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determining $n$ in a given sequence $\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $ Given that: $$\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $$
Determine $n$.
The memorandum says the answer is 2011 but how is that so? Where did I go wrong?
| $1 + 3 + 5 + .. + 2n - 1 = n^2$, and $2 + 4 + ... + 2n = n(n + 1) = n^2 + n$. So:
$\dfrac{n^2}{n^2 + n} = \dfrac{2011}{2012}$. So: $2012n^2 = 2011n^2 + 2011n$, and $n^2 = 2011n$. So $n = 2011$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help with integral from Apostol I've been working on this all day and I'm still stumped. To state the problem (ref: Apostol Section 5.11, Question 29):
Show
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{a(2n+3)}(x^n(ax+b)^{3/2} - nb\int x^{n-1}\sqrt{ax+b}\,\,dx) + c$$
My attempts so far have been as follows (NB: Im avoiding using integrating with partial fractions because Im not up to that stuff in Apostol):
*
*Let $w = \sqrt{ax+b}$ which lead to
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{a^{n+1}}\int w^2 (w^2 - b)^n\, dw
$$
and then use integration by parts with $u = w$ and $dv = (\frac{w^2-b}{a})^{n} \frac{2w}{a}$ to get
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{1}{n+1}x^{n+1}\sqrt{ax+b} - \frac{1}{n+1}\int x^{n+1}\frac{a}{2\sqrt{ax+b}} dx
$$
*Use integration by parts directly with $u = x^n$ and $dv =\sqrt{ax+b}\,dx$ to get
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{3a}(x^n(ax+b)^{3/2} - n\int\,x^{n-1}(ax+b)^{3/2}\,dx
) + c
$$
*Same as (1) but with the integration by parts, let $dv = w^2$ and $u = (\frac{w^2-b}{a})^{n}$ to get
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{3a}(x^n(ax+b)^{3/2} - n\int\,x^{n-1}(ax+b)^{3/2}\,dx
) + c
$$
*Using integration by parts using $u = x^n\sqrt{ax+b}$ which lead to
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{x^{n+1}}{n+1}\sqrt{ax+b} - \frac{a}{2(n+1)}\int x^{n+1}(ax+b)^{-1/2}\, dx
$$
which doesn't help because evaluating the integral on the RHS would cancel each side out.
As always, not looking for an answer but some broad hints (so I learn more!)
| I would start with integration by parts with the choice $$u = x^n, \quad dv = (ax+b)^{1/2} \, dx,$$ which gives $$du = nx^{n-1}, \quad v = \frac{2}{3a}(ax+b)^{3/2}.$$ Thus if $I$ is the given integral, then $$\begin{align*} I &= \int u \, dv = uv - \int v \, du \\ &= \frac{2}{3a}x^n (ax+b)^{3/2} - \frac{2n}{3a} \int x^{n-1} (ax+b)^{3/2} \, dx \\ &= \frac{2}{3a} x^n (ax+b)^{3/2} - \frac{2n}{3a} \int ax^n (ax+b)^{1/2} + bx^{n-1} (ax+b)^{1/2} \, dx \\ &= \frac{2}{3a} x^n(ax+b)^{3/2} - \frac{2n}{3} I - \frac{2nb}{3a} \int x^{n-1} (ax+b)^{1/2} \, dx . \end{align*}$$ Therefore, $$\Bigl(\frac{2n}{3}+1\Bigr)I = \frac{2}{3a}x^n(ax+b)^{3/2} - \frac{2nb}{3a} \int x^{n-1} (ax+b)^{1/2} \, dx.$$ Dividing both sides by the constant gives the desired identity.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maclaurin series: $\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\frac{x^{12}}{12!}+...$ The Taylor series of a real or complex-valued function ƒ(x) that is infinitely differentiable at a real or complex number a is the power series
$$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots. $$
I have a curiosity concerning the following polynomial:
$$\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\frac{x^{12}}{12!}+...$$
Is this a known development in Maclaurin series?
| Note that
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.\tag{1}$$
A little less familiar is
$$\sinh x=\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots.\tag{2}$$
Subtract (1) from (2) and divide by $2$. We get
$$\frac{1}{2}(\sinh x-\sin x)=\frac{x^3}{3!}+\frac{x^7}{7!}+\frac{x^{11}}{11!}+\cdots.$$
That takes care of the odd guys. Now play a similar game with $\cos x$ and $\cosh x$. Or integrate the closed form for the sum of the odds from $0$ to $x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $b-d$ when $\log_ab={3\over2}$ and $\log_cd={5\over4}$ $a,b,c$ are three natural numbers such that $\log_ab={3\over2}$ and $\log_cd={5\over4}$.
Given: $a-c=9$
Find $b-d$
| Hint: $$\log_ab={3\over2} \Rightarrow b=a^{\frac32}\Rightarrow a \text{ is a perfect square, } a= \alpha^2 \to b=\alpha^3$$
$$\log_cd={5\over4} \Rightarrow d=c^{\frac54} =(a-9)^{\frac54} \Rightarrow a-9 \text{ is a perfect 4-th power} \to \alpha^2-9=\beta^4 \to \alpha^2-\beta^4=(\alpha+\beta^2)(\alpha-\beta^2)=9$$
Since $\alpha+\beta^2 \gt 0$ the other factor of $9$ ($\alpha-\beta^2$) must be $\gt0\Rightarrow$
\begin{cases}
\alpha+\beta^2=1, \alpha - \beta^2 =9\\
\alpha+\beta^2=3, \alpha - \beta^2 =3\\
\alpha+\beta^2=9, \alpha - \beta^2 =1
\end{cases}
\begin{cases}
\text{Impossible since } \alpha+\beta^2 \ge \alpha-\beta^2\\
\alpha=3, \beta=0\\
\alpha=5, \beta =2
\end{cases}
\begin{cases}
a=3^2=9, b=3^3=27, c=9-9=0, d= 0^{\frac54}=0 \text{, impossible since } d \gt 0\\
a=5^2=25, b=5^3=125, c= 25-9=16,d=16^{\frac54}=2^5=32
\end{cases}
The only solution is $$a=25\qquad b=125\qquad c=16\qquad d=32$$
$$\Rightarrow b-d=125-32=93$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Using the method of Lagrange multipliers, find the extreme values of a function Using the method of Lagrange multipliers, find the extreme values of the function
$f(x,y)= \frac{2y^3}{3} + 2x^2 +1$ on the ellipse $5x^2 + y^2 = 1/9$
. Identify the (absolute) maximal and minimal values of f
taken on the ellipse.
currently I have that $\nabla f(x,y)=L*\nabla g(x,y)$ where $g(x,y)=5x^2 + y^2-1/9$
this leads to: $4x-10Lx=0$, so $x=0$ or $L=2/5$
and:$2y^2-2Ly=0$, so $y=0$ or $y=L$.
where do I go from here? andy help would be greatly appreciated, thank you in advance.
| $\dfrac{\partial f}{\partial x} = 4x= \gamma\cdot 10x$, and $\dfrac{\partial f}{\partial y} = 2y^2 = \gamma\cdot 2y$. So: $2x(2 - 5\gamma) = 0$, and $2y(y - \gamma) = 0$. So
case 1: $x = 0$, so $y^2 = \dfrac{1}{9}$. So $y = \dfrac{1}{3}$ or $-\dfrac{1}{3}$. So:
$f(0,\frac{1}{3}) = \dfrac{83}{81}$, and $f(0,-\frac{1}{3}) = \dfrac{79}{81}$
case 2: $x \neq 0$, and $\gamma = \dfrac{2}{5}$, and $y = 0$ or $y = \dfrac{2}{5}$. If $y = 0$, then $x = \dfrac{1}{3\sqrt{5}}$ or $x = -\dfrac{1}{3\sqrt{5}}$, and in both cases $f(\frac{1}{3\sqrt{5}}, 0) = \dfrac{47}{45}$. If $y = \dfrac{2}{5}$, then $5x^2 = -\dfrac{12}{225} < 0$, and there is no solution in this case.
So: $f_{min} = \dfrac{79}{81} = 0.975$, and $f_{max} = \dfrac{47}{45} = 1.044$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770411",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Which conic is represented by $r = a \cos \theta$ The polar equation $r = a \cos \theta$ represents which conic?
| Multiply both sides by $r$ to get $r^2 = a r \cos \theta$. Then we use the substitution $r^2 = x^2 + y^2$ and $r \cos \theta = x$ to get $x^2 + y^2 = ax$. Then we get $\displaystyle \left(x - \frac{a}{2}\right)^2 + y^2 = \frac{a^2}{4}$. Looks like it's a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771829",
"timestamp": "2023-03-29T00:00:00",
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How prove this $100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$
let $a,b,c\ge 0$, and such $$a+b+c=6$$
show that
$$100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$$
My idea: since
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=36-2(ab+bc+ac)$$
$$a^2b^2+b^2c^2+a^2c^2=(ab+ca+bc)^2-2abc(a+b+c)=(ab+bc+ac)^2-12abc$$
so let $p=a+b+c=6, q=ab+bc+ac,r=abc$,so
$$\Longleftrightarrow 100+5[36-2q]-2[q^2-12r]-r^2\ge 0$$
$$\Longleftrightarrow r^2+2q^2-24r+10q\le 280$$
then I can't.Thank you
| Let $G(a,b,c)=100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2$. We will
show that $G(a,b,c) \geq G(2,2,2)=0$.
From the identity
$$
4\frac{G(a,b,c)-G(a,\frac{b+c}{2},\frac{b+c}{2})}{(c-b)^2}
+\frac{G(a,b,c)-G(a,b+c,0)}{bc}=\frac{(a^2+2)(b+c)^2}{4}
$$
we see that at least one of the two numbers
$G(a,b,c)-G(a,\frac{b+c}{2},\frac{b+c}{2})$ or
$G(a,b,c)-G(a,b+c,0)$ is nonnegative, so that it suffices to treat the cases when
one variable is zero or when two variables are equal. This is what we do in the next
two lemmas.
Lemma 1. $G(a,b+c,0) \geq G(3,3,0)=28$.
Proof of lemma 1 Putting $d=b+c=6-a$, and $x=a-3$ we have
$$
\begin{array}{lcl}
G(a,d,0) &=& 100+5(a^2+d^2)-2a^2d^2 \\
&=& 100+5((3+x)^2+(3-x)^2)-2((9-x^2)^2) \\
&=& 28+46x^2-2x^4 \geq 28=G(3,3,0).
\end{array}
$$
Lemma 2. $G(a,\frac{b+c}{2},\frac{b+c}{2}) \geq G(2,2,2)=0$.
Proof of lemma 2 Putting $e=\frac{b+c}{2}=\frac{6-a}{2}$, we have
$$
\begin{array}{lcl}
G(a,e,e) &=& 100+5(a^2+2e^2)-2(2a^2e^2+e^4)-a^2e^4 \\
&=& 100+5\bigg(a^2+2\big(\frac{6-a}{2}\big)^2\bigg)
-2\bigg(2a^2\bigg(\frac{6-a}{2}\bigg)^2+\bigg(\frac{6-a}{2}\bigg)^4\bigg)
-a^2\bigg(\frac{6-a}{2}\bigg)^4 \\
&=& (a-2)^2\Bigg(7+a\bigg(\frac{7}{4}+\frac{9(6-a)}{8}+\frac{(6-a)^2}{8}+\frac{(6-a)^3}{16}\bigg)
\bigg) \Bigg) \geq 0.
\end{array}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $\cos x+8\sin x-7=0$
Solve $\cos x+8\sin x-7=0$
My attempt:
\begin{align}
&8\sin x=7-\cos x\\
&\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\
&\implies 8\sin \frac{x}{2}\cos \frac{x}{2}=3+\sin^2 \frac{x}{2}\\
&\implies 0=\sin^2 \frac{x}{2}-8\sin \frac{x}{2}\cos \frac{x}{2}+3\\
&\implies 0=\sin \frac{x}{2}\left(\sin \frac{x}{2}-8\cos \frac{x}{2}\right)+3
\end{align}
I'm not sure how to proceed from here (if this process is even right at all?) . Any help would be appreciated. Thanks.
| $1.) \; \cos x+8\sin x-7=0; \tag{1}$
$2.) \; \cos^2 x = (7- 8\sin x)^2; \tag{2}$
$3.) \; \cos^2 x = 1 - \sin^2 x; \tag{3}$
$4.) \; 1 - \sin^2 x = (7- 8\sin x)^2; \tag{4}$
$5.) \; 1 - \sin^2 x = 49 - 112 \sin x + 64 \sin^2 x; \tag{5}$
$6.) \; 65 \sin^2 x -112\sin x + 48 = 0; \tag{6}$
$7.) \; \sin x = \dfrac{112 \pm \sqrt{(112)^2 - 4(48)(65)}}{130}; \tag{7}$
$8.) \; \sin x = \dfrac{112 \pm \sqrt{12,544 -12,480}}{130}; \tag{8}$
$9.) \sin x = \dfrac{112 \pm 8}{130} = \dfrac{104}{130}, \dfrac{120}{130} = \dfrac{4}{5}, \dfrac{12}{13}; \tag{9}$
$10.) \; \text{Note there exist unique} \; \alpha, \beta \in [0, \dfrac{\pi}{2}] \; \text{with} \; \sin \alpha = \dfrac{4}{5}, \sin \beta = \dfrac{12}{13}; \tag{10}$
$11.) \; \text{Note} \; \cos \alpha = \dfrac{3}{5}, \cos \beta = \dfrac{5}{13}; \tag{11}$
$12.) \; \text{Note} \; \alpha \; \text{alone satisfies (1)}; \tag{12}$
$13.) \; \text{Note there exist unique} \; \gamma, \delta \in [\dfrac{\pi}{2}, \pi] \; \text{with} \; \sin \gamma = \dfrac{4}{5}, \sin \delta = \dfrac{12}{13}; \tag{13}$
$14.) \; \text{Note that} \; \cos \gamma = -\dfrac{3}{5}, \cos \delta = -\dfrac{5}{13}; \tag{14}$
$15.) \; \text{Note that} \; \delta \; \text{alone satisfies (1)}; \tag{15}$
$16.) \; \text{Note by periodicity} \; \alpha \pm 2n\pi, \delta \pm 2n\pi \; \text{satisfy (1) for} \; n \in \Bbb Z; \tag{16}$
$17.) \; \text{Conclude that all solutions are given by item (16) above}; \tag{17}$
$18.) \; \text{Kick back for a minute; you deserve it!} \tag{18}$
Hope this helps! Cheerio,
and as always,
Fiat Lux!!!
| {
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"timestamp": "2023-03-29T00:00:00",
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Algebra question: Finding inverse function This question is about finding the inverse function of $f(x)=-\sqrt{9-x^2}$
I seem to be making an error with one of the manipulations. Here is my attempt.
$$x=-\sqrt{9-y^2}$$
$$x^2=(-\sqrt{9-y^2})^2$$
$$x^2= -(9-y^2)$$
$$x^2=y^2-9$$
$$x^2+9=y^2$$
$$\sqrt{x^2+9}=\sqrt{y^2}$$
$$y=\sqrt{x^2+9}$$
The answer is $f^{-1} (x)=-\sqrt{9-x^2},-3≤x ≤0$
In which step did I make a mistake?
Thank you for your time.
| \begin{align}
y=-\sqrt{9-x^2}
\end{align}
Change $x$ to $y$ and vice versa.
\begin{align}
x&=-\sqrt{9-y^2}\\
-x&=\sqrt{9-y^2}\\
(-x)^2&=\left(\sqrt{9-y^2}\right)^2\\
x^2&=9-y^2\\
y^2&=9-x^2\\
y&=\pm\sqrt{9-x^2}
\end{align}
The last step, change $y$ to $f^{-1}(x)$.
\begin{align}
f^{-1}(x)&=\pm\sqrt{9-x^2}
\end{align}
Since there is a symmetry between a function and its inverse. Specifically, if $f(x)$ is an invertible function with domain $x$ and range $y$, then its inverse $f^{-1}(x)$ has domain $y$ and range $x$. Therefore, the inverse function of $f(x)=-\sqrt{9-x^2}$ is
\begin{align}
f^{-1}(x)&=-\sqrt{9-x^2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Elegant proof of $\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$? Let $a, b > 0$ satisfy $a^2-4b^2 \geq 0$. Then:
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$$
One way to calculate this is by computing the residues at the poles in the upper half-plane and integrating around the standard semicircle. However, the sum of the two residues becomes a complicated expression involving nested square roots, which magically simplifies to the concise expression above.
Sometimes such 'magical' cancellations indicate that there is a faster, more elegant method to reach the same result.
Is there a faster or more insightful way to compute the above integral?
| Let $b=\frac{a}{2}\sin2\theta$ ($0<\theta<\frac{\pi}{2}$), then
\begin{eqnarray}
\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + \frac14a^2\sin^22\theta}&=&\int_{-\infty}^{\infty}\frac{dx}{(x^2+a\sin^2\theta)(x^2+a\cos^2\theta)}\\
&=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\int_{-\infty}^{\infty}\left(\frac{1}{x^2+a\sin^2\theta}-\frac{1}{x^2+a\cos^2\theta}\right)dx\\
&=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\left(\frac{1}{\sqrt{a}\sin\theta}\arctan\frac{1}{\sqrt{a}\sin\theta}x\right.\\
&&-\left.\frac{1}{\sqrt{a}\cos\theta}\arctan\frac{1}{\sqrt{a}\cos\theta}x\right)\bigg|_{-\infty}^\infty\\
&=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\left(\frac{1}{\sqrt{a}\sin\theta}-\frac{1}{\sqrt{a}\cos\theta}\right)\pi\\
&=&\frac{\pi}{a\sqrt{a}(\cos\theta+\sin\theta)\sin\theta\cos\theta}\\
&=&\frac{\pi}{b\sqrt{2b+a}}.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{a}{(a-b)(a-c)(x-a)}+\frac{b}{(b-c)(b-a)(x-b)}+\frac{c}{(c-a)(c-b)(x-c)}$ We need to simplify $$\dfrac{a}{(a-b)(a-c)(x-a)}+\dfrac{b}{(b-c)(b-a)(x-b)}+\dfrac{c}{(c-a)(c-b)(x-c)}$$
The biggest problem is that the above expression has four variables.I transformed the expression into $$\dfrac{a}{-(a-b)(c-a)(x-a)}+\dfrac{b}{-(b-c)(a-b)(x-b)}+\dfrac{c}{-(c-a)(b-c)(x-c)}$$
and then tried to add them up,but as you can probably guess,it yielded something almost impossible to handle.I tried to factorize it using standard techniques,but that was fruitless too.A hint will be appreciated.
N.B: The chapter deals with Remainder-factor theorem,partial fractions,factoring cyclic homogenous polynomials and manipulating algebraic expressions.
| This answer is to show that this can be done by hand without any particular clever trick, but by being clearheaded and organised. With the original expression being symmetric, you should expect the appearance of symmetric expressions in the expansion, and often a fair amount of cancellation.
If we put everything over a common denominator we get $$\frac {-a(b-c)(x-b)(x-c)-b(c-a)(x-a)(x-c)-c(a-b)(x-a)(x-b)} {(a-b)(b-c)(c-a)(x-a)(x-b)(x-c)}$$
Now we notice that the coefficient of $x^2$ in the numerator is $-a(b-c)-b(c-a)-c(a-b)=0$ and the constant coefficient is $-abc(b-c)-abc(c-a)-abc(a-b)=0$.
The coefficient of $x$ is $$a(b-c)(b+c)+b(c-a)(c+a)+c(a-b)(a+b)$$$$=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2$$
One way to factorise such an expression is to note that it is zero if $a=b$, or $b=c$ or $c=a$ so must have a factor $d(a-b)(b-c)(c-a)$ where d is a constant obtained by equating the coefficient of $ab^2$, for example.
Another way (if you don't have a better idea) is to treat it as a quadratic in $c$ so that $$(b-a)c^2+(a^2-b^2)c+ab(b-a)=(b-a)\left(c^2-(a+b)c+ab\right)=(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)$$
A further note - the original expression was symmetric. The common denominator had a factor $(a-b)(b-c)(c-a)$ which is skew symmetric, and this can only happen if this same factor, which is the basic skew symmetric expression in three variables, also appears as a factor in the numerator (otherwise the whole expression will fail to be symmetric). The factor has to cancel - and this helps us to know what we are looking for, and to imagine short cuts.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\cdots$ Evaluate
$$
\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\frac{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}{4}\cdots .
$$
First, it is clear that terms tend to $1$.
It seems that the infinity product is not 0. This is related to the post Sequence $x_{n+1}=\sqrt{x_n+a(a+1)}$.
| Define the sequence $x_n$ by $x_0=\dfrac{1}{2}$ and $x_{n+1}=\dfrac{1}{2}+\dfrac{\sqrt{x_n}}{2}$,
and let $y_n=x_0x_1\cdots x_{n}$. The question is to evaluate
$\lim\limits_{n\to\infty}y_n$.
It is easy to see by induction that $0<x_n<1$ for every $n$, so we can define
$$\theta_n=\arccos(\sqrt{x_n})$$
So that
$$\cos^2(\theta_{n+1})=x_{n+1}=\frac{1+\cos\theta_n}{2}=\cos^2\left(\frac{\theta_n}{2}\right).$$
Thus
$\theta_{n+1}=\dfrac{\theta_n}{2}$. This shows that $\theta_n=2^{-n}\theta_0=\dfrac{\pi}{2^{n+2}}$.
Now, noting that $\cos(\theta_{k+1})=\dfrac{\sin(2\theta_{k+1})}{2\sin\theta_{k+1}}
=\dfrac{\sin\theta_{k}}{2\sin\theta_{k+1}}$, we conclude that
$$
x_{k}=\frac{1}{4}\frac{\sin^2\theta_{k-1}}{\sin^2\theta_{k}}
$$
Thus
$$
y_n=\prod_{k=0}^{n}x_k=\frac{1}{2}\prod_{k=1}^{n}\left(\frac{1}{4}\frac{\sin^2\theta_{k-1}}{\sin^2\theta_{k}}\right)=\frac{1}{2^{2n+1}} \frac{\sin^2\theta_{0}}{\sin^2\theta_{n}}
$$
Finally,
$$
y_n=\frac{1}{2^{2n+2}\sin^2(2^{-n-2}\pi)}
$$
So, $$\lim_{n\to\infty}y_n=\frac{4}{\pi^2 },$$
which is the desired limit.$\qquad \square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the sum $1! + 2! + 3! + \cdots + n! \le ?? < (2n)!$ How to evaluate the sum below as close as possible?
$$1! + 2! + 3! + \cdots + n! \le ??? $$
Is the next evaluation $ 1! + 2! + 3! + \cdots + n! \le n n! < (2n)! $ correct?
| $1!+2!+3!+\ldots+n!\le 2n!$ follows by induction (using $1+\ldots+n!+(n+1)!\le 2n!+(n+1)n!=(n+2)n!\le2(n+1)n!=2(n+1)!$). Of course, for $n\ge 2$, we have $2n!\le nn!<(n+1)!<(2n)!$.
A somewhat better (because $\frac{n+1}{n-1}\to1$) approximation is given by
$$ 1!+2!+3!+\ldots+n!\le\frac{n+1}{n-1}n!\qquad\text{for }n\ge2$$
which is again shown by induction: For $n=2$ the claim just says $1!+2!=3\le \frac{3}{1}2!=6$. The induction step
$$ 1!+2!+3!+\ldots+n!+(n+1)!\le\frac {n+1}{n-1}n!+(n+1)!=\left(\frac{1}{n-1}+1\right)(n+1)!\le\frac{n+2}{n}(n+1)!$$
works because for $n\ge 2$
$$\frac1{n-1}+1\le\frac{n+2}{n} \iff n^2\le (n+2)(n-1)\iff 0\le n-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all solutions for $\cos(2x)\cos x-\sin(2x)\sin x=\frac{1}{\sqrt{2}}$ if $0\leq x<\pi$ Find all solutions for $\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$ if $0\leq x< \pi$
Can you verify my work? Thanks!
$$\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$$
$$\cos(2x+x)=\frac{1}{\sqrt{2}}$$
$$\cos(3x)=\frac{1}{\sqrt{2}}$$
$$\cos^{-1}(\frac{1}{\sqrt{2}})$$
Reference angle: $\frac{\pi}{4}$
$$3x=\frac{\pi}{4}+2k\pi$$
$$x=\frac{\pi}{12}+\frac{2k\pi}{3}$$
$$3x=\frac{7\pi}{4}+2k\pi$$
$$x=\frac{7\pi}{12}+\frac{2k\pi}{3}$$
$$x=\frac{\pi}{12},\frac{3\pi}{4},\frac{17\pi}{12}, \frac{7\pi}{12},\frac{5\pi}{4},\frac{23\pi}{12} $$
| The solution $x=\frac{\pi}{12} + \frac{2k\pi}{3}$ gives, in $[0, \pi[$, the angles $\frac{\pi}{12}$ and $\frac{9\pi}{12}=\frac{3\pi}{4}$.
