this inequality using only the inequality between the arithmetic mean and the geometric mean.
Does anyone succeed? Thanks!
| Without full expanding:
If $abc = 0$, clearly the inequality is true.
In the following, WLOG, assume that $a \ge b \ge c > 0$.
We split into two cases:
Case 1: $b^2 \ge ca$
Using Holder, we have
\begin{align*}
\mathrm{RHS} &\ge 27\left(\sqrt[3]{bc \cdot ca \cdot c^2} + \sqrt[3]{a^2\cdot b^2\cdot ab}\right)^3\\
&= 27\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to? If $S_n=\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to?
So, the most obvious course of action in my mind is to find a closed form for the partial summations, but alas, this task eludes me. I started doing this by hand.... | this is a telescopic series, this means that each term cancel part of other term, in your case you first need to apply partial fraction on the given expression:
$\sum_{k=1}^n \frac{1}{k(k+1)} =\sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} $
this means that the first terms are:
$ k=1: 1-\frac{1}{2} $
$ k=2: \frac{1}{2}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Directional derivatives of $f(x,y) := \frac{2x^2y}{x^4 + y^2}$ in $(0,0)$ Given is the function $f: \mathbb{R^2} \to \mathbb{R}$ with
$ f(x,y) := \frac{2x^2y}{x^4 + y^2} \text{ for all } (x,y) \neq (0,0) \text{ and } 0 \text{ for all } (x,y) = (0,0)$
How can one prove that in all (bilateral) directional derivatives of... | According to the definition of directional derivative at the origin, if we let $v = (a,b)$, we get that:
\begin{align*}
D_{v}f(0,0) & = \lim_{t\to 0}\frac{f((0,0) + t(a,b)) - f(0,0)}{t}\\\\
& = \lim_{t\to 0}\frac{2t^{3}a^{2}b}{t^{3}(t^{2}a^{4} + b^{2})}\\\\
& = \lim_{t\to 0}\frac{2a^{2}b}{t^{2}a^{4} + b^{2}}\\\\
& = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Effective method to solve $\frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $ I want to know, is there an easy method to solve below equation $$\frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $$ I tried it by plotting and find the so... | $$ \frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $$
Use componendo-dividendo:
$$ \frac{\sqrt[3]{1+x} -\sqrt[3]{1-x} + (\sqrt[3]{1+x} +\sqrt[3]{1-x})}{\sqrt[3]{1+x} -\sqrt[3]{1-x} - (\sqrt[3]{1+x} +\sqrt[3]{1-x})} = \frac{x(x^2+3) + 3x^2+1}{x(x^2+3) -(3x^2+1)} $$
$$ \frac{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove by induction $\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- \frac {1}{2^n} $ I'm trying to solve this problem about mathematical induction but every time I try to solve it I end up with an incorrect answer.
The problem is as follows:
Prove by induction : $$\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- ... | You went wrong when you combined the fractions. You should have $$1- \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{2^{k+1} - 2^k}{2^{k+1}2^k} = 1 - \frac{2 - 1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Finding the number of $2\times2$ matrices whose sum of element is zero If the number of non skew symmetric matrices $A = [a_{ji}],$ where $a_{ji}\in\{-2,-1,0,1,2\} $ and the sum of elements is zero then find the number of such matrices.
I tried setting up a recursion using (-n....0....n) and adding -n-1 and n+1 but not... | split into cases based on the number of zeros in your 4 elements, there can only be 0, 1, or 2.
The biggest case will be for 0 zeros, for which the elements would be from one of 3 sets,
{1, 1, -1, -1}, {2, 2, -2, -2} or {1, -1, 2, -2}
To find the number of matrices you just need the permutations of each of the sets
$$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the following definite integral exist? I encounter a problem in which I would need to deal with the folloing definite integral
$$I(t)=\int_{0}^{\infty} \mathrm dp \frac{p^2}{\omega^5} \sin^2\left(\frac{\omega t}{2}\right)$$
in which $$\omega=\sqrt{m^2+p^2}$$
where m is just some positive constant, t is also a pa... | With Mathematica I have:
$$\int_0^{\infty } \frac{p^2 \sin ^2\left(\frac{1}{2} \sqrt{m^2+p^2} t\right)}{\left(m^2+p^2\right)^{5/2}} \, dp=\int_0^{\infty } \left(\frac{p^2}{2 \left(m^2+p^2\right)^{5/2}}-\frac{p^2 \cos \left(\sqrt{m^2+p^2} t\right)}{2 \left(m^2+p^2\right)^{5/2}}\right) \, dp=\\\frac{1}{6 m^2}-\frac{\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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MCQ about the product of $4$ consecutive odd numbers The product of four consecutive odd numbers must be …
(A) A multiple of 3, but not necessarily of 9.
