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How to prove this trig. identity???(question 1 : solved)(question 2 : solved) question 1:
We know :
$$
\left\{
\begin{array}{ll}
\alpha \ + \beta\ +\theta\ = \pi\\
\alpha\ = \frac{\pi}{2} \\
\end{array}
\right.
$$
How to prove this :
$$
\sin(\alpha)\sin(\beta)\sin(\alpha\ -\ \beta) + \sin(\beta)\sin(\theta)\sin(\beta\ -\ \theta) + \sin(\theta)\sin(\alpha)\sin(\theta\ -\ \alpha)+ \\
\sin(\alpha\ -\ \beta)\sin(\beta\ -\ \theta)\sin(\theta\ -\ \alpha) = 0
$$
question 2:
And if we know that :
$$
\left\{
\begin{array}{ll}
\alpha \ + \beta\ +\theta\ = \pi\\
\cot(\alpha) \ + \cot(\beta)\ + \cot(\theta) \ = 2 \\
\end{array}
\right.
$$
How to prove this:
$$1+\cos(\alpha)\cos(\beta)\cos(\theta) = 2\times \sin(\alpha)\sin(\beta)\sin(\theta)$$
Thanks a lot.
|
Question 2 :
The relation $\alpha + \beta + \theta = \pi$ implies that $\cos (\alpha + \beta + \theta) = -1$. But
\begin{align}
\cos(\alpha + \beta + \theta) &= \cos\alpha \cos(\beta + \theta) - \sin \alpha\sin(\beta + \theta) \\
&= \cos(\alpha)(\cos\beta\cos\theta - \sin\beta \sin\theta) - \sin\alpha(\sin\beta \cos\theta + \sin\theta \cos\beta) \\
&= \cos\alpha \cos\beta \cos\theta - \cos\alpha\sin\beta\sin\theta - \sin\alpha\sin\beta\cos\theta - \sin\alpha\sin\theta\cos\beta,
\end{align}
so
$$1 +\cos\alpha\cos\beta\cos\theta = \cos\alpha\sin\beta\sin\theta + \sin\alpha\sin\beta\cos\theta + \sin\alpha\sin\theta\cos\beta.$$
Therefore
$$\frac{1}{\sin\alpha\sin\beta\sin\theta} + \cot\alpha\cot\beta\cot\theta = \cot\alpha + \cot\beta + \cot\theta = 2$$
and the result follows.
|
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"url": "https://math.stackexchange.com/questions/1682586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solve the Integral $\int \frac{dx}{\left(x-2\right)^3\sqrt{3x^2-8x+5}}$ $$\int \frac{dx}{\left(x-2\right)^3\sqrt{3x^2-8x+5}}$$
I am pretty sure that a certain substitution should work.
I tried using $x-2=\frac{1}{t}$
And got the integral:
$$\int \:\frac{-\frac{dt}{t^2}}{\frac{1}{t^3}\sqrt{3\left(\frac{1}{t}+2\right)^2-8\left(\frac{1}{t}+2\right)^2+5}}$$
After some simplification:
$$\int \:\frac{tdt}{\sqrt{3\left(\frac{1}{t}+2\right)^2-8\left(\frac{1}{t}+2\right)^2+5}}$$
And at this point I'm lost.
|
From the point that you stopped:
$$\int \:\frac{tdt}{\sqrt{3\left(\frac{1}{t}+2\right)^2-8\left(\frac{1}{t}+2\right)^2+5}}$$
$$=\int\frac{tdt}{\sqrt{5-5\left(\frac 1 t+2\right)^2}}=\int\frac{tdt}{\sqrt{-\frac{5}{t^2}-\frac{20}{t}-15}}=-\frac{i}{\sqrt 5}\int\frac{t^2}{\sqrt{3t^2+4t+1}}dt$$
$$\overset{\text{complete the square }}{=}-\frac{i}{\sqrt 5}\int\frac{t^2}{\sqrt{\left(\sqrt 3 t+\frac{2}{\sqrt 3}\right)^2-\frac 1 3}}dt$$
Set $u=\sqrt t+\frac{2}{\sqrt 3}$ and $du=\sqrt 3 dt$
$$=-\frac{i}{\sqrt 5}\int\frac{(2\sqrt 3-3u)^2}{27\sqrt{u^2-\frac 1 3}}du$$
substitute $u=\frac{\sec s}{\sqrt 3}$ and $du=\frac{\tan s \sec s}{\sqrt 3}ds$ then $\sqrt{u^2-\frac 1 3}=\sqrt{\frac{\sec^2 s}{3}-\frac 1 3}=\frac{\tan s}{\sqrt 3}$ and $s=\sec^{-1}(\sqrt 3 u)$
$$=-\frac{i}{81\sqrt 5}\int \sqrt 3 \sec s(2\sqrt 3-\sqrt 3\sec s)^2$$
$$=-\frac{i}{9\sqrt{15}}\int \sec^3 s ds+\frac{4i}{9\sqrt{15}}\int \sec^2 s ds-\frac{4i}{9\sqrt{15}}\int \sec s ds=\dots$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1684228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Symmetric matrix with nonnegative determinant Given positive reals $a,b$, and $c$, is it true that the matrix
$$
\begin{pmatrix} a^4+b^4+c^4 & a^3+b^3+c^3 & a^2+b^2+c^2 \\ a^3+b^3+c^3 & a^2+b^2+c^2 & a+b+c \\ a^2+b^2+c^2 & a+b+c & 3\\ \end{pmatrix}
$$
has nonnegative determinant?
I conjecture the answer is affirmative; on the other hand, at the moment I didn't try with numerical evidences..
|
Note that
$$
\begin{pmatrix} a^4+b^4+c^4 & a^3+b^3+c^3 & a^2+b^2+c^2 \\ a^3+b^3+c^3 & a^2+b^2+c^2 & a+b+c \\ a^2+b^2+c^2 & a+b+c & 3\\ \end{pmatrix}=\begin{pmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1\\ \end{pmatrix}\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1\\ \end{pmatrix}
$$
The determinant thus is the square of the determinant of
$$\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1\\ \end{pmatrix}
$$ which is $(a-b)^2(b-c)^2(c-a)^2$ which is non negative for all real $a,b,c$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$ Calculate:
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$
I don't know how to use L'Hôpital's Rule.
I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}$.
|
$$\begin{align}
\lim_{x\to0^{+}}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}
&=\lim_{x\to0^{+}}\frac{\tan^{3/2}x-\sin^{3/2} x}{x^3 \sqrt{x}}\\
&=\lim_{x\to0^{+}}\frac{\sin^{3/2}x}{x^{3/2}}\frac{\sec^{3/2}x-1}{x^2}\\
&=\lim_{x\to0^{+}}\left(\frac{\sin x}{x}\right)^{3/2}\cdot \lim_{n\to\infty}\frac{\sec^{3/2}x-1}{x^2}\\
&=\lim_{x\to0^{+}}\frac{\sec^{3/2}x-1}{x^2}\\
&=\lim_{x\to0^{+}}\frac{\left(1+\frac12x^2+\frac5{24}x^4+\cdots\right)^{3/2}-1}{x^2}\\
&=\lim_{x\to0^{+}}\frac{\left(1+\frac34x^2+\cdots\right)-1}{x^2}\\
&=\frac34
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1687677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.
|
"Completing the square" would be my choice.
$x^2+ (2cos(b)cos(c))x+ cos^2(b)cos^2(c)- cos^2(b)cos^2(c)+ cos^2(b)+ cos^2(c)- 1= 0$
$(x+ cos(b)cos(c))^2= 1- cos^2(b)- cos^2(c)+ cos^2(b)cos^2(c)$
You say that "$a+ b+ c= \pi$ but there is NO "a" in your equation.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve in positive integers the equation $a^3+b^3=9ab$ Solve in positive integers the equation:
$$a^3+b^3=9ab$$
I try to:
$$\dfrac{a^2}{b}+\dfrac{b^2}{a}=9\Longrightarrow a^2<9b,b^2<9a$$
Of course, I can't solve it. Can anyone help?
|
By the Arithmetic Mean Geometric Mean Inequality, we have
$$\frac{9}{2}=\frac{1}{2}\left(\frac{a^2}{b}+\frac{b^2}{a}\right)\ge \sqrt{ab}.$$
Now we only have a smallish number of possibilities to examine.
A simple congruential observation then dramatically cuts down the number of cases. For $a$ and $b$ must both be even. (If they are both odd, the left-hand side is even and the right-hand side is odd. If they are of different parity, the left-hand side is odd and the right-hand side is even.)
Thus $a^3+b^3$ is divisible by $8$, and one of $a$ and $b$ is divisible by $2$, and the other is divisible by $4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1691052",
"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$ How to evaluate
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$$
I completely have no idea how to find the result.Mathematic gave me the following answer part of the integral
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=-\frac{1}{4}\mathbf{G}\pi +\frac{\pi ^{2}}{16}\ln 2+\frac{21}{64}\zeta \left ( 3 \right )$$
where $\mathbf{G}$ donates the Catalan's Constant.
But it can't evaluate the other part.So I'd like to know how to evaluate the original integral or the above integral.
|
Let $t=\arctan x$. Then
\begin{eqnarray}
&&\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x\\
&=&-\int_{0}^{\frac{\pi}{4}}t\ln[\cos t(\cos t+\sin t))]\mathrm{d}t\\
&=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[\cos t^2(\cos t+\sin t)^2]\mathrm{d}t\\
&=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[\frac{1+\cos 2t}{2}(1+\sin 2t)]\mathrm{d}t\\
&=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[(1+\cos 2t)(1+\sin 2t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\
&=&-\frac{1}{8}\int_{0}^{\frac{\pi}{2}}t\ln[(1+\cos t)(1+\sin t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\
&=&-\frac{\pi}{32}\int_{0}^{\frac{\pi}{2}}\ln[(1+\cos t)(1+\sin t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\
&=&-\frac{\pi}{16}\int_{0}^{\frac{\pi}{2}}\ln(1+\cos t)\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\
\end{eqnarray}
Noting
\begin{eqnarray}
\int_{0}^{\frac{\pi}{2}}\ln(1+\cos t)\mathrm{d}t&=&\int_{0}^{\frac{\pi}{2}}\ln(2\cos^2\frac{t}{2})\mathrm{d}t\\
&=&\frac{\pi}{2}\ln 2+2\int_{0}^{\frac{\pi}{2}}\ln(\cos\frac{t}{2})\mathrm{d}t\\
&=&\frac{\pi}{2}\ln 2+4\int_{0}^{\frac{\pi}{4}}\ln(\cos t)\mathrm{d}t\\
&=&2G-\frac{\pi}{2}\ln 2
\end{eqnarray}
and hence
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x=\frac{1}{64} \pi (3\pi \ln2-8 G).$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$
Prove that $$1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$$
$\bf{My\; Try::}$ Using Binomial expansion of $$(1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+.......$$
So we get $$nx=\frac{1}{3}$$ and $$\frac{nx(nx+x)}{2}=\frac{1}{3}\cdot \frac{3}{6}$$
We get $$\frac{1}{3}\left(\frac{1}{3}+x\right)=\frac{1}{3}\Rightarrow x=\frac{2}{3}$$
So we get $$n=\frac{1}{2}$$
So our series sum is $$(1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$
Although I know that this is the simplest proof, can we solve it any other way someting like defining $a_{n}$ and then use Telescopic sum.
Thanks.
|
Here is a variation to obtain $\sqrt{3}$ based upon the generating function of the Central binomial coefficients
\begin{align*}
\sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad\qquad |z|<\frac{1}{4}
\end{align*}
We obtain
\begin{align*}
1&+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+\cdots\\
&=1+\frac{1!!}{3^11!}+\frac{3!!}{3^22!}+\frac{5!!}{3^33!}+\cdots\tag{1}\\
&=1+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n!}\frac{1}{3^n}\\
&=1+\sum_{n=1}^{\infty}\frac{(2n)!}{n!(2n)!!}\frac{1}{3^n}\tag{2}\\
&=1+\sum_{n=1}^{\infty}\frac{(2n)!}{n!n!}\frac{1}{6^n}\tag{3}\\
&=\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{6^n}\\
&=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=\frac{1}{6}}\\
&=\frac{1}{\sqrt{1-\frac{2}{3}}}\\
&=\sqrt{3}
\end{align*}
Comment:
*
*In (1) we use double factorial for odd values $$(2n-1)!!=(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1$$
*In (2) we use the identity
\begin{align*}
(2n)!=(2n)!!(2n-1)!!
\end{align*}
*In (3) we use the identity
\begin{align*}
(2n)!!=2^nn!
\end{align*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1692678",
"timestamp": "2023-03-29T00:00:00",
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|
If $a^{2}+84a+2008=b^{2}$ what is $a+b$ let $a, b$ are two positive integer satisfy the condition $a^{2}+84a+2008=b^{2}$. Find out $a+b$
My Solution
$a^{2}+84a+2008=b^{2} \implies (a+42)^{2}+244=b^{2} \implies (b+a+42)(b-a-42)=2^{2}61$. Considering (244,1) , (122,2) ,(61,4) we observe only (244,1) give integer $(a,b)=(18,62) \implies a+b=80$
My Question
Is there any other way to solve this problem?
|
With the quadratic formula you get
$$a=\frac{-84\pm \sqrt{84^2-4(2008-b^2)}}{2}=-42\pm \sqrt{b^2-244}$$
Now with this, it is possible to obtain the root since $b^2-244=x^2$, and as $b$ and $a$ must be positive integers, that root must be too, so neccesarily $x$ too. This lets
$$(b+x)(b-x)=244$$
And that implies that:
$$b+x=244\ \ \ \ \ \ \ \ \ b-x=1 \ \ \Rightarrow \ \ 2b=245 \mbox{ not integer}$$
$$b+x=122\ \ \ \ \ \ \ \ \ b-x=2 \ \ \Rightarrow \ \ 2b=124 \Rightarrow b=62$$
$$b+x=61\ \ \ \ \ \ \ \ \ b-x=4 \ \ \Rightarrow \ \ 2b=65 \mbox{ not integer}$$
$$b+x=1\ \ \ \ \ \ \ \ \ b-x=244 \ \ \Rightarrow \ \ 2b=245 \mbox{ not integer}$$
Since $b>0$, this lets $b=62$, so $a=-42\pm 2(62)$ but only is possible $a=-42+124=18$, so $a=18$ and $a+b=80$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If a family of straight lines is $\lambda^2 P+\lambda Q+R=0$ ,then the family of lines will be tangent to the curve $Q^2=4PR.$ I have read this theorem in my book but i do not know how to prove it.