And the solution $x=-\frac{\pi}{12} + \frac{2k\pi}{3}$ gives, in $[0, \pi[$, the angle $\frac{7\pi}{12}$.
And that's all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/784603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Limit of a sequence involving roots How can I find the limit of the following sequence:
$b_n = \sqrt[4]{n^4+n^3} - \sqrt[4]{n^4+1}$?
I succeeded in finding the limits of other sequences involving roots using the binomic formulas but I don't know what to apply to fourth roots.
Thanks for your help!
| This calls for multiplication with the conjugate: $\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b}$.
$$ \begin{align}b_n&=\sqrt[4]{n^4+n^3}-\sqrt[4]{n^4+1}\\
&=\frac{\sqrt{n^4+n^3}-\sqrt{n^4+1}}{\sqrt[4]{n^4+n^3}+\sqrt[4]{n^4+1}}\\
&=\frac{(n^4+n^3)-(n^4+1)}{(\sqrt[4]{n^4+n^3}+\sqrt[4]{n^4+1})(\sqrt{n^4+n^3}+\sqrt{n^4+1})}\\
&=\frac{n^3}{(\sqrt[4]{n^4+n^3}+\sqrt[4]{n^4+1})(\sqrt{n^4+n^3}+\sqrt{n^4+1})}\\
&=\frac{1}{(\sqrt[4]{1+n^{-1}}+\sqrt[4]{1+n^{-4}})(\sqrt{1+n^{-1}}+\sqrt{1+n^{-4}})}\\&\to \frac1{(1+1)(1+1)}=\frac14\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of the integral Find the value of the following integral conatining a term with natural logarithm$$\int_0^1 (1-y) \ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\, dy.$$
| $$I=\int_{0}^{1}(1-y)\log{\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)}\,dy$$
Substitute $x=1-y$.
$$I=\int_{0}^{1}(1-y)\log{\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)}\,dy=\int_{0}^{1}x\,\log{\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}\right)}\,dx.$$
This can be tackled by integration by parts. First we'll need the derivative of the logarithmic factor:
$$\frac{d}{dx}\log{\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}\right)}=\left(\frac{2-\sqrt{x}}{2+\sqrt{x}}\right)\cdot\left(\frac{(2-\sqrt{x})\frac{1}{2\sqrt{x}}+(2+\sqrt{x})\frac{1}{2\sqrt{x}}}{(2-\sqrt{x})^2}\right)=\frac{2}{(4-x)\sqrt{x}}.$$
Then,
$$I=\frac12x^2\log{\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}\right)}\big |_{0}^{1}-\int_{0}^{1}\frac{x^{3/2}}{4-x}\,dx\\
=\frac12\log{3}-\int_{0}^{1}\frac{x^{3/2}}{4-x}\,dx.$$
To find the integral $\int_{0}^{1}\frac{x^{3/2}}{4-x}\,dx$, substitute $x=4u^2$:
$$\int_{0}^{1}\frac{x^{3/2}}{4-x}\,dx=16\int_{0}^{\frac12}\frac{u^4}{1-u^2}\,du.$$
This last integral can readily be computed by expanding the integrand by partial fractions* and integrating term by term.
*Partial fraction expansion of $\frac{u^4}{1-u^2}$:
$$\frac{u^4}{1-u^2}=-u^2+\frac{u^2}{1-u^2}=-u^2-1+\frac{1}{1-u^2}=-u^2-1+\frac{1}{2(1+u)}+\frac{1}{2(1-u)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Solving $315 x \equiv 5 \pmod {11}$ I have to solve this: $$315 x \equiv 5 \pmod {11}$$
Isn't it like that?
$$315 \equiv (22+8) \cdot 10+15 \equiv 8 \cdot 3+4 \equiv 5+8+4 \equiv 6$$
Or have I done something wrong?
| If we use Fermat's little theorem and the fact that $315 \equiv 7\text{ (mod 11)}$, then $7^{11 - 1}\equiv 1 \text{ (mod } 11) \implies 7^{11 - 1} \cdot 5 \equiv 5 \text{ (mod } 11) \implies 7 \cdot 7^{11-2} \cdot 5\equiv 5\text{ (mod 11)} \implies 7 \cdot 8 \cdot 5 \equiv 5 \text{ (mod 11)} \implies x \equiv 40 \text{ (mod 11)} \implies x \equiv 7 \text{ (mod 11)}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Maximum and Minimum value of an inverse function
Find the maximum and minimum value of
$\arcsin \left(x\right)^3+\arccos \left(x\right)^3$.
given that $-1\le x\le 1$
I have solved the problem but i am just curious to know if there are any other ways to solve this particular problem other than the method i used below.
By using the fact that $\arcsin \left(x\right)+\arccos \left(x\right)$ =$\frac{\pi }{2}$
i found that
$\arcsin \left(x\right)^3+\arccos \left(x\right)^3$=$3\left(\frac{\pi }{2}\right)^2\left(\left\{\arcsin \left(x\right)-\frac{\pi }{4}\right\}^2+\frac{\left(8\pi -3\pi ^2\right)}{48}\right)$
so it is minimum when $\left\{\arcsin \left(x\right)-\frac{\pi }{4}\right\}^2$$=0$
or $x=\sin \left(\frac{\pi }{4}\right)$
Therefore minimum value$=\frac{1}{32}\pi ^3$
and it is maximum when $\arcsin \left(x\right)=-\frac{\pi }{2}$
Therefore maximum value=$\frac{7}{8}\pi ^3$
| Help:
a^3+b^3=(a+b)((a+b)^2-3ab)
Let p=Pi
ASinx+ACosx=p/2
f(x)=(ASinx)^3+(ACosx)^3
=p/2 (p^2/4-3.ASinx.ACosx)
=p/2 (p^2/4-3.(p/2).ASinx+3.(ASinx)^2)
This shall be maximum for ASinx= -(p/2)
Hence max value of the function will be:
f (x)max=(7/8) (p^2)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int \sqrt{x^2-3x+2}\ dx$ How to evaluate $$\int_3^{17} \sqrt{x^2-3x+2}\ dx \ ?$$ I tried Euler's substitution, that is $$\sqrt{x^2-3x+2}=x+t \Longleftrightarrow \frac{t^2-2}{-3-2t}+t=\frac{t^2+3t+2}{2t+3}\ ,$$ which I obtained from $$x^2+2tx+t^2=x^2-3x+2\Longleftrightarrow x=\frac{t^2-2}{-3-2t}$$ $$dx=\frac{2t(-3-2t)-(t^2-2)(-2)}{(-3-2t)^2}\ dt=-\frac{2t^2+6t+4}{(3+2t)^2} \ dt\ .$$ $x\in[3,17]$, so $t\in[\sqrt{2}-3,\sqrt{240}-17]$ (right?). Then we got $$\int_{\sqrt{2}-3}^{\sqrt{240}-17} \frac{t^2+3t+2}{2t+3}(-\frac{2t^2+6t+4}{(2t+3)^2})\ dt=-2\int_{\sqrt{2}-3}^{\sqrt{240}-17} \frac{(t^2+3t+2)^2}{(2t+3)^3}\ dt$$ and it's where I stuck...
EDIT
Ok I finished it my way, but I've heard there's a possibility to do it using $(x-\frac{3}{2})=\tan (t)$ substitution. How exactly?
| Hint
I think that you made your life very complicated.
If I may suggest, start completing the square $$x^2-3x+2=\Big(x-\frac{3}{2}\Big)^2-\frac {9}{4}+2=\Big(x-\frac{3}{2}\Big)^2-\frac {1}{4}$$ and now change variable such $$x-\frac{3}{2}= \frac{1}{2}\cosh(y)$$ You will arrive to something very simple for the integrand and then for the antiderivative.
I am sure that you can take from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Cubic trig equation I'm trying to solve the following trig equation:
$\cos^3(x)-\sin^3(x)=1$
I set up the substitutions $a=\cos(x)$ and $b=\sin(x)$ and, playing with trig identities, got as far as $a^3+a^2b-b-1=0$, but not sure how to continue. Is there a way to factor this so I can use the zero product property to solve?
Thanks for any help/guidance!
P
| $$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x\cos x)$$
If we set $\displaystyle\cos x-\sin x=u, u^2=1-2\sin x\cos x\implies (\cos x-\sin x)(1+\sin x\cos x)=u\left(1+\frac{1-u^2}2\right)=\frac{3u-u^3}2$
So, the problem reduces to $\displaystyle u^3-3u+2=0$
Clearly, $1$ is a root and $\displaystyle\displaystyle\frac{u^3-3u+2}{u-1}=u^2+u-2=(u+2)(u-1)$
$\displaystyle\implies u^3-3u+2=(u+2)(u-1)^2$
If $\displaystyle u+2=0, u=-2,\cos x-\sin x=-2,$ $\displaystyle\cos x-\sin x=\sqrt2\cos\left(x+\frac\pi4\right)$
$\displaystyle\implies-\sqrt2\le\cos x-\sin x\le\sqrt2$
So, $u=\cos x-\sin x$ must be $\displaystyle=1\implies \cos x-\sin x=1\iff \sqrt2\cos\left(x+\frac\pi4\right)=1$
$\displaystyle\implies\cos\left(x+\frac\pi4\right)=\frac1{\sqrt2}=\cos\frac\pi4$
$\displaystyle\implies x+\frac\pi4=2m\pi\pm \frac\pi4$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/800320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Double integral for $\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$ I'm trying to evaluate this
$$\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$$
tried substition
$$ u = {(x^2+y^2+1)}^{-1} \ \ du = \ln {(x^2+y^2+1)}$$
but du is not found in the given equation. I have the feeling that I should use arctan but I do not know how to apply that in two variables.
| $\int_{0}^{1}\int_{-1}^{0}\dfrac{xy}{1 + x^2 + y^2}dydx = \int_{0}^{1}\dfrac{x}{2}\int_{-1}^{0}\dfrac{2ydy}{1+x^2+y^2}dx = \int_{0}^{1}\dfrac{x}{2}\int_{-1}^{0}\dfrac{d(1 + x^2 + y^2)}{1 + x^2 + y^2}dx$. Thus,
$\int_{0}^{1}\int_{-1}^{0}\dfrac{xy}{1 + x^2 + y^2}dydx = \int_{0}^{1}\dfrac{x}{2}\ln(1 + x^2 + y^2)\mid_{-1}^{0}dx = -\int_{0}^{1}\dfrac{x}{2}\ln\biggl(1 + \dfrac{1}{1 + x^2}\biggr)dx$
If $u = 1 + x^2$, then $du = 2xdx$ and
$-\int_{0}^{1}\dfrac{x}{2}\ln\biggl(1 + \dfrac{1}{1 + x^2}\biggr)dx = -\dfrac{1}{4}\int_{1}^{2}\ln(1 + 1/u)du = -\dfrac{1}{4}\int_{1}^{2}[\ln(1 + u) - \ln u]du$
But, $\int \ln x dx = x\ln x - x + C$. Therefore,
$-\dfrac{1}{4}\int_{1}^{2}[\ln(1 + u) - \ln u]du = -\dfrac{1}{4}\biggl[(1 + u)\ln(1 + u) - 1 - u - u\ln u + u\biggr]_{1}^{2} = -\dfrac{1}{4}[3\ln 3 - 2\ln 2 - 1 - 2\ln 2 + 1] = -\dfrac{3}{4}\ln 3 + \ln 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding generators for an ideal of $\Bbb{Z}[x]$ We know that $\Bbb{Z}$ is Noetherian. Hence, we can conclude that $\Bbb{Z}[x]$ is Noetherian, too. Consider the ideal generated by $\langle 2x^2+2,3x^3+3,5x^5+5,…,px^p+p,…\rangle$ for all prime natural numbers $p$. How can I determine a finite number of elements which generate this ideal? Clearly the Euclidean algorithm is not valid in $\Bbb{Z}[x]$.
Thank you
|
The arguments of the following proof relies on computations within $\mathbb Z[x]$ without any reference to bigger rings.
Let $I=\langle 2x^2+2,3x^3+3,5x^5+5,\dots,px^p+p,\dots\rangle$. Since $$px^p+p=p(x^p+1)=p(x+1)(x^{p-1}-\cdots+1)$$ for any prime $p\ge 3$, we have $$I\subseteq \langle 2x^2+2,x+1\rangle.$$ For $2x^2+2=2(x+1)(x-1)+4$ we get $\langle 2x^2+2,x+1\rangle=\langle 4,x+1\rangle$. Now let's prove that $4,x+1\in I$ and then conclude
$$I=\langle 4,x+1\rangle.$$
Note that $$3(x^3+1)=3(x+1)(x^2-x+1)$$ $$5(x^5+1)=5(x+1)(x^4-x^3+x^2-x+1)=5(x+1)[x^2(x^2-x+1)-x+1],$$ and therefore $15x^2(x^3+1)-15(x^5+1)=15(x^2-1)$. Since $15\cdot2(x^2+1)-30(x^2-1)=60$ we get $60=4\cdot3\cdot 5\in I$. Repeating the same procedure with $11$ and $13$ instead of $3$ and $5$ we get $4\cdot11\cdot13\in I$, and since $(3\cdot5,11\cdot13)=1$ we find $4\in I$.
It remains to prove $x+1\in I$: since $2x^2+2\in I$ and $4\in I$ we get $2x^2-2\in I$ and summing up $4x^2\in I$. Since $3x^3+3\in I$ and $4x^3\in I$ we get (by subtraction) $x^3-3\in I$, so $x^3+1\in I$. From $5x^5+5=5x^2(x^3+1)-5x^2+5\in I$ we get $5x^2-1=(5x^2-5)+4\in I$. Now recall that $4x^2\in I$, so $x^2-1\in I$ and using $x^3+1\in I$ we get $x+1\in I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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System of linear first order DEs Question:
$$3\dot{x} + \dot{y} + 5x - y = 2e^{-t}+4e^{-3t}$$
$$\dot{x} + 4\dot{y} - 2x + 7y = -3e^{-t}+5e^{-3t}$$
Subject to:
$$x(0)=y(0)=0$$
Attempt at a solution:
I have gotten to:
$$\left(\begin{array}{cc}1&-1&3&-3\\ 1&1&1&1\end{array}\right)\left(\begin{array}{cc}\dot{x}\\ \dot{y}\\ x \\ y \end{array}\right) = 2 \left( \begin{array}{cc}e^{-t}\\ e^{-3t}\end{array}\right) $$
But I am not sure what to do next.
| One way to rewrite:
$$
\left[\begin{array}{cc} 3 & 1 \\ 1 & 4\end{array}\right]
\frac{d}{dt}\left[\begin{array}{c} x \\ y\end{array}\right]+
\left[\begin{array}{cc} 5 & -1 \\ -2 & 7\end{array}\right]
\left[\begin{array}{c} x \\ y\end{array}\right]=
\left[\begin{array}{cc} 2 & 4 \\ -3 & 5\end{array}\right]
\left[\begin{array}{c} e^{-t} \\ e^{-3t}\end{array}\right].
$$
Multiplying by the inverse
$$
\left[\begin{array}{cc} 3 & 1 \\ 1 & 4\end{array}\right]^{-1}=
\frac{1}{11}\left[\begin{array}{cc} 4 & -1 \\ -1 & 3\end{array}\right]^{-1}
$$
gives
$$
\frac{d}{dt}\left[\begin{array}{c}x \\ y\end{array}\right]+
\left[\begin{array}{cc} 2 & -1 \\ -1 & 2\end{array}\right]
\left[\begin{array}{c} x \\ y\end{array}\right]=
\left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
\left[\begin{array}{c} e^{-t} \\ e^{-3t}\end{array}\right].
$$
The coefficient matrix $C$ on the left has characteristic polynomial $(\lambda-2)^{2}-1=(\lambda-3)(\lambda-1)$. So the coefficient matrix has diagonal representation:
$$
C=\left[\begin{array}{cc} 2 & -1 \\ -1 & 2\end{array}\right] =
U^{T}\left[\begin{array}{cc}3 & 0 \\ 0 & 1 \end{array}\right]U\;\;\;
\mbox{ where } U = \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right].
$$
Therefore,
$$
e^{tC} = U^{T}\left[\begin{array}{c}e^{3t} & 0\\0 & e^{t}\end{array}\right]U
= \frac{1}{2}\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]
\left[\begin{array}{c}e^{3t} & 0\\0 & e^{t}\end{array}\right]
\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]
$$
Multiplying both sides of the differential equation by the matrix $e^{tC}$ gives
$$
\frac{d}{dt}\left(e^{tC}\left[\begin{array}{c}x \\ y\end{array}\right]\right)
= e^{tC}\left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
\left[\begin{array}{c} e^{-t} \\ e^{-3t}\end{array}\right]
$$
Integrating,
$$
e^{tC}\left[\begin{array}{c}x \\ y\end{array}\right]
-\left[\begin{array}{c}x(0) \\ y(0)\end{array}\right]
= \int_{0}^{t} e^{sC}\left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
\left[\begin{array}{c} e^{-s} \\ e^{-3s}\end{array}\right]\,ds
$$
Applying $x(0)=y(0)=0$ and the specific form of $e^{sC}$ yields the final answer
$$
\begin{align}
\left[\begin{array}{c} x \\ y\end{array}\right] & = e^{-tC}\int_{0}^{t}
\left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
\left[\begin{array}{cc} e^{2s}\\e^{-2s}\end{array}\right]\,ds \\
%% & = \left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
%% \left[\begin{array}{cc} e^{-3t} & 0 \\ 0 & e^{-t}\end{array}\right]
%% \int_{0}^{t}\left[\begin{array}{cc} e^{2s}\\e^{-2s}\end{array}\right]\,ds \\
%% & = \left[\begin{array}{cc}e^{-3t} & e^{-t} \\ -e^{-3t} & e^{-t}\end{array}\right]
%% \frac{1}{2}\left[\begin{array}{c}e^{2t}-1 \\ 1-e^{-2t}\end{array}\right] \\
%% & = \frac{1}{2}\left[\begin{array}{cc}e^{-t}-e^{-3t}+e^{-t}-e^{-3t} \\
%% -e^{-t}+e^{-3t}+e^{-t}-e^{-3t}\end{array}\right] \\
& = \left[\begin{array}{c}e^{-t}-e^{-3t} \\ 0\end{array}\right].
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}$ I am trying to prove this interesting integral
$$
I:=\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}.
$$
I tried using $y=1+x^3$ but that didn't help.
We can possibly try
$$
I=\int_0^\infty \frac{\log(1+x^3) x}{1+x^3} \,dx-\int_0^\infty \frac{\log(x^3) x}{1+x^3}\,dx.
$$ These integrals would be much easier had the bounds been from $0 $\ to $\infty$, however they are not. Perhaps partial integration will work but I didn't find the way if we try
$$
dv=\frac{x}{1+x^3}, \quad u= \log(1+x^3)
$$
but I ran into a divergent integral. Thanks how can we prove I?
| Letting $x\mapsto \frac{1}{x}$ simplify the integral as
$$
I=\int_0^{\infty} \ln \left(\frac{1+x^3}{x^3}\right) \frac{x}{1+x^3} d x =\int_0^{\infty} \frac{\ln \left(x^3+1\right)}{x^3+1} d x
$$
Next consider another integral
$$
I(a)=\int_0^{\infty}\left(x^3+1\right)^a d x
$$
and transform $I(a)$, by putting $y=\frac{1}{x^3+1}$, into a beta function
$$
\begin{aligned}
I(a) &=\frac{1}{3} \int_0^1 y^{-a-\frac{4}{3}}(1-y)^{-\frac{2}{3}} d y \\
&=\frac{1}{3} B\left(-a-\frac{1}{3}, \frac{1}{3}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ yields
$$
I^{\prime}(a)=\frac{1}{3} B\left(-a-\frac{1}{3}, \frac{1}{3}\right)\left(\psi\left(-a-\frac{1}{3}\right)-\psi(-a)\right)
$$
Then putting $a=1$ gives our integral$$
\begin{aligned}
I&=I^{\prime}(-1) \\&=\frac{1}{3} B\left(\frac{2}{3}, \frac{1}{3}\right)\left(\psi\left(\frac{2}{3}\right)-\psi(1)\right) \\
&=\frac{1}{3} \cdot \frac{2 \pi}{\sqrt{3}}\left(\gamma-\frac{\pi}{2 \sqrt{3}}+\frac{3 \ln 3}{2}-\gamma\right) \\
&=\frac{\pi}{\sqrt{3}} \ln 3-\frac{\pi^2}{9}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Compute $\det(A^n+B^n)$
Let $A, B $ be two real $3\times 3 $ matrices, $AB=BA$, and $ \det(A-B)=\det(A^2+B^2)=1,\det(A+B)=3, \det(B)=0 $, then, what is ?
$$\det(A^n+B^n)$$
here $n$ is a positive integer.
The problem looks very horrible. Any help will be appreciated
| Although I don't know the real tricks behind the problem, the problem is not as terrible as it appears.
As $A$ and $B$ commute, they are simultanesouly triangularisable over $\mathbb C$. Given that $\det(B)=0$, we may assume that $A$ and $B$ are triangular matrices whose diagonal entries are respectively $a_1,a_2,a_3$ and $b_1,b_2,0$. So, the other three determinant conditions amount to
\begin{align}
(a_1-b_1)(a_2-b_2)a_3&=1,\tag{1}\\
(a_1+b_1)(a_2+b_2)a_3&=3,\tag{2}\\
(a_1^2+b_1^2)(a_2^2+b_2^2)a_3^2&=1.\tag{3}
\end{align}
Hence $a_3$ is nonzero. Now $A$ must be invertible. Otherwise, if, say, $a_2=0$, then the sum and difference of $(1)$ and $(2)$ would give $a_1b_2a_3=1$ and $b_1b_2a_3=2$, which contradict $(3)$.
Let $C=A^{-1}B$. Then $C$ is singular (because $B$ is). Let its complex eigenvalues be $x,y,0$. By the given conditions, we have
\begin{align}
\det(I+C)&=3\det(I-C),\\
(1+x)(1+y)&=3(1-x)(1-y),\\
2(x+y)&= xy + 1,\tag{4}
\end{align}
and also
\begin{align}
\det(I-C)^2&=\det(I+C^2),\\
(1-x)^2(1-y)^2 &= (1+x^2)(1+y^2),\\
4xy - 2(x+y)(xy+1) &= 0,\\
4xy - (xy+1)^2 &= 0 \quad\text{ by } (4),\\
xy &= 1.\tag{5}
\end{align}
Now equations $(4)$ and $(5)$ show that $x$ and $y$ are roots of the equations $xy=1$ and $x+y=1$. As $C$ is a real matrix, nonreal eigenvalues must occur in a conjugate pair, so $\{x,y\}=\{e^{i\pi/3},e^{-i\pi/3}\}$. Substitute $b_1=e^{i\pi/3}a_1$ and $b_2=e^{-i\pi/3}a_2$ into $(1)$, we get $\det(A)=1$. Therefore
$$
\det(A^n+B^n)=\det(I+C^n)\det(A)^n=|1+e^{in\pi/3}|^2=2 + 2\cos\left(\frac{n\pi}{3}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Prove that if x and y are odd natural numbers, then $x^2+y^2$ is never a perfect square. Prove that if x and y are odd natural numbers, then $x^2+y^2$ is never a perfect square.
