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Evaluating $\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx$
How am I supposed to solve the following definite integral?
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx
$$
This definite integral is solved if the minus sign is replaced by a plus sign, and it yields $\pi^2$.
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} + \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx \text{
— (I)} \\
\implies \mathcal{I} = \int_0^\pi (\pi - x) \cdot \frac{\cos{\frac{x}{2} + \sin{\frac{x}{2}}}}{\sqrt{\sin{x}}} dx \text{
— (II)}
$$
On (I) + (II), we have,
$$
\mathcal{I} = \frac{\pi}{2}\int_0^\pi \frac{\sin{\frac{x}{2}} + \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx = \frac{\pi}{2} \int_0^\pi \frac{\sin{\frac{x}{2}}+\cos{\frac{x}{2}}}{\sqrt{1 - (\sin{\frac{x}{2}-\cos{\frac{x}{2}}})^2}} dx
$$
On substitution,
$$
\sin{\frac{x}{2}} - \cos{\frac{x}{2}} = u \implies \left(\sin{\frac{x}{2}} + \cos{\frac{x} {2}}\right) dx = 2 \cdot du
$$
The upper and lower limits changes to 1 and -1. Now, we have
$$
\mathcal{I} = \frac{\pi}{2} \int_{-1}^1 \frac{2 \cdot du}{\sqrt{1 - u^2}} du = \pi \cdot \left[\arcsin{u}\right]_{-1}^1 = \pi^2
$$
But...
*
*The sign was not supposed to be changed. We get $(2x - \pi)$ instead of $\pi$ in the nominator when adding both integrals. It complicates the problem.
*Using integral-calculator.com or a scientific calculator is helpless.
*The answer to the original problem should be $2\pi \cdot \ln{2}$ (approx 4.35.)
| $$I=\int_0^\pi x \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx\overset{IBP}=4\int_0^\frac{\pi}{2}\ln(\sqrt{\sin x}+\sqrt{1+\sin x})dx$$
$$\overset{\large x\to \frac{\pi}{2}-2x}=8\int_0^\frac{\pi}{4}\ln\left(\sqrt{\cos(2x)}+\sqrt 2 \cos x\right)dx\overset{\cos x\to x}=8\int_\frac{1}{\sqrt 2}^1\frac{\operatorname{arccosh}(\sqrt 2x)}{\sqrt{1-x^2}}dx$$
$$\require{cancel}
\mathcal J(a)=\int_\frac{1}{a}^1\frac{\operatorname{arccosh}(ax)}{\sqrt{1-x^2}}dx\Rightarrow \mathcal J'(a)=\frac{1}{a^2}\frac{\cancelto{0}{\operatorname{arccosh}\left(a\frac{1}{a}\right)}}{\sqrt{1-\frac{1}{a^2}}}+\int_\frac{1}{a}^1\frac{x}{\sqrt{1-a^2x^2}\sqrt{1-x^2}}dx$$
$$\overset{1-x^2\to x^2}=\frac1a\int_0^{\sqrt{1-\frac{1}{a^2}}}\frac{1}{\sqrt{1-\frac{1}{a^2}-x^2}}dx=\frac1a\arcsin\left(\frac{x}{\sqrt{1-\frac{1}{a^2}}}\right)\bigg|_0^\sqrt{1-\frac1{a^2}}=\frac{\pi}{2a}$$
$$I=8\left(\mathcal J(\sqrt 2)-\mathcal J(1)\right)=4\pi\int_1^\sqrt 2\frac{1}{a}da=2\pi \ln 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
How to compute $\lim\limits_{x\to 0} (-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$? Consider the function
$$f(x)=(-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$$
How do we compute $\lim\limits_{x\to 0} f(x)$?
Here is my attempt
If $t\geq 0$ then $\frac{t^{2n+2}}{1+t^2} \leq t^{2n+2}$ and if $t\leq 0$ then $t^{2n+2}\leq\frac{t^{2n+2}}{1+t^2}$. That is, for any $t$
$$-|t|^{2n+2}\leq\frac{t^{2n+2}}{1+t^2}\leq |t|^{2n+2}$$
Thus
$$-\int_0^x|t|^{2n+2}\leq\int_0^x\frac{t^{2n+2}}{1+t^2}\leq \int_0^x|t|^{2n+2}$$
$$-\frac{|x|^{2n+3}}{2n+3}\leq\int_0^x\frac{t^{2n+2}}{1+t^2}\leq \frac{|x|^{2n+3}}{2n+3}$$
$$-\lim\limits_{x\to 0}\frac{|x|^{2n+3}}{2n+3}\leq\lim\limits_{x\to 0}\int_0^x\frac{t^{2n+2}}{1+t^2}\leq \lim\limits_{x\to 0}\frac{|x|^{2n+3}}{2n+3}$$
$$0\leq\lim\limits_{x\to 0}\int_0^x\frac{t^{2n+2}}{1+t^2}\leq 0$$
$$\lim\limits_{x\to 0}\int_0^x\frac{t^{2n+2}}{1+t^2}=0$$
With this result in hand, we go back to the original limit.
$$\lim\limits_{x\to 0}(-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$$
$$=(-1)^{n+1}\lim\limits_{x\to 0}\frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$$
$$=\frac{0}{0}$$
By L'Hopital's Rule
$$=(-1)^{n+1}\lim\limits_{x\to 0}\frac{\frac{x^{2n+2}}{1+x^2}}{(2n+1)x^{2n}}$$
$$=(-1)^{n+1}\lim\limits_{x\to 0} \frac{x^2}{(2n+1)(1+x^2)}$$
$$=0$$
is this calculation correct? I know the result is correct, but I am interested in the steps.
For context, $f(x)$ is the remainder in a Taylor Polynomial of the $\arctan$ function. The book I am reading (Spivak's Calculus) skipped over the details of this calculation.
| You can expand ${1\over 1+t^2}$ as a geometric series as $x \rightarrow 0$.
$$\lim_{x \rightarrow 0} {1\over x^{2n+1}}{\int_{0}^{x}{t^{2n+2} \over {1+t^2}}dt}\space =\space\lim_{x \rightarrow 0} {1\over x^{2n+1}}\sum_{k=0}^{\infty}{(-1)^k}\int_{0}^{x}{t^{2n+2k+2}dt}$$
$$\lim_{x \rightarrow 0} {1\over x^{2n+1}}\sum_{k=0}^{\infty}{(-1)^k\cdot x^{2n+2k+3}\over(2n+2k+3)}=\space\lim_{x \rightarrow 0} \sum_{k=0}^{\infty}{(-1)^k\cdot x^{2k+2}\over(2n+2k+3)}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\int_{0}^{\pi} \ln (\sin x+2) d x$ and $\int_{0}^{\pi} \ln (2-\sin x) d x$ I first encountered this integral
$$
I=\int_{0}^{\pi} \ln (\sin x+2) d x
$$
several months ago without any idea and had tried many methods such as integration by parts, substitution and Fourier series etc. but all are in vain. Today, I tried the tangent substitution and succeeded. Now I want to share with you and seek any other alternatives.
For simplicity, I convert, by symmetry, the integral into
$$
I=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x+2) d x
$$
and substitute $t=\tan x$, and get
$$
\begin{aligned}
I &=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x+2) d x \\
&=2 \int_{0}^{\infty} \ln \left(\frac{2 t}{1+t^{2}}+2\right) \frac{d t}{1+t^{2}} \\
&=2 \int_{0}^{\infty} \frac{\ln 2+\ln \left(t^{2}+t+1\right)-\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \int_{0}^{\infty} \frac{\ln \left(t^{2}+t+1\right)}{1+t^{2}}-2 \int_{0}^{\infty} \frac{\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \underbrace{ \left[\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3} G\right]}_{(*)} -2 \underbrace{\pi \ln 2}_{(**)} \\
&=-\pi \ln 2+\frac{2 \pi}{3} \ln (2+\sqrt{3})+\frac{8}{3} G
\end{aligned}
$$
Note:
(*): post 1, (**):post 2
For the second integral, we use the similar technique and arrive at
\begin{aligned}
J&=2 \int_{0}^{\infty} \frac{\ln 2+\ln \left(1-t+t^{2}\right)-\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=2 \int_{0}^{\infty} \frac{\ln 2}{1+t^{2}} d t+2 \int_{0}^{\infty} \frac{\ln \left(1-t+t^{2}\right)}{1+t^{2}} d t-2 \int \frac{\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \underbrace{\left(\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G\right)}_{(***)}-2 \pi \ln 2 \\
&=-\pi \ln 2+\frac{4 \pi}{3} \ln (2+\sqrt{3})-\frac{8}{3} G
\end{aligned}
Note:(***) post 3
Eager to know whether there are any alternatives!
Comments and alternative solutions are highly appreciated.
| To evaluate the second integral, we may use their product.
$$\int_{0}^{\pi} \ln (\sin x+2) d x+ \int_{0}^{\pi} \ln (2-\sin x) d x = 2\int_{0}^{\pi/2} \ln (4-\sin^2 x) d x $$
Using my post,
$$\begin{aligned}\int_{0}^{\pi} \ln (2-\sin x) d x &=2 \pi \ln \left(\frac{2+\sqrt{3}}{2}\right) -\left(-\pi \ln 2+\frac{2 \pi}{3} \ln (2+\sqrt{3})+\frac{8}{3} G \right)\\&= -\pi \ln 2+\frac{4 \pi}{3} \ln (2+\sqrt{3})-\frac{8}{3} G \end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $f(x) =ax^3+bx^2+cx+d$ is a cubic equation with roots $\alpha,\beta,\gamma.$ Is there a way to find $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha?$ Suppose $f(x) = ax^3 + bx^2 + cx + d$ is a cubic equation with roots $\alpha, \beta, \gamma.$ Then we have:
$\alpha + \beta + \gamma= -\frac{b}{a}\quad (1)$
$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\quad (2)$
$\alpha\beta\gamma = -\frac{d}{a}\quad (3)$
We can find $\alpha^2\beta + \beta^2\gamma + \gamma^2\alpha + \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha$ in terms of $a,b,c,d$ with the formula:
$$ \alpha^2\beta + \beta^2\gamma + \gamma^2\alpha + \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha = (\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma) - 3\alpha\beta\gamma $$
$$=\left(\frac{-b}{a}\right) \left(\frac{c}{a}\right) - 3\left(-\frac{d}{a}\right).$$
But I was wondering if there was some way to find $ \alpha^2\beta + \beta^2\gamma + \gamma^2\alpha\ $ and therefore also $\ \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha\ $ in terms of $a,b,c,d,\ $ with some algebraic manipulation, i.e. without finding the roots with a cubic formula?
Notice that there are two possible values of $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha,$ namely $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha = \beta^2\gamma+\gamma^2\alpha+\alpha^2\beta = \gamma^2\alpha+\alpha^2\beta+\beta^2\gamma$ and $\alpha^2\gamma+\gamma^2\beta+\beta^2\alpha = \gamma^2\beta+\beta^2\alpha+\alpha^2\gamma = \beta^2\alpha + \alpha^2\gamma+\gamma^2\beta.$
| Quantity is not-symmetric. This results in rather complex formula:
$$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=t_{1}\,t_{3}^2-{{b\,t_{3}^2}\over{3\,a}}+t_{2}^2\,t_{3}-{{2\,b\,
t_{2}\,t_{3}}\over{3\,a}}-{{2\,b\,t_{1}\,t_{3}}\over{3\,a}}+\\{{b^2\,
t_{3}}\over{3\,a^2}}-{{b\,t_{2}^2}\over{3\,a}}+t_{1}^2\,t_{2}-{{2\,b
\,t_{1}\,t_{2}}\over{3\,a}}+{{b^2\,t_{2}}\over{3\,a^2}}-{{b\,t_{1}^2
}\over{3\,a}}+{{b^2\,t_{1}}\over{3\,a^2}}-{{b^3}\over{9\,a^3}}$$
$$t_i=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4}-\frac{p^3}{27}}}\,e^{k_i\frac{2i\pi}{3}}+\frac{p}{3\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4}-\frac{p^3}{27}}}}e^{-k_i\frac{2i\pi}{3}}, i\in\{1,2,3\}$$
$$p={{b^2}\over{3\,a^2}}-{{c}\over{a}}, q={{b\,c
}\over{3\,a^2}}-{{d}\over{a}}-{{2\,b^3}\over{27\,a^3}}$$
$\{k_1,k_2,k_3\}=\{0,1,2\}$ determines order of roots taken as $\alpha$, $\beta$, $\gamma$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
\dfrac{5}{2} & \dfrac{3}{2}\\
-\dfrac{3}{2} & -\dfrac{1}{2}
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it comes out to be something like this,
$$A^2 = \begin{bmatrix}
4 & 3\\
-3 & -2
\end{bmatrix}$$
$$ A^3 = \begin{bmatrix}
\dfrac{11}{2} & \dfrac{9}{2}\\
-\dfrac{9}{2} & -\dfrac{7}{2}
\end{bmatrix}$$
$$A^4 = \begin{bmatrix}
7 & 6\\
-6 & -5
\end{bmatrix}$$
so $$ A^n = \begin{bmatrix}
\dfrac{3n}{2}+1 & \dfrac{3n}{2}\\
-\dfrac{3n}{2} & -\dfrac{3n}{2}+1
\end{bmatrix} \tag{1}\label{eq1}$$
thus after substituting $n=2022$ we get $ A^{2022} = \begin{bmatrix}
3034 & 3033\\
-3033 & -3032
\end{bmatrix}$
$$\\$$
Is there is any other way to find the value of $A^{2022}$ without finding out the pattern and then general equation \eqref{eq1} like I did above?
| The characteristic polynomial of $A$ is
\begin{align}
\det(A-\lambda I)=\left|\begin{array}{c}
\frac{5}{2}-\lambda & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}-\lambda
\end{array}\right|\\
= (5/2-\lambda)(-1/2-\lambda)+9/4 \\
= -5/4-2\lambda+\lambda^2+9/4 \\
= (\lambda-1)^2
\end{align}
Cayley-Hamilton gives $(A-I)^2=0$.
So, a binomial expansion works great because only two terms are non-zero:
\begin{align}
A^{2022} &= (I+(A-I))^{2022} \\
&=I+2022 (A-I) \\
&=2022 A - 2021 I \\
&=2022\left[\begin{array}{rr}
\frac{5}{2} & \frac{3}{2} \\
-\frac{3}{2} & -\frac{1}{2}
\end{array}\right]
-\left[\begin{array}{rr}2021 & 0 \\ 0 & 2021\end{array}\right] \\
&=\left[\begin{array}{rr}
3034 & 3033 \\
-3033 & -3032
\end{array}\right].
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Calculate $\frac{4-5\sin\alpha}{2+3\cos\alpha}$
Calculate $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}$$ if $\cot\dfrac{\alpha}{2}=-\dfrac32$.
My first approach was to somehow write the given expression only in terms of the given $\cot\frac{\alpha}{2}$ and just put in the value $\left(-\dfrac{3}{2}\right)$. Now I don't think that's possible because we have constants (4 and 2). My try, though: $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}=\dfrac{4-5.2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\left(\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}\right)}=\dfrac{4-10\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\cos^2\frac{\alpha}{2}-3\sin^2\frac{\alpha}{2}}$$ My second idea was to find the value of the trig functions of $\alpha$. I don't know if this is the most straight-forward approach, but $$\cot\alpha=\dfrac{\cot^2\frac{\alpha}{2}-1}{2\cot\frac{\alpha}{2}}=\dfrac{\frac94-1}{-2.\frac32}=-\dfrac{5}{12}.$$ Am I now supposed just to find the values of $\sin\alpha$ and $\cos\alpha$? Nothing more elegant? We would have $\dfrac{\cos\alpha}{\sin\alpha}=-\dfrac{5}{12}\Rightarrow\cos\alpha=-\dfrac{5}{12}\sin\alpha$ and putting into $\sin^2\alpha+\cos^2\alpha=1$ we'd get $\cos\alpha=\pm\dfrac{12}{13}$.
| You just use the trigonometrical formulae: $sina=\dfrac{2tan(a/2)}{1+tan^{2}(a/2)}$ and
$cosa=\dfrac{1-tan^{2}(a/2)}{1+tan^{2}(a/2)}$ and set $\,\,\,tan(a/2)=\dfrac{1}{cot(a/2)}=-\dfrac{2}{3}$.
If we put this value in the original fraction we obtain :
value=$\dfrac{102}{41}$. If my numerical calculations are correct!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
What is wrong in this solution of $\sec x + \csc x = 2 \sqrt{2}$
Find number of solutions in the interval $[0,2\pi]$ of the equation - $$\csc x + \sec x = 2 \sqrt{2}.$$
$⇒ \dfrac{1}{\sin x}+ \dfrac{1}{\cos x }= 2 \sqrt2$
$⇒ \dfrac{\sin x +\cos x}{\sin x \cos x} = 2 \sqrt 2 ⇒ (\sin x+\cos x)^2=8\sin^2x\cos^2x⇒1+2\sin x \cos x=8\sin^2x \cos^2x$
$⇒\sin x \cos x=-1/4 ,\sin x \cos x=1/2⇒2\sin x \cos x=-1/2 , 2\sin x\cos x=1$
$⇒\sin2x =-1/2 , \sin2x=1.$ But if we check manually , $\sin2x=-1/2$ is not a solution.
$⇒\sin 2x =1⇒2x=(-1)^{n}\dfrac{\pi}{2}+n\pi⇒x=(-1)^{n}\dfrac{\pi}{4}+\dfrac{n\pi}{2}⇒$ For $[0,2\pi], \boxed{x=\dfrac{\pi}{4},\dfrac{5 \pi}{4}}$ are solutions.
But if we check the graph neither are the solutions . But $\pi /4 $ is a solution but $5\pi /4 $ is not a solution if we plug values manually.
Can someone point out the mistake? Thanks
$\text{References:}$ Link for above graph : https://www.desmos.com/calculator/eaakqvd9y5
| Hint:
Instead of squaring $$\dfrac {\sin x + \cos x}{\sin x \cos x} = 2\sqrt{2}$$ clear fractions to get $$\sin x + \cos x = 2\sqrt{2}\sin x \cos x$$ and note that you have a double angle identity hidden on the RHS. Then you can use another identity to get the LHS and find the relationship between the two.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Triangle ABC with $\angle{ACB} = 3\angle{ABC }$ and $AB = \frac{10}{3}BC$. Find $\cos{A}\cos{B}\cos{C}$ assume that $\angle{ABC}=y, \angle{ACB}=3y, BC=3x,$ and $AB=10x$ then by sine rule, i obtain following
$ \frac{10x}{\sin{3y}}=\frac{AC}{\sin{y}}=\frac{3x}{\sin{4y}}$
by cosine rule in try to figure out AC
$AC=\sqrt{109x^2-60x^2\cos{y}}$
i have no idea how to combine these two informations in order to solve the problem
| Set $\cos y=p$, then you can get this equation in sine rule you wrote:
$$
10\sin4y=3\sin 3y
\\
40\sin y\cos y\cos2y=9\sin y-12\sin^3 y
\\
40p(2p^2-1)=9-12(1-p^2)=12p^2-3
\\
80p^3-12p^2-40p+3=(4p-3)(20p^2+12p-1)=0
$$
Then, you can see $\cos B=\frac34$, and the only thing left is the calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the all real roots of the polynomial $x^6+3 x^5+3 x-1=0$ in closed form
Find the all real roots of the polynomial
$$x^6+3 x^5+3 x-1=0$$
in exact form.
WolframAlpha gives only numerical results. I've asked a few similar questions before. The source of the problem comes from the algebra precalculus workbook (but not homework workbook).
The rational root theorem doesn't work. Because, $x=1$ and $x=-1$ are not roots.
Wolfram says , we have only $2$ real roots. What kind of factorization should I try?Can a factorization of the form $P(x)=(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)$ work?
I also tried the trick $\frac {P(x)}{x^n}$, but I failed.
| This problem is not easy, but you can simplify the work a bit by observing that you can divide both sides by $x^3$. It can be shown that $x=0$ is not a solution to the original polynomial, hence this operation is allowed. We get
$$x^3-\frac 1{x^3}+3\left(x^2+\frac 1{x^2}\right)=0$$
Now, use the two identities below
\begin{align*}
x^3-\frac 1{x^3} & =\left(x-\frac 1x\right)^3+3\left(x-\frac 1x\right)\\
x^2+\frac 1{x^2} & =\left(x-\frac 1x\right)^2+2
\end{align*}
Our polynomial now becomes
$$\left(x-\frac 1x\right)^3+3\left(x-\frac 1x\right)+3\left(x-\frac 1x\right)^2+6=0$$
Make the substitution $x-\frac 1x=z$ and the equation is now a cubic in $z$.
$$z^3+3z^2+3z+6=0$$
Use Cardano's method to solve this cubic for $z$ and calculate $x$ by using
$$x-\frac 1x=z$$
The resulting six solutions are the roots of your original polynomial.
EDIT: Here's an actual attempt at solving the problem. Starting with the cubic in $z$, then we get
$$(z+1)^3=-5\qquad\implies\qquad z_0=-1-\sqrt[3]{5}$$
Since the OP is only interested in the real roots, we can neglect the two other complex roots. However, for thoroughness, the two other complex roots would be
\begin{align*}
z_1 & =-1-\omega\sqrt[3]{5}\\
z_2 & =-1-\omega^2\sqrt[3]{5}
\end{align*}
Where $\omega$ is the cube root of unity
$$\omega=\frac {-1+i\sqrt3}2\qquad\qquad\qquad\omega^2=-\frac {1+i\sqrt3}2$$
Returning to the substitution in $x$, we have that
$$x^2-z_0x-1=0$$
A direct application of the quadratic formula gives the two real roots as
\begin{align*}
x_+ & \color{blue}{=\frac 12\left(-1-\sqrt[3]{5}+\sqrt{5+2\sqrt[3]{5}+\sqrt[3]{25}}\right)}\\
x_- & \color{blue}{=\frac 12\left(-1-\sqrt[3]{5}-\sqrt{5+2\sqrt[3]{5}+\sqrt[3]{25}}\right)}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\lim _{x \to \infty }\frac{1 - 2x^2}{x^2 + 3}\ = -2$ I solved it using the definition of limit to the $\left|\frac{-7}{x^2 +3}\right| < \varepsilon$
Then $\varepsilon > \frac{7}{x + 3} > \frac{7}{x^2 +3}$
Then $x > \frac{7}{\varepsilon} - 3$
But my teacher said this was wrong and I have to fix it to
$x > \sqrt{\left|\frac{7}{\varepsilon} - 3\right|}$
Can anyone please explain why am I wrong here? Thank you very much
| We have that for $0<\varepsilon <\frac 7 3$
$$\left|\frac{1 - 2x^2}{x^2 + 3}\ -(-2)\right|<\varepsilon \iff \left|\frac{7}{x^2 + 3}\ \right|<\varepsilon \iff x^2>\frac 7 \varepsilon -3$$
which leads to
$$x>\sqrt{\frac 7 \varepsilon -3}$$
or also, as you teacher suggests, for any $\varepsilon>0$
$$x>\sqrt{\left|\frac 7 \varepsilon -3\right|}$$
Your way is not wrong when we deal with $x$ sufficiently large (i.e. $x>1$), such that $x<x^2$, and for any fixed $0<\varepsilon<\frac 7 4$, indeed by the definition of limit if we have
$$ x>x_0=\sqrt{\frac 7 \varepsilon-3}>1 \implies \frac{7}{x^2 +3}<\varepsilon $$
by
$$\frac{7}{x +3}<\varepsilon \implies x\ge x_1=\frac 7 \varepsilon-3>1$$
we also have that
$$x_1=\frac 7 \varepsilon-3>\sqrt{\frac 7 \varepsilon-3}=x_0$$
and therefore assuming $x>x_1$ suffices.
For $\varepsilon\ge \frac 7 4$ we just need $x>1$.
| {
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"answer_count": 3,
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} |
How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$? So this was part of a bigger induction proof problem, but I'm stuck at what I think is the last step. I need to show that
$\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$, for $n \ge 1$
Would it be enough to show that the limit as it approaches positive infinity is both 1, and that since the inequality is true for $n=1$, and $1/2 < \sqrt{1/2}$, the function on the right is always greater? I'm pretty sure this isn't the way to go about it but I'm really stuck.
| There are lots of ways to prove this.
One way is to square both sides and
cross-multiply.
Another way is to write
$\dfrac{2n+1}{2n+2}
=1-\dfrac{1}{2n+2}
$
and
$\dfrac{\sqrt{n+1}}{\sqrt{n+2}}
=\sqrt{\dfrac{n+1}{n+2}}
=\sqrt{1-\dfrac{1}{n+2}}
$.
Squaring both sides,
you want
$1-\dfrac{1}{n+2}
\ge (1-\dfrac{1}{2n+2})^2
=1-\dfrac1{n+1}+\dfrac1{(2n+2)^2}
$
or
$\dfrac1{(2n+2)^2}
\le \dfrac1{n+1}-\dfrac1{n+2}
=\dfrac1{(n+1)(n+2)}
$
and this is true
for all $n \ge 0$.
Another way is write it as
$\dfrac{2n+2}{2n+1}
\ge \dfrac{\sqrt{n+2}}{\sqrt{n+1}}
$
or
$1+\dfrac{1}{2n+1}
\ge \sqrt{1+\dfrac{1}{n+1}}
$.
Squaring,
this is
$1+\dfrac{2}{2n+1}+\dfrac{1}{(2n+1)^2}
\ge 1+\dfrac{1}{n+1}
$
or
$\begin{array}\\
0
&\le \dfrac{2}{2n+1}-\dfrac{1}{n+1}+\dfrac{1}{(2n+1)^2}\\
&= \dfrac{2(n+1)-(2n+1)}{(2n+1)(n+1)}+\dfrac{1}{(2n+1)^2}\\
&= \dfrac{1}{(2n+1)(n+1)}+\dfrac{1}{(2n+1)^2}\\
\end{array}
$
which is always true.
| {
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"answer_count": 6,
"answer_id": 1
} |
$\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi}\dfrac{1}{(\sin\theta - 2\cos\theta)^2}\,d\theta$: singularity in Weierstrass sub Could I use the integration by Weierstrass substitution or tangent-half angle substitution for this case? Because I see on Wikipedia, when the lower limit was $0$, they made the definite integral split into two things, where the first integral is from $0$ to $\pi$ and the second one is $\pi$ to $2 \pi$. Because the article said that "In the first line, one cannot simply substitute $t=0$ for both limits of integration."
This is what the article work like this
Blockquote
$$\begin{align}
\int_0^{2\pi}\frac{dx}{2+\cos x}
&= \int_0^\pi \frac{dx}{2+\cos x} + \int_\pi^{2\pi} \frac{dx}{2+\cos x} \\[6pt]
&=\int_0^\infty \frac{2\,dt}{3 + t^2} + \int_{-\infty}^0 \frac{2\,dt}{3 + t^2} & t &= \tan\tfrac x2 \\[6pt]
&=\int_{-\infty}^\infty \frac{2\,dt}{3+t^2} \\[6pt]
&=\frac{2}{\sqrt 3}\int_{-\infty}^\infty \frac{du}{1+u^2} & t &= u\sqrt 3 \\[6pt]
&=\frac{2\pi}{\sqrt 3}.