If a family of straight lines can be represented by an equation $\lambda^2 P+\lambda Q+R=0$ where $\lambda $ is a parameter and $P,Q,R$ are linear functions of $x$ and $y$ then the family of lines will be tangent to the curve $Q^2=4PR.$
My try:
Let $P:a_1x+b_1y+c_1=0$
$Q:a_2x+b_2y+c_2=0$
$R:a_3x+b_3y+c_3=0$
Then the family of straight lines can be represented by an equation $\lambda^2 (a_1x+b_1y+c_1)+\lambda (a_2x+b_2y+c_2)+(a_3x+b_3y+c_3)=0$
$x(\lambda^2a_1+\lambda a_2+a_3)+y(\lambda^2b_1+\lambda b_2+b_3)+(\lambda^2c_1+\lambda c_2+c_3)=0$
But i do not know how to prove that the family of lines will be tangent to the curve $Q^2=4PR.$
|
This is a pretty brute-force method but it works.
Test
To figure out what was going on, I tried a simple case: $P=x$, $Q=y$, $R=1$. The family of lines is $$\lambda^2x +\lambda y + 1 = 0$$ and the quadratic is $$y^2 = 4x$$ To say that the family of lines is tangent to the curve means that wherever they intersect, their derivatives agree.
So first, we look at the intersections. If $\lambda^2 x + \lambda y + 1 = 0$, then $y = -\lambda x -\frac{1}{\lambda}$ ($\lambda \neq 0$ because then the line's equation becomes $1=0$). So $y^2 = \lambda^2 x^2 + 2x + \frac{1}{\lambda^2}$. On the other hand, if $y^2 = 4x$, then
\begin{align*}
\lambda^2 x^2 + 2x + \frac{1}{\lambda^2} &= 4x \\
\implies
\lambda^2 x^2 - 2x + \frac{1}{\lambda^2} &= 0 \\
\implies \left(\lambda x - \frac{1}{\lambda}\right)^2 &= 0\\
\implies x &= \frac{1}{\lambda^2}
\end{align*}
If follows that $y = -\lambda \cdot \frac{1}{\lambda^2} - \frac{1}{\lambda} = -\frac{2}{\lambda}$. This jibes with the second equation $y^2 = 4x$.
The derivative $\frac{dy}{dx}$ along the line $\lambda^2 x + \lambda y + 1 = 0$ is clearly the slope $-\lambda$. The derivative along the curve $y^2=4x$ is
$$
2y \frac{dy}{dx} = 4\implies \frac{dy}{dx} = \frac{2}{y}
$$
If $y=-\frac{2}{\lambda}$ then $\frac{2}{y} = -\lambda$, as desired.
The general case
Now I'm going to be careful not to divide by zero. Let $F = \lambda^2 P + \lambda Q + R$ and $G = Q^2-4PR$. We want to show
$$
\begin{vmatrix} F_x & F_y \\ G_x & G_y \end{vmatrix} =0
$$
at the intersections $F=G=0$.
First, we reduce the equations for the intersections. If $\lambda^2 P + \lambda Q + R = 0$, then $\lambda Q = - \lambda^2 P - R$, so
$$
\lambda^2 Q^2 = \lambda^4 P^2 + 2\lambda^2 PR + R^2
$$
On the other hand, if $Q^2 =4PR$, then $\lambda^2 Q^2 = 4\lambda PR$, and we have
\begin{align*}
\lambda^4 P^2 + 2\lambda^2 PR + R^2 &= 4\lambda PR \\
\implies
\lambda^4 P^2 - 2\lambda^2 PR + R^2 &= 0 \\
\implies
\left(\lambda^2 P-R\right)^2 &=0 \\
\implies
\lambda^2 P-R &=0 \tag{1}\\
\end{align*}
Substituting $R=\lambda ^2P$ into the first equation gives
$$
2\lambda^2 P + \lambda Q = 0 \implies \lambda(2\lambda P + Q) = 0 \implies 2\lambda P + Q=0 \tag{2}
$$
Now we compute the determinant of the derivatives:
$$
\begin{vmatrix} F_x & F_y \\ G_x & G_y \end{vmatrix}
= \begin{vmatrix}
\lambda^2 P_x + \lambda Q_x + R_x &
\lambda^2 P_y + \lambda Q_y + R_y \\
2QQ_x - 4P_xR - 4PR_x &
2QQ_y - 4P_yR - 4PR_y \\
\end{vmatrix}
$$
Add $4P$ times the first line to the second line.
The determinant is
$$
\begin{vmatrix}
\lambda^2 P_x + \lambda Q_x + R_x &
\lambda^2 P_y + \lambda Q_y + R_y \\
2(2\lambda P+Q)Q_x +4(\lambda^2P- R)P_x &
2(2\lambda P+Q)Q_y +4(\lambda^2P- R)P_y
\end{vmatrix}
$$
In light of (1) and (2), the second line is zero.
|
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|
Find the Cartesian equation of the locus described by $\arg \left(\frac{z-2}{z+5} \right)= \frac{\pi}{4}$ My working:
$$ \frac{x + iy - 2}{x + iy + 5} $$
$$ \frac{(x - 2 + iy)(x+5-iy)}{(x + 5 + iy)(x+5-iy)} $$
$$ \frac{x^2+5x-ixy-2x-10+2iy+ixy+5iy+y^2}{x^2+5x-ixy+5x+25-5iy+ixy+5iy+y^2} $$
$$ \frac{x^2+3x-10+y^2+7iy}{x^2+10x+25+y^2}$$
$$ \frac {\Im(z)}{\Re(z)} = \tan \frac{\pi}{4} = 1$$
$$ x^2 + 3x + y^2 - 10 = 7y $$
$$ x^2 + 3x + y^2 - 7y = 10 $$
$$ \left(x+ \frac 32\right)^2 + \left(y- \frac 72\right)^2 = 10 + \frac 94 + \frac {49}{4} $$
$$ \left(x- -\frac 32\right)^2 + \left(y- \frac 72\right)^2 = \left(\frac {7 \sqrt{2}}{2}\right)^2 $$
Is this correct? Is there a better method than what I did here?
|
Hint for another method:
Translate this problem into pure geometry: denote $A$ the point with affix $-5$, $B$ the point with affix $2$, $M$ the point with affix $z$. The condition translates into $\angle AMB=\frac\pi4$.
Now use the inscribed angle theorem: the locus of $M$ is a circular arc. The centre $O$ of this arc is such that $\angle AOB=\frac\pi2$. Hence $O$ is the intersection point of the upper semi-circle with diameter $[AB]$ and the perpendicular bisector of $[AB]$.
|
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|
indefinite integral of $\cos(x) / x$? How can I evaluate ;
$$\int \frac{\cos(x)}{x} \, dx $$
I tried to do partial integration but it became to an infinite loop of partial integrations
|
Recall the Taylor series expansion for $\cos x$
\begin{align*}\cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \\
\implies \frac{\cos x}{x} &= \frac{1}{x} - \frac{x}{2!} - \frac{x^3}{4!} - \frac{x^5}{6!} + \cdots \\
\therefore \int \frac{\cos x}{x} &= ln\ x - \frac{x^2}{2 \times 2!} + \frac{x^4}{4 \times 4!} - \frac{x^6}{6 \times 6!} + \cdots + \frac{x^{2k}}{(-1)^{k}\times2k \times (2k)!} + \cdots
\end{align*}
This sum approaches zero so that the indefinite integral is $ln(x)$ up to an integration constant. Moreover, if the terminals of integration are say $a$ and $b$ (not zero or infinity), the definite integral would be $ln(a)-ln(b)$. Where the terminals include zero or infinity, the Trigonometric Integral $Ci(x)$ or $Cin(x)$ need to be used.
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|
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0<x<1$, then find $\int f(x)\mathrm{d}x$
$$f(x)=\lim_{n\to\infty}2x[1+2x^2+\cdots+nx^{2n}]$$
$$S=\frac{f(x)}{2x}=1+2x^2+\cdots+nx^{2n}$$
$S$ is an AGP. I used the general method for finding $S$.
$$x^2S=x^2+2x^4+\cdots+nx^{2n+2}$$
$$(1-x^2)S=1+(x^2+x^4+\cdots+x^{2n})+nx^{2n+2}$$
$$(1-x^2)S=\frac{1-x^{2n+1}}{1-x^2}+nx^{2n+2}$$
$$S=\frac{1-x^{2n+1}}{(1-x^2)^2}+\frac{nx^{2n+2}}{1-x^2}$$
$$f(x)=\lim_{n\to\infty}\left(2x\frac{1-x^{2n+1}}{(1-x^2)^2}+2x\frac{nx^{2n+2}}{1-x^2}\right)$$
I got stuck here.
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You can turn that long expression into a infinite series.
$$ f(x) = \sum_{n=1}^\infty 2n x^{2n-1}
$$
Let $g = \int f(x)\,dx$. By termwise integration, (This is justified by differentiating $g$ and noticing that $g$ has the same radius of convergence as $f$. Then it follows from the uniqueness of antiderivative up to a constant.)
$$ g(x) = \sum_{n=1}^\infty 2n \frac{x^{2n}}{2n} = \sum_{n=1}^\infty x^{2n} = \sum_{n=1}^\infty (x^2)^n = \frac{x^2}{1-x^2}
$$
and finally add the constant of integration to the end.
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System of inequalities $x^2+2x+\alpha\leq0$ and $x^2-4x-6\alpha\leq 0$ has unique solution.
Find all values of $\alpha$ for which the system of inequalities $x^2+2x+\alpha\leq0$
and $x^2-4x-6\alpha\leq 0$
has unique solution.
My Try: We can write it as $$\frac{x^2-4x}{6}\leq \alpha \leq -x^2-2x$$
So we get $$\frac{(x-2)^2-4}{6}\leq \alpha \leq -(x+1)^2+1$$
So for the existence of solution, Here $-\frac{2}{3}\leq \alpha \leq 1$.
Now how can I solve for unique solution?
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As both the Quadratic inequalities is less than zero, so we can say that their discriminant must be greater than or equal to Zero ,i.e., real roots must exists.
So, we will get $$4-4\alpha \ge 0$$ $$16 + 24\alpha \ge 0$$On solving these eqations, we will get $$\frac{-2}3 \le \alpha \le 1$$ Now, we can determine the roots of these quadratic equations as, $$-1 \pm \sqrt{1-\alpha}$$ $$2 \pm \sqrt{4 + 6\alpha}$$ Now, we can see that both the roots of first equation are negative and one of the roots of the other equation is positive.
Therefore for a unique solution, the other root of the second equation must be negative. Thus, $$2 - \sqrt{4 + 6\alpha} \lt 0$$ On solving this, we get$$\alpha \gt 0$$ and we know that, $$\frac{-2}3 \le \alpha \le 1$$ Therefore, the required set of values of $\alpha$ is $$0 \lt \alpha \le 1$$
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|
Find a plane perpendicular to $yz$, passing by a point and making an angle with another plane The problem is to find the equation of a plane (let's call it $A$) that is perpendicular to the $yz$ plane, containing the point $P(2,1,1)$, and making an angle of $\cos^{-1} \frac{2}{3}$ with the plane $2x - y + 2z - 3 = 0$ (let's call it $B$).
First, I know that the equation perpendicular to the $yz$ plane will have $x=0$. Second, I think that the angle between planes $A$ and $B$ will still be $\cos^{-1} \frac{2}{3}$ where $x=0$, ie within the $yz$ plane. So, the equation of the line intersecting planes $B$ and $yz$ is
$$
-y + 2z - 3 = 0
$$
And let the line intersecting planes $A$ and $yz$ be
$$
b(y-1) + c(z-1) = 0
$$
Third, I also think that the angle between the normal lines perpendicular to those equations should still be $\cos^{-1} \frac{2}{3}$. So, I use this formula and make a few calculations
$$
\vec{B}\cdot \vec{A} = \|\vec{B}\|\|\vec{A}\|\cos \theta \\
\langle 0,-1,2\rangle \cdot \langle 0,b,c\rangle = \|\langle 0,-1,2 \rangle \|\|\langle 0,b,c \rangle \| \frac{2}{3} \\
0\cdot 0 + (-1)\cdot b + 2\cdot c = \sqrt{0^2+(-1)^2+2^2}\sqrt{0^2+b^2+c^2}\frac{2}{3} \\
-b + 2c = 2 \sqrt{b^2+c^2} \\
\frac{b^2}{4} - bc + c^2 = b^2 + c^2 \\
4c = -3b \\
$$
This looks close to a solution but I can't seem to reach the actual solutions, because there are actually two, of $4y-3z-1=0$ and $z=1$.
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With the relationship you found between $b$ and $c$, as well as $x=0$, you have
$$
ax + by + cz + d = 0 \\
a(0) + by + (\frac{-3b}{4})z + d = 0 \\
4by - 3bz + 4d = 0 \\
$$
With the components of $(2,1,1)$, you then have
$$
4b(1) - 3b(1) + 4d = 0 \\
b = -4d \\
$$
With this relationship between $b$ and $d$, you finally have
$$
4by - 3bz + 4(\frac{-b}{4}) = 0 \\
4y - 3z -1 = 0 \\
$$
Note that you made a mistake when calculating $\| \vec{B} \|$ as $\sqrt{0^2+(-1)^2+2^2} = \sqrt{5}$. Instead, you should have $\sqrt{2^2+(-1)^2+2^2} = \sqrt{9} = 3$ which cancels nicely with the denominator. You did this anyway which is a case where two wrongs make a right.
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If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then show that $\tan \beta=\tan\alpha(4\sec^2\alpha-1)$.
The chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is given by $x\cos\frac{\alpha-\beta}{2}-y\sin\frac{\alpha+\beta}{2}=a\cos\frac{\alpha+\beta}{2}$....................$(1)$
And the normal to the hyperbola at $P(a\sec\alpha,a\tan\alpha)$ is given by
$\frac{x}{\sec\alpha}+\frac{y}{\tan\alpha}=2a$.............$(2)$
equation $(1)$ and $(2)$ are the same lines
Comparing them,gives
$\frac{\cos\frac{\alpha-\beta}{2}}{\frac{1}{\sec\alpha}}=\frac{-\sin\frac{\alpha+\beta}{2}}{\frac{1}{\tan\alpha}}=\frac{a\cos\frac{\alpha+\beta}{2}}{2a}$
$\sec\alpha=\frac{\cos\frac{\alpha+\beta}{2}}{2\cos\frac{\alpha-\beta}{2}}$ and $\tan\alpha=\frac{\cos\frac{\alpha+\beta}{2}}{-2\sin\frac{\alpha+\beta}{2}}$
I am stuck here.I do not know how to proceed further.