(B) A multiple of 5 .
(C) A multiple of 7.
(D) A multiple of 9.
(E) A multiple of 3×5×7×9= 945.
Let $n\in \mathbb{Z}$, then
$(2n-1)(2n+1)(2n+3)(2n-3)$
$= (... | Let $n$ be the first of the consecutive odd integers, with
$$f(n) = n(n+2)(n+4)(n+6)$$
If $n \equiv 0 \pmod{3}$, then $3 \mid f(n)$ (i.e., $f(n)$ is an integral multiple of $3$). If $n \equiv 1 \pmod{3}$, then $n + 2 \equiv 0 \pmod{3}$ so $3 \mid f(n)$. Finally, if $n \equiv 2 \pmod{3}$, then $n + 4 \equiv 0 \pmod{3}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combinatorial proof of $\sum_{k=0}^{n} k \binom{n+1}{k+1} n^{n-k} = n^{n+1}$ Show :
$$\sum_{k=0}^{n} k \binom{n+1}{k+1} n^{n-k} = n^{n+1}$$
for natural number $n$. I randomly discovered this identity, and managed to prove it using simple algebra. I tried a combinatorial proof of this, but it seems too difficult for me.... | $\underline{\text{Preliminary Results}}$
PR-1
$\displaystyle\sum_{i=0}^r \binom{r}{i}r^{r-i} = (1 + r)^r.$
This is directly from binomial expansion.
PR-2
$\displaystyle\sum_{k=0}^n \binom{n+1}{k+1}n^{n-k} = (1 + n)^{n+1} - n^{n+1}.$
Proof
First, re-index it to
$\displaystyle\sum_{k=1}^{n+1} \binom{n+1}{k}n^{n+1-k}.$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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If $P_n=\alpha^n+\beta^n\;, \alpha+\beta=1, \;\alpha \cdot \beta=-1,\;P_{n-1}=11,\; P_{n+1}=29$ Find $(P_n)^2,\;$ where $n\in \mathbb N$
If $P_n=\alpha^n+\beta^n\;, \alpha+\beta=1, \;\alpha \cdot \beta=-1,\;P_{n-1}=11,\; P_{n+1}=29$, $\alpha$ and $\beta$ are real numbers. Find $(P_n)^2,\;$ where $n\in \mathbb N$
My... | Since $P_n = \alpha^n + \beta^n $, then $P_n$ satisfies the difference equation
$ (E - \alpha)(E - \beta) P_n = 0 $
where $E$ is the advance operator, i.e. $E\left(P_n\right) = P_{n+1}$
Hence, multiplying the operators, we get the difference equation
$P_{n+2} - (\alpha + \beta) P_{n+1} + \alpha \beta P_{n} = 0 $
Substi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Double integration in polar coordinates: $\int_{-\pi /4}^{\pi /4} \int_0^{a \sqrt{2 \cos (2 \theta )}} \sqrt{2a^2-r^2} r dr d\theta $ This is exercise 14 in Section 9.3 of Serge Lang's Calculus of Several Variables.
Question statement: The base of a solid is the area of one loop in Exercise 13(b)* and the top is bounde... | You wrote
$$\int^{2a^2}_{4a^2\sin^22θ}
\frac{\sqrt u}{2}\,du=\color{red}{\frac{2a^3}{3}(\sqrt 2−4\sin^3θ)}
$$ in your arguments.
However, it should be $$\frac{2a^3}3(\sqrt 2−4\left|\sin^32θ\right|)
.$$
| {
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"url": "https://math.stackexchange.com/questions/4426895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem with $\int_0^{\infty} \frac{x^a}{(1-x)^2} \ \mathrm dx$ I want to evaluate:
$\displaystyle \tag*{} I = \int_0^{\infty} \frac{x^a}{(1-x)^2} \ \mathrm dx$
Where $-1<a<1$. I tried to split the integral at $x=1$
$\displaystyle \tag*{}I= \int_0^1 \frac{x^a}{(1-x)^2} \ \mathrm dx + \int_1^\infty \frac{x^a}{(1-x)^2... | New Answer. Let $0 < \varepsilon < 1$, and consider the integral
\begin{align*}
I_{\varepsilon} = \int_{0}^{1-\varepsilon} \frac{x^a}{(x-1)^2} \, \mathrm{d}x + \int_{1+\varepsilon}^{\infty} \frac{x^a}{(x-1)^2} \, \mathrm{d}x.
