Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Given a sequence $a_n =\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$, prove that $\lim_{n \to \infty} a_{n} = g$. The given Sequence is $a_n = \frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$.
I showed that the Sequence $a_n$ converges towards a Value $g = \frac{1}{3}$.
How do I determine for each $\epsilon > 0$ an $n_0$ so that: $... | What you have done is correct. You need $\frac {2n+8} {3n(3n-7)+15} <\epsilon$.
One way to achieve this is to make $3n-7 >\frac 1 {\epsilon}$. $\cdots$ (1).
Now we need $2n+8 <3n+15\epsilon$. For this we need
$n>8-15\epsilon$. Hence $n >8-15\epsilon$ and $n>\frac 1 3 (7+\frac 1 {\epsilon})$ is good enough. Choose $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Proving that $\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$ using geometry I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result.
$$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$
My Attempt:
Let ... | $\text{We are going to prove that}$
$\displaystyle \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}. \tag*{}$
$\textrm{Considering the roots of the polynomial equation }\\ \text{of degree 2m}$
$\displaystyle (z+1)^{2 m+1}-(z-1)^{2 m+1}=0 \cdots(*)\tag*{} $
$\displaystyle \left(\frac{z-1}{z+1}\right)^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove that $3^{4n-1}-2$ is always a multiple of 5. I am trying to prove the following statement: $f(n) =3^{4n-1}-2 $ is always a multiple of 5, for $n\in \mathbb Z^+$. Using proof by induction:
Base case: $f(1)=25$, which is a multiple of 5 and hence holds for $n=1$.
Assumption step: Assume $f(k) =3^{4k-1}-2 $ is a mul... | So we need to prove that \begin{equation}
5 | 3^{4 n-1}-2 \quad \forall n \in \mathbb Z^{+}
\end{equation}
For n=1, we have \begin{equation}
5 | 25
\end{equation}. We assume it is true for some $n\in\mathbb Z^+$
Now we prove \begin{equation}
\begin{array}{l}
{3^{4(n+1)-1}-2} \\
{3^{4 n+4-1}-2} \\
\end{array}
\end{equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Prove that $\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n + 2)2^{n-1}$. How can I prove the following identity?
$$\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n +
2)2^{n-1}$$
I thought about differentiating this:
$$(1 + x) ^ n = \sum_{k = 0} ^ {n} \binom{n}{k}x^k$$
and then evaluating it at $x =... | Another way.
Multiply
$(1+x)^n=\sum_{k=0}^n{n\choose k}x^k
$
by $x$
to get
$x(1+x)^{n}=\sum_{k=0}^n{n\choose k}x^{k+1}
$.
Now when you differentiate
you get
$\begin{array}\\
\sum_{k=0}^n(k+1){n\choose k}x^{k}
&=(x(1+x)^{n})'\\
&=(1+x)^n+xn(1+x)^{n-1}\\
&=(1+x)^{n-1}(1+x+nx))\\
&=(1+x)^{n-1}(1+x(n+1))\\
\end{array}
$
Se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Area bounded by a hyperbola Find the area "A" of the given graph
Answers should be in terms of a, b and h
My attempt:
(1) Look for an equation in terms of y (or maybe in terms of x since the region is bounded between $y=0$ and $y=h$?)
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \\
b^2x^2-a^2y^2 = a^2b^2 \\
a^2y^2 = b^2x^2-... | Integrate over $y$, it will be easier to work out.
$$A = 2a\int_0^h \sqrt{1+\frac{y^2}{b^2}}dy = 2ab \int_0^{h/b} \sqrt{1+u^2}du $$
The primitive of $\sqrt{1+u^2}$ can be found by substitution. I suggest $u=\sinh t$, then
$$\int \sqrt{1+u^2}du = \int \cosh^2 t dt = \frac{1}{4} \sinh(2t) + \frac{t}{2} = \frac{1}{2} u\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^{\infty } \frac{\tan ^{-1}\left(\sqrt{a^2+x^2}\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx$ J. Borwein's review on experimental mathematics gives the following
$$\int_0^{\infty } \frac{\tan ^{-1}\left(\sqrt{a^2+x^2}\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx=\frac{\pi \left(2 \tan ^{-1}\left(... | As suggested by Ali Shather in the comments, we can write the inverse tangent as an integral to obtain
\begin{align}
\int \limits_0^\infty \frac{\arctan\left(\sqrt{a^2+x^2}\right)}{(1+x^2)\sqrt{a^2+x^2}} \, \mathrm{d} x &= \int \limits_0^\infty \int \limits_0^1 \frac{\mathrm{d} y}{1+y^2 (a^2+x^2)} \,\frac{\mathrm{d} x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Finding the value of $\sum_{n=1}^\infty\frac{n^2}{(n+1)(n+2)(n+3)(n+4)}$ Problem_
Find the value of $$\sum_{n=1}^\infty\frac{n^2}{(n+1)(n+2)(n+3)(n+4)}$$
It seems like I have to use the partial sum in order to get the exact value. But making it into the partial fractions was not that easy to me because of the numerat... | First add and subtract 1 to the numerator
it becomes $\sum_{n=1}^\infty \frac{(n+4-5)}{(n+2)(n+3)(n+4)} + \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)(n+3)(n+4)}$
$=\sum_{n=1}^\infty\frac{1}{(n+2)(n+3)} -\sum_{n=1}^\infty\frac{5}{(n+2)(n+3)(n+4)}+\sum_{n=1}^\infty\frac{1}{(n+1)(n+2)(n+3)(n+4)}$.(try evaluating this yourself)
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find circumradius of an octagon inscribable into a circle with side lengths $1,1,1,1,3\sqrt2,3\sqrt2,3\sqrt2$ and $3\sqrt2$ An octagon has side lengths $1,1,1,1,3\sqrt2,3\sqrt2,3\sqrt2$ and $3\sqrt2.$ What should be the length of its circumradius?
I tried solving it using elementary geometry, but that was of no use. I ... | The circumradius does not depend on the order of the sides. To show this, draw the isosceles triangle from the center to the vertices. The angle subtended by each of the $1$ sides is the same, as is the angle subtended by each of the $3\sqrt 2$ sides. These angles must sum to $\frac \pi 2$. Draw the octagon with th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Integral $\int \arcsin \left(\sqrt{\frac{x}{1-x}}\right)dx$ $$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$
I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)
I tried substituting $x$ for $t^2$:
$$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) ... | Substitute
$$u = \sqrt{\frac{x}{1-x}} \implies \frac{du}{dx} = \frac{\sqrt{x}}{(1-x)^{\frac{3}{2}}} + \frac{1}{2\sqrt{1-x}\sqrt{x}}$$
Then the integral become:
$$2 \int \frac{u\arcsin(u)}{(u^2 + 1)^2} \,du$$
Integrating by parts $\left(f = \arcsin(u), g'= \frac{u}{(u^2 + 1)^2} \implies f'= \frac{1}{\sqrt{1 - u^2}}, g ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Prove that $\frac{\sum \sqrt{xy}}{\sqrt{xyz}}-2(\sum \sqrt{x})\ge \sum \frac{\sqrt{xy}}{y}$ Let $x,y,z>0$ such that $x+y+z+2\sqrt{xyz}\le 1$. Prove that $$\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{\sqrt{xyz}}-2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\ge \frac{\sqrt{xy}}{y}+\frac{\sqrt{yz}}{z}+\frac{\sqrt{xz}}{x}$$
My try: L... | The inequality can be written as:
$$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq 2(\sqrt{x}+\sqrt{y}+\sqrt{z})+\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}}$$
From the condition and using AM-GM:
$$
\begin{aligned}
(1-\sqrt{x})(1+\sqrt{x})&=1-x\\
&\geq y+z+2\sqrt{xyz}\\
&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $k$ in this question Suppose $x_1$ and $x_2$ are solutions to $x^2+x+k=0$, if $x_1^2+x_1x_2+x_2^2=2k^2$, find the value of $k$.
$x_1+x_2 = -1$
$x_1x_2 = k$
$(x_1+x_2)^2 = x_1^2+2(x_1x_2)+x_2^2=2k^2$
$=\frac {x_1^2}{2}+x_1x_2+\frac {x_2^2}{2}=2k^2=\frac{1}{2}$
$2k^2=\frac{1}{2}$
$k^2=\frac{1}{4}$
$k... | $$(x_1^2+x_2^2+x_1x_2)=(x_1+x_2)^2-x_1x_2=1-k=2k^2\implies 2k^2+k-1 \implies k=-1,1/2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Perron vector of the adjacency matrix of complete bipartite graph $K_{p,q}$. We know that the spectral radius of the adjacency matrix of complete bipartite graph $K_{p,q}$ is $\sqrt{pq}$. Also note that the perron vector corresponding to the spectral radius has positive entries and is of the form $$(a,a,\cdots,a,b,b,\c... | I think that $a= \sqrt{q}$, $b= \sqrt{p}$ works:
\begin{align*}
A_{p,q}v_{\sqrt{pq}} &= \phantom{\sqrt{pq}} \left[\begin{array}{cccccc}
0& \dots & 0 & 1 & \dots& 1 \\
\vdots& & & && \vdots \\
0& \dots & 0 & 1 & \dots& 1 \\
1& \dots & 1 & 0 & \dots& 0 \\
\vdots& & & && \vdots \\
1& \dots & 1 & 0 & \dots& 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence In L2 norm Given that
\begin{align*}
&u_n(x) = \frac{x}{\sqrt{x^2+1/n}}&\\
&\text{and }&\\
&v(x) = \left\{ \begin{array}{cc}
\frac{|x|}{x} & \hspace{5mm} x\neq 0 \\
0 & \hspace{5mm} x=0 \\
\end{array} \right.&
\end{align*}
Show that $u_n \to v$ with respect ... | No, since
$$ \frac{1}{x + \sqrt{x^2 + 1/n}} > 1 $$
for $n$ large and $x \in [0, 1]$ small. I think the easiest proof is probably by the dominated convergence theorem, as $u_n \rightarrow v$ pointwise and $|u_n| \leq 1$. But if you want to proceed in this way, you can. Since
$$ \frac{1}{x + \sqrt{x^2 + 1/n}} \leq \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve modular equivalence with polynomial I'm finding the solutions for the modular equation $x^2 -3x+2 \equiv 0\pmod{14}$. This is what I've done so far:
\begin{align}
0 & \equiv x^2 -3x+2 \pmod{14}\\
& \equiv (x-1)(x-2) \pmod{14} \\
\end{align}
This implies that two of the solutions are the polynomial's usual ... | If $14|(x-1)(x-2)$, we cannot conclude $14|x-1$ or $14|x-2$, because $14$ is not prime.