\end{align}$$
In the first line, one cannot simply substitute
${\textstyle t=0}$
for both limits of integration. The singularity (in this case, a vertical asymptote) of
${\textstyle t=\tan {\tfrac {x}{2}}}$
at
${\textstyle x=\pi }$
must be taken into account. Alternatively, first evaluate the indefinite integral, then apply the boundary values.
$${\displaystyle {\begin{aligned}\int {\frac {dx}{2+\cos x}}&=\int {\frac {1}{2+{\frac {1-t^{2}}{1+t^{2}}}}}{\frac {2\,dt}{t^{2}+1}}&&t=\tan {\tfrac {x}{2}}\\[6pt]&=\int {\frac {2\,dt}{2(t^{2}+1)+(1-t^{2})}}=\int {\frac {2\,dt}{t^{2}+3}}\\[6pt]&={\frac {2}{3}}\int {\frac {dt}{{\bigl (}t{\big /}{\sqrt {3}}{\bigr )}^{2}+1}}&&u=t{\big /}{\sqrt {3}}\\[6pt]&={\frac {2}{\sqrt {3}}}\int {\frac {du}{u^{2}+1}}&&\tan \theta =u\\[6pt]&={\frac {2}{\sqrt {3}}}\int \cos ^{2}\theta \sec ^{2}\theta \,d\theta ={\frac {2}{\sqrt {3}}}\int d\theta \\[6pt]&={\frac {2}{\sqrt {3}}}\theta +C={\frac {2}{\sqrt {3}}}\arctan \left({\frac {t}{\sqrt {3}}}\right)+C\\[6pt]&={\frac {2}{\sqrt {3}}}\arctan \left({\frac {\tan {\tfrac {x}{2}}}{\sqrt {3}}}\right)+C.\end{aligned}}} $$
By symmetry,
$$\begin{align}
\int_{0}^{2\pi} \frac{dx}{2 + \cos x} &= 2 \int_{0}^{\pi} \frac{dx}{2 + \cos x} = \lim_{b \rightarrow \pi} \frac{4}{\sqrt3} \arctan \left( \frac{\tan\tfrac x2}{\sqrt3}\right) \Biggl|_{0}^{b}\\[6pt]
&= \frac{4}{\sqrt3} \Biggl[ \lim_{b \rightarrow \pi} \arctan \left(\frac{\tan\tfrac b2}{\sqrt3}\right) - \arctan (0) \Biggl] = \frac{4}{\sqrt 3} \left( \frac{\pi}{2} - 0\right) = \frac{2\pi}{\sqrt 3},
\end{align}$$
From: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution
| THASsing, as I like to call the process, is perfectly fine here as there are no poles (which solve $\sin x=2\cos x$ or $\tan x=2$) on the interval of integration. After simplifying you get
$$\int_1^\infty\frac{1+t^2}{2(t^2+t-1)^2}\,dt=\left[-\frac t{2(t^2+t-1)}\right]_1^\infty=\frac12$$
| {
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Squeeze Theorem for $(2^n + 3^n + 4^n)^{\frac{1}{n}}$ How to find this limit when $n \rightarrow \infty$.
$$(2^n + 3^n + 4^n)^{\frac{1}{n}}.$$
My Idea:
We need to squeeze this sequence.
$$ (4^n)^{\frac{1}{n}}\le (2^n + 3^n + 4^n)^{\frac{1}{n}} \le (3\cdot4^n)^{\frac{1}{n}}$$
So that we obtain
$$ 4\le (2^n + 3^n + 4^n)^{\frac{1}{n}} \le (3)^{\frac{1}{n}} 4$$
By taking the limit $n \rightarrow \infty$
$$4\le (2^n + 3^n + 4^n)^{\frac{1}{n}} \le 4 $$
So, by squeeze theorem
$$\lim_{n \to \infty} (2^n + 3^n + 4^n)^{\frac{1}{n}} = 4.$$
Actually, I still doubt about my idea.
Is my idea correct? Thanks in advance.
| I have another idea that might be useful:
$$
(2^n+3^n+4^n)^\frac{1}{n}=\frac{4}{4}(2^n+3^n+4^n)^\frac{1}{n}=4\left(\frac{2^n+3^n+4^n}{4^n}\right)^\frac{1}{n}=4\left(\left(\frac{2}{4}\right)^n+\left(\frac{3}{4}\right)^n+1\right)^\frac{1}{n}
$$
Taking limits you get $4$
| {
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} |
$\epsilon$-$\delta$ argument Given $x \in \left ( \frac{1}{10}, \frac{3}{10} \right)$. Find $\delta > 0$ so that
$$0< \left| x- \frac{1}{5} \right| < \delta \Rightarrow \left| \frac{1}{x} - 5\right| < \frac{1}{10}.$$
How can I answer that question?
My Idea
Let's do a little observation:
$$\left| \frac{1}{x} - 5\right| < \frac{1}{10} \Longleftrightarrow -\frac{1}{10} < \frac{1}{x}-5 <\frac{1}{10} \Longleftrightarrow \frac{49}{10}< \frac{1}{x}<\frac{51}{10}.$$
We get
$$\frac{10}{51} < x < \frac{10}{49} \Longleftrightarrow \frac{10}{51}-\frac{1}{5}< x - \frac{1}{5} < \frac{10}{49} - \frac{1}{5}.$$
As a consequence
$$\frac{-1}{51 \cdot 5} < x - \frac{1}{5}< \frac{1}{49 \cdot 5} \Longleftrightarrow
\frac{-1}{255} < x - \frac{1}{5} < \frac{1}{255}< \frac{1}{245}.$$
We obtain
$$\left| x - \frac{1}{5}\right| < \frac{1}{255}.$$
Choose $\delta = \frac{1}{255}$, so that the implication above is true (?)
How can I fix this problem? Is my idea correct?.
| Your idea is correct. As you may directly verify, you have the implication $$0< \left| x- \frac{1}{5} \right| < \frac1{255} \;\;\Rightarrow\;\; \left| \frac{1}{x} - 5\right| < \frac{1}{10}.$$ Sometimes I prefer to avoid double inequalities (i.e. $A<x<B$), since they can be difficult to look at if there is a longer sequence of steps. Here we can get a hint by looking at the graph of $f(x)=1/x.$ For any $\delta$-neighborhood of $1/5$ consisting of strictly positive inputs $x$, the graph will always be steeper throughout the left-hand half of the interval, compared with all the right-hand half. So we suspect the tougher requirement on $\delta$ will be found for $x\in(-\delta+1/5,\;1/5)$.
Indeed, if $0<x<1/5,$ then $\frac1x-5>0$, and $|\frac1x-5|=\frac1x-5.$ Now we require $$\frac1x-5 \;<\; \frac1{10}.$$
But you can readily check that this occurs iff:
$$\frac15-x \;<\;\frac15 - \frac{10}{51} \;\quad\left(=\;\frac1{255}\right).$$
On the other hand, you can repeat the above for the right-hand half (where $f$ is comparatively flat) and obtain
$$5-\frac1x \;<\; \frac1{10} \;\;\iff\;\; x-\frac15 \;<\;\frac{10}{49} - \frac{1}{5} \;\quad\left(=\;\frac1{245}\right).$$
| {
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To prove: $e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\gt e^2$ Question:
Statement I: If $0\lt\theta\lt\frac\pi2$, then $e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\gt e^2$
Statement II: AM of positive numbers is always greater than or equal to their GM.
A) Both the statements are true and Statement II is correct explanation of Statement I.
B) Both the Statements are true and Statement II is not the correct explanation of Statement I.
C) Statement I is true and Statement II is false.
D) Statement I is false and Statement II is true.
My Attempt: I can see that both the statements are true. I wonder how to prove Statement I.
Answer given is option A.
$\frac{e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta}2\ge\sqrt{e^{\csc^2\theta}\sin^2\theta e^{\sec^2\theta}\cos^2\theta}$
Thus, $$RHS=\sqrt{e^{\frac1{\sin^2\theta}+\frac1{\cos^2\theta}}\sin^2\theta\cos^2\theta}\\ =e^{\frac1{\sin\theta\cos\theta}}\sin\theta\cos\theta$$
Not able to conclude.
| The inequality is false as stated. However, we can show that $$e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\geq e^2 \tag{1}$$
for $0 < \theta < \frac\pi 2$. Using the AM-GM inequality,
$$e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\geq 2 e^{\frac{1}{2\sin^2\theta \cos^2\theta}} \sin\theta\cos\theta = 2 \phi(\sin\theta\cos\theta)$$
where $$\phi(x):= xe^{\frac{1}{2x^2}}$$
for $-\frac 12 \le x\le \frac12$. Of course,
$$2\phi(\sin\theta\cos\theta) \ge 2\inf_{\theta\in \left(0, \frac\pi 2 \right)} \phi(\sin\theta\cos\theta)$$ for every $\theta\in \left(0, \frac\pi 2 \right)$. To compute the infimum on the right-hand side, we shall analyze the behavior of $\phi(x)$ for $-\frac 12 \le x\le \frac12$ (since $-\frac 12 \le \sin\theta\cos\theta\le \frac12$). Differentiating $\phi$, we have
$$\phi'(x) = e^{\frac{1}{2x^2}}\left(1 - \frac{1}{x^2} \right)$$
Since $-\frac 12 \le x\le \frac12$, we have $-\infty < 1 - \frac{1}{x^2} \le -3$. Therefore, $\phi'(x) < 0$ for $-\frac 12 \le x\le \frac12$.
With this information, it is clear that $\phi$ is strictly decreasing, and so it attains its minimum value at $x = \frac12$, or $\theta = \frac{\pi}{4}$. Putting everything together, we obtain
$$e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\geq 2 \phi(\sin\theta\cos\theta) \ge 2\phi\left(\frac12 \right) = 2 \cdot \frac12 \cdot e^{2} = e^2$$
as desired.
| {
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} |
how to find $ x^3+y^3$ if $(\sin^{-1}x)^2+(\cos^{-1}y)^2 = \frac{\pi^2}{2}$ I have to find $ x^3+y^3$ if $$(\sin^{-1}x)^2+(\cos^{-1}x)^2 = \frac{\pi^2}{2}$$
My book has arbitrarily declared that $x=y$, I have an idea, but I have no way to check if it's conceptually right as we just assumed $x=y$ even when we did this in class a few months ago.
I'd appreciate any help
Idea:- let $x= \sin(\theta)$
so
$$\theta^2 +y^2= \frac{\pi^2}{2}$$
which leaves us with two variables, beyond which, I'm stuck
what do I do from here
| You're idia would only work if it's sin(x) and cos(x)
but there are arcsin(x) and arccos(x)...
Trigonometric Way
We can use the trigonometric identity that relates arcsin and arccos:
$$ \arccos(x) = \frac{\pi}{2} - \arcsin(x) $$
Now we can say:
$$
\begin{align*}
\sin^{-1}(x)^{2} + \cos^{-1}(x)^{2} &= \frac{\pi^{2}}{2}\\
\arcsin(x)^{2} + \arccos(x)^{2} &= \frac{\pi^{2}}{2}\\
\arcsin(x)^{2} + (\frac{\pi}{2} - \arcsin(x))^{2} &= \frac{\pi^{2}}{2}\\
\arcsin(x)^{2} + (\frac{\pi}{2})^{2} - 2 \cdot \frac{\pi}{2} \cdot \arcsin(x) + \arcsin(x)^{2} &= \frac{\pi^{2}}{2}\\
\arcsin(x)^{2} + (\frac{\pi}{2} - \arcsin(x))^{2} &= \frac{\pi^{2}}{2}\\
2 \cdot \arcsin(x)^{2} + \frac{\pi^{2}}{4} - \pi \cdot \arcsin(x) &= \frac{\pi^{2}}{2} \quad\mid\quad y := \arcsin(x)\\
2 \cdot y^{2} + \frac{\pi^{2}}{4} - \pi \cdot y &= \frac{\pi^{2}}{2} \quad\mid\quad -(\frac{\pi^{2}}{2})\\
2 \cdot y^{2} - \pi \cdot y + \frac{\pi^{2}}{4} -\frac{\pi^{2}}{2} &= 0 \quad\mid\quad \div 2\\
y^{2} - \frac{\pi}{2} \cdot y + \frac{\pi^{2}}{8} -\frac{\pi^{2}}{4} &= 0 \quad\mid\quad \text{pq-Formula}\\
p = -\frac{\pi}{2} ~&\text{ and }~ q = \frac{\pi^{2}}{8} -\frac{\pi^{2}}{4}\\
y_{1, ~2} &= -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^{2} - q}\\
y_{1, ~2} &= -\frac{-\frac{\pi}{2}} \pm \sqrt{(\frac{-\frac{\pi}{2}}{2})^{2} - (\frac{\pi^{2}}{8} -\frac{\pi^{2}}{4})}\\
y_{1, ~2} &= \frac{\pi}{4} \pm \sqrt{(\frac{\pi}{4})^{2} - \frac{\pi^{2}}{8} + \frac{\pi^{2}}{4}} \quad\mid\quad y := \arcsin(x)\\
\arcsin(x) &= \frac{\frac{\pi}{2}}{2} \pm \sqrt{(\frac{\pi}{4})^{2} - \frac{\pi^{2}}{8} + \frac{\pi^{2}}{4}} \quad\mid\quad \sin()\\
x &= \sin(\frac{\frac{\pi}{2}}{2} \pm \sqrt{(\frac{\pi}{4})^{2} - \frac{\pi^{2}}{8} + \frac{\pi^{2}}{4}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{(\frac{\pi}{4})^{2} + \frac{-\pi^{2} + 2 \cdot \pi^{2}}{8}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{(\frac{\pi}{4})^{2} + \frac{\pi^{2}}{8}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{\frac{\pi^{2}}{16} + \frac{\pi^{2}}{8}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{\frac{\pi^{2} + 2 \cdot \pi^{2}}{16}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{\frac{3 \cdot \pi^{2}}{16}})\\
&\Rightarrow x^{3} = \sin(\frac{\pi}{4} \pm \sqrt{\frac{3 \cdot \pi^{2}}{16}})^{3}
\end{align*}
$$
Complex Way
I would try to solve the problem using the complex logarithm forms of the arc functions:
$$
\begin{align*}
\sin^{-1}(x) = \arcsin(x) &= -\mathrm{i} \cdot \ln(x \cdot \mathrm{i} + \sqrt{1 - x^{2}})\\
\cos^{-1}(x) = \arccos(x) &= -\mathrm{i} \cdot \ln(x + \sqrt{1 - x^{2}} \cdot \mathrm{i})\\
\end{align*}
$$
Now we can say:
$$
\begin{align*}
\sin^{-1}(x)^{2} + \cos^{-1}(x)^{2} &= \frac{\pi^{2}}{2}\\
(-\mathrm{i} \cdot \ln(x \cdot \mathrm{i} + \sqrt{1 - x^{2}}))^{2} + (-\mathrm{i} \cdot \ln(x + \sqrt{1 - x^{2}} \cdot \mathrm{i}))^{2} &= \frac{\pi^{2}}{2}\\
\ln(x \cdot \mathrm{i} + \sqrt{1 - x^{2}})^{2} + \ln(x + \sqrt{1 - x^{2}} \cdot \mathrm{i})^{2} &= \frac{\pi^{2}}{2}\\
...
\end{align*}
$$
But that won't work that fast...
| {
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determining the start velocity and angle based on the goals(End point) angle and coordinates. ========
I'm trying to figure out the start angle and velocity of an parabola, When i know only the start coordinates and the end point coordinates and the angle which it reaches the endpoint.
I'm tasked with figuring this out by using derivative method.
So the known parameters are:
Start point (0,0)
end point (6,4)
end point angle (-18)
unknown are:
velocity = unknown.
start angle =unkown.
This is a homework assiment thats been bothering me for months now.
Here is a picture of what i've been working with:
| The equation of motion of a projectile starting from the origin, and shot at an angle $\theta $ above the horizon, at an initial velocity $\mathbf{v_0}$ is
$\mathbf{P}(t) = \mathbf{v_0} t + \dfrac{1}{2} \mathbf{g} t^2 $
Now, $\mathbf{P}(t) = ( x(t), y(t) ) $ and $\mathbf{v_0} = v_0 (\cos \theta, \sin \theta) $, and $ \mathbf{g} = (0, - g ) $ , where $g = 9.81 \ \text{m/s}^2$
Hence,
$ x(t) = v_0 \ cos (\theta) \ t $
$ y(t) = v_0 \ \sin(\theta) \ t - \dfrac{1}{2} \ g \ t^2 $
Solving for $t$ from the first equation for $x(t)$,
$ t = \dfrac{x}{v_0 \cos \theta} $
Substitute this into the equation for $y(t)$
$ y = \tan(\theta) \ x - \dfrac{g \ x^2}{2 v_0^2 \cos^2(\theta) }\hspace{25pt} (1) $
Substitute the point $(6, 4)$ into equation $(1)$
$ 4 = 6 \tan(\theta) - \dfrac{ 18 g }{ v_0^2 \cos^2 \theta } \hspace{25pt} (2)$
Differentiate equation $(1)$ with respect to $x$ to get the slope of the tangent to the projectile trajectory,
$ \dfrac{dy}{dx} = \tan(\phi) = \tan(\theta) - \dfrac{ g x }{ v_0^2 \cos^2 \theta } \hspace{25pt} (3) $
Since at the end point (which is $(6, 4)$) the angle is $\phi = -18^\circ$, then substituting this into $(3)$, gives us
$ \tan(-18^\circ) = \tan(\theta) - \dfrac{ 6 g }{ v_0^2 \cos^2 \theta } \hspace{25pt}(4)$
Eliminating the second term from equations $(2)$ and $(4)$, we get
$ 4 - 3 \tan(-18^\circ) = 3 \tan(\theta) $
Hence, $ \theta = \tan^{-1}\bigg( \dfrac{ 4 - 3 \tan(-18^\circ) }{3} \bigg) = 58.9^\circ$
Substitute this into $(2)$ and solve for $v_0$,
$ v_0 = \sqrt{ \dfrac{18 g }{ (6 \tan(\theta) - 4) \cos^2(\theta) } } = 10.55 \ \text{m/s}$
| {
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Finding the value of given integral The given integral is :$$\displaystyle\int_0^1 \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} \, d\theta$$
I tried solving it with some trigonometric identities but comes out it does not work here.
However solution given starts with these equations mentioned down below and later they substituted $z= \cos{\theta}+ i\sin(\theta)$
$$\begin{array}{r}
z^{10}-1=\left(z^2-1\right)\left[z^2-2 \cos \left(\frac{\pi}{5}\right) z+1\right]\left[z^2-2\left(\cos \frac{2 \pi}{5}\right) z+1\right] \\
\quad \times\left(z^2-2 \cos \frac{4 \pi}{5} z+1\right)\left(z^2-2 \cos \frac{6 \pi}{5} z+1\right)
\end{array}$$
$$\begin{aligned}
z^5-\frac{1}{z^5}=\left(z-\frac{1}{z}\right) &\left(z-2 \cos \frac{\pi}{5}+\frac{1}{z}\right)\left(z-2 \cos \frac{2 \pi}{5}+\frac{1}{z}\right) \\
& \times\left(z-2 \cos \frac{4 \pi}{5}+\frac{1}{z}\right)\left(z-2 \cos \frac{6 \pi}{5}+\frac{1}{z}\right)
\end{aligned}$$
Can anyone help me understand how I can derive these equations and use them here in this integral?
Thank you for your help.
| First you should simplify the term and then bild the antiderivative, what is really easy after simplifying the term:
$$
\begin{align*}
y &= \int_{0}^{1} \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} ~\operatorname{d}\theta\\
F(\theta) &= \int_{0}^{\theta} \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} ~\operatorname{d}\theta = \int_{0}^{\theta} f(\theta) ~\operatorname{d}\theta\\
f(\theta) &= \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}}\\
f(\theta) &= \dfrac{1}{16}\\
F(\theta) &= \int_{0}^{\theta} \dfrac{1}{16} ~\operatorname{d}\theta\\
F(\theta) &= \dfrac{\theta}{16}\\
y &= \int_{0}^{1} \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} ~\operatorname{d}\theta = F(1) - F(0)\\
F(1) - F(0) &= \dfrac{1}{16} - \dfrac{0}{16}\\
F(1) - F(0) &= \dfrac{1}{16} - 0\\
F(1) - F(0) &= \dfrac{1}{16}\\
F(1) - F(0) &= 0.0625\\
\end{align*}
$$
$$
\int_{0}^{1} \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} ~\operatorname{d}\theta = 0.0625\\ $$
If you can't simplefy it, you also would be able to integrate it without simplefying it but that's way more complex and unambiguously.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Prove that it converges to zero. Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Use an appropriate theorem and the fact that $\lim_{n\to\infty} (\frac{1}{2})^n = 0$ to prove that $\lim_{n\to\infty} a_n = 0$.
My attempt: $a_n$ is a decreasing sequence because $a_n < \frac{1}{2}a_{n-1}$, for all $n \ge 2$. And we can continue: $\frac{1}{2}a_{n+1} < \frac{1}{2}a_{n}$; $\frac{1}{2}a_{n+1} < \frac{1}{2}a_{n} <\frac{1}{2}a_{n-1}$. Then, $(\frac{1}{2})^na_{n+1} < (\frac{1}{2})^na_{n} <(\frac{1}{2})^na_{n-1}$ is also true. I see that the squeeze theorem is applicable but I am not sure how to proceed from here. Please advise.
| EDIT
One possible approach consists in noticing that:
\begin{align*}
0 < a_{n} = \frac{a_{n}}{a_{n-1}}\times\frac{a_{n-1}}{a_{n-2}}\times\ldots\frac{a_{2}}{a_{1}}\times a_{1} < a_{1}\times\left(\frac{1}{2}\right)^{n-1}
\end{align*}
Consequently, due to the arithmetic properties of numerical sequences, one gets the desired result:
\begin{align*}
0 < a_{n} < a_{1}\times \left(\frac{1}{2}\right)^{n-1} \Rightarrow 0\leq\lim_{n\to\infty}a_{n} \leq \lim_{n\to\infty}a_{1}\times\left(\frac{1}{2}\right)^{n-1} \Rightarrow \lim_{n\to\infty}a_{n} = 0.
\end{align*}
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show a function is continuous on $\mathbb{R}$ by any method Let $f(x)=\frac{Kx}{K^2+x^2}$ where K is some constant, show this is continuous on $\mathbb{R}$.
Here are my scratch work in looking for a delta. let $x,y\in \mathbb{R} $ WTS $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|<\epsilon,\forall \epsilon>0$ whenever $|x-y|<\delta$. So, $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|=|K||\frac{x(K^2+y^2)-y(K^2+x^2)}{(K^2+y^2)(K^2+x^2)}|\leq |\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}|\leq|\frac{xK^2-yK^2+xy^2}{(K^2+y^2)(K^2+x^2)}|$ I'm a bit stuck here on how to relate this inequality to $|x-y|$. Can someone help out? Thanks!
| Though you agreed to use a more appropriate general argument (see comments), let us finish your $\epsilon-\delta$-proof (assuming $K\ne0$).
You were stuck on how to relate the inequality
$$\left|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}\right|\le|K|\left|\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}\right|$$
to $|x−y|.$ Just factorize it in the numerator:
$xK^2-yK^2+xy^2-yx^2=(x-y)(K^2-xy)$
hence for $|x-y|<\delta$,
$$\left|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}\right|\le\delta|K|M\quad\text{where}\quad M:=\frac{K^2+|xy|}{(K^2+y^2)(K^2+x^2)}$$
and you just need to bound $M$.
If you only want to prove the continuity at any fixed point $x$, you can end up like this: wlog $\delta<1$, so that
$$M\le\frac{K^2+|x|(|x|+1)}{K^4}.$$
But you can even prove that $f$ is uniformly continuous:
$$M\le\frac{K^2}{K^4}+\frac{|xy|}{K^2(x^2+y^2)}\le\frac1{K^2}+\frac1{2K^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Whats the probability for person A to win the dice game? The rules: Two people rolls a single dice. If the dice rolls 1,2,3 or 4, person A gets a point. For the rolls 5 and 6, person B gets a point. One person needs a 2 point lead to win the game.
This is a question taken from my math book. The answer says the probability is $\frac{4}{5}$ for person A to win the game. Which I dont understand.
My thought process: Lets look at all the four possible outcomes with the two first roles. These would be AA, BB, AB or BA. AA means person A gets a point two times a row. Under is the probability for all these scenarios:
$P(AA)=(\frac{2}{3})^2=\frac{4}{9}$
$P(BB)=(\frac{1}{3})^2=\frac{1}{9}$
$P(AB)=\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$
$P(BA)=\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$
If AB or BA happens they have an equal amount of points again, no matter how far they are into the game. The probability of this would then be $2\cdot\frac{2}{9}=\frac{4}{9}$. Since they have an equal amount of points, you can look at that as the game has restarted.
Meaning person A has to get two points a row to win no matter what. Would that not mean the probability is $\frac{4}{9}$ for person A to win? Can someone tell me where my logic is flawed and what the correct logic would be?
| As you showed, person $A$ can win the first round of two rolls with probability $4/9$. However, as Fishbane pointed out in the comments, person $A$ can also win if both person $A$ and person $B$ fail to win in the first round and person $A$ obtains two points in a subsequent round before person $B$ does. Hence, the probability that person $A$ wins should be higher than $4/9$ since person $A$ does not have to win the game in the first round of two rolls.
Method 1: We add the probabilities that person $A$ wins in the $k$th round of two rolls.
Since the probability neither $A$ nor $B$ wins a round is $4/9$, the probability neither $A$ nor $B$ wins in any of the first $k - 1$ rounds of two rolls is $(4/9)^{k - 1}$. The probability that person $A$ then wins the $k$th round of two rolls is $4/9$. Hence, the probability that person $A$ wins in the $k$th round is $(4/9)^k$. Therefore, the probability that person $A$ wins is
\begin{align*}
\sum_{k = 1}^{\infty} \left(\frac{4}{9}\right)^k & = \frac{4}{9}\sum_{k = 1}^{\infty} \left(\frac{4}{9}\right)^{k - 1}\\
& = \frac{4}{9} \cdot \frac{1}{1 - \frac{4}{9}}\\
& = \frac{4}{9} \cdot \frac{1}{\frac{5}{9}}\\
& = \frac{4}{9} \cdot \frac{9}{5}\\
& = \frac{4}{5}
\end{align*}
Method 2: Let $p$ be the probability that $A$ wins the game. You showed that the probability $A$ wins in the first round is $4/9$. You also showed that the probability that neither $A$ nor $B$ wins the first round is $4/9$, at which point the game restarts, so $A$ again has probability $p$ of winning. Hence,
$$p = \frac{4}{9} + \frac{4}{9}p$$
Solving for $p$ yields $p = 4/5$, as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of the areas of all n-sided polygons inscribed within each other Imagine you take an $n$-sided regular polygon with side length $x$, and connect the midpoints of each of its sides to construct a smaller but identical $n$-sided regular polygon within it.
Now perform the same process with the smaller polygon, and then with the next one, and the next and so on.
If you continue this, endlessly inscribing smaller and smaller regular $n$-sided polygons, what is the sum of all of their individual areas in terms of $x$ and $n$?
Below is an image describing this question when n = 6:
I know there must be a common ratio between each of the areas, so i'm thinking about turning this into a geometric series of the form $\frac{a}{1-r}$ (since $r$ < 1), where $a$ = the area of the first polygon, however I do not know how to find the specific values for $a$ and $r$ in terms of $x$ and $n$.