Let $x=2m+1$ and $y=2l+1$ where m,l are integers.
$x^2+y^2=(2m+1)^2+(2l+1)^2=4(m^2+m+l^2+l)+2$
Where do I go from here?
| You can now look at all the natural numbers modulo $4$. We know that numbers must either be even or odd, hence they have the form
$$2n\text{ or } 2n+1.$$
In modulo $4$, they are just $2n\text{ or } 2n+1\mod4$. Now, look at the squares of these numbers, we have that
$$(2n)^2\equiv 4n^2\equiv0\mod4.$$
Also, the odds give
$$(2n+1)^2\equiv 4n^2+4n+1\equiv1\mod4.$$
Now we know that squares MUST be either $0$ or $1$ modulo $4$. What is your number modulo $4$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove this inequality $\frac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\frac{\pi}{n+1}}$
Let $a_{1},a_{2},\cdots,a_{n},n\ge 2$ be real numbers,show that
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\dfrac{\pi}{n+1}}$$
I think this result is interesting.
When $n=2$, clearly
$$\dfrac{a_{1}a_{2}}{a^2_{1}+a^2_{2}}\le\dfrac{1}{2}=\cos{\dfrac{\pi}{3}}=\dfrac{1}{2}$$
When $n=3$,
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}}{a^2_{1}+a^2_{2}+a^2_{3}}\le\dfrac{\sqrt{2}}{2}.$$
This is true, because
$$a^2_{1}+\dfrac{1}{2}a^2_{2}\ge \sqrt{2}a_{1}a_{2},$$
$$\dfrac{1}{2}a^2_{2}+a^2_{3}\ge\sqrt{2}a_{2}a_{3}.$$
But for general $n$ I cannot prove it.
| Let $M_1 = \dfrac{1}{4\cos\dfrac{\pi}{n+1}}$ and $M_{k+1} = \dfrac{1}{4[\cos\dfrac{\pi}{n+1} - M_k]}$. If we can prove $0 <M_k < \cos\dfrac{\pi}{n+1}$ for $k < n -1$, then we have
\begin{align}
a_1a_2 \leq \cos\dfrac{\pi}{n+1}a_1^2 + M_1 a_2^2 \\
a_2 a_3 \leq (\cos\dfrac{\pi}{n+1} - M_1)a_2^2 + M_2 a_3^2 \\
\vdots \\
a_{n-1}a_n \leq (\cos\dfrac{\pi}{n+1} - M_{n-2})a_{n-1}^2 + M_{n-1}a_n^2
\end{align}
thus all we need to prove is $0 <M_k < \cos\dfrac{\pi}{n+1}$ for all $ k < n-1$ and $M_{n-1} = \cos\dfrac{\pi}{n+1}$
To do this, we aim at finding constants $x$ and $y$, such that $\exists c$ with
\begin{align}
\frac{M_{k+1} + x}{M_{k+1} + y} = c \frac{M_{k} + x}{M_{k} + y}
\end{align}
After simplication, we find we can take $x$ and $y$ as different solutions of $$4\lambda^2 + 4\cos\dfrac{\pi}{n+1}\lambda + 1 = 0$$ and $c = \frac{x}{y}$, i.e.:
\begin{align}
x = -\frac{1}{2}(\cos\dfrac{\pi}{n+1} + i \sin\dfrac{\pi}{n+1})= -\frac{1}{2}e^{i\theta}\\
y = -\frac{1}{2}(\cos\dfrac{\pi}{n+1} - i \sin\dfrac{\pi}{n+1}) = -\frac{1}{2}e^{-i\theta}\\
c = (\cos\dfrac{\pi}{n+1} + i \sin\dfrac{\pi}{n+1})^2 = e^{i2\theta}
\end{align}
wiht $\theta = \frac{\pi}{n+1}$
Then we have
\begin{align}
\frac{M_{n-1} - \frac{1}{2}e^{i\theta}}{M_{n-1} - \frac{1}{2}e^{-i\theta}} = e^{2(n-2)i\theta} \frac{M_{1} - \frac{1}{2}e^{i\theta}}{M_{1} - \frac{1}{2}e^{-i\theta}}
\end{align}
Since $M_1 = \dfrac{1}{2(e^{i\theta} + e^{-i\theta})}$, we get
\begin{align}
\frac{M_{1} - \frac{1}{2}e^{i\theta}}{M_{1} - \frac{1}{2}e^{-i\theta}} = e^{4i\theta}
\end{align}
Thus
\begin{align}
\frac{M_{n-1} - \frac{1}{2}e^{i\theta}}{M_{n-1} - \frac{1}{2}e^{-i\theta}} = e^{2ni\theta} = e^{-2i\theta}
\end{align}
Then we get easily $M_{n-1} = \dfrac{e^{i\theta} + e^{-i\theta}}{2}=\cos\frac{\pi}{n+1}$
Finally note that if $M_k < \cos\frac{\pi}{n+1} - \frac{1}{4\cos\frac{\pi}{n+1}}$
$$M_{k+1} - \cos\frac{\pi}{n+1} = \dfrac{1 - 4\cos^2\frac{\pi}{n+1} + 4M_k \cos\frac{\pi}{n+1}}{4(\cos\frac{\pi}{n+1} - M_k)} < 0$$
Note also that $M_{k+1} - M_{k} = \dfrac{1 - 4\cos\frac{\pi}{n+1}M_k + 4M_k^2}{4(\cos\frac{\pi}{n+1} - M_k)} > 0$ when $M_k < \cos\frac{\pi}{n+1} $
since we know that $M_{n-2} = \cos\frac{\pi}{n+1} - \frac{1}{4\cos\frac{\pi}{n+1}}$, we can conclude $M_k$ is increasing for $k < n-1$, therefore $0<M_k < \cos\frac{\pi}{n+1}$ for $k < n-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/807326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 2
} |
Asymptotic expansions for the roots of $\epsilon^2x^4-\epsilon x^3-2x^2+2=0$ I'm trying to compute the asymptotic expansion for each of the four roots to the following equation, as $\epsilon \rightarrow 0$:
$\epsilon^2x^4-\epsilon x^3-2x^2+2=0$
I'd like my expansions to go up through terms of size $O(\epsilon^2)$.
I´ve made the change of variables $x=\delta y$, performed dominant balance and found out that the only two valid options are: (rescaled eq: $\epsilon^2\delta^4y^4-\epsilon \delta^3 y^3-2\delta^2 y^2+2=0$)
*
*$\delta^2 \sim 2 \Rightarrow \delta=O(1)$
*$\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta=O(\epsilon^{-1})$
How do I proceed?
| Your analysis is correct.
Let's look at the original equation
$${\epsilon ^2}{x^4} - \epsilon {x^3} - 2{x^2} + 2 = 0$$
and plot the Newton-Kruskal diagram:
There are only two possible placements of straight lines passing through two or more points with all remaining points "above the line"; those two lines correspond exactly to
*
*$\delta^2 \sim 2 \Rightarrow \delta \sim 1$, and
*$\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta \sim \epsilon^{-1}.$
Before we continue, let's rescale the original equation by making the change of variables $x=\delta y$ to arrive at
$${\epsilon ^2}{\delta ^4}{y^4} - \epsilon {\delta ^3}{y^3} - 2{\delta ^2}{y^2} + 2 = 0.$$
From the first balance, $\delta^2 \sim 2 \Rightarrow \delta \sim 1$, we immediately recover the original equation:
$$\tag{1} {\epsilon ^2}{y^4} - \epsilon {y^3} - 2{y^2} + 2 = 0.$$
Letting $\epsilon \rightarrow 0$, we get
$$2{y^2} + 2 = 0 \Rightarrow y = \pm 1$$
Let´s check if there are roots close to $y=-1$ and $y=1$.
Plugging an asymptotical expansion of the form $y \sim \pm 1 + {a_1}\epsilon + {a_2}{\epsilon ^2} + {a_3}{\epsilon ^3} + {a_4}{\epsilon ^4} + \ldots $ into (1), and then matching coefficients, we arrive at
$${x_a} \sim {y_a} \sim - 1 - \frac{\epsilon }{4} - \frac{{13}}{{32}}{\epsilon ^2} + O\left( {{\epsilon ^3}} \right),$$
$${x_b} \sim {y_b} \sim 1 - \frac{\epsilon }{4} + \frac{{13}}{{32}}{\epsilon ^2} + O\left( {{\epsilon ^3}} \right).$$
Regarding the second possibility of balancing, $\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta \sim \epsilon^{-1}$, we get the following equation
$$\tag{2}{y^4} - {y^3} - 2{y^2} + 2{\epsilon ^2} = 0$$
Letting $\epsilon \rightarrow 0$, we get
$${y^4} - {y^3} - 2{y^2} = 0 \Rightarrow {y^2}\left( {{y^2} - y - 2} \right) = 0 \Rightarrow y = 0 \vee y = - 1 \vee y = 2.$$
Let´s check if there are roots close to $y=-1$ and $y=2$ ($y=0$ is of no interest, as we just recover one of the roots previously found). Plugging an asymptotical expansion of the form $y \sim {a_0} + {a_1}\epsilon + {a_2}{\epsilon ^2} + {a_3}{\epsilon ^3} + {a_4}{\epsilon ^4} + \ldots $ (with ${a_0} = - 1 \vee {a_0} = 2$) into (2), and then matching coefficients, we arrive at
$${x_c} \sim {\epsilon ^{ - 1}}{y_c} \sim - {\epsilon ^{ - 1}} + \frac{2}{3}\epsilon + O\left( {{\epsilon ^3}} \right),$$
$${x_d} \sim {\epsilon ^{ - 1}}{y_d} \sim 2{\epsilon ^{ - 1}} - \frac{1}{6}\epsilon + O\left( {{\epsilon ^3}} \right).$$
Thus, we've determined all the four roots of the original equation.
-EDIT-
Here's the Mathematica code I used to do my calculations, more specifically, the last one.
n = 4;
a[0] = 2;
x = a[0] + Sum[a[i] \[Epsilon]^i, {i, 1, n}] + O[\[Epsilon]]^(n + 1);
x^4 - x^3 - 2 x^2 + 2 \[Epsilon]^2 == 0;
LogicalExpand[%]
Solve[%]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it down to $ \tan^2\theta.$
HELP!!!!
I also need help proving that $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$
| $$\large\frac{1+\tan^2\theta}{1+\cot^2\theta}$$
$$\implies\large \frac {\frac{cos^2\theta+sin^2\theta}{cos^2\theta}}{\frac{sin^2\theta+cos^2\theta}{sin^2\theta}}$$
$$\implies \large \frac {\frac{1}{cos^2\theta}}{\frac{1}{sin^2\theta}} \\$$
$$\implies \large \frac {sin^2\theta}{cos^2\theta} $$
$$\implies \large \tan^2\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 5
} |
proving $\tan^{-1} \left( \frac{x}{1-x^2} \right) = \tan^{-1}x + \tan^{-1}(x^3)$ Two related questions, one easy, one just a bit harder:
1) Prove the identity
$$
\tan^{-1} \left( \frac{x}{1-x^2} \right) =
\tan^{-1}x + \tan^{-1}(x^3)
$$
2) Now try to find a geometric or trigonometric proof of that same geometry, without resorting to calculus.
I'll post an answer to both questions in a couple of days if nobody has one yet.
| Part I)
Taking the tan of the RHS, we have a form $tan(A+B)$ where $A = tan^{-1}(x)$, $B = tan^{-1}(x^3)$.
$$
tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)*tan(B)}
= \frac{x + x^3}{1 - x^4}
= \frac{x(1 + x^2)}{(1-x^2)(1+x^2)}
= \frac{x}{1-x^2}
$$
Now the tan of LHS is also $\frac{x}{1-x^2}$.
Therefore the identity is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/812123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute and find 2009th decimal(2009th digit after the point), without automation, the following sum Compute and find 2009th decimal of (2009th digit after the point), without automation, the following sum
$$\frac{10}{11}+\frac{10^2}{1221}+\frac{10^3}{123321}+ \cdots +\frac{10^9}{123456789987654321}$$
| You can write your series as
$$f(x)=\sum_{1}^{x}\frac{81\times10^k}{(10^k-1)(10^{k+1}-1)}=9\sum_{1}^{x}\frac{1}{10^k-1}-\frac{1}{10^{k+1}-1}$$
Where $x=9$. By telescopy we can show:
$$f(x)=\frac{10^{x+1}-10}{10^{x+1}-1}=1-\frac{9}{10^{x+1}-1}$$
So your sum is $S=1-1111111111^{-1}=0.\overline{9999999990}$. So since $2009\equiv 9\pmod{10}$, the digit is $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
| \begin{align}
x^2+x-6 = x^2 + \underbrace{3x - 2x} - 6 & = \underbrace{x^2+3x}+\underbrace{{}-2x-6} \\[8pt]
& = x(x+3) + (-2)(x+3) \\[8pt]
& = x(\cdots\cdots) + (-2)(\cdots\cdots) \\[8pt]
& = x(\cdots\cdots) -2(\cdots\cdots) \\[8pt]
& = (x-2)(\cdots\cdots) \\[8pt]
& = (x-2)(x+3).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Cannot solve by hand:$ x + y = 2; 4x^2 + y^2 = 5(2x - y)(xy)^{\frac12}$ Firstly, this is not my homework. I am well past high school (finished graduate school several years ago) but I am mentoring a high schooler and I want to explain how to solve this by hand using just pen and paper.
A more presentable form + solution is here
My own attempts could only simplify it to a fourth degree polynomial in $x$ and equate it to $0$. There's gotta be a better way but it is beyond me and I feel embarrassed!
| since $xy>0,x+y=2>0, \implies x>0,y>0$
let $x=a^2,y=b^2,a>0,b>0 \implies 4a^4+b^4=5(2a^2-b^2)ab \iff \\ 4a^4-10a^3b+5ab^3+b^4=0$
by observation, $a=b$ is a solution as $4-10+5+1=0$
so we get $(a-b)(4a^3-6a^2b-6ab^2-b^3)=0$
now to check $4a^3-6a^2b-6ab^2-b^3=0$,a factor $b=-2a$ is existed. $\implies$
$(2a+b)(2a^2-4ab-b^2)=0 \implies 2a^2-4ab-b^2=0 $
with $a^2+b^2=2 $ we get $3a^2-4ab=2 \implies b=\dfrac{3a^2-2}{4a} \implies x+\dfrac{(3x-2)^2}{16x}=2$
$25x^2-44x+4=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach :
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(8\pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(\pi + \pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)-\sin(\pi/7) $
I am not getting any clue how to proceed further or whether it is correct or not. Please help thanks..
| Let $\Phi_7(x)$ the minimal polynomial of $\xi=\exp\left(\frac{2\pi i}{7}\right)$. Its Galois group over $\mathbb{Q}$ is cyclic and generated by $\xi\mapsto\xi^3$, since $3$ is a generator of $\mathbb{Z}/(7\mathbb{Z})^*$. The quadratic residues $\!\!\pmod{7}$ are $1,2,4$ and the quadratic non-residues are $3,5,6$, hence:
$$ \alpha = \xi+\xi^2+\xi^4,\qquad \beta=\xi^3+\xi^5+\xi^6 $$
are two conjugated algebraic numbers with degree $\frac{6}{3}=2$. In particular, both $\alpha+\beta$ and $\alpha\beta$ are rational numbers:
$$ \alpha+\beta=\sum_{k=1}^{6}\xi^k=-1,\qquad \alpha\beta=3+\sum_{k=1}^{6}\xi^k=2 $$
and the minimal polynomial of $\alpha$ and $\beta$ is given by $x^2+x+2$.
It is straightforward to check that $\text{Im}(\alpha)>0$ and $\text{Im}(\beta)<0$, hence by computing the discriminant of $x^2+x+2$ we have:
$$ \alpha-\beta = \sqrt{-7} $$
and the claim follows by De Moivre's formula, since, for instance,
$$ \xi-\xi^6 = \exp\left(\frac{2\pi i}{7}\right)-\exp\left(-\frac{2\pi i}{7}\right)=2i\sin\left(\frac{2\pi}{7}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Area of triangle having an inscribed circle
The radius of an inscribed circle in a triangle is $2 cm$. A point of tangency divides a side into $3 cm$ and $4 cm$. Find the area of the triangle.
We know that a side is $7 cm$, the others are $4+x$, $3+x$. I tried finding $2$ equations of the area, $S=pr \Rightarrow S=2x+14$ and by herone and equalizing, but that didn't give me a good answer.
|
$AR=AP=3$, $CR=CQ=4$, $BP=BQ=x$, $OP=OQ=OR=2$.
$AO=\sqrt{3^2+2^2}=\sqrt{13}$;
$CO=\sqrt{4^2+2^2}=\sqrt{20}$.
$$S = \dfrac{1}{2} \cdot AB\cdot AC \cdot \sin \angle BAC = \dfrac{1}{2} \cdot CA\cdot CB \cdot \sin \angle ACB$$
Using formula $\sin 2\alpha = 2 \sin \alpha \cos \alpha$, we get
$$
S=AB\cdot AC\cdot \sin \angle OAR \cdot \cos \angle OAR = CA\cdot CB \cdot \sin \angle OCR \cdot \cos \angle OCR
$$
$$
S=(3+x)\cdot 7 \cdot \dfrac{2}{\sqrt{13}}\cdot \dfrac{3}{\sqrt{13}} = (4+x)\cdot 7 \cdot \dfrac{2}{\sqrt{20}}\cdot \dfrac{4}{\sqrt{20}};
$$
$$
S=\dfrac{42}{13}(3+x) = \dfrac{56}{20}(4+x);
$$
$$
840(3+x)=728(4+x);
$$
$$
112 x = 392;
$$
$$
x=\dfrac{7}{2};
$$
$$
S=21.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Answer to simple algebraic formula manipulation I have to show that
$2y+(x + 1) = 3 \cdot 2^{x+1}− (x + 1) − 2$
is equal to
$y=3⋅2^x−x−2$
I can get this far:
$2y+(x+1)=3⋅2^{x+1}−(x+1)−2$
$2y+(x+1)=3⋅2^{x+1}−x−1−2$
$2y+(x+1)=3⋅2^{x+1}−x−3$
$2y+x+1=3⋅2^{x+1}−x−3$
$2y=3⋅2^{x+1}−x−3−1−x$
$2y=3⋅2^{x+1}−2x−4$
Now I should divide by 2, but the 3 on the right side throws me off. Some help would be appreciated!
| If you divide $3 \cdot 2^{x+1}$ by $2$ you obtain $3 \cdot 2^x$, since $2^{x+1}=2 \cdot 2^{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the number of values of $x$ in $(\sqrt{2})^x+(\sqrt{3})^x = (\sqrt{13})^{\frac{x}{2}}$. Question:
Find the number of values of $x$ with $$(\sqrt{2})^x+(\sqrt{3})^x = (\sqrt{13})^{\frac{x}{2}}.$$
My attempt:
I tried setting $p = \sqrt{2}$, $q = \sqrt{3}$, and resetting the equation to:
$$p^x + q^x = \sqrt{p^4+q^4}^{\frac{x}{2}}$$
But I don't still think that expansion would probably solve it. Anyone has a better way ?
| Divide by $(\sqrt{13})^{\large\frac{x}{2}}$, and write $x = 4y$. You obtain
$$\left(\frac{4}{13}\right)^y + \left(\frac{9}{13}\right)^y = 1.$$
For $0 < a < 1$, the sole solution of $a^y + (1-a)^y = 1$ is $y = 1$, since $y\mapsto a^y$ and $y\mapsto (1-a)^y$ are both strictly decreasing.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$
But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
| HINT :
Rewrite the integrand
$$
\frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}
$$
as
$$
\frac{x\cos x}{x\sin x +\cos x}+\frac{x\sin x}{x\cos x -\sin x}
$$
then
$$
\frac{\color{red}{\sin x}+x\cos x-\color{red}{\sin x}}{x\sin x +\cos x}+\frac{\color{blue}{\cos x}+x\sin x-\color{blue}{\cos x}}{x\cos x -\sin x}.
$$
Now let $u=x\sin x +\cos x$ and $v=x\cos x -\sin x$.
| {
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"url": "https://math.stackexchange.com/questions/821862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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} |
minimum and maximum of $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$ we are asked to find the minimum and maximum of the function$f:A \to A$ $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$
Where $A$ is the triangle bound by $x=0$,$y=0$ and $y=-x+2\pi$
I'd like someone to review my answer.
What I did:
$A$ is a closed and bounded set, $f(x,y)$ is continuous, so according to Weierstrass theorem, $f$ receives maximal / minimal values, either on the boundary, or an internal point $(a,b)$ where $\triangle f(a,b)=(0,0)$,($\triangle$ represents gradient.)
First let's find points where the gradient is zero:
$\triangle f(x,y)=(\cos(x)-\cos(x+y),\cos(y)-\cos(x+y))=0$
this implies $\cos(x)=\cos(y)=\cos(x+y)$.
On the triangle we were given, this can only happen at $(0,2\pi)$ or $(2\pi,0)$, otherwise we are outside the boundaries of the triangle. and $f(0,2\pi)=f(2\pi,0)=0$.
Let's see what happens on the boundary, assume first $x=0$:
$f(0,y)=\sin(0)+\sin(y)-\sin(0+y)=0$
Same thing happens when $y=0$.
Now let's see $y=-x+2\pi$:
$f(x,-x+2\pi)=\sin(x)+\sin(-x+2\pi)-\sin(x+x-2\pi)=\sin(x)+\sin(-x)-\sin(2x)=\sin(x)-\sin(x)-\sin(2x)=-\sin(2x)$
Let's denote $g(x)=-\sin(2x)$, then $g'(x)=-2\cos(2x)$. $g'(x)=0$ implies $\cos(2x)=0$, which implies $2x=\frac{\pi}{2}+k\pi$, then $x=\frac{\pi}{4}+k\frac{\pi}{2}$
The only such valid point on our triangle would be the point $(\frac{\pi}{4},\frac{7\pi}{4})$. and at that point: $f(\frac{\pi}{4},\frac{7\pi}{4})=\sin(\frac{\pi}{4})+\sin(\frac{7\pi}{4})-\sin(2\pi)=0$
At all the potentially extreme points, we got $f=0$. this makes me believe that $f(A)=\{0\}$. There is no point on the triangle where $f$ is not zero. Is this true? Is it possible to verify this result with trigonometric identities? to simplify $\sin(x)+\sin(y)-\sin(x+y)$ and eventually hope to reach zero?
| You are correct in saying that if $f$ attains a minimum/maximum at $(x,y)$ in the interior of $A$, then $(x,y)$ must satisfy $\cos x = \cos y = \cos(x+y)$.
However, $(0,2\pi)$ and $(2\pi,0)$ are not the only points in $A$ at which this condition can be met.
Suppose $(x,y) \in A$ satisfies $\cos x = \cos y = \cos(x+y)$.
Then, $0 = \cos x - \cos(x+y) = 2\sin \dfrac{y}{2} \sin\left(x + \dfrac{y}{2}\right)$.
Hence, $\sin \dfrac{y}{2} = 0$ or $\sin\left(x + \dfrac{y}{2}\right) = 0$.
Thus, we must have either $y = 2\pi m$, or $2x+y = 2\pi m$ for some integer $m$.
Similarly, we must have either $x = 2\pi n$, or $x+2y = 2\pi n$ for some integer $n$.
With a bit of casework, we can see that the only $(x,y)$ in the interior of $A$ is $(x,y) = \left(\dfrac{2\pi}{3},\dfrac{2\pi}{3}\right)$ (the solution to $2x+y = x+2y = 2\pi$).
So you need to test this point as well. (As it turns out, this ends up being the maximum).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Intersection of ellipse and hyperbola at a right angle Need to show that two functions intersect at a right angle.
Show that the ellipse
$$ \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1
$$
and the hyperbola
$$
\frac{x^2}{α^2} −\frac{y^2}{β^2} = 1
$$
will intersect at a right angle if
$$α^2 ≤ a^2 \quad \text{and}\quad a^2 − b^2 = α^2 + β^2$$
Not sure how to tackle this question, graphing didn't help.
| Apparently, if $\alpha^2>a^2$, the two curves never intersect in the first place. Now let's just find the derivatives at the point of intersection and show that their product is $-1$.