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As $(2)$ passes through $Q(a\sec\beta,a\tan\beta),$
$$\dfrac{\sec\beta}{\sec\alpha}=2-\dfrac{\tan\beta}{\tan\alpha}$$
Squaring and writing $\tan\alpha=p,\tan\beta=q,$
$$\dfrac{1+q^2}{1+p^2}=4+\dfrac{q^2}{p^2}-\dfrac{4q}p$$
$$\iff4-\dfrac{4q}p=\dfrac{1+q^2}{1+p^2}-\dfrac{q^2}{p^2}$$
$$\iff\dfrac{4(p-q)}p=\dfrac{p^2-q^2}{(1+p^2)p^2}$$
If $p=q$ i.e., $\tan\alpha=\tan\beta,(2)\implies\sec\beta=\sec\alpha\implies P=Q$
So, cancelling $p-q\ne0$ and assuming $p\ne0,$
$$\iff4=\dfrac{p+q}{(1+p^2)p}\iff q=4p^3+3p=p\{4(p^2+1)-1\}$$
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|
Given $f(z)=(z^2-1)^{\frac{1}{2}}$ find the first three terms of the Laurent expansion Given
$$f(z)=(z^2-1)^{\frac{1}{2}}$$
which has a branch cut for $|z|<1$. Find the first three terms of the Laurent expansion.
I proceed by factoring out a $z^2$.
$$z(1-\frac{1}{z^2})^{\frac{1}{2}}$$
where I know the following expansion
$$\frac{1}{z^2}=\sum_{n=0}^{\infty} (n+1)(z+1)^n=1+2(z+1)+3(z+1)^3+ \ldots$$
However, although plugging this in results in a cancellation of the 1's, I do not know how to deal with the 1/2 power.
$$z(2(z+1)+3(z+2)^3+\ldots)^{\frac{1}{2}}$$
Advice, and also how might I deal with the branch cut?
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This function has branch points at $z=+1$ and at $z=-1$ and a branch line connecting these two points. Because it is a square-root singularity, the branch line for $|z|>1$ cancels, as may be seen by considering the net phase change when both branch points are encircled:
\begin{align*}
\left.\arg\left(\sqrt{z^2-1}\right)\right|_{\arg z =0}^{\arg z=2\pi}=0\qquad \mod 2\pi
\end{align*}
So, the Laurent expansion for $|z|>1$ is using the binomial series expansion
\begin{align*}
\left(z^2-1\right)^{\frac{1}{2}}&=z\left(1-\frac{1}{z^2}\right)^{\frac{1}{2}}\\
&=z\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}\left(-\frac{1}{z^2}\right)^n\\
&=z\left[\binom{\frac{1}{2}}{0}-\binom{\frac{1}{2}}{1}\frac{1}{z^2}+\binom{\frac{1}{2}}{2}\frac{1}{z^2}\mp\cdots\right]\\
&=z-\frac{1}{2z}-\frac{1}{8z^3}\mp\cdots
\end{align*}
Note: This is example 6.3.1 found here.
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The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is
A. 0;
B. 3;
C. 5;
D. 1.
I don't know how to solve this.
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Consider the polynomial $\mod{x^3 +1} $
$x^5 + 2x^3 + x^2 + 2 \equiv -x^2 -2 + x^2 + 2 \equiv 0$
Thus, the polynomial has a factor of $x^3+1$
The remainder of the question can be done as above.
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How do i find eigenvectors for a $3\times 3$-matrix when eigenvalues are mixed complex or real? My task is this;
Find the eigenvalues and eigenvectors for the matrix: $A = \begin{pmatrix}2 & 1 & -3\\4 & 2 & 3\\0 & 0 & 1\end{pmatrix}$.
My work so far is this:
Apply the lemma that states $\lambda$ is and eigenvalue for $n\times n$-matrix M iff $det(\lambda I_n - M) = 0$.
We get (developing $A$ along row 3 in the determinant step):
$\lambda I_3 - A = \begin{pmatrix}\lambda - 2& -1 & 3\\-4&\lambda - 2&-3\\0& 0&\lambda - 1\end{pmatrix} \to det(A) = (\lambda - 1)\begin{vmatrix}\lambda - 2& -1\\-4 & \lambda - 2\end{vmatrix} =\\[-10mm]$
$(\lambda - 1)((\lambda - 2)^2 - 4)= \lambda(\lambda - 1)(\lambda - 4)$.
Which gives us $\lambda = \{0, 1, 4\}$ which in this case were real, but i'm supposed to solve for mixed complex solutions aswell (recall that complex solutions occur in pairs and that we will get three solutions for a third degree polynomial from the determinant of a $3\times 3$).
We should get three equations: $$\begin{equation} Av_1 = 0v_1\\ Av_2 = v_2\\ Av_3 = 4v_3\end{equation}$$
Now i'm not sure how to determine the eigenvectors from these equations? This time i would actually like someone to show and explain the reasoning behind the missing step from these equations to the set of eigenvectors. Don't get me wrong, i can setup the system of equations, but from there on i'm confused on how to determine the eigenvectors in terms of real numbers and not in terms of variables.
Thanks in advance!
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Your eigenvalues are correct. Let's step through one eigenvector example and you can do the other two. We will use $\lambda = 0$. Note that there are other approaches to finding eigenvectors, this is just one approach.
To find the eigenvectors, we want to find $v$ such that $Av = \lambda v$ or $(A- \lambda I) v = 0$.
For $\lambda_1 = 0$,
$(A- \lambda_1 I)v_1 =
\begin{pmatrix}2 & 1 & -3\\4 & 2 & 3\\0 & 0 & 1\end{pmatrix} v_1 = \begin{pmatrix}
0 \\
0 \\
0
\end{pmatrix}$.
Row reduction yields
$$\begin{pmatrix}1 & \dfrac{1}{2} & 0\\0 &0 & 1\\0 & 0 & 0\end{pmatrix} v_1 = \begin{pmatrix}1 & \dfrac{1}{2} & 0\\0 &0 & 1\\0 & 0 & 0\end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$
In other words, for the eigenvector $v_1 = (a, b, c)$,
\begin{align*}
a &= -\dfrac{1}{2} b \\
c &= 0
\end{align*}
This gives us a free variable, $b$, that we can set to anything we'd like (in other words, eigenvectors are not unique). To make the eigenvector easy to write, let's choose $b = 2$, giving us the eigenvector $$v_1 = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}$$
Next, try finding the eigenvectors for $\lambda_2 = 1$ and $\lambda_3 = 4$.
One solution for these eigenvectors can be:
$$v_2 = \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}, v_3 \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$$
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solve $\int \frac{dx}{3+\sqrt{x+2}}$
$$\int \frac{dx}{3+\sqrt{x+2}}$$
How should I approach this? U-substituion does not seem to work
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Let $y = x + 2, dy = dx$.
$$ \int \frac{dx}{3+\sqrt{x+2}} = \int \frac{dy}{\sqrt{y} + 3}
$$
Let $z = \sqrt{y}, dz = dy/(2\sqrt{y})$.
$$ \int \frac{dy}{\sqrt{y} + 3} = \int \frac{2\sqrt{y} dz}{z+3} = 2\int \frac{z}{z+3} dz
$$
Then do partial fraction decomposition.
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|
Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is $P(|Y| > 1)$?
Suppose that $X_1,X_2,\ldots,X_{25}$ are independent random variables from $\mathcal{N}(1, 4)$. Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is the probability that $|Y| > 1$?
We know that $Y$ is a standard normal random variable.
I keep trying to transform it into normal distribution, but I keep getting the wrong answer. The answer is $0.6826$. How do I solve this?
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Assuming "$4$" is the variance, ...
$X$ is a linear function of i.i.d. r.v.s. It's mean is $\frac{1}{25} ( 25 \cdot 1) = 1$. It's variance is $\left( \frac{1}{25} \right)^2 (25 \cdot 4) = 4/25$. (This is a standard result. There are several ways to get this. Google "sums of normally distributed variables" for a plethora.)
$Y$ is a linear function of $X$. It's mean is $\frac{5}{2} \cdot 1 - \frac{2}{5} = \frac{21}{10}$. It's variance is $\left( \frac{5}{2} \right)^2 \cdot \frac{4}{25} = 1$. (This is another standard result. Ask Professor Google for "linear transform of normally distributed variable".)
Note that $Y - \frac{21}{10}$ is a $\mathcal{N}(0,1)$ distributed random variable, so we can use standard tables to look up \begin{align*}
|Y| &> 1 \\
Y < -1 &\text{ or } Y > 1 \\
Y - \frac{21}{10} < \frac{-31}{10} &\text{ or } Y - \frac{21}{10} > \frac{-11}{10} \text{.}
\end{align*} This is $\frac{1}{2} \left(\text{erf}\left(\frac{11}{10
\sqrt{2}}\right)-\text{erf}\left(\frac{31}{10 \sqrt{2}}\right)\right) + 1 = 0.865302\dots$.
If, on the other hand, "$4$" is the standard deviation, the same process leads to the probability being $0.769411\dots$.
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The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Find the sum of the distances of these four points The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Compute the sum of the distances of these four points from the point $(-3,2).$
$x^2+y^2+6x-24y+72=0$ is a circle and $x^2-y^2+6x+16y-46=0$ is a hyperbola.These cut at four points.
$x^2+y^2+6x-24y+72=0$ becomes $(x+3)^2+(y-12)=9^2$.
$x^2-y^2+6x+16y-46=0$ becomes $(x+3)^2-(y-8)^2=-9$.
When i solved these two equations to find the points of intesection,$y=10\pm \sqrt{41}$ and $x$ is $-3\pm\sqrt{36\pm 4\sqrt{41}}$
Now it is very difficult to find the sum of distances of these points from $(-3,2)$.Answer is $40$ given in the book.
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note: for any point of 4, the distance to $(-3,2)$ is $\sqrt{(-3\pm\sqrt{36\pm 4\sqrt{41}}-(-3))^2+(10\pm \sqrt{41}-2)^2}=\sqrt{36\pm 4\sqrt{41}+(8\pm \sqrt{41})^2}=\sqrt{(10\pm\sqrt{41})^2}=10\pm\sqrt{41}$
so it is easy to see the final answer.
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|
How can I evaluate $\int_{0}^{1}\frac{(\arctan x)^2}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x$ How to calculate this relation?
$$I=\int_{0}^{1}\frac{(\arctan x)^2}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=\frac{\pi^3}{96}\ln{2}-\frac{3\pi\zeta{(3)}}{128}-\frac{\pi^2G}{16}+\frac{\beta{(4)}}{2}$$
Where G is the Catalan's constant, and $$\beta(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^{x}}$$ is the Dirichlet's beta function.
Integrate by parts $$u=(\arctan{\,x})^2\ln{(1+x^2)}$$ $$v=\arctan{\,x}$$
We have $$3I=\frac{\pi^3}{64}\ln{2-2\int_0^{1}x(\arctan{\,x})^3}\frac{dx}{1+x^2}$$
But how to calculate the latter integral?Could anybody please help by offering useful hints or solutions?Ithing very difficult to prove.
|
Take $\arctan\left(x\right)=u,\,\frac{dx}{1+x^{2}}=du
$. Then $$I=\int_{0}^{1}\frac{\left(\arctan\left(x\right)\right)^{2}}{1+x^{2}}\log\left(1+x^{2}\right)dx=\int_{0}^{\pi/4}u^{2}\log\left(1+\tan^{2}\left(u\right)\right)du
$$ $$=-2\int_{0}^{\pi/4}u^{2}\log\left(\cos\left(u\right)\right)du
$$ and now using the Fourier series $$\log\left(\cos\left(u\right)\right)=-\log\left(2\right)-\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(2ku\right)}{k},\,0\leq x<\frac{\pi}{2}
$$ we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}+2\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}\int_{0}^{\pi/4}u^{2}\cos\left(2ku\right)du
$$ and the last integral is trivial to estimate $$\int_{0}^{\pi/4}u^{2}\cos\left(2ku\right)du=\frac{\pi^{2}\sin\left(\frac{\pi k}{2}\right)}{32k}-\frac{\sin\left(\frac{\pi k}{2}\right)}{4k^{3}}+\frac{\pi\cos\left(\frac{\pi k}{2}\right)}{8k^{2}}
$$ so we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}+\pi^{2}\sum_{k\geq1}\frac{\left(-1\right)^{k}\sin\left(\frac{\pi k}{2}\right)}{16k^{2}}-\sum_{k\geq1}\frac{\left(-1\right)^{k}\sin\left(\frac{\pi k}{2}\right)}{2k^{4}}+\pi\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(\frac{\pi k}{2}\right)}{4k^{3}}
$$ and now observing that $$\cos\left(\frac{\pi k}{2}\right)=\begin{cases}
-1, & k\equiv2\,\mod\,4\\
1, & k\equiv0\,\mod\,4\\
0, & \textrm{otherwise}
\end{cases}
$$ and $$ \sin\left(\frac{\pi k}{2}\right)=\begin{cases}
-1, & k\equiv3\,\mod\,4\\
1, & k\equiv1\,\mod\,4\\
0, & \textrm{otherwise}
\end{cases}
$$ we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}-\frac{\pi^{2}}{16}K+\frac{\beta\left(4\right)}{2}-\frac{3\pi\zeta\left(3\right)}{128}\approx 0.064824$$ where the last sum is obtained using the relation between Dirichlet eta function and Riemann zeta function.
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What is $\frac{{\partial f(\left| z \right|)}}{{\partial x}} = $? Let $f(z)$ is polynomial and $z=x+iy$ , $x,y\in R$.