Any help with this question would be greatly appreciated!
| Let us first find $a$.
Let $O$ be the center of the regular polygon. Then, a circle centered at $O$ circumscribes the polygon. Each side is a chord of the circle and hence the perpendicular from the center bisects the side. Also, $\triangle AOM\cong\triangle BOM$ so $\angle AOM=\angle BOM=\dfrac{\pi}{n}$. Now, in $\triangle OAM $,$$\frac{AM}{MO}=\frac{x/2}{r}=\tan\frac{\pi}{n}$$ so $$r=\frac{x}{2\tan\frac{\pi}{n}}$$ so the area of triangle $OAB$ is $\dfrac12OM\times AB=\dfrac{x^2}{4\tan\frac{\pi}{n}}$. There are exactly $n$ such triangles so the total area is, $$\fbox{$a= \dfrac{nx^2}{4\tan\frac{\pi}{n}}$}.$$
Now,
to find the side length of the smaller second polygon, join BC. By the midpoint theorem on $\triangle ABC$, the side length of the second polygon is $\dfrac{BC}{2}$. Now, to find $BC$, note that $\angle CAB=\dfrac{(n-2)\pi}{n}$. By the cosine rule in $\triangle ABC$,$$\frac{x^2+x^2-BC^2}{2x^2}=\cos\angle CAB$$$$\implies 1-\frac{BC^2}{2x^2}=\cos\frac{(n-2)\pi}{n}=\cos\left(\pi-\frac{2\pi}{n}\right)$$$$=-\cos\frac{2\pi}{n}$$ so that $$\frac{BC^2}{2x^2}=1+\cos\frac{2\pi}{n}=2\cos^2\frac{\pi}{n}$$ which implies that $$BC=2x\cos\frac{\pi}{n}$$ or that the side of the second polygon is $\dfrac{BC}{2}= x\cos\dfracπn.$ Thus, its area will be (by similarity) $a\cos^2\dfrac{\pi}{n}$, which implies that the common ratio for the geometric series is $\cos^2\dfrac{\pi}{n}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Find the smallest possible value for $x^2-3x+2y^2+4y+2$ Find the smallest possible value for $x^2-3x+2y^2+4y+2$.
I know this: $x^2-3x+2y^2+4y+2=(x-\frac{3}{2})^2+2(y+1)^2-\frac{9}{4}$ so the smallest value is $-\frac{9}{4}$.
Now we also know this: $x^2-3x+2y^2+4y+2=(x-1)^2-1-x+2(y+1)^2=(*)(x-1)^2+2(y+1)^2-(1+x)$
My question is: why we cannot say that the smallest value possible is $-(1+x)$ when $(x-1)^2=0$ (i.e. when $x=1$) and $(y+1)^2=0$, and because $x=1$ so the smallest value is $-2$?
| $x^2-3x+2y^2+4y+2=0$ is an ellipse, so what do you mean by "the smallest possible value of $x^2-3x+2y^2+4y+2$"?
Anyway, since you have two distinct values of $y=y(x)$ for the given $x: x=\frac{3}{2}$ such that one of them (the negative one) is minimum, I will write only the positive value of $y(x) : x=\frac{3}{2}$, leaving the final step (and its proof) as an exercise.
Well, $y\left(\frac{3}{2}\right) : y>0$ is equal to $\left(\frac{3}{2\cdot \sqrt{2}} - 1\right)$ so that you have to show why $y\left(\frac{3}{2}\right) : y<0$ is the minimum and then find the value.
P.S. If we are just interested in the global minimum of $x^2-3x+2y^2+4y+2$ in $\mathbb{R}^2$, it is at $(x,y)=(\frac{3}{2},-1)$ and it is $-\frac{9}{4}$ indeed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove or disprove $ab+bc+cd+da\leq1$ if $a+b+c+d=2$ Non-negative real numbers $a,b,c,d$ are such that $a+b+c+d=2$. Prove or disprove that
$$ab+bc+cd+da\leq1$$
I see there are multiple equality cases, where $(a,b,c,d)$ is for example $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$, $(\frac{3}{4},\frac{1}{2},\frac{1}{4},\frac{1}{2})$, $(\frac{3}{4},\frac{5}{8},\frac{1}{4},\frac{3}{8})$.
I suspect it's true, and maybe it can be proven with rearrangement, but I have not found a way.
It's reminiscent of Chebyshev's inequality, since the desired inequality is equivalent to
$$4(ab+bc+cd+da)\leq(a+b+c+d)(b+c+d+a)$$
But we cannot assume that $(a,b,c,d)$ and $(b,c,d,a)$ are oppositely ordered.
| Let $x = a+c$ and $y=b+d$. Then, we have $xy = (a+c)(b+d) = ab + ad + bc + cd$ which is the quantity we want to bound. So, we want to show that if $x+y = 2$, then $xy \leq 1$.
We just use basic algebra now: $x=2-y$, so $xy = y(2-y)$. Now you can rearrange, $y(2-y) = -(y-1)^2 +1$. Since $(y-1)^2 \geq 0$, it follows that $-(y-1)^2\leq 0$ and so $xy = -(y-1)^2+1 \leq 1$.
You could also just differentiate $-y^2 + 2y$, to see that you have a local maximum when $y=1$.
| {
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"url": "https://math.stackexchange.com/questions/4557878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}$ without using derivatives This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$
where $a>0$, $b>0$ are some constants.
I know how to calculate this limit using the L'Hopital's rule:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)};
$$
$$
\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)=
\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)}{x}$$
$$
=
\lim\limits_{x\to 0} \frac{2}{a^{\sin x}+b^{\sin x}}\cdot\frac12\cdot
\left( a^{\sin x}\cos x \ln a+b^{\sin x}\cos x \ln b \right)=
\frac12\left( \ln a+ \ln b \right)
$$
$$
\Rightarrow
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.
$$
I'm allowed to use the limits $\lim_{x\to0}\frac{\sin x}{x}=1$,
$\lim_{x\to0}\frac{a^x-1}{x}=\ln a$,
$\lim_{x\to0}\frac{\log_a(1+x)}{x}=\log_a e$ and $\lim_{x\to0} (1+x)^{1/x}=e$.
| Taking logarithm, then using $\sin x\sim x$ and $\ln(1+x)\sim x$ when $x\to0$, we have
\begin{align*}
\frac1x\cdot\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)&\sim\frac1{\sin x}\cdot\left(\frac{a^{\sin x}+b^{\sin x} }{2}-1\right)\\
&=\frac12\frac{a^{\sin x}-1}{\sin x}+\frac12\frac{b^{\sin x}-1}{\sin x}\\
&\rightarrow\frac12(\ln a+\ln b),\qquad x\to0,
\end{align*}
where in the last line we've used
$$\lim_{x\to0}\frac{a^{\sin x}-1}{\sin x}=\lim_{t\to0}\frac{a^{t}-1}{t}=\ln a. $$
Therefore,
$$\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.$$
| {
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"url": "https://math.stackexchange.com/questions/4561863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$3$-var inequality: $\frac{bc}{\sqrt{a}+3}+\frac{ca}{\sqrt{b}+3}+\frac{ab}{\sqrt{c}+3} \leq \frac{3}{4}$ for $a+b+c=3$. Problem: Let $a,b,c$ be positive numbers satisfied $a+b+c=3$. Prove that $$\dfrac{bc}{\sqrt{a}+3}+\dfrac{ca}{\sqrt{b}+3}+\dfrac{ab}{\sqrt{c}+3} \leq \dfrac{3}{4}$$
I've tried U.C.T method but it doesn't reach the solution. I also thought of $p,q,r$ method, but I think it will end up a ugly solution. The only useful thing I get is $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab+bc+ca$. Anyone?
| We need to prove that $$\sum_{cyc}\frac{a^2b^2}{3+c}\leq\frac{3}{4},$$ where $a$, $b$ and $c$ are positives such that $a^2+b^2+c^2=3$.
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the condition does not depend on $w^3$ and we need to prove that
$$4\sum_{cyc}a^2b^2(a+3)(b+3)\leq3\prod_{cyc}(a+3)$$ and since
$$\sum_{cyc}a^3b^3=27v^6-27uv^2w^3+3w^6$$ and the rest terms are less degree, we need to prove that $f(w^3)\geq0,$ where $f$ is a concave function.
But a concave function gets a minimal value for an extremal value of $w^3$, which by $uvw$(see here: https://artofproblemsolving.com/community/c6h278791) happens in the following cases.
1.$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$.
Thus, we need to prove that $$\frac{a^2b^2}{3}\leq\frac{3}{4}$$ or $$(a^2-b^2)^2\geq0.$$
2.Two variables are equal.
Let $b=a$. Thus, $c=\sqrt{3-2a^2},$ where $0<a<\sqrt{\frac{3}{2}}$ and we need to prove that
$$\frac{a^4}{3+\sqrt{3-2a^2}}+\frac{2a^2(3-2a^2)}{3+a}\leq\frac{3}{4}.$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
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How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$ I am trying to find a closed form for
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}.
$$
I have done trig substitution and it results in
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,} =
\int_{0}^{\pi/3}\sec\left(\theta\right)
\ln\left(\frac{\sec\left(\theta\right)}
{2\left[\sec\left(\theta\right) - 1\right]^{\,2}} \right){\rm d}\theta
$$
which doesn't help.
By part integration with
$\displaystyle u = \ln\left(\frac{x + 1}{2x^{2}} \right)$, $\displaystyle\,\,{\rm d}v=\frac{\displaystyle\,\,{\rm d}x}{\,\sqrt{\,{x^{2} + 2x}\,}\,}$ also makes it more complicated.
I appreciate any help on this problem.
| Let
$$x=y^2,\quad y^2+1=t,\tag1$$
then
$$I=\int\limits_0^1 \dfrac{\ln\left(\dfrac{x+1}{2x^2}\right)}{\sqrt{x^2+2x}}dx
=\int\limits_0^1 \dfrac{\ln\left(\dfrac{y^2+1}{2y^4}\right)}{\sqrt{y^4+2y^2}}\, 2y\,\text dy
=\int\limits_1^2 \dfrac{\ln\left(\dfrac{t}{2(t-1)^2}\right)}{\sqrt{t^2-1}}\,\text dt,$$
$$I=I_1-2I_2-I_3,$$
where
\begin{cases}
I_1=\int\limits_1^2 \dfrac{\ln t\,\text dt}{\sqrt{t^2-1}}
= \dfrac{\pi^2}{24} -\dfrac12 \operatorname{Li_2}\left(\dfrac{2 - \sqrt3}4\right) - \ln^2 2-\dfrac14\arccos^2 2\\
I_2=\int\limits_1^2 \dfrac{\ln(t-1)\,\text dt}{\sqrt{t^2-1}}
=-\dfrac{\pi^2}3+2\operatorname{Li}_2(2-\sqrt3) +6\operatorname{arcsinh}^2\dfrac1{\sqrt2}
-4 \operatorname{arcsinh}\dfrac 1{\sqrt2}\,\log(1+\sqrt3)\\
I_3=\int\limits_1^2 \dfrac{\ln 2\text dt}{\sqrt{t^2-1}} = \dfrac12\ln2 \ln\left(4\sqrt{3}+7\right) = \ln2 \ln\left(2+\sqrt{3}\right)\\
\arccos 2 = i\ln(2+\sqrt3),
\end{cases}
Therefore,
$$I=\dfrac{17\pi^2}{24}-\dfrac12 \operatorname{Li_2}\left(\dfrac{2 - \sqrt3}4\right)-4\operatorname{Li}_2(2-\sqrt3)-\ln^2 2 -\ln\left(2+\sqrt{3}\right)\ln 2$$
$$+\dfrac1{4}\ln^2(2+\sqrt3)
-12 \operatorname{arcsinh}^2 \dfrac1{\sqrt2}+8\operatorname{arcsinh}\dfrac 1{\sqrt2}\,\ln(1+\sqrt3),$$
$$I\approx4.9348022005446793094$$
(see also Wolfram Alpha calculations),
in accordance with numeric calculations.
| {
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"url": "https://math.stackexchange.com/questions/4568778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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Use the substitution $u=\frac{1}{\sqrt{3}}\tan x$ to find $\int\frac{dx}{3-2\sin^2x}$ Question:
Use the substitution $u=\frac{1}{\sqrt3}\tan x$ to find $$\int\frac{dx}{3-2\sin^2x}$$
My Working:
Let $u=\frac{1}{\sqrt3}\tan x$, then
\begin{align}
\frac{du}{dx}&=\frac{1}{\sqrt3}\cdot\sec^2x+0=\frac{\sec^2x}{\sqrt3}\\
du&=\frac{\sec^2x}{\sqrt3}\cdot dx
\end{align}
Unfortunately, after that, I do not know how to proceed. Could anyone please help? Thank you!
| Your substitution $u=\frac{1}{\sqrt 3}\tan x$ works well as below:
$$
\begin{aligned}
\int \frac{dx}{3-2 \sin ^2 x} =& \int \frac{\frac{\sqrt{3} d u}{3 u^2+1}}{3-2\left(\frac{3 u^2}{3 u^2+1}\right)} \\
=& \sqrt{3} \int \frac{d u}{3 u^2+3} \\
=& \frac{1}{\sqrt{3}} \tan ^{-1} u+C \\
=&\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan x}{\sqrt{3}}\right)+C
\end{aligned}
$$
Alternatively, multiplying both denominator and numerator by $\sec^2x$ gives
$$
\begin{aligned}
\int \frac{dx}{3-2 \sin ^2 x} &=\int \frac{\sec ^2 x}{3 \sec ^2 x-2 \tan ^2 x} d x \\
&=\int \frac{d(\tan x)}{\tan ^2 x+3} \\
&=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan x}{\sqrt{3}}\right)+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
In $\triangle ABC$, $BD$ is a median, $\angle DAB=15$ and $\angle ABD=30$. Find $\angle ACB$. As title implies, the goal is to find the measure of the missing angle in the following figure. While I have solved this, which I'll show below as an answer, I'm not quite sure if my answer is accurate, so I'm posting this here to see if my answer is correct and if there are any alternative approaches. Please post your own approaches to this!
|
WLOG, let $AD=DC=1$
Using sine rule yields
$$
\begin{aligned}
\frac{A B}{\sin 135^{\circ}}&=\frac{1}{\sin 30^{\circ}}
\Rightarrow A B =\sqrt{2}
\end{aligned}
$$
Using cosine rule gives
$$
\begin{aligned}
BC^2 &=2^2+\sqrt{2}^2-2(2) \sqrt{2} \cos 15^{\circ} \\
&=6-4 \sqrt{2} \cos 15^{\circ} \\
&=6-2(\sqrt{3}+1) \\
B C &=\sqrt{4-2 \sqrt{3}}=\sqrt{3}-1
\end{aligned}
$$
By sine rule again, we have
$$
\begin{aligned}
\frac{\sin C}{A B} &=\frac{\sin 15^{\circ}}{BC} \\
\sin C &=\frac{\sqrt{2} \sin 15^{\circ}}{\sqrt{3}-1}=\frac{\frac{\sqrt{3}-1}{2}}{\sqrt{3}-1}=\frac{1}{2} \\
C &=30^{\circ}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 6,
"answer_id": 4
} |
Volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2$ This is question 13 on page 294 of Vector Calculus by Marsden and Tromba.
Find the volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2.$
I have attempted it as follows.
The region can be described by using spherical polar coordinates $(r,\theta,\phi)$ and we have $0 \leq r \leq \sqrt{10}$, $0 \leq \theta \leq 2 \pi$.
For the limits for $\phi$, I think that we can find it by saying that $z=r \cos \phi = \sqrt{10} \cos \phi \geq 2$ using that $r = \sqrt{10}$ on the surface. This gives $0 \leq \phi \leq \cos^{-1}(2/\sqrt{10}).$ Hence I get
\begin{align} \iiint_V dV &= \int_0^{2\pi} \int_0^{\cos^{-1}(2/\sqrt{10})}\int_0^{\sqrt{10}} r^2 \sin \phi drd\phi d\theta \\ &= 2\pi \int_0^{\cos^{-1}(2/\sqrt{10})} 10 \dfrac{\sqrt{3}}{3} \sin \phi d\phi \\&= 20\pi \dfrac{\sqrt{10}}{3} - 40\dfrac{\pi}{3}. \end{align}
However, the answer at the back of the book is
$20\pi \dfrac{\sqrt{10}}{3} - 52\dfrac{\pi}{3}$
but I have been unable to identify the mistake. Please, could someone help me? Thank you very much.
| What you have computed is the volume of a larger region, which is the region that you have been given plus the cone whose vertex is located at the origin and whose base is the the intersection of the solid sphere centered at the origin with radius $\sqrt{10}$ and the plane $z=2$.
I suggest the use of cylindrical coordinates:\begin{align}\int_0^{2\pi}\int_2^{\sqrt{10}}\int_0^{\sqrt{10-z^2}}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta&=2\pi\int_2^{\sqrt{10}}5-\frac{z^2}2\,\mathrm dz\\&=20\pi\frac{\sqrt{10}}3-52\frac\pi3.\end{align}
| {
"language": "en",
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"answer_count": 2,
"answer_id": 0
} |
Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$ (Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right).$$
(Attempt) Let $H_n$ denote the n-th harmonic number $\displaystyle H_{n}=\sum_{k=1}^{n}\frac{1}{k}$. I will warn this process gets ugly, but it is the best I have so far. Expanding the integrand as a series, we get:
$$
\eqalign{
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+2}}dx+\sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+4}}dx, \cr
}
$$
which simplifies down to
$$\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-1}}{-4n-1}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-1}}{-4n-1}+\frac{4\sqrt{3}^{-4n-1}}{\left(-4n-1\right)^{2}}\right)+\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-3}}{-4n-3}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-3}}{-4n-3}+\frac{4\sqrt{3}^{-4n-3}}{\left(-4n-3\right)^{2}}\right).$$
Since both series converge, we can split up the first series like
$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-1}}{4n+1}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{\left(4n+1\right)^{2}}$$
and the second series like
$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-3}}{4n+3}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{\left(4n+3\right)^{2}}.$$
Next, I found that
$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}=\frac{\pi}{12}+\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)$$
and
$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}=\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{12}.$$
However, after trying for a while, I am out of ideas for evaluating the other sums.
I realize I skipped a lot of steps, but that is because I don't want this question to be too long. So for your convenience, I put all of these into Desmos, so I believe my process is correct so far based on numerical approximations.
(Question) Does anyone have an idea of how to evaluate the integral in question, or how to evaluate the sums I am stuck on? Any hints and ideas are appreciated.
(Miscellaneous) Here are some other ideas I have:
$$
\eqalign{
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{\operatorname{arcsec}\left(\sqrt{3}\right)}^{\frac{\pi}{2}}\frac{\ln\left(\left(\sec x\right)^{4}-1\right)}{\sec^{2}x-1}\sec\left(x\right)\tan\left(x\right)dx \cr
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{0}^{\infty}\frac{\ln\left(\left(x+\sqrt{3}\right)^{4}-1\right)}{\left(x+\sqrt{3}\right)^{2}-1}dx.
}
$$
Maybe I could construct a keyhole contour for the last integral?
| Mathematica 13.1 produces
Integrate[Log[x^4 - 1]/(x^2 - 1), {x, Sqrt[3], Infinity}]//FullSimplify
$$\frac{1}{32} \left(16 \left(\text{Li}_2\left(\frac{1}{2} \left(1-\sqrt{3}\right)\right)+\text{Li}_2\left(\sqrt{3}-1\right)+\text{Li}_2\left(\left(-\frac{1}{2}+\frac{i}{2}\right) \left(\sqrt{3}-1\right)\right)+\text{Li}_2\left(\left(-\frac{1}{2}-\frac{i}{2}\right) \left(\sqrt{3}-1\right)\right)+\text{Li}_2\left(\frac{1-i}{\sqrt{3}+1}\right)+\text{Li}_2\left(\frac{1+i}{\sqrt{3}+1}\right)+\log ^2\left(\sqrt{3}+1\right)\right)+7 \pi ^2+8 \log ^2\left(\sqrt{3}-1\right)+4 \log (2) \left(\log (8)+4 \log \left(2 \sqrt{3}+\frac{7}{2}\right)\right)\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Laurent Series of $(z+1)^2\sin(z)$ centred at $0$ How am I to write the Laurent series of $(z+1)^2\sin(z)$ centred at $0$? I know
$$(z+1)^2 \sin(z) = (z^2+2z+1) \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1)!} \\ \Rightarrow (z+1)^2 \sin(z) = \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+3}}{(2k+1)!} + \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+2}}{(2k+1)!} + \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1)!},$$
but how am I to appropriately re-index is summation such that I might have a (closed-form) expression, containing only a single sum, as in, say,
$$(z+1)^2 \sin(z) = \sum_{k = 0}^{\infty} a_k z^{k}?$$
| We derive a series representation of $(z+1)^2\sin(z)$ evaluated at $z=0$ by calculating the coefficients $a_n$ in
\begin{align*}
\color{blue}{(z+1)^2\sin(z)}&=(z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&\,\,\color{blue}{=\sum_{n=0}^{\infty}a_nz^n}\tag{1}\\
&=\sum_{n=0}^{\infty}a_{2n}z^{2n}+\sum_{n=0}^{\infty}a_{2n+1}z^{2n+1}\tag{2}\\
\end{align*}
At first we derive a representation (2) and then we merge the sums to obtain (1). We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. We obtain for $n\geq 1$:
\begin{align*}
\color{blue}{a_{2n}}&=[z^{2n}](z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&=\left([z^{2n-2}]+2[z^{2n-1}]+[z^{2n}]\right)\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&\,\,\color{blue}{=\frac{2(-1)^{n-1}}{(2n-1)!}}\tag{3}\\
\color{blue}{a_{2n+1}}&=[z^{2n+1}](z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&=\left([z^{2n-1}]+2[z^{2n}]+[z^{2n+1}]\right)\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&\,\,\color{blue}{=\frac{(-1)^{n-1}}{(2n-1)!}+\frac{(-1)^n}{(2n+1)!}}\tag{4}
\end{align*}
We derive from (3) and (4) for $n\geq 1$:
\begin{align*}
a_n&=\frac{1+(-1)^n}{2}\,\color{blue}{\frac{2(-1)^{\frac{n}{2}-1}}{(n-1)!}}
+\frac{1-(-1)^n}{2}\left(\color{blue}{\frac{(-1)^{\frac{n-1}{2}-1}}{(n-2)!}+\frac{(-1)^{\frac{n-1}{2}}}{n!}}\right)\\
&=\frac{\left(1+(-1)^n\right)(-1)^{\frac{n}{2}-1}}{(n-1)!}
+\frac{\left(1-(-1)^n\right)(-1)^{\frac{n-1}{2}}}{2n!}\left(1+n-n^2\right)\tag{5}\\
\end{align*}
Comment:
*
*In (3) we select the coefficient of $z^{2n-1}$. The other coefficients with even power of $z$ are zero.
*In (4) we select the coefficients of $z^{2n-1}$ and $z^{2n+1}$. The coefficient with even power of $z$ is zero.
*In (5), we combine even and odd indexed coefficients $a_n$ by multiplying them by $\frac{1+(-1)^n}{2}$ and $\frac{1-(-1)^n}{2}$, respectively, and adding them.
We finally derive from (5)
\begin{align*}
&\color{blue}{(z+1)^2\sin(z)}\\
&\qquad=1+\sum_{n=1}^{\infty}\left(\left(1+(-1)^n\right)(-1)^{\frac{n}{2}-1}n\right.\\
&\qquad\qquad\qquad\qquad\left.+\left(1-(-1)^n\right)(-1)^{\frac{n-1}{2}}\frac{1+n-n^2}{2}\right)\frac{z^n}{n!}\\
&\qquad\color{blue}{=1+\sum_{n=1}^{\infty}\left(\left((-1)^{n-1}-1\right)n
-i\left(1-(-1)^n\right)\frac{1+n-n^2}{2}\right)\frac{\left(iz\right)^n}{n!}}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that: $\frac{1}{c}\sqrt{h} \Big\|\begin{pmatrix} f(0) \\f(h) \end{pmatrix} \Big\|_2 \le \Big\|f\Big\|_{L^2} \le c\sqrt{h}\ldots$ Let $\mathcal P:=\left\{f:\Omega \rightarrow \mathbb R \space \vert \space f(x)=ax+b, a,b \in \mathbb R \right\}$ with $\Omega:=[0,h], h\in \mathbb R_+$. I want to show this inequality holds for some suitable constant $c\in \mathbb R$ and I want to determine this constant $c$. I have seen this before in equivalent norms but I am having some trouble with the arithmetic.
$$\frac{1}{c}\sqrt{h} \Big\|\begin{pmatrix} f(0) \\f(h) \end{pmatrix} \Big\|_2 \le \Big\|f\Big\|_{L^2\left( \Omega\right)} \le c\sqrt{h}\Big\|\begin{pmatrix} f(0) \\f(h) \end{pmatrix} \Big\|_2$$
My attempt:
$$\begin{aligned}&\frac{1}{c} \sqrt{h}\sqrt{\left(f(0)\right)^2+\left(f(h)\right)^2}\le\int_0^h (ax+b)^2 dx\le c\sqrt{h} \sqrt{\left(f(0)\right)^2+\left(f(h)\right)^2} \\ &\iff \frac{1}{c}\sqrt{h}\sqrt{b^2+(ah+b)^2} \le \int_0^h (a^2x^2+2abx+b^2) dx \le c\sqrt{h}\sqrt{b^2+(ah+b)^2} \\ &\iff \frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le \int_0^h (a^2x^2+2abx+b^2) dx \le c\sqrt{a^2h^3+2bh^2+2b^2h} \\ & \iff \frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le a^2\left[\frac{x^3}{3}\right]_0^h+ ab\left[x^2\right]_0^h+b^2 \left[x\right]_0^h \le c\sqrt{a^2h^3+2bh^2+2b^2h} \\ & \iff \frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le \frac{a^2h^3}{3}+abh^2+b^2h \le c\sqrt{a^2h^3+2bh^2+2b^2h} \\ & \iff \ldots?\end{aligned}$$
I feel like I am very close but I am not sure how to proceed. Can $c$ be any expression involving $a,b,h$? Getting rid of the the square roots seems to be the problem. Is this even the right approach or should I have used some inequality (Cauchy-Schwarz, etc.) earlier to simplify the $L^2$-norm after squaring the entire inequality?
| I made a mistake writing down the $L^2$-Norm. We have:
$$\vert \vert f \vert\vert_{L^2} = \left( \int_0^h \vert f(x) \vert^2dx \right)^{\frac{1}{2}} $$
Therefore the inequality becomes:
$$\frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le \sqrt{\frac{a^2h^3}{3}+abh^2+b^2h} \le c\sqrt{a^2h^3+2bh^2+2b^2h}$$
Let $c=\sqrt{4}=2$
$$\implies \sqrt{\frac{a^2h^3}{4}+\frac{bh^2}{2}+\frac{b^2h}{2}}\le \sqrt{\frac{a^2h^3}{3}+abh^2+b^2h} \le \sqrt{4a^2h^3+8bh^2+8b^2h}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580951",
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
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Verify that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$ in a general triangle $\triangle ABC$ So, this problem is inspired by a contest preparation problem I saw back in Japan, and it is as follows:
In a general triangle $\triangle ABC$, show that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$
Now, while I still haven't figured out a geometric interpretation of this inequality, here is my attempt to prove this:
Recall that:
$$\sin^2(\frac{A}{2})=\frac{(1-\cos(A))}{2}$$
$$\sin^2(\frac{A}{2})=\frac{1}{2}(1-\frac{b^2+c^2-a^2}{2bc})$$
$$\sin^2(\frac{A}{2})=\frac{1}{2}(\frac{a^2-b^2-c^2+2bc}{2bc})$$
$$\sin^2(\frac{A}{2})=\frac{a^2-(b-c)^2}{4bc}$$
Now, obviously $\frac{a^2-(b-c)^2}{4bc} \le \frac{a^2}{4bc}$, therefore:
$$\sin^2(\frac{A}{2}) \le \frac{a^2}{4bc}$$
$$\sin(\frac{A}{2}) \le \frac{a}{2\sqrt{bc}}$$
This can be done for $\sin(\frac{A}{2}), \sin(\frac{B}{2})$ and $\sin(\frac{C}{2})$.
$$\sin(\frac{A}{2}) \le \frac{a}{2\sqrt{bc}}$$
$$\sin(\frac{B}{2}) \le \frac{b}{2\sqrt{ac}}$$
$$\sin(\frac{C}{2}) \le \frac{c}{2\sqrt{ab}}$$
Therefore:
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le (\frac{a}{2\sqrt{bc}})(\frac{b}{2\sqrt{ac}})(\frac{c}{2\sqrt{ab}})$$
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \frac{abc}{8abc}$$
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \frac{1}{8}$$
However I'm not sure if this is correct or if it is, I don't think this brute force approach is good. Are there any better options to prove this inequality? Please share your answers!
| By your work
$$\prod_{cyc}\sin\frac{\alpha}{2}=\prod_{cyc}\sqrt{\frac{(a-b+c)(a+c-b)}{4bc}}=\frac{\prod\limits_{cyc}(a+b-c)}{8abc}\leq\frac{1}{8}$$
because $$abc-\prod\limits_{cyc}(a+b-c)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$
$$=\sum_{cyc}(a^3-abc-b^2c-a^2c+2abc)=\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-c\right)=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b-c)\geq0.$$
By the way, last inequality is true for any non-negatives $a$, $b$ and $c$, but it's another problem.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Why can't I use the form $\frac1{x^2+1}$ from the derivative of $\arctan(x)$ to convert the integral form in this situation? In the question that solves the integral $\displaystyle\int\frac1{6x^2 + 36x + 78} \,\mathrm{d}x$, I first tried to solve it by changing the denominator in a form of $\dfrac1{x^2 + 1}$ to apply $\;\arctan(x)$.