\end{align*}
To analyze the behavior of this integral, we consider the counter-clockwise conto... | {
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"timestamp": "2023-03-29T00:00:00",
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Suppose that $a_n\neq0$ for all $n\in\mathbb{N}$ and $L=\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$. Prove that $a_{n}\to 0$. If $L<1$, how do I prove that $\lim_{n\rightarrow\infty}a_n=0$ ?
I tried setting $L$ as $1-\delta$, because
$$L<1\Longleftrightarrow L=1-\delta,\, \delta>0$$
So, by definition:
$$... | Let $\epsilon > 0$. Let $N$ be large enough so that $\big| \frac {a_{n+1}} {a_n} \big| < 1 - \epsilon$ for all $n \ge N$. Then,
$$
\frac {1} {a_n} =
\frac {a_{n+1}} {a_n} \cdot
\frac {a_{n+2}} {a_{n+1}} \cdot
\frac {a_{n+3}} {a_{n+2}} \cdot
\frac {a_{n+4}} {a_{n+3}} \cdots
\frac {a_{n+k}} {a_{n+k-1}} \cdot
\frac {1} {a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4431609",
"timestamp": "2023-03-29T00:00:00",
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proving or disproving that $\sum_{k=0}^{N-1} {N-1\choose k}\epsilon^k - 1 < 1.01 N\epsilon$
Prove or disprove the following: if N is a positive integer and $\epsilon > 0$ is such that $N < \frac{.01}{\epsilon}$, then $\sum_{k=0}^{N-1} {N-1\choose k}\epsilon^k - 1 < 1.01 N\epsilon$.
Clearly from the question, we must ... | Fact 1: Let $c, x$ be real numbers such that $0 < c < 1/100$ and $1 \le x \le \frac{1}{100c}$. Then
$$(1 + c)^{x - 1} - 1 < \frac{101}{100} x c.$$
(The proof is given at the end.)
According to Fact 1, the desired result follows.
Proof of Fact 1:
Denote $q = 101/100$.
Taking logarithm, it suffices to prove that
$$f(x) ... | {
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If I define summation recursively, how do I prove formally that both of the following definitions are equivalent? First definition:
$\sum_{x=a}^{b} f(x)=f(b)+\sum_{x=a}^{b-1} f(x)$ if $a\leq b$, and equals $0$ otherwise
Second definition:
$\sum_{x=a}^{b} f(x)=f(a)+\sum_{x=a+1}^{b} f(x)$ if $a\leq b$, and equals $0$ oth... | First, we assume $a,b$ are integers (and so must be finite). If $a > b$, both sums yield $0$ (and thus agree), so assume $a \le b$.
A, intuitive direct approach is to apply the recursion. You have with the first definition:
$$
\begin{split}
\sum_{x=a}^b f(x)
&= f(a) + \sum_{x=a+1}^b f(x) \\
&= f(a) + f(a+1) + \sum_{... | {
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Inverse of a function related area problem
My solution is based on the following diagram which I will illustrate in details
Now ${A_1} = \frac{5}{4};{A_3} = 1$
$g\left( {2x} \right) = 2f\left( x \right)$
$g\left( 2 \right) = 2f\left( 1 \right) = 2 \times 1 = 2$
$g\left( 4 \right) = 2f\left( 2 \right)$
$\int\limits_1^... |
Lemma: If $f(x)$ is a continuous and increasing function and $a<b$, then:
$$ \int_{a}^{b} f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = b\, f(b)-a\, f(a). $$
To prove it, you just need to draw a picture or look at this Wikipedia page.
Now, following my comment, required integral = $ \displaystyle \int_{1}^8 x f'(x) dx... | {
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"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Maclaurin series of $1-\left(\frac{\sin x}{x}\right) ^{\frac{2}{5}}$ has all coefficients positive I had a similar problem posted here, and after experimentation with WA I noticed that the function $1-\left(\frac{\sin x}{x}\right) ^{\frac{2}{5}}$ has a Maclaurin expansion with all coefficients positive.