Ellipse:
$$y=b\sqrt{1-\left({x\over a}\right)^2}\\
y'=b\cdot{-2x/a^2\over2\sqrt{1-\left({x\over a}\right)^2}}=-{b^2x\over a^2y}\tag{1}$$
Hyperbola:
$$y=\beta\sqrt{\left({x\over\alpha}\right)^2-1}\\
y'=\beta\cdot{2x/\alpha^2\over2\sqrt{\left({x\over\alpha}\right)^2-1}}={\beta^2x\over\alpha^2y}\tag{2}$$
Point of intersection:
$${x^2\over\alpha^2} − {y^2\over\beta^2} = 1\\
{x^2\over a^2} + {y^2\over b^2} = 1$$
Let's multiply the first equation by $\beta^2$, the second by $b^2$, and add them together.
$${\beta^2x^2\over\alpha^2}+{b^2x^2\over a^2} = \beta^2+b^2\\
x^2={\beta^2+b^2\over{\beta^2\over\alpha^2}+{b^2\over a^2}}=a^2\alpha^2{\beta^2+b^2\over a^2\beta^2+\alpha^2b^2}\\
y^2=b^2\left(1-{x^2\over a^2}\right)=b^2\cdot{a^2\beta^2+\alpha^2b^2-\alpha^2\beta^2-\alpha^2b^2\over a^2\beta^2+\alpha^2b^2}=b^2\beta^2\cdot{a^2-\alpha^2\over a^2\beta^2+\alpha^2b^2}\tag{3}$$
Now back to the derivatives:
$$-{b^2x\over a^2y}\cdot{\beta^2x\over\alpha^2y}=-1\\
{b^2\beta^2x^2\over a^2\alpha^2y^2}=1
$$
Plug in $x^2$ and $y^2$ from (3). Then nearly everything magically cancels out, and we're left with...
$${\beta^2+b^2
\over
a^2-\alpha^2}
=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/829621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to:
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.
Prove...
$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$
I keep trying to expand to $6(2k+2)^2$ and factorising but I end up being short on one factor, e.g., I end up with $\frac{(k+1)(2k+3)(7k+6)}{6}$.
| Check this out
#
$(k+1)^2 +(k+2)^2+ ............ (k+k)^2 = \frac{k(2k+1)(7k+1)}{6}$
#
$p(k+1): (k+1)^2+(k+2)^2...........(2k)^2+(2k+1)^2+.(2k+2)^2-(k+1)^2= \frac{(k+1)(2k+3)(7k+8)}{6}$
#
$p(k+1): \frac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2$=RHS
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How can I find the roots of a quartic equation, knowing one of its roots? I need to decompose (in $\Bbb{C}[x]$) the function
$$
f(x) = x^4 + 4x^3 - 4x^2 + 24x + 15
$$
in its simplest form, knowing that $1 - 2i$ is one of its roots. Any ideas?
| The answer has already been given in the comments, so I'll make a community answer out of it.
So we have the function $$f(x) = x^4 + 4x^3 - 4x^2 + 24x + 15$$
One root has been found at $x = 1-2i$.
Since the coefficients are real, there must be another root at $x = 1+2i$. We can now reduce this equation to a quadratic one, by dividing those out.
$$\begin{align}
& \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{(x-(1-2i))(x-(1+2i))} \\
= \ & \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{(x-1+2i)(x-1-2i)} \\
= \ & \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{x^2-2x+5} \\
= \ & x^2+6x+3
\end{align}$$
The roots for that quadratic equation can be found at $x = -3-\sqrt{6}$ and $x = -3 + \sqrt{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the limit $\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$ I need to evaluate the limit without using l'Hopital's rule.
$$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$
| $\frac{\sqrt{1+x\sin(x)}-\sqrt{\cos(2x)}}{\tan^{2}(\frac{x}{2})}=\bigg(\frac{1-\cos(2x)}{\tan^{2}(\frac{x}{2})}+\frac{x\sin(x)}{\tan^{2}(\frac{x}{2})}\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}=\bigg(\frac{1-\cos(2x)}{(2x)^{2}}\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}(16\cos^{2}(\frac{x}{2}))+\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}\frac{\sin(x)}{x}(4\cos^{2}(\frac{x}{2}))\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}$
Letting $x$ go to $0$ in the last formulation of the limit gives:
$(\frac{1}{2}\cdot1\cdot16+1\cdot1\cdot4)\frac{1}{2}=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Use induction to show that $a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) .$
Let $a_0$ and $a_1$ be distinct real numbers. Define
$a_n=\frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $n\geq 2$. Prove that $$a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) $$
This is what I have so far:
Let $P(n)$ be the statement: For all $n\geq 2$, $n\in\mathbb{N}$, $$
a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0). $$ We must show
1. $P(2)$ holds, 2. For all $k\geq 2$, $k\in\mathbb{N}$, if $P(k)$ holds, then $P(k+1)$ holds. Let $n=2$. Then $$ \begin{aligned}
a_{n+1}-a_n &=a_3-a_2\\ &=\frac{a_2+a_1}{2}-\frac{a_1+a_0}{2}\\
&=\frac{a_1+a_0+2a_1}{4}-\frac{a_1+a_0}{2}\\
&=\frac{3a_1+a_0}{4}+\frac{-2a_1-2a_0}{4}\\ &=\frac{a_1-a_0}{4}\\
&=\biggl(-\frac{1}{2}\biggr)^2(a_1-a_0). \end{aligned} $$ Hence,
$P(2)$ holds. Suppose that for all $k\geq 2$, $k\in\mathbb{N}$,
$P(k)$ holds, that is $$ \begin{aligned} a_{k+1}-a_k
&=\frac{a_k+a_{k-1}}{2}-\frac{a_{k-1}+a_{k-2}}{2}\\
&=\biggl(-\frac{1}{2} \biggr)^k (a_1-a_0). \end{aligned} $$ Then
$$ \begin{aligned} a_{(k+1)+1}-a_{k+1} &=a_{k+2}-a_{k+1}\\
&=\frac{a_{k+1}+a_k}{2}-\frac{a_k+a_{k-1}}{2}\\
&=\frac{a_k+a_{k-1}+a_{k-1}+a_{k-2}}{4}-\frac{a_{k-1}+a_{k-2}+a_{k-2}+a_{k-3}}{4}\\ &=\biggl(\frac{a_k+a_{k-1}-a_{k-2}-a_{k-3}}{2}\biggr)^2\\
&=\frac{a_{k-1}+a_{k-2}+a_{k-2}+a_{k-3}-a_{k-3}-a_{k-4}-a_{k-4}-a_{k-5}}{8}\\
&=\biggl(\frac{a_{k-1}+2a_{k-2}-2a_{k-4}-a_{k-5}}{2}\biggr)^3
\end{aligned} $$
Ok, so here it seems that I will just end up substituting forever. I'm sure that I'm missing something obvious. Any help would be greatly appreciated. Thanks.
| You should use the induction hypothesis (with two base cases) immediately after your second equality when computing $a_{k + 1} - a_{k + 1}$, for we then have
\begin{align*}
\frac{a_{k + 1} + a_k}{2} - \frac{a_k + a_{k - 1}}{2} &= \frac 1 2\left(- \frac 1 2\right)^k (a_1 - a_0) - \frac 1 2\left(-\frac 1 2\right)^{k - 1} (a_1 - a_0) \\
&= (-1)^k \left(\frac{1}{2^{k + 1}} - \frac 1 {2^k}\right) (a_1 - a_0) \\
&= (-1)^k \left(-\frac{1}{2^{k + 1}}\right) (a_1 - a_0) \\
&= \left(-\frac 1 2 \right)^{k + 1} (a_1 - a_0)
\end{align*}
as desired.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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GCD of polynomials over $\mathbb{Z}_3$ $f$ and $g$ are polynomials over field $\space \mathbb{Z}_3$. $f=X^4+X^3+X+2, \space g=X^4+2X^3+2X+2$. And I been asked to find the GCD of them.
What I have done is using Euclidean algorithm. After long division I get,
$$X^4+2X^3+2X+2=(1)(X^4+X^3+X+2)+(X^3+X)$$
So GCD$(f,g)=X^3+X$, however the answer to this is $X^2+1$. I want to know what I did wrong.
| This is for the completeness, ever step is done by long division.
$$X^4+2X^3+2X+2=(1)(X^4+X^3+X+2)+(X^3+X) \\
X^4+X^3+X+2 = (X+1)(X^3+X)+(X^2+1)(2) \\
X^3+X = (2X)(2X^2+2)+0$$
Therefore, the GCD of $f$ ad $g$ is $X^2+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $n^2(n^2+16)$ is divisible by 720 Given that $n+1$ and $n-1$ are prime, we need to show that $n^2(n^2+16)$ is divisible by 720 for $n>6$.
My attempt:
We know that neither $n-1$ nor $n+1$ is divisible by $2$ or by $3$, therefore $n$ must be divisible by both $2$ and $3$ which means it must be divisible by $6$.
So, $n = 6k$ and since $n>6$ we must have $k>3$. So the expression becomes $36k^2(36k^2+16) = 144k^2(9k^2+4)$
...
But then I get stuck.
Could someone please guide me towards a solution.Thanks.
| HINT (since this is worth solving yourself): You need to use that $n-1, n+1$ are prime. If either of these are equal to $5$, you have the excluded values $n=6$ or $n=4$.
Otherwise, note that the product of any five successive integers is divisible by $5$. You are given $n+1$ and $n-1$ - can you see which five successive integers you might use?
Further to comments below, and to fill out a solution. We note that the conditions mean that $(n+1)$ and $(n-1)$ have no factors $2,3,5$ and that $720 = 16\times 9 \times 5$.
The product of three successive integers $(n-1)n(n+1)$ is divisible by $3$, so $n$ is divisible by $3$ and $n^2$ is divisible by $9$.
Also $n$ must be even, so $n^2$ and $n^2+16$ are both divisible by $4$ and we have a factor $16$.
One of the five consecutive numbers $n-2, n-1, n, n+1, n+2$ is divisible by $5$. It is not $n-1$ or $n+1$. If it is $n$ we are done - we have the factor $5$ we need. Else $(n+2)(n-2)=n^2-4=n^2+16-20$ is divisible by $5$, whence the same is true of $n^2+16$ and again we are done.
It is also possible to do this with congruences, of course, but I have always liked this kind of method, so I look out for it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Type of this Conic section I want to determine, to which type the following Conic sections belong to:
$$
\begin{align}
\textrm{(i)}&\quad-8x^2+12xy-6x+8y^2-18y+8=0\\
\textrm{(ii)}&\quad5x^2-8xy+2x+5y^2+2y+1=0
\end{align}
$$
To (i)
Matrix notation:
$$
\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}-8&6\\6&8\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+2\begin{pmatrix}-3&-9\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+8=0\\
$$
$A=\begin{pmatrix}-8&6\\6&8\end{pmatrix}$
Eigenvalues are $\lambda_1=10,\lambda_2=-10$
Eigenvectors: $v_{\lambda_1}=\begin{pmatrix}1\\3\end{pmatrix}, v_{\lambda_2}=\begin{pmatrix}-3\\1\end{pmatrix}$
diagonal matrix $D=\begin{pmatrix}10&0\\0&-10\end{pmatrix}$
rotation matrix $B=\begin{pmatrix}\frac{1}{\sqrt{10}}&\frac{-3}{\sqrt{10}}\\\frac{3}{\sqrt{10}}&\frac{1}{\sqrt{10}}\end{pmatrix}$
Let $x=Bx'$
$\Rightarrow x^TAx+u^Tx=x'^TB^TABx'+u^TBx'=x'^TDx'+u^TBx'$
Calculating $u'^T=u^TB$
$u'^T=2(\begin{pmatrix}-3&-9\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{10}}&\frac{-3}{\sqrt{10}}\\\frac{3}{\sqrt{10}}&\frac{1}{\sqrt{10})}\end{pmatrix}
)=\begin{pmatrix}-6\sqrt{10}&0\end{pmatrix}$
So the transformed quadric is:
$10x'^2-6\sqrt{10}x'-10y'^2+8=0$
I know, that the solution must be $10x^2-10y^2=1$, but completing the square didn't get me there... Is my work so far correct? Where is my mistake? Can somebody please help me?
| Well the second equation
$$ 5 x^2 - 8 x y + 5 y^2 + 2 x + 2 y + 1 = 0 $$
can be written as
$$
5 \Big( \big[ x - x_o \big] + x_o \Big)^2
- 8 \Big( \big[ x - x_o \big] + x_o \Big) \Big( \big[ y - y_o \big]
+ y_o \Big) + 5 \Big( \big[ y - y_o \big] + y_o \Big)^2\\
+ 2 \Big( \big[ x - x_o \big] + x_o \Big)
+ 2 \Big( \big[ y - y_o \big] + y_o \Big) + 1 = 0
$$
so
$$
5 \big[ x - x_o \big]^2
- 8 \big[ x - x_o \big] \big[ y - y_o \big]
+ 5 \big[ y - y_o \big]\\
+ \Big( 10 x_o - 8 y_o + 2 \Big) \big[ x - x_o \big]
+ \Big( 10 y_o - 8 x_o + 2 \Big) \big[ y - y_o \big]\\
+ 5 x_o^2 - 8 x_o y_o + 5 y_o^2 + 2 x_o + 2 y_o + 1 = 0
$$
So $x_o = -1$ and $y_o = -1$ gives
$$
5 \big[ x + 1 \big]^2
- 8 \big[ x + 1 \big] \big[ y + 1 \big]
+ 5 \big[ y + 1 \big]^2 = 1\\
$$
This can be written as
$$
9 \big[ x + 1 \big]^2
- 18 \big[ x + 1 \big] \big[ y + 1 \big]
+ 9 \big[ y + 1 \big]^2\\
+ \big[ x + 1 \big]^2
+ 2 \big[ x + 1 \big] \big[ y + 1 \big]
+ \big[ y + 1 \big]^2 = 2\\
$$
so
$$
9 \big( x - y \big)^2 + \big( x + y + 2 \big)^2 = 2
$$
which is an ellipse with the axis $x-y=0$ and $x+y+2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate the following indefinite integral? $\int\frac{1}{x(x^2-1)}dx.$ I need the step by step solution of this integral
please help me!
I can't solve it!
$$\int\frac{1}{x(x^2-1)}dx.$$
| We use partial fraction decomposition:
$$\int\frac{1}{x(x^2-1)}dx = \int \frac 1{x(x-1)(x+1)}\,dx = \int \left(\frac A{x} + \frac{B}{x - 1} + \frac C{x+1}\right)\,dx$$
Solving for $A, B, C$:
$$A(x-1)(x+1) + Bx(x+1) + Cx(x-1) = 1$$
When $x = 1 \implies 2B = 1 \implies B = \frac 12$
$x = -1 \implies 2C = 1 \iff C = \frac 12$
$x = 0 \implies -A = 1 \iff A = -1$.
That gives us: $$\int \left(\frac {-1}{x} + \frac{1}{2(x - 1)} + \frac 1{2(x+1)}\right)\,dx$$
Now use the fact that $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/839510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Find the last two digits of $9^{{9}^{9}}$ I have to find the last two decimal digits of the number $9^{{9}^{9}}$.
That's what I did:
$$m=100 , \phi(m)=40, a=9$$
$$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$
$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \cdot 9^2 \cdot 9} \equiv 9^{81 \cdot 81 \cdot 81 \cdot 81 \cdot 9} \equiv 9^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv (9^{(40+40+1)})^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv 9^9 \equiv 9^4 \cdot 9^4 \cdot 9 \equiv 6561 \cdot 6561 \cdot 9 \equiv 3721 \cdot 9 \\ \equiv 21 \cdot 9 \equiv 89 \pmod{100}$$
So,the last two digits are $8 \text{ and } 9$.
$$$$But,is there also an other way two calculate the last two digits of $9^{{9}^{9}}$ or is the above the only one?
| Let $k$ be the order of $9$ mod $100$. Then
$$1\equiv 9^k=(10-1)^k \equiv 10\binom{k}{1}\cdot 10(-1)^{k-1}+(-1)^k\pmod{100}$$
It implies that $1=\pm (10k-1)\pmod{100}$. The minimal integer which satisfies this condition is $k=10$. It follows that $$9^{9^9} \equiv 9^9 \equiv 9^{-1}\equiv 89\pmod{100}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Solve for x if $z$ is a complex number such that $z^2+z+1=0$ I was given a task to solve this equation for $x$:
$$\frac{x-1}{x+1}=z\frac{1+i}{1-i}$$
for a complex number $z$ such that $z^2+z+1=0$.
Solving this for $x$ is trivial but simplifying solution using the given condition is what's bothering me. Thanks ;)
| You said that solving for $x$ was trivial, so I imagine you were able to get to:
$$x=\frac{1+zi}{1-zi}$$
This is equal to:
$$x=\frac{\left(1+zi\right)^2}{(1-zi)(1+zi)}=\frac{1+2zi-z^2}{1+z^2}$$
Since $1+z+z^2=0$ we can conlude that $1+z^2=-z$ and also that $1=-z-z^2$. We substitute that into the expression to get:
$$x=\frac{(-z-z^2)+2zi-z^2}{-z}=\frac{-z-2z^2+2zi}{-z}=1+2z-2i$$
$$x=1+2z-2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/843719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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analytical solution for linear 1st order PDE using laplace and seperation of variables I am looking for the solution of the following pde:
$\frac{\partial y(x,t)}{\partial t} = a* \frac{\partial y(z,t)}{\partial x} + b* y(x,t) + c$
and need help with the boundary and initial conditions:
$y(x=0,t)=0 $ bzw $y(x=0,t)=const. $ and $y(x,t=0)=f(x)$
a, b are negative constants and c is a positive constant.
I used Laplace Transform to derive a solution as follows:
$L^{-1}(\frac{\partial y(x,t)}{\partial t}) = L^{-1}(a \frac{\partial y(z,t)}{\partial x} + b y(x,t) + c) $
$s*Y(x,s) - y(x,0)=a* \frac{\partial Y(z,s)}{\partial x} +b* Y(x,s) + \frac{c}{s} $
$ \frac{\partial Y(z,s)}{\partial x} + \frac{b-s}{a} * Y(x,s)= -\frac{1}{a} * (y(x,0)+\frac{c}{s})$
which results in
$\int d(exp(\frac{b-s}{a}*x)*Y(x,s)) =\int -\frac{1}{a} * (y(x,0)+\frac{c}{s}) dx$
with $F(x) = \int f(x) dx= \int y(x,0)dx$ and $y(x=0,s)=0$ this results in
$Y(x,s)=- \frac{1}{a}*exp(\frac{s-b}{a}*x)*(F(x)-F(0)+\frac{c}{s}*x)$
Transforming this result back into time domain gives
$y(x,t) = -\frac{1}{a}*exp(\frac{-b}{a}*x)*((F(x)-F(0))* \delta (t+\frac{x}{a}) +c*x* H(t+\frac{x}{a})) $
with $\delta$ beeing the dirac delta function and $H$ beeing the heaviside step function.
This analytical equation cannot reproduce my numeric solution, in my opinion because the initial conditions are only multiplied with a dirac impulse, so is the transformation correct?
I also looked at the solution seperating the variables as shown in Analytical Solution of a PDE
$y(x,t)=C*exp(kt) * exp (\frac{k-b}{a}*x) - \frac{c}{b} $, however when I use the boundary condition, this results in a time independend solution, since the only way for the solution to be $0$ at $x=0$ is $y(x,t)=\frac{c}{b}*(exp(\frac{k-b}{a}*x)-1)$
Am I overseeing something?
Thanks for your help
| The solution is
$$ y(x,t) = \left( e^{-\frac{b}{a}x}y_l + e^{-\frac{b}{a}x}\frac{c}{b}\left( 1 - e^{b(t+\frac{x}{a})} \right)\right)\sigma\left(t + \frac{x}{a}\right) - \frac{c}{b}\left(1-e^{bt}\right) - \mathcal{L}^{-1}\{ F(x,s) \}, $$
where
$$F(x,s) = \frac{1}{as}e^{\frac{s-b}{a}x} \int\limits_0^x e^{\frac{b-s}{a}\xi}f(\xi) \text{d}\xi, $$
with $y_l := y(0,t)$ and $f(x) := y(x,0)$ the boundary and initial condition, respectively.
To verify this for $f(x)\equiv 0$ note that $y(x,t)$ satisfies the boundary condition
$$y(0,t) = \left( y_l + \frac{c}{b}\left( 1 - e^{bt} \right) \right) \sigma(t) - \frac{c}{b}\left( 1 - e^{bt} \right) \sigma(t) = y_l \sigma(t),$$
and the initial condition
$$ y(x,0) = \left( e^{-\frac{b}{a}x}y_l + e^{-\frac{b}{a}x}\frac{c}{b}\left( 1-e^{\frac{b}{a}x} \right) \right)\sigma\left(\frac{x}{a}\right) - \frac{c}{b}\left( 1 - e^{b\cdot 0}\right) = 0 \equiv f(x).$$
Since $\frac{x}{a} < 0$, we have $\sigma\left(\frac{x}{a}\right) = 0$.
To show that $y(x,t)$ satisfies the pde, we derive over $t$ and $x$
$$
\begin{align}
y_t(x,t) = & \left( e^{-\frac{b}{a}x}y_l + e^{-\frac{b}{a}x}\frac{c}{b}\left( 1 - e^{b\left( t + \frac{x}{a} \right)} \right) \right) \delta\left( t + \frac{x}{a} \right) \\
& - e^{-\frac{b}{a}x} \frac{c}{b} e^{b+\left( t + \frac{x}{a} \right)}b \sigma\left( t + \frac{x}{a} + ce^{bt} \right), \\
y_x(x,t) = & \left( -\frac{b}{a}e^{-\frac{b}{a}x}y_l + \frac{c}{b}\left( -\frac{b}{a}e^{-\frac{b}{a}x} \left( 1 - e^{b\left( t + \frac{x}{a} \right)} \right) + e^{-\frac{b}{a}x} \left( -e^{b\left( t + \frac{x}{a} \right)} \right) \right) \right) \sigma\left( t + \frac{x}{a} \right) \\
& + \left( e^{-\frac{b}{a}x} y_l + e^{-\frac{b}{a}x}\frac{c}{b} \left( 1 - e^{b\left( t + \frac{x}{a} \right)} \right) \right) \delta\left( t + \frac{x}{a} \right) \frac{1}{a},
\end{align} $$
such that $y_t - ay_x - by - c = 0$ for $t < \left|\frac{x}{a}\right|$, $t > \left|\frac{x}{a}\right|$ and $t = \left|\frac{x}{a}\right|$.