$\dfrac1{6\!\cdot\!\left(x^2 + 6x + 13\right)}=\dfrac16\!\cdot\!\dfrac1{(x + 3)^2 + 2^2}$
Now, in order to make it in a form of $\;\dfrac1{x^2 + 1}\;,\;$ I divide everything by $\,2^2$:
$=\dfrac16\!\cdot\!\dfrac1{\left(\frac{x+3}2\right)^2+1}\!\cdot\!\dfrac1{2^2}
=\dfrac1{24}\!\cdot\!\dfrac1{\left(\frac{x+3}2\right)^2+1}$
Then assume that $\;u=\dfrac{x+3}2\;,\;$ I thought I can apply arctan to get rid of the integral form:
$\dfrac1{24}\!\cdot\!\arctan\left(\dfrac{x+3}2\right)+c\;.$
But the correct answer is $\;\dfrac1{12}\!\cdot\!\arctan\left(\dfrac{x+3}2\right) + c\;,\;$ not $\;\dfrac1{24}\,.$
The answer also explained to use $\;\dfrac1{x^2 + k^2} = \left[\dfrac1k\!\cdot\!\arctan\left(\dfrac xk\right)\right]’$ (derivative), but I wonder why I cannot use the form of $\;\dfrac1{x^2 + 1}\;,\;$ which is the only equation I have known.
| Basically, you messed up in your substitution:
$$\int \frac{1}{6x^2+36x+78}dx = \frac{1}{6} \int \frac{1}{(x+3)^2+2^2}dx = \frac{1}{24} \int \frac{1}{(\frac{x+3}{2})^2+1}dx$$
Let $u=\frac{x+3}{2}$, Then $du = \frac{dx}{2}$ So:
$$\frac{1}{24} \int \frac{1}{(\frac{x+3}{2})^2+1}dx = \frac{1}{24} \int \frac{1}{(u)^2+1} \cdot 2 du = \frac{1}{12} \int \frac{1}{(u)^2+1} du $$
$$= \frac{1}{12}\arctan(\frac{x+3}{2})$$
It seems you instead used $du=dx$
Note that you can use whatever form you'd like to evaluate your integral but what the question tells you to use requires less working so may aswell.
| {
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"question_score": "2",
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How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)^n$
$\log L=n\log\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)$
Is this equal to 0? Then the answer would be $e^0=1$.
The answer was given as $e^4$ and I have no idea how to get to that.
| I've also posted this answer with the thought that some readers might find it useful.
Using high school algebra trick, we have:
$$
\begin{align}&\lim_{n\to\infty}\left(\frac{n^2+5n+3}{n^2+n+2}\right)^n\\
=~&\lim_{n\to\infty}\left(1+\frac{4n+1}{n^2+n+2}\right)^n\\
=~&\lim_{n\to\infty}\left(\left(1+\frac {1}{\frac{n^2+n+2}{4n+1}}\right)^{\frac {n^2+n+2}{4n+1}}\right)^{\frac {4n^2+n}{n^2+n+2}}\\
=~&\lim_{n\to\infty}\left(\left(1+\frac {1}{\frac{n+1+2/n}{4+1/n}}\right)^{\frac {n+1+2/n}{4+1/n}}\right)^{\frac {4+1/n}{1+1/n+2/n^2}}\\
=~&~e^4~.\end{align}
$$
That's all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Prove that $2\cdot 3^x +1= p^y$ has no solution Prove that the Diophantine equation of $3$ variables $(x,y,p)$
$$2\cdot 3^x +1= p^y$$
has no solution where $x,y\in\mathbb{N}_+$, $x\ge2, y\ge2$ and $p$ is a prime number.
I found that $y$ cannot be even, if $y =2k$ then $4| (p^k -1)(p^k+1) =2\cdot3^x$ which is a contradiction. But I cannot prove there is no solution for the case $y = 2k+1$.
I try to apply the technique in this answer (equation $7^x = 3 \cdot 2^y +1$) but it doesn't work as $p$ is not known.
Any help would be greatly appreciated!
| Note that
$$2\cdot 3^x = p^y - 1 = (p - 1)(p^{y-1} + p^{y-2} + \ldots + p + 1) \tag{1}\label{eq1A}$$
With $p$ being odd and $p \gt 3$, then
$$p - 1 = 2(3^{a}) \; \; \to \; \; p = 2(3^{a}) + 1 \tag{2}\label{eq2A}$$
for some integer $a \ge 1$. Next, using the lifting-the-exponent (LTE) lemma, we have
$$x = \nu_3(p^y - 1) = \nu_3(p - 1) + \nu_3(y) = a + \nu_3(y) \tag{3}\label{eq3A}$$
Letting $\nu_3(y) = b$, there's then an integer $c \ge 1$ where $3 \nmid c$ and
$$y = c(3^b) \tag{4}\label{eq4A}$$
We next have
$$\begin{equation}\begin{aligned}
p^y & = 2^y(3^{ay}) + y(2(3^{a}))^{y-1} + \ldots + y(2(3^{a})) + 1 \\
p^y - 1 & \gt 2^y(3^{ay}) \\
2(3^x) & \gt 2^y(3^{ay}) \\
3^{x} & \gt 3^{ay} \\
3^{a+b} & \gt 3^{ac(3^b)} \\
a + b & \gt a(3^b)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
If $b = 0$, we get $a \gt a$, which is not true, so $b \gt 0$. Next, we have from \eqref{eq5A} that
$$a + b \gt a(3^b) \; \; \to \; \; b \gt a(3^b - 1) \ge 3^b - 1 \tag{6}\label{eq6A}$$
which is not possible. Thus, there's no solution to \eqref{eq1A} with $x \ge 2$, $y \ge 2$ and $p$ being a prime number (actually, no integer $p \gt 3$ works).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Indefinite Integral of ${\tan(x)}^{p/q}$ I have seen a lot of videos in which people integrate functions like $\sqrt{\tan(x)}$, $\sqrt[3]{\tan^2(x)}$, etc. I was wondering if there was a closed-form expression for
$$\int (\tan{x})^{\frac{p}{q}} dx$$
Where $p,q$ are integers, and $q \neq 0$. Eventually, I would like to generalize the result to an integrand ${\tan(x)}^r$, for $r \in \mathbb{R}$, but for now we can stay with the rationals.
My attempt at solving this:
Let $u = \tan(x) \implies x = \arctan(u), dx = \frac{du}{1+u^2}$. This yields
$$\int \frac{u^{\frac{p}{q}}}{1+u^2}du$$
But I'm not sure where to go from here. My first thought was long division, but that can't work here. I also considered treating $\frac{1}{1+u^2}$ like a geometric series, but that has a finite radius of convergence.
| Here is a more general result. Let $a \in \mathbb{R}\backslash\left\{-1,-3\right\}$ and let the integral in question be $I$. For this answer, we will suppose the following:
$$\cos^2(x) =\text{ } _2F_1\left(1, \frac{1+a}{2}; \frac{3+a}{2};-\tan^2(x)\right)-\frac{2\tan^2(x)}{3+a}\text{ }_2F_1\left(2, \frac{3+a}{2}; \frac{5+a}{2};-\tan^2(x)\right)$$
(which can be seen here (I have tried multiple real values of $a$)) where $\text{ } _2F_1 (a,b;c;z)$ is the Hypergeometric Function. Applying this, we get
$$
\eqalign{
I =& \int\tan^{a}\left(x\right)dx \cr
=& \int\frac{\tan^{a}\left(x\right)}{\cos^{2}\left(x\right)}\cos^{2}\left(x\right)dx \cr
=& \int\frac{\tan^{a}\left(x\right)}{\cos^{2}\left(x\right)}\text{ } _2F_1\left(1, \frac{1+a}{2}; \frac{3+a}{2};-\tan^2(x)\right)dx \cr
&- \frac{2}{3+a}\int\frac{\tan^{2+a}\left(x\right)}{\cos^{2}\left(x\right)}\text{ }_2F_1\left(2, \frac{3+a}{2}; \frac{5+a}{2};-\tan^2(x)\right) \cr
}
$$
Next, we will suppose
$$\frac{d}{dz} \text{ } _2F_1 (a,b;c;f(z)) = \frac{ab}{c}\text{ } _2F_1 (a+1,b+1;c+1;f(z))f'(z).$$
Integrating by parts on the first integral, we let $u =\text{ }_2F_1\left(1, \frac{1+a}{2}; \frac{3+a}{2};-\tan^2(x)\right)$ and $dv = \dfrac{\tan^{a}\left(x\right)}{\cos^{2}\left(x\right)}dx$ so that $du = -\dfrac{2+2a}{3+a} \text{ } _2F_1\left(2, \frac{3+a}{2}; \frac{5+a}{2}; -\tan^2(x)\right)\tan(x)\sec^2(x)$ and $v = \dfrac{\tan^{1+a}(x)}{1+a}$. Then
$$
\eqalign{
I =& \frac{\tan^{1+a}\left(x\right)}{1+a} \text{ }_2F_1\left(1, \frac{1+a}{2}; \frac{3+a}{2};-\tan^2(x)\right) \cr
&- \frac{1}{1+a}\int\tan^{1+a}\left(x\right)\left(-\frac{1+a}{3+a}\text{ } _2F_1\left(2, \frac{3+a}{2}; \frac{5+a}{2};-\tan^2(x)\right)\right)2\tan\left(x\right)\sec^{2}\left(x\right)dx \cr
&- \frac{2}{3+a}\int\frac{\tan^{2+a}\left(x\right)}{\cos^{2}\left(x\right)} \text{ } _2F_1\left(2, \frac{3+a}{2}; \frac{5+a}{2};-\tan^2(x)\right). \cr
}
$$
Therefore,
$$\int\tan^{a}\left(x\right)dx = \frac{\tan^{1+a}\left(x\right)}{1+a} \text{ }_2F_1\left(1, \frac{1+a}{2}; \frac{3+a}{2};-\tan^2(x)\right) + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Fair Value of a Dice Game 1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.
What is the fair price of this game?
Strategy
Since the expected value of a single roll is 3.5, I will stop on any roll if I get a 4, 5, or 6.
If Fair Price = Expected Value
And Expected Value = P(i_th roll) x E(i_th roll) i=1 to 3
Then Fair Value = P(1 roll) E(1st roll) + P(2 rolls) E(2nd roll) + P(3 rolls) * E(3rd roll)
E(i_th roll) = 3.5 for any roll
P(only 1 roll) = 1/2 :: must roll 4|5|6
P(only 2 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 4|5|6
P(3 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 1|2|3
$\frac{1}{2} * 3.5 + \frac{1}{4} * 3.5 + \frac{1}{4} * 3.5 = $
$\frac{1}{2}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} = $
$\frac{7}{4} + \frac{7}{8} + \frac{7}{8} = $
$\frac{14}{8} + \frac{7}{8} + \frac{7}{8} = $
$\frac{28}{8} = 3.50$
Whn trying to solve this before, I think I made a mistake and end up with something like $4\frac{5}{8} = 4.625$
Should 3.5 be the answer, or should the 1st and 2nd rolls be "wieghted" by what values you must roll to move on (1,2 or 3) instead of the general 3.5 expected value for a single roll?
| The value of a single roll is $3.5$
With two rolls, keep the first if it is more than $3.5$, so the value is $$\frac12\times5+\frac12\times3.5=4.25$$
With three rolls, keep the first if it is more than $4.25$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving the Diophantine equation $k^2 = a^2 - 4b + 8$ for $k,a,b$ We have the system of Diophantine equations in 4 variables
$$\begin{align}
a & = p+q \\
b-2 &= pq \tag 1
\end{align}$$
We have
$$\begin{align}
(p-q)^2 & = (p+q)^2 - 4pq \\
& = a^2 - 4(b-2) \tag 2
\end{align}$$
Putting $p-q = k$, we need to solve the 3-variable Diophantine equation
$$
k^2 = a^2 - 4b + 8 \tag 3
$$
Assuming we have a solution for Eqn. (3), we can then solve
$$
p+q = a \\
p-q = k \tag 4
$$
I can get particular solutions for eg., if I set $b = 0$, then $k-a = \text{$d_i$, a divisor of 8}$ and $k+a = 8/d_i$.
How do I solve Eqn. (3) to obtain the general solutions?
If it helps to provide context, $a, b$ are the coefficients in a monic palindromic polynomial of degree 4.
$$
\begin{align}
f(x) & = x^4 + ax^3 + bx^2 + ax + 1 \\
& = (x^2 + px + 1)(x^2 + qx + 1) \\
& = x^4 + (p+q)x^3 + (pq+2)x^2 + (p+q)x + 1
\end{align} \tag 5
$$
Equating the coefficients, we get the original Eqn. (1).
Note: WolframAlpha gives the following parametric solutions for Eqn (3):
$$
\begin{align}
a & = 2 c_1, b = {c_1}^2 - {c_2}^2 + 2, k = 2 c_2, & \text{ where } c_1, c_2 \in \mathbb{Z} \\
a & = 2 c_1 + 1, b = {c_1}^2 + c_1 - {c_2}^2 + c_2 + 2, k = 2 c_2 - 1, & \text{ where } c_1, c_2 \in \mathbb{Z}
\end{align}
$$
Are these the only solutions?
| If $a$ and $b$ are integer constants in your question, then $p$ and $q$ are obviously restricted and the pairs $(p,q)$ or $(q,p)$ are the roots of the quadratic equation $x^2-ax+(b-2)=0$ therefore $(p,q)$ is not always a pair of integers.
Thus, $a$ and $b$ cannot be considered as constants.
If $a,b,p,q$ are all integer variables, then you can construct the general solution set as follows:
$$\begin{align}&p=c_1\\
&q=c_2\\
&a=c_1+c_2\\
&b=c_1c_2+2\end{align}$$
where $c_1,c_2\in\mathbb Z$.
Thus, you can easily obtain the general solution you are looking for depending on the variables $a,b,k\,:$
$$
\begin{align}&a=c_1+c_2\\
&b=c_1c_2+2\\
&k=±(c_1-c_2)\end{align}
$$
where $c_1,c_2\in\mathbb Z$.
Indeed, we see that:
$$
\begin{align}(c_1-c_2)^2=(c_1+c_2)^2-4(c_1c_2+2)+8\end{align}
$$
Finally, observe that the solutions that Wolfram Alpha or other software generates for the equation $k^2=a^2-4b+8$ over integers, are a subset of the solution set we found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Parabola passing through TWO points with known tangents directions Given the points $ P_1 (0, 5), P_2 (0, 0), P_3 (3, 0) $, a parabola passes through $P_1$ and is tangent to the segment $P_1 P_2$ and passes through $P_3$ and is tangent to the segment $P_2 P_3$. Determine the vertex, and axis of symmetry of the parabola, as well the quadratic term coefficient.
My Attempt:
Based on the Bezier quadratic curve, the above parabola can be written as
$P(t) = P_1 (1 - t)^2 + 2 t (1 - t) P_2 + t^2 P_3 $
Expanding this, we get
$P(t) = t^2 (P_1 - 2 P_2 + P_3) + t ( - 2 P_1 + 2 P_2 ) + P_1 $
Note that $P(0) = P_1 $ and $P(1) = P_3$, and that the tangent vector
$P'(t) = (2 t ) (P_1 - 2 P_2 + P_3) + (-2 P_1 + 2 P_2 ) $
so that
$ P'(0) = 2 (P_2 - P_1) $ which is along the segment $P_1 P_2$, and
$ P'(1) = 2 ( P_3 - P_2 ) $ which is along the segment $P_2 P3$
The vertex of of this parabola is the point $P(t_1)$ where $ P'(t_1) \cdot P''(t_1) = 0 $
I still have to find a way to compute the axis of symmetry and the quadratic term coefficient.
EDIT:
The above parabola parametrization can be written as follows
$P(t) = A t^2 + B t + C $
with $A = P_1 - 2 P_2 + P_3 , \ B = -2 P_1 + 2 P_2, \ C = P_1 $
Completing the square in $t$, this becomes
$ P(t) = A' (t - \tau)^2 + B' (t - \tau) + C' $
where $ \tau $ is the value of $t$ where the vertex $C'$ occurs.
Comparing the two expressions which must be equal for all $t$, we have
$ A' = A $
$ - 2 \tau A' + B' = B $
$ A' \tau^2 - B' \tau + C' = C $
which can be solved for $A', B', C'$ if $\tau$ is known. To determine $\tau$, we must use the requirement that $B'$ is perpendicular to $A'$, and this gives
$ A \cdot (B + 2 A \tau) = 0 $
Note that this equation is exaclty the same as $P'(\tau) \cdot P''(\tau) = 0 $
Having found $\tau$ we can readily solve for $A', B', C'$.
Clearly, the axis of symmetry points along vector $A'$ (which is the same as $A$).
If we now choose a unit vector along $B'$ to be our $x'$ axis and a unit vector along vector $A'$, and let the origin $O'$ of this coordinate system be the vector $C'$, then
$P(t) = (x(t), y(t) ) = A' (t - \tau)^2 + B' (t - \tau) + C' $
Define the unit vectors $\hat{u} = \dfrac{B'}{\| B' \|}$ and $\hat{v} = \dfrac{A'}{\| A' \| } $, then
$ \vec{P(t)} = x' \hat{u} + y' \hat{v} + C' $
where $x' = \| B' \| (t - \tau ) $
and $ y' = \| A' \| (t - \tau )^2 $
Hence, $y' = \bigg( \dfrac{\| A' \|}{ \| B' \|^2 } \bigg) x' \ ^2 $
So that the quadratic term coefficient is
$ a = \dfrac{\| A' \|}{ \| B' \|^2 } $
| Following the method outlined in my OP, we have the parametric equation of the parabola,
$P(t) = A t^2 + B t + C = A' (t - \tau)^2 + B' (t - \tau) + C' $
where
$A = P_1 - 2 P_2 + P_3 = (0, 5) - 2 (0, 0) + (3, 0) = (3, 5) $
$B = -2 P_1 + 2 P_2 = - 2 (0, 5) + 2 (0, 0) = (0, -10) $
$C = P_1 = (0, 5) $
And the constant $\tau$ is computed from
$ A \cdot (2 A \tau + B) = 0 $, hence $ (3, 5) \cdot ( (6, 10) \tau + (0, -10) ) = 0 $, which gives $\tau = \dfrac{50}{68} = \dfrac{25}{34} $
In addition, we have
$A' = A = (3, 5) $
$B' = B + 2 A' \tau = (0, -10) + \dfrac{25}{17} (3, 5) = \dfrac{1}{17} (75, -45) $
$C' = C + B' \tau - A' \tau^2 = (0, 5) + \dfrac{25}{578} (75, -45) - \dfrac{625}{1156} (3, 5) = \dfrac{1}{1156} ( 1875, 405 ) $
$C'$ is the vertex of the parabola.
The unit vectors $\hat{u}$ and $\hat{v}$ are defined as follows
$ \| B' \| = \dfrac{1}{17} \sqrt{7650} = \dfrac{ 15 \sqrt{34} }{17} = \dfrac{30}{\sqrt{34} } $
$ \| A' \| = \sqrt{34} $
$ \hat{u} = \dfrac{B'}{\| B' \|} = \dfrac{(75, -45) }{\sqrt{7650}} $
$ \hat{v} = \dfrac{A'}{\| A' \|} = \dfrac{ (3, 5) }{\sqrt{34} } $
The quadratic term coefficient is
$ a = \dfrac{ \| A' \|}{\| B' \|^2 } = \dfrac{34 \sqrt{34}}{900} $
Hence, if $p$ is the semi-latus rectum, then
$ 4 p = \dfrac{1}{a} $
Hence,
$ p = \dfrac{ 225 }{ 34 \sqrt{34} } $
The focus is given by
$F = C' + p \hat{v} = \dfrac{1}{1156} ( 1875, 405 ) + \bigg( \dfrac{ 225 }{ 34 \sqrt{34} } \bigg) \bigg( \dfrac{ (3, 5) }{\sqrt{34} } \bigg) = \bigg(\dfrac{2550}{1156}, \dfrac{1530}{1156} \bigg) = \bigg( \dfrac{75}{34}, \dfrac{45}{34} \bigg) $
Here's a plot of the parabola showing the vertex, the focus, and the axis of symmetry, as well as the tangency points with the coordinate axes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Attempt to solve $\lim_{x\to +\infty} \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}$ I tried to solve this limit. Is it correct? There exists a more straightforward way?
$$\lim_{x\to +\infty} \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}$$
$$\lim_{x\to +\infty} \exp{\left(\frac{x^2-1+1}{1+x}\right)} -\exp\left(x^2\ln\left(1+\frac1x\right)\right)$$
$$\lim_{x\to +\infty} \exp(x-1)\cdot\exp{\left(\frac{1}{1+x}\right)} -\exp\left(x^2\left(\frac1x-\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)\right)\right)$$
$$\lim_{x\to +\infty} \exp(x-1)\cdot{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)} -\exp\left(x-\frac{1}{2}+o(1)\right)$$
$$\lim_{x\to +\infty} \exp\left(-\frac12\right)\cdot\exp\left(x-\frac12\right)\cdot{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)} -\exp\left(x-\frac{1}{2}+o(1)\right)$$
$$\lim_{x\to +\infty}\underbrace{\exp\left(x-\frac12\right)}_{\to+\infty}\left[\exp\left(-\frac12\right)\cdot{\underbrace{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)}_{\to0}} -\underbrace{\exp\left(o\left(1\right)\right)}_{\to 1}\right] $$
$$= +\infty\cdot\left[0-1\right] = -\infty$$
| I always try to find a solution without using $o(f(x))$ (yes, I know that De L'Hopital is equivalent to $o(f(x))$)
\begin{align*}
& \lim_{x\to +\infty}\left[ \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}\right] = \\
& \lim_{x\to +\infty}\left[ \exp{\left(\frac{x^2}{1+x}\right)} - \exp\left(x^2\log\left(1+\frac1x\right)\right)\right] = \\
& \lim_{y\to 0^+}\left[ \exp{\left(\frac{1}{y+y^2}\right)} - \exp\left(\frac{\log\left(1+y\right)}{y^2}\right)\right] = \\
& \lim_{y\to 0^+}\left[ \exp{\left(\frac{1}{y}-\frac{1}{1+y}\right)} - \exp\left(\frac{1}{y}+\frac{\log\left(1+y\right)-y}{y^2}\right)\right] = \\
& \lim_{y\to 0^+} \exp\left(\frac{1}{y}\right) \left[ \exp{\left(-\frac{1}{1+y}\right)} - \exp\left(\frac{\log\left(1+y\right)-y}{y^2}\right)\right] = (*)
\end{align*}
and given that
$$
\lim_{y\to0}\frac{\log\left(1+y\right)-y}{y^2}=\lim_{y\to0}\frac{\frac{1}{1+y}-1}{2y}=-\frac{1}{2}
$$
we have
$$
(*) = \lim_{y\to0^+}\exp\left(\frac{1}{y}\right)[e^{-1}-e^{-1/2}] = -\infty
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the (1,1) entry in this dynamical system? Here's the question:
Consider the dynamical system $V_{k+1}$ = $AV_k$ where
$$
A=\begin{pmatrix}
0 & 1 \\
-13 & 4\\
\end{pmatrix}
\quad \text{and} \quad
V_0=\begin{pmatrix}
1 \\
3
\end{pmatrix}
$$
Find a formula in terms of k for the $(1,1)$-entry $x_k$ of $V_k$.
Can anyone help point me in the right direction for this? my first though to Diagonalize A and use the formula $V_k = PD^kP^-1V_0$, but all I got wax this insanely janky looking vector that I know is wrong. How are you supposed to find the formula for k in this instance?
| Obviously,
$ V_k = A^k V_0 $
Now,
$ A^k = \alpha_0 I + \alpha_1 A \hspace{30pt} (*)$
To find $\alpha_0$ and $\alpha_1$, we use the eigenvalues of $A$.
The characteristic polynomial of $A$ is $ \lambda^2 - 4 \lambda + 13 = 0 $
whose roots (the eigenvalues) are $\lambda_1= 2 - 3 i $ and $ \lambda_2 = 2 + 3 i $
Substitute this in the above equation, i.e.