What is $\frac{{\partial f(\left| z \right|)}}{{\partial x}} = $?
|
$$\begin{align} & \frac{{\partial f(\left| z \right|)}}{{\partial x}}
\\ & =\frac{{\partial f(\sqrt{x^2+y^2})}}{{\partial x}}
\\ & =\frac{\partial f(\sqrt{x^2+y^2})}{\partial (\sqrt{x^2+y^2})}\cdot \frac{\partial (\sqrt{x^2+y^2})}{\partial x}
\\ & =\frac{d\{f(\sqrt{x^2+y^2})\}}{d(\sqrt{x^2+y^2})}\cdot \frac{\partial (\sqrt{x^2+y^2})}{\partial x}
\\ & =\color{red}{f'(\sqrt{x^2+y^2})\cdot \frac{x}{\sqrt{x^2+y^2}}}\end{align}$$
Hope this helps.
|
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|
Series of squares of n integers - where is the mistake? Given the following two series:
$$1^3 + 2^3 + ... + n^3$$
$$0^3 + 1^3 + .... + (n-1)^3$$
I take the difference vertically of the two:
$$\left(1^3-0^3\right) + \left(2^3-1^3\right) + .... + \left(n^3-(n-1)^3\right)$$
This equals to $n^3$
If I now express this in sum notation:
$$\sum_{i=1}^n\left(i^3-(i-1)^3\right) = n^3$$
If I expand: $(i-1)^3 = i^3 - 3i^2 + 3i - 1$
Thus
$$\left(i^3 - (i-1)^3\right) = 3i^2 - 3i +1$$
My sum is now:
$$3\sum_{i=1}^n i^2 -3 \sum_{i=1}^n i + n = n^3$$
$$\sum_{i=1}^n i^2 = \frac{1}{3} \left(n^3 + 3 \frac{n(n-1)}{2} -n\right)$$
And the expression on the RHS above is not $\frac{1}{6} n (n+1) (2n+1)$
I don't want to solve the above using forward difference, I want to keep it backward.
|
$$\sum_{i=1}^n i={n(n+1)\over2}\quad\text{ not }\quad {n(n-1)\over2}$$
|
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|
Closed form for $\sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2} \,dx$ I am trying to get this to equal $\displaystyle-\frac {\pi^a}{b}$ for some positive integers $a$ and $b$ . My efforts so far give:
$\displaystyle \sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2} \,dx=A-B+C$
where $A=\displaystyle \sum_{n=1}^\infty\frac{2\zeta(3)}{n^3}=2\zeta(3)^2$
and $C=\displaystyle \sum_{n=1}^\infty\frac{\pi^2}{3n^4}=\frac{\pi^6}{270}$
and $B=\displaystyle \sum_{n=1}^\infty\frac{f(n)}{n^6}$ with f(1)=6, f(2)=34, f(3)=393/4, f(4)=5750/27, f(5)=339059/864, f(6)=325493/500, f(7)=36107863/36000, ... ,f(11)=29071954407257/8001504000, ...
I can not figure out what $f(n)$ might be. The sequences of numerators and denominators are unrecognized by OEIS. Maybe I am off the track.
Perhaps a different approach from the beginning would be better. The results for A and C were interesting anyway.
|
Hint. One may use some MZVs algebra.
For $n\geq1$, set
$$
I_n:=\frac1{n^2}\int_0^1 x^{n-1}\ln^2(x)\ln(1-x)\:dx,
$$ then, differentiating the Euler beta integral three times, one gets
$$
I_n=-\frac4{n^6}-2\frac{H_{n-1}}{n^5}+2\frac{\frac{\pi^2}6-H_{n-1,2}}{n^4}+2\frac{\zeta(3)-H_{n-1,3}}{n^3}
$$ summing with respect to $n$ gives
$$
\sum_{n=1}^\infty I_n=-4\zeta(6)-2\zeta(5,1)+\frac{\pi^2}3\zeta(4)-2\zeta(4,2)+2\zeta(3)^2-2\zeta(3,3),
$$ using
$$
\begin{align}
\zeta(5,1)&=-\frac12\zeta(3)^2 +\frac{\pi^6}{1260}
\\\\ \zeta(4,2)&= \zeta(3)^2 -\frac{4\pi^6}{2835}
\\\\ \zeta(3,3)&=\frac12\zeta(3)^2 -\frac{\pi^6}{1890}
\end{align}
$$ leads to
$$
\sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2}=-\frac{\pi^6}{2835}
$$
or
$$
\int_0^1 \frac{\text{Li}_2(x)\ln^2(x)\ln(1-x)}x=-\frac{\pi^6}{2835}.
$$
Probably there is a direct path using $\rm{Li}_2(\cdot)$ properties.
|
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|
Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reasoning?
Question
Calculate the limit of:
$$x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
when $n$ goes to infinity.
Answer
This can also be written as:
$$\lim_{n \to \infty} \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
The denominator can be written as:
$$\ln(1 + \sqrt[3]{n} + \sqrt[4]{n}) = \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) + \ln(\sqrt[3]{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) \rightarrow 0$$
The numerator can be written as:
$$\ln(1 + \sqrt{n} + \sqrt[3]{n}) = \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) + \ln(\sqrt{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) \rightarrow 0$$
This means that we have the following limit:
$$\lim_{n \to \infty} \frac{\ln(\sqrt{n})}{\ln(\sqrt[3]{n})} = \lim_{n \to \infty} \frac{\ln(n^{\frac{1}{2}})}{\ln(n^{\frac{1}{3}})} = \lim_{n \to \infty} \frac{\frac{1}{2}\ln(n)}{\frac{1}{3}\ln(n)} = \lim_{n \to \infty} \frac{3 \ln(n)}{2\ln(n)} \rightarrow \frac{3}{2}$$
The limit converges towards $\frac{3}{2}.$
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With all due respect to the answers already posted, I'm not in favor of using "equivalents" without enough explanation, or writing $\lim_{n\to \infty}$ before the limit has been shown to exist. Instead, I would do this: The expression for $n>1$ is less than
$$\frac{\ln (3n^{1/2})}{\ln n^{1/3}} = \frac{\ln 3 + (1/2)\ln n}{(1/3)\ln n} = \frac{\ln 3}{(1/3)\ln n} + \frac{3}{2} \to \frac{3}{2}.$$
Similarly, the expression is greater than
$$\frac{\ln n^{1/2}}{\ln (3n^{1/3})} = \frac{(1/2)\ln n}{\ln 3 +(1/3)\ln n} = \frac{(1/2)}{(\ln 3)/\ln n +(1/3)} \to \frac{3}{2}.$$
By the squeeze theorem, the limit is $3/2.$
|
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|
What is the Permutation Matrix in FFT DFT Factorization? Given:
$$F_N = \frac{1}{\sqrt{N}} \begin{bmatrix}
1&1&1&1&\cdots &1 \\
1&\omega&\omega^2&\omega^3&\cdots&\omega^{N-1} \\
1&\omega^2&\omega^4&\omega^6&\cdots&\omega^{2(N-1)}\\ 1&\omega^3&\omega^6&\omega^9&\cdots&\omega^{3(N-1)}\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&\omega^{N-1}&\omega^{2(N-1)}&\omega^{3(N-1)}&\cdots&\omega^{(N-1)(N-1)}\\
\end{bmatrix}, $$
where $D = $ diag($1,\omega,\omega^2,...,\omega^N$). We note that $\omega^N = 1$, so $\omega^{N/2} = -1 $.
What is meant by odd-even permutation in this context?
Available notes on FFT online are generally highly condensed, thus they are very difficult to decipher. Using $F_4$ as a template, i've found that permutation matrices $P$ such that $P F$ is row sorted into odd powers of $\omega$ and then even powers of $\omega$, and visa versa, don't ostensibly work.
I additionally am not sure how you can convert the consecutive $F_{N/2}$ matrices on the top of the second matrix into rows that span to all necessary powers of $\omega$.
Thank you.
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The permutation matrix P for FFT of $F_8$ is
$
P=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&0&0&0&1&0\\
0&1&0&0&0&0&0&0\\
0&0&0&1&0&0&0&0\\
0&0&0&0&0&1&0&0\\
0&0&0&0&0&0&0&1
\end{bmatrix}
$
|
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|
Prove that $\lim_{x\to\infty} \frac{1}{x}\int_0^xf(t)dt$ exists and find it, where $f(t)$ is an alternating function. The problem: function $f$ alternates between $1$ and $-1$ on $[0^2; 1^2), [1^2; 2^2), ... [(n-1)^2; n^2)$, so on $[0^2; 1^2], \;f(x) = 1;$ on $[1^2; 2^2],\;f(x) = -1$ etc.
Prove that $\;\;\lim_{x\to\infty} \frac{1}{x}\int_0^xf(t)dt\;\;$ exists and find it.
What I did:
Questions: Where did I go wrong? I honestly don't believe this limit exists but it says so in the problem statement so I don't even know. I tried doing it without the $x \in \mathbb N$, but I get the same result: The limit can't exist because the value alternates between $1$ and $-1$. Thank you!
|
Let $x \in \mathbf R$. Then there is a unique integer $n \in \mathbf N$ such that $n^2 \le x < (n+1)^2$. We have
\begin{align*}
\int_0^x f(t)\, dt &= \sum_{k=0}^{n-1} (-1)^k (2k+1) + (-1)^n(x-n^2)\\
&= 2\sum_{k=0}^{n-1} (-1)^k k + \frac 12\bigl(1 + (-1)^{n}\bigr) + (-1)^n(x-n^2)\\
&= \frac 12\bigl((-1)^{n-1}2n - 1\bigr) + \frac 12\bigl(1+(-1)^n\bigr) + (-1)^n(x-n^2)\\
\end{align*}
Dividing by $x$ now (note that $x \sim n^2$, this was your main mistake, if you let $x \in \mathbf N$, that is fine, but the sum goes up to $\sqrt x$ then, only), we have
\begin{align*}
\def\abs#1{\left|#1\right|}\abs{\frac 1x \int_0^x f(t)\, dt}
&\le \frac 12 \frac{2n + 1}x + \frac 1x + \frac{x-n^2}x
\end{align*}
Now $x-n^2 \le (n+1)^2 - n^2 = 2n+1$ and $\frac 1x \le \frac 1{n^2}$ as $n^2\le x$. Hence
\begin{align*}
\abs{\frac 1x \int_0^x f(t)\, dt} &\le \frac 32 \frac{2n +1}{n^2} + \frac 1{n^2}
\end{align*}
Now $n \to \infty$ for $x \to \infty$, that is the above converges to $0$, therefore
$$ \frac 1x \int_0^x f(t)\, dt \to 0, \qquad x \to \infty $$
|
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|
Induction for divisibility: $3\mid 12^n -7^n -4^n -1$ I must use mathematical induction to show that
$a_{n} = 12^n −7^n −4^n −1$ is divisible by 3 for all positive integers n.
Assume true for $n=k$
$a_{k} = 12^k -7^k -4^k -1$
Prove true for $n=k+1$
$a_{k} = 12^{k+1} -7^{k+1} -4^{k+1} -1$
$ = (12^k)(12) - (7^k)(7) - (4^k)(4) -1$
$ = (12^k)(12) - (7^k)(3+4) - (4^k)(3+1) -1$
I'm not really sure about the last step, as someone just told me to do it. Am I supposed to find the right addends to use and then distribute the exponent terms until I get a multiple of the original $a_{k}$? Because I can't get it to work out evenly, and the -1 at the end gives me trouble. Also, I know that $12^n$ is a multiple of three already, but I don't know how to implement that fact to my advantage. Can I prove that $7^{n}-4^{n}-1$ is also a multiple of three and go from there?
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Since $(x-12)(x-7)(x-4)(x-1)=x^4-24x^3+183x^2-496x+336$, we have that
$a_n=12^n-7^n-4^n-1$ satisfies
$$
a_n=24a_{n-1}-183a_{n-2}+496a_{n-3}-336a_{n-4}\tag{1}
$$
Since the first $4$ values
$$
a_1=0,a_2=78,a_3=1320,a_4=18078\tag{2}
$$
are all divisible by $6$, induction with $(1)$ and $(2)$ insure that $6\mid a_n$ for all $n\ge1$.
|
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|
Perimeter and area of a regular n-gon. A friend of mine asked me how to derive the area and perimeter of a regular $n$-gon with a radius $r$ for a design project he is working on. I came up with this, but I want to make sure I didn't make any errors before giving it to him.
First, I assumed that the $n$-gon was inscribed in a circle of radius r centered at the origin, with the first vertex of the circle being at the point $(r,0)$.
The vertices of the $n$-gon will divide the circle into $n$ equal sections. Because the total angle of a circle is $2\pi$, then the angle between the $x$-axis and the second vertex is $\frac{2\pi}{n}$. Using trigonometry, the coordinates of this vertex are $\left(r\cos\left(\frac{2\pi}{n}\right), r\sin\left(\frac{2\pi}{n}\right)\right)$.
Now, the origin, the first vertex, and the second vertex form a triangle. The edge of this triangle which touches the circle in two places, using the distance formula, will have a length of $r\sqrt{\left(\cos\left(\frac{2\pi}{n}\right)-1\right)^2 + \left(\sin\left(\frac{2\pi}{n}\right)\right)^2}$.
Now, the $n$-gon will be made up of $n$ of these triangles, and so the perimeter is: $nr\sqrt{\left(\cos\left(\frac{2\pi}{n}\right)-1\right)^2 + \left(\sin\left(\frac{2\pi}{n}\right)\right)^2}$.
Now, the triangle has a base of $r$ and a height of $r\sin(\frac{2\pi}{n})$. There area of a triangle is half the product of its base and height, so the area of the triangle is $\frac{r^2\sin\left(\frac{2\pi}{n}\right)}{2}$.
Again, the $n$-gon is made up of $n$ of these triangles, so its area is: $\frac{nr^2\sin\left(\frac{2\pi}{n}\right)}{2}$
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Consider a regular polygon with side length $s$ inscribed in a circle with radius $r$. Let $\theta$ be the measure of a central angle subtended by a side of the regular polygon as shown in the figure below.
As you observed, since a full revolution is $2\pi$ radians, each central angle that subtends a side of an inscribed regular polygon with $n$ sides has measure
$$\theta = \frac{2\pi}{n}$$
Each triangle that is formed by connecting the center of the circle to adjacent vertices of the inscribed regular polygon is isosceles since the segments connecting the center to the vertices are radii of the circle.
Let's look more carefully at a triangle formed by connecting the center of the circle to adjacent vertices of the regular polygon. If we draw an altitude from the vertex angle to the base of an isosceles triangle, it bisects both the vertex angle and the base, as shown in the Figure below.
The perimeter of a regular polygon with $n$ sides of side length $s$ is $P = ns$. Since
$$\frac{s}{2} = r\sin\left(\frac{\theta}{2}\right)$$
and
$$\frac{\theta}{2} = \frac{1}{2} \cdot \frac{2\pi}{n} = \frac{\pi}{n}$$
we have
$$\frac{s}{2} = r\sin\left(\frac{\pi}{n}\right) \implies s = 2r\sin\left(\frac{\pi}{n}\right)$$
Hence, the perimeter of the regular polygon is
$$P = ns = n\left[2r\sin\left(\frac{\pi}{n}\right)\right] = 2nr\sin\left(\frac{\pi}{n}\right)$$
Note that the length of the altitude of the triangle is
$$a = r\cos\left(\frac{\theta}{2}\right) = r\cos\left(\frac{\pi}{n}\right)$$
Hence, the area enclosed by the triangle is
\begin{align*}
A_{\triangle} & = \frac{1}{2}sa\\
& = \frac{1}{2}\left[2r\sin\left(\frac{\pi}{n}\right)\right]\left[r\cos\left(\frac{\pi}{n}\right)\right]\\
& = \frac{1}{2}r^2\left[2\sin\left(\frac{\pi}{n}\right)\cos\left(\frac{\pi}{n}\right)\right]\\
& = \frac{1}{2}r^2\sin\left(\frac{2\pi}{n}\right)
\end{align*}
Since the area enclosed by the regular polygon is comprised of $n$ such triangular regions, the area enclosed by the regular polygon is
$$A = \frac{1}{2}nr^2\sin\left(\frac{2\pi}{n}\right)$$
which agrees with the answer you obtained by taking one of the legs as the base of the triangle.
|
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|
Invert $\overline{x+1}$ in $\mathbb Q[x]/(x^2+x+1)$
Invert $\overline{x+1}$ in $\mathbb Q[x]/(x^2+x+1)$.