It seems to ho... | Using the same approach as earlier
$$1-\left(\frac{\sin (x)}{x}\right)^a= \frac{a x^2}{6}\Bigg[1+\frac 1{60} \sum_{n=1}^\infty (-1)^n\frac{P_n(a)}{b_n }x^{2n}\Bigg]$$ where the first $b_n$ are
$$\{1,126,15120,997920,16345929600,\cdots\}$$ and the first $P_n(a)$ are
$$\left(
\begin{array}{cc}
n & P_n(a) \\
1 & 5 a-2 \\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Simplify $\frac{1+\sin\alpha-2\sin^2\left(45^\circ-\frac{\alpha}{2}\right)}{4\cos\frac{\alpha}{2}}$ Simplify $$\dfrac{1+\sin\alpha-2\sin^2\left(45^\circ-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}$$
I am reading the solution of the authors and I really don't see how $$\dfrac{1+\sin\alpha-2\sin^2\left(45^\circ-\df... | We start from the classical formula
$$\cos(a+b)=$$
$$\cos(a)\cos(b)-\sin(a)\sin(b)$$
which gives
$$\cos(2a)=\cos^2(a)-\sin^2(a)$$
$$=1-2\sin^2(a)$$
or
$$2\sin^2(a)=1-\cos(2a)$$
thus
$$2\sin^2(45-\frac{\alpha}{2})=$$
$$1-\cos(90-\alpha)=1-\sin(\alpha)$$
your expression becomes
$$\frac{1+\sin(\alpha)-(1-\sin(\alpha))}{4\... | {
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"source": "stackexchange",
"question_score": "1",
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Solve $\frac{d^{2} y}{d x^{2}}+(\tan x-3 \cos x) \frac{d y}{d x}+2 y \cos ^{2} x=\cos ^{4} x$ Solve the differential equation $$\frac{d^{2} y}{d x^{2}}+(\tan x-3 \cos x) \frac{d y}{d x}+2 y \cos ^{2} x=\cos ^{4} x$$
My try:
I multiplied throughout with $\sec^2 x$, getting
$$\sec^{2} x \frac{d^{2} y}{d x}+\sec^{2} x \ta... | The equation has trigonometric coefficients and singularities at the roots of the cosine. On the intervals between these roots, $\sin x$ is an equivalent parametrization to $x$, so try if $y(x)=u(\sin x)$ gives a more simple equation.
$$
y(x)=u(\sin x)\\
y'(x)=\cos x\,u'(\sin x)\\
y''(x)=\cos^2x\,u''(\sin x)-\sin x\, u... | {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\... | What you have shown is that $|y| \leq \dfrac{2}{\sqrt{3}}$. While that does imply that $-\dfrac{2}{\sqrt{3}} \leq y \leq \dfrac{2}{\sqrt{3}}$, you cannot necessarily conclude that the range is $\left[-\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}\right]$ since you obtained that result by squaring both sides of the equation... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Calculate the power series centered in $z_0=1$ I did this, I don't know if is correct
$$
\begin{gathered}
h(z)=\left(\frac{z}{z+1}\right)^{2} \\
u=z-1,z_0=1 \\
h(z)=\left(\frac{z}{z+1}\right)^{2}=z^{2} \frac{1}{(z+1)^{2}}=z^{2} \frac{1}{(1-(-z))^{2}} \rightarrow \text { geometric series } \\
\sum_{n=0}^{\infty} u^{n}=\... | Comment on your approach:
It is not very clear how each step performs, in addition the changes of variables are strange in the part that tries to adjust to the convergence radius. We can use only geometric series if we first make use of partial fractions and then study each series representation for each function obta... | {
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"source": "stackexchange",
"question_score": "3",
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Inequality with interesting independent constants Let $b_1,\dots,b_{n-1}$ be integers satisfying $0 \le b_i \le n-i$ for each $i \in [n-1]$ such that $\sum_{i=1}^{n-1} b_i = \alpha \binom{n}{2}$ where $\alpha$ is constant strictly between $0$ and $1$. Prove that $\sum_{i=1}^{n-1} ib_i \ge c\alpha^2(n-1)\binom{n}{2}$ fo... | Trivially the constant $c=0$ will satisfy the inequality for any $\alpha$ and $n$. For the following, I will pick $c = \frac 14$.
For the proof, I will relax the requirement of $\alpha$ to $0\le \alpha \le 1$. Hence every $b_1, \ldots, b_{n-1}$ may satisfy $0\le b_i \le n-i$ without further global constraints.
I will a... | {
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"source": "stackexchange",
"question_score": "3",
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Find all polynomials $P$ such that $P(P(x))=P(x^n)+P(x)-1$
Find all polynomials $P$ of degree $n$ such that $P(P(x))=P(x^n)+P(x)-1$.