$ \lambda_1^k = \alpha_0 + \alpha_1 \lambda_1 $
$ \lambda_2^k = \alpha_0 + \alpha_1 \lambda_2 $
Solving this linear system in $\alpha_0, \alpha_1 $ we get
$\alpha_0 = \dfrac{ \lambda_2 \lambda_1^{k} - \lambda_1 \lambda_2^k }{ \lambda_2 - \lambda_1 } $
$\alpha_1 = \dfrac{ \lambda_2^k - \lambda_1^k } {\lambda_2 - \lambda_1} $
Now, we have
$ \lambda_2 - \lambda_1 = 6 i $
$ \lambda_1^k = \sqrt{13} e^{-i k \phi} , \lambda_2^k = \sqrt{13} e^{i k \phi} $
where $ \phi = \tan^{-1} \dfrac{3}{2} $
Substituting this, and simplifying,
$ \alpha_0 = - \dfrac{13^{(k+1)/2}}{3} \sin( ( k - 1) \phi ) $
$ \alpha_1 = \dfrac{13^{k/2}}{3} \sin (k \phi ) $
Substituting these in equation (*) above, then
$ A^k = \begin{bmatrix} - \dfrac{13^{(k+1)/2}}{3} \sin( ( k - 1) \phi ) && \dfrac{13^{k/2}}{3} \sin (k \phi ) \\ -13 \dfrac{13^{k/2}}{3} \sin (k \phi ) && - \dfrac{13^{(k+1)/2}}{3} \sin( ( k - 1) \phi ) + 4 \dfrac{13^{k/2}}{3} \sin (k \phi ) \end{bmatrix} $
Hence,
$ V_k = A^k V_0 = \begin{bmatrix}- \dfrac{13^{(k+1)/2}}{3} \sin( ( k - 1) \phi ) + 13^{k/2} \sin (k \phi ) \\ - \dfrac{13^{k/2}}{3} \sin (k \phi ) - 13^{(k+1)/2} \sin( ( k - 1) \phi ) \end{bmatrix} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $\int_0^1f(x)dx$
If $$f(x)=\binom{n}{1}(x-1)^2-\binom{n}{2}(x-2)^2+\cdots+(-1)^{n-1}\binom{n}{n}(x-n)^2$$
Find the value of $$\int_0^1f(x)dx$$
I rewrote this into a compact form.
$$\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}$$
Now,
$$\int_0^1\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}dx$$
$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}-\sum_{k=1}^n\binom{n}{k}\frac{(-k)^3}{3}(-1)^{k-1}$$
$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}+\sum_{k=1}^n\binom{n}{k}\frac{k^3}{3}(-1)^{k-1}$$
After this, I took $\dfrac13$ common and did some simplifications but nothing useful came out.
Any help is greatly appreciated.
| A devious little problem indeed! I am interested in where you found it. We in fact have a very nice formula:
$$\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 =x^2$$
It follows from:
$$F(x)=\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 \\ =\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}\big(x^2-2kx+k^2\big) \\ =x^2\color{red}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}}-2x\color{green}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}k}+\color{blue}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}k^2}$$
Now, we need to somehow show
$$\color{red}{\blacksquare}=1 \\ \color{green}{\blacksquare}=\color{blue}{\blacksquare}=0$$
These sums have been studied before:
$\color{red}{\blacksquare}$ : Alternating sum of binomial coefficients equal to $1$
Follows from doing a binomial expansion of $(1-1)^n$.
$\color{green}{\blacksquare}$ : Binomial coefficient series $\sum\limits_{k=1}^n (-1)^{k+1} k \binom nk=0$
Follows from the recurrence $k\binom{n}{k}=n\binom{n-1}{k-1}$.
$\color{blue}{\blacksquare}$ : This is the hard one. We proceed as follows:
$$\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}=n\sum_{k=1}^n (-1)^{k-1}k \binom{n-1}{k-1} \tag{1}$$ $$ =n\sum_{l=0}^{n-1}(-1)^l(l+1)\binom{n-1}{l} \\ =n\sum_{l=0}^{n-1}(-1)^l l\binom{n-1}{l}+n\underbrace{\sum_{l=0}^{n-1}(-1)^l \binom{n-1}{l}}_{=(1-1)^{n-1}=0} $$
Finally, since $\binom{n-1}{n}=0$, and since the $l=0$ summand is zero, we can remove the $l=0$ index and add a $l=n$ index, and then rename the index back to $k$:
$$n\sum_{l=0}^{n-1}(-1)^l l\binom{n-1}{l}=n\sum_{k=1}^n(-1)^k k\binom{n-1}{k}$$
Now, using the recurrence relation for the binomials,
$$n\sum_{k=1}^n(-1)^k k\binom{n-1}{k}=n\underbrace{\sum_{k=1}^n k(-1)^k \binom{n}{k}}_{=\color{green}{\blacksquare}=0}-n\sum_{k=1}^n (-1)^kk \binom{n-1}{k-1}$$
Again, $k\binom{n-1}{k-1}=k^2\binom{n}{k}$ and hence
$$-n\sum_{k=1}^n (-1)^kk \binom{n-1}{k-1}=-n\sum_{k=1}^n (-1)^kk^2 \binom{n}{k}=n\sum_{k=1}^n (-1)^{k-1}k^2 \binom{n}{k}\tag{2}$$
But, retracing our steps from $(1)$ to $(2)$, we have just proved that
$$\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}=~\boldsymbol{n}~\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}$$
This can only be true for general $n$ if the sum is zero. Hence,
$$\boxed{\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2=x^2\color{red}{\blacksquare}-2x\color{green}{\blacksquare}+\color{blue}{\blacksquare} \\ =x^2}$$
QED!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
How to find the constant $C$ such that $f(x)\geq Cx$ Problem :
Define for strictly positive $x$ :
$$f\left(x\right)=\left(\prod_{k=1}^{\operatorname{floor}\left(x\right)}\left(1+\sum_{n=1}^{k}\frac{1}{k\cdot2^{n}}\right)\right)$$
Does there exists a constant $C$ such that :
$$f(x)\geq Cx$$
I think definitely yes and $C\simeq 0.6516...$
Currently Symbolic calculator don't find it .
As side notes we have :
$$-\ln\left(2\right)+\sum_{k=1}^{\infty}\frac{1}{k\cdot2^{k}}=0$$
$$\frac{1}{2}\left(1-3\ln\left(\frac{4}{3}\right)\right)+\left(+\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)\cdot2^{1+2k}}\right)=0$$
and so on .
Edit 26/12/2022 :
we have with a computer :
$$\left|\exp\left(1-\prod_{k=1}^{1000}\left(1-\frac{1}{2^{k}(k+1)}\right)\right)-\sqrt{2}-\frac{1}{2^{9}}-\frac{1}{2^{11}}-\frac{1}{2^{14}}-\frac{1}{2^{16}}+\frac{1}{2^{18}}-\frac{1}{2^{20}}-\frac{1}{2^{21}}-\frac{1}{2^{24}}+\frac{1}{2^{27}}+\frac{1}{2^{30}}-\frac{1}{2^{33}}+\frac{1}{2^{36}}+\frac{1}{2^{38}}+\frac{1}{2^{39}}-\frac{1}{2^{41}}\right|<4*10^{-13}$$
I don't know if we can go like this at infinity .
Question :
How to show the existence and does $C$ admits a closed form ?
Thanks a lot .
| Note that
\begin{equation}
f(x)=\prod_{k=1}^{\left \lfloor x \right \rfloor} \left(1+ \frac{1}{k}\sum_{j=1}^k \frac{1}{2^j}\right)= \prod_{k=1}^{\left \lfloor x \right \rfloor} \left( 1+ \frac{1-2^{-k}}{k} \right)
\end{equation}
Therefore,
\begin{align}
\ln f(x) =& \sum_{k=1}^{\left \lfloor x \right \rfloor} \ln \left(1+ \frac{1-2^{-k}}{k} \right) \approx \sum_{k=1}^{\left \lfloor x \right \rfloor} \frac{1-2^{-k}}{k}= H_{\left \lfloor x \right \rfloor}-\sum_{k=1}^{\left \lfloor x \right \rfloor} \frac{1}{k 2^k}\\ &\geq H_{\left \lfloor x \right \rfloor}- \sum_{k=1}^{\infty} \frac{1}{2^k}= H_{\left \lfloor x \right \rfloor}-1
\end{align}
where $H_n$ denotes the harmonic series, and the approximation follows from a Taylor expansion. Next as $H_n\approx \ln n +\gamma$, where $\gamma\approx 0.5772$ is the Euler–Mascheroni constant,
\begin{equation}
f(x)\geq \frac{1}{e} e^{H_{\left \lfloor x \right \rfloor}} \approx e^{\gamma-1} \left \lfloor x \right \rfloor \geq e^{\gamma-1} (x-1) \approx 0.6552 (x-1)
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
What particular solution should I guess for $y'' - 4y' + 4y = (6x - 2)e^{2x}$? I have the following initial value problem
$$ y'' - 4y' + 4y = (6x-2)e^{2x},\ y(0)=2,\ y'(0)=1 $$
to which I have obtained the following solution for the homogenous ODE
$$ y_{h} = e^{2x}(2 - 3x) $$
Now I am trying to figure out the particular solution but am lost as to what I should guess. My initial thought was to simply guess $y_{p} = Ce^{2x}$ but this quickly leads to nothing. I then guessed $y_{p} = (Ax+B)e^{2x}$ and it too lead to nothing. How do I continue?
EDIT: I have solved the problem with the help of the comments, see my solution down below.
| After advice from the comments I managed to solve the problem.
\begin{split}
y'' - 4y' + 4y &= 2(3x-1)e^{2x} \\
y(0) &= 2\\
y'(0) &= 1
\end{split}
$$ y'' - 4y' + 4y = 0 $$
$y_{h} = e^{rx}$, $y'_{h} = re^{rx}$, $y''_{h} = r^{2}e^{rx}$.
\begin{split}
r^{2}e^{rx} - 4re^{rx} + 4e^{rx} &= 0 \\
e^{rx}(r^{2} - 4r + 4) &= 0
\end{split}
Solving for the roots of the characteristic equation
$$ r^{2} - 4r + 4 = 0 $$
yields a double root $r=2$. General solution becomes
$$ y_{h} = C_{1}xe^{2x} + C_{2}e^{2x} $$
Now for the particular solution I make the assumption
$$y_{p} = e^{2x}(Ax^{3} + Bx^{2}) $$
Finding expressions for $y'_{p}$ and$y''_{p}$ and then substituting them into the ODE we obtain
\begin{split}
2e^{2x}(Ax(2x^{2}+6x+3) + B(2x^{2}+4x+1)) + 4e^{2x}(Ax^{3}+Bx^{2}) - 4e^{2x}x(Ax(2x+3)+2B(x+1)) &= (6x-2)e^{2x} \\
e^{2x}(4Ax^{3} + 12Ax^{2} + 6Ax + 4Bx^{2} + 8Bx + 2B + 4Ax^{3} + 4Bx^{2}) - 4e^{2x}x(2Ax^{2}+3Ax + 2Bx + 2B) &= (6x-2)e^{2x} \\
e^{2x}(4Ax^{3} + 12Ax^{2} + 6Ax + 4Bx^{2} + 8Bx + 2B + 4Ax^{3} + 4Bx^{2} - 8Ax^{3} - 12Ax^{2} - 8Bx^{2} - 8Bx) &= (6x-2)e^{2x} \\
e^{2x}( 6A + 2B ) &= (6x-2)e^{2x} \\
\end{split}
\begin{split}
6A = 6 &\implies A = 1 \\
2B = -2 &\implies B = -1
\end{split}
$$ y_{p} = e^{2x}(x^{3} - x^{2}) $$
Now for the final solution we get
$$ y = y_{h} + y_{p} = C_{1}xe^{2x} + C_{2}e^{2x} + e^{2x}(x^{3} - x^{2})$$
What is left is to determine the coefficients $C_{1}$ and $C_{2}$ which is easily done by applying our initial values.
\begin{split}
y &= C_{1}xe^{2x} + C_{2}e^{2x} + e^{2x}(x^{3} - x^{2}) \\
y' &= e^{2x}(2C_{1}x + C_{1} + 2C_{2} + x(2x^{2} + x -2))
\end{split}
Solving for the coefficients yields $C_{1} = -3$ and $C_{2} = 2$ which gives us the final complete solution
$$ y(x) = e^{2x}(x^{3} - x^{2} -3x + 2) $$
Edit: This way of solving this initial value problem seems highly tedious and annoying, is there any better faster way to do it?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is wrong with the calculation $ \lim\limits_{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $? We are given
$$ \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $$
We can write
$$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} = \lim _{x \to 0} \frac {x (cos x - \sin x/x)} {x^2 \sin x} $$
It can be written as
$$ L = \lim _{x \to 0} \frac { \cos x - \sin x/x} {x^2 \sin x/x} $$
We know
$$ \lim _{x \to 0} \frac {\sin x} {x} = 1 $$
The original limit can be written as
$$ L = \lim _{x \to 0} \frac { \cos x - 1} {x^2} $$
Now using $ ( \cos x ) $ expansion formula, the answer will be $ L = -\frac12$ , but the answer is $ L = -\frac13$ , which is not matching.
I want to know what I did wrong in this method. I know the answer using other methods, but I want to know what went wrong with this one.
| The thing is you can take a partial limit when it only involves products (or quotients). Here there is an additive $\cos(x)$ term so you cannot do that.
Remember that $\cos(x) = 1 - \frac{x^2}{2} + o(x^2)$ and $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + o(x^2)$ therefore $\cos x - \frac{\sin x}{x} = -\frac{x^2}{3} + o(x^2)$ which means that $ \frac { \cos x - \sin x/x} {x^2 }\to -\frac{1}{3}. $
Therefore $$\lim_{x \to 0} \frac { \cos x - \sin x/x} {x^2 \sin x/x} = \lim_{x \to 0} \frac { -\frac{1}{3} } {\sin x/x} = -\frac{1}{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{c^2-s^2\over s^4+c^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{(1-s^2)-s^2\over s^4+(c^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(1-s^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(s^2-1)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+s^4-2s^2+1}\mathrm d\theta\\
&=\int_{0}^{2\pi}\underbrace{\color{red}{\left({1-2s^2\over 2s^4-2s^2+1}\right)}}_{\text{I got stuck here}}\mathrm d\theta\\
\end{align}$$
I need your help.
| Method 1
Note that by defining $u=\theta-\dfrac{\pi}{2}$, we have
$$
\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
=
\int_{\dfrac{\pi}{2}}^{2\pi+\dfrac{\pi}{2}}{\sin^2u-\cos^2u\over\cos^4 u+\sin^4 u}\mathrm du.
$$
Now, since the integrand is periodic with a period of $2\pi$, we obtain:
$$
\int_{\dfrac{\pi}{2}}^{2\pi+\dfrac{\pi}{2}}{\sin^2u-\cos^2u\over\cos^4 u+\sin^4 u}\mathrm du
=\int_{0}^{2\pi}{\sin^2u-\cos^2u\over\cos^4 u+\sin^4 u}\mathrm du=-I,
$$
hence, $I=-I=0$.
Method 2
The indefinite integral can be found as follows:
$$
\int{\cos^2\theta-\sin^2\theta\over\sin^4\theta+\cos^4\theta} d\theta
{=
\int{\cos2\theta\over\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta-2\sin^2\theta\cos^2\theta} d\theta
\\=
\int{\cos2\theta\over(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta}d\theta
\\=\int \dfrac{\cos2\theta}{1-\dfrac{1}{2}\sin^22\theta}d\theta
\\=\int \dfrac{2\cos2\theta}{2-\sin^22\theta}d\theta
}
$$
where by considering $u=\sin 2\theta$ we achieve
$$
\int\dfrac{du}{2-u^2}{=
\int\dfrac{\dfrac{1}{2\sqrt 2}}{\sqrt2-u}+\dfrac{\dfrac{1}{2\sqrt 2}}{\sqrt2+u}du
\\=\dfrac{1}{2\sqrt 2}\ln\left|\dfrac{\sqrt 2+u}{\sqrt 2-u}\right|+C
\\=\dfrac{1}{2\sqrt 2}\ln\dfrac{\sqrt 2+\sin 2\theta}{\sqrt 2-\sin 2\theta}+C.
}
$$
Now, the value of the definite integral becomes $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Solution of an equation with $3$ variables I've been thinking about a solution for the following equation for integers $x, y, z$:
$$6 = (x-y)^2 + (y-z)^2 + (x-z)^2$$
A possible approach would probably be to set $(x-y)^2 = a$, $(y-z)^2 = b$ and $(x-z)^2 = c$.
Now, the euqation woud look like this:
$$6 = a + b + c$$
Hence, the only possible solutions would be $$(a, b, c) = (4, 1, 1)$$ $$(a, b, c) = (1, 4, 1)$$ $$(a, b, c) = (1, 1, 4)$$
Now one could formulate equation for $x, y$ and $z$. However, this approach would be very time consuming. Is there a simpler solution to the problem?
| Hint:
Let $x-y=a$ etc.
$\implies a+b+c=0\iff c=-?$
$$6=a^2+b^2+c^2=2a^2+2ab+2b^2\iff2a^2+a(2b)+2b^2-6=0$$
$$a=\dfrac{-2b\pm\sqrt{12(4-b^2)}}4=\dfrac{-b\pm\sqrt{3(4-b^2)}}2$$
For integer $a,$ $$b^2=4,1\implies b=\pm2,\pm1$$
We shall get the same set of values for $c,$ if we start with $b=-(c+a)$
So, $(a,b,c)\in(1,1,-2)$ or $\in(-1,-1,2)$ as $a+b+c=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4606594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Where did I go wrong with this integration? $$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2-3x^2}}$$
$$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2(1-\frac32x^2)}}$$
$$\frac 1{\sqrt 2}\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {1-\frac32x^2}}$$
$$\bigg(\frac 1{\sqrt 2}\sin^{-1}{\sqrt {\frac 32 x}}\bigg)\bigg|_0^\frac 1{\sqrt 3}$$
$$\frac 1{\sqrt 2} \times \frac {\pi}{4}$$
However, the answer is $\frac 1{\sqrt 3} \times \frac {\pi}{4}$. Where did I go wrong?
| Let $x=\frac{\sqrt{2} \sin \theta}{\sqrt{3}}$, then $dx=\frac{\sqrt{2}}{\sqrt{3}} \cos \theta d \theta$
$$
\begin{aligned} \\
I & =\int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{2-2 \sin ^2 \theta}} \frac{\sqrt{2}}{\sqrt{3}} \cos \theta d \theta \\
& =\frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{4}} d \theta \\
& =\frac{\pi}{4 \sqrt{3}}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\log 40\sqrt{3}}{\log 4n} = \frac{\log 45}{\log 3n} \\
\log 40\sqrt{3}\log 3n = \log 45\log 4n\\
\log 40\sqrt{3} \cdot (\log 3 + \log n) = \log 45 \cdot (\log 4 + \log n)\\
\log n \cdot (\log 40\sqrt{3} - \log 45) = \log 45\log 4 - \log 40\sqrt{3}\log 3
\end{align*}
Dividing through and putting the coefficients as powers, we have:
\begin{align*}
\log n &= \frac{\log 45^{\log 4} - \log \left[(40\sqrt{3})^{\log 3}\right]}{\log\left(\frac{40\sqrt{3}}{45}\right)}
=\frac{\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right) }{\log\left(\frac{40\sqrt{3}}{45}\right)} \\
&=\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
which shows that
\begin{align*}
n^3 = \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{3\cdot\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
Somehow it feels like this answer may be simplified further. Are the steps shown so far correct and can the answer be expressed in a better way?
| I would go the following way.
You have :
$$\begin{align}&\log_{4n} 40\sqrt{3} = \log_{3n} 45\\
\implies &\log_{4n} \left(3\cdot 40^2\right) = \log_{3n} 45^2=k
\end{align}$$
This leads to :
$$\begin{align}&\begin{cases}(4n)^k=3\cdot 40^2\\
(3n)^k=45^2 \end{cases}\\
\implies &\left(\frac 43\right)^k=\frac {3\cdot 40^2}{45^2}=\left(\frac {4}{3}\right)^3\\
\implies &k=3\\
\implies &n^3=\frac {45^2}{3^3}=75\thinspace.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
How to prove $\sum_{k=1}^n{n\choose k}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^n\frac{1}{k}$? I think by induction we can do it. Let
$I(n)=\sum_{k=1}^n{n\choose k}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^n\frac{1}{k}.$ Then, we must show that $I(n+1)-I(n)=\frac{1}{n+1}$.
$\begin{align}
I(n+1)-I(n)&=\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n\left({n+1\choose k}-{n\choose k}\right)\frac{(-1)^{k+1}}{k}\\
&=I(n+1)-I(n)=\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n{n\choose k-1}\frac{(-1)^{k+1}}{k}\\
&=\frac{(-1)^{n+1}}{n+1}-\frac{1}{n+1}\sum_{k=1}^n{n+1\choose k}(-1)^{k}\\
&=\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^n+1}{n+1}\\
&=\frac{1}{n+1}
\end{align}$
I saw these steps now. If you don't write you can't see anything. Thanks for help and elementary proofs.
WA says it is about Digamma function. What is the connection? These are very hot topics for students like me.
Thanks in advance.
| Let $n\geq 1$ and define the sequence,
$$ a_n = \sum_{k=1}^n {n\choose k} \frac{1}{k} (-1)^{k+1} = \sum_{k=1}^{\infty} {n\choose k} \frac{1}{k} (-1)^{k+1} $$
(If $k>n$ then the binomial coefficient is zero, not changing the sum).
The generating function of this sequence is equal to,
$$ A(x) = \sum_{n=1}^{\infty} a_n x^n = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} {n\choose k} \frac{1}{k} (-1)^{k+1} x^n = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \left\{ \sum_{n=1}^{\infty} {n\choose k} x^n \right\}$$
The series in the bracket is a variant of the geometric series and is well-known to sum to,
$$ A(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \left\{ \frac{x^k}{(1-x)^{k+1}} \right\} = \frac{1}{1-x}\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \left( \frac{x}{1-x} \right)^k $$
The remaining series is again well-known to sum of the logarithmic series, therefore,
$$ A(x) = \frac{1}{1-x} \log \left( 1 + \frac{x}{1-x} \right) = \frac{1}{1-x} \log \left( \frac{1}{1-x} \right) $$
Now recall that $\log(\tfrac{1}{\theta}) = -\log(\theta)$, therefore, by using logarithmic series again,
$$ A(x) = \frac{1}{1-x} \bigg( -\log(1-x) \bigg) = \frac{1}{1-x}\left( \sum_{k=1}^{\infty} \frac{x^k}{k} \right) = \left( \sum_{k=0}^{\infty} x^k \right) \left( \sum_{k=1}^{\infty} \frac{x^k}{k} \right)$$
Now it is easy to find the $x^n$ coefficient of $A(x)$, it is equal to,
$$ a_n = \frac{1}{n} + \frac{1}{n-1} + ... + \frac{1}{1} = \sum_{k=1}^n \frac{1}{k} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the right solution, as $-\frac{\sqrt{3}}{2}$ returns $-1$, not $1$. My question is where did I miss a condition that excludes the negative "solution"? I expect somewhere along the line I squared where I wasn't allowed to square without an additional condition, hoping that I don't have to check these solutions every time.
| Yet another way :
Let $\sqrt{1+x}=a, \sqrt{1-x}=b$
For real $x, a\ge0, b\ge0 $
By the given condition,
$$a-b=1$$ and $$a^2+b^2=2\implies2=(b+1)^2+b^2\iff2b^2+2b-1=0$$
$\implies(2b+1)^2=3\implies2b+1=+\sqrt3$ as $b\ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
Maximizing/minimizing $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta)$
Assume that $f : [0, 2\pi]\rightarrow \mathbb{R}$ is a function such that $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta)$. Then, how can we maximize/minimize $f$?
We can re-parametrize our function $f$ by defining another function $g : [-1, 1]\rightarrow \mathbb{R}$ function such that for every $t\in [-1, 1]$,
$$g(t) = \sqrt{2}\sqrt{1-t^2}-4t$$
$$\frac{dg}{dt} = \frac{d}{dt}\left(\sqrt{2}\sqrt{1-t^2}-4t\right) = \frac{\sqrt{2}t}{\sqrt{1-t^2}} + 4 = 0$$
From which we conclude that $g$ attains its maximum/minimum at $\left(-\frac{2\sqrt{2}}{3}, g\left(-\frac{2\sqrt{2}}{3}\right)\right), (1, g(1))\in \mathbb{R}^2$ respectively.
| $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta) = \sqrt{18}( \sin\alpha \cos \theta - \cos \alpha \sin \theta)$ where $\sin \alpha = \frac{\sqrt{2}}{\sqrt{18}}$ and $\cos \alpha = \frac{4}{\sqrt{18}}$ and hence $\alpha = arctan \frac{\sqrt{2}}{4}$. Thus
$$f(\theta) = \sqrt{18}\sin(\alpha - \theta) \le \sqrt{18}$$ and the estimate is best possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate this double Integral in polar coordinates. $$ \iint y^2(a^2-x^2)^{0.5}dxdy $$ over $x^2+y^2\le a^2$
I have evaluated this in $x-y$ plane and got $32a^5/45$. Please help in evaluating the same in polar coordinates.
Ive tried putting $x=r\cos\theta$ and $y=r\sin\theta$ and After considering the Jacobian I am getting this expression
$$ \int\limits_0^a\int\limits_0^{2π}r^3\sin^2θ(a^2-r^2cos^2θ)^{0.5}\,drdθ $$
Please help me evaluate this.
| Since the integrand and the region $x^2 + y^2 \le a^2$ is symmetric, it suffices to calculate the quarter.
$$
\begin{align}
\iint_{x^2 + y^2 \le a^2} y^2\sqrt{a^2 - x^2} dx dy &= 4 \iint_{0 \le x,y \land x^2+y^2 \le a^2} y^2\sqrt{a^2 - x^2} dx dy \\
&=4 \int_{0}^{a} \sqrt{a^2 - x^2} dx \int_0^{\sqrt{a^2 - x^2}} y^2 dy \\
&= \frac43 \int_0^a (a^2 - x^2)^2 dx \\
&= \frac43 \int_0^a (x^4 - 2a^2 x^2 + a^4) dx \\
&= \frac 43 (\frac15 a^5 - \frac23 a^5 + a^5) \\
&= \frac{32}{45}a^5
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Simplifying $\frac{x^4(y-z)+y^4(z-x)+z^4(x-y)}{(y+z)^2+(z+x)^2+(x+y)^2}$ I need help to simplify this expression.
$$\frac{x^4(y-z)+y^4(z-x)+z^4(x-y)}{(y+z)^2+(z+x)^2+(x+y)^2}$$
I only know to start, I extracted the $(x-y)$ from top one, and then I don't know what's next. Extracting $(x-y)$ from the numerator:
$$(x^4y - y^4x) - (zy^4-zx^4)+z^4(x-y)$$
$$ = xy(x^3-y^3)- z(x^4-y^4)+z^4(x-y)$$
$$ = xy(x-y)(x^2+xy+y^2)-z(x-y)(x+y)(x^2+y^2)+z^4(x-y)$$
$$ = (x-y)\left[ xy(x^2+xy+y^2) - z(x+y)(x^2+y^2) +z^4 \right]$$
$$ = (x-y)(x^3y+x^2y^2+xy^3-zx^3-zy^3+z^4-zxy^2-zx^2y)$$
| The nominator is a degree 5 polynomial and the denominator degree 2 polynomial, so if the fraction should simplify into a polynomial, it would have to be of degree 3, symmetric, and with the same number of positive and negative terms. My first guess is to consider the symmetric polynomial
$$
(x-y)(y-z)(z-x)
= y (z^2-x^2) + z (x^2-y^2) + x (y^2-z^2).
$$
Let's try when we multiply it by the denominator of the fraction:
\begin{align}
&\frac 1 2((x-y)(y-z)(z-x)
\Big((y+z)^2 + (z+x)^2 + (x+y)^2\Big) \\
&= \Big(y (z^2-x^2) + z (x^2-y^2) + x (y^2-z^2)\Big) \Big(x^2 + y^2 + z^2 + xy + yz + zx\Big)\\
&= \Big(y (z^2-x^2) + \ldots \Big) \Big(x^2 + y^2 + z^2 + xy + yz + zx\Big)\\
&= \Big(y (z^2-x^2) + \ldots\Big) \Big(x^2 + y^2 + z^2 \Big) \quad \color{red}{=: A}\\
& \ + \Big(y (z^2-x^2) + \ldots\Big) \Big(xy + yz + zx\Big) \quad \color{red}{=: B},
\end{align}
where I use the notation $(\it \text{expression}+\ldots)$ to represent the cyclical completion of the expression $x\to y \to z (\to x)$. For example $(xy+\ldots)$ represents the expression $xy + yz + zx$, which could equivalently be represented by $(yz+\ldots)$ or $(yz+\ldots)$.