But $14=2\times7$, so $14|(x-1)(x-2)$ means $2|(x-1)(x-2)$ and $7|(x-1)(x-2)$.
Since $2$ and $7$ are prime, this means ($2|x-1 $ or $2|x-2$) and ($7|x-1$ or $7|x-2$).
Thus, there are four solutions (mod $14$). Can you take it fr... | {
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"url": "https://math.stackexchange.com/questions/3568373",
"timestamp": "2023-03-29T00:00:00",
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Does $\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $ converge? Does the following integral converge? I will post my solution, but I am unsure if it is true.
$$\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $$
My solution
Let
$$ g(x) = \frac{2x}{\sqrt {x^3}}$$
$$ f(x) = \frac{2x +3}{\sqrt {x^3... | Perhaps redundant:
Let $x \ge 2$.
$\dfrac{x}{\sqrt{3x^3}}\lt \dfrac{2x+3}{\sqrt{x^3+2x+5}}$, or
$0<g(x):=(1/√3)x^{-1/2} <$
$\dfrac{2x+3}{\sqrt{x^3+2x+5}}=:f(x);$
$\int_{2}^{\infty} g(x)dx$ diverges, so does $\int_{2}^{\infty}f(x)dx$ (Monotonicity of integral).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and... | $$4\cos x\cos3x\cos2x=2\cos2x(\cos2x+\cos4x)=2\cos2x(\cos2x+2\cos^22x-1)$$
$$\implies0=4\cos^32x+2\cos^22x-2\cos2x-1$$
$$=2\cos^22x(2\cos2x+1)-(2\cos2x+1)$$
$$=(2\cos2x+1)(2\cos^22x-1)$$
If $2\cos2x+1=0,$
$$\cos2x=-\cos\dfrac\pi3=\cos\left(\pi-\dfrac\pi3\right)$$
Else $0=2\cos^22x-1=\cos4x,4x=(2n+1)\dfrac\pi2$ where $n... | {
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"timestamp": "2023-03-29T00:00:00",
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Given that $3^{15a} = 5^{5b} = 15^{3c}$, show that $5ab-bc-3ac=0$ Given that $3^{15a} = 5^{5b} = 15^{3c}$, show that $5ab-bc-3ac=0$
The only thing I can do is:
$3^{5a} = 5^{b} = 15^{\frac{3}{5}c}$
and then i am stuck, I figure that there must be a relationship between 3 and 5? should i utilise $5ab-bc-3ac=0$?
| To get the variables directly take logarithms. (Doesn't matter which base).
$\log 3^{15a} = \log 5^{5b} = \log 15^{3c}$ so
$15a \log 3 = 5b \log 5 = 3c \log 15$.
So flip a coin and choose which one we should express the others in. I pick ... $b$.
$a = \frac {\log 5}{3\log 3} b$ and $c = \frac {5\log 5}{3\log 15} b$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Find $f(x)$ such that: $ f'(x) + f(x^2) = 2x + 1 $ How to find the function $f(x)$ that is derivable on $\mathbb{R}$ and satisfies the equation:
$$ f'(x) + f(x^2) = 2x + 1 \text{ } \text{ } \forall x \in \mathbb{R}$$
My attempt:
Substitute $-x$ for $x$ in the equation, we have a system of 2 equations :
\begin{cases}... | We can write a power series for $f(x)$ as $$f(x) = \sum_{n=0}^\infty a_nx^n$$
Then we get that $$f'(x) + f(x^2) = \sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n} + \sum_{n=0}^{\infty} a_nx^{2n} = 2x+1$$
Breaking this into odd and even powers of $x$, this can be rewritten as $$\sum_{n=0}^{\infty}((2n+1)a_{2n+1} + a_n)x^{2n} + \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\c... | You work is correct, You may further simplify
$$I=-9\sqrt{9-x^2}+\frac{1}{3}\left(\sqrt{9-x^2}\right)^3 +C= \sqrt{9-x^2} \left (-9+\frac{9-x^2}{3}\right)+C$$
$$\implies I=-\frac{1}{3}\sqrt{9-x^2}~~(18+x^2)+C,$$
which is the final correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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prove that: $\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}$ Pythagoras's constant in Fibonacci number!
How do I show that?
$$\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}\tag1$$
Where $F_n$ is Fibonacci sequence.
$$2(F_n^4+F_{n+1}^4+F_{n+2}^4)^2=F_n^2+F_{n+1}... | Using the same notation as Matteo $x=F_{n}$ and $y=F_{n+1}$ we have \begin{eqnarray}F_n^4+F_{n+1}^4+F_{n+2}^4&=&x^4+y^4+(x+y)^4 \\
&=& 2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\\
&=&2(x^4+2x^3y+3x^2y^2+2xy^3+y^4)\\
&=&2(x^2+y^2+xy)^2
\end{eqnarray}
On the other hand we have \begin{eqnarray}F_n^2+F_{n+1}^2+F_{n+2}^2&=& x^2+y^2+(x+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Isosceles triangle $ABC$ with an inside point $M$, find $\angle BMC$ We have an isosceles $\triangle ABC, AC=BC, \measuredangle ACB=40^\circ$ and a point $M$ such that $\measuredangle MAB=30^\circ$, $\measuredangle MBA=50^\circ$.
Find $\measuredangle BMC$.
Starting with $\angle ABC=\angle BAC=70^\circ \Rightarrow \ang... | Here's a trigonometric approach. Let $\angle BCM=\varphi\Rightarrow \angle ACM=40^{\circ}-\varphi$. Apply the law of sines in $\triangle AMC$ and $\triangle BMC$:
$$\frac{AC}{CM}=\frac{\sin(80^{\circ}-\varphi)}{\sin(40^\circ)} \\ \frac{BC}{CM}=\frac{\sin(20^{\circ}+\varphi)}{\sin(20^\circ)} $$
Since $AC=BC$, the two ra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be? If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be?
$$y^2(x^2-5x+4)+y(x^2-5x+2)+(x^2-5x+4)<0$$
As it is true for all real y, hence $D<0$
$$(x^2-5x+2)^2-4(x^2-5x+4... | It is only necessary that the discriminant is less than $0$; not sufficient. We also need that the leading coefficient is $<0$ (otherwise the quadratic is $>0$ for all $y$). Thus, we also require
$$x^2-5x+4<0\implies 1<x<4.$$
Intersecting this with what you got gives the interval $(2,3)$.
| {
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"answer_id": 0
} |
Compute $\left[\begin{smallmatrix}1-a & a \\ b & 1-b\end{smallmatrix}\right]^n$ Compute $\begin{bmatrix}1-a & a \\ b & 1-b\end{bmatrix}^n$, where the power of $n\in\mathbb N$ denotes multiplying the matrix by itself $n$ times; $a,b\in[0,1]$.
Edit:
I considered using induction, computed the desired matrix:
$$\begin{bm... | Note that the characteristic polynomial of $A$ is given by
$$\begin{align}
\chi_A(\lambda) &= \lambda^2 - (2-a-b)\lambda + 1-a-b \\
&= \left( \lambda - \left( 1 - \frac{a+b}{2} \right) \right)^2 - \left( \frac{a+b}{2} \right)^2
\end{align}$$
so the eigenvalues of $A$ are $1-\dfrac{a+b}{2} \pm \dfrac{a+b}{2}$, i.e. $1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators Suppose that we decompose $1$ as a sum of Egyptian fractions with odd denominators.
I noticed (from a cursory view) that the fraction
$$\frac{1}{3}$$
appears in each of such decompositions.