I stumbled upon this problem while solving functional equations in polynomials. However, even after trying most of the methods I know, I'm still yet to solve it. Here's what I tried:
*... | $\bullet\ $ If $n=0$, then $P=a$ for a constant $a$. Replacing in the equation, you get $a=a+a-1$, so $a=1$, so you get thet solution
$$P(X)=1$$
$\bullet\ $ If $n=1$, then $P(X)=aX+b$ for some constants $a\neq 0$ and $b$. Replacing in the equation, you get
$$a(aX+b)+b=aX+b+aX+b-1$$
so $a^2=2a$ and $ab+b=2b-1$. You get ... | {
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"source": "stackexchange",
"question_score": "2",
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Minimum possible sum of squares of two numbers with sum $k$? If the sum of two numbers is k. Find the minimum value of the sum of their squares.
This is my calculations so far.
a + b = k <-- google says that I should put x + y = k rather than a + b = k
a² + b² = y (but I don't know why s... | I focus on a calculus method given your tags on the question. We are given $a + b = k$. The sum of their squares is given by $a^2 + b^2$. Since $a + b = k$, we hat that $k - b = a$, hence
$$g(b) = a^2 + b^2 = (k-b)^2 + b^2 = k^2-2kb+b^2+b^2 = k^2 - 2kb + 2b^2\text{.}$$
$k$ is a fixed value that is known, so $b$ is the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\iiint\frac{xyz\,dx\,dy\,dz}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}$ over $\{(x,\,y,\,z)\in[0,\,\infty)^3\mid x^2+y^2+z^2\le R^2\}$
Evaluate $$\iiint\frac{xyz\,dx\,dy\,dz}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}$$ for $a>b>c>0$ where $\Omega:=\{(x,\,y,\,z)\in[0,\,\infty)^3\mid x^2+y^2+z^2\le R^2\}$
What do we do with the deno... | Integrate in spherical coordinates
\begin{align}
&\iiint\frac{xyz}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}\,dx\,dy\,dz\\
=& \int_0^{\pi/2} \int_0^{\pi/2}\int_0^R
\frac{r^4\sin \theta \cos\theta \sin^3\phi \cos\phi}
{\sqrt{ a^2\cos^2\theta\sin^2\phi+b^2\sin^2\theta\sin^2\phi +c^2\cos^2\phi}}dr d\theta d\phi\\
=& \int_0^{\pi/2} \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Integer part equation with sum and radicals
For a fixed $n$, natural number greather or equal than $2$, solve the following equation for real $x$.
$$\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor = (n - 1) \lfloor \sqrt{x^2 + 1} \rfloor + \sqrt{|x^2 - 1|}$$
I have noticed that all but the last terms i... | As you noted, we must have $|x^2-1|=a^2$ for some integer $a$.
Consider two cases . . .
Case $(1)$:$\;|x|\le 1$.
From $|x|\le 1$ we get $|x^2-1|=1-x^2$, hence
\begin{align*}
&
|x^2-1|=a^2
\\[4pt]
\implies\;\;&
1-x^2=a^2
\\[4pt]
\implies\;\;&
1-a^2=x^2
\\[4pt]
\implies\;\;&
1-a^2\ge 0
\\[4pt]
\implies\;\;&
a^2\le 1
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\int_{0}^{\infty} \frac{1}{x^{n}+1} d x?$ I first investigate the integral
$$\int_{0}^{\infty} \frac{1}{x^{6}+1} d x$$
using contour integration along the semicircle $\gamma=\gamma_{1} \cup \gamma_{2},$
$\textrm{ where }$ $$ \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} ... | Let $\frac{1}{t}=1+x^{n} \Rightarrow \textrm{ then } x= \left(\frac{1}{t}-1\right)^{\frac{1}{n}}\textrm{ and } \displaystyle dx= \frac{1}{n}\left(\frac{1}{t}-1\right)^{\frac{1}{n}-1}\left(-\frac{d t}{t^{2}}\right)$
$$
\begin{aligned}
I &=\int_{1}^{0} t\cdot\frac{1}{n}\left(\frac{1}{t}-1\right)^{\frac{1}{n}-1}\left(-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Series representation of GCD(x, y) While playing around with WolframAlpha, I discovered that apparently the GCD of any integers x, y is equal to the following sum:
$$
x + y - xy + 2\sum_{k = 1}^{x-1} \lfloor \frac{ky}{x}\rfloor
$$
I've personally verified that this formula works.
My question is: where does this even co... | Let $x,y \in \mathbb{N}_0$, and let $d= \gcd(x,y)$. Let also $\displaystyle{S = \sum_{k = 1}^{x-1} \left\lfloor \frac{ky}{x}\right\rfloor}$.