Taking into account that the expression $\big(x^2 + y^2 + z^2\big)$ is symmetric, we can write
\begin{align}
A &= \Big( y (z^2-x^2) \big(x^2 + y^2 + z^2 \big) + \ldots \Big) \\
&=\Big( y \big(z^2x^2- x^4 + z^2 y^2 -x^2 y^2 + z^4 - x^2 z^2\big) + \ldots \Big) \\
&= \Big(2y (z^2 y^2 -x^2 y^2 + z^4-x^4) + \ldots \Big) \\
&= \Big(y^3 z^2 - x^2 y^3 + \ldots \Big) + 2\Big(y(z^4-x^4) + \ldots \Big) \\
&= \Big(x^3 y^2 - x^2 y^3 +\ldots \Big)
-\Big(x^4(y-z) + y^4(z-x) + z^4 (x-y) \Big),
\end{align}
where in the last line we took into the fact that the cyclical term $(y^3z^2+\ldots) = (x^3y^2+\ldots)$.
Taking into account that the expression $\big(xy + yz + zx\big)$ is symmetric, we can write
\begin{align}
B &= \Big(y (z^2-x^2) \big(xy + yz + zx\big) + \ldots \Big) \\
&\Big(
xy^2z^2 - x^3 y^2 + y^2 z^3 - x^2 y^2 z + xyz^3 - x^3 y z + \ldots \Big) \\
&=xyz \Big(yz - xy + z^2 - x^2 + \ldots \Big)+
\Big(- x^3 y^2 + y^2 z^3 + \ldots \Big) \\
&= \Big(- x^3 y^2 + y^2 z^3 + \ldots \Big) \\
&= \Big(- x^3 y^2 + x^2 y^3 + \ldots \Big)
\end{align}
We conclude that
$A + B = -\Big(x^4(y-z) + y^4(z-x) + z^4 (x-y) \Big),$
and so
$$
\frac{x^4(y-z) + y^4(z-x) + z^4 (x-y)}
{(y+z)^2 + (z+x)^2 + (x+y)^2} =
- \frac 1 2 (x-y)(y-z)(z-x).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Can $\,9\!\cdot\!10^n+4\,$ be a perfect square? I think $\,9\!\cdot\!10^n+4\,$ can be a perfect square, since it is $0 \pmod 4$ (a quadratic residue modulo $4$), and $1 \pmod 3$ (also a quadratic residue modulo $3$).
But when I tried to find if $\;9\!\cdot\!10^n+4\,$ is a perfect square, I didn’t succeed. Can someone help me see if $\;9\!\cdot\!10^n+4\,$ can be a perfect square ?
| Assume that $9\cdot10^n+4\equiv4$ is a perfect square.
$9\cdot10^n+4\equiv4\pmod9$, so $9\cdot10^n+4$ can be represented as $(9m-2)^2=81m^2-18m+4$ or $(9m+2)^2=81m^2+18m+4$, where $m\in\mathbb{N}$.
If $9\cdot10^n+4=81m^2-18m+4$, $$10^n=9m^2-2m=m(9m-2)$$
This means that $m$ and $9m-2$ must be powers of $10$.
Clearly, $9m-2>m$ because $m>0$. If $m≠1$, $m\equiv0\pmod{10}$, but then $9m-2\equiv8\pmod{10}$, which doesn't work. If $m=1$, $9m-2=7$, which also doesn't work.
If $9\cdot10^n+4=81m^2+18m+4$, $$10^n=9m^2+2m=m(9m-2)$$
This means that $m$ and $9m+2$ must be powers of $10$.
Clearly, $9m+2>m$ because $m>0$. If $m≠1$, $m\equiv0\pmod{10}$, but then $9m+2\equiv2\pmod{10}$, which doesn't work. If $m=1$, $9m+2=11$, which also doesn't work.
Therefore, there is a contradiction and so $9\cdot10^n+4$ cannot be a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving system of equations by echelon form. I have an example of a solution for a system of equations and there is a step I do not understand:
Could anyone offer some help please?
Kind regards
Stany
| $$
\begin{aligned}
\left(\begin{array}{cccc}
1 & 1 & 1 & 0 \\
3 & -1 & 1 & 20 \\
6 & 2 & -1 & 40
\end{array}\right) & \rightarrow\left(\begin{array}{cccc}
1 & 1 & 1 & 0 \\
0 & -4 & -2 & 20 \\
0 & -4 & -7 & 4
\end{array}\right) \\
& \rightarrow\left(\begin{array}{cccc}
1 & 1 & 1 & 0 \\
0 & -4 & -2 & 20 \\
0 & 0 & -5 & 20
\end{array}\right) \\
\therefore-5 z=20 & \Rightarrow z=-4 \\
-4 y-2 x(-4)=20 & \Rightarrow y=-3 \\
x-3-4=0 & \Rightarrow x=7
\end{aligned}
$$
Hence $(x,y,z)=(7,-3,-4).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$
If $g\left( x \right) = ax + c$, $f\left( x \right) = {x^b} + 3$, and $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$ what is the value of $a+b+c$?
My approach is as follows
Given $g\left( x \right) = ax + c$ & $f\left( x \right) = {x^b} + 3 \Rightarrow {\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = x$
$\frac{{g\left( x \right) - c}}{a} = x$
${\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = \frac{{g\left( x \right) - c}}{a}$
$g\left( x \right) = g\left( x \right) \Rightarrow {g^{ - 1}}\left( {g\left( x \right)} \right) = x$
I am not able to proceed.
| You can start by noting that
$$
(g \circ f)(x) = a(x^b+3)+c
$$
and, inverting this expression, get to
$$
(g\circ f)^{-1}(x) = \left(
\frac{x-c-3a}{a} \right)^{\frac 13}
$$
Identifying coefficients, you'll see that $a=2, b=3, c=1$, and consequentely, $a+b+c = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $c>b>a$ with $c,b,a\in\Bbb N,c-b=b-a=1$, prove $\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\ge\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}$
Given $c>b>a$, also $c-b=b-a=1$ where $c,b,a$ are Natural numbers, prove that
$$
\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b} \geqslant \frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}
$$
The LHS becomes $$
\frac{a^{b+c} c^b+b^{b+c} a^b+c^{b+c} \cdot b^b}{(a b c)^b},
$$
and the RHS becomes
$$
\frac{a^{b+a} c^a+b^{a+b} a^a+c^{a+b} b^a}{(a b c)^a}.
$$
Now, $a^{b+c}>a^{b+a}$,$c^b>c^a$, $b^{b+c}>b^{b+a}$,$a^b>a^a$,
$c^{b+c}>c^{b+a}$,$b^b>b^a$. So the numerator on LHS is clearly bigger than the numerator on RHS. But $(abc)^{b}>(abc)^{a}$. How do I proceed.
$\textbf{Edit}$: added an important detail.
| For $(a, b, c) = (x, x+1, x+2)$ and positive real $x$ is your inequality equivalent to
$$
2 \frac{(x+2)^{x+1}}{x^{x+1}} - \frac{x^{x+1}}{(x+1)^{x+1}} - \frac{(x+1)^{x+1}}{(x+2)^{x+1}} \ge 0
$$
or
$$
2 \ge \left( \frac{x^2}{(x+1)(x+2)}\right)^{x+1} + \left( \frac{x(x+1)}{(x+2)^2}\right)^{x+1}
$$
and that is true because both terms on the right are less than one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Averaging neighbours in a list In a list of twelve numbers, the first number is 1 and the last number is 12.
1, a, b, c, d, e, f, g, h, i, j, 12 (assume the letters represent unknown numbers)
Each other number in the list is calculated by taking the average of its neighbours and adding 1. The question is asking to find the largest number in this list.
I have assumed the third number of the list to be x, then the second number can be calculated in terms of x:
1, $\frac{x+1}{2} + 1$, x
Continuing this process I got:
$1$, $\frac{x+3}{2}$, $x$, $\frac{3x-7}{2}$, $2x - 9$, $\frac{5x-33}2$, $3x - 26$, $\frac{7x-75}2$, $4x - 51$, $\frac{9x-133}2$, $5x - 84$, $12$
I found that x is equal to 21 and the biggest number in the list is 37.
This method took a long time and is inefficient. Is there any better way to do this?
| if $x, y, z$ are three adjacent numbers, then $y = 1 + \frac{x+z}2$, which solves to $$z = 2y - x - 2$$
Also, let me call the values as $$1, a_1 = t, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, 12$$
Thus we can calculate
*
*$a_0 = 1 = 0t + 1$
*$a_1 = t - 0$
*$a_2 = 2t - 3$
*$a_3 = 2a_2 - a_1 - 2 = 3t-8$
*$a_4 = 2a_3 - a_2 - 2 = 4t - 15$
*$a_5 = 2a_4 - a_3 - 2 = 5t -24$
From which it appears that $$a_n = nt - n^2 +1$$
If this holds $a_n$ and $a_{n-1}$ then
$$\begin{align}a_{n+1} &= 2a_n - a_{n-1} - 2\\
&=2(nt - n^2 +1) - ((n-1)t - (n-1)^2 + 1) - 2\\
&=[2nt-(n-1)t] - [2n^2 - (n-1)^2 + 1] +[2 - 2]\\
&=[(n+1)t] - [n^2 + 2n]\\
&=(n+1)t - (n+1)^2 + 1\end{align}$$
And thus by induction, this holds for the remaining values, including $a_{11} = 12$.
So $$11t - 11^2 + 1 = 12\\t - 11 = 1\\t = 12$$
And we can calculate
$$a_1 = 12 - 0 = 12\\
a_2 = 24 - 3 = 21\\
a_3 = 36 - 8 = 28\\
a_4 = 48 - 15 = 33\\
a_5 = 60 - 24 = 36\\
a_6 = 72 - 35 = 37\\
a_7 = 84 - 48 = 36\\
a_8 = 96 - 63 = 33\\
a_9 = 108 - 80 = 28\\
a_{10} = 120 - 99 = 21\\
a_{11} = 132 - 120 = 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Polynomial solutions to $ax^4+bx^2+c^2 = y^2$, with $a,b,c$ as polynomials While investigating a system involving "equal sums of like powers", an elliptic curve popped up,
$$9 (1 + 4 n^4)^2 + 30 (4 + n^2 - 24 n^4 + 4 n^6) x^2 + 5 (32 - 40 n^2 + 53 n^4) x^4 = y^2$$
Some easy rational points are,
$$x = (0,\; 1,\; n)$$
Using the tangent-chord method, from the last two, we get,
$$\; x = \frac{6 (1 + 4 n^4) (17 + 12 n^2)}{151 - 49 n^2 + 144 n^4 + 144 n^6}$$
$$x = \frac{\; 6 (1 + 4 n^4) (3n + 23 n^3)}{9 + 9 n^2 - 79 n^4 + 121 n^6}$$
Question: But are there rational points where the numerator and denominator are polynomials of degree less than 6, preferably only quadratics?
| Looking at a bunch of examples, it appears that your curve (which is isomorphic to an elliptic curve) has rank $2$ over $\mathbb{Q}(n)$. Using this, one can find some other rational points of degree $\leq 6$, but not where the numerator or denominator is a quadratic. In particular, there are solutions when
$$
x = \frac{12 n^{4} + 3}{12 n^{3} \pm 23n^{2} + 17n \pm 3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ How do I evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ ?
Note that this is a Q&A post and I've presented my solution below.
| Another way to evaluate the OP’s integral is using Euler’s substitutions.
By letting $\;t=x^2\,,\;$ we get that
$\displaystyle\int\frac{2x}{\left(1-x^2\right)\sqrt{x^4-1}}\,\mathrm dx=\int\frac1{(1-t)\sqrt{t^2-1}}\,\mathrm dt\;.$
Now, we will apply Euler’s first substitution that is $\,\sqrt{t^2-1}=-t+u\,$ and obtain that
$\displaystyle\int\frac1{(1-t)\sqrt{t^2-1}}\,\mathrm dt=\int\frac{-2}{(u-1)^2}\,\mathrm du=\frac2{u-1}+C\;.$
Hence ,
$\displaystyle\int\frac{2x}{\left(1-x^2\right)\sqrt{x^4-1}}\,\mathrm dx=\frac2{x^2-1+\sqrt{x^4-1}}+C$
Addendum:
Note that
$\begin{align}\dfrac2{x^2-1+\sqrt{x^4-1}}+C&= \dfrac2{x^2-1+\sqrt{x^4-1}}+1+C-1=\\&=\dfrac{x^2+1+\sqrt{x^4-1}}{x^2-1+\sqrt{x^4-1}}+C-1=\\&=\dfrac{\sqrt{x^2+1}\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}{\sqrt{x^2-1}\left(\sqrt{x^2-1}+\sqrt{x^2+1}\right)}+C-1=\\&=\dfrac{\sqrt{x^2+1}}{\sqrt{x^2-1}}+C-1=\\&=\sqrt{\dfrac{x^2+1}{x^2-1}}+C^*\,.\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving a trigonometric finite sum $\sum_{k=1}^N(-1)^k(\cos \frac{k\pi}{N})^{N-m}(\sin\frac{k\pi}{N})^m=(-1)^{m/2}\frac{N}{2^{N-1}}$ How to prove this following formula?
$\sum_{k=1}^N(-1)^k(\cos \frac{k\pi}{N})^{N-m}(\sin\frac{k\pi}{N})^m=(-1)^{m/2}\frac{N}{2^{N-1}}$
for m is even, and $0$ for m is odd.
If we know $\sum_{k=1}^N(-1)^k(\cos\frac{k\pi}{N})^m=N/2^{N-1}$ for $m=N$, and $=-1/2(1-(-1)^{N+m})$ for $m<N$.
I have some ideas for this problem, but all fail. I tried Abel transformation or recursion. Maybe others ways can solve, like residual theory, but I'm not familiar with it. So I ask for help about this question.
From the given series, Abel transformation is easy to take. That is, one series is $(-1)^k*(\cos\frac{k\pi}{N})^{N-m}$ and the other is $(\sin\frac{k\pi}{N})^m$. But it doesn't work, because, I don't know the truncated sum for the first series, which is a necessary condition for Abel transformation calculation.
(Update on 20230224) I want to add some notes on the origin of these formulas. I encounter this identity when I want to calculate the witness for GHZ state as shown in doi:10.1103/PhysRevA.76.030305 and the following picture:
I hope this note would be helpful for future readers.
| We seek to show that
$$S_{N,m} = \sum_{k=1}^N (-1)^k (\cos \frac{k\pi}{N})^{N-m}
(\sin\frac{k\pi}{N})^m =
\frac{1+(-1)^m}{2} (-1)^{m/2} \frac{N}{2^{N-1}}.$$
Observe that
$$\sum_{k=N+1}^{2N} (-1)^k (\cos \frac{k\pi}{N})^{N-m}
(\sin\frac{k\pi}{N})^m
\\ = \sum_{k=1}^N (-1)^{N+k}
(-1)^{N-m} (\cos \frac{k\pi}{N})^{N-m}
(-1)^m (\sin\frac{k\pi}{N})^m = S_{N,m}$$
so that
$$S_{N,m} = \frac{1}{2}
\sum_{k=1}^{2N} (-1)^k (\cos \frac{k\pi}{N})^{N-m}
(\sin\frac{k\pi}{N})^m.$$
With $\rho_k =\exp(k i\pi/N)$ the roots of $z^{2N}-1=0$ this becomes
$$\frac{1}{2}
\sum_{k=1}^{2N} (-1)^k \frac{1}{2^{N-m}} (\rho_k+1/\rho_k)^{N-m}
\frac{1}{2^m i^m} (\rho_k-1/\rho_k)^m.$$
Now introduce
$$f(z) = \frac{1}{2^{N+1} i^m} z^N
(z+1/z)^{N-m} (z-1/z)^m
\frac{2N/z}{z^{2N}-1}.$$
We then have by inspection that the sum is given by the residues due
to the rational term at $\rho_k = \exp(k i\pi/N)$ with $1\le k\le 2N.$
Here we use that
$$\lim_{z\rightarrow\rho_k} \frac{z-\rho_k}{z^{2N}-1}
= \frac{1}{2N \rho_k^{2N-1}}.$$
Note that should $z+1/z$ or $z-1/z$ be zero the corresponding
trigonometric sum term would have been zero as well. In that case the
simple pole from the rational term is canceled, making for a zero
contribution and everything is in order. Continuing,
$$f(z) = \frac{N}{2^N i^m} \frac{1}{z}
(z^2+1)^{N-m} (z^2-1)^m
\frac{1}{z^{2N}-1}.$$
With residues adding to zero we must compute minus the residue at zero
and minus the residue at infinity. We get for the former (including the
switched sign)
$$\frac{N}{2^N i^m} (-1)^m.$$
and the latter (the residue at infinity is
$-\mathrm{Res}_{z=0}\frac{1}{z^2} f\left(\frac{1}{z}\right)$)
$$\frac{N}{2^N i^m}
\mathrm{Res}_{z=0} \frac{1}{z^2} z (1/z^2+1)^{N-m}
(1/z^2-1)^m \frac{1}{1/z^{2N}-1}
\\ = \frac{N}{2^N i^m}
\mathrm{Res}_{z=0} \frac{1}{z}
\frac{(z^2+1)^{N-m}}{z^{2N-2m}}
\frac{(1-z^2)^m}{z^{2m}}
\frac{z^{2N}}{1-z^{2N}}
\\ = \frac{N}{2^N i^m}
\mathrm{Res}_{z=0} \frac{1}{z}
(z^2+1)^{N-m} (1-z^2)^m \frac{1}{1-z^{2N}}
= \frac{N}{2^N i^m}.$$
Therefore we have
$$S_{N,m} = \frac{N}{2^N i^m} (-1)^m
+ \frac{N}{2^N i^m}
= \frac{1+(-1)^m}{2} \frac{1}{i^m} \frac{N}{2^{N-1}}.$$
This is zero when $m$ is odd as claimed. When $m$ is even the term
$1/i^m$ simplfies and we have at last
$$\frac{1+(-1)^m}{2} (-1)^{m/2} \frac{N}{2^{N-1}}$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Computing the limit of $ \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $ I had this exercise:
Compute the limit
$$ \lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $$
I tried two different approaches and got different answers.
Approach 1:
$$\begin{split}
\lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{n^4\sqrt{4 +1/n^8}}&=\lim_{n\to\infty} \frac{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\cdots+\frac{n^3}{n^4}}{\sqrt{4 +1/n^8}}\\
&=\lim_{n\to\infty} \frac{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\cdots+\frac{1}{n}}{\sqrt{4 +1/n^8}}\\
&= \frac{0}{\sqrt{4+0}}\\
&= 0\\
\end{split}$$
Approach 2:
We substitute sum of cubes of $n$ natural numbers in the numerator and get
$$\begin{split}
\lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^4\sqrt{4 +1/n^8}}&= \lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^4\sqrt{4 +1/n^8}}\\
&= \lim_{n\to\infty} \frac{n^4(1+\frac{1}{n})^2}{4n^4\sqrt{4 +1/n^8}}\\
&= \lim_{n\to\infty} \frac{(1+\frac{1}{n})^2}{4\sqrt{4 +1/n^8}}\\
&= \lim_{n\to\infty} \frac{1}{4\sqrt{4}}\\
&= \frac{1}{8}\\
\end{split}$$
I'm not sure why the two methods are giving me different answers and which one is correct?
| In your first attempt, $n$ is also present as the number of terms. You cannot use "the limit of the sum is the sum of the limit" in that case. The same way you cannot say
$$
\lim_{n\to\infty}\sum_{k=1}^n\frac1n=\sum_k0=0.
$$
In this example the sum is $1$ for all $n$, and so the limit is $1$.
| {
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"answer_count": 2,
"answer_id": 0
} |
Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$ Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$. It seems to be true for all $n \ge 11$, but I only need it to be true for $n \ge 347$. I tried to manipulate the equation in all sorts of ways but I couldn't get it to work. These are the difference of prime bounds if that helps. $\dfrac{x}{\ln x-1} < \pi(x)$ for $x \ge 5393$ and $\pi(x) < \dfrac{x}{\ln x-1.1}$ for $x \ge 60184$ where $\pi(x)$ is counts the number of primes less than or equal to $x$.
| $$\frac{n^2}{\ln n^2-1}-\frac{\frac{n^2}{2}}{\ln\frac{n^2}{2}-1.1} \gt n \; \iff \;
\frac{1}{\ln n^2-1}-\frac{1}{2(\ln\frac{n^2}{2}-1.1)} \gt \frac{1}{n} \tag{1}\label{eq1A}$$
For $n \ge 347$, we have $\ln(n) \gt 2(\ln(2) + 1.1) \; \to \; \ln(n) - \ln(2) - 1.1 \gt \frac{1}{3}\left(2\ln(n) - \ln(2) - 1.1\right)$ so, using this below, we get
$$\begin{equation}\begin{aligned}
\frac{1}{\ln n^2-1}-\frac{1}{2(\ln\frac{n^2}{2}-1.1)} & = \frac{1}{2\ln(n)-1}-\frac{1}{2(2\ln(n) - \ln(2)-1.1)} \\
& = \frac{2(2\ln(n) - \ln(2)-1.1) - (2\ln(n)-1)}{(2\ln(n)-1)2(2\ln(n) - \ln(2)-1.1)} \\
& = \frac{2\ln(n) - 2\ln(2)- 1.2}{(2\ln(n)-1)2(2\ln(n) - \ln(2)-1.1)} \\
& \gt \frac{2\ln(n) - 2\ln(2)- 2(1.1)}{(2\ln(n)-1)2(2\ln(n) - \ln(2)-1.1)} \\
& = \frac{\ln(n) - \ln(2)- 1.1}{(2\ln(n)-1)(2\ln(n) - \ln(2)-1.1)} \\
& \gt \frac{2\ln(n) - \ln(2)- 1.1}{3(2\ln(n)-1)(2\ln(n) - \ln(2)-1.1)} \\
& = \frac{1}{3(2\ln(n)-1)}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note that
$$\frac{1}{3(2\ln(n)-1)} \gt \frac{1}{n} \iff 3(2\ln(n)-1) \lt n \iff n - 6\ln(n) + 3 \gt 0 \tag{3}\label{eq3A}$$
Thus, to prove \eqref{eq1A}, it's sufficient to prove the RHS above. Define
$$f(n) = n - 6\ln(n) + 3 \tag{4}\label{eq4A}$$
We have $f(347) \approx 314.9$, and $f'(n) = 1 - \frac{6}{n} \gt 0$, so $f(n) \gt 0$ for all $n \ge 347$.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Distance between the point of touching in three touching circles $\textbf{Question : }$ Say you have three touching circles $\Gamma_1,\Gamma_2,\Gamma_3$ with radii $x,y,z$ and centers $A,B,C$ as per the diagram, then prove the following $$|DE|=\frac{2}{\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfrac{1}{z}\right)}}$$
$\textbf{My Attempt}$
Let $\angle{DAE} = \alpha, \angle{FBE}=\beta , \angle{FCD}=\pi-\alpha-\beta$ then through sine rule in the $\Delta ABC$ we can say
$$\frac{\sin{\alpha}}{y+z}=\frac{\sin{\beta}}{z+x}=\frac{\sin{(\alpha+\beta)}}{x+y} $$
As we know
$$|DE|=2x\sin\frac{\alpha}{2}$$
We just need to prove
$$\sin\frac{\alpha}{2} = \dfrac{1}{x}\dfrac{1}{\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfrac{1}{z}\right)}}$$
So using the sine rule relation
$$\sin\beta = \frac{z+x}{y+z} \sin \alpha$$
And again from the sine rule relation
$$\sin \alpha = \frac{y+z}{x+y} \sin (\alpha+\beta)$$
Now substituting $\sin \beta$ in terms of $\sin \alpha$
$$\sin \alpha = \sin \alpha \cdot \sqrt{1-\left( \frac{z+x}{y+z}\sin \alpha \right)^2}+\sqrt{1-\sin^2 \alpha}\cdot \frac{z+x}{y+z}\sin \alpha $$
$$1 = \sqrt{1-\left( \frac{z+x}{y+z}\sin \alpha \right)^2}+\sqrt{1-\sin^2 \alpha}\cdot \frac{z+x}{y+z}$$
Now this is the step where I get stuck, it is simply too complicated to solve by hand for me at least. Maybe someone can suggest a way to solve this in reasonable time or even better a different approach...
| From double angle formulae, we have $$2\sin^2\frac{\alpha}2=1-\cos\alpha.$$
Using cosine rule,
$$\cdots=1-\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-(b-c)^2}{2bc}=\frac{(a-b+c)(a+b-c)}{2bc}$$
where $a=y+z,\ b=x+z,\ c=x+y$ in your diagram.
Thus,
$$\begin{align}2\sin^2\frac{\alpha}2&=\frac{4yz}{2(x+z)(x+y)}\\ \sin\dfrac{\alpha}2&=\dfrac1{\sqrt{\dfrac{(x+z)(x+y)}{yz}}}\\&=\frac1x\frac1{\sqrt{\dfrac{(x+z)}{xz}\dfrac{(x+y)}{xy}}}\\&=\frac1x\frac1{\sqrt{\left(\dfrac1x+\dfrac1z\right)\left(\dfrac1x+\dfrac1y\right)}}\end{align}$$
$\square$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proving that $0 \leq k \sin\left(\frac{2\pi}{n}\right) - \sin\left(\frac{2\pi k}{n}\right)$ I need help proving that
$$
0 \leq k \sin\left(\frac{2\pi}{n}\right) - \sin\left(\frac{2\pi k}{n}\right)
$$
For $n > 2$ and positive $k$.
I've tried all sorts of identities, and nothing have worked. Any help would be greatly appreciated.
The things I have tried are the following:
Using a identity for multiples angle formula:
$$
k\sin \left(\frac{2 \pi}{n}\right) >\sin\left(k\cdot\frac{2 \pi }{n}\right)=2^{k -1}\prod_{i=0}^{k-1}\sin \left(\frac{\pi i}{k}+\frac{2 \pi }{n}\right)
$$
$$
k \geq 2^{k -1}\prod_{i=1}^{k-1}\sin \left(\frac{\pi i}{k}+\frac{2 \pi }{n}\right)
$$
Which I could not make sense of.