Questions
Must the fraction $1/3$ appear in ea... | $1 = \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} + \frac{1}{17} + \frac{1}{19} + \frac{1}{21} + \frac{1}{23} + \frac{1}{25} + \frac{1}{27} + \frac{1}{33} + \frac{1}{611} + \frac{1}{265525} + \frac{1}{97544139723} + \frac{1}{8457652617058141652925}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stuck in a finite summation in a physics problem.r³/(r-1)² Actually it was not a hard problem as seen from physics point of view, just some application of Coulomb's law (multiple charge system) but it finally ended up as the summation of:-
$$
\sum_{r=2}^n \frac{(r+1)^3}{r^2}
$$
I tried to open things up but it finally... | We have:
$$\frac{(r+1)^3}{r^2}=\frac{r^3+3r^2+3r+1}{r^2}=r+3+\frac{3}{r}+\frac{1}{r^2}$$
Thus:
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\sum_{r=2}^n \bigg( r+3+\frac{3}{r} \frac{1}{r^2} \bigg)$$
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\sum_{r=2}^n r +\sum_{r=2}^n 3 +3 \sum_{r=2}^n \frac{1}{r} + \sum_{r=2}^n \frac{1}{r^2}$$
$$\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On the integral $\int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$ I'm having a difficult time evaluating the integral
$$\mathcal{J} = \int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$$
This is integral arose after simplifying the integral $\displaystyle \int_{0}^{\pi/4 } \a... | This answer is based on Feynman's trick. Put
\begin{equation*}
I(a) = \int_{0}^{\pi/4}\arctan\left(a\sqrt{\dfrac{1-\tan^2 x}{2}}\right)\, dx .
\end{equation*}
Then
\begin{gather*}
I'(a) = \int_{0}^{\pi/4}\dfrac{1}{1+a^2\dfrac{1-\tan^2 x}{2}}\cdot \sqrt{\dfrac{1-\tan^2 x}{2}} \, dx = \\[2ex]\int_{0}^{\pi/4}\dfrac{1}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the locus of the complex number from the given data If $|\sqrt 2 z-3+2i|=|z||\sin (\pi/4+\arg z_1) + \cos (3\pi/4-\arg z_1)|$, where $z_1=1+\frac{1}{\sqrt 3} i$, then the locus of $z$ is ...
$$\arg~z_1 =\frac{\pi}{6}$$
So RHS of the equation is $\;|z|\dfrac{1}{\sqrt 2}$.
Then
$\;|\sqrt 2 z-3+2i|=|z|\dfrac{1}{\sqr... | Continue with $|\sqrt 2 z-3+2i|=\frac{1}{\sqrt 2}|z|$ to get,
$$(\sqrt 2 z-3+2i) (\sqrt 2 \bar z-3-2i) =\frac{1}{2}|z|^2$$
or,
$$3|z|^2 -2\sqrt2 (3-2i) \bar z -2\sqrt2 (3+2i) z +26=0$$
which can be written in the form of a circle, i.e.
$$\bigg| z - \frac{2\sqrt2}3(3-2i) \bigg|^2 = \frac{26}9$$
Thus, the locus is a circ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$ For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$
ATTEMPT
We transform the equation into $xy^2+12y-3xy-3y^2\le 12$
I noticed that for $x=0, y=2$ equality is achieved, but I am lost here.
| I am interested in elementary solution
$$
\begin{aligned}
(y-2)^{2}&\geq 0\\
y^{2}-4y+4&\geq 0\\
4-y&\geq y(3-y)\\
\\
x&\geq 0\\
3&\geq 3-x
\end{aligned}
$$
Only one of $y(3-y)$ and $3-x$ can be negative, thus $3(4-y)\geq y(3-x)(3-y)$. from the inequalities, equality is when $y=2$ and $x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$
Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.
Is ... | Here is a proof without using Menelaus's thm. Let $A_2$, $B_2$, and $C_2$ be the midpoints of $CA_1$, $AB_1$, and $BC_1$, resp. Draw $MA_1$, $KA_2$, $KB_1$, $LB_2$, $LC_1$, $LC_2$, $MA_2$, $KB_2$, and $LC_2$. In this proof, $[\mathcal{P}]$ is the area of a polygon $\mathcal{P}$.
Due to symmetry $\triangle KLM$ is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Convergence of $S(x,y)=\sum_{n=1}^{\infty}\left(\frac{x+2n+1}{(x+2n+1)^2+y^2}+\frac{x+2n}{(x+2n)^2+y^2}-\frac{x+n}{(x+n)^2+y^2}\right)$ I know that the following sum is convergent and equal to $\ln2$ for all $x\not\in\mathbb{Z}$
$$\sum_{n=0}^{\infty}\left(\frac{1}{x+2n+1}+\frac{1}{x+2n}-\frac{1}{x+n}\right)-\ln2=0$$
Ca... | According to Wolfy,
each term is
$
\dfrac{\left(8 n^4 x - 4 n^4 + 20 n^3 x^2 - 4 n^3 x + 12 n^3 y^2 - 2 n^3 + 18 n^2 x^3 + 3 n^2 x^2 + 18 n^2 x y^2 - 3 n^2 x + 9 n^2 y^2 + 7 n x^4 + 4 n x^3 + 10 n x^2 y^2 - n x^2 + 8 n x y^2 + 3 n y^4 + n y^2 + x^5 + x^4 + 2 x^3 y^2 + 2 x^2 y^2 + x y^4 + y^4\right)}
{(n^2 + 2 n x + x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given that $a+b+c=0$, show that $2(a^4+b^4+c^4)$ is a perfect square Given that $a+b+c=0$. Show that: $2(a^4+b^4+c^4)$ is a perfect square
MY ATTEMPTS:
I found that when $a+b+c=0$, $a^3+b^3+c^3=3abc$
So I did:
$(a^3+b^3+c^3)(a+b+c)$ -- $a^4+b^4+c^4=-(a^3c+a^3b+ab^3+b^3c+c^3a+c^3b)$
And then I tried to substitute $a^4+b... | I don't pretend that my solution is shorter, but for such questions, I try to avoid "redicovering all from scratch" ; I systematically apply the Newton-Girard identities one finds here, in particular for the present case :
$$p_4=\underbrace{e_1^4-4e_1 e_2+4e_1 e_3}_{= \ 0 \ \text{because} \ e_1= \ 0}+2e_2^2-\underbrace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks
| $P(k)$ is the statement $ 2n^3<3^n $. Do not write "$P(k)=.....$"; $P(k)$ is not a mathematical value.
If we assume that then we have $2n^3 < 3^n$
So $2(n+1)^3 = 2n^3 + 6n^2 + 6n + 1$ And we have $2n^3 < 3^n$ so
$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + 6n^2 + 6n + 1$
And $n\ge 6$ so $6n^2 \le n*n^2 =n^3$ and $6n+1 < 6n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How to find $\int \frac {(1+x^2)(2+x^2)}{(x \cos x+\sin x)^4}dx$? I came across this integral while studying indefinite integrals.
So far I have had many unsuccessful attempts which include - trying by parts - but I could not find a way to proceed with it.
I even tried dividing both numerator and denominator by $x^4$ ... | Let $f_n(x) = \left( \frac {\cos x - x \sin x}{x\cos x + \sin x} \right)^n$ and note that,
$$\frac{df_n(x)}{dx}
= -\frac{n(\cos x - x \sin x)^{n-1}(x^2+2)}{(x\cos x + \sin x)^{n+1}}$$
Then, for $n=1$ and $n=3$, we have respectively,
$$\frac{df_1(x)}{dx}
= -\frac{x^2+2}{(x\cos x + \sin x)^{2}}$$
$$\frac{df_3(x)}{dx}=- \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What are the integer solutions to $a^{b^2} = b^a$ with $a, b \ge 2$ I saw this in quora.
What are all the
integer solutions to
$a^{b^2} = b^a$
with $a, b \ge 2$?
Solutions I have found so far:
$a = 2^4 = 16, b = 2,
a^{b^2}
= 2^{4\cdot 4}
=2^{16},
b^a = 2^{16}
$.
$a = 3^3, b = 3,
a^{b^2} = 3^{3\cdot 9}
=3^{27},
b^a = 3... | Render $a=tb^2$. Then
$(tb^2)^{b^2}=b^{tb^2}$
$tb^2=b^t$ ($b$ is assumed nonzero)
$t=b^{t-2}$ ($t$ is nonzero bdcause $a$ is assumed nonzero)
If $b$ is to be a positive integer and $t$ is rational, then $t$ will be a positive integer or $b$ will be a perfect power. But if $t>4$ then $b=t^{1/(t-2)}$ will lie strictly ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this geometry problem without trigonometry?
Let $ABC$ be a triangle with $D$ on the side $AC$ such that $\angle DBC=42^\circ$ and $\angle DCB=84^\circ$. If $AD = BC$, find $x = \angle DAB$.
It's supposed to be solved with constructions, but I couldn't figure out. See the trigonometric version here.... |
Here is a mostly geometric solution. First of all, let $C'$ be another point on $AC$ such that $BC'=BC$. Observe that the triangle $BC'D$ has $\angle C'BD=30^\circ$ and $\angle BDC'=\angle BDC=54^\circ$. Thus, by the Law of Sines on the triangle $BC'D$, we obtain
$$\frac{C'D}{C'B}=\frac{\sin(\angle C'BD)}{\sin(\ang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$ Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck.
Show that $1+2^n+2^{2n}$ is divisible b... | $n=3k+1$, $1+2^{3k+1}+2^{2(3k+1)}$, $2^3=1$ mod $7$ implies that
$1+2^{3k+1}+2^{2(3k+1)}=1+2+4$ mod $7$.