How to get there:
The Laplace transform of the pde yields
$$ sY(x,s) - Y(x,0) = aY_x(x,s) + bY(x,s) + \frac{c}{s}, $$
or, equivalently,
$$ Y_x(x,s) = Y(x,s) \frac{s-b}{a} - \frac{1}{a} \left( Y(x,0) + \frac{c}{s} \right) $$
The general solution to this linear, inhomogeneous 1$^\text{st}$-order ode in $x$ is
$$ Y(x,s) = e^{\frac{s-b}{a}x}Y(0,s) - \frac{1}{a}\int\limits_0^x e^{\frac{s-b}{a}(x-\xi)}\left( Y(\xi,0) + \frac{c}{s} \right) \text{d}\xi. $$
With $Y(0,s) = \mathcal{L}\{ y(0,t) \} = \frac{y_l}{s}$ and $Y(x,0) = \mathcal{L}\{ y(x,0) \} = \frac{f(x)}{s}$, we arrive at the solution $Y(x,s)$ in the frequency domain:
$$ Y(x,s) = e^{s\frac{x}{a}} \left( \frac{1}{s}e^{-\frac{b}{a}x}y_l - \frac{c}{s(s-b)}e^{-\frac{b}{a}x} \right) + \frac{c}{s(s-b)} - F(x,s), $$
which can then be back-transformed by using the following correspondences:
$$ \frac{a}{s(s+a)} = \mathcal{L}\{(1-e^{-at})\} $$
$$ e^{-as}F(s) = \mathcal{L}\{ f(t-a) \}$$
$$ \frac{1}{s} = \mathcal{L}\{ \sigma(t) \} $$
$$ 1 = \mathcal{L}\{ \delta(t) \}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of infinite series $\sum_{n=0}^{\infty}=\frac{n^2}{4n^2-1}t^n$ I have this problem, finding infinite sum of this series:
$$\sum_{n=0}^{\infty}\frac{n^2}{4n^2-1}t^n$$
It should be done using derivatives and integrals, like for example:
$$\sum_{n=1}^{\infty}\frac{t^n}{n}=\sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1}=\sum_{n=0}^{\infty}\int_{0}^{t}s^nds=\int_{0}^{t}\sum_{n=0}^{\infty}s^nds=\int_{0}^{t}\frac{1}{1-s}ds=-ln(1-t)$$
I could think about doing this:
$$\sum_{n=0}^{\infty}\frac{n^2}{4n^2-1}t^n=\sum_{n=0}^{\infty}\frac{n^2}{(2n-1)(2n+1)}t^n=\frac{1}{2}\sum_{n=0}^{\infty}\frac{n^2}{(2n-1)}t^n-\frac{1}{2}\sum_{n=0}^{\infty}\frac{n^2}{(2n+1)}t^n=\ldots$$
but then again, I don't know how could I make it to the end.
Any help would be very appreciated. Thanks!
| The series
\begin{align}
S = \sum_{n=0}^{\infty} \frac{ n^{2} \ t^{n} }{ 4n^{2}-1}
\end{align}
can be reduced as follows.
\begin{align}
S &= \frac{1}{4} \sum_{n=0}^{\infty} \frac{(4n^{2}-1) + 1}{4n^{2}-1} \ t^{n} \\
&= \frac{1}{4} \sum_{n=0}^{\infty} t^{n} + \frac{1}{8} \sum_{n=0}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \ t^{n} \\
&= \frac{1}{4(1-t)} + \frac{1}{8} \left[ \left( \sqrt{t} + \frac{1}{\sqrt{t}} \right) \tanh^{-1}(\sqrt{t}) - 1 \right],
\end{align}
where
\begin{align}
\sum_{n=0}^{\infty} \frac{t^{n}}{2n-1} = \sqrt{t} \tanh^{-1}(\sqrt{t}) -1
\end{align}
and
\begin{align}
\sum_{n=0}^{\infty} \frac{t^{n}}{2n+1} = \frac{1}{\sqrt{t}} \tanh^{-1}(\sqrt{t})
\end{align}
was used.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that all real roots of the polynomial $P (x) = x^5 − 10x + 35$ are negative. I got this problem out of Andreescu's Putnam and Beyond. I solved it differently from the given solution and could not understand the solution. Can you explain what is happening in the last step of the solution?
Because P (x) has odd degree, it has a real zero r. If r > 0, then by the AM–GM
inequality
$P (r)$
$ = r^5 + 1 + 1 + 1 + 2^5 − 5 · 2 · r $
$≥ 0.$ (why?)
And the inequality is strict since $1 \neq 2$. Hence r < 0, as desired.
Also, here is my own (edit: incorrect) solution:
Let p be a positive root. Then $p^5-10p+35=0$
$10p > p^5 + 35 \implies 10 > p^4 + 35/p > 2 \sqrt{35p^3} > 10p \implies p < 1$
$10p > 35 \implies p > 3.5$
Contradiction!
| To see what can be done without calculus, we can use a little bit from the "theory of equations" to show that $ \ x^5 \ - \ 10x \ - \ 35 \ $ has either five, three, or one real roots (complex conjugate zeroes) and that either two or none of those are positive and just one is negative (Descartes' Law of Signs).
If we just work with $ \ x^5 \ - \ 10x \ = \ x \ (x^2 \ - \ 10^{1/2} ) \ (x^2 \ + \ 10^{1/2} ) $ , we have an even function with three real zeroes. From what we know about the form of such a curve ("negative tail" goes to negative infinity, "positive tail" to positive infinity), between $ \ x \ = \ -(10^{1/4}) \ $ and $ \ x \ = \ 0 \ $ lies the local maximum for this function and the local minimum lies between $ \ x \ = \ 0 \ $ and $ \ x \ = \ +(10^{1/4}) \ $ This latter value of $ \ x \ $ is about $ \ 3.2^{1/2} \ \approx \ 1.8 \ $ (since $ \ \sqrt{10} \ \approx \ 3.2 \ $ and $ \ \sqrt{3.24} \ = \ 1.8 \ $ ) .
We can now estimate the minimal value of the function, using values of $ \ x \ $ between $ \ 1 \ $ and $ \ 1.8 \ $ , by noting that
$$ P \ (1) \ = \ 1^5 \ - \ 10 \ = \ -9 \ \ , $$
$$ P \ ( \ \frac{3}{2} \ ) \ = \ \left( \frac{3}{2} \right)^5 \ - \ 10 \ \left( \frac{3}{2} \right) \ = \ \left( \frac{243}{32} \right) \ - \ 15 \ \sim \ 8 \ - \ 15 \ \sim \ -7 \ \ , $$
$$ P \ ( \ \frac{4}{3} \ ) \ = \ \left( \frac{4}{3} \right)^5 \ - \ 10 \ \left( \frac{4}{3} \right) \ = \ \left( \frac{1024}{243} \right) \ - \ \frac{40}{3} \ \sim \ 4 \ - \ 13 \ \sim \ -9 \ \ , $$
where small integer ratios are chosen to keep the calculations simple. It is clear that the local minimal value is probably not significantly different from $ \ -10 \ $ (in fact, it's about $ \ -9.51 \ $ ) .
Shifting the curve for $ \ x^5 \ - \ 10x \ $ "vertically" by 35 units then guarantees that $ \ P \ (0) \ $ and the local minimum of $ \ P(x) \ $ are positive. So only the single negative $ \ x-$ intercept remains. (The other four zeroes are in complex conjugate pairs.)
| {
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"url": "https://math.stackexchange.com/questions/849977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying polynomial fraction Working through an old book I got and am at this problem: Simplify:
$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$
The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answer. :/
| $$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4} = \frac{3(x^2 + x -2)}{2(x^2 + 3x + 2)}= \frac{3\require{cancel}\cancel{(x+2)}(x-1)}{2(x+1)\cancel{(x + 2)}}=\frac{3(x-1)}{2(x+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Newtonian potential at (0, 0, – a) I found this problem in the book Advanced Calculus, written by Friedman.
"Newtonian potential at (0, 0, – a) due to a mass with constant densinty $\sigma$ on the hemisphere S: $x^2 + y^2 + z^2 = a^2$, $z \geq 0$, is
$$\int\int_S \frac{\sigma}{[x^2+y^2+(z+a)^2]^{1/2}}dS$$
Compute U."
I wrote $z=\sqrt{a^2-x^2-y^2}$, and the hemisphere as $(x, y, \sqrt{a^2-x^2-y^2})$, then $dS=\sqrt{1+{z_x}^2+{z_y}^2}dxdy=\dfrac{a}{\sqrt{a^2-x^2-y^2}}dxdy$.
So, my initial integral became
$$\int_{0}^{a}\int_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} \frac{\sigma a}{\sqrt{a^2-x^2-y^2}[2a^2+2a\sqrt{a^2-x^2-y^2}]^{1/2}}dxdy$$
I don't even know how to start integrating this thing. So I decided to use polar coordinates, so I got:
$$\int_{0}^{2\pi}\int_{0}^{a} \frac{\sigma ar}{\sqrt{a^2-r^2}[2a^2+2a\sqrt{a^2-r^2}]^{1/2}}drd\theta,$$
which wasn't so much better...
If someone could tell me where I went wrong, or give a full solution, or just a hint, I'd be really grateful.
Thanks!
| The surface integral,
$$\Phi(0,0,-a)=\iint_{\Sigma}\frac{\sigma}{\left[x^2+y^2+(z+a)^2\right]^{1/2}}\mathrm{d}S,$$
is most easily done using spherical coordinates: $\langle x,y,z\rangle=\langle a\cos{\varphi}\sin{\theta}, a\sin{\varphi}\sin{\theta}, a\cos{\theta}\rangle$, where $\varphi\in[0,2\pi]$ and $\theta\in[0,\frac{\pi}{2}]$. With these coordinates,
$$\begin{align}
x^2+y^2+(z+a)^2&=a^2\cos^2{\varphi}\sin^2{\theta}+a^2\sin^2{\varphi}\sin^2{\theta}+(a\cos{\theta}+a)^2\\
&=a^2\sin^2{\theta}+a^2(\cos{\theta}+1)^2\\
&=a^2\sin^2{\theta}+a^2(\cos^2{\theta}+2\cos{\theta}+1)\\
&=a^2(\sin^2{\theta}+\cos^2{\theta}+2\cos{\theta}+1)\\
&=2a^2(\cos{\theta}+1),
\end{align}$$
which implies, $\left[x^2+y^2+(z+a)^2\right]^{1/2}=\sqrt{2}\,a\sqrt{\cos{\theta}+1}$.
The surface element on the surface of the hemisphere is $\mathrm{d}S=a^2\sin{\theta}\,\mathrm{d}\theta\mathrm{d}\varphi$, so putting it all together the surface integral is:
$$\begin{align}
\Phi(0,0,-a)&=\iint_{\Sigma}\frac{\sigma}{\left[x^2+y^2+(z+a)^2\right]^{1/2}}\mathrm{d}S\\
&=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\frac{\sigma a^2\sin{\theta}}{\sqrt{2}\,a\sqrt{\cos{\theta}+1}}\mathrm{d}\theta\mathrm{d}\varphi\\
&=\frac{a\sigma}{\sqrt{2}}\int_{0}^{2\pi}\mathrm{d}\varphi\int_{0}^{\frac{\pi}{2}}\frac{\sin{\theta}}{\sqrt{\cos{\theta}+1}}\mathrm{d}\theta\\
&=\sqrt{2}\,\pi a\sigma\int_{0}^{\frac{\pi}{2}}\frac{\sin{\theta}}{\sqrt{\cos{\theta}+1}}\mathrm{d}\theta\\
&=\sqrt{2}\,\pi a\sigma\left[2(\sqrt{2}-1)\right]\\
&=(4-2\sqrt{2})\pi a\sigma.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Exercise about linear independence A) Let $V$ a vector space over $\mathbb{R}$ and $T: V \rightarrow V$ a linear mapping, such that $T^n=0$ for a natural $n \geq 2$. If $x \in V$, then the set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly independent if and only if $T^{n-1}x \neq 0$.
B) Let $V$ a vector space over $\mathbb{R}$ and $n \in \mathbb{N}$ odd. If the vectors $x_1, \dots,x_n$are linearly independent, then the same stands also for the vectors $x_1+x_2, x_2+x_3, \dots, x_{n-1}+x_n, x_n+x_1$.
$$$$
I have done the following:
For $(A)$:
for the direction $\Rightarrow$
The set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly independent:
$c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \Rightarrow c_0=c_1=c_2= \dots=c_{n-1}=0$
$c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \overset{\cdot T}{\Rightarrow} \\ c_0Tx +c_1 T^2x +c_2 T^3x+ \dots+c_{n-1} T^{n}x=0 \Rightarrow \\ c_0Tx +c_1 T^2x +c_2 T^3x+ \dots+c_{n-2}T^{n-1} =0 \overset{\cdot T}{\Rightarrow} \\ c_0T^2x +c_1 T^3x +c_2 T^4x+ \dots+c_{n-2}T^{n} =0 \Rightarrow \\ c_0T^2x +c_1 T^3x +c_2 T^4x+ \dots+c_{n-3}T_{n-1} =0 \overset{\cdot T}{\Rightarrow} \\ c_0T^3x +c_1 T^4x +c_2 T^5x+ \dots+c_{n-3}T_{n} =0 \Rightarrow \\ c_0T^3x +c_1 T^4x +c_2 T^5x+ \dots+c_{n-4}T^{n-1}=0 \overset{\cdot T}{\Rightarrow} \dots \Rightarrow c_0T^{n-2}x+c_1T^{n-1}=0 \overset{\cdot T}{\Rightarrow} \\ c_0T^{n-1}x+c_1T^{n}=0 \Rightarrow \\ c_0T^{n-1}x=0 \overset{c_0=0}{\Rightarrow} T^{n-1}x \neq 0$
Is this correct??
for the direction $\Leftarrow$
Knowing that $T^{n-1}x \neq 0$ we have to show that
$c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \Rightarrow c_1=c_2=\dots c_{n-1}=0$, right? But how could I do that??
For $(B)$:
$x_1, \dots,x_n$are linearly independent:
$c_1 x_1+c_2x_2+ \dots c_nx_n=0 \Rightarrow c_1=c_2= \dots =c_n=0 \ \ \ (*)$
$a_1(x_1+x_2)+a_2(x_2+x_3)+\dots+a_{n-1}(x_{n-1}+x_n)+a_n(x_n+x_1)=0 \Rightarrow \\ (a_1+a_n)x_1+(a_1+a_2)x_2+\dots(a_{n-1}+a_n)x_n=0 \overset{(*)}{\Rightarrow} \\ a_1+a_n=a_1+a_2= \dots =a_{n-1}+a_n=0$
Can we conclude from here that $a_1=a_2= \dots =a_{n-1}=a_n=0$ ??
| To prove (A), it is clear that if $T^{n-1} x = 0$, then the set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly dependent.
Conversely, suppose the set is linearly dependent, and let
$$
c_{0} x + c_{1} Tx + \dots + c_{n-1} T^{n-1} x = 0\tag{eq}
$$
be a relation where not all coefficients are zero.
If $c_{0} = c_{1} = \dots = c_{k-1} = 0$, and $c_{k} \ne 0$, apply $T^{n-1-k}$ to (eq) to get
$$
c_{k} T^{n-1} x = 0,
$$
so that $T^{n-1} x = 0$ as $c_{k} \ne 0$.
As to (B), note that if
$$
a_{1} (x_{1} + x_{2}) + a_{2} (x_{2} + x_{3}) + \dots + a_{n} (x_{n} + x_{1}) = 0,
$$
then
$$
(a_{n} + a_{1}) x_{1} + (a_{1} + a_{2} ) x_{2} + \dots + (a_{n-1} + a_{n}) x_{n} = 0,
$$
so that
$$
a_{n} + a_{1} = a_{1} + a_{2} = \dots = a_{n-1} + a_{n} = 0.\tag{eq2}
$$
Now note that since $n$ is odd
$$
0 = (a_{1} + a_{2}) - (a_{2} + a_{3}) + \dots - (a_{n-1} + a_{n}) + (a_{n} + a_{1}) = 2 a_{1},
$$
so $a_{1} = 0$ and thus, by (eq2), all $a_{i}$ are zero.
| {
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} |
Calculation of Trigonometric Limit with Summation. If $\displaystyle f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$ Then value of $f(x)$ is
$\bf{My\; Try::}$ Let $\displaystyle \left(\frac{x}{2^{r+1}}\right)=y$. So Trigonometric Expression is
$\displaystyle \frac{\tan y\cdot \left(1+\tan^2 y\right)}{\left(1-\tan^2 y\right)}=\frac{\tan y}{\cos 2y}=\frac{\sin y}{\cos 2y \cdot \cos y} = \frac{1}{2}\left\{\frac{\sin (2y-y)}{\cos 2y \cdot \cos y}\right\} = \frac{1}{2}\left\{\tan (2y)-\tan (y)\right\}$
Now How Can I solve after that
Help me
Thanks
| Continuing on from your answer,
$$\eqalign{\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}
&=\frac{1}{2}\sum_{r=0}^{n}\Bigl(\tan\Bigl(\frac{x}{2^r}\Bigr)-\tan\Bigl(\frac{x}{2^{r+1}}\Bigr)\Bigr)\cr}$$
which is a telescoping sum, giving
$$\frac{1}{2}\left(\tan x-\tan\Bigl(\frac{x}{2}\Bigr)+\tan\Bigl(\frac{x}{2}\Bigr)-\tan\Bigl(\frac{x}{2^2}\Bigr)+\cdots-\tan\Bigl(\frac{x}{2^{n+1}}\Bigr)\right)\ .$$
The sum is
$$\frac{1}{2}\left(\tan x-\tan\Bigl(\frac{x}{2^{n+1}}\Bigr)\right)$$
which tends to $\frac{1}{2}\tan x$ as $n\to\infty$.
| {
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"source": "stackexchange",
"question_score": "1",
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How to solve this linear equation? which has an x on each side I have made this equation.
$5x + 8 = 10x + \dfrac{3}{6}$
And I have achieved this result:
$x = 9$
Is my result correct?
I have already posted two other questions related to this topic, I'm a programmer and am learning Math out of my interest, this is not homework.
My steps taken:
10x - 5x = 48 - 3
5x = 45
5x/5 = 45/5
x = 9
This is my steps which led to a wrong answer.
| A good tip would be to put all of the variables on the left hand side of the equation and all of the numbers on the right side equation.
$5x + 8 = 10x + \dfrac{3}{6}$
First we would reduce the fraction
$5x + 8 = 10x + \dfrac{1}{2}$
and then we would subtract $10x $ from both sides, so we will have
$ 5x-10x + 8 = \dfrac{1}{2}$
Then we would subtract $8$ from both sides and we will have
$ 5x-10x = \dfrac{1}{2} - 8$
We know that $ 5x - 10x = -5x$ but what about the right hand side of the equation? We need to multiply $8$ by $\frac{2}{2}$
$ 5x-10x = \dfrac{1}{2} - 8 \times \dfrac{2}{2}$
Now, we must simplify the terms.
$-5x = \dfrac{1-16}{2}$
$-5x = \dfrac{-15}{2}$
Then, we divide $-5$ from both sides and get
$ x = \frac{\dfrac{-15}{2}}{-5}$
Flipping the second fraction, we have
$ x = \dfrac{-15}{2}\frac{-1}{5}$
So the final answer is $x = \frac{-15}{10}$ which reduces to $ x = \frac{3}{2}$
Now we shall substitute $ x = \frac{3}{2}$ back into the original equation
$5 \times \dfrac{3}{2} + 8 = 10 \times \dfrac{3}{2} + \dfrac{1}{2}$
$ \dfrac{15}{2}+ 8 = \dfrac{30}{2} + \dfrac{1}{2}$
$ \dfrac{15}{2}+ 8 \times \dfrac{2}{2} = \dfrac{30}{2} + \dfrac{1}{2}$
$ \dfrac{15+16}{2} = \dfrac{30+1}{2}$
$ \dfrac{31}{2}=\dfrac{31}{2}$
Since the left hand side equals the right hand side, we have found our x.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Eliminate $\theta$ from the equations $\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$
Eliminate $\theta$ from the equations
$$\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$$
Ans: $m^2+m\cos\alpha-2=0$.
I tried using the following two identities:
$$\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$$
$$\sin(3\theta)=3\sin(\theta)-4\sin^3(\theta)$$
but these didn't help much. I am sure that this is a simple problem but I am unable to figure out the right approach to solve it. :(
Any help is appreciated. Thanks!
| We have, $$\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$$
Then, $$m=\frac{\sin\theta\cos(\alpha-3\theta)+\cos\theta\sin(\alpha-3\theta)}{\sin\theta\cos^3\theta+\cos\theta\sin^3\theta}=\frac{\sin(\alpha-2\theta)}{\sin\theta\cos\theta}$$
Expanding $\sin(\alpha-2\theta)$ and using $2\sin\theta\cos\theta=\sin2\theta$, we get, $$\frac m2=\sin\alpha\cot2\theta-\cos\alpha\tag1$$
Similarly, $$m=\frac{\cos(\alpha-2\theta)}{\cos^4\theta-\sin^4\theta}=\frac{\cos(\alpha-2\theta)}{\cos2\theta}$$
which gives, $$m=\cos\alpha+\sin\alpha\tan2\theta\tag2$$
From $(1)$ and $(2)$,
$$\left(\frac m2+\cos\alpha\right)\left(m-\cos\alpha\right)=\sin^2\alpha$$ $$\frac{m^2}2+\frac m2\cos\alpha-\cos^2\alpha=\sin^2\alpha$$ $$m^2+m\cos\alpha-2=0.$$
$\square$
| {
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Is this procedure for $5^{300} \bmod 11$ correct? I'm new to modular exponentiation. Is this procecdure correct?
$$5^{300} \bmod 11$$
$$5^{1} \bmod 11 = 5\\
5^{2} \bmod 11 = 3\\
5^{4} \bmod 11 = 3^2 \bmod 11 = 9\\
5^{8} \bmod 11 = 9^2\bmod 11 = 4\\
5^{16} \bmod 11 = 4^2 \bmod 11 = 5\\
5^{32} \bmod 11 = 5^2 \bmod 11 = 3$$
$$5^{300} = 3 + 3 + 3 + 3 +3 +3 + 3 + 3 +3 +4 + 9$$
| You could also notice $11$ is prime, so by Fermat's Little Theorem $ 5^{10}\equiv 1 \bmod 11$
from here we get $5^{300}=(5^{10})^{30}\equiv1^{30}\equiv 1 \bmod 11$
| {
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I need help solving these limits I've been struggling to solve the following limits for about an hour. I've tried using conjugates as well as common factors, but it gets me nowhere. Wolfram Alpha does not provide steps for these limits. I could really use some help.
(x greater than 1) (SOLVED)
(SOLVED)
Thanks for taking your time!
| For question 1: You can factorize $x^2+2x-3 = (x+3)(x-1)$ to get $\frac{1}{\sqrt{x-1}}(\frac{1}{2}-\frac{1}{\sqrt{x+3}})$. After that, I'd suggest calculating Taylor expansions at x=1 for $\frac{1}{\sqrt{x-1}}$ and $\frac{1}{2}-\frac{1}{\sqrt{x+3}}$ (you can do this with Wolfram Alpha). Btw, the limit is zero, I think (used a calculator entering x slightly >1).
I'd post this as a comment, if possible, but I don't have enough reputation, so sorry that the answer is incomplete :(
EDIT:
For the second question, I have this idea:
First (inspired from André Nicolas' comment), we multiply by $\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$, so that we have
$$\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$$
Next, we devide numerator and denominator by $\sqrt{x}$:
$$=\frac{\sqrt{\frac{x+\sqrt{x}}{x}}}{\sqrt{\frac{x+\sqrt{x+\sqrt{x}}}{x}}+1}
=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{x+\sqrt{x}}{x^2}}}+1}
=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}^3}}}+1}$$
Now we can see that $\frac{1}{x}$, $\frac{1}{\sqrt{x}}$ and $\frac{1}{\sqrt{x}^3}$ will all become zero for $x\rightarrow\infty$, so what remains is $$\frac{\sqrt{1}}{\sqrt{1}+1}=\frac{1}{1+1} = 0.5$$
| {
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If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?
My Attempt:
By factoring out powers of $2$, we can write
$$
\begin{align}
K &= 2^{50}\times \left(\frac{1}
{1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\
&= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\
&= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99}
\end{align}
$$
How can I solve for $K$ from here?
| I'm sure this answer is not rigorous, but perhaps someone can make it so.
By manipulating factorials we get
$$\frac{1}{K}=\frac{1}{2^{100}}\binom{100}{50}=P(X\,{=}\,50)\ ,$$
where $X$ is a binomial random variable with $n=100$ and $p=\frac{1}{2}$. Approximating $X$ by a normal random variable $Y$ in the usual way, we have $Y\sim N(50,5^2)$ and so
$$\frac{1}{K}\approx P(49.5\,{<}\,Y\,{<}\,50.5)=P(-0.1\,{<}\,Z\,{<}\,0.1)
\approx0.0796\ .$$
Hence
$$K\approx12.56$$
and so $\lfloor K\rfloor=12$.
| {
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Commutant Algebra of Matrix Representation I am currently working on Bruce Sagan's The Symmetric Group.