So i know that the coset representatives for $\mathbb Q[x]/(x^2+x+1) = \{a+bx : a,b \in \mathbb{Q}\}$. But I am unsure as to how to invert this. Any help?
would $\overline{x+1} = \overline{(x^2+x+1) - x^2} = \overline{0} - \overline{x^2} = \overline{-x^2} $ work? I am not sure where to go from here.
|
We need to find some polynomial $P(x)$ so that $P(x)\cdot (x+1) = 1$. In $\Bbb Q[x]/\langle x^2+x+1\rangle$ we have $x^2 = -x-1$, so there are no polynomials of degree 2 or higher. The polynomial $P$ that we seek therefore has the form $q_1x+q_0$. So we want $$(q_1x+q_0)(x+1) = 1\qquad \pmod{x^2+x+1}.$$
Expanding the left-hand side we get $$q_1x^2+(q_1+q_0)x+ q_0 = 1$$
and then because $x^2 = -x-1$ we can reduce this to
$$\begin{align}q_1(-x-1)+(q_1+q_0)x+ q_0 &= \\
q_0x+ q_0-q_1 &= 1
\end{align}$$
Equating coefficients on both sides, we find $q_0=0$ and $q_1 = -1$, so $$P(x) = -x.$$
To check, multiply $(x+1)\cdot -x = -x^2 - x = (x+1) - x = 1$ which is what we wanted.
|
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|
Derive $\tan(3x)$ in terms of $\tan(x)$ using De Moivre's theorem
Derive the following identity: $$\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$
The way I approached the questions is that I first derived $\sin(3x)$ and $\cos(3x)$ because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Then substituting:
$$\tan(3x)=\frac{\sin(3x)}{\cos(3x)}=\frac{-4\sin^3(x)+3\sin(x)}{4\cos^3(x)-3\cos(x)}$$
I transformed it into $$\tan(3x)=\frac{-4\sin^2(x)\tan(x)+3\tan(x)}{4\cos^2(x)-3}$$ and also to $$\tan(3x)=-\tan(x)\frac{4\sin^2(x)-3}{-4\sin^2(x)+1}$$ but I'm stuck on either of these forms. Any help?
|
Wthout de Moivre:
$$\tan(2x+x)=\frac{\tan2x+\tan x}{1-\tan2x\tan x}=\frac{\frac{2\tan x}{1-\tan^2x}+\tan x}{1-\frac{2\tan x}{1-\tan^2x}\tan x}=$$
$$=\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan x\frac{(\tan x-\sqrt3)(\tan x+\sqrt3)}{(\sqrt3\,\tan x-1)(\sqrt3\,\tan x+1)}$$
|
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|
Prove: $2^k$ is the sum of two perfect squares If $k$ is a nonnegative integer, prove that $2^k$ can be represented as a sum of two perfect squares in exactly one way. (For example, the unique representation of $10$ is $3^2+1^2$; we do not count $1^2+3^2$ as different.)
I understand that $2^{2n}=0+2^{2n}$ and $2^{2n+1}=2^{2n}+2^{2n}$. But how can we prove that $2^k$ can be represented as two perfect squares in exactly one way?
|
Use induction.
Note that if $$x^2+y^2=2^k\ (\text{where } k \geq 2)$$
If $x \equiv 1 \pmod 2$, then from $x^2+y^2 \equiv 0 \pmod 2$ we get $y \equiv 1 \pmod 2$.
However, then $$x^2+y^2 \equiv 1+1 \equiv 2 \pmod 4$$ A contradiction.
Thus, $x=2x_{1}, y=2y_{2}$. Thus $x_1^2+y_1^2 =2^{k-2}$.
|
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|
Why $xy+yz+xz=1$ is two-sheeted hyperboloid? I can't see why $xy+yz+xz=1$ is two-sheeted hyperboloid.
I know that the equation for two-sheeted hyperboloid is: $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=-1$.
|
You need to perform Gauß' reduction to see it:
\begin{align*}
xy+yz+zx&=(x+z)(y+z)-z^2=\frac14\Bigl((x+z+y+z)^2-(x+z-y-z)^2\Bigr)-z^2\\
&=\Bigl(\frac{x+y+2z}2\Bigr)^2-\Bigl(\frac{x-y}2\Bigr)^2-z^2.
\end{align*}
Hence the equation of the quadric can be written as
$$\Bigl(\frac{x-y}2\Bigr)^2+z^2-\Bigl(\frac{x+y+2z}2\Bigr)^2=-1.$$
|
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|
Plane $3x + y - z= 4$ touches the ellipsoid $2z^2 = \sqrt7(1 - 2x^2 -y^2)$ Show that the Plane $3x + y - z= 4$ touches the ellipsoid $2z^2 = \sqrt7(1 - 2x^2 -y^2)$
My attempt: First I tried to convert the equation of ellipsoid in general form and then further applying the condition of plane as tangent which is $$ \frac{u^2}{a} + \frac{v^2}{b} + \frac{w^2}{c} = p^2$$
Rewriting the equation of conic we get $$2x^2 + y^2 + \frac{2}{\sqrt7}z^2 = 1$$
Using this we get $a = 2 , b = 3$ and $c = \frac{2}{\sqrt7}$
and we using equation of plane we get $u = 3 , v = 1 , w = -1$ and $p = 4$
The condition is not satisfied. What am I doing wrong here? I seem to do everything correct and tried this question five times.
I'm really stuck at this problem. Kindly help.
Note: Please ignore the B3P36 in the image that was written by me as a code to something. It has no significance to the problem.
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The equation of the plane in general form is
$$3 x + y - z - 4 = 0$$
and therefore its normal is $(3,1,-1)$. Its minimum distance $D$ to origin is
$$D = \frac{\lvert -4 \rvert}{\sqrt{3^2 + 1^2 + (-1)^2 }} = \frac{4}{\sqrt{11}} \approx 1.206$$
and the unit normal vector is $\hat n = (3/\sqrt{11},1/\sqrt{11},-1/\sqrt{11})$. Distance $t$ from origin along the plane normal is therefore
$$\vec n (t) = \left ( \frac{3 t}{\sqrt{11}}, \frac{t}{\sqrt{11}}, -\frac{t}{\sqrt{11}} \right )\tag{1}\label{1}$$
Divide both sides of the formula for the ellipsoid
$$2 z^2 = \sqrt{7} ( 1 - 2 x^2 - y^2 )$$
by $\sqrt{7}$ and move $-2 x^2 - y^2$ to the other side:
$$\frac{2}{\sqrt{7}} z^2 + 2 x^2 + y^2 = 1\tag{2}\label{2}$$
If we move the coefficients to denominators, we can see the general form:
$$\frac{x^2}{(\sqrt{1/2})^2} + \frac{y^2}{1^2} + \frac{z^2}{(\sqrt{\sqrt{7}/2})^2} = 1$$
and we see it is an axis-aligned ellipsoid with semi-principal axes $a = \sqrt{1/2} \approx 0.7071$, $b = 1$, and $c = \sqrt{\sqrt{7}/2} \approx 1.150$.
(At this point, we notice that all the semi-principal axes are smaller than the minimum distance from the plane to the center of the ellipsoid, origin, which means the two surfaces cannot come into contact. The ellipsoid is just too small.)
The closest the ellipsoid comes to the plane (if they do not touch), the point that contacts the plane (if they contact at exactly one point), or the point on the ellipsoid furthest on the other side of the plane (if they intersect in a ring), is on the line parallel to the plane normal that passes through the center of the ellipsoid (here, origin).
So, to find that point on the ellipsoid we insert equation $\eqref{1}$ into $\eqref{2}$:
$$\frac{2}{\sqrt{7}} \left ( -\frac{t}{\sqrt{11}} \right )^2 + 2 \left ( \frac{3 t}{\sqrt{11}} \right )^2 + \left ( \frac{t}{\sqrt{11}} \right )^2 = 1$$
$$\frac{2}{11 \sqrt{7}} t^2 + \frac{18}{11} t^2 + \frac{1}{11} t^2 - 1 = 0$$
and solve for $t$, the (signed) distance from origin (negative referring to opposite direction from plane surface normal):
$$t = \pm \frac{\sqrt{10241 + 154\sqrt{7}}}{2\sqrt{7} + 133} \approx \pm 0.7462$$
Because the closest point to origin the plane comes to is $\vec n(D)$, i.e. at distance $D \approx 1.206$, the plane never comes closer than approximately $1.206 - 0.7462 = 0.4598$ to the ellipsoid.
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Trigonometric substitution $\tan{\frac{x}{2}}=t$. What is $\cos{x}$ then? For example, the integral is:
$$\int \frac{\sin{x}}{3\sin{x}+4\cos{x}}dx$$
And we use the substitution: $\tan{\frac{x}{2}}=t$
Now, to get $\cos{x}$ in terms of $\tan\frac{x}{2}$, I first expressed $\cos^2\frac{x}{2}$ and $\sin^2\frac{x}{2}$ in temrs of $\tan\frac{x}{2}$:
$$\cos^2\frac{x}{2}=\frac{1}{\frac{1}{\cos^2\frac{x}{2}}}=\frac{1}{\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}}{\cos^2\frac{x}{2}}}=\frac{1}{1+\tan^2\frac{x}{2}}=\frac{1}{1+t^2}$$
$$\sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}=1-\frac{1}{1+t^2}=\frac{t^2}{1+t^2}$$
Now, using trig idendity: $\cos{2x}=\cos^2{x}-\sin^2{x}$ we have:
$$\cos{x}=\bigg(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\bigg)=\frac{1}{1+t^2}-\frac{t^2}{1+t^2}=\frac{1-t^2}{1+t^2}$$
which is good, but, I don't know what is wrong with this next procedure (first, getting $\sin^2{x}$, then using trig identity $\sin^2{x}+\cos^2{x}=1$ getting $\cos{x}$).
So, first, expressing $\sin{x}$ in terms of $\tan\frac{x}{2}$ using trig identity $\sin{2x}=2\sin{x}\cos{x}$:
$$\sin^2{x}=4\sin^2\frac{x}{2}\cos^2\frac{x}{2}=4\cdot\frac{t^2}{1+t^2}\cdot\frac{1}{1+t^2}=\frac{4t^2}{(1+t^2)^2}$$
So:
$$\cos^2{x}=1-\sin^2{x}=1-\frac{4t^2}{(1+t^2)^2}=\frac{t^4-2t^2+1-4t^2}{(1+t^2)^2}=\frac{(t^2-1)^2}{(t^2+1)^2}$$
And, $\cos{x}$ is then:
$$\cos{x}=\pm\frac{t^2-1}{t^2+1}$$
Why do I get $\pm$ here? Did I make a mistake somewhere? Thank you for your time.
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I'd rather have a more elegant way to do this, but this way at least works:
You've shown that $\cos x = \dfrac{1-t^2}{1+t^2}$, and then that $\sin^2 x = \dfrac{4t^2}{(1+t^2)^2}$.
Notice that if you start with $a=b=3$ and then square both sides of $a=b$ to get $a^2=b^2$, you can deduce that $a = \pm b$. That just means that either $a=b$ or $a=-b$; it doesn't mean $a$ is equal to both $b$ and $-b$. So you must conclude that for each value of $x$, either $\sin x = \dfrac{2t}{1+t^2}$ or $\sin x = \dfrac{-2t}{1+t^2}$. Might it be equal to that first expression for some values of $x$ and the second for others? If so, things would get somewhat more complicated. But now look at the graphs of $x\mapsto\sin x$ and $x\mapsto \tan\dfrac x 2$ on the interval $-\pi\le x\le\pi$. What you see is that $\sin x\ge0$ if $\tan\dfrac x 2\ge0$ and $\sin \le 0$ if $\tan\dfrac x 2\le 0$. Consequently you have $\sin x = \dfrac{2t}{1+t^2}$ for all values of $x$. Now compare the graphs of $x\mapsto\cos x$ and $x\mapsto \tan\dfrac x 2$ on that same interval, and observe that $\cos x\ge 0$ if $-1\le\tan \dfrac x 2\le 1$ and $\cos x\le 0$ if $\tan\dfrac x 2\ge 1$ or $\tan\dfrac x 2\le -1$. That is the behavior of $\dfrac{1-t^2}{1+t^2}$ and not of $-\dfrac{1-t^2}{1+t^2}$. That decides between $\text{“}{+}\text{''}$ and $\text{“}{-}\text{''}$.
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How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:
$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$
Wolframalpha shows this, but I can't prove it.
http://www.wolframalpha.com/input/?i=1%2Bcos(2pi%2F7)-4cos%5E2(2pi%2F7)-8cos%5E3(2pi%2F7)%3D0
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We prove a closely related result, which in particular shows there is a typo in the given equation.
The number $e^{2\pi i/7}$ is a root of $x^7=1$, and therefore of $x^6+x^5+\cdots+x+1=0$, or equivalently of
$$(x^3+x^{-3})+(x^2+x^{-2})+(x+x^{-1})+1=0.\tag{1}$$
(We divided through by $x^3$.) Let $w=\frac{1}{2}(x+x^{-1})$.
Note that $x^3 +x^{-3}=8w^3-6w$ and $x^2+x^{-2}=4w^2-2$ and $x+x^{-1}=2w$ So our equation can be rewritten as
$$8w^3+4w^2-4w-1=0.\tag{2}$$
Since $e^{2\pi i/7}$ is a root of (1), it follows that $\cos(2\pi/7)$ is a root of (2).
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Determine the first three non-zero terms in the Taylor polynomial approximation for the initial value problem: $y''+\sin(y)=0$ Having trouble understanding how to solve this problem. Did I at least set it up correctly?
$y''+\sin(y)=0,\;y(0)=1,\;y'(0)=0$
So assuming $y(x)=\sum_{n=0}^{\infty}a_nx^n$ then $y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$
And since $\sin(y)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$ rewriting the original equation I get:
$$y''+\sin(y) = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}+\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$$
I tried combining terms and simplifying so that I can solve for $a_{n+2}:
$$\sum_{n=0}^{\infty}x^n\left((n+2)(n+1)a_{n+2}+\frac{(-1)^nx^2}{(2n!)}\right)$$
But I don't know how to get rid of the factor of $x$ inside the parenthesis so that I can solve for the coefficient. Am I going down the wrong path?