$n=3k+2$, $1+2^{3k+2}+2^{2(3k+2)}=1+4+16$ mod $7$ and $1+4+16=0$ mod $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\fra... | $$2x^3+x^2+x-1=0`~~~(1)$$
$a,b, c$ are roots of the cubic, let $$s=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{a^3b^3+b^3c^3+c^3a^3}{(abc)^3}=0$$ Let one root od the new cubic be $$y=-\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=s-\frac{2}{a^3}=s-\frac{2}{x^3} \implies x=(\frac{2}{-y})^{1/3}~~~~(2)$$
From (1) we can w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$ Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$
Take $x \in \mathbb{Q}[\sqrt{3}+ \sqrt{5}]. x = a_x + b_x( \sqrt{3} + \sqrt{5}) = a_x + b_x\sqrt{3} + b_x\sqrt{5} \in \mathbb{Q}[\sqrt{3}, \sqrt{5}] \Rightarrow \ma... | While it is possible to show the other inclusion, I think it is far easier to show that the dimensions of each field extension (as vector spaces over $\mathbb{Q}$) are the same (which is $4$ in this case). Then since one vector space is contained within the other, they must be the same if they have the same dimension. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Verifying that Kummer hypergeometric function is a solution to $xy''+(b-x)y'-ay=0$ The following second order differential equation (see YouTube link)
$$xy''+(b-x)y'-ay=0$$
has two solutions, one of them resmenble Kummer function of the first kind:
$$y=M(a,b,x)=\sum_{n=0}^\infty \frac{a^{(n)}x^n}{b^{(n)}n!},$$
where $a... | $\def\d{\mathrm{d}}$If $y(x) = \sum\limits_{n = 0}^∞ \dfrac{a^{(n)} x^n}{b^{(n)} n!} = 1 + \sum\limits_{n = 1}^∞ \dfrac{a^{(n)} x^n}{b^{(n)} n!}$, then$$
ay(x) = a + \sum_{n = 1}^∞ \frac{a · a^{(n)} x^n}{b^{(n)} n!},
$$\begin{gather*}
(b - x) y'(x) = (b - x) \sum_{n = 1}^∞ \frac{a^{(n)} nx^{n - 1}}{b^{(n)} n!} = \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Precalculus linear combination problem What is the maximum possible length of the vector resulting from the following linear combination?
$$
\frac{1}{\| \mathbf{v_1} \|} \,\mathbf{v_1} + \frac{1}{\|\mathbf{v_2}\|} \,\mathbf{v_2} + \cdots + \frac{1}{\| \mathbf{v_n} \|} \,\mathbf{v_n}
$$
| Let $\mathbf{u_i} = \frac{1}{\| \mathbf{v_i} \|} \,\mathbf{v_i}$, then $\| \mathbf{u_i} \| = 1.$
$
\Big\|\frac{1}{\| \mathbf{v_1} \|} \,\mathbf{v_1} + \frac{1}{\|\mathbf{v_2}\|} \,\mathbf{v_2} + \cdots + \frac{1}{\| \mathbf{v_n} \|} \,\mathbf{v_n}\Big\| =
\| \mathbf{u_1} + \mathbf{u_2} + \dots + \mathbf{u_n}\| \le \|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3622452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Area and center location of an ellipse generated by the intersection of an ellipsoid and plane I am working on a model that requires that I know the area and center coordinates of the ellipse that is created by the intersection of an ellipsoid and a plane.
Specifically, for the location of the center of the ellipse I w... | Without loss of generality, you can translate the ellipsoid to origin, rotate it so that its semiaxes are parallel to the Cartesian coordinate axes, and reorder the axes so that the $z$ component of the intersection plane normal in the rotated coordinates has the largest magnitude. Then, the points on the ellipsoid fu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3622888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Fastest way to solve $x^3\equiv x \pmod{105}$ $$x^3\equiv x \pmod{105}$$
I'm trying to solve this equation. Here's what I tried so far:
$$x^3\equiv x \pmod{105} \iff x^2\equiv 1 \pmod{105}$$
Then, applying the Chinese remainder theorem, I got the system:
$$\cases{x^2 \equiv 1 \pmod{5}\\x^2 \equiv 1 \pmod{7}\\x^2 \equiv... | As $x^3-x=(x-1)x(x+1)$ is a product of three consecutive integers
$3$ must divide $x^3-x$
So, we need $$x^3\equiv x\pmod{5\cdot7}$$
If $(x-1)x(x+1)\equiv0\pmod 5$
$\implies x\equiv0\ \ \ \ (1), x\equiv-1\ \ \ \ (2), x\equiv1\pmod5\ \ \ \ (3)$
Similarly, $x\equiv0\ \ \ \ (4), x\equiv-1\ \ \ \ (5), x\equiv1\pmod7\ \ \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+... | Use the (not so) popular algebraic limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ Dividing both the numerator and denominator of the expression under limit by $x-2$ we can see that the desired limit equals $$\left. \frac{1}{3}\cdot 2^{-2/3}\middle/\frac{1}{2}\cdot 2^{-1/2}\right.=\frac {2^{5/6}}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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A question about a function $q(b)$ User Mathphile invented the function $\ q(b)\ $ which is defined as the sum of the values $ \lfloor \frac bj \rfloor$ for $j=1,\cdots ,b$ without duplicates.
Example : The values for $b=11$ are $[11, 5, 3, 2, 2, 1, 1, 1, 1, 1, 1]$ , without duplicates we have $[1, 2, 3, 5, 11]$ , the... | If $b$ is composite integer then for $ 1\leq j\leq [\sqrt{b}]$ we have that $[\frac{b}{j}] > [\frac{b}{j+1}]$ because $\frac{b}{j}-\frac{b}{j+1}\geq 1$ which is true for all $1 \leq j \leq [-0.5+0.5 \sqrt{1+4b}] \leq [\sqrt{b}]$, so for $j=1,2,3,\cdots, [\sqrt{b}]$ and we have that $[\frac{b}{j}]\geq [\frac{b-1}{j}]$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Show that $\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}$ I've been trying to develop this integral in parts twice, but I can't come up with anything meaningful,
Could you give me some hints on how to develop this exercise?
Prove t... | Hint
By differentiating the equality, we should prove $$\frac{1}{(a+b\cos x)^n}={d\over dx}\frac{A\sin x}{(a+b\cos x)^{n-1}}+{B \over (a+b\cos x)^{n-1}}+{C \over (a+b\cos x)^{n-2}}, \hspace{1cm}$$ with $|a|\neq |b|$. Also $${d\over dx}\frac{A\sin x}{(a+b\cos x)^{n-1}}{={A\cos x(a+b\cos x)^{n-1}+(n-1)Ab\sin^2 x(a+b\cos ... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying ... | Assume $u=\sqrt{a^+b^2+2ab\cos x} $ so that $u\to a+b$ as $x\to 0$. The expression under limit is $$\frac{ab\sin x} {2u\sqrt{a+b-u}}$$ whose limit is same as that of $$\frac{ab} {2(a+b)}\frac{x}{\sqrt{a+b-u}}$$ Next we can observe that $$\frac{a+b-u} {x^2}=\frac{(a+b) ^2-u^2}{x^2(a+b+u)}=\frac{2ab}{a+b+u}\cdot\frac{1-\... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$ without L'Hôpital This limit is one of the "Problems Plus" from Stewart Calculus:
$$\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$$
Note that the limit is of the indetermi... | Without heavy machinery, only that $\lim_{x \to 0} \sin x / x = 1$, you can use the sum-to-product formulas, as follows.
\begin{align*}
\sin(a+2x) - 2 \sin(a+x) + \sin a & = \sin(a+2x) - \sin(a+x) + \sin a - \sin(a+x) \\
& = 2 \sin(x/2) \cos(a + 3x/2) - 2 \sin (x/2) \cos(a + x/2) \\
& = 2 \sin(x/2) \left (\cos(a + 3x/2... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Floor functions in a set of numbers Find the number of distinct numbers in the list
$$\left\lfloor \frac{1^2}{1000} \right\rfloor, \ \left\lfloor \frac{2^2}{1000} \right\rfloor, \ \left\lfloor \frac{3^2}{1000} \right\rfloor, \ \dots, \ \left\lfloor \frac{1000^2}{1000} \right\rfloor.$$
The floor of all numbers till $\le... | Consider $\lfloor \frac{x^2}{1000} \rfloor = k$ for some integer $k$. This means:
$$k\leq \frac{x^2}{1000}<k+1$$
$$1000k\leq x^2 \leq 1000k+999$$
It is clear that when $x\geq 500$, $(x+1)^2-x^2=2x+1>1000$ so each value of $\lfloor\frac{x^2}{1000}\rfloor$ is distinct. That's $\fbox{501}$ distinct values from $x=500,501,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the polynomial $P(x)$ s.t. for any $a,b\in\Bbb C$ s.t. $a^2+b^2=ab$, $P(a+b)=6\big(P(a)+P(b)\big)+15a^2b^2(a+b)$.
Suppose that $P(x)$ is a polynomial such that
$$P(a+b)=6\big(P(a)+P(b)\big)+15a^2b^2(a+b)$$
for any complex numbers $a$ and $b$ satisfying $a^2+b^2=ab$, then find the polynomial $P(x)$.
I don't ... | In this solution, $P$ is assumed to be an entire function (which is more generalised than just being a polynomial). Let $\omega_1,\omega_2$ be roots of $x^2-x+1=0$.