In the following example they show that for a representation that contains 2 different subrepresentations the commutative algebra Com$X$ has the form
\begin{equation} \text{Com}X = \left\{ c_1I_{d_1} \oplus c_2I_{d_2}: c_1, c_2 \in \mathbb{C} \right\}. \end{equation}
Suppose that $X$ is a matrix representation such that
\begin{equation} X = \begin{pmatrix}X^{(1)}&0\\0&X^{(2)}\end{pmatrix} = X^{(1)} \oplus X^{(2)}, \end{equation}
where $X^{(1)}, X^{(2)}$ are inequivalent and irreducible of degrees $d_1, d_2,$ respectively. What does Com$X$ look like?
Suppose that
\begin{equation} T = \begin{pmatrix}T_{1,1}&T_{1,2}\\T_{2,1}&T_{2,2}\end{pmatrix} \end{equation}
is a matrix partitioned in the same way as $X$. If $TX = XT$, then we can multiply out each side to obtain
\begin{equation} \begin{pmatrix}T_{1,1}X^{(1)}&T_{1,2}X^{(2)}\\T_{2,1}X^{(1)}&T_{2,2}X^{(2)}\end{pmatrix} = \begin{pmatrix}X^{(1)}T_{1,1}&X^{(1)}T_{1,2}\\X^{(2)}T_{2,1}&X^{(2)}T_{2,2}\end{pmatrix}. \end{equation}
Equating corresponding blocks we get
\begin{equation} T_{1,1}X^{(1)} = X^{(1)}T_{1,1}, \end{equation}
\begin{equation} T_{1,2}X^{(2)} = X^{(1)}T_{1,2}, \end{equation}
\begin{equation} T_{2,1}X^{(1)} = X^{(2)}T_{2,1}, \end{equation}
\begin{equation} T_{2,2}X^{(2)} = X^{(2)}T_{2,2}. \end{equation}
Using Corollaries 1.6.6 and 1.6.8 along with the fact that $X^{(1)}$ and $X^{(2)}$ are inequivalent, these equations can be solved to yield
\begin{equation} T_{1,1} = c_1I_{d_1}, T_{1,2} = T_{2,1} = 0, T_{2,2} = c_2I_{d_2}, \end{equation}
where $c_1, c_2 \in \mathbb{C}$ and $I_{d_1}, I_{d_2}$ are identity matrices of degrees $d_1, d_2$. Thus
\begin{equation} T = \begin{pmatrix} c_1I_{d_1}&0\\0&c_2I_{d_2} \end{pmatrix}. \end{equation}
We have shown that when $X = X^{(1)} \oplus X^{(2)}$ with $X^{(1)} \neq X^{(2)}$ and irreducible, then
\begin{equation} \text{Com}X = \left\{ c_1I_{d_1} \oplus c_2I_{d_2}: c_1, c_2 \in \mathbb{C} \right\}, \end{equation}
where $d_1 = \text{deg}X^{(1)}, d_2 = \text{deg}X^{(2)}$.
Corollary 1.6.6 is the matrix version of Schur's Lemma and Corollary 1.6.8 says: "Let $X$ be an irreducible matrix representation of $G$ over the complex numbers. Then the only matrices $T$ that commute with $X(g)$ for all $g \in G$ are those of the form $T = cI$ - i.e., scalar multiples of the identity matrix."
There are three things that I do not understand in this example:
$1$. How can we assume that $T$ has the block form
\begin{equation} T = \begin{pmatrix}T_{1,1}&T_{1,2}\\T_{2,1}&T_{2,2}\end{pmatrix}? \end{equation}
Would it not be possible that $T$ does not have such a block form?
$2$. How do we know that $T_{1,2} = T_{2,1} = 0$? Schur's Lemma says that either $T_{1,2}, T_{2,1}$ are invertible or they are zero matrices. How do we know that they are not invertible?
Thank you very much for your help!
| Write $V=V_1\oplus V_2$, a direct sum of two irreducible representations of $G$ with dimensions $d_1,d_2$, respectively. We are showing the intertwiners in ${\rm End}_G(V)$ look like $c_1I_{V_1}\oplus c_2 I_{V_2}$.
(1) Every $(n+m)\times(n+m)$ matrix can be written as a block matrix $(\begin{smallmatrix} A & B \\ C & D \end{smallmatrix})$ with the appropriate matrices $A\in M_{n\times n}$, $B\in M_{n\times m}$, $C\in M_{m\times n}$, $D\in M_{m\times m}$. This is actually trivial. What do you think the matrices $A,B,C,D$ are? They're already right there in the matrix already! We just need to put parentheses (or brackets if that's your thing) around them. For instance:
$$\quad \begin{pmatrix} \color{Blue}{1} & \color{Blue}{2} & \color{Blue}{3} & \color{Purple}{4} & \color{Purple}{5} \\ \color{Blue}{6} & \color{Blue}{7} & \color{Blue}{8} & \color{Purple}{9} & \color{Purple}{10} \\ \color{Blue}{11} & \color{Blue}{12} & \color{Blue}{13} & \color{Purple}{14} & \color{Purple}{15} \\ \color{Magenta}{16} & \color{Magenta}{17} & \color{Magenta}{18} & \color{Red}{19} & \color{Red}{20} \\ \color{Magenta}{21} & \color{Magenta}{22} & \color{Magenta}{23} & \color{Red}{24} & \color{Red}{25}\end{pmatrix}\longrightarrow \begin{pmatrix} \color{Blue}{\begin{pmatrix} 1 & 2 & 3 \\ 6 & 7 & 8 \\ 11 & 12 & 13\end{pmatrix}} & \color{Purple}{\begin{pmatrix} 4 & 5 \\ 9 & 10 \\ 14 & 15\end{pmatrix}} \\ \color{Magenta}{\begin{pmatrix} 16 & 17 & 18 \\ 21 & 22 & 23 \end{pmatrix}} & \color{Red}{\begin{pmatrix}19 & 20 \\ 24 & 25\end{pmatrix}} \end{pmatrix}. $$
(2) We're not using Schur's lemma to prove $T_{1,2}$ and $T_{2,1}$ are zero. Instead, as the text explicitly says, we're using that fact that $V_1$ and $V_2$ are inequivalent. Look at the first equation, which reads as $T_{1,2}X^{(2)}=X^{(1)}T_{1,2}$. This tells us $T_{1,2}$ is a morphism $V_2\to V_1$ of representations. Can there be any nonzero morphisms of representations between inequivalent irreducible representations?
| {
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How to find orthogonal projection of vector on a subspace? Well, I have this subspace: $V = \operatorname{span}\left\{ \begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix},\begin{pmatrix}1 \\3 \\4\end{pmatrix}\right\}$
and the vector $v = \begin{pmatrix}9 \\0 \\0\end{pmatrix}$
How can I find the orthogonal projection of $v$ on $V$?
This is what I did so far:
\begin{align}&P_v(v)=\langle v,v_1\rangle v_1+\langle v,v_2\rangle v_2 =\\=& \left\langle\begin{pmatrix}9 \\0 \\0\end{pmatrix},\begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix}\right\rangle\begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix}+\left<\begin{pmatrix}9 \\0 \\0\end{pmatrix},\begin{pmatrix}1 \\3 \\4\end{pmatrix}\right>\begin{pmatrix}1 \\3 \\4\end{pmatrix} = \begin{pmatrix}10 \\29 \\38\end{pmatrix}\end{align}
Is this the right method to compute this?
| That would be the correct method...if $v_1$ and $v_2$ were orthogonal and unit length. Unfortunately, they're not.
Three alternatives:
*
*Compute $w = v_1 \times v_2$, and the projection of $v$ onto $w$ -- call it $q$. Then compute $v - q$, which will be the desired projection.
*Orthgonalize $v_1$ and $v_2$ using the gram-schmidt process, and then apply your method.
*Write $q = av_1 + bv_2$ as the proposed projection vector. You then want $v - q$ to the orthogonal to both $v_1$ and $v_2$. This gives you two equations in the unknowns $a$ adn $b$, which you can solve.
| {
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Calculate $\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$ I have homework questions to calculate infinity sum, and when I write it into wolfram, it knows to calculate partial sum...
So... How can I calculate this:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$$
| Lets solve the problem generally;
$$\begin{array}{l}\sum\limits_{k = i}^\infty {\frac{1}{{\left( {k + a} \right)\left( {k + b} \right)}}} = \\\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = i}^n {\frac{1}{{\left( {k + a} \right)\left( {k + b} \right)}}} = \\\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = i}^n {\frac{1}{{b - a}}\left[ {\frac{1}{{k + a}} - \frac{1}{{k + b}}} \right]} = \\\frac{1}{{b - a}}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = i}^n {\left[ {\frac{1}{{k + a}} - \frac{1}{{k + b}}} \right]} = \\\frac{1}{{b - a}}\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{i + a}} - \frac{1}{{n + b}}} \right]\end{array}
$$
So the solution is:
$$\begin{array}{l}\sum\limits_{k = i}^\infty {\frac{1}{{\left( {k + a} \right)\left( {k + b} \right)}}} = \frac{1}{{b - a}}\frac{1}{{i + a}}\end{array}
$$
And going back to your questions:
$$\begin{array}{l}\sum\limits_{k = 1}^\infty {\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}} = \frac{1}{{2 - 1}}\frac{1}{{1 + 1}} = \frac{1}{2}\end{array}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tridiagonal Symmetric Matrix Could anyone help me to find the determinant of a $N\times N$ tri-diagonal symmetric matrix, named "$A[i,j]$" with $i,j \le N$, that has all the elements in the super-diagonal and sub-diagonal equal, and in the main diagonal all the elements are also equal, except for $A[1,N]$ and $A[N,N]$, that are different of others but are equal to each other.
| For a general $N \times N$ tridiagonal matrix $A$, define:
$$d_i = A_{i,i}, \, i=1 \dots N \\
u_i = A_{i,i+1}, \, i=1 \dots N-1 \\
l_i = A_{i+1,i}, \, i=1 \dots N-1 \\
f_0 = 1 \\
f_1 = d_1 \\
f_n = d_n f_{n-1} - u_{n-1} l_{n-1} f_{n-2}, \, n = 2 \dots N$$
Then $\text{det}(A)=f_N$.
In your situation, $u_i \equiv s$, $l_i \equiv s$, and $d_i \equiv d$ for $i=2,\dots,N-1$ and $d_1=d_N=c$. So you solve the constant coefficient recursion relation
$$f_0 = 1 \\
f_1 = c \\
f_n = d f_{n-1} - s^2 f_{n-2}, \, n \geq 2$$
up to $n=N-1$, then finally
$$f_N = c f_{N-1} - s^2 f_{N-2}$$
Solving these, we get two cases. If $d^2 \neq 4s^2$:
$$\text{det}(A) = c \left ( \alpha \left ( \frac{d + \sqrt{d^2-4s^2}}{2} \right )^{N-1} + \beta \left ( \frac{d - \sqrt{d^2-4s^2}}{2} \right )^{N-1} \right ) - s^2 \left ( \alpha \left ( \frac{d + \sqrt{d^2-4s^2}}{2} \right )^{N-2} + \beta \left ( \frac{d - \sqrt{d^2-4s^2}}{2} \right )^{N-2} \right ) $$
If $d^2 = 4s^2$, then
$$\text{det}(A) = c \left ( \alpha \left ( \frac{d}{2} \right )^{N-1} + \beta (N-1) \left ( \frac{d}{2} \right )^{N-1} \right ) - s^2 \left ( \alpha \left ( \frac{d}{2} \right )^{N-2} + \beta (N-2) \left ( \frac{d}{2} \right )^{N-2} \right )$$
In both cases one finds $\alpha$ and $\beta$ from the conditions $f_0=1$, $f_1=c$.
This was taken essentially from http://en.wikipedia.org/wiki/Tridiagonal_matrix#Determinant.
| {
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Fastest way to integrate $\int_0^1 x^{2}\sqrt{x+x^2} \hspace{2mm}dx $ This integral looks simple, but it appears that its not so.
All Ideas are welcome, no Idea is bad, it may not work in this problem, but may be useful in some other case some other day ! :)
| Note that
$$x^2\sqrt{x+x^2}=\frac{x^3+x^4}{\sqrt{x(1+x)}}$$
So, let us look for a polynomial $P(x)=a x^3+bx^2+cx+d$ such that the derivative
$(P(x)\sqrt{x(1+x)})^\prime$ is as close as we can to this function. An easy calculation shows that
$$
\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{4 a x^4+(\frac{7 a }{2}+3 b)
x^3+(\frac{5 b }{2}+2 c)
x^2+(\frac{3 c }{2}+d )x+\frac{d}{2}}{\sqrt{x(1+x)}}
$$
So, choosing $a=\frac{1}{4}$, $b=\frac{1}{24}$, $c=-\frac{5}{96}$ and $d=\frac{5}{64}$ we see that
$$
\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{ x^4+
x^3}{\sqrt{x(1+x)}}+\frac{5}{64}\cdot\frac{1}{2\sqrt{x(1+x)}}
$$
This reduces the considered integral to a simple one:
$$\int_0^1x^2\sqrt{x(1+x)}dx=\Big[P(x)\sqrt{x(1+x)}\Big]_0^1-\frac{5}{64}\int_0^1\frac{dx}{2\sqrt{x(1+x)}}$$
The last integral is easy since $\log(\sqrt{x}+\sqrt{1+x})$ is a primitive of the integrand.
Thus
$$\int_0^1x^2\sqrt{x(1+x)}dx=\Big[P(x)\sqrt{x(1+x)}-\frac{5}{64}\log(\sqrt{x}+\sqrt{1+x})\Big]_0^1$$
Finally,
$$\int_0^1x^2\sqrt{x(1+x)}dx=
\frac{61}{96 \sqrt{2}}-\frac{5}{64}
\log \left(1+\sqrt{2}\right).
$$
| {
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The number of distinct multiples of composites greater than $n$ that can be factored into two naturals less than or equal to $n$ Given a list of composites between $n$ and $\lfloor \frac{n^2}{2} \rfloor$:
What would be the most efficient way to count, for each composite, the number of its distinct multiples that can be factored into two naturals less than or equal to $n$?
For example, $n = 6$:
For the composite 8, there are 3:
$8 \times 1$, factors $4 \times 2$
$8 \times 2$, factors $4 \times 4$
$8 \times 3$, factors $4 \times 6$
The multiple $8 \times 4$ cannot be factored into two naturals less than or equal to 6.
For the composite 9, there are 3:
$9 \times 1$, factors $3 \times 3$
$9 \times 2$, factors $6 \times 3$
$9 \times 4$, factors $6 \times 6$
The multiple $9 \times 3$ cannot be factored into two naturals less than or equal to 6.
| Let the composite be $m$. Factor $m$, then find all factorizations with no factor greater than $n$. For each one, see how much room you have. I'll take the larger example $n=25, m=144=2^43^2$ From the prime factorization we can find $144=6\cdot 24=8 \cdot 18 = 9 \cdot 16 = 12 \cdot 12$ are possible, while $1 \cdot 144, 2 \cdot 72, 3 \cdot 48, 4 \cdot 36$ have a factor too large. Now find what you can multiply each factor by while staying below $25$. The $6$ can be multiplied by $1,2,3,4$, while the $24$ cannot be multiplied by anything. You can get this by computing $\lfloor \frac {25}6 \rfloor$ In this case, that is all you can get. Both of the $12$'s can be multiplied by $1,2$, but that doesn't give anything new, so there you are.
| {
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How find this P(x) if $ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $ Let $m \neq 0 $ be an integer. Find all polynomials $P(x) $ with real coefficients such that
$$ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $$
This problem is IMO Shortlist 2013
let $$P(x)=\sum_{i=0}^{n}a_{i}x^i,a_{i}\in R$$
then
$$\sum_{i=0}^{n}[a_{i}x^3(x+1)^i-mx^2(x+1)^i+(x+1)^i]+\sum_{i=0}^{n}[x^3(x-1)^i+mx^2(x-1)^i+(x-1)^i]=2\sum_{i=0}^{n}[x^{3+i}-mx^{i+1}+x^i]$$
then I can't
| Note that, $P(x)=cx$ is a trivial solution, we will prove that it is the only one.
Suppose that degree of this polynomial is $>1$.
From your work, comparing coefficients of $x^{n+1}$ gives , $n=2m$, and so $m>0$.
Now,after doing some manipulations(multiply by $x$ and then regroup) , we get,
$$ (x^3-mx^2+1)F(x)= (x^3+mx^2+1)F(x-1) \cdots (1)$$
Where, $F(x) = xP(x+1)-(x+1)P(x)$ .
Comparing, degree of $F(x)$ with $P(x)$ we get that they both have same degree($n=2m$).
Consider $G(x)= x^3+mx^2+1$ Since it is cubic it has either $1$ or $3$ real roots. Now, $G'(x) = 3x^2+2mx$ . $G'(x)=0 \implies x = -2m/3, x=0$ So local extrema at $x=0$ or $x=-2m/3$ . $G(0)=1>0, G(-2m/3) = 1+\frac{4m^3}{27} $
So three real roots if $ m \le -2 $ or, one real root if $m \ge -1$
Let $H(x) = x^3-mx^2+1 $. Suppose that , $G(x)$ and $H(x)$ have unequal number of roots. This is impossible since, in $(1)$ $LHS/RHS$ will have one root more than $RHS/LHS$ which is absurd.
So $G(x)$ and $H(x)$ have same number of real roots. $m \le -2, m \ge 2 \implies ABSURD $. $m \ge -1, m \le 1 \implies m=1 \implies n=2 $(since m is $>0$)
Now, it is not hard to check that $n=2$ has no solutions.(Since you are attempting $IMOSL$ , i don't think it will be hard to prove this part)
| {
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Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative Hope someone can help on this inequality using nonanalytical method (i.e. simple elementary method leveraging basic inequalities are prefered).
Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative.
Here is what I already got.
First of all, one should notice equality holds not when $a=b=c$ as one might initially thought. Rather, the equality holds when $(a,b,c)$ is a permutation of $({1\over2},{1\over2},0)$.
Secondly and obviously, this is cyclical and homgeneous and hence we can apply the EMV theorem developed by a IMO golden medalist (reference here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1130901) and then the original inequality can be easily proved by assuming $$f(a,b,c)={\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\over {a+b+c}},$$and then prove $$f(1,1,1)\leq{3\over2},\\f(a,b,0)\leq{3\over2},\forall a,b\geq0.$$
But this kind of proof does not fit my appetite as it not only involves some additional theorem but also not very nice as the simple nice form of the question of itself.
So can people help on some elementary proof that might only taking use of the basic inequalities like AM-GM, Jensen's inequality, etc.
| Proof.
(i) if abc=0, W.L.O.G assume c=0, $\Longleftrightarrow a+b \geq 2\sqrt{ab}$, it is clearly true.
(ii) in the following, assume $abc > 0$.
Case 1: $a^2+b^2+c^2 \leq 2(ab+bc+ca)$
according to Cauchy-Schwarz inequality, we have:
$$LHS \leq \sqrt{3(a^2+b^2+c^2+ab+bc+ca)}$$
$$\Longleftrightarrow \sqrt{3(a^2+b^2+c^2+ab+bc+ca)} \leq {3\over2}(a+b+c)$$
$$\Longleftrightarrow 3(a^2+b^2+c^2+ab+bc+ca) \leq {9\over4}(a+b+c)^2$$
$$\Longleftrightarrow a^2+b^2+c^2 \leq 2(ab+bc+ca)$$
it is clearly true.
Case 2: $a^2+b^2+c^2 \geq 2(ab+bc+ca)$
W.L.O.G. assume c=max{a,b,c},
according to Cauchy-Schwarz inequality, $$\sqrt{a^2+bc} + \sqrt{b^2+ac} \leq \sqrt{2(a^2+b^2+ac+bc)}$$
and $$\sqrt{c^2+ab}=\sqrt{c^2+ab}-c+c=\frac{ab}{\sqrt{c^2+ab}+c} +c \leq \frac{ab}{2c} +c$$
$LHS \leq \sqrt{2(a^2+b^2+ac+bc)} + \frac{ab}{2c} +c \leq {3\over2}(a+b+c)$
$$\Longleftrightarrow \sqrt{2(a^2+b^2+ac+bc)} \leq {3\over2}(a+b+c) - c - \frac{ab}{2c}$$
$$\Longleftrightarrow \sqrt{2(a^2+b^2+ac+bc)} \leq {3\over2}(a+b) + {1\over2}c - \frac{ab}{2c}$$
$$\Longleftrightarrow 2(a^2+b^2+ac+bc) \leq ({3\over2}(a+b) + {1\over2}c - \frac{ab}{2c})^2$$
$$\Longleftrightarrow \frac{a^2b^2-6a^2bc-6ab^2c+a^2c^2+16abc^2+b^2c^2-2ac^3-2bc^3+c^4}{4c^2} \geq 0$$
$$\Longleftrightarrow \frac{a^2b^2+abc(16c-6a-6b)+c^2(a^2+b^2+c^2-2ac-2bc)}{4c^2} \geq 0$$
it is also true.
| {
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Find the derivative of $y=\cos(x) - 2\sin(x),$ when the gradient is $1$ I need to find the smallest positive value of $x$ for which the gradient of the curve has value 1.
For this equation:
$$
y =\cos(x)-2\sin(x)
$$
The answer is 2.5c grad.
The following is my working out but I’ve stopped at the factor formula, how can I change the equation having just $\sin()$ functions? Or is there another neat way?
$$
y = \cos(x)-2\sin(x) \\
\frac{d}{dx} \cos(x)-2\sin(x) = 1 \\
\frac{dy}{dx} =-\sin(x)-2\cos(x) = 1 \\
\frac{dy}{dx} =\sin(x)+2\cos(x) = -1 \\
\frac{dy}{dx} = \sin^2(x)+2\cos(x)\sin(x)=-\sin(x)
$$
| $$\sin x + 2\cos x + 1 = 0\tag{1}$$ Now use the Weierstrass substitution $y = \tan\left(\frac x2\right).$ From this, it follows that $$\sin x = \frac {2y}{y^2 + 1},\;\cos x = \frac{1-y^2}{y^2 + 1}\tag{2}$$
Then, substituting the values from $(2)$ into $(1)$ gives us $$ \frac {2y}{y^2 + 1} + \frac{2(1-y^2)}{y^2 + 1} + 1=0$$
Now, solving for $y$, $$\begin{align} &\quad \frac {2y}{y^2 + 1} + \frac{2(1-y^2)}{y^2 + 1} + 1=0 \\ \\
&\iff \dfrac{2y+2(1-y^2)+ y^2 + 1}{y^2 + 1} = 0\\ \\
&\iff \frac{y^2 - 2y - 3}{y^2 +1} = 0\\ \\
&\iff y^2 - 2y - 3 = (y - 3)(y+1) = 0\\ \\
& \iff y = 3\;\text{ or }\;y = -1
\end{align}$$
Now simply recall that $\;y = \tan\left(\frac x2\right) \implies \tan^{-1}(y) = \frac x2 \iff x = 2\tan^{-1}(y).$
So use $x = 2\tan^{-1}(y)$ to solve for $x$. One of your two $y$ values will yield the desired conclusion.
| {
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$ \int \frac{1}{(x-a)(x+b)} dx $ Could you please explain how to integrate this integral:
$$ \int \frac{1}{(x-a)(x+b)} dx $$
| $$\int\frac1{(x-a)(x+b)}dx$$
Use partial fractions
$$\int\frac1{x(a+b)-a(a+b)}+\frac1{-x(a+b)-b(a+b)}dx$$
Integrate the sum by term
$$\int\frac1{x(a+b)-a(a+b)}dx+\int\frac1{-x(a+b)-b(a+b)}dx$$
Substitute $u=-x(a+b)-b(a+b)$ and $du=(-a-b)dx$
$$\int\frac1{x(a+b)-a(a+b)}dx + \frac1{-a-b}\int\frac1u du$$
Since $\int\frac1u du=\log u$
$$\int\frac1{x(a+b)-a(a+b)}dx + \frac{\log u}{-a-b}$$
Substitute $s=x(a+b)-a(a+b)$ and $ds=(a+b)dx$
$$\frac1{a+b}\int\frac1sds + \frac{\log u}{-a-b}$$
Since $\int\frac1s ds=\log s$
$$\frac{\log s}{a+b} + \frac{\log u}{-a-b}+C$$
Substitute back for $s$ and $u$
$$\frac{\log ((a+b)(x-a))}{a+b} + \frac{\log (-(a+b)(b+x))}{-a-b}+C$$
This is equivalent to
$$\frac{\log(x-a)-\log(b+x)}{a+b}$$
Assuming $x-a>0$ and $b+x>0$ this is the same as
$$\frac{\log\frac{x-a}{b+x}}{a+b}$$
| {
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Sum of series with triangular numbers Can you please tell me the sum of the seires
$ \frac {1}{10} + \frac {3}{100} + \frac {6}{1000} + \frac {10}{10000} + \frac {15}{100000} + \cdots $
where the numerator is the series of triangular numbers?