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$y = a_0 + a_1 x + a_2 x^2 + a_3 x^3$
$y(0) = 1, y'(0) = 0$
$y = 1 + a_2 x^2 + a_3 x^3$
$y'' = 2 a_2 + 6 a_3 x + 12 a_4 x^2$
$\sin y = y - y^3 / 6 + y^5/5!...$
$y^3 = 1 + 3 a_2 x^2 + 3 a_3 x^3 + 3 a_2x^4 + 3a_4 x^4...$
$y^5 = 1 + 5 a_2 x^2 + 5 a_3 x^3 + 10 a_2x^4 + 5a_4 x^4...$
$\sin y = (1-1/6+1/120...) + (1-1/2+1/24...) a_2 x^2$
$\sin y = \sin 1 + (cos 1) a_2 x^2.. $
$y'' + \sin y = 0$
$2a_2 + \sin 1 + 6a_3 x + (12 a_4 + \cos 1 a_2) x^2 = 0$
$a_2 = (-1/2) (\sin 1), a_3 = 0, a_4 = (-1/12)(\cos 1)(-1/2)(\sin 1)$
$y = \sin 1 - (1/2) (\sin 1) x^2 + (1/48) (\sin 2) x^4$
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Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$
$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$
$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$
$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$
Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:
$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\
z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$
I obtain $x$ from $z$ by taking the inverse cosine.
The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).
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Hint: I think the approach of @JanEerland is instructive. Here some thoughts how we could find this kind of substitution.
When looking at
\begin{align*}
\sin x+\cos x=\frac{1}{3}
\end{align*}
and we think of the trigonometric addition formulas we know that
\begin{align*}
\sin(x+a)=\sin x \cos a+\cos x\sin a
\end{align*}
It would be convenient if $\cos a=\sin a$, so that we can separate them. This is the case if $a=\frac{\pi}{4}$ and we obtain
\begin{align*}
\sin\left(x+\frac{\pi}{4}\right)
&=\sin x\sin \frac{\pi}{4}+\cos x\cos \frac{\pi}{4}\\
&=\frac{\sqrt{2}}{2}\left(\sin x+\cos x\right)
\end{align*}
It follows
\begin{align*}
\sin x+\cos x&=\frac{1}{3}\\
\sin\left(x+\frac{\pi}{4}\right)&=\frac{1}{3\sqrt{2}}
\end{align*}
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Finding equation of sphere, $2|PB|=|PA|, A(-2,5,2)$ and $B(5,2,-1)$ I have question asking to find an equation for a sphere with points $P$ such that the distance from $P$ to $A$ is twice the distance from $P$ to $B$ with $A(-2,5,2)$ and $B(5,2,-1)$.
I have:
$$4[(x-5)^2+(y-2)^2+(z+1)^2] = [(x+2)^2+(y-5)^2+(z-2)^2]$$
$$=> 4[(x^2-10x+25)+(y^2-4y+4)+(z^2+2z+1)] - (x^2+4x+4) - (y^2-10y+25)-(z^2-4z+4)=0$$
$$=>3x^2-44x+96+3y^2+6y-9+3z^2+12z=0$$
$$=>3x^2-44x+3y^2+6y+3z^2+12z=(-87)$$
$$=>x^2-22x+y^2+2y+z^2+4z=(-29)$$
$$=>(x^2-11x+121)+(y^2+y+1)+(z^2+2z+4)=(97)$$
$$=>(x+11)^2+(y-1)^2+(z-2)^2=(97)$$
I looked it over a few times and found one error I had originally made but it's still not coming out right and I can't find where else I messed up.
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The first error is in the coefficient of $y$ :
$$3x^2-44x+3y^2\color{red}{-}6y+3z^2+12z=-87$$
The second error is in the coefficient of $x$ :
$$x^2-\color{red}{\frac{44}{3}}x+y^2-2y+z^2+4z=-29$$
Now, add $(-22/3)^2+(-1)^2+2^2$ to the both sides.
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integrate $\int \frac{\tan^4x}{4}\cos^3x$
$$\int \frac{\tan^4x}{4}\cos^3x$$
$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac{\sin^2x}{\cos x}-\frac{\cos^2x}{\cos x}=\frac{1}{4}\int \frac{1-\cos^2x}{\cos x}-\cos x=\frac{1}{4} \int \frac{1}{\cos x}-2\cos x=\frac{1}{4}(\ln(\tan x+\sec x)+2\sin x$$
Is it correct?
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\begin{align}
\frac{1}{4}\int \tan^4 x \cos^3 x \, dx
&= \frac{1}{4} \int \frac{\sin^4 x}{\cos^4 x}\cos^3 x \, dx\\[0.3cm]
&= \frac{1}{4} \int \frac{\sin^4 x}{\cos^2 x}\cos x \, dx\\[0.3cm]
&= \frac{1}{4} \int \frac{\sin^4 x}{1-\sin^2x} \cos x \, dx
\end{align}
Now do a $u$-sub with $u = \sin x$.
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How to compute such a limit? Knowing $f(x,y) = 2x^2 +3y^2 -7x +15y$, one simply proves $$|f(x,y)|\leq 5(x^2+y^2)+22 \sqrt{x^2 + y^2}$$ How can I use this info to compute
$$ \lim_{(x,y)\to(0,0)} \frac{f(x,y) - 2(x^2+y^2)^{1/4}}{(x^2+y^2)^{1/4}}\;\;\; ?$$
Thanks!
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Approach limit with $y = mx$
then
$$
\lim_{(x,y)\to(0,0)} \frac{f(x,y) - 2(x^2+y^2)^{1/4}}{(x^2+y^2)^{1/4}} =
$$
$$
\lim_{x\to0} \frac{x(2x +3m^2x -7 +15m)}{x^{1/2}(1+m^2)^{1/4}} - 2 =
$$
$$
\lim_{x\to0} \frac{x^{1/2}( -7 +15m)}{(1+m^2)^{1/4}} - 2 = -2
$$
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In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$
In $\triangle ABC$, if
$$\cos A \cos B \cos C=\frac{1}{3}$$
then can we find value of
$$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$
Please give some hint. I am not sure if $\tan A \tan B+\tan B \tan C+\tan C \tan A$ will be constant under given condition.
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Let $$S=\tan A\tan B+\tan B\tan C+\tan C\tan A$$
Multiplying by $\cos A \cos B \cos C=\frac 13$, we get $$\frac 13S=\sin A\sin B\cos C+\cos A\sin B\sin C+\sin A\cos B \sin C$$
However,$$\cos(A+B+C)=-1=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C$$
Therefore, $$\frac 13S=\cos A\cos B\cos C+1=\frac 43\Rightarrow S=4$$
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Minimum point of the function $f (x) = 2 ^{- x-\sqrt {x ^ 2 + 1}} + 2 ^ {2x-2\sqrt {1-x ^ 2}}$ I would like to find without the use of derivatives of the minimum point of the function $$f (x) = 2 ^{- x-\sqrt {x ^ 2 + 1}} + 2 ^ {2x-2\sqrt {1-x ^ 2}}$$
(In fact, the point minimum is known: $x = 0$ and the minimum value of the function is equal to $3/4$).
Let $x=\sqrt{\cos 2y}, 0 \le 2y \le \frac{\pi}2$
$$f (x) = 2 ^{-\sqrt{\cos 2y} -\sqrt {1+\cos 2y}} + 2 ^ {2\sqrt{\cos 2y}-2\sqrt {1-{\cos 2y}}}$$
What's next?
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Let $a,b > 0$. Then for any positive $t$ inequality
$$at^2+\frac{b}t=at^2+\frac{b}{2t}+\frac{b}{2t}\ge3\sqrt[3]{\frac{ab^2}4}.$$
In our case we set $t=2^x$, $a=2^{-2\sqrt{1-x^2}}$, $b=2^{-\sqrt{1+x^2}}$. Then $ab^2=2^{-2(\sqrt{1-x^2}+\sqrt{1+x^2})}\ge\frac1{16}$ due to inequality $\sqrt{1-x^2}+\sqrt{1+x^2}\le2$, which obviously is checked by squaring.
Here $f(x)=at^2+\frac{b}t\ge3\sqrt[3]{\frac{ab^2}4}\ge\frac34$ and equality holds only at the origin.
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Using squeeze thorem find $ \lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$ It is already solved here at Math.stackexchange, but we haven't learned Stirling's approximation (at our school), so can it be solved using only squeeze theorem?
$$\lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$$
My attempt,
Let $y_n = \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}$, we see that $ \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6\cdot ...\cdot 2n} > \frac{1 \cdot 2 \cdot 2 \cdot ... \cdot 2}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n} = \frac{1}{1 \cdot 2 \cdots .. \cdot (n-1)\cdot2n} = x_n$
$$\lim_{n \to \infty}{x_n} = 0 $$
Now I just need to find a sequnce $z_n>y_n$, so, $\lim_{n \to \infty} z_n = 0$.
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Hint multiply both numerator,denominator by $2.4.6....2n$ so you get $$\frac{(2n)!}{4^n(n!)^2}$$ where i get $4^n$ by taking $2$ common so we can write it as $$\frac{{2n\choose n}}{4^n}$$ thus we know $4^n\geq {2n\choose n}$. So limit is $0$.
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proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$ Question: proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$
Hi... I am stumped on an intro analysis problem.
so it is stated Show $lim \frac{1}{x^{2}-1}$ = -1 as $x_{0} \to 0$.
here is my work, and I can't quite simplify the $|f(x) - L | < \epsilon$ to get an expression for $\delta$.
|$\frac{1}{x^{2}-1} + 1| \leq |\frac{1}{(x+1)(x-1)} + 1 | \leq | \frac{1}{x-1} - \frac{1}{x+1} + 2| \leq \epsilon $ not sure if I should use partial fractions here.
I am not sure the neighborhood to select for $x_0$ , so I'm guessing $|x| < 1$
this choice yields $1 < \frac{1}{x+1} < \frac{1}{x} < \frac{1}{x-1} $
so the expression of $\epsilon$ can be written:
$|\frac{1}{x-1} - \frac{1}{x+1} + 2 | < |\frac{2}{x-1} + 2 | $ this can be simplified to
$| x - 0 | < \frac{\epsilon}{2}(x-1)$ where given our choice of $x_{0}$ (x-1) $< 2$
so I chose $\delta = min( 1, \epsilon)$.
but I am not sure this work is any good, and am having trouble writing the proof.
sincerely thank you!
Here is the Proof
let $\delta = min\{\frac{1}{2},\frac{5\epsilon}{2} \}$
let $\delta = \frac{1}{2} \implies x^{2} < \frac{|x|}{2} <\frac{1}{4}$
where $|x| < \frac{1}{2}$
then $\frac{-5}{4} < x^{2} - 1 < \frac{-3}{4} \implies \frac{5}{4} < |x^{2}-1| < \frac{3}{4} \implies \frac{1}{x^{2}-1}<\frac{4}{5}$
then $\frac{x^{2}}{x^{2}-1} < \frac{4|x|}{2*5} < \frac{4*1}{4*5} = \frac{1}{5} < \epsilon$
where if $\delta = \frac{1}{2} \implies \frac{1}{2} < \frac{5\epsilon}{2} \implies \frac{1}{5}<\epsilon$
for $\delta = \frac{5\epsilon}{2} $
let $|x - 0| < \frac{5\epsilon}{2} $ and for $|\frac{x}{2}|<\frac{5 \epsilon}{4}$ then
$x^{2}<|\frac{x}{2}| \implies |\frac{x^{2}}{x^{2}-1}| \leq \frac{2|x|}{5}$
so $x^{2} < \frac{|x|}{2} < \frac{5\epsilon}{4}$
then
$|\frac{x^{2}}{x^{2}-1}| < \frac{4|x|}{5} <\frac{4}{5}*\frac{5\epsilon}{4} = \epsilon$
|
Note that in our case,
$$|f(x)-L|=\frac{x^2}{|x^2-1|}.$$
If $|x|\lt 1/2$, then $|x^2-1|\gt \frac{3}{4}$ and therefore
$$|f(x)-L|\le \frac{4x^2}{3}.$$
Now it should not be hard to find an appropriate $\delta$. To make things even smoother, note that if $|x|\lt \frac{1}{2}$ then $x^2\le \frac{|x|}{2}$.
Added: We wanted to show that given $\epsilon\gt 0$, we can find a $\delta$ such that if $|x-0|\lt \delta$, then $\frac{x^2}{|x^2-1|}\lt \epsilon$. The "top" can be easily made small by choosing $|x|$ small enough. But the bottom could spoil things, because $\frac{1}{|x^2-1|}$ can be very large if $x$ is close to $1$ or $-1$. So the first thing to do is to neutralize the bottom by specifying that $\delta$ will be $\le 1/2$. Instead of $1/2$ we could have chosen $1/10$, or $8/10$.
We concluded that if $|x|\le 1/2$ then the expression we are trying to make small is $\le 4x^2/3$, which, since $|x|\le 1/2$, is $\le 2|x|/3$. This in turn is $\le |x|$. So if we make $\delta=\min(1/2,\epsilon)$, then if $|x|\lt \delta$ we will have $\frac{x^2}{|x^2-1|}\lt \epsilon$.
Note that we expended no effort whatsoever to choose the "largest" $\delta$ that would do the job. Choosing an optimal or near optimal $\delta$ might be of great importance if we were doing numerical analysis, but is of no importance here.
|
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|
Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one
Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one for all $k,m,n\in \mathbb N$. where $m\geqslant n$ and $p$ is a prime.
What I did:
$\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m}{p^k}-\frac{n}{p^k}\bigg\rfloor$
Let $x:=\frac{m}{p^k}$ and let $y:=\frac{n}{p^k}$
we get $\lfloor x\rfloor-\lfloor y\rfloor-\lfloor x-y\rfloor$
What should I do now?
|
Since
$x = \left\lfloor x \right\rfloor + \left\{ x \right\}$
then we have
$$
\eqalign{
& \left\lfloor {x - y} \right\rfloor = - \left\lceil {y - x} \right\rceil = \cr
& = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} - \left\{ y \right\}} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor - \left[ {\left\{ x \right\} < \left\{ y \right\}} \right] \cr}
$$
where
$$
\left[ P \right] = {\rm Iverson bracket} = \left\{ \matrix{
1\;{\rm if}\;P = TRUE \hfill \cr
0\;{\rm if}\;P = FALSE \hfill \cr} \right.
$$
Thus
$$
\left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor - \left\lfloor {x - y} \right\rfloor = \left[ {\left\{ x \right\} < \left\{ y \right\}} \right] = 0\;OR\;1\quad \left| {\;\forall x,y\;{\rm real}} \right.
$$
|
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|
Simpler Way to Write a sum of factorial I have a fairly simple question. I was wondering if there is a simpler way to write the following: $$\sum_{k=0}^{x-1}\frac{(n+k+1)!}{k!}$$
So for example, given $n=4$ and $x=3$, we would have $$1\cdot 2\cdot 3\cdot 4\cdot 5+2\cdot 3\cdot 4\cdot 5\cdot 6+3\cdot 4\cdot 5\cdot 6\cdot 7=3360$$
Is there a simpler way to write this or do this calculation?