We note that when $a=\omega_jt$ and $b=t$, we have $a^2+b^2=ab$, so that
$$P\big((\omega_j+1)t\big)-6\big(P(\omega_j t)-6P(t)\big)-15\omega_j^2(\omega_j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3647743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Linear Algebra matrix calculation help please A = $$
\begin{pmatrix}
2 & 1 \\
3 & 4 \\
\end{pmatrix}
$$
How can i find matrix B s.t B^3 = A?
| You can define $B$ as
$$B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
And solve the equation following equation system:
$$B^3 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}*\begin{bmatrix} a & b \\ c & d \end{bmatrix}*\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix},$$
wh... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\righ... | A NOTE
The given inequality
$$
\left(\sum_{cyc}\frac{a+b}{c}\right)^2\geq 4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}\frac{1}{a^2}\right)\textrm{, }a,b,c>0\tag 1
$$
is written as
$$
\left(\sum_{cyc}ab(a+b)\right)^2\geq4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2b^2\right).\tag 2
$$
Hence if $s=a+b+c$ and $p=abc$ and $s... | {
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"url": "https://math.stackexchange.com/questions/3649363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Solving $\displaystyle \int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}\log(2\cos \theta)d\theta$ $$\int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}\int_{\frac{1}{a}}^{2\cos\theta}\frac{1}{x}dxd\theta=\int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}(\log(2\cos \theta)\,\mathrm +\log(a))d\theta=$$
$... | Note that, for $a\to \infty$, $\tan^{-1}\sqrt{2a^2-1} \to \frac\pi2$. Then, the Integral
$$\int_{-\frac\pi2}^{\frac\pi2}\log(2\cos \theta)d\theta=0$$
However, the first term $2\log(a)(\tan^{-1}\sqrt{4a^2-1})$
diverges; so does the original integral.
————-
Edit:
\begin{align}
I&=\int_0^{\frac{\pi}{2}}\log(2\cos x)dx=\i... | {
"language": "en",
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"source": "stackexchange",
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Prove that $\frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1$ Question -
Let $x, y, z$ be distinct real numbers. Prove that
$$
\frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1
$$
My work -
first i apply directly C-S and after simplification i have to prove... | Let $a = \dfrac{x}{x-y}, b = \dfrac{y}{y-z}, c = \dfrac{z}{z-x}$, and note
$a+b+c = ab+bc+ca+1$. We need to show $a^2+b^2+c^2 \geqslant 1$. But this is now just equivalent to $(a+b+c-1)^2 \geqslant 0$
Alternately, the original inequality $\iff \dfrac{(x^2y+y^2z+z^2x-3xyz)^2}{(x-y)^2(y-z)^2(z-x)^2}\geqslant 0$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$
Let $x, y \in \mathbb R$ s. t . $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2-\sqrt{32}$
I know this is a duplicate of another question, but that question has solutions involving calculus and geometry,... | Yet another approach: Let $S=\{(x,y) : x^2+y^2=2x-2y+2\}$ be the algebraic curve of the constraint and let $A=\{(x,y): x+y=0\}$ be the anti-diagonal. We have the basic inequality: $(x-y)^2 \leq 2(x^2+y^2)$ with equality iff $(x,y)\in A$. Thus, for points in $S$ we have:
$$ (x-y)^2 \leq 4(x-y) + 4 $$
with equality iff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to compute $\int_0^{\frac{\pi}{2}}\frac{\arctan(\sqrt{\tan(x)})}{\tan(x)}dx$ I have been asked to compute the integral $$\int_0^{\frac{\pi}{2}}\frac{\arctan(\sqrt{\tan(x)})}{\tan(x)}dx$$
I have been told that it converges, and we only need its value.
I tried the substitutions $ u=\tan(x)$, $\;\; v^2=\tan(x) $ .
I t... | $$\begin{align}
\int_{0}^{\frac{\pi}{2}}\frac{\arctan\left(\sqrt{\tan\left(x\right)}\right)}{\tan\left(x\right)}dx &= \frac{1}{2}\int_{0}^{\pi}\frac{\arctan\left(\sqrt{\tan\left(\frac{x}{2}\right)}\right)}{\tan\left(\frac{x}{2}\right)}dx
\\&\stackrel{(1)}{=}\int_{0}^{\infty}\frac{\arctan\left(\sqrt{t}\right)}{t\left(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove $\cos 20^\circ \cos 40^\circ \cos 80^\circ= \frac18$ geometrically I understand the wizardry that $\cos 20^\circ \cos 40^\circ \cos 80^\circ=\dfrac18$
Proving it isn't that hard. Taking the left hand side and multiplying it up and down by $\sin 20$ yields:
\begin{align}
& \dfrac{\sin 20^\circ}{\sin 20^\circ} \... |
Configure the angles in the diagram. Then
\begin{align}
& \triangle ABE: \>\>\> \cos 20 = \frac{\frac y2}x = \frac y{2x}\\
& \triangle BGE: \>\>\> \cos 40 = \frac{\frac x2}1 = \frac x2\\
& \triangle ABC: \>\>\> \cos 80 = \frac{\frac12}y = \frac1{2y} \\
\end{align}
Thus,
$$\cos 20\cos 40 \cos 80 = \frac y{2x}\frac x2\... | {
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"source": "stackexchange",
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Prove that $\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$ Question -
Prove that for all non-negative real numbers a,b, c, we have
$$
\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a... | By C-S twice we obtain: $$\sum_{cyc}\sqrt{\frac{2a^2+bc}{a^2+2bc}}-2\sqrt2=\sum_{cyc}\frac{\sqrt{(2a^2+bc)(a^2+2bc)}}{a^2+2bc}-2\sqrt2\geq$$
$$\geq\sum_{cyc}\frac{\sqrt2(a^2+bc)}{a^2+2bc}-2\sqrt2=\sqrt2\left(\sum_{cyc}\left(\frac{a^2+bc}{a^2+2bc}-\frac{1}{2}\right)-\frac{1}{2}\right)=$$
$$=\sqrt2\left(\sum_{cyc}\frac{a... | {
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"url": "https://math.stackexchange.com/questions/3659642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Finding $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$
If $a, b$ and $c$ (all distinct) are the sides of a triangle ABC opposite to the angles $A, B$ and $C$, respectively, then $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$ is equal to $?$
By opening, $\sin(A-B)$ as $\sin A\cos B-\cos A\sin B$ and ... | Assume that $a>b$.
There exists a point $D$ on line segment $AB$ such that $CD=b$, Produce $CD$ to meet the circumscribed circle of $\Delta ABC$ at point $E$. (See the figure below.)
Since $AC=CD$, $\angle CDA=\angle CAD=A$. Hence, $\angle BCD=A-B$.
Since $\angle BAE$ and $\angle BCE$ are angles in the same segment, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\sum _{n=1}^{\infty }\sum _{i=1}^{n }\frac{\left(-1\right)^n}{i\cdot n}$ have a finite value? If so, evaluate its closed form. Does $\sum _{n=1}^{\infty }\sum _{i=1}^{n }\frac{\left(-1\right)^n}{i\cdot n}$ have a finite value? If so, evaluate its closed form.
I'm pretty sure its related to $\begin{array}{l}\zeta ... | Consider $$\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty\right)^2=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...\infty+2\left(1\left(-\frac{1}{2}\right)+1\left(\frac{1}{3}\right)+...\ +-\frac{1}{2}\left(\frac{1}{3}\right)+-\frac{1}{2}\left(-\frac{1}{4}\right)...\infty\right)$$
The numbers in the brackets are the s... | {
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"source": "stackexchange",
"question_score": "1",
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Integral $ \int_{-\infty}^{+\infty} \frac{ \sin^2(\sqrt{(x-a)^2 + b^2}\,\,t)}{(x-a)^2 + b^2} dx$ I am trying to solve the following definite integral:
$$ \int_{-\infty}^{+\infty} \frac{ \sin^2(\sqrt{(x-a)^2 + b^2}\,\,t)}{(x-a)^2 + b^2} dx$$
with $a$ and $b$ being real constants. Notice that the integrand is non-negati... | Thanks Gary. Your solution was quite impressive.
Let me complicate the integral to a level higher and see if we can proceed similarly to get an analytical solution. I now write the integral to be solved as:
$$ \int_{-\infty}^{+\infty} \frac{\sin^2 \sqrt{(x^2-a^2)^2 + b^4 }\, t}{(x^2-a^2)^2 + b^4 } \, \, dx $$
You wi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Can the expression $a^2 + b^2 - c^2$ be factored as a product of two quaternions, where $a$, $b$, $c$ are real numbers? I'm trying to factor the expression
$$a^2+b^2-c^2$$
as a product of two quaternions, where $a,b,c$ are reals. Can anyone give me the answer?
I think it can't be done.
I started with multiplying
$$(Aa... | It cannot be written as the product of two $\mathbb{H}$-linear combinations of variables $a,b,c$. (We assume the formal variables commute with all quaternion scalars, of course.)
Suppose it were, say $(ap+q)(ar+s)$ where $p,q\in\mathbb{H}$ and $r,s$ are $\mathbb{H}$-linear combinations of $b$ and $c$. The leading term ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Euclidean geometry : find out the area.
${\angle}ABC=45°$, $AD{\bot}AE$, $AD=AE$, $AC{\bot}BE$, $F$ is the intersection of $AC$ and $BE$, $BF=9$, so what is the area of ${\square}ABCE$?
| Let $F(0,0), A(0,a), B(-b,0), C(0,-c), E(e,0)$, then $D=tB+(1-t)C=
(-tb,-(1-t)c)$ for some $t$ ($t\in(0;1)$ if we believe the drawing, but I won't use it).
The desired area is then $\frac12(ae+ec+bc+ba)$.