Is there a simple way to find the sum?
Thank you.
| I thought I might add another derivation (devised by me). This one is long and involves dissecting the sequence into its simplest terms.
$1/10 + 3/100 + 6/1000 + \ldots$
$= 1/10 + (1+2)/100 + (1+2+3)/1000 + \ldots$ (from the definition of
triangular numbers.)
$= 1/10 + 1/100 + 2/100 + 1/1000 + 2/1000 + 3/1000 + \ldots$
(by grouping terms with similar numerator together)
$= (1/10 + 1/100 + 1/1000 + \ldots) + (2/100 + 2/1000 + \ldots) + (3/1000 +
\ldots) + \ldots$ $= 1/9 + 2/90 + 3/900 + \ldots$
($1/9$ is a common factor)
$= 1/9 [ 1 + 2/10 + 3/100 + \ldots]$ $= 1/9 [ 1 + 1/10 + 1/10 + 1/100
+ 1/100 + 1/100 + \ldots ]$
(after rearranging the terms)
$= 1/9 [ 1 + (1/10 + 1/100 + 1/100 + \ldots) + (1/10 + 1/100 + 1/100 +
\ldots) + (1/100 + \ldots) + \ldots ] $ $= 1/9 [ 1 + 1/9 + (1/9 + 1/90
+ 1/900 + 1/900 + \ldots) ]$
(the terms between the parentheses represent a geometric series
whose sum is $10/81$)
$= 1/9 [ 1 + 1/9 + 10/81 ]$
$= 1/9 \times 100/81$
$= 100/729$
| {
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Homework - Resolve the recurrence relation What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$
| The best course of action for simple inhomogeneous recurrences is to make use of a smart "change of variables" (read: substitute with another recurrence relation) to turn it into a homogeneous recurrence. A nice observation here is to notice that $2^n$ is itself a recurrence relation, namely:
$$y_n=2y_{n-1},\,\,(y_0=1)$$
Let's use this to our advantage:
$$\begin{align}
a_n=a_{n-1}+2a_{n-2}+2^n &\Leftrightarrow 2^n=a_n-a_{n-1}-2a_{n-2} \\
&\Leftrightarrow 2\cdot2^{n-1}=a_n-a_{n-1}-2a_{n-2} \\
&\Leftrightarrow 2\cdot (a_{n-1}-a_{n-2}-2a_{n-3})=a_n-a_{n-1}-2a_{n-2} \\
&\Leftrightarrow a_n-3a_{n-1}+4a_{n-3}=0
\end{align}$$
Bam, homogeneous recurrence. Can apply the characteristic equation and finish it by yourself?
Edit: The characteristic equation for the last recurrence is $x^3-3x+4=0$, whose roots are $2,2$ and $-1$. It is well-known then, that given these roots, the solution is of the form:
$$a_n=(An+B)2^n+C(-1)^n$$
For some constants $A,B,C$. To find out what the constants are, we plug in values of $n$ for which the value of $a_n$ is known. The values of $a_0,a_1$ were given, but we need three to solve a linear system with three variables. We can easily calculate $a_2$: $$a_2=a_1+2a_0+2^2=8$$
Now we solve the system:
$$\begin{cases}
(A\cdot0+B)2^0+C(-1)^0=1 \\
(A\cdot1+B)2^1+C(-1)^1=2 \Rightarrow \\
(A\cdot2+B)2^2+C(-1)^2=8
\end{cases}$$
$$\Rightarrow \begin{cases}
B+C=1 \\
2(A+B)-C=2 \\
4(2A+B)+C=8
\end{cases}$$
Solving this will yield $A=\frac{2}{3}$,$B=\frac{5}{9}$ and $C=\frac{4}{9}$. Therefore, the closed form is:
$$a_n=\frac{(6n+5)2^n+4(-1)^n}{9}$$
| {
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Calculating the area For the two graphs $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x+9}{x+4} $, calculate the area which is confined by them;
Attempt to solve: Limits of the integral are $1$ and $-3$, so I took the definite integral of the diffrence between $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x+9}{x+4} $ since the latter is a graph of a lower-degree polynomial than the first one, but as a result I got a strange negative area result, which is quite funny since the offical answer is $12 -5 \ln5$...
What am I doing wrong?
Note that in this exercise you need to somehow imagine the two graphs by yourself without the use of any comupters programs or whatever. Hence it is not as trivial to evalutae the position of each graph to the other...
| Note that
$$
\left|\frac{x^2+2x-3}{x+4}\right| = -x + 2 - \frac{5}{x+4},\ \textrm{for}\ x \in [-1,3].
$$
So we obtain
\begin{eqnarray}
\int_{-3}^1 dx \left| \frac{x^2+2x-3}{x+4} \right|
&=& \int_{-3}^1 dx \Big( -x + 2 - \frac{5}{x+4} \Big)\\
&=& \left[ - \frac{1}{2} x^2 + 2 x - 5 \ln(x+4) \right]_{-3}^1\\
&=& \Big( - \frac{1}{2} + 2 - 5 \ln(5) \Big) - \Big( - \frac{9}{2} - 6 - 5 \ln(1) \Big)\\
&=& 12 - 5 \ln(5).
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Nth value of Function Given x and y we define a function as follow :
f(1)=x
f(2)=y
f(i)=f(i-1) + f(i+1) for i>2
Now given x and y, how to calculate f(n)
Example : If x=2 and y=3 and n=3 then answer is 1
as f(2) = f(1) + f(3), 3 = 2 + f(3), f(3) = 1.
Constraints are : x,y,n all can go upto 10^9.
| The general way to solve $$F(i+1)= F(i) - F(i - 1)$$
Is to first to see that if $F(i + 1)$ can be written as the sum of 2 geometric series, so that $$F(n) = ar^n + bs^n$$
So $$ar^{n+1} + bs^{n+1} = ar^{n} + bs^n - ar^{n-1} - bs^{n-1}$$
$$ar^{n-1}(1 - r + r^2) = -bs^{n-1}(1 - s + s^2)$$
Since $r\ne s$, it follows that both $r$ and $s$ must satisfy the quadratic equation $$1-x+x^2 = 0$$
So $$r = \frac{1 + \sqrt{3}i}{2}, s = \frac{1 - \sqrt{3}i}{2}$$
Now we find $a$ and $b$, using $F(1)$ and $F(2)$
$$F(1) = x = a\left(\frac{1 + \sqrt{3}i}{2}\right)^1 + b\left(\frac{1 - \sqrt{3}i}{2}\right)^1$$
$$F(2) = y = a\left(\frac{1 + \sqrt{3}i}{2}\right)^2 + b\left(\frac{1 - \sqrt{3}i}{2}\right)^2$$
And then you find the closed form of your recurrence series.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that this expression is a perfect square? Show that this expression is a perfect square?
$(b^2 + 3a^2 )^2 - 4 ab*(2b^2 - ab - 6a^2)$
| Once you factored it out and got $9a^4+24a^3b+10a^2b^2−8ab^3+b^4$, you'd suspect that this is the square of a sum. What are the parts of the sum?
If we let b = 0, the sum is $9a^4$ so there should be a term $±3a^2$. If we let a = 0, the sum is $b^4$, so there should be a term $±b^2$.
We know that $(x+y)^2=x^2 + 2xy + y^2$. Guessing that $x=3a^2$ and $24a^3b=2xy=6a^2y$, that would make y = 4ab. From the other side, if $x=±b^2$ and $2xy=-8ab^3=-/+2yb^2$, that makes $x = -b^2$ and $y = 4ab$.
So an educated guess would be $(3a^2+4ab-b^2)^2$, which lucky enough produces the requested result.
| {
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Evaluate $\int_0^1 \sqrt{2x-1} - \sqrt{x}$ $dx$ I'm trying to calculate the area between the curves $y = \sqrt{x}$ and $y= \sqrt{2x-1}$
Here's the graph:
I've already tried calculating the area with respect to $y$, i.e.
$\int_0^1 (\frac{y^2+1}{2} - y^2)$ $ dx$
[since $y^2=x$ for the first curve and $\frac{y^2+1}{2}=x$ for the second curve]
And the result for that integral is $1/3$ which should be the same for the result of the integral in question
$\int_0^1 (\sqrt{2x-1} - \sqrt{x})$ $dx$
But, $\int_0^1 (\sqrt{2x-1} - \sqrt{x})$ $dx$ = $\int_0^1 (2x-1)^{1/2}$ $dx$ - $\int_0^1 {x}^{1/2} $ $dx$
[let $2x-1 = u$ and $2 dx = du$]
$\Rightarrow$ $\frac 12 \int_0^1 (u)^{1/2}$ $du$ - $\frac{2x^{3/2}}{3}$
$=\frac{u^{3/2}}{3}$ - $\frac{2x^{3/2}}{3}$
$=\frac{(2x-1)^{3/2}}{3}$ - $\frac{2x^{3/2}}{3}$
Evaluated at $x=1$ and $x=0$ and subtracting:
$(\frac{-1}{3})-(\frac{-1}{3})$
Why?
| As noted in the comments the correct way is
$$
A
= \int_0^{1/2} \color{blue}{\sqrt{x}} \, \mathrm{d}x +
\int_{1/2}^1 \color{red}{\sqrt{x} - \sqrt{2x-1}} \,\mathrm{d}x
= \int_0^1 \frac{y^2+1}{2} - y^2 \mathrm{d}y
= 1/3
$$
Since $\sqrt{2x-1}$ is not defined for $x<1/2$. Note that you could also have done the integral as
$$
A = \int_0^1 \sqrt{x} \,\mathrm{d}x - \int_{1/2}^1 \sqrt{2x-1} \,\mathrm{d}x
$$
(why?) without it making the computation any easier.
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of a triangle adjacent to two similar triangles This GMAT problem states:
In the figure above $AD = 4$, $AB = 3$ and $CD = 9$. What is the area of triangle $AEC$
The solution states:
to find the base we need to see that triangles $AEB$ and $CDE$ are similar. The ratio $AB: CD$, is therefore equal to the ratio $AE: ED$. The given information shows that the ratio is $3:9$, or $1:3$. Now dividing $AD$ ($4$) in this ratio gives us $AE$ as $1$.
The area of $AEC = 1/2\times\text{base} \times \text{height}=1/2 \times 9 = 4.5$
If $AD$ is $4$, and the ratio is $1:3$, how did they get $AE$ to be $1$ and not $4/3$?
| Apply Componendo Dividendo on fraction from similar triangles
$$ \frac{AE}{ED}= \frac{AB}{CD}=\dfrac{3}{9}= \dfrac{1}{3}$$
$$\dfrac{AE}{AE+ED}= \dfrac{1}{1+3}= \dfrac{1}{4} $$
$$ =\dfrac{AE}{AD}= \dfrac{1}{4} $$
Since $AD=4$
$$AE=1$$
| {
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How prove that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $ How check that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $?
| use three equations:
$$a^3+b^3=(a+b)(a^2-ab+b^2)\quad (1)$$
$$a^3-b^3=(a-b)(a^2+ab+b^2)\quad (2)$$
$$(a+b)^3=a^3+3a^2b+3ab^2+b^3\quad (3)$$
for your problem:
$$left\\=(\sqrt[3]{\frac{1}{3}})^2-\sqrt[3]{\frac{1}{3}}\sqrt[3]{\frac{2}{3}}+(\sqrt[3]{\frac{2}{3}})^2\\=\frac{\frac{1}{3}+\frac{2}{3}}{\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}}}\quad using(1)\\=\sqrt[3]{\frac{1}{(\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}})^3}}\\=\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}}\\=\sqrt[3]{\frac{3}{(1+\sqrt[3]{2})^3}}\\=\sqrt[3]{\frac{3}{1+3\sqrt[3]{2}+3\sqrt[3]{2^2}+2}}\quad using(3)\\=\sqrt[3]{\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{2^2}}}\\=\sqrt[3]{\sqrt[3]{2}-1}\quad using(2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Longest path in a grid I recently saw a computer programming question that asked for the longest path that one can build in a $3\times3$ unit grid connecting the vertexes, with the following rules(the same rules of a pattern password):
*
*Each vertex may be used at most once.
*A vertex cannot be skipped in a line if it is not used previously.
*If a vertex is used then it must be skipped if a line crosses it.
Now, since $9!=362880$ the problem is trivial by bruteforce. But, is there an analytical way of solving the problem?
| I suggest to start with longest distances and then add shorter.
You have the following vertex sequence
$$
\begin{array}{cccc}
1 & \cdots & 6 & \cdots & 9 & \cdots & 3 \\
\vdots & & \vdots & & \vdots & & \vdots \\
10 & \cdots & 13 & \cdots & 16 & \cdots & 8 \\
\vdots & & \vdots & & \vdots & & \vdots \\
4 & \cdots & 15 & \cdots & 14 & \cdots & 12 \\
\vdots & & \vdots & & \vdots & & \vdots \\
7 & \cdots & 11 & \cdots & 2 & \cdots & 5 \\
\end{array}
$$
And we obtain the distance
$$
\begin{array}{ccccccccccccccc}
1 & [\sqrt{13}] & 2 & [\sqrt{10}] &
3 & [\sqrt{13}] & 2 & [\sqrt{10}] &
5 & [\sqrt{13}] & 6 & [\sqrt{10}] &
7 & [\sqrt{13}] & 8
\end{array}\\
\begin{array}{ccccccccccc}
8 & [\sqrt{2}] &
9 & [\sqrt{5}] &
10 & [\sqrt{5}] &
11 & [\sqrt{5}] &
12 & [\sqrt{5}] &
13
\end{array}\\
\begin{array}{ccccccccccc}
13 & [\sqrt{2}] &
14 & [1] &
15 & [\sqrt{2}] &
16
\end{array},\\
$$
(if I did understood the rules...)
Totaly
$$
1 + 3 \sqrt{2} + 4 \sqrt{5} + 3 \sqrt{10} + 4 \sqrt{13},
$$
or
$$
38.09595...
$$
| {
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How find $x$ in a right triangle $ABC$ (${\measuredangle}A=90^\circ$) where ${\measuredangle}DBC={\measuredangle}DCA=x$,${\measuredangle}BAD=5x$? In a right triangle $ABC$ (${\measuredangle}A=90^\circ$) taken in point $D$ such that $BD=AC$, ${\measuredangle}DBC={\measuredangle}DCA=x$,${\measuredangle}BAD=5x$. How find $x$?
| W.l.o.g. you may assume the following coordinates:
$$A=(0,0)\quad B=(a,0)\quad C=(0,1)\quad D=(b,c)$$
If you define $d:=\cos(x)$ then you can compute the cosines of multiple angles using a Chebyshev polynomial as
$$\cos(5x)=T_5(d)=16d^5 - 20d^3 + 5d\;.$$
The dot product of two vectors is proportional to the cosine, so you can write the first angle equality as
$$\frac{\langle D-B,C-B\rangle^2}{\lVert D-B\rVert^2\cdot\lVert C-B\rVert^2} = d^2$$
and likewise for the other two angle conditions, the last of which will make use of $\cos^2(5x)$ on its right hand side. The equal length condition is simply expressed as
$$\lVert B-D\rVert^2 = \lVert A-C\rVert^2\;.$$
By using squares in all steps involving the lengths of some vector, we managed to avoid introducing square roots into our equations. So now you have four polynomial equations in four variables, which can be rewritten as
\begin{align*}{}
- a^4 d^2 + a^4 + 2 a^3 b d^2 - 2 a^3 b - a^2 b^2 d^2 + a^2 b^2 - a^2 c^2 d^2 \\{}
+ 2 a^2 c - a^2 d^2 - 2 a b c + 2 a b d^2 - b^2 d^2 - c^2 d^2 + c^2 &= 0
\\[2ex]{}
- b^2 d^2 - c^2 d^2 + c^2 + 2 c d^2 - 2 c - d^2 + 1 &= 0
\\[2ex]{}
-256 a^2 b^2 d^{10} + 640 a^2 b^2 d^8 - 560 a^2 b^2 d^6 + 200 a^2 b^2 d^4 - 25 a^2 b^2 d^2 \\{}
+ a^2 b^2 - 256 a^2 c^2 d^{10} + 640 a^2 c^2 d^8 - 560 a^2 c^2 d^6 + 200 a^2 c^2 d^4 - 25 a^2 c^2 d^2 &= 0
\\[2ex]{}
a^2 - 2 a b + b^2 + c^2 - 1 &= 0
\end{align*}
Now in theory you could use your favorite computer algebra system to compute solutions for these. Perform some extra checks, like ensuring that the orientations match your expectations, and you can obtain possible $x=\arccos(d)$ from the remaining solutions. In practice, my sage is still busy trying to solve this system. I'll update this answer if it manages to come up with a solution.
While I still don't have a list of all solutions, I can confirm that you have a solution of this system for
$$a=1\qquad b=1-\tfrac12\sqrt3\qquad c=\tfrac12\qquad
d=\tfrac12\sqrt{\sqrt3 + 2}\qquad x=15°$$
as your comment indicated. However, strictly speaking you have $\angle DBC=\angle DCA=-15°$ here, but $\angle BAD=+75°$. So this is a solution except for the orientation of the angles.
| {
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What is the value of this double integral? Let $C$ be the subset of the plane given by
$$ C \colon= \{ \ (x,y) \in \mathbb{R}^2 \ | \ 0 \leq x^2 + y^2 \leq 1 \}.$$
Then what is the value of the double integral
$$ \int_{C} \int (x^2 + y^2) \ dx \ dy?$$
My work:
In $C$, we have $-1 \leq x \leq 1$ and $-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}$. So we have
$$ \int_C \int (x^2 + y^2) \ dx \ dy = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \ (x^2 + y^2) \ dy \ dx = 2 \int_{-1}^1 (x^2 \sqrt{1-x^2} + \frac{ (\sqrt{1-x^2})^3}{3} ) \ dx = \frac{4}{3} \int_0^1 (2x^2 + 1) \sqrt{1-x^2} \ dx. $$ Now let $x= \sin t$. Then we obtain
$$ \int_C \int (x^2 + y^2) \ dx \ dy = \frac{4}{3} \int_0^1 (2x^2 + 1) \sqrt{1-x^2} \ dx = \frac{4}{3} \int_{t = 0}^{\pi/2} (2\sin^2 t + 1) \cos^2 t \ dt \ \ = \frac{1}{3} \int_{t=0}^{\pi/2} (2\sin^2 2t + 4 \cos^2 t) \ dt = \frac{4}{3} \int_{t=0}^{\pi/2} ( 1 - \cos 4t + 2(1+ \cos 2t) ) \ dt = \frac{1}{3} \int_{t=0}^{\pi/2} (3 + 2 \cos 2t - \cos 4t) \ dt = \frac{\pi}{2}. $$ Am I right?
| Use polar coordinates,
We know that
$$r^2 = x^2 + y^2$$
So our double integral becomes
$$\int_{0}^{2\pi} \int_0^{1}r^2\cdot rdrd\theta$$
Now solve.
EDIT
I see that your computation is correct, I am simply offering another alternative and more easier way to solve this double integral.
| {
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Evaluate $\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$ I'm trying to solve
$$\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$$
My first approach was to factorise and then do a partial integration. However the factorisation $(u+1)\left(u+\frac{1}{2}-\frac{\sqrt{5}}{2} \right)\left(u+\frac{1}{2}+\frac{\sqrt{5}}{2} \right)$ leads me to heavy calculations. How would you proceed to solve this?
This is a continued calculation of Solving $\int_0^1 \frac{dx}{e^x-e^{-2x}+2}$ with substitution
| As $$u^3+2u^2-1=(u+1)(u^2+u-1)$$
Write,
$$\frac u{u^3+2u^2-1}=\frac A{u+1}+B\frac{\dfrac{d(u^2+u-1)}{du}}{u^2+u-1}+\frac C{u^2+u-1}$$
$$u=A(u^2+u-1)+(2Bu+B)(u+1)+C(u+1)$$
Set $\displaystyle u+1=0$ to find $A$
Now comparing the coefficients of $u^2$, $0=A+2B\iff B=?$
Again comparing the constants $0=-A+B+C\implies C=?$
Observe that the second integral $\displaystyle\int B\frac{\dfrac{d(u^2+u-1)}{du}}{u^2+u-1}du=B\int\frac{d(u^2+u-1)}{u^2+u-1}=?$
For the third, $\displaystyle\int\frac C{u^2+u-1}=4C\int\frac{du}{(2u+1)^2-(\sqrt5)^2}$
use $\#1$ of this
Or set $\displaystyle 2u+1=\sqrt5\sec\theta$
| {
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Euler's proof of divergence of sum of reciprocals of primes On Wikipedia at link currently is:
\begin{align}
\ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right)
= -\sum_p \ln \left( 1-\frac{1}{p}\right) \\
& {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\
& {} = \left( \sum_{p}\frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{3p} + \frac{1}{4p^2} + \cdots \right) \\
& {} < \left( \sum_p \frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( 1 + \frac{1}{p} + \frac{1}{p^2} + \cdots \right) \\
& {} = \left( \sum_p \frac{1}{p} \right) + \left( \sum_p \frac{1}{p(p-1)} \right) \\
& {} = \left( \sum_p \frac{1}{p} \right) + C
\end{align}
and since $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, so must $\sum_{p} \frac{1}{p}.$
Currently there is language like "audacious leaps of logic" and "correct result by questionable means".
Wouldn't this be a valid proof if we reversed it, though? If we assume $\sum_p \frac{1}{p}$ converges, can't we just go through the steps backwards and find $\sum_{n=1}^{\infty} \frac{1}{n}$ converges, contradiction? Euler's work seems reasonable to me.
| $
\ln \left( \sum_{n=1}^\infty \frac{1}{n}\right)$ has no meaning.
It is like you say $$\ln (\infty)=...$$
But $\ln x$ is defined on real numbers and not on "anything we understand".
But everyone understands that Euler "understood" that the method was correct.
This is why the proof is attributed to him.
| {
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Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rarely.
Things I have done so far: The inequality look is similar to Nesbitt's inequality.
We could re-write it as: $$\sum \limits_{cyc}\frac {a}{(b+c)^2}(2(a+b+c)) \geq \frac{9}{2}$$
Re-write it again:$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \frac{9}{2}$$
Cauchy appears:
$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2$$
So, if I prove $\left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2 \geq \frac {9}{2}$ then problem is solved.
Re-write in semi expanded form:$$2\left(\sum \limits_{cyc}\frac{a}{b+c}+2\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)}\right) \geq 9$$
We know that $\sum \limits_{cyc}\frac{a}{b+c} \geq \frac {3}{2}$.So$$4\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq 6$$
So the problem simplifies to proving this $$\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq \frac{3}{2}$$
And I'm stuck here.
| $\sum \limits_{cyc}\dfrac {a}{(b+c)^2}=\sum \limits_{cyc}\dfrac {2a^2}{2a(b+c)^2} \ge \dfrac{2(a+b+c)^2}{2a(b+c)^2+2b(a+c)^2+2c(a+b)^2}\ge \dfrac{2(a+b+c)^2}{3\times \left(\dfrac{2a+2b+2c}{3}\right)^3}=\dfrac {9}{4(a+b+c)}$
| {
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How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$? I was going through answers on this question and came across this answer and I was wondering how the user arrived at the first line where they state:
$$f(x) \equiv x^3 - \sin^3 x = x^3 + {1 \over 4} \,\sin {3x} - {3 \over 4}\,\sin x$$
How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$?