I also tried rewriting the inequality
$$
k\sin \left(\frac{2 \pi}{n}\right) +\sin \left(\frac{2 \pi (n -k)}{n}\right)>0
$$
Which got me here:
$$\begin{split}
&=(k-1)\sin \left(\frac{2 \pi}{n}\right)+\sin \left(\frac{2 \pi}{n}\right)+\sin \left(\frac{2 \pi (n -k)}{n}\right)\\
&=(k-1)\sin \left(\frac{2 \pi}{n}\right)+2\sin \left(\frac12\left[\frac{2 \pi}{n}+\frac{2 \pi (n -k)}{n}\right]\right)\sin \left(\frac12\left[\frac{2 \pi}{n}-\frac{2 \pi (n -k)}{n}\right]\right)
\end{split}$$
But I could not make that work either.
| Assume $0 < k \le 1$ and $n \ge 4$. Put $x = \dfrac{2\pi k}{n} \implies k = \dfrac{nx}{2\pi}$. Thus you prove: $\dfrac{\sin\left(\frac{2\pi}{n}\right)}{\frac{2\pi}{n}} \le \dfrac{\sin x}{x}$. But the function $ f(x) = \dfrac{\sin x}{x}$ on $\left(0,\frac{2\pi}{n}\right]$ has $f’(x) = \dfrac{x\cdot \cos x - \sin x}{x^2} < 0$ since $\tan x > x$ on this same domain of $x$. Thus $f$ is a decreasing function and you have $f(x) \ge f\left(\frac{2\pi}{n}\right)$ since $0 < x < \frac{2\pi}{n}$. Done.
Note:The inequality reverses when $k < 1$.
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
} |
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
|
Above is an image representing Pascals Triangle. What I want to draw attention to is the hockey stick formation, particularly, the blue hockey stick. Notice how the entries in the stick of the blue hockey stick are in arithmetic progression, and that the entry in the blade represents the sum of the entries in the stick.
To prove this inductively we have as a bootstrap condition
$$1=\frac{1(1+1)}{2}=\binom{1+1}{2} = \sum\limits_{i=1}^1\binom{i}{1}=1$$
and for the general case
$$\begin{array}{lll}
\sum\limits_{i=1}^{n+1}&=&(n+1)+\sum\limits_{i=1}^{n}i\\
&=&\binom{n+1}{1}+\binom{n+1}{2}\\
&=&\binom{n+2}{2}\\
&=&\binom{(n+1)+1}{2}\\
&=&\frac{(n+1)((n+1)+1)}{2}
\end{array}$$
Of course, we assumed that $\binom{n}{k}+\binom{n}{k+1} = \binom{n+1}{k+1}$ holds.
$$\begin{array}{lll}
\binom{n}{k}+\binom{n}{k+1}&=&\frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-(k+1))!}\\
&=&\frac{n!(k+1)}{k!(k+1)(n-k)!} + \frac{n!(n-k)}{(k+1)!(n-(k+1))!(n-k)}\\
&=&\frac{n!k+n!+n!n-n!k}{(k+1)!(n-k)!}\\
&=&\frac{n!+n!n}{(k+1)!((n+1)-(k+1))!}\\
&=&\frac{n!(n+1)}{(k+1)!((n+1)-(k+1))!}\\
&=&\frac{(n+1)!}{(k+1)!((n+1)-(k+1))!}\\
&=&\binom{n+1}{k+1}\\
\end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
"answer_count": 36,
"answer_id": 18
} |
Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$ Consider the sequence defined as
$x_1 = 1$
$x_{n+1} = \sin x_n$
I think I was able to show that the sequence $\sqrt{n} x_{n}$ converges to $\sqrt{3}$ by a tedious elementary method which I wasn't too happy about.
(I think I did this by showing that $\sqrt{\frac{3}{n+1}} < x_{n} < \sqrt{\frac{3}{n}}$, don't remember exactly)
This looks like it should be a standard problem.
Does anyone know a simple (and preferably elementary) proof for the fact that the sequence $\sqrt{n}x_{n}$ converges to $\sqrt{3}$?
| If your sequence is of the form
$$x_{n+1} = f(x_n)$$
with
$$f(x) = x( 1 - c x^{\alpha} + \textrm{h. o. t} )$$
then notice ( e.g. ) that
$$\frac{1}{f(x)^{\alpha}} = \frac{1}{x^{\alpha}} + \alpha c+ \textrm{h. o. t.}$$
and so
$$\frac{1}{x_{n+1}^{\alpha}} - \frac{1}{x_n^{\alpha}} \to \alpha c$$
and with Cesaro-Stolz we get
$$\frac{1}{n x_n^{\alpha} } \to \alpha c$$
In the particular case $f(x) = \sin x= x( 1 - \frac{x^2}{6} + \cdots)$, we get
$$\frac{1}{n x_n^2} \to 2 \cdot \frac{1}{6} = \frac{1}{3}$$
$\bf{Added}$ We can get effective estimates for the sequence
$x_{n+1} = \sin x_n$ as follows
Consider the Taylor expansion
$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} - 1}{x^2} = \frac{1}{5} + \frac{2 x^2}{63} + \cdots$$
where all the remaining coefficients are positive. We conclude that for $0<|x|< \pi$ we get
$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} > \frac{1}{5}$$
and for $0 < |x|< 1$ we have
$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} <
\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} \ _{x=1}=0.2368\ldots < \frac{1}{4}$$
Therefore we have for $0< |x|\le 1$
$$\frac{3}{x^2} + 1 + \frac{x^2}{5}<\frac{3}{\sin^2 x} < \frac{3}{x^2} + 1 + \frac{x^2}{4}$$
Now for the sequence $y_n= \frac{3}{x_n^2}$ we get the inequalities
$$y_n+ 1 + \frac{3}{5 y_n} < y_{n+1} < y_n + 1 + \frac{3}{4 y_n}$$
and we can use this to get estimates for $y_n$ and so $x_n$.
| {
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"source": "stackexchange",
"question_score": "81",
"answer_count": 3,
"answer_id": 0
} |
How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$? Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
| Assuming you mean $e^{ix}=\cos x+i\sin x$, one way is to use the MacLaurin series for sine and cosine, which are known to converge for all real $x$ in a first-year calculus context, and the MacLaurin series for $e^z$, trusting that it converges for pure-imaginary $z$ since this result requires complex analysis.
The MacLaurin series:
\begin{align}
\sin x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots
\\\\
\cos x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots
\\\\
e^z&=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots
\end{align}
Substitute $z=ix$ in the last series:
\begin{align}
e^{ix}&=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots
\\\\
&=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots
\\\\
&=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)
\\\\
&=\cos x+i\sin x
\end{align}
| {
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"source": "stackexchange",
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"answer_count": 17,
"answer_id": 12
} |
Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$ Mathematica tells me that
$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$
Although I have not been able to come up with a proof.
Proofs, hints, or references are all welcome.
| $$S_n=\sum\limits_{k=0}^n \dfrac{(-1)^k \binom{n}{k}}{2k+1}$$
We have:
$$\begin{align}S_n&=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1} \dfrac{(-1)^k \binom{n}{k}}{2k+1}\\
&=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1}\left[\dfrac{n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\
&=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{n(2n+1)}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\
&=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk+2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\
&=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\&+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\
&=\dfrac{(-1)^n}{2n+1}+\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \dfrac{n(-1)^k \binom{n-1}{k}}{n-k}\\
&=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \left[(-1)^k \binom{n}{k}\right]+\dfrac{(-1)^n}{2n+1}\\
&=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^n \left[(-1)^k \binom{n}{k}\right]\end{align}$$
Therefore,
$$S_n=\dfrac{2n}{2n+1}S_{n-1}+0
\Rightarrow S_{n-1}=\dfrac{2n-2}{2n-1}S_{n-2}
...
\Rightarrow S_1=\dfrac{2}{3}S_0$$
and $S_0=1$
Hence,
$$S_n=\dfrac{2n(2n-2)...2}{(2n+1)(2n-1)...3.1}=\dfrac{(2n)!!}{(2n+1)!!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 1
} |
$\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} > \sqrt{a+b+c}$ is true for positive a,b,c How do you prove that for all positive $a,b,c$ this formula holds true:
\begin{equation*}
\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} > \sqrt{a+b+c}?
\end{equation*}
Any help will be invaluable.
| Since
$$a+b+c=\sqrt{(a+b+c)^2}=\sqrt{a+b+c}\ \sqrt{a+b+c}\iff \dfrac{a+b+c}{\sqrt{a+b+c}}=\sqrt{a+b+c}$$
and
$$\dfrac{a}{\sqrt{a+b}}>\dfrac{a}{\sqrt{a+b+c}}$$
$$\dfrac{b}{\sqrt{b+c}}>\dfrac{b}{\sqrt{a+b+c}}$$
$$\dfrac{c}{\sqrt{c+a}}>\dfrac{c}{\sqrt{a+b+c}}$$
we have
$$\dfrac{a}{\sqrt{a+b}}+\dfrac{b}{\sqrt{b+c}}+\dfrac{c}{\sqrt{c+a}}>\dfrac{a+b+c}{%
\sqrt{a+b+c}}=\sqrt{a+b+c}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/6950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find an isomorphism from the octic group $G$ to the group $G'$ Find an isomorphism from the octic group $G$ to the group $G'$:
$G = \{e, s, s^2, s^3, b, g, d, t\}$
$$e = (1)\quad s = (1234)\quad s^2= (13)(24)\quad s^3= (1432)\quad b = (14)(23)\quad g = (24)\quad d = (12)(34)\quad t = (13)$$
$G' = \{I_2, R, R^2, R^3, H, D, V, T\}$
$$\begin{align*}I_2&=\begin{pmatrix}1 & 0\\ 0& 1 \end{pmatrix}& R&=\begin{pmatrix} 0& -1\\1 &0 \end{pmatrix}& R^2&=\begin{pmatrix}-1 & 0\\ 0&-1 \end{pmatrix}& R^3&=\begin{pmatrix} 0& 1\\-1 & 0\end{pmatrix}\\\\
H&=\begin{pmatrix} 1& 0\\0 & -1 \end{pmatrix}& V&=\begin{pmatrix} -1& 0\\0 & 1 \end{pmatrix}& D&=\begin{pmatrix} 0& 1\\1& 0\end{pmatrix}& T&=\begin{pmatrix} 0& -1\\ -1& 0\end{pmatrix}\end{align*}$$
| Jessica: Preserving the order of all elements isn't sufficient; you have to show that it preserves the group operation.
Here's a hint though. $G^\prime$ is the set of symmetries of a square, right? Try labelling the square's corners. Then each symmetry is just a permutation of its corners...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| The following proof is by Khalaf Ruhemi ( he is not a MSE member)
By partial fraction decomposition, we have
$$\frac{y}{(1+y^2)(y^2+x^2)}=\frac{1}{x^2-1}\left(\frac{y}{1+y^2}-\frac{y}{y^2+x^2}\right).$$
Integrate both sides from $y=0$ to $y=\infty$,
\begin{gather*}
\int_0^\infty\frac{y}{(1+y^2)(y^2+x^2)}\mathrm{d}y=\frac{1}{x^2-1}\int_0^\infty\left[\frac{y}{1+y^2}-\frac{y}{y^2+x^2}\right]\mathrm{d}y\\
=\frac{1}{x^2-1}\left[\frac12\ln(1+y^2)-\frac12\ln(y^2+x^2)\right]_0^\infty=\frac{1}{2(x^2-1)}\left[\ln\left(\frac{1+y^2}{y^2+x^2}\right)\right]_0^\infty\\
=\frac{1}{2(x^2-1)}\left[\ln(1)-\ln\left(\frac{1}{x^2}\right)\right]=\frac{1}{2(x^2-1)}\left[2\ln(x)\right]=\frac{\ln(x)}{x^2-1}.
\end{gather*}
Next, integrate both sides from $x=0$ to $x=\infty$
\begin{gather*}
\int_0^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x=\int_0^\infty\int_0^\infty\frac{y}{(1+y^2)(y^2+x^2)}\mathrm{d}y\,\mathrm{d}x\\
\{\text{change the order of integration}\}\\
=\int_0^\infty\frac{1}{1+y^2}\left[\int_0^\infty\frac{y\,\mathrm{d}x}{y^2+x^2}\right]\mathrm{d}y\\
=\int_0^\infty\frac{1}{1+y^2}\left[\arctan\left(\frac{x}{y}\right)\right]_0^\infty dy=\int_0^\infty\frac{1}{1+y^2}\left[\frac{\pi}{2}-0\right] \mathrm{d}y\\
=\frac{\pi}{2}\int_0^\infty\frac{1}{1+y^2} dy=\frac{\pi}{2}\arctan(y)\bigg|_0^\infty=\frac{\pi}{2}\cdot\frac{\pi}{2}=\frac{\pi^2}{4}.
\end{gather*}
Thus,
\begin{gather*}
\frac{\pi^2}{4}=\int_0^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x=\left(\int_0^1+\int_1^\infty\right)\frac{\ln(x)}{x^2-1}\mathrm{d}x\\
=\int_0^1\frac{\ln(x)}{x^2-1}\mathrm{d}x+\underbrace{\int_1^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x}_{x\to1/x}\\
=2\int_0^1\frac{\ln(x)}{x^2-1}\mathrm{d}x=-\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x-\int_0^1\frac{\ln(x)}{1+x}\mathrm{d}x\\
\left\{\text{use $\frac{1}{1+x}=\frac{1}{1-x}-\frac{2x}{1-x^2}$ in the second integral}\right\}\\
=-2\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x+2\underbrace{\int_0^1\frac{x\ln(x)}{1-x^2}\mathrm{d}x}_{x^2\to x}\\
=-2\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x+\frac12\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x\\
=-\frac32\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x\overset{1-x\to y}{=}-\frac32\int_0^1\frac{\ln(1-y)}{y}\mathrm{d}y\\
\{\text{expand $\ln(1-y)$ in series}\}\\
=\frac32\sum_{n=1}^\infty \frac{1}{n}\int_0^1 y^{n-1}\mathrm{d}y=\frac32\sum_{n=1}^\infty\frac{1}{n^2}=\frac{3}{2}\zeta(2).
\end{gather*}
So we have
$$\frac{\pi^2}{4}=\frac{3}{2}\zeta(2)\Longrightarrow \zeta(2)=\frac{\pi^2}{6}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 29
} |
Explain this step How the following conversion is justified ?
$$\cos \biggl( \theta + \frac{2\pi}{3} \biggr) + \cos \biggl( \theta + \frac{4\pi}{3} \biggr) = \cos \biggl( \frac{\pi}{2} - \theta \biggr) - \cos \biggl( \frac{\pi}{3} +\theta \biggr) $$
| First, notice that
$$\cos(x+\pi) = \cos(x)\cos(\pi) - \sin(x)\sin(\pi) = \cos(x)(-1)-\sin(x)(0) = -\cos(x).$$
So, rewriting, using the above, and using that $\cos(-x)=\cos x$ you have:
\begin{align*}
\cos\left(\theta+\frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right) &= \cos\left(\left(\theta-\frac{\pi}{3}\right)+\pi\right) + \cos\left(\left(\theta+\frac{\pi}{3}\right)+\pi\right)\\
&= -\cos\left(\theta-\frac{\pi}{3}\right) - \cos\left(\theta+\frac{\pi}{3}\right)\\
&= -\cos\left(\frac{\pi}{3}-\theta\right) - \cos\left(\frac{\pi}{3}+\theta\right).
\end{align*}
So: the equality as written cannot hold: you would need $\cos(\frac{\pi}{2}+\theta) = - \cos(\frac{\pi}{3}-\theta)$ for all $\theta$, and plugging in $\theta=0$ shows this cannot hold. I suspect your $\frac{\pi}{2}$ should have been a $\frac{\pi}{3}$, and that you are missing a minus sign before the first term. If this is not the case, then please edit the question to give the equality you are actually trying to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$
Possible Duplicate:
Value of $\sum x^n$
Proof to the formula
$$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$
| For a more mechanical proof, you could use induction. The proof then boils down to finding a common denominator:
$\frac{x^{n+1}-1}{x-1} + x^{n+1} = \frac{x^{n+1}-1+(x-1)x^{n+1}}{x-1} = \frac{x^{n+2}-1}{x-1}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Algebra Problem The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer.
Please give an explanation of how the answer
| $ax^3+bx^2+25=(x^2-4x+5)(ax+k)$ , $k$ = constant
$ax(x^2-4x+5) + k(x^2-4x+5) =0$
$ax^3 + x^2(k-4a)+x(5a-4k)+5k=0$
compare to original then $5k=25 \implies k=5$
and $(5a-4k)=0$ so $a =4$
and $b= k-4a = 5 -16 =-11$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Bounding a series from above using the integral test If $a_{n}$ is a non-negative, decreasing sequence, we know from the integral test that if $f(n)=a_{n}$ is an integrable function, then $\sum_{n=1}^{\infty} a_{n}$ and $\int_{1}^{\infty} f(x)dx$ converge/diverge together.
When the theorem is proven, it is shown that:
$$\forall k \in \mathbb{N}, a_{k+1} \leq \int_{k}^{k+1} f(x)dx \leq a_{k}$$
Which gives us an upper bound of $a_{n}$. Can I use this fact and conclude the following:
$$\sum_{n=1}^{\infty} a_{n} \overset{?}{\leq} \int_{1}^{\infty} f(x)dx + a_{1}$$
What I'm trying to do is check if $\sum\limits_{n=1}^{\infty} \frac{a}{a^2+n^2} \lt \frac{\pi}{4}+\frac{1}{2}$ when $0\lt a\lt 1$. Using the above gives:
$$\int_{1}^{\infty} \frac{a}{a^2+x^2} dx = \frac{1}{a} \int_{1}^{\infty} \frac{1}{1+(\frac{x}{a})^2} dx=\int_{1/a}^{\infty} \frac{1}{1+t^2} dt=$$
$$=\lim_{t \to \infty} \arctan t - \arctan 1/a \lt \frac{\pi}{2} - 1$$
And then
$$\sum_{n=1}^{\infty} \frac{a}{a^2+n^2} \lt \frac{\pi}{2} - 1 + \frac{a}{a^2+1} \lt \frac{\pi}{2}$$
(For some reason the end result is not what I wanted since $\frac{\pi}{2} \gt \frac{\pi}{4} + 1/2$ but I may have made an error along the way).
| Given $\forall k \in \mathbb{N}, a_{k+1} \leq \int_{k}^{k+1} f(x)dx \leq a_{k}$,
$$\sum_{k=1}^{\infty}a_{k+1} \leq \sum_{k=1}^{\infty}\int_{k}^{k+1} f(x)dx \leq \sum_{k=1}^{\infty}a_{k}$$
so
$$\sum_{k=2}^{\infty}a_{k} \leq \int_{1}^{\infty} f(x)dx \leq \sum_{k=1}^{\infty}a_{k}$$
and, using the left part,
$$\sum_{k=1}^{\infty}a_{k}=a_1+\sum_{k=2}^{\infty}a_{k} \leq a_1+\int_{1}^{\infty} f(x)dx.$$
Now, to $\int_{1}^{\infty} \frac{a}{a^2+x^2} dx=\frac{\pi}{2}-\arctan\frac{1}{a}$. As in Jack Schmidt's comment, $\arctan\frac{1}{a}>\frac{\pi}{4}$ when $0<a<1$, so $\frac{\pi}{2}-\arctan\frac{1}{a}<\frac{\pi}{4}$. So,
$$\sum_{k=1}^{\infty}a_{k}\leq a_1+\int_{1}^{\infty} f(x)dx<\frac{a}{a^2+1}+\frac{\pi}{4}<\frac{1}{2}+\frac{\pi}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/16478",
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"source": "stackexchange",
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Solve $\cos(\theta) + \sin(\theta) = x$ for known $x$, unknown $\theta$? After looking at the list of trigonometric identities, I can't seem to find a way to solve this. Is it solvable?
$$\cos(\theta) + \sin(\theta) = x.$$
What if I added another equation to the problem:
$$-\sin(\theta) + \cos(\theta) = y,$$
where $\theta$ is the same and $y$ is also known?
Thanks.
EDIT:
OK, so using the linear combinations I was able to whip out:
$$a \sin(\theta) + b \cos(\theta) = x = \sqrt{a^2 + b^2} \sin(\theta + \phi),$$
where $\phi = \arcsin \left( \frac{b}{\sqrt{a^2 + b^2}} \right) = \frac{\pi}{4}$ (as long as $a\geq 0$)
Giving me:
$$x = \sin(\theta + \frac{\pi}{4}) \text{ and } \arcsin(x) - \frac{\pi}{4} = \theta.$$
All set! Thanks!
| Linear equations in $\sin \theta $ and $\cos \theta $ can be solved by a
resolvent quadratic equation, using the two identities (also here):
$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2%
}}$$
and
$$\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}}.$$
In this case, we have
$$\begin{eqnarray*}
\cos \theta +\sin \theta &=&x \\
&\Leftrightarrow &\left( 1-\tan ^{2}\frac{\theta }{2}\right) +2\tan \frac{%
\theta }{2}=x\left( 1+\tan ^{2}\frac{\theta }{2}\right) \\
&\Leftrightarrow &(x+1)\tan ^{2}\frac{\theta }{2}-2\tan \frac{\theta }{2}%
+x-1=0 \\
&\Leftrightarrow &\tan \frac{\theta }{2}=\frac{2\pm \sqrt{4-4(x+1)(x-1)}}{%
2(x+1)} \\
&\Leftrightarrow &\tan \frac{\theta }{2}=\frac{1\pm \sqrt{2-x^{2}}}{x+1} \\
&\Leftrightarrow &\theta =2\arctan \frac{1\pm \sqrt{2-x^{2}}}{x+1}.
\end{eqnarray*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
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Limits and Measure Theory How would I evaluate these limits?
$$ \lim_{n \to \infty} \int_0^\infty \frac{n}{1+(nx)^2} \ dx$$
and
$$ \lim_{n \to \infty} \int_0^\infty \frac{(1+(nx)^2)}{(1+nx^2)^n} \ dx$$
| From $(1+a)^n\geq1+na\;\;\;\forall a\geq0$ with $a=nx^2$, you have:
$\frac{1+n^2x^2}{(1+nx^2)^n}\leq1$.
Furthermore, you have:$\frac{1+n^2x^2}{(1+nx^2)^n}<\frac{1+n^2x^2}{(nx^2)^n}=\frac{1}{n^nx^{2n}}+\frac{1}{n^{n-2}x^{2(n-1)}}<\frac{2}{x^2}$ if $n\geq2$ and $x>0$.
Hence, by the Lebesgue dominated convergence theorem, you need only consider the integral of the limit function to evaluate the limit
(let $g(x)=\begin{cases}
1 & \text{ if } 0\leq x\leq1 \\
\frac{2}{x^2}& \text{ if } x>1
\end{cases}$ ).
Now $0\leq\frac{1+n^2x^2}{(1+nx^2)^n}\leq\frac{n^2+n^2x^2}{(1+x^2)^n}=\frac{n^2}{(1+x^2)^{n-1}}$.
Hence $lim_{n\rightarrow\infty}\frac{1+n^2x^2}{(1+nx^2)^n}=0$ by the squeeze theorem (we can neglect $x=0$ since it constitutes a set of measure zero).
Hence the desired integral is equal to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/20615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Evaluate $\int \cos^3 x\;\sin^2 xdx$ Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.
$$\int (\cos^3x)(\sin^2x)dx = \int(\cos x)(\cos^2x)(\sin^2x)dx
= \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$
let $u = \sin x$, $du = \cos xdx$
$$\int(1-u^2)u^2du = \int(u^2-u^4)du
= \frac{u^3}{3} - \frac{u^5}{5} +C$$
Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C
| Just for the heck of it, another substitution could have been $u=\sin^3 x,$ in which case, $du = 3\sin^2x\cos x \ dx.$ Now, $1-u^{2/3} = 1-\sin^2x = \cos^2x.$ Thus,
$$
\begin{align*}
\int \cos^3x\sin^2x \ dx &= \frac{1}{3}\int \cos^2x(3\sin^2x\cos x) dx \\
&= \frac{1}{3} \int(1-u^{2/3})du \\
&= \frac{u}{3} - \frac{u^{5/3}}{5} + C \\
&= \frac{\sin^3x}{3} - \frac{\sin^5x}{5} + C \ .
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/21589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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How to solve this recurrence relation What are some ways to solve this recurrence relation:
$a(n+1)=2 a(n) - a(n-1) -1, \text{ with }a(0)=0, a(10)=0?$
I tried to first convert this inhomogeneous equation into a homogeneous one following Wikipedia:
$b_{n}=Ab_{n-1}+Bb_{n-2}+K \,$ can be
converted into homogeneous form as
follows: The steady state is found by
setting $b_n = b_{n−1} = b_{n−2} = b^{*}$ to
obtain $b^{*} = \frac{K}{1-A-B}. \, $.
Then the non-homogeneous recurrence
can be rewritten in homogeneous form
as $[b_n -b^{*}]=A[b_{n-1}-b^{*}]+B[b_{n-2}-b^{*}], \, $
But in the recurrence relation given at the beginning of this post, the denominator in $b^{*} = \frac{K}{1-A-B} \, $ is $1-2+1=0$. So how can I solve the recurrence relation? Thanks in advance!
PS: Are there some books or websites with some summary of various ways to solve recurrence relation?
| Generating functions to the rescue. Define $A(z) = \sum_{n \ge 0} a(n) z^n$,
multiply the shifted by 2 recurrence by $z^n$ and sum over $n \ge 0$. Recognizing:
\begin{align}
\sum_{n \ge 0} a(n + 1) z^n &= \frac{A(z) - a(0)}{z} \\
\sum_{n \ge 0} a(n + 2) z^n &= \frac{A(z) - a(0) - a(1) z}{z^2} \\
\sum_{n \ge 0} z^n &= \frac{1}{1 -z}
\end{align}
gives, using the unknown $a(1)$:
$$
\frac{A(z) - a(1) z}{z^2} = 2 \frac{A(z)}{z} - A(z) - \frac{1}{1 - z}
$$
From this maxima gets the partial fraction expansion:
$$
A(z) = - \frac{1 + a(1)}{1 - z} + \frac{2 + a(1)}{(1 - z)^2} - \frac{1}{(1 - z)^3}
$$
From here, using the generalized binomial theorem:
$$
a(n) = - (1 + a(1)) + (n + 1)(2 + a(1)) - \frac{(n + 1) (n + 2)}{2}
= - \frac{n^2 - (2 a(1) + 1) n}{2}
$$
As $a(10) = 0$, this gives $a(1) = 9 / 2$, and thus:
$$
a(n) = \frac{n (10 - n)}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Find the length of the curve from x=-2 to x=1 of $y=\sqrt{a^2-x^2}$ Homework problem I recently got:
Find the length of the curve of $y=\sqrt{a^2-x^2}$
from $x=-2$ to $x=1$. Also find the area of the surface of revolution about the $x$-axis.
Having trouble understanding this. I'm not even sure what it would look like. I assume a is supposed to be a constant. Is it just a semi-circle with radius $a$. $X$-intercepts of $(-a,0)$ and $(a,0)$ and a max at $(0,a)$?
I'm a little iffy about how to first find the length.
I do know that I should start with $\int_{-2}^{1} 2\pi y ds$ since I am revolving about the x-axis.
Thanks for anyone who can help me out.
| The graph, as you surmised, of $y=\sqrt{a^2-x^2}$ is the upper half of the semicircle with center at $(0,0)$ and radius $a$. (Square both sides to get $y^2 = a^2-x^2$, or $x^2+y^2=a^2$, the equation of the circle described; since $y\geq 0$, it's the upper half). (Of course, you need $|a|\geq 2$ for your limits to make sense).