We have:
$AD\perp AE\Leftrightarrow (D-A).(E-A)=0 \Leftrightarrow (-tb,-(1-t)c-a).(e,-a)=0 \Leftrightarrow -tbe+(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Determining the period of $\sin 2x +\sin\frac{x}{2}$ without using the LCM of periods
I tried to calculate period of function described as:
$$y=\sin 2x +\sin\frac{x}{2}$$
but without using LCM of periods.
From definition of periodic function we have:
$$\begin{align}
0 &= \sin(2x+2T)+\sin\left(\frac{x+t}{2}\righ... | Let $$\sin2(x+T)+\sin\frac{x+T}{2}=\sin2x+\sin\frac{x}{2},$$ where $T>0$.
Thus, for $x=0$ we obtain:
$$\sin2T+\sin\frac{T}{2}=0$$ and for $x=2\pi$ we obtain:
$$\sin2T-\sin\frac{T}{2}=0,$$
which gives $$\sin2T=\sin\frac{T}{2}=0$$ or
$$T=2\pi k,$$ where $k$ is a positive integer number.
We see that $k=1$ is not valid, bu... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $z^2+z+2=0$, then $z^2 + \frac{4}{z^2} = -3$
If $z^2+z+2=0$, for $z\in\mathbb C$, demonstrate:
$$z^2 + \frac{4}{z^2} = -3$$
$z^2 = - 2 - z$, but it didn't help me.
Is there any other elegant solution?
| $$z^2+\frac{4}{z^2}+3$$
$$=\frac{z^4+3z^2+4}{z^2}$$
$$=\frac{(z^2+z+2)(z^2-z+2)}{z^2}$$
$$=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove this inequality with $xyz=1$ let $x,y,z>0$ and such $xyz=1$,show that
$$f(x)+f(y)+f(z)\le\dfrac{1}{8}$$
where $f(x)=\dfrac{x}{2x^{x+1}+11x^2+10x+1}$
I try use this $2x^x\ge x^2+1$,so we have
$$2x^{x+1}+11x^2+10x+1\ge x^3+11x^2+11x+1=(x+1)(x^2+10+1)$$
It need to prove
$$\sum_{cyc}\dfrac{x}{(x+1)(x^2+10x+1)}\le\dfr... | The TL method helps!
Since $$x^x\geq\frac{x^3-x^2+x+1}{2},$$ it's enough to prove that
$$\sum_{cyc}\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{1}{8}$$ or
$$\sum_{cyc}\left(\frac{1}{24}-\frac{x}{x^4-x^3+12x^2+11x+1}\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{24}-\frac{x}{x^4-x^3+12x^2+11x+1}-\frac{1}{48}\ln{x}\right)\geq0.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674817",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Convergence of the recursive sequence $x_n=\frac{x_{n-1}}2+\frac3{x_{n-1}},n\ge 1$ Consider the iterative scheme :
$x_n=\frac{x_{n-1}}2+\frac3{x_{n-1}},n\ge 1$
with the initial point $x_0 \gt 0$. Then the sequence $\{x_n\}$
$(a)$ converges only if $x_o \lt 3$
$(b)$converges for any $x_0$
$(c)$does not converge for any... | There is a nice substitution and technique applied here. It enables us to find a closed form of $x_n$ from which the correct alternative can be easily derived. Given
\begin{align*}
x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\qquad\qquad n\geq 1
\end{align*}
we substitute
$\color{blue}{x_n=\sqrt{6}\,y_n}$
and we obtain
\beg... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding pdfs of $\frac1{X^2}$ and $\frac{1}{4}\left(\frac1{X^2}+\frac1{W^2}\right)$ where $X,W$ are independent $N(0,1)$ $X,W$ are independent random variables, both $N(0,1)$, i.e. $f_X(x)=f_W(x)= \frac{1}{{\sqrt{2\pi}}}e^{-\frac{x^2}{2}}$.
*
*Find PDF of $Y:=\frac{1}{X^2}$
*Find PDF of $\frac{1}{4}\left(\frac{1}{X... | Your work for the first problem is correct.
Let $\Phi(\cdot)$ and $\phi(\cdot)$ be the cdf and pdf of standard normal distribution, as usual.
Since $X$ is standard normal, you have for every $y>0$,
$$P\left(\frac1{X^2}\le y\right)=\cdots=1-\Phi\left(\frac1{\sqrt y}\right)+\Phi\left(-\frac1{\sqrt y}\right)=2\left(1-\Phi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are there formulas for $\sin\left(\frac{\theta}{3}\right)$ and $\cos\left(\frac{\theta}{3}\right)$? So far, I have seen the half angle formula (i.e.)
$$\sin^2\left(\frac{\theta}{2}\right) = \frac{1-\cos\theta}{2}$$
Is there a way to find $\sin\left(\frac{\theta}{3}\right)$ and $\cos\left(\frac{\theta}{3}\right)$ in ter... | There is no analytic formula for $\sin\frac{\theta}{3}$ & $\cos\frac{\theta}{3}$ . But the value of $\sin\frac{\theta}{3}$ & $\cos\frac{\theta}{3}$ can be derived in terms of $\sin\theta $ & $\cos\theta$ using triple angle formula (which may also give complex roots) as follows
$$\sin3x=3\sin x-4\sin^3x$$
setting $3x=\t... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability when putting balls from one box into another and then draw one ball from it I am struggling with the following problem:
Consider two boxes with several blue and red colored balls in.
In the right box there are 4 red and 5 blue balls.
In the left one are 7 red and 3 blue balls.
Now you blindly draw 4 ball... | Let $R$ be the event that after the transfer, you draw a red ball. Let $T(n)$ be the event that you transfer $n$ red balls from the right bin to the left bin (and $4-n$ blue balls).
Then the probability you are looking for is:
$$P(R) = P(R|T(0))P(T(0))+P(R|T(1))P(T(1))+P(R|T(2))P(T(2))+P(R|T(3))P(T(3))+P(R|T(4))P(T(4))... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the radius of the circle given in the picture below. This is the image of the question. I am not able to get how to find the radius. Please help with that.
This is my try. I can't proceed now after it.
Thanks
|
$$PT^2=PC\cdot PA\implies PT^2=2\cdot 5=10$$
Using Pythagorean theorem in right $\Delta PTO$, $PO=\sqrt{OT^2+PT^2}=\sqrt{R^2+10}$
Using cosine rule in $\Delta ABP$, $$\cos\angle APB=\frac{PB^2+PA^2-AB^2}{2(PB)(PA)}=\frac{(\sqrt{R^2+10}+R)^2+5^2-(\sqrt{11})^2}{2(\sqrt{R^2+10}+R)(5)}\tag1$$
Similarly, using cosine r... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Spivak calculus chapter 1 consequence of property 9 Spivak state's that one of the consequence of the distributive property is being able to solve multiplication of Arabic numerals.
1 3
x 2 4
_______
3 1 2
is arranged as:
$13 . 24$
$=13 . (2 . 10 + 4)$
$= 13 . 2 . 10 + 13 . 4$
$= 26 . 10 + 52$
$= 312$
and
... | He is showing us how did he obtain $13.4=52$ in the first working using $P9$ property explicitly.
\begin{align}13\cdot 24 &= 13 \cdot (2\cdot 10 + 4) \\
&=13 \cdot 2 \cdot 10 + \color{blue}{13 \cdot 4} \\
&= 26 \cdot 10 + \color{blue}{52}
\end{align}
The multiplication $13\cdot 4=52$ uses $P9$ also:
| {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2
= 8 - 4 \sqrt{3}
\end{alig... | Note the denesting formula
$$ \sqrt{a-\sqrt c} = \sqrt{\frac{a+\sqrt {a^2-c}}2}- \sqrt{\frac{a-\sqrt {a^2-c}}2}
$$
which can be verified by squaring both sides, and apply it to
$$\sqrt{8-4\sqrt3}=2\cdot \sqrt{2-\sqrt3}= 2\left(\sqrt{\frac32} -\sqrt{\frac12}\right)=\sqrt6-\sqrt2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Why does Stolz- Cesaro fail to evaluate the limit of $\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$, I need to find the limit of the sequence
$\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,
My strategy is to use Stolz's Cesaro theorem for this seque... | Note that, as mentioned in the comments below, your computation of the ratio is incorrect. Regardless, the hypothesis of Stolz-Cesaro assumes that the limit $\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}$ exists. If it doesn't exist, it does not imply that the original limit does not exist.
A better way to ap... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Maximize $\sum\limits_{k =1}^n x_k (1 - x_k)^2$ Given problem for maximizing
\begin{align}
&\sum_{k =1}^n x_k (1 - x_k)^2\rightarrow \max\\
&\sum_{k =1}^n x_k = 1,\\
&x_k \ge 0, \; \forall k \in 1:n.
\end{align}
My attempt: first of all i tried AM-GM, or we can just say, that $x_k (1 - x_k)^2 \le x_kx_k^2 =x_k^3$, but ... | For $x_1=x_2=...=x_n=\frac{1}{n}$ we get a value $\frac{(n-1)^2}{n^2}.$
We'll prove that it's a maximal value.