Are they using some simple identity or is there some other observation happening?
Thanks!
| Hint:
$$\sin x =\frac{opposite}{hypotenuse}$$
| {
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A certain “harmonic” sum Is there a simple, elementary proof of the fact that:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
I have thought of a very simple notation for "harmonic" sums like these: just write down the numerators. So, for example:
$[\overline{1}]=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots=\infty\;$ is the harmonic series
$[\overline{1,-1}]=\frac{1}{1}+\frac{-1}{2}+\frac{1}{3}+\dots=\ln2\;$ is well known
$[\overline{1,1,-2}]=\frac{1}{1}+\frac{1}{2}+\frac{-2}{3}+\dots=\ln3\;$ is slightly less well known (I think)
$[\overline{1,0,-1,0}]=\frac{1}{1}+\frac{0}{2}+\frac{-1}{3}+\dots=\frac{\pi}{4}\;$ is the Gregory-Leibniz series for $\pi$
What I claim is that $[\overline{1,-1,-2,-1,1,2}]$ is equal to $0$. I wonder if there are any simple proofs of this (i.e. definitely without using calculus, preferably without appealing to complex numbers/taylor series/etc.)
P.S. I know a method that doesn't use any integrals or derivatives, but requires knowledge of the taylor series for $\ln(x)$ and the Euler formula for $e^{ix}$.
The reason I believe that there should be an elementary proof is that the sum, $0$, is a very simple number.
| I think we can "squeeze" something out of this:
$$0=\sum_{n=0}^\infty\left(\frac{1}{6n+6}+\frac{-1}{6n+6}+\frac{-2}{6n+6}+\frac{-1}{6n+6}+\frac{1}{6n+6}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+1}+\frac{-2}{6n+1}+\frac{-1}{6n+1}+\frac{1}{6n+1}+\frac{2}{6n+1}\right)=0$$
| {
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"url": "https://math.stackexchange.com/questions/883233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Integration of $x\cos(x)/(5+2\cos^2 x)$ on the interval from $0$ to $2\pi$
Compute the integral
$$\int_{0}^{2\pi}\frac{x\cos(x)}{5+2\cos^2(x)}dx$$
My Try: I substitute $$\cos(x)=u$$
but it did not help. Please help me to solve this.Thanks
| This is not an answer to the post but a reply to David's comment
The antiderivative does not express in terms of elementary functions. For your curiosity, I write it down, but, as said, it looks like a nightmare.
$$4 \sqrt{14}\int\frac{x\cos(x)}{5+2\cos^2(x)}dx=-2 i \text{Li}_2\left(-\frac{i \left(-7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)+2 i \text{Li}_2\left(\frac{i \left(-7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)+2 i \text{Li}_2\left(-\frac{i \left(7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)-2 i \text{Li}_2\left(\frac{i \left(7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)+2 x \log \left(1-\frac{i \left(\sqrt{35}-7\right) e^{-i
x}}{\sqrt{14}}\right)-\pi \log \left(1-\frac{i \left(\sqrt{35}-7\right) e^{-i
x}}{\sqrt{14}}\right)-2 x \log \left(1+\frac{i \left(\sqrt{35}-7\right) e^{-i
x}}{\sqrt{14}}\right)+\pi \log \left(1+\frac{i \left(\sqrt{35}-7\right) e^{-i
x}}{\sqrt{14}}\right)-2 x \log \left(1-\frac{i \left(7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)+\pi \log \left(1-\frac{i \left(7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)+2 x \log \left(1+\frac{i \left(7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)-\pi \log \left(1+\frac{i \left(7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)-4 \sin ^{-1}\left(\frac{\sqrt{7+\sqrt{14}}}{2^{3/4}
\sqrt[4]{7}}\right) \log \left(1-\frac{i \left(\sqrt{35}-7\right) e^{-i
x}}{\sqrt{14}}\right)+4 \sin ^{-1}\left(\frac{\sqrt{7+\sqrt{14}}}{2^{3/4}
\sqrt[4]{7}}\right) \log \left(1+\frac{i \left(7+\sqrt{35}\right) e^{-i
x}}{\sqrt{14}}\right)-\pi \log \left(\sqrt{14} \sin (x)-7\right)+\pi \log
\left(\sqrt{14} \sin (x)+7\right)-4 i \sinh
^{-1}\left(\frac{\sqrt{7-\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \log
\left(1+\frac{i \left(\sqrt{35}-7\right) e^{-i x}}{\sqrt{14}}\right)+4 i \sinh
^{-1}\left(\frac{\sqrt{7-\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \log
\left(1-\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)+8 i \sin
^{-1}\left(\frac{\sqrt{7+\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \tan
^{-1}\left(\frac{\left(\sqrt{14}-7\right) \cot \left(\frac{1}{4} (2 x+\pi
)\right)}{\sqrt{35}}\right)+8 \sinh ^{-1}\left(\frac{\sqrt{7-\sqrt{14}}}{2^{3/4}
\sqrt[4]{7}}\right) \tan ^{-1}\left(\frac{\left(7+\sqrt{14}\right) \cot
\left(\frac{1}{4} (2 x+\pi )\right)}{\sqrt{35}}\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of a definite integral We need to calculate $$\lim_{x \to 0}\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}$$
Integral itself doesn't seem to be the problem here. When making a substitution $\sqrt{t}=u$, we get $$\lim_{x \to 0}2\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5(1+u)}=2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5+u^6}$$
Then by partial fractions, which I did manually and chcecked with WolframAlpha afterwards, it becomes $$\begin {align}
2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\left(\frac{1}{u^5}-\frac{1}{u^4}+\frac{1}{u^3}-\frac{1}{u^2}+\frac{1}{u}-\frac{1}{1+u}\right) du =\\
\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}2\left(\log{u}-\log{(1+u)}+\frac{1}{u}-\frac{1}{2u^2}+\frac{1}{3u^3}-\frac{1}{4u^4}\right) du =\\
\lim_{x \to 0}\int_{\sin x}^{x}\left(\log{t}-2\log{(1+\sqrt{t})}+\frac{2}{\sqrt{t}}-\frac{1}{t}+\frac{2}{3t^{3/2}}-\frac{1}{2t^2}\right) dt
\\\end{align}$$
Fianlly we obtain the following limit:
$$\lim_{x \to 0}\left(\log {x}-\log {\sin x}+2\log {(1+\sqrt{x})}-2\log {(1+\sqrt{\sin x})}+\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}-\frac{1}{x}+\frac{1}{\sin x}+\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}-\frac{1}{2x^2}+\frac{1}{\sin^2 x}\right)$$
Here's where I stuck. It gets messy when I try to calculate $\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}$ and $\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}$. The rest is rather doable - de l'Hospital's rule is useful with $-\frac{1}{x}+\frac{1}{\sin x}$ which is $0$ in limit, so as logarithm expressions (obviously) and Taylor expansion helps with $-\frac{1}{2x^2}+\frac{1}{\sin^2 x}$ which, in turn, equals $1/6$ when $x$ approaches $0$.
Did I make any mistakes? I hope not, but even so I'm not certain what to do with this horrible limit. I'd be glad if anyone could point out what to do.
| This is not much different from Sami Ben Romdhane's answer.
Since the function $g(t)=\frac{1}{t^3(1+\sqrt{t})}$ is decreasing over $\mathbb{R}^+$, by the mean value theorem:
$$(x-\sin x) \frac{1}{\sin^3 x\,(1+\sqrt{\sin x})}\leq\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}\leq (x-\sin x)\frac{1}{x^3(1+\sqrt{x})},$$
but when $x$ approaches $0$ both the LHS and the RHS approaches $\frac{1}{6}$.
| {
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"url": "https://math.stackexchange.com/questions/885458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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prove by induction $7 \mid 3^{3^n}+8$ Okay so ive been trying to prove this for about 5 hours...
really need salvation from the geniouses around here.
prove by induction
$7\mid 3^{3^n}+8$
i really need some directions on what to do here...
| First note that this is equivalent to $7$ dividing $3^{3^n}+1$. Clearly if $n=1$, then $3^{3^1}+1 = 28$ and $7$ divides this.
Let's suppose it's true for $n$. Then $3^{3^{n+1}} = 3^{3\cdot3^n} = (3^{3^n})^3$. We want then to show that $7$ divides $(3^{3^n})^3+1$. A nice factorization is that $a^3+b^3 = (a+b)(a^2-ab+b^2)$. So then we can write $(3^{3^n})^3+1$ as $(3^{3^n}+1)((3^{3^n})^2-3^{3^n}+1)$. Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
} |
Maximum and minimum of $z=\frac{1+x-y}{\sqrt{1+x^2+y^2}}$ Find the maximum and minimum of the function
$$z=\frac{1+x-y}{\sqrt{1+x^2+y^2}}$$
I have calculated
$\frac{\partial z}{\partial x}=\frac{1+y^2+xy-x}{(1+x^2+y^2)^{\frac{3}{2}}}$
$\frac{\partial z}{\partial y}=\frac{-1-x^2-xy-y}{(1+x^2+y^2)^{\frac{3}{2}}}$
I have problem to solve the system of equations:
$$1+y^2+xy-x=0$$
$$1+x^2+xy+y=0$$
| $z^2 = \dfrac{(1+x+(-y))^2}{1+x^2+(-y)^2} \leq 3$ because $(1-x)^2 + (x+y)^2 +(y+1)^2 \geq 0$.
The second inequality becomes $ = $ when $x = 1, y = -1$. From this we have: $z_\text{min} = -\sqrt{3}$, and $z_\text{max} = \sqrt{3}$.
Note: The CS inequality is: $(a_1b_1 + a_2b_2 +a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$ with equality when $\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} = \dfrac{a_3}{b_3}$. Apply it with $a_1 = a_2 = a_3 =1$, and $b_1 = 1, b_2 = x, b_3 = -y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $\sum_{n=0}^\infty \frac{1}{2^n}$ and $\sum_{n=0}^\infty \frac{1}{4^n}$, what is the Cauchy product? Given $\sum_{n=0}^\infty \frac{1}{2^n}$ and $\sum_{n=0}^\infty \frac{1}{4^n}$, what is the Cauchy product?
The definition of the Cauchy product is for two given series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$: $$c_n = \sum_{j=0}^n a_j b_{n-j}$$.
With the two series I would get $c_n = \sum_{j=0}^n \frac{1}{2^j} \frac{1}{4^{n-j}} = \sum_{j=0}^n \frac{1}{2^{-j}2^n}$ How can I simplify this to a closed form?
| I think you mean
$c_n=\sum\limits_{j=0}^n\frac{1}{2^j}\frac{1}{4^{n-j}}=\sum\limits_{j=0}^n\frac{1}{2^j}\frac{1}{2^{2n-2j}}=\sum\limits_{j=0}^n\frac{1}{2^{2n-j}}=\sum\limits_{j=0}^n2^{j-2n}=2^{-2n}\sum\limits_{j=0}^n2^j$
The sum on the right hand side is a geometric series, which has the formula
$\sum\limits_{j=0}^nar^j=\frac{a(1-r^{n+1})}{1-r}$
Plugging in $a=1,r=2$, we have
$c_n=2^{-2n}\sum\limits_{j=0}^n2^j=2^{-2n}(\frac{1-2^{n+1}}{1-2})=\frac{2^{n+1}-1}{2^{2n}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $OD^2+OG^2=3AB^2$
$AOB$ is cicular sector of $90^{\circ}$.
$C$ is a point on $\stackrel \frown {AB}$.
$ACDE$ and $CBFG$ are squares.
Prove $OD^2+OG^2=3AB^2$
My attempt :
$OA=OC=OB=r$
$O$ is central angle so :
$$\angle O_{1}+\angle O_{2}=\stackrel \frown {BC}=m \hspace{20pt}\angle O_{3}+\angle O_{4}=\stackrel \frown {AC}=n$$
$\angle A_{2}$ and $\angle B_{2}$ are inscribed angles so :
$$\angle A_{2}=\frac{\stackrel \frown {BC}}{2}=\frac{m}{2} \hspace{20pt}\angle B_{2}=\frac{\stackrel \frown {AC}}{2}=\frac{n}{2}$$
in $\triangle OCB$ :
$$\angle OCB=180-(\angle O_{1}+\angle O_{2})-(\angle B_{1}+\angle B_{2})=180-(m)-(45+\frac{n}{2})=135-(m+\frac{n}{2})$$
in $\triangle OCA$ :
$$\angle OCA=180-(\angle O_{3}+\angle O_{4})-(\angle A_{1}+\angle A_{2})=180-(n)-(45+\frac{m}{2})=135-(n+\frac{m}{2})$$
so :
$$\angle OCG=\angle OCB+90=225-(m+\frac{n}{2})=x$$
$$\angle OCD=\angle OCA+90=225-(n+\frac{m}{2})=y$$
after that I applied Law of cosines :
$$OG^2=r^2+CG^2-2r*CG*\cos{x}$$
$$OD^2=r^2+CD^2-2r*CD*\cos{y}$$
And I stuck here.
| Hint: Note that $CG = CB$ and $CD = AC$ so:
$CG^2 + CD^2 = CB^2 + AC^2 = AB^2 +2CB·AC·\cos 135 = r^2 + 2CB·AC·\cos 135$.
All you have to do now is to prove that:
$r·CB\cos x + r·AC\cos y = AC·CB·\cos 135$.
Note also that $x = 90 + \angle OCA$ so, applying the formula of cosine of sum of angles: $\cos x = -\sin \angle OCA$. Equivalently, $\cos y = -\sin \angle OCB$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Find a reduced echelon basis from a reduced echelon matrix. The reduced row matrix was this ---> $\begin{pmatrix}1&2&0&1&0\\0&0&1&3&0\\0&0&0&0&1\\0&0&0&0&0&\end{pmatrix} = 0$
So i computed the basis to be such that
$\begin{pmatrix}x\\y\\z\\t\\u\end{pmatrix} = y\begin{pmatrix}-2\\1\\0\\0\\0\end{pmatrix} + t\begin{pmatrix}-1\\0\\-3\\1\\0\end{pmatrix}$
So there is a basis $(f_1, f_2)$ where $f_1$ = $(-2, 1, 0, 0, 0)$ and $f_2$ = $(-1, 0, -3, 1, 0)$. I did it up to there but in addition there is a note saying that a reduced echelon basis is therefore $(−f_2, 2f_2 + f_1)$. Where does this come from? How did they come about this reduced echelon basis?
| I suppose they simply made a rref from this:
$$
\begin{pmatrix}
-1 & 0 &-3 & 1 & 0 \\
-2 & 1 & 0 & 0 & 0 \\
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 3 &-1 & 0 \\
-2 & 1 & 0 & 0 & 0 \\
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 3 &-1 & 0 \\
0 & 1 & 6 &-2 & 0 \\
\end{pmatrix}
$$
i.e.
$$
\begin{pmatrix}
f_2 \\
f_1 \\
\end{pmatrix}\sim
\begin{pmatrix}
-f_2 \\
f_1 \\
\end{pmatrix}\sim
\begin{pmatrix}
-f_2 \\
f_1-2f_2 \\
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can we expect to find some constant $C$; so that, $\sum_{n\in \mathbb Z} \frac{1}{1+(n-y)^{2}} Fix $y\in \mathbb R;$ and consider the series:
$$\sum_{n\in \mathbb Z}\frac{1}{1+(n-y)^{2}}.$$
My Question is: Can we expect to find some constant $C$; so that,
$$\sum_{n\in \mathbb Z} \frac{1}{1+(n-y)^{2}} <C$$ for all $y\in \mathbb R;$ If yes, How ?
Thanks,
| \begin{align}
\sum_{n=-\infty}^{[y]} \frac{1}{1+(n-y)^2} = \sum_{n=-\infty}^{[y]-1} \frac{1}{1+(n-y)^2} + \frac{1}{1+([y]-y)^2} \leq \sum_{n=-\infty}^{1} \frac{1}{1+n^2} + 1
\end{align}
\begin{align}
\sum_{[y]+1}^{n=+\infty} \frac{1}{1+(n-y)^2} = \sum_{[y]+2}^{n=+\infty} \frac{1}{1+(n-y)^2} + \frac{1}{1+([y]+1 -y)^2} \leq \sum_{1}^{+\infty} \frac{1}{1+n^2} + 1
\end{align}
Thus $\sum_{n=-\infty}^{+\infty} \frac{1}{1+(n-y)^2} \leq 2 \sum_{n=1}^{+\infty}\frac{1}{n^2} + 2 = \frac{\pi^2}{3}+2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Elementary ways to calculate the arc length of the Cantor function (and singular function in general) Cantor's function: http://en.wikipedia.org/wiki/Cantor_function
There is an elementary way to prove that the arc length of the Cantor function is 2?
In this article (http://www.math.helsinki.fi/analysis/seminar/esitelmat/stat0312.pdf) they use the following result:
If $f:[a,b] \rightarrow \mathbb{R}$ is a continuous monotone function, then $f$ is singular if and only if
$$L_a^b = |f(a)-f(b)|+|a-b|$$
But, there is a way for calculate the arc length of singular function without using this property? like using the arc length definition
If $X$ is a metric space with metric $d$, then we can define the ''length'' of a curve $\!\,\gamma : [a, b] \rightarrow X$ by $$\text{length} (\gamma)=\sup \left\{ \sum_{i=1}^n d(\gamma(t_i),\gamma(t_{i-1})) : n \in \mathbb{N} \text{ and } a = t_0 < t_1 < \cdots < t_n = b \right\}. $$
where the sup is over all $n$ and all partitions $t_0 < t_1 < \cdots < t_n$ of $[a, b]$.
| Here is an elementary way to derive the arc length of the Cantor Function:
We know that the following sequence of functions converges uniformly to the Cantor Function. Letting $c_0(x) = x$ and $n\ge 0$, we define
$$c_{n+1}(x)= \begin{cases}
\frac{1}{2}c_n(3x) & 0 \le x \le \frac{1}{3},\\
\frac{1}{2} & \frac{1}{3} \le x \le \frac{2}{3},\\
\frac{1}{2}c_n(3x-2)+\frac{1}{2} & \frac{2}{3}\le x \le 1.
\end{cases}
$$
We also know that once the sequence of functions goes constant on an interval, it stays constant on that interval. Then from the definition, we see that at $c_n$, $n\ge1$, we add $2^{n-1}$ constant portions of length $3^{-n}$ each. So for the totally horizontal (i.e. constant) portions of $c_n$, we have that they have total length $\displaystyle \sum_{i=0}^n \frac{2^k}{3^{k+1}}$. To find the lengths of the positively sloped portions for a given $c_n$, we note that from the definition, there are $2^n$ sloped portions, which increase the function $2^{-n}$ over a length of $3^{-n}$. Then we have that the total length of the positively sloped portions is
$$2^n\sqrt{(\frac{1}{2^n})^2 + \frac{1}{3^n})^2} = \sqrt{\frac{(2^n)^2((2^n)^2 + (3^n)^2)}{(2^n)^2(3^n)^2}} = \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}.$$
So then $L(c_n) = \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}+ \displaystyle \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k}$, and so
$$\lim_{n \to \infty}L(c_n) = \lim_{n \to \infty} \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}+ \displaystyle \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k}$$
$$ = \sqrt{\lim_{n\to \infty}\frac{2^{2n}+3^{2n}}{3^{2n}}} + \lim_{n \to \infty} \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k} = \sqrt{\lim_{n \to \infty}(\frac{2}{3})^n +1} + \frac{1}{3}\left(\frac{1}{1-\frac{2}{3}}\right)$$
$$= 1 + \frac{1}{3}\left(\frac{1}{\frac{1}{3}}\right) = 2.$$
Letting $c$ be the Cantor function, we have $L(c) \ge 2$, since $L(c) \ge L(c_n)$ for all $n$. To see this, note that on any constant portion of $c_n$, $c$ agrees with $c_n$. But on the diagonal portions of $c_n$, the graph of $c$ is broken up into more and more horizontal and diagonal portions. In essence, if on any $(a,b)$ on which $c_n$ is diagonal, $c$ is less diagonal. Then since the shortest distance between any two points is given by the straight line between them, we have that the length of $c$ is greater than the length of $c_n$ on that portion. Then we have that $L(c) \ge L(c_n)$ for all $n$, so $L(c) \ge 2$. To see that $L(c) \le 2$, note that we have $L(c) = \displaystyle \sup_{P} \sum_{i=1}^n \sqrt{(\triangle x_i)^2 + (\triangle c(x_i))^2} \le \sup_{P}\sum_{i=1}^n |\Delta x_i| + |\Delta c(x_i)| = \sup_{P}\sum_{i=1}^n \Delta x_i + \Delta c(x_i) =2$, since the total length of the horizontal segments is $1$, and since $c$ is monotonically increasing the total length of the vertical growth is $1$. Thus $L(c) =2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Throwing dice twice, with unlike probability of occourence? A loaded dice has the property that when the dice is thrown the probability of showing a given number is proportional to the number. For example $2$ is twice as likely to show up compared to $1$ and $3$ is thrice as likely to show up compared to $1$, And so on. What is the probability that when the dice is thrown twice the sum is $4$ or less.
$$
P = \frac 36 \cdot \frac 16 +\frac 16 \cdot \frac 36 + \frac 26 \cdot \frac 26 + \frac 16 \cdot \frac 26 + \frac 26 \cdot \frac 16 + \frac 16 \cdot \frac 16 =
\frac{15}{36}
$$
Where I am getting it wrong?
| Since $1+2+\cdots+6=21$, the probabilities of $1,2,3,4,5,6$ in $1$ toss are respectively $\frac{1}{21}$, $\frac{2}{21}$, and so on up to $\frac{6}{21}$.
The probability of a sum of $2$ in $2$ throws is $\frac{1}{21}\cdot\frac{1}{21}$.
The probability of a sum of $3$ is $2\cdot \frac{1}{21}\cdot\frac{2}{21}$.
The probability of a sum of $4$ is $2\cdot \frac{1}{21}\cdot\frac{3}{21}+\frac{2}{21}\cdot\frac{2}{21}$.
Add up.
Remark: Your numerator, and therefore the basic analysis, was right. The denominators were not.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$ Question:
$$\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$$
Prove that L.H.S.=R.H.S.
My Efforts:
L.H.S.$$=\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta}\times\frac{\csc\theta+1}{\csc\theta+1}$$
$$=\frac{\tan^2\theta(1-\cos\theta)(\csc^2\theta-1)}{(1-\cos^2\theta)(\csc\theta+1)}$$
| You are almost there.
Hint:
$$\sin^2\theta+\cos^2\theta=1\implies \left(\frac{1}{\sin^2\theta}\right)(\sin^2\theta+\cos^2\theta)=\frac{1}{\sin^2\theta}$$
Also note that in the calculation you have reproduced, there is an error in the final denominator. It should be $(\csc\theta +1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this function decreased with $x$? Given three positive integers $a,b,c\ge 1$, I am wondering if the following $f(x)$ is decreased with $x$ ?
$$f(x)=\frac{c+2x}{(a+x)(b+x)}, \quad x \in Z^+ \cup \{0\}$$
where $1\le c \le ab$.
| since my calculations are error-prone, i will merely show some working which someone might check. if this result is correct then at present i cannot see see how it can be non-trivially related to the given condition $1 \le c \le ab$. might one expect a condition involving $c^2$ rather than $c$? apologies in advance
$$
\frac{df}{dx} = \frac{2(a+x)(b+x) - (c+2x)(a+b+2x)}{(a+x)^2(b+x)^2}
$$
the numerator is
$$
2(ab+(a+b)x+x^2) - c(a+b) - 2cx -2x(a+b)-4x^2 \\
= 2ab - 2x^2 -2cx - c(a+b)\\
= - ( 2x^2+2cx + [c(a+b)-2ab])
$$
the critical values of $x$ are thus the solutions, if any, of:
$$
x = \frac{-2c \pm \sqrt{(4c^2-8c(a+b)+16ab)}}{4}\\
= -\frac12 \left(c \pm \sqrt{(c-2a)(c-2b)} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
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