Edit: Ok I have simplified the sum above to the following: $$\frac{x(n+x+1)!}{(n+2)(x!)}$$
Can I simplify this further?
|
Note that we can write
$$\frac{(n+k+1)!}{k!}=\frac{1}{n+2}\left(\frac{(k+n+2)!}{k!}-\frac{(k+n+1)!}{(k-1)!}\right)$$
Therefore, evaluating the telescoping sum is trivial and yields
$$\begin{align}
\sum_{k=0}^{x-1}\frac{(n+k+1)!}{k!}&=(n+1)!+\sum_{k=1}^{x-1}\frac{(n+k+1)!}{k!}\\\\
&=(n+1)!+\frac{1}{n+2}\left(\frac{(x+n+1)!}{(x-1)!}-\frac{(n+2)!}{1!}\right)\\\\
&=\frac{1}{n+2}\frac{(x+n+1)!}{(x-1)!}\\\\
&=\frac{1}{n+2}\prod_{\ell =0}^{n+1}(x+\ell)\\\\
&=(n+1)!\binom{x+n+1}{x-1}
\end{align}$$
|
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|
Prove the convergence and find the sum of series $\sum\limits_{n=1}^{\infty}\left(n^3\sin\frac{\pi}{3^n}\right)$. We know that $0<\sin\frac{\pi}{3^n}\le\frac{\sqrt 3}{2},\forall n\ge 1$. How to find the boundary for $n^3\sin\frac{\pi}{3^n}$ (how to use comparison test here)?
I tried using the ratio test, but the limit $$\lim\limits_{n\to\infty}\left((n+1)^3\sin\frac{\pi}{3^{n+1}}\cdot \frac{1}{n^3\sin\frac{\pi}{3^n}}\right)$$ isn't that easy to evaluate.
|
You can evaluate the limit in this way
$$\begin{align}
\lim_{n\to\infty} \left( \frac{(n+1)^3}{n^3} \cdot \frac{\sin\frac{\pi}{3^{n+1}}}{\sin\frac{\pi}{3^{n}}}\right) &= \lim_{n \to \infty} \frac{(n+1)^3}{n^3} \lim_{n \to \infty} \frac{\sin\frac{\pi}{3^{n+1}}}{\sin\frac{\pi}{3^{n}}} \\
&= 1 \cdot \lim_{n \to \infty} \frac{\frac{\pi}{3^{n+1}}}{\frac{\pi}{3^{n}}} \\
&= 1 \cdot \frac{1}{3} \\
&= \frac{1}{3}
\end{align}$$
and since the result is less than $1$ the series will converge.
|
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|
Similarity of Linear Transformations in $\mathbb{R}^4$ Define $\mathbb{R}^4 \rightarrow \mathbb{R}^4$ by $T(x,y,z,w) = (x+z, y-z,z,w-z)$.
Let $$B = \{(1,2,3,4), (1,2,3,0), (1,2,0,0), (1,0,0,0)\}$$
$$B' = \{(1,0,-1,0), (0,2,3,0), (1,0,0,0), (1,1,1,1)\}$$
Find the matrix $[T]_B$ of the linear transformation $T$ with respect to basis B, the matrix $[T]_{B'}$ of the linear transformation $T$ with respect to the basis $B'$, the transition matrix $P$ from $B$ to $B'$ and verify that $[T]_B$ and $[T]_{B'}$ are similar matrices.
*
*I can easily find the transition matrix. Where I'm having issues is finding $[T]_B$ and $[T]_{B'}$
*Can I simply plug the vectors into the operation?
$$T \begin{pmatrix} \begin{bmatrix} 1\\2\\3\\4 \end{bmatrix} \end{pmatrix} = \begin{bmatrix} 1+3\\2-3\\3\\4-3 \end{bmatrix} = \begin{bmatrix} 4\\-1\\3\\1 \end{bmatrix} $$
Repeat with all other vectors...
$$[T]_B = \begin{bmatrix} 4&4&1&1\\-1&-1&2&0\\3&3&0&0\\1&-3&0&0 \end{bmatrix}$$
Similar process with $[T]_{B'}$
Once I find transition matrix $P$, checking the similarity of $B$ and $B'$ is a simply a matter of checking if:
$$P^{-1}BP = B'$$
Is this correct? Are there any other requirements that I have to check for similarity between matrices?
|
You have calculated is the matrix $[T]_{B,E}$, where $E$ is the standard basis.
In general if $\mathscr{A}, \mathscr{B}$ are two bases then, the n th column of $[T]_{\mathscr{A}, \mathscr{B}}$ is $ \begin{bmatrix} \lambda_1\\ \vdots\\ \lambda_n \end{bmatrix}$,
where $T(a_n) = \Sigma_ i \lambda_i b_i $ and $a_i, \in \mathscr{A}$,$b_i, \in \mathscr{B}$.
So for $[T]_B = [T]_{B,B}$ compute:
$$
\begin{align}
T \begin{pmatrix} \begin{bmatrix} 1\\2\\3\\4 \end{bmatrix} \end{pmatrix} & = \begin{bmatrix} 4\\-1\\3\\1 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 1\\2\\3\\4 \end{bmatrix} + \frac{3}{4} \begin{bmatrix} 1\\2\\3\\0 \end{bmatrix} + \frac{-3}{2} \begin{bmatrix} 1\\2\\0\\0 \end{bmatrix} + \frac{9}{2} \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}
\end{align}
$$
Thus first column of $[T]_B$ is $ \begin{bmatrix} \frac{1}{4}\\ \frac{3}{4}\\ \frac{-3}{2}\\ \frac{9}{2}\\ \end{bmatrix}$.
Repeating for other basis vectors will yield $[T]_B$ .
Your calculation for verifying similar matrices is correct.
|
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|
Why can't I solve for this second derivative? Here's the equation I have to find the second derivative point for.
$$f(x)=\frac{x+2}{x^{\frac{1}{2}}}$$
$$f'(x) = \frac{x-2}{2x^{\frac{3}{2}}}$$
From here I then calculate the second derivative and set it equal to 0.
But it doesn't work.. Take a look:
$$f''(x)=\frac{-x^\frac{3}{2} + 6x^{\frac{1}{2}}}{4x^{3}} = 0$$
FROM first DERIVATIVE TO second:
$$\frac{2x^{\frac{3}{2}}[x-2]'-((x-2)[2x^{\frac{3}{2}}]'}{4x^{3}}$$
$$\frac{2x^{\frac{3}{2}} -3x^{}\frac{3}{2}+6x^{\frac{1}{2}}}{4x^{3}}=0$$
$$-x^{\frac{3}{2}} + 6x^{\frac{1}{2}} = 0$$
$$(-x^{\frac{3}{2}})^{2} + (6x^{\frac{1}{2}})^{2}$$
$$x^{3} + 36x = 0$$
$$x (x^{2} + 36) = 0$$
$$x = 0 \text{ or } x^{2} = - 36 \text{ no solution..}$$
This doesn't seem right.. Yet I have no idea why. It's easier to write the function as a product but I want to solve it using the quotient rule..
What's going on?
|
We proceed via the quotient rule:
$$f''(x)=\frac{2x^{3/2}-(x-2)3x^{1/2}}{4x^3}=\frac{2x^{3/2}-3x^{3/2}+6x^{1/2}}{4x^3}=x^{1/2}\left(\frac{6-x}{4x^3}\right)=\frac{6-x}{4x^{5/2}}$$
We can set $f''(x)=0$ and find that $f''(6)=0$. To find the points of inflection, we simply need to test a point less than and greater than $x=6$.
|
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|
Finding series solution about zero $y''+x^2y'+4y=1-x^2$
To find a power series, one substitutes in $y= \sum_0^\infty a_nx^n$. So after substitution, I've gotten
$\sum_0^\infty (n+1)(n+2)a_{n+2}x^n + \sum_1^\infty (n-1)a_{n-1}x^n + 4\sum_0^\infty a_nx^n$
The recurrence relation would then be in terms of $a_{n-2}$ since it's the lowest term. But how do I solve this if there are two other unknown terms? And I've never worked with a power series set to equal anything other than zero. I am just lost.
|
For the equation
$$ y'' + x^2 \, y' + 4 \, y = 1 - x^2$$
use the solution
$$y = \sum_{n=0}^{\infty} a_{n} \, x^n$$
to obtain:
\begin{align}
\sum_{n} (n)(n-1) \, a_{n} \, x^{n-2} + \sum_{n} n \, a_{n} \, x^{n+1} + 4 \, \sum_{n} a_{n} &= 1 - x^2 \\
\sum_{n=0} [(n+1)(n+2) \, a_{n+2} + 4 \, a_{n}] \, x^{n} + \sum_{n=1} n \, a_{n} \, x^{n+1} &= 1 - x^2 \\
\sum_{n=0} [(n+1)(n+2) \, a_{n+2} + 4 \, a_{n}] \, x^{n} + \sum_{n=0} (n+1) \, a_{n+1} \, x^{n+2} &= 1 - x^2 \\
\end{align}
Now equating coefficients leads to:
\begin{align}
2 \, a_{2} + 4 \, a_{0} &= 1 \\
6 \, a_{3} + 4 \, a_{1} &= 0 \\
12 \, a_{4} + a_{1} + 4 \, a_{2} &= -1 \\
(n+4)(n+5) \, a_{n+5} + (n+2) \, a_{n+2} + 4 \, a_{n+3} &= 0.
\end{align}
From these equations it is determined that:
\begin{align}
a_{2} &= \frac{1}{2} \, (1- 4 \, a_{0}) \\
a_{3} &= - \frac{4}{6} \, a_{1} \\
a_{4} &= \frac{1}{12} \, ( -3 + 8 \, a_{0} - a_{1} ) \\
a_{5} &= \frac{1}{60} \, (-3 + 12 \, a_{0} + 8 \, a_{1} ),
\end{align}
where $a_{0}$ and $a_{1}$ are determined by the initial conditions.
|
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|
$\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$ $\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f'(x)=\cos x(\cos x+\frac{\sin x\cos x}{\sqrt{\sin^2x+\sin^2\alpha}})-\sin x(\sin x+\sqrt{\sin^2x+\sin^2\alpha})$
I am stuck here and could not find the minimum and maximum values of $f(x),$The answer given is $-\sqrt{1+\sin^2\alpha}\leq f(x)\leq\sqrt{1+\sin^2\alpha}$.
|
You're derivative need work. $$f'(x)=-\sin(\sin(x)+\sqrt{\sin^2 x+\sin^2\alpha})\left(\cos x+\frac{2\sin x\cos x}{\sqrt{\sin^2 x+\sin^2\alpha}}\right).$$
Now determine when this is zero, which is precisely when $$\sin(\sin(x)+\sqrt{\sin^2 x+\sin^2\alpha})=0,$$ or $$\cos x+\frac{2\sin x\cos x}{\sqrt{\sin^2 x+\sin^2\alpha}}=0.$$
|
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|
curve of $(x^2+y^2)^2=2(x^2-y^2)$ The diagram shows the curve $(x^2+y^2)^2=2(x^2-y^2)$ and one of its maximum points $M$. Find the coordinates of $M$.
My attempt.
Differentiate the equation and I got
$\frac{dy}{dx}=-\frac{x(x^2+y^2-1)}{y(x^2+y^2+1)}$
How should I proceed?
|
HINT.-You have a closed curve, symmetrical with respect to the two axes (a lemniscate indeed) and you have the alternative of take the explicit form for $y$ solving a quadratic equation; you have
$$y^4+2(x^2+1)y^2+(x^4-2x^2)=0$$ By symmetry you can take only the positive signs in solving so you have
$$y=\left(-x^2-1+\sqrt{4x^2+1}\right)^{\frac 12}$$
$$y'=\frac{-x+\frac{2x}{\sqrt{4x^2+1}}}{\sqrt y}$$
Hence you get for the maximun $$x=\frac{\sqrt 3}{2}$$ Thus $$y=\frac 12$$
|
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|
Find the first two terms in the perturbation expansion of the solution I want to find the $\mathcal{O}(1)$ and $\mathcal{O}(\epsilon)$ terms in the pedestrian expansion $y = y_0 + \epsilon y_1 + \epsilon ^2 y_2 + \dots$, where $y$ satisfies the following second order ODE:
$\frac{d^2y}{dx^2} + (1-\epsilon x)y = 0$,
with $|{\epsilon}|\ll 1$ and subject to $y(0) = 1$, $y'(0) = 0$.
I have already found that $y_0 = \cos x$, but I'm stuck with finding $y_1$.
Any help would be much appreciated.
|
You're right that $y_0 = \cos x$. To find $y_1$, we need to plug our expansion for $y$ into the ODE and collect the $\mathcal{O} (\epsilon)$ terms:
$\frac{d^2}{dx^2} (y_0 + \epsilon y_1 + \dots) + (1-\epsilon x)(y_0 + \epsilon y_1 + \dots) = 0$.
Taking order $\epsilon$ terms:
$\frac{d^2y_1}{dx^2} + y_1 - xy_0 = 0$.
Subbing $y_0 = \cos x$:
$\frac{d^2y_1}{dx^2} + y_1 - x \cos x = 0$.
This is a second order nonhomogeneous ODE, which we can solve by finding the complementary and particular solutions and adding them together.
First we need to work out our conditions on $y_1$. We know that $y(0) = y_0(0) +\epsilon y_1(0) + \dots = 1$. Since $1$ is $\mathcal{O}(1)$, we can only have $y_0(0) = 1$, $y_i(0) = 0 \ \ \forall \ \ i \neq 1$. Also, since $y'(0) = y_0' + \epsilon y_1' + \dots = 0$, we must have $y_i'(0) = 0 \ \ \forall \ \ i = 0,1,2,\dots$
So we've now got the conditions $y_1(0)=y_1'(0)=0$.
The complementary solution is $y_1^p = c_1 \sin x + c_2 \cos x$ for some constants $c_1,c_2$. (Solution to $\frac{d^2y_1}{dx^2} + y_1 = 0$).
For the particular solution, we can use the trial function: $y_1 = (Ax^2+Bx)\sin x + (Cx^2+Dx)\cos x$, with $A,B,C,D$ constants to be determined.