Then one has
\begin{align*} S = \sum_{k = 1}^{x-1} \left\lfloor \frac{ky}{x}\right\rfloor & = \sum_{k = 1}^{x-1} \left\lfloor \frac{(x-k)y}{x}\right\rfloor\\
& = \sum_{k = 1}^{x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Using change of coordinates to find the exact value of an integral Use an appropriate change of coordinates to find the exact value of the integral
$$\int_{-\sqrt{3}}^{\sqrt{3}}\int_{-\sqrt{3-x^2}}^{\sqrt{3-x^2}}\int_{-3+x^2+y^2}^{3-x^2-y^2}x^2dzdydx$$
My work so far:
$-3+x^2+y^2\leq z\leq 3-x^2-y^2$
$-\sqrt{3-x^2}\leq... | Yes, I confirm that your calculation is correct.
By symmetries, one can either use the cylindrical coordinates for the triple integral or the polar coordinates after integrating with the variable $z$.
The iterated triple integral can be written as
$$
J=\iint_D (6-2x^2-2y^2)x^2\;dxdy
$$
where $D=\{(x,y)\mid x^2+y^2\le ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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Why is the triangle inequality equivalent to $a^4+b^4+c^4\leq 2(a^2b^2+b^2c^2+c^2a^2)$? Consider the existential problem of a triangle with side lengths $a,b,c\geq0$. Such a triangle exists if and only if the three triangle inequalities
$$a+b\geq c,\quad b+c\geq a\quad\text{and}\quad c+a\geq b\tag{0}$$
are all satisfie... | The equations in (0) can be equivalently expressed as
$$a + b - c \ge 0$$
$$a - b + c \ge 0$$
$$-a + b + c \ge 0$$
Since the product of non-negative factors must also be non-negative, we must have:
$$(a + b - c)(a - b + c)(-a + b + c) \ge 0$$
Of course, this product would also be positive if exactly two of the original... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove or disprove $\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18$ for $\sum\limits_{i=1}^n x_i = \frac12$($x_i\ge 0, \forall i$)
Problem 1: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove or disprove that
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac1... | An incomplete approach:
$\sum\limits_{1\le i < j \le n} A_i A_j=\frac{1}{2}[(\sum_{k=1}^{n} A_k )^2=\sum_{k=1}^{n} A_k^2]=$
Using integral representation, we can write
$\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} =\sum\limits_{1\le i < j \le n} \int_{0}^{1} x_i x_j t^{-x_i-x_j} dt=\int_{0}^{1} dt\sum\limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Product of $n$ terms of sequence where the $n^{th}$ term is of the form $(x^{a^n}+1)$ While practicing from a book I found a product in the form $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)\cdot(x^{a^4}+1)$$ and was immediately curious if I could a formula to solve the product for $n$ terms, that is, a single formula ... | For the specific value of $a=1/2$ (as mentioned in the comments), one has
$$\prod_{j=1}^n \left(x^{2^{-j}}+1\right)=\frac{x-1}{x^{2^{-n}}-1},$$
which can be easily proven by induction on $n$: if this holds for $n$, then
$$\prod_{j=1}^{n+1}\left(x^{2^{-j}}+1\right)=\frac{x-1}{x^{2^{-n}}-1}\left(x^{2^{-(n+1)}}+1\right)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is there any method other than Feynman’s Integration Technique to find $ \int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x?$ We are going to find the formula, by Feynman’s Integration Technique, for
$$\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x,$$
where $a+c$ $\... | We can write the initial integral in the form
$\displaystyle I=\int_0^\frac{\pi}{2}\ln(a\cos^2x+b\sin^2x+c)dx=\int_0^\frac{\pi}{2}\ln\big((a-b)\cos^2x+b+c\big)dx\tag*{}$
Using $1+\tan^2x=\frac{1}{\cos^2x}$
$\displaystyle I=\int_0^\frac{\pi}{2}\ln\Big((a-b)+(b+c)(1+\tan^2x)\Big)dx+\int_0^\frac{\pi}{2}\ln(\cos^2x)dx\tag*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472100",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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$t^{n}-1 = \prod^{n-1}_{k=0}(t-C^k)$ where $C=e^{\frac{2\pi i}{n}}$ in the ring $\mathbb{C}[t]$. $t^{n}-1 = \prod^{n-1}_{k=0}(t-C^k)$ where $C=e^{\frac{2\pi i}{n}}$ in the ring $\mathbb{C}[t]$.