The formula for the length of the graph of $y=f(x)$ from $x=r$ to $x=s$ is
$$\mathrm{Arc\ Length} = \int_r^s \sqrt{1 + \left(\frac{dy}{dx}\right)^2}
\,dx.$$
So here, you have $y=\sqrt{a^2-x^2}$, so take the derivative, square, plug it into the formula, that will give the arc length.
For the area of the surface, I don't know which formula you know. The one I know is that the area is given by
$$\mathrm{Area} = \int_r^s 2\pi f(x)\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.$$
Added. So, since you've essentially worked it out: if $|a|\gt 2$ (so we don't have to worry about the derivative and the curve being defined over the entire interval $[-2,1]$), we have
$$\frac{dy}{dx} = -\frac{x}{\sqrt{a^2-x^2}}.$$
So
$$ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^2}{a^2-x^2} = \frac{a^2}{a^2-x^2}.$$
So for arc length, we have:
\begin{align*}
\mathrm{Arc\ Length} &= \int_{-2}^1 \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx \\
&= \int_{-2}^1 \sqrt{\frac{a^2}{a^2-x^2}}\,dx\\
&= \int_{-2}^1 \frac{a}{\sqrt{a^2-x^2}}\,dx\\
&= a\int_{-2}^1\frac{dx}{\sqrt{a^2-x^2}}\\
&= a\arcsin\left(\frac{x}{a}\right)\Biggm|_{-2}^1\\
&= a\arcsin\left(\frac{1}{a}\right) - a\arcsin\left(\frac{-2}{a}\right)\\
&= a\arcsin\left(\frac{1}{a}\right) + a\arcsin\left(\frac{2}{a}\right).
\end{align*}
For the surface area, you have:
\begin{align*}
\mathrm{Surface\ Area} &= \int_{-2}^1 2\pi f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\\
&= 2\pi \int_{-2}^1 \sqrt{a^2-x^2}\left(\frac{a}{\sqrt{a^2-x^2}}\right)\,dx\\
&= 2\pi a \int_{-2}^1 \,dx = 2\pi a(1-(-2)) = 6\pi a.
\end{align*}
If $a=2$, then the two integrals are improper and you need to use limits, but you'll end up with the same answer in both cases. If $0\lt a\lt 2$, the integrals are not well-defined (but then neither is the problem).
| {
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"question_score": "1",
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Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$
I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.
I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.
| by using Fourier series of $f(x)=1, x\in[0,1]$
$$1=\sum_{n=1,3,5,..}^\infty\frac{4}{n\pi}\sin (n\pi x)$$
take triple integrals as follows
$$\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}1.dxdxdx=\int_{0}^{1}\int_{0}^{x}\int_{0}^{x} \sum_{n=1,3,5,..}^\infty\frac{4}{n\pi}\sin (n\pi x) dxdxdx$$
$$\frac{1}{6}=\sum_{n=1,3,5,..}^\infty\frac{2}{(n\pi)^2}-\frac{8}{(n\pi)^4}$$
$$\frac{\pi^4}{6}=\sum_{n=1,3,5,..}^\infty\frac{2\pi^2}{n^2}-\frac{8}{n^4}$$
We know that
$$\sum_{n=1,3,5,..}^\infty\frac{1}{n^2}=\frac{\pi^2}{8}$$
so
$$\frac{\pi^4}{96}=\sum_{n=1,3,5,..}^\infty\frac{1}{n^4}$$
then we use the inequality series
$$\sum_{n=1}^\infty\frac{1}{n^4}=\sum_{n=1}^\infty\frac{1}{(2n-1)^4}+\sum_{n=1}^\infty\frac{1}{(2n)^4}$$
simplify it to get
$$\sum_{n=1}^\infty\frac{1}{n^4}=\frac{16}{15}\sum_{n=1}^\infty\frac{1}{(2n-1)^4}$$
so,
$$\sum_{n=1}^\infty\frac{1}{n^4}=\frac{16}{15}\frac{\pi^4}{96}=\frac{\pi^4}{90}$$
| {
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"source": "stackexchange",
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} |
Proof by induction: $\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$ How can the following be proved by induction?
$$\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$$
I am out of ideas after practicing for a while:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)
$$
Does this involve telescoping series?
| $n=1$, check. note that
$$
\frac{1}{2(n+1)-1}-\frac{1}{2(n+1)}=\frac{(2n+2)-(2n+1)}{(2n+1)(2n+2)}=\frac{1}{(2n+1)(2n+2)}
$$
(this is what changes on the rhs for $n+1$).
on the left hand side the change is
$$
\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\frac{(2n+2)+(2n+1)-2(2n+1)}{(2n+1)(2n+2)}=\frac{1}{(2n+1)(2n+2)}
$$
as desired.
| {
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Where are these additional solutions coming from? Solve for $x$: $2\sin(2x)-\sqrt{2} = 0$ in interval $[0,2\pi)$
Step $1$: Add $\sqrt{2}$ and divide by $2$ to get $\sin(2x) = \dfrac{\sqrt{2}}{2}$
Step $2$: Set $2x$ equal to the angles where $\sin(x) = \dfrac{\sqrt{2}}{2}$:
$2x = \dfrac{\pi}{4}$ and $2x = \dfrac{3\pi}{4}$
Step $3$: Solve for $x$ by dividing by $2$: $x = \dfrac{\pi}{8}$ and $x = \dfrac{3\pi}{8}$
My textbook also lists $\dfrac{9\pi}{8}$ and $\dfrac{11\pi}{8}$ as additional solutions, anyone know where they may have came from? thanks
| $\sin(\theta) = \sqrt(2)/2$ iff $\theta = \pi/4 + 2k \pi$ or $\theta =3\pi/4 + 2k \pi$.
| {
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"source": "stackexchange",
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} |
How to find roots of $X^5 - 1$? How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)
| It is easy to write down the solution in trigonometric functions, namely, $x=e^{2\pi i n/5}$ for $n=0,\ldots, 4$.
Here is an algebraic solution. First factor the polynomial as $$x^5-1=(x-1)(x^4+x^3+x^2+x+1).$$
The first factor gives you the obvious solution $x=1$. To find other roots, we need to solve the equation
$$ x^4+x^3+x^2+x+1=0.$$
Divides both sides by $x^2$, and use the fact that $(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$, we get
$$ (x+\frac{1}{x})^2 +(x+\frac{1}{x})-1=0$$
Set $y=x+\frac{1}{x}$, and we get a quadratic equation
$$ y^2+y-1=0,$$
which one may easily solve using quadratic formulas. Once we have solved for $y=y_0, y_1$, the equation reduces into two other quadratic equations
$$ x+ \frac{1}{x}= y_i, \qquad i=0, 1 $$
or equivalently,
$$ x^2-y_i x+1=0.$$
We just apply the quadratic formula again.
| {
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"source": "stackexchange",
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For what $a$ and $b$ is $9x^4-12x^3+28x^2+ax+b$ a perfect square? If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$.
This is one of my past year examination's questions, some help on it?
(The answer for this problem is $a=-16$, $b=16$.)
| $9x^4-12x^3+28x^2+ax+b=(3x^2+cx+d)^2=9x^4+6cx^3+(c^2+6d)x^2+2cdx+d^2$
So $c=-2, d=4$ and $a=2cd=-16, b=d^2=16$
| {
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Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 &= x^2 - 3x - 4 \\
0 &= (x-4)(x+1) \\
\end{align}
$$
Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.
But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:
$$
\begin{align}
\sqrt{4 + 5} &= 4 - 1 \\
\sqrt{9} &= 3 \\
3 &= 3
\end{align}
$$
So using 4 works as expected, but when using $-1$:
$$
\begin{align}
\sqrt{-1 + 5} &= -1 - 1 \\
\sqrt{4} &= -2 \\
2 &\ne -2
\end{align}
$$
At what stage am I going wrong?
And according the WolframAlpha, the solution is $x = 4$.
| $\sqrt{x+5} = x - 1$
Because, you need to put condition $x\geq 1$
So, If you tried to plug $x = -1 < 1$, which not satified.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "20",
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Confused by textbook solution to trig problem The following is part of a question and the solution from my textbook:
Question
Given that $2 \sin{2\theta} = \cos{2\theta}$, show that $\tan{2\theta} = 0.5$
Solution
$2 \sin{2\theta} = \cos{2\theta}$
$\Rightarrow \frac{2 \sin{2\theta}}{\cos{2\theta}} = 1$
$\Rightarrow 2 \tan{2\theta} = 1~~~\tan{2\theta} = \frac{\sin{2\theta}}{\cos{2\theta}}$
I am unsure about this part: Since it is possible that $\cos(2\theta) = 0$, isn't it bad form to divide by $\cos(2\theta)$?
| If $\cos(2\theta) = 0$ then $\sin(2\theta)$ should be $1$ or $-1$.
So $2\sin(2\theta)$ could not be equal to $\cos(2\theta)$.
| {
"language": "en",
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Exercising divergent summations: $\lim 1-2+4-6+9-12+16-20+\ldots-\ldots$ I'm trying to make sense of some (assumed to be) simple exercises in divergent summation. One example I cannot resolve.
First I assume the sequence of binomialcoefficients $ \{ b_k = \binom k2 \}_{k=2\to\infty}=\{1,3,6,10,15,...\}$
Then to assign a meaningful value to the alternating sum $ S= 1-3+6+10.... $ I compute the Abel-sum
$ \qquad S = \lim_{x \to 1} 1-3x+6x^2-10x^3+...-... = {1 \over (1+x)^3 } = 1/8 $ (Abel)
But I want to proceed one more step. The sequence of partial sums may be denoted as
$ \{ c_k \}_{k=0\to\infty}=\{1,-2,4,-6,9,-12,16,-20,...\} $
Q - my question is: what is the sum $ T = 1-2+4-6+9... $ ?
I tried two approaches, but I'm lost.
The generating function is simply $ 1/(1+x)^3/(1-x) $ but here I cannot let x approach 1, so the simple application of the Abel-sum is impossible. Also the Euler-sum seems to not to converge; instead I get increasing partial sums like k/16 for the k'th partial sum .
On the other hand, I observe, that the coefficients are near the squares, so if I consider the alternating $\zeta$ (usually called Dirichlet-$\eta$)
$ \qquad \qquad \begin{eqnarray} X &=& 1 - 2.25 + 4 - 6.25 + 9 - 12.25 + ... \\ &=&( 4 - 9 + 16 - 25 + ... - ...)/4 \\ &=& (1-\eta(-2))/4 \\ &=& (1-0)/4 = {1 \over 4} \end{eqnarray}$
then in T I had just each second coefficient $1/4$ above that of the $\eta$-series in X and possibly could go along something like
$ \qquad \qquad \begin{eqnarray} T &=& 1 - (2.25-0.25) + 4 - (6.25-0.25) + ... - ... \\
&=& X + 0.25*\zeta(0) \\
&=& X-1/8 = {1 \over 8} \end{eqnarray}$
But surely this is only an outline how I could come nearer to a solution. How could I actually proceed here?
[update]: One more idea was to make use of reordering summation. The coefficients $c_k$ can be seen as rowsums of the following matrix:
$\qquad \small
\begin{array} {rrrrr}
1 & . & . & . & . & . &\ldots\\
-3 & 1 & . & . & . & . \\
6 & -3 & 1 & . & . & . \\
-10 & 6 & -3 & 1 & . & . \\
15 & -10 & 6 & -3 & 1 & . \\
-21 & 15 & -10 & 6 & -3 & 1 & \ldots \\
... & ... \\
\end{array}
$
so that all columns evaluate to $1/8$ due to the Abel-summation. If I add all that columnsums, I should have to write $\zeta(0)*1/8 $ and evaluate $T=-1/16$ (Q&D) now. But this is all fumbling, because I not even reflect the infinite application of downshifting by rows when evaluating the columns...
[update2]: Hmm. I played with the reciprocals of the series. I just did the "paper&pen" divisions and got:
$ \small \qquad \qquad 1 \qquad : 1-3x+6x^2-10x^3+15x^4-21x^5+...-...=1+3x+3x^2+1x^3=(1+x)^3=|_{x=1}8 $
and
$ \small \qquad \qquad 1 \qquad : 1-2x+4x^2-6x^3+9x^4-12x^5+...-...=1+2x-2x^3-1x^4=(1+x)^3(1-x)=|_{x=1}0 $
So this is also immediately what Lubos pointed out. Good for my intuition, I hope this model is not too much misleading in other obvious cases....
| The elementary Ramanujan sum of $1-2+4-6+9-\cdots$ is 1/4 and the series belongs to the class $R=2$.
In general, for $f(x)=b(x)/(1-x)$ and $b(x)$ Abel summable, the sum of $f(1)$
is $-b'(1) + b(1)/2$, with $b'(x)=xdb(x)/dx$.
Edit 1. Verification.
$$
\frac{1}{(1+x)^3(1-x)}=\frac{x}{8(1-x)}+\frac{8+7x+4x^2+x^3}{8(1+x)^3}~.
$$
The first term on the right-hand side corresponds to
$$
\frac{1}{8}(1+1+1+\cdots)=-\frac{1}{16}
$$
and the second term is Abel summable to $5/16$ (the value at $x=1$).
Linearity implies that the sum is $-1/16 + 5/16 = 1/4$.
Edit 2. The shifted series $0+1-2+4-\cdots=1/8$.
The shifted series corresponds to $x$ times the previous series, thus
$$
\frac{x}{(1+x)^3(1-x)}=\frac{x}{8(1-x)}-\frac{x}{8}+\frac{8x+7x^2+4x^3+x^4}{8(1+x)^3}~.
$$
As before, linearity implies that the sum is $-1/16-1/8+5/16=1/8$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to calculate $\int \sqrt{(\cos{x})^2-a^2} \, dx$ How to calculate:
$$\int \sqrt{(\cos{x})^2-a^2} \, dx$$
| In SWP (Scientific WorkPlace), with Local MAPLE kernel, I got the following evaluation
$$\begin{eqnarray*}
I &:&=\int \sqrt{\cos ^{2}x-a^{2}}dx \\
&=&-\frac{\sqrt{\sin ^{2}x}}{\sin x}a^{2}\text{EllipticF}\left( \left( \cos
x\right) \frac{\text{csgn}\left( a^{\ast }\right) }{a},\text{csgn}\left(
a\right) a\right) \\
&&-\text{EllipticF}\left( \left( \cos x\right) \frac{\text{csgn}\left(
a^{\ast }\right) }{a},\text{csgn}\left( a\right) a\right) \\
&&+\text{EllipticE}\left( \left( \cos x\right) \frac{\text{csgn}\left(
a^{\ast }\right) }{a},\text{csgn}\left( a\right) a\right) F
\end{eqnarray*}$$
where
$$F=\sqrt{\frac{-\cos ^{2}x+a^{2}}{a^{2}}}\sqrt{\cos ^{2}x-a^{2}}\text{csgn}\left( a^{\ast }\right) \frac{a}{-\cos ^{2}x+a^{2}}$$
As an example:
$$\begin{eqnarray*}
\int \sqrt{\cos ^{2}x-2^{2}}dx &=&\frac{\sqrt{\sin ^{2}x}}{\sin x}3\text{EllipticF}\left( \frac{1}{2}\cos x,2\right) \\
&&+\text{EllipticE}\left( \frac{1}{2}\cos x,2\right) \frac{\sqrt{-\cos
^{2}x+4}}{\sqrt{\cos ^{2}x-4}}
\end{eqnarray*}$$
| {
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"url": "https://math.stackexchange.com/questions/48024",
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"source": "stackexchange",
"question_score": "3",
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Find the rotation matrix for +15° out of the rotation matrix +60° without using trigonometric functions Find the rotation matrix for +15° out of the rotation matrix +60° without using trigonometric functions. And as much as I would love to tell you what I did so far. I don't even know where to start with this exercise..
The matrix for +60° is this.
$$\left[
\begin{matrix}
1/2 & -\sqrt 3/2\\
\sqrt 3/2 & 1/2 \\
\end{matrix} \right]
$$
How do I even start?
| The only approach I can see is to take the fourth root of the matrix, that is, find a $2 \times 2$ matrix $A$, such that $A^4=\left(\begin{matrix}
1/2 & -\sqrt 3/2\\
\sqrt 3/2 & 1/2 \\
\end{matrix}\right)$. If you expect that it is of the form $\left(\begin{matrix}
a & -b\\
b & a \\
\end{matrix}\right)$ with $a^2+b^2=1$, it should go through.
Added: If you first take a square root, you get $a^2-b^2=1/2, -2ab=\sqrt{3}/2$, which gives $b^2=\frac{3}{16a^2}, a^2-\frac{3}{16a^2}=\frac{1}{2}, a=\frac{\sqrt{3}}{2}, b=\frac{1}{2}$ (taking the positive root). This give the $30^{\circ}$ matrix as $A^2=\left(\begin{matrix}
\sqrt 3/2 & -1/2\\
1/2 & \sqrt 3/2 \\
\end{matrix}\right)$.
Now the same approach gives $a^2-b^2=\sqrt 3/2, -2ab=1/2$ and you can get a quadratic in $a^2$ again.
| {
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$
Multiplying both sides by $(a + b + c)$:
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?
| An elementary approach, without $\text{AM} \ge \text{GM}$ is to use the identity
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left(\frac{(x-y)^2 + (y-z)^2 + (z-x)^2}{2}\right)$$
Thus $$\text{if } \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge 0\ \text{then } (a+b+c)^3 \ge 27abc$$
by setting $x = \sqrt[3]{a}$ etc
| {
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Simplifying trigonometric expressions, is there a unified theory? $\frac{1}{3}\cos^3 x \cos(2x)+\frac{1}{12}\sin(2x)(\sin(3x)+3\sin x)=\frac{1}{3} \cos x$
I got this as the result of a differential equation that I solved. The answer in the book is (1/3) cos(x), but after applying variation of parameters I got the expression on the left. To my delight Wolfram Alpha tells me that they are equal! (yay!!)
But, without cheating and using the computer, how would I ever know that? Nothing about my expression screams "simplify me" unless I'm missing something.
Perhaps I'd notice the graph looked like cosine if I happened to graph it.
I know many trig identities, but I have never heard of a formal procedure that always works to simplify. Is there such a thing? How would you approach this messy expression?
How can I get better at this important skill?
I have more problems to solve and it feels cheap to keep plugging my answers in to W.alpha to see if they are right.
Ps. Is there a widget to convert thing formatted for mathematical to latex and vice versa?
| Try to express everything on the $\text{l.h.s}$ in terms of $\cos$ and then see what happens. Use the following identities:
*
*$\sin{2x} = 2\cdot \sin{x} \cdot \cos{x}$
*$\sin{3x} = 3\sin{x} - 4 \sin^{3}{x}$
*$1-\sin^{2}{x}= \cos^{2}{x}$
So you have $$\frac{1}{3} \cos^{3}{x} \cdot \bigl[ 2 \cos^{2}{x}-1 \bigr]=\frac{2}{3}\cos^{5}{x} -\frac{1}{3}\cos^{3}{x} \qquad (\text{I})$$ and $$\frac{1}{12} \cdot 2\cdot \sin{x} \cos{x} \cdot \Bigl[ 6\sin{x} -4\sin^{3}{x}\Bigr]=\sin^{2}{x}\cdot \cos{x} -\frac{2}{3}\sin^{4}{x}\cdot\cos{x} \qquad (\text{II})$$
Now write $\sin^{2}{x}$ as $1-\cos^{2}{x}$ and $\sin^{4}{x}$ as $(1-\cos^{2}{x})^{2}$ and add equations $\text{I}$ and $\text{II}$ to get the final answer.
Added. Equation $\text{II}$ becomes $$(1-\cos^{2}{x}) \cdot \cos{x} - \frac{2}{3}(1-\cos^{2}{x})^{2}\cdot\cos{x}$$ $$ = \cos{x}-\cos^{3}{x} -\frac{2}{3} \cdot \Bigl[ \cos{x} - 2\cdot\cos^{3}{x} +\cos^{5}{x}\Bigr]$$ $$= \frac{1}{3}\cdot \cos{x} + \frac{1}{3}\cos^{3}{x} -\frac{2}{3}\cdot \cos^{5}{x} $$
The basic idea for simplifying such expressions is to observe the RHS and the LHS and then use double angle, half-angle formulas to simplify the equations. I don't think there are any such theoreies behind them.
| {
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Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an easy proof?
| You can express the inequality as
$$ \frac{a^3+b^3+c^3}{3} > \frac{a+b+c}{3} \frac{a^2+b^2+c^2}{3}$$
or
$$ [m_3(a,b,c)]^3 > [m_1(a,b,c)] \; [m_2(a,b,c)]^2 $$
where $m_p$ is the generalized power (Hölder) $p$-mean. Applying the (important) property that $m_p > m_q$ whenever $p>q$ (and the values are not all equal), you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/49211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all pairs of positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$? How to find ALL pairs of positive integers $(x,y)$ such that the difference in their squares is a perfect cube and the difference in their cubes is a perfect square.
i.e.,
Positive integers $(x,y)$ such that
$x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$?
Finding infinite number of pairs is no problem, as in:
$( 2^{6j+1} \cdot 3^{6k} \cdot 5 , 2^{6j+1} \cdot 3^{6k+1} )$ for any integral $j,k \geq 0$
But how would you determine the exhaustive list?
| $ x^2 - y^2 = a^3 $ is fairly trivial, because clearly for every integer solution there are integers $ p, q, r, s $ with $p$ and $q$ squarefree and coprime such that $ x + y = p q^2 r^3 $ and $ x - y = p^2 q s^3 $, so that:
$ 2 x = p q (q r^3 + p s^3) $
$ 2 y = p q (q r^3 - p s^3) $
$ a = p q r s $
and these satisfy the equation identically, with $x, y$ integers iff either $p q$ is even or $p, q, r, s$ all odd.
(or $r, s$ both even, although for "reduced" solutions with no integer $t > 1$ such that t^3 divides $x, y$ and t^2 divides $a$ we can assume that $gcd(r, s) = 1$, which we do hereafter ..)
Plugging these relations into $(2 x)^3 - (2 y)^3 = 8 b^2 $ gives $ p^4 q^3 (3 q^2 r^6 s^3 + p^2 s^9) = 4 b^2 $.
Noting that $p$ and $q$ are coprime and squarefree we then conclude $2 b = p^2 q^2 s B$ for some integer $B$, so that $ s (3 q^2 r^6 + p^2 s^6) = q B^2 $.
From this we see that $q$ divides $p^2 s^7$ and hence, since $gcd(p, q) = 1$ and $q$ is squarefree, that $q$ divides $s$.
Thus with $s = q S$ and $B = q C$ we obtain $ S (3 r^6 + p^2 q^4 S^6) = C^2 $.
Then, taking $S = u v^2$ with $u$ squarefree, we conclude that $C = u v D$ for some integer $D$ so that finally $3 r^6 + (p q^2 u^3 v^6)^2 = u D^2 $
In this $u$ divides $3 r^6$; but since $gcd(r, s) = 1$ and $u$ divides $s$ we conclude that $u$ divides 3.
Thus, since $u$ > 0, we must have $u = 1$ or $u = 3$, which reduces the problem to one of the following respectively:
$3 r^6 + M^2 = N^2$ ($u = 1$)
$ r^6 + 3 M^2 = N^2$ ($u = 3$)
Now $X^2 + 3 Y^2 = Z^2$ has general integer solution $X, Y, Z = k(m^2 - 3 n^2), 2 k m n, k(m^2 + 3 n^2)$ with $gcd(m, n) = 1$ and $m + n$ odd.
So the cases require either $2 k m n = r^3$ or $k(m^2 - 3 n^2) = r^3$, each of which is trivial by suitable choice of $k$ (although arguably if you want explicit values of m, n the second is not so trivial).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/53292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Get number of elements of a square matrix given a vector that has upper right elements of that matrix Suppose I have some vectors:
*
*$[1,2,1]$ its length is $3$ wich represents a matrix like $\begin{pmatrix}
1 & 2\\
-1 & 1
\end{pmatrix}$ the complete matrix would have $4$ elements
*$[1,2,1,3,4,1]$ its length is $6$ wich represents $\begin{pmatrix}
1& 2& 3\\
-1& 1 & 4\\
-1& -1 &1
\end{pmatrix}$ the complete matrix would have $9$ elements
*$[1,2,1,3,4,1,5,6,7,1]$ its length is $10$ wich represents $\begin{pmatrix}
1 & 2& 3& 5\\
-1 &1 & 4& 6\\
-1 &-1 &1 & 7\\
-1 &-1 &-1 &1
\end{pmatrix}$ the complete matrix would have $16$ elements
*$[1,2,1,3,4,1,5,6,7,1,8,9,10,11,1]$ its length is $15$ wich represents
$\begin{pmatrix}
1 &2 &3 &5 &8 \\
-1 &1 &4 &6 &9 \\
-1 &-1 &1 &7 &10 \\
-1 &-1 &-1 &1 &11 \\
-1 &-1 &-1 &-1 &-1
\end{pmatrix}$ the complete matrix would have $25$ elements
How do I find a general formula to get all elements of a square matrix from the number of elements in a vector?
So
*
*$3\to 4$
*$6\to 9$
*$10\to 16$
*$15\to 25$
*etc...
also: How would you know the matrix size $2\times 2$, $3\times 3$,...$n\times n$?
| The function
$n\mapsto \frac12 ( 1 + 4 n - \sqrt{1+8n})$
does the job.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/55859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A "fast" way to find the sum of the sequence $5,5.5,5.55,5.555,5.5555,\ldots $ (20 terms) My initial approach is diving the whole sum by $9$ and taking the common $5$ out which gives $$\frac{5}{9}[(10-1)+(10-0.1)+(10-0.01)+\cdots + (10-10^{-19})]$$ after some algebra this could be reduced to $$\frac{5}{9} \times [200-\frac{10}{9} \times (1-10^{-20})]$$
after this I am not sure how to show that is almost equal to $110.5$? Also if any body wants to suggest any other tricky/fast way I will appreciate it.
| A slower way. It solves the general case with $n$ terms. I denote your sum as
$$S_{20}=5+5.5+5.55+5.555+\ldots +5.\underset{19}{\underbrace{555\ldots 5}}$$
and the general case for $n$ as
$$S_{n}=5+5.5+5.55+5.555+\ldots +5.\underset{n-1}{\underbrace{555\ldots 5}}.$$
Since $\sum_{j=1}^{k-1}10^{j-1}=\frac{1}{90}10^{k}-\frac{1}{9}$, we have in general
$$\begin{eqnarray*}
S_{n} &=&5n+5\sum_{k=1}^{n}\frac{\sum_{j=1}^{k-1}10^{j-1}}{10^{k-1}} =5n+5\sum_{k=1}^{n}\frac{\frac{1}{90}10^{k}-\frac{1}{9}}{10^{k-1}}
=\frac{1000}{9}-\frac{50}{81}+\frac{50}{81}10^{-n},
\end{eqnarray*}$$
and in the present case
$$\begin{eqnarray*}
S_{20} &=&\frac{1000}{9} -\frac{50}{81}+\frac{50}{81}10^{-20} \\
&=&\frac{220987654320987654321}{2000\,000\,000000\,000\,000}\approx 110.49.
\end{eqnarray*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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