Indeed,
\begin{align}
\frac{(n-1)^2}{n^2}-\sum_{k=1}^nx_k(1-x_k)^2&=\sum_{k=1}^n\left(\frac{(n-1)^2}{n^3}-x_k(1-x_k)^2\right)\\
&=\frac{1}{n^3}\sum_{k=1}^n(1-nx_k)(n^2x_k^2-n(2n-1)x_k+(n-1)^2)
\end{align}
\b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Use differentiation under integral sign to prove $\int_{0}^{\infty} e^{-(x^2+\frac {a^2}{x^2})b^2} dx=\frac {\sqrt {\pi}}{2b} e^{-2ab^2}$ Using differentiation under integral sign, prove that $$\int_{0}^{\infty} e^{-\Big(x^2+\frac {a^2}{x^2}\Big)b^2} dx=\frac {\sqrt {\pi}}{2b}e^{-2ab^2}$$
My Attempt:
Let,
$$F(a)= \int_... | Proceed by averaging the two expressions as follows
\begin{align}
F’(a)&= -b^2 \int^{\infty}_{0} e^{-\Big(x^2+\frac {a^2}{x^2} \Big)b^2} \left(1+\frac a{x^2}\right)dy\\
&=- b^2 e^{-2ab^2}\int^{\infty}_{0} e^{-\left(x-\frac {a}{x} \right)^2 b^2} d\left(x-\frac a{x}\right)\\
&=- b \>e^{-2ab^2}\int^{\infty}_{-\infty} e^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solution for the roots of $x^4+x^2+1=0$ Is this solution to find the roots of $x^4+x^2+1=0$ correct?
$x^4+x^2+1=0$
$x^4+2x^2+1-x^2=0$
$(x^2+1)^2-x^2=0$
$[(x^2+1)-x][(x^2+1)+x]=0$
$(x^2-x+1)(x^2+x+1)=0$
For this equation to be true, either $(x^2-x+1)=0$ or/and $(x^2+x+1)=0$.
Using the quadratic formula, I got
$x=\frac... | Given equation
\begin{eqnarray}
x^4+x^2+1=0
\end{eqnarray}
Suppose $x^2=y$ then $y^2+y+1=0$
using quadratic formula we have
\begin{eqnarray}
y=\frac{-1\pm \sqrt{-3}}{2}\\
\end{eqnarray}
\begin{eqnarray}
Thus~~~~~~~~ x^2=\frac{-1\pm \sqrt{-3}}{2}\\
\Rightarrow ~~x=\pm\sqrt\frac{-1\pm \sqrt{-3}}{2}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$ Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following eq... | $$\frac{x}{1+x^2}=\frac{1}{1-(\color{red}{1-x-\frac 1x})}$$
$$=\frac{1}{1-u}$$
When $ x \to 0 $, $ u =\color{red}{1-x-\frac 1x} $ goes to $ \infty $, so you are not allowed to repalce $ X $ by $ u $ in the expansion
$$\frac{1}{1-X}=1+X+X^2+..$$
In the first part, $ u=x(2-x)$ goes to zero.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$ For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$
$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117... | Because $(a+b+c)(ab+bc+ca) \geqslant 9abc,$ so we will prove stronger inequality
$$(a^2+b^2+c^2)^3 \geqslant (a+b+c)(ab+bc+ca)(a^3+b^3+c^3).$$
or
$$(a^2+b^2+c^2-ab-bc-ca)^2\sum (a^2+bc)+ \frac{ab+bc+ca}{2} \sum a^2(b-c)^2 \geqslant 0.$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Fourier series for function $f(x) = \arccos (\cos x), \ \ x\in ]-\pi, \pi[$ $f(x) = \arccos (\cos x), \ \ x\in [-\pi; \pi]$
Here is what I did:
$$a_0 = \frac{1}{2\pi} \int^{\pi}_{-\pi} \arccos (\cos x)dx = \frac{\pi}{2} \\ a_n = \frac{1}{\pi} \int^{\pi}_{-\pi} \arccos (\cos x) \cos \left(\frac{\pi n x}{\pi} \right)dx =... | I am a bit puzzled as to why you write $f(x) = \arccos(\cos x))$. This is identical to $f(x) = \lvert x \rvert$ which seems to me a simpler form, and I think is the form you used to evaluate the integrals.
Then the calculation for $a_0 = \pi / 2$ is fine. $b_n = 0$ because $f(x) \sin nx$ is an odd function. For $a_... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the minimun of $MN+\frac{3}{5}MP$, $MN$ and $MP$ is two sides of a quadrilateral. In a quadrilateral $OPMN$ ，$\angle NOP=90^\circ$,$ON=1$,$OP=3$, and $M$ satisfy $\vec{MO}\cdot\vec{MP}=4$, find the minimum of
$MP+\frac{3}{5}MN$
I choose the vertex $O$ of $OPMN$ as the origin of the coordinate system, and ON as th... | Now, $f'(\theta)=0$ gives $$(17\cos\theta-3\sin\theta-5)(15\sin2\theta+10\cos2\theta-6\sin\theta-16\cos\theta)=0$$ and
$$17\cos\theta-3\sin\theta-5=0$$ gives a minimal value for
$$\frac{17}{\sqrt{298}}\cos\theta-\frac{3}{\sqrt{298}}\sin\theta=\frac{5}{\sqrt{298}},$$
which gives $$\theta_{min}=-\arccos\frac{17}{\sqrt{29... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$ for real $t$ and $0
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$
where $0<x<1$ and $t\in\mathbb{R}$. Prove that $x=1/2$.
It is evident that $x=1/2$ satisfies the above equation. Please help.
| Well, define $y \in \mathbb{R}$ such that $x=1-y$ (noting that also $0<y<1$), we get
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$
$$\implies \frac{t^2-(1-y)^2}{(t^2+(1-y)^2)^2}+\frac{y^2-t^2}{(y^2+t^2)^2}=0$$
Note that the first equation implies
$$\frac{t^2-x^2}{(t^2+x^2)^2}=\frac{t^2-(1-x)^2}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Combinations series: $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$
Evaluate $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$ for $n=10$.
Attempt: I'll deal with the case n being even, as we need to evaluate for n=10.
the numerator is
$${n \choose 1}(n-1)^3+{... | Here's an alternative approach that does not depend on the hint.
Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=\binom{n}{k+1}(k+1)^3$ to obtain
\begin{align}
&\sum_{k\ge 0} \bi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Explain the value of $(0,0)$ for $f(x,y)=1+x\frac{x^2-y^2}{x^2+y^2}$
Let $f:\mathbb R^2\to\mathbb R$ (where $\mathbb R$ is the real numbers) continuous in $\mathbb R^2$, such that
$$f(x,y)=1+x\frac{x^2-y^2}{x^2+y^2}$$ for $(x,y)\neq(0,0)$
Justify the value of $f(0,0)$.
The function is continuous in its domain, and... | We have: $\left|x\cdot \dfrac{x^2-y^2}{x^2+y^2}\right| \le |x|\cdot \left|\dfrac{x^2}{x^2+y^2}+\dfrac{y^2}{x^2+y^2}\right| = |x| \implies \displaystyle \lim_{(x,y) \to (0,0)} f(x,y) = 1+0 = 1 \implies f(0,0) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calcuating an Integral via Residues I am trying to solve the following integral:
$$\int_0^\infty\frac{dx*x}{(x^2+1)(2+x)}$$
What I have done is an analytical extenstion to the upper half plane only by recognising that the above integral is just 1/2 the integral over the entire real line, then only considered poles that... | Consider $\displaystyle \oint_C \frac{z \, \textrm{Log }z \, dz}{(z^2+1)(z+2)}$ where $C$ is the keyhole contour that goes from $\epsilon$ to $R$ then around a large circle of radius $R$ counterclockwise, from $R$ to $\epsilon$ (below the positive real axis) and around a circle of radius $\epsilon$ clockwise. There ar... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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When the quotient $\frac {13n^2+n}{2n+2}$ is an integer? Find the nonzero values of integer $n$ for which the quotient $\frac {13n^2+n}{2n+2}$ is an integer?
My Attempt
I assumed $(2n+2 )| (13n^2+n)$ implies existence of integer $k$ such that $$13n^2+n=k(2n+2)$$
$\implies\ 13n^2+(1-2k)n-2k=0$
$\implies\ n=\frac{(2k-1)\... | You made a mistake in the last line of your post. The $n^2$ shouldn’t be there. You need: $(1-2k)^2+104=h^2\implies h^2-(1-2k)^2=104\implies (h+1-2k)(h-1+2k)=104=1\cdot 104= 2\cdot 52 = 4\cdot 26= 8\cdot 13= (-1)(-104) = (-2)(-52) = (-4)(-26) = (-8)(-13)$. Those are the possibilities for them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Rearranging these formulas The entropy of mixing formula is $\Delta S_{\mathrm{mix}} = R(x_1\ln(x_1) + x_2\ln(x_2))$ where $x$ is the mole fraction $x_1 = \frac{n_1}{n_1+n_2}$ and $x_2 = \frac{n_2}{n_2+n_1}$ where $n$ is the number of moles and the subscript denotes which compound it is. I am asked to express this entr... | Given
\begin{align}
\Delta S_{\mathrm{mix}}
&=
R(x_1\ln(x_1) + x_2\ln(x_2))
\tag{1}\label{1}
,\\
x_1 &= \frac{n_1}{n_1+n_2}
\tag{2}\label{2}
,\\
x_2 &= \frac{n_2}{n_1+n_2}
\tag{3}\label{3}
,\\
y_1 &= \frac{m_1}{m_1+m_2}
\tag{4}\label{4}
,\\
m_1 &= n_1M_1
\tag{5}\label{5}
,\\
m_2 &= n_2M_2
\tag{6}\label{6}
.