I wanted to make a proof by induction. I've checked it for $n = 3$:
\begin{equation}
\begin{split}
\prod^{3-1}_{k=0}(t-C^k) & ... | You are overcomplicating the question. Just note that the roots of $t^n-1$ are exactly $C^k$ with $k$ between $0$ and $n-1$. Just factorice the polynomial and get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2... | As said in the comments, your answer looks good. If you wanted to generalize to the case $D\in\mathbb{R}$, which would be natural since your integral is invariant under $D\to -D$, your answer would then be $\frac{\pi}{2|D|^3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$ How can I prove the following sequence converges?
$$\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$$
I tried everything.
Could not find any candidates for comparison test, and failed to find an upper bound.
Any hints will be app... | By Cauchy-Schwarz inequality:$$\begin{align}a_n&= \dfrac{1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots + \dfrac{1}{\sqrt{n}}}{n^2}\\&\le \dfrac{\sqrt{n}\sqrt{1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}}}{n^2}\\&\le \dfrac{\sqrt{\ln(n+1)}}{n^{\frac{3}{2}}}\\&\le \dfrac{(n+1)^c}{n^{\frac{3}{2}}}\\&= \left(1+\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summing even and odd terms of binomial coefficients, with increasing powers of 2 and alternating signs For a complex number $z$, let $(1+z)^{15}=a_{0}+a_{1} z+\cdots+a_{15} z^{15}$. The value of
$$
\left(a_{0}-4 a_{2}+16 a_{4}+\cdots-2^{14} a_{14}\right)^{2}+\left(2 a_{1}-8 a_{3}+32 a_{5} \cdots-2^{15} a_{15}\right)^{2... | Let
$$F= \left(\sum_{k=0}^{7} 2^{2 k}({ }^{15} C_{2 k})(-1)^k\right)^{2}+\left(\sum_{k=0}^{7} 2^{2 k+1}\left({ }^{15} C_{2 k+1}\right)(-1)^k\right)^{2} $$
From binomial expansion we have
$S_1=\sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k} x^{2k}=\frac{1}{2}[(1+ix)^{2n+1}+(1-ix)^{2n+1}]-(1+x^2)^{n+1/2} \cos (2n+1)t$,
$t=\tan^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int\frac{3x}{x^5+x^4+1}dx$ I have an integral which I solved. But, I am not sure whether my answer is right or not. The integral is $$\int\frac{3x}{x^5+x^4+1}dx$$ My answer $$3\left(-\dfrac{\displaystyle\sum_{\left\{Z:\>Z^3-Z+1=0\right\}}\frac{\left(2Z^2-3Z-1\right)\ln\left(\left|x-Z\right|\right)}{3Z^2-1}}... | Maple agrees with your answer, except that $\ln(|x-Z|)$ is replaced by $\ln(x-Z)$. Maple does this because Maple considers $x$ to be a complex variable. Two out of three of your zeros $Z$ are complex numbers anyway, so $\ln(|x-Z|)$ would be wrong even if $x$ is a real variable.
A simpler example. Factor $(x^2+1) =... | {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Numbers of $2\times 2$ matrices $A$ with elements from the set $\{-1, 0, 1\}$ such that $A^2 =I$ where $I$ is an identity matrix of order $2$. Numbers of $2\times 2$ matrices $A$ with elements from the set $\{-1, 0, 1\}$ such that $A^2 =I$ where $I$ is an identity matrix of order $2$.
My Approach: Let $\;A=\begin{bmat... | Some simplifications can be made. First, $\det(A)=ad-bc=\pm1$, so $ad=0$ and $bc=\pm1$ or $ad=\pm1$ and $bc=0$.
If $bc=0$, then since replacing $A$ with $A^t$ doesn't affect anything, we might as well assume $c=0$. If $ad=-1$ then all possibilities for $b$ work ($A$ has eigenvalues $\pm1$ so is conjugate to $\begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Solve $\frac{dx}{dt} = \frac{1}{y}$ and $\frac{dy}{dt} = \frac{1}{x}$ by integrable combinations method? I have a such system to solve using integrable combinations method:
$$
\frac{dx}{dt} = \frac{1}{y}
$$
$$
\frac{dy}{dt} = \frac{1}{x}
$$
And the right answer for it is:
$$
C_1 x^2 = 2t + C_2
$$
$$
y^2 = C_1 (2t + C_2... | \begin{align*}
\frac{dx}{dt} & =\frac{1}{y}\tag{1}\\
\frac{dy}{dt} & =\frac{1}{x}\tag{2}
\end{align*}
From (1) $dt=ydx$ and from (2) $dt=xdy$. Hence $ydx=xdy$ or $ydx-xdy=0$. But
$d\left( \frac{x}{y}\right) =\frac{ydx-x dy}{y^{2}}$. This shows that
$d\left( \frac{x}{y}\right) =0$ or $\frac{x}{y}=c$. Where $c$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all primes x such that f(x) is also a prime number. There is a function
\begin{align*}
f(x)=x^3 + x^2 +11x +2 \\
\end{align*}
Find all prime $x$ such that $f(x)$ is also a prime number.
I found that this is satisfied with an x value of 3 then the function is equal to 71, so both are primes, but I am unsure how to ... | Mod 3: $\mod 3$ we know by Fermat's little theorem $x^3 \equiv x\pmod 2$ and so $f(x) = x^3 + x^2 + 11x + 2 \equiv x + x^2 -x -1 \equiv x^2 - 1\equiv (x+1)(x-1)\pmod 3$.
What does this tell us?
It tells us that if $x$ is divisible by $3$ then $f(x)\equiv -1 \pmod 3$. But if $x$ is not divisible by $3$ (i.e. if $x \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Evaluating $\int_{-\infty}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^n \ \mathrm{d}x$ I am trying to find a general form of
$$I_n = \int_{-\infty}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^n \ \mathrm{d}x$$
as well as methods of solving it (preferably elementary). This is easy enough to solve for in... | A systemic procedure is to utilize the reduction formula
$$\int_0^\infty \frac{x^{2m}}{(1+x^2+x^4)^n}dx =I_{m,n}=-2I_{m-1,n}+ \frac{4n-2m-3}{2(n-1)}I_{m-1, n-1}$$
Then, given the initial values
$$I_{0,1}=I_{1,1}=\frac\pi{2\sqrt3}, \>\>\>I_{0,2}=\frac\pi{3\sqrt3}, \>\>\> I_{0,3}=\frac{13\pi}{48\sqrt3}, \>\>\>\cdots$$
we... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Can this property of certain pythagorean triples in relation to their inner circle be generalized for other values of $n$? This question was raised in comments of
Is the $(3,4,5)$ triangle the only rectangular triangle with this property?
and I was suggested to ask it as a separate one.
First some notation, let's write... | Let $(a,b,c)$ be a primitive Pythagorean triple with $a<b<c$. Then $\{a,b\}=\{2mn,m^2-n^2\}$ and $c=m^2+n^2$ for some coprime integers $m>n>0$ of different parity. If $2mn<m^2-n^2$, then $c-b=2n^2$ is twice a square. Moreover, if $m>3n$, then $2mn<(2/3)m^2<(8/9)m^2<m^2-n^2$, so for every $n$ there exists $(a,b,c)$ such... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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A closed form for integral $\int_0^\infty \frac{\ln(1-e^{-\pi x})}{1+x^2} dx$ $$\int_0^\infty \frac{\ln(1-e^{-\pi x})}{1+x^2} dx$$
let $u=e^{-\pi x}$, the integral goes to:
$$\pi \int_0^1 \frac{1}{u}\cdot\frac{\ln(1-u)}{\pi^2+\ln^2(u)}du$$
This looks a little like Gregory's coefficient, but not the same, because the in... | Similar to the evaluation of $\int_{0}^{\infty} \operatorname{Li}_{2}(e^{-\pi x}) \arctan(x) \, \mathrm dx$ here, we have $$ \begin{align} \int_{0}^{\infty} \frac{\ln(1-e^{-\pi x})}{1+x^{2}} \, \mathrm dx &= -\int_{0}^{\infty} \frac{1}{1+x^{2}} \sum_{n=1}^{\infty} \frac{e^{- \pi n x}}{n} \, \mathrm dx\\ &= -\sum_{n=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
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How to find the sum of constants given the following systems of equations? Given that $q,r,s$, and $t$ are different constant values in the following systems of equations containing $a,b,c$, and $d$. Find the sum $q+r+s+t$.
$\frac{1}{qa+1} + \frac{1}{qb+1} + \frac{1}{qc+1}+ \frac{1}{qd+1} = 1$
$\frac{1}{ra+1} + \frac{1... | Begin by noting that $q,r,s,t$ are the four roots of the equation $$\frac{1}{ax+1}+ \frac{1}{bx+1} + \frac{1}{cx+1} + \frac{1}{dx+1} =1$$
Simplifying, we get $$(bx+1)(cx+1)(dx+1)+(ax+1)(bx+1)(cx+1)+(cx+1)(dx+1)(ax+1)+(dx+1)(ax+1)(bx+1)=(ax+1)(bx+1)(cx+1)(dx+1).$$ Now, we know that for a general cubic equation $ux^4+vx^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Fractions in Questions and Answers
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