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$ I'm having trouble proving that for any $x,y,z>0$ such that $x+y+z=1$ the following inequality is true:
$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$
It seems to me that Jensen's inequality could do the trick, but I'm having tro... | Another way.
We need to prove that
$$\sum_{cyc}\left(\frac{3}{2}-\frac{3x+1}{x+1}\right)\geq0$$ or
$$\sum_{cyc}\frac{1-3x}{x+1}\geq0$$ or
$$\sum_{cyc}\left(\frac{1-3x}{x+1}+\frac{3}{4}(3x-1)\right)\geq0$$ or
$$\sum_{cyc}(3x-1)\left(\frac{3}{4}-\frac{1}{x+1}\right)\geq0$$ or
$$\sum_{cyc}\frac{(3x-1)^2}{x+1}\geq0$$ and w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given polynomial $P(x) = x^2 + ax + b$ and that there only exists one $c$ such that $P^2(c) = c$. Calculate the minimum value of $a + b + c$.
Given polynomial $P(x) = x^2 + ax + b$. Knowing that there only exists one value of real $c$ such that $P^2(c) = c$, calculate the minimum value of $a + b + c$.
Notation: $P^2... | Just a less computational argument for the first part of the solution: if $P(x) = x$ had no real solutions, then $P(x)>x$ for all real $x$ and hence $P(P(x)) > P(x) > x$, contradicting the existence of $c$. Any solution of $P(x) = x$ is also a solution of $P(P(x)) =x$, hence $P(x) = x$ must have exactly one solution, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the sum of the series with terms given by ${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$ The given series has general term as $${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$$
I have tried to approach this problem by making a telescopic series as follows, but I end up cancelling $r$ in the numerator, $$\frac{1}{(r+1)(r+3)}-\frac{1}{(r+3... | In general you can guess (in general more hope) a decomposition $\frac{r}{(r+1)(r+3)(3+4)} = \frac{a}{r+1}+\frac{b}{r+3} + \frac{c}{r+4}$ for certain $a,b,c \in \mathbb{R}$.
So what you have to do is to solve the system given by $ \frac{r}{(r+1)(r+3)(3+4)} = \frac{a}{r+1}+\frac{b}{r+3} + \frac{c}{r+4} = \frac{a(r+3)(r+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:- Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long... | Use $x^8+x^4+1=(x^4+x^2+1)(x^4-x^2+1)$ to partial decompose
\begin{align} \int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx
&=\frac12\int_{0}^{\infty}\frac {x^2+1}{x^4+x^2+1}dx
- \frac12\int_{0}^{\infty}\frac {x^2-1}{x^4-x^2+1}dx\\
&= \frac12\int_{0}^{\infty}\frac {d(1-\frac1x)}{(x-\frac1x)^2+3}
- \frac12\int_{0}^{\infty}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Diophantine equation : $6^m+2^n+2=x^2$
Find $m,n,x\in\mathbb{N}$ such that $6^m+2^n+2=x^2$.
My first approach is to show that for $m,n\geq2$, there exist no solution for $x$ by using modulo $4$.
Case $1$ : $m=1$, $x^2=2^n+8$.
As $n\geq1\implies2\mid RHS\implies2\mid x^2\implies4\mid x^2\implies4\mid LHS\implies 4\mi... | $\textbf{Hint:}$Write the last equality as,
$2^{m-2}3^m=(\bar x-1)(\bar x+1)$.
Here, $gcd(\bar x+1,\bar x-1)=2$ unless $m=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+... | $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}=\int \dfrac{1-\dfrac{1}{x^2}}{x^3 \sqrt{2-\dfrac{2}{x^2}+\dfrac{1}{x^4}}}dx$$$$=\int \frac{\left(1-\dfrac{1}{x^2}\right)dx}{x^3 \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}}$$
$$=\frac14\int \frac{2\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{-2}{x^3}\right)dx}{ \sqrt{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
$\frac{1}{n}+\ln {n}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$ for $n\ge2$. How to prove the following bound:
$\frac{1}{n}+\ln {n}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$ for $n\ge2$.
My Attempt:
We have:$$\int_1^n\frac1xdx=\ln n$$So,$$\sum_{i=1}^n\frac{1}{i+1}\le\int_1^n\frac1xdx=\ln n\le... | Since $e^x>1+x$ for all $x>0$, we have for $k\geq 1$ that
$$
e^{\frac{1}{k}} > 1 + \frac{1}{k} = \frac{{k + 1}}{k}.
$$
Thus
$$
e^{\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} } > \prod\limits_{k = 1}^{n - 1} {\frac{{k + 1}}{k}} = n,
$$
$$
\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} > \log n,
$$
$$
\sum\limits_{k = 1}^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Residue of $f(z)=\frac{1}{z^2(e^z-1)}$ I want to find the residue of the function $f(z)=\frac{1}{z^2(e^z-1)}$. I have tried something that I am 99 precent sure about, but still I would appreciate some feedback. What I have done is that:
$$\frac{1}{z^2(e^z-1)}=\frac{1}{z^2}\cdot\frac{1}{1+z+\frac{z}{2}+o(z^2)-1}=\frac{1... | The value found is not correct. You need to expand the exponential at order $3$:
\begin{align}
\frac{1}{z^2(e^z-1)}&=\frac{1}{z^2\bigl(z+\tfrac{z^2}2+\tfrac{z^3}{6}+o(z^3)\bigr)}=\frac{1}{z^3\bigl(1+\tfrac{z}2+\tfrac{z^2}{6}+o(z^2)\bigr)}\\
&=\frac{1}{z^3}\Bigl(1-\Bigl(\frac{z}2+\frac{z^2}6\Bigr)+\Bigl(\frac{z}2+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$
Simplify:
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
After the substitution as $\cos(x)=a$ and $\s... | So we have :
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
First of all, by using double-angles formulas we can get this:
$$\cos ^2\left(x\right) = \frac{1+\cos\left(2x\right)}{2}$$
$$\sin ^2\left(x\right) = \frac{1-\cos\left(2x\right)}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to calculate the integral $I\left(a,b,c\right)=\int_{a}^{b}exp\left(u^{2}\right)\times\mbox{erfi}\left(\sqrt{\frac{u}{c}}\right)du$? I want to compute the following integral depending on a,b and c all strictly positive real numbers:
$$I\left(a,b,c\right)=\int_{a}^{b}exp\left(u^{2}\right)\times\mbox{erfi}\left(\sqrt... | Let's begin by an expansion of the $\mbox{erfi}(x)$
function:
$$I\left(a,b,c\right)=\int_{a}^{b}exp\left(u^{2}\right)\times\mbox{erfi}\left(\sqrt{\frac{u}{c}}\right)du$$
$$\Leftrightarrow I\left(a,b,c\right)=\int_{a}^{b}exp\left(u^{2}\right)\times\frac{2}{\sqrt{\pi}}\sum_{j=0}^{+\infty}\frac{\sqrt{\frac{u}{c}}^{2j+1}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 1... | $\frac{1}{4}(k-5)^2(k-4)^2$ = $\frac{1}{4}(k^4-18k^3+121k^2-360k+400)$=$pk+17$ for some integer $p$. So, $\frac{1}{4}(k^4-18k^3+121k^2-360k+332)$ = $pk$ => $k$ should divide $332$ . So we get possible values of $k$ as $1,83,166,332$. As $k>4$, $k=1$ is not possible. Similarly for $k=332$, $(k^3-18k^2+121k-360+1)$ is cl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$... | Another way.
Let $a^2+b^2-c^2=z$, $a^2+c^2-b^2=y$ and $b^2+c^2-a^2=x$.
Thus, $x$, $y$ and $z$ are positives, $\cos\alpha=\frac{x}{\sqrt{(x+y)(x+z)}},$ $\cos\beta=\frac{y}{\sqrt{(x+y)(y+z)}}$, $\cos\gamma=\frac{z}{\sqrt{(x+z)(y+z)}}$
and we need to prove that
$$2\sum_{cyc}\frac{x^3}{\sqrt{(x+y)^3(x+z)^3}}+\frac{128x^3y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Doubt about how to compute $\sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$ In the book I am reading, I have encountered the following sum.
$$S = \sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$$
From here, I factored the denominator since it does not seem to be dependent on $n,$ and I rewr... | Let's replace $2019$ by $T$. In the numerator:
$$ \sum_{n=1}^T \sum_{k=1}^n k = \sum_{n=1}^T \frac{n(n+1)}{2} = \frac{(T+1)^3 - (T+1)}{6}$$
In the denominator:
$$ \sum_{k=1}^T k^3 = \frac{(T+1)^4 - 2 (T+1)^3 + (T+1)^2}{4} $$
So:
$$ \frac{ \sum_{n=1}^T \sum_{k=1}^n k }{\sum_{k=1}^T k^3} = \frac{4}{6} \frac{(T+1)^3 - (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
prove formula by induction I need to prove by induction that $\sum_{n=2}^{m} \frac{1}{n^2 - 1} = \frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})$.
I've seen some similar questions about the convergence of the infinite series, however none about the finite case, I've tried in several ways to factorize or add things... | Here is an explicit evaluation which may be useful. This is focused on the $m+1$ portion of the inductive proof:
$P\left(m+1\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{m+1}-\frac{1}{m+2}\right)$
$P\left(m+1\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{m+1}\right)-\frac{1}{2}\frac{1}{m+2}$
In order for the induct... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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Fractions in Questions and Answers
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