Subbing this in and equating coefficients, we end up with particular solution $y_1^p = \frac{x^2}{4}\sin x + \frac{x}{4} \cos x$.
So:
$y_1 = y_1^c +y_1^p = c_1\sin x + c_2 \cos x + \frac{x^2}{4}\sin x + \frac{x}{4} \cos x$
Applying the conditions, we find that $c_1=c_2=0$, so that:
$y_1 = \frac{x^2}{4}\sin x + \frac{x}{4} \cos x$.
Hence, $y \sim \cos x + \epsilon\left(\frac{x^2}{4}\sin x + \frac{x}{4} \cos x \right)$.
$x$." />
The dotted line is the asymptotic approximation, the solid line is the solution to the ODE. You can see the approximation is pretty much exact for small $x$.This is for $\epsilon = 0.01$, the approximation would get better and better the smaller $\epsilon$ becomes.
|
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|
Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $ Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $
My attempt : I recognised that this series can be decomposed into the taylor expansion of $ e $ around 0.
So I thought of writing the series as
$ 1 + \frac{1}{2!} + \frac{1}{3!} ...$ + $ 2[ \frac{1}{2!} + \frac{1}{3!}...]$ $ +$ $ 3[ \frac{1}{3!} + \frac{1}{4!} ...] + ...$
However I got stuck here and couldn't proceed further.
Any hints on how to proceed further , or a better method to solve the question would be appreciated.
|
The general term is $$a_n=\frac{1}{n!}\sum_{i=1}^n i=\frac 12\frac{n (n+1)}{ n!}$$ So, now consider $$S=\frac 12\sum_{n=1}^\infty \frac{n (n+1)}{ n!}x^n=\frac 12\sum_{n=1}^\infty \frac{n (n-1)+2n}{ n!}x^n=\frac 12\sum_{n=1}^\infty \frac{n (n-1)}{ n!}x^n+\sum_{n=1}^\infty \frac{n}{ n!}x^n$$ $$S=\frac {x^2}2\sum_{n=1}^\infty \frac{n (n-1)}{ n!}x^{n-2}+x\sum_{n=1}^\infty \frac{n}{ n!}x^{n-1}$$ where you should recognize derivatives of $e^x$. At the end, set $x=1$.
|
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|
Show that there are no solutions to $x^2 + y^2 = 3z^2$ in $\Bbb{Z}$ I'm attempting to work through some of the questions in Whitelaw's "Introduction to Abstract Algebra" but am having some difficulty.
The question is as follows
Show that $\forall n \in \Bbb{Z},$ $n^2 \equiv 0\ \text{or}\ 1 \mod{3}.$ Show further that $$3\rvert(x^2 + y^2) \implies 3\rvert x\ \text{and}\ 3\rvert y.$$
Deduce that there is no solution in $\Bbb{Z}$ to the equation
$$x^2 + y^2 = 3z^2$$
except $x = y = z = 0.$
For the first part, I simply let $n = 3k,$ $3k+1$ or $3k + 2$ with $k \in \Bbb{Z}$. Hence,
\begin{align}
(3k)^2 &\equiv 0 \mod{3}\\
(3k+1)^2 = 9k^2 + 6k + 1 &\equiv 1\mod{3}\\
(3k+2)^2 = 9k^2 + 12k + 3 + 1 &\equiv 1\mod{3}.
\end{align}
For the second part, since $n^2 \equiv 0$ or $1 \mod{3}$, and that the only case in which $n^2 \equiv 0 \mod{3}$ was when $3\rvert n$, it must then be the case that both $x^2$ and $y^2$ are divisible by $3$ for $3$ to divide their sum (else we get that $x^2 + y^2 \equiv 1$ or $2 \mod{3}$). We also know that if $3\rvert n$ then $3\rvert n^2$. Hence, $3\rvert x$ and $3\rvert y$.
I am, however, unsure how to approach the third part.
Thanks for any help!
|
Let's suppose that we have a solution $(a,b,c)$ where $c\neq 0$ is the smallest positive integer such that the equality holds.
$$a^2 + b^2= 3c^2 \Rightarrow 3|(a^2 + b^2)$$
But we know that this means $3|a$ and $3|b$. However, on squaring we get that $9$ divides the LHS, and so it must also divide the RHS:
$$a'^2 + b'^2 = \frac{c^2}{3}$$
Where $3a' = a, 3b' = b$. But $3$ is prime, so if it divides the product it must divide $c$. Factor that out as $3c'$ and divide:
$$a'^2 + b'^2 = 3c'^2$$
But this is another solution triple with $c' < c$ (and also $c' \neq 0$), and we chose the smallest possible $c$, so we have a contradiction.
|
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|
Find the sum to $n$ terms of the series $10+84+734+....$ Find the sum to n terms of the series $10+84+734+....$
*
*$\frac{9(9^n+1)}{10} + 1$
*$\frac{9(9^n-1)}{8} + 1 $
*$\frac{9(9^n-1)}{8} + n $
*$ \frac{9(9^n-1)}{8} + n^2$
My attempt:
I'm getting option $(4)$.
For $n=1$, all options are right,
For $n=2$, sum must be $10+84=94$.
Can you explain in formal way? Please.
|
Note: I think formal issues are not that essential. But you have to reasonably show that you have checked all variants and derive the appropriate conclusion. Here is one variation of the theme.
We have four options each providing a sequence
\begin{align*}
(a_n)_{n\geq 1}=(a_1,a_2,a_3,\ldots)
\end{align*}
The sequence against we have to check is
\begin{align*}
\left(10,10+84,10+84+734,\ldots\right)=\left(10,94,828,\ldots\right)
\end{align*}
We do so by listing each of the four alternatives as far as necessary:
\begin{align*}
\left(\frac{9(9^n+1)}{10} + 1\right)_{n\geq 1}&=(10,74.1,\ldots)\tag{1}\\
\left(\frac{9(9^n-1)}{8} + 1\right)_{n\geq 1}&=(10,91,\ldots)\tag{2}\\
\left(\frac{9(9^n-1)}{8} + n\right)_{n\geq 1}&=(10,92,\ldots)\tag{3}\\
\left(\frac{9(9^n-1)}{8} + n^2\right)_{n\geq 1}&=(10,94,828,\ldots)\tag{4}\\
\end{align*}
We observe alternatives (1) to (3) are no proper solution, since the second element of each of these three alternatives is not equal
\begin{align*}
10+84=94
\end{align*}
The last option (4) coincides in all three values $(10,10+84,10+84+734)=(10,94,828)$ with the series which is to generate.
and conclude the sequence
\begin{align*}
\left(\frac{9(9^n-1)}{8} + n^2\right)_{n\geq 1}=(10,94,828,\ldots)
\end{align*}
is the solution.
|
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|
Let $x \ge 0$. Determine a condtion on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$ I'm trying to understand the logic of this proof.
Let $x \ge 0$. Determine a condition on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$
Proof
$|\sqrt{x} - 2| = \frac{|(\sqrt{x} - 2)(\sqrt{x} + 2)|}{\sqrt{x} + 2} = \frac{x-4}{\sqrt{x} + 2} \le \frac{|x-4|}{2}$
For $x$ satisfying $|x-4| = 2 \cdot 10^{-2}$
we have $|\sqrt{2} -2| \le \frac{2 \cdot 10^{-2}}{2}=10^{-2}$
Does the $|\sqrt{2} -2|$ just come from the 2 in $|x-4| = 2 \cdot 10^{-2}$? Is there more too it? Or is there a better, more complete proof?
Thank you, in advance.
|
When you say $$|\sqrt{x} - 2| = \frac{|(\sqrt{x} - 2)(\sqrt{x} + 2)|}{\sqrt{x} + 2} = \frac{|x-4|}{\sqrt{x} + 2} \le \frac{|x-4|}{2}$$ The final $\le$ comes from replacing $\sqrt x +2$ with $2$. As $\sqrt x \ge 0$, this is decreasing the denominator, which increases the fraction. The overall inequality
$$|\sqrt{x} - 2|\le \frac{|x-4|}{2}$$ is then used. We want to require that $|\sqrt x - 2| \lt 0.02$, so we demand that $$\frac{|x-4|}{2}\lt 0.02$$, which gives $$|x-4| \lt 0.01$$
|
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|
How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$ Let $a, b, c$ be positive real numbers, show that
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$
I think this is likely to turn out to be proved by Hölder, but I can't see how. Any hints will be appreciated.
|
W.L.O.G., let $a\ge b\ge c$, then by the rearrangement inequality, we have
\begin{align*}
\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}
&\ge \frac{b^6}{b^2+c^2}+\frac{c^6}{a^2+c^2}+\frac{a^6}{a^2+b^2},\tag{1}\\
\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}
&\ge \frac{c^6}{b^2+c^2}+\frac{a^6}{a^2+c^2}+\frac{b^6}{a^2+b^2}.\tag{2}
\end{align*}
Thus combine $(1)$ and $(2)$,
\begin{align*}
2&\left(\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}\right)\\
&=\frac{b^6+c^6}{b^2+c^2}+\frac{a^6+c^6}{a^2+c^2}+\frac{a^6+b^6}{a^2+b^2}\\
&=(b^4-b^2c^2+c^4)+(a^4-a^2c^2+c^4)+(a^4-a^2b^2+b^4)\\
&=(a^4+b^4+c^4)+(a^4+b^4+c^4-a^2b^2-a^2c^2-b^2c^2)\\
&\ge a^4+b^4+c^4\tag{3}\\
&=\frac{a^4+b^4}{2}+\frac{b^4+c^4}{2}+\frac{c^4+a^4}{2}\\
&\ge \frac{a^4+b^2c^2}{2}+\frac{b^4+a^2c^2}{2}+\frac{c^4+a^2b^2}{2}\tag{4}\\
&\ge a^2bc+b^2ac+c^2ab\\
&=abc(a+b+c),
\end{align*}
where $(3)$ and $(4)$ are also used the rearrangement inequality. So the result follows.
|
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|
In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these
The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$
I am stuck here.
|
Since $0=a^2+b^2+c^2-ac-ab\sqrt{3}=\left(b-\frac{\sqrt{3}}2a\right)^2+\left(c-\frac12a\right)^2\,,$ we have $b=\frac{\sqrt{3}}2a$ and $c=\frac{1}{2}a$. Thus, $\angle A=\frac{\pi}{2}$, $\angle B=\frac{\pi}{3}$, and $\angle C=\frac{\pi}{6}$.
It can be shown that, if $\alpha,\beta,\gamma\in(0,\pi)$ and $\alpha+\beta+\gamma=\pi$, then there is a unique triangle $ABC$, up to scaling, with $BC=a$, $CA=b$, and $AB=c$ such that $$a^2+b^2+c^2=2bc\cos(\alpha)+2ca\cos(\beta)+2ab\cos(\gamma)\,.$$
To show this, one observes that the matrix $$\textbf{X}:=\begin{bmatrix}1&-\cos(\gamma)&-\cos(\beta)\\-\cos(\gamma)&1&-\cos(\alpha)\\-\cos(\beta)&-\cos(\alpha)&1\end{bmatrix}$$ is positive-semidefinite and the eigenspace associated with the eigenvalue $0$ of $\textbf{X}$ is spanned by $\begin{bmatrix}\sin(\alpha)\\\sin(\beta)\\\sin(\gamma)\end{bmatrix}$.
|
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|
Obtaining the Rodrigues formula On $So(3)$ the algebra of a $3 \times 3$ skew symmetric matrices define Lie bracket $[A,B]=AB-BA$
Consider the exponential map $$EXP: So(3) \to So(3)$$.
We have the $So(3)$ matrix $$A=\begin{bmatrix} 0 & -c & b \\c & 0 & -a\\
-b & a & 0\end{bmatrix}$$
Upon letting $\theta=\sqrt{a^2 + b^2 +c^2}$ , show that we obtain the identity (which is the Rodrigues formula) $$EXP (A)=I_3 + \frac{sin \theta}{\theta} A+ \frac{I-cos \theta}{\theta^2} A^2$$
I am not sure how we get the expressions $$A^{2n}=(-1)^{n+1} \theta^{2(n+1)}\begin{bmatrix} -(b^2+c^2) & ab & ac \\ab & -(a^2+c^2) & bc\\
ac & bc & -(a^2+b^2)\end{bmatrix}$$
and $$A^{2n+1}=(-1)^n \theta^{2n}\begin{bmatrix} 0 & -c & b \\c & 0 & -a\\
-b & a & 0\end{bmatrix}$$
I understand that you look at the powers of $A$, $A^2$, $A^3$ and so on.
I also understand that you get $A^{2n}$ for even powers of n and $A^{2n+1}$ for odd powers of n.
I work out $$A^2= \begin{bmatrix} -b^2-c^2 & ab & ac \\ab & -a^2-c^2 & bc\\
ac & bc & -a^2-b^2\end{bmatrix}$$
$$A^3= \begin{bmatrix} 0 & a^2c-c(-b^2-c^2) & -a^2b+b(-b^2-c^2)
\\-b^2 c+c(-a^2-c^2) & 0 & ab^2-a(-a^2-c^2)\\
bc^2-b(-a^2-b^2) & -ac^2+a(-a^2-b^2) & 0\end{bmatrix}$$
From $A^2$ how do you get the expression for $A^{2n}$?
From $A^3$ how do you get the expression for $A^{2n+1}$?
For example how is $\theta$ incorporated?
|
"Rodrigues formula" will be easier to prove with $A/\theta=B$, thus the result we have to prove is under the form:
$$Exp(\theta B)=I_3 + (\sin \theta) B+ (1-\cos \theta) B^2 \ \ \ (0)$$
by taking this definition:
$$B:=\begin{bmatrix} 0 & -c & b \\c & 0 & -a\\
-b & a & 0\end{bmatrix} \ \ (1a) \ \ \text{with} \ \ \sqrt{a^2+b^2+c^2}=1 \ \ \ (1b)$$
(your presentation has its own merits, but its drawback is that it is un-natural for an angle to have the dimension of a length).
Using (1a) and (1b), the characteristic polynomial of $B$ is found to be $$det(B-\lambda I_3)=-\lambda^3-\lambda \ \ \ $$
whose roots (eigenvalues of $B$) are $\{0,i,-i\}$.
As a consequence, Cayley-Hamilton's theorem gives
$$B^3=-B \ \ \ \text{and, consequently} \ \ \ B^4=-B^2, B^5=-B^3=B, \ \ (2) \ \ \text{etc.}$$
(cycling: $-B, -B^2, B, B^2, -B...$). Use now the definition
$$Exp(\theta B)=\sum_{k=0}^{\infty}\dfrac{1}{k!}\theta^k B^k \ \ \ (3)$$
Transform (3) by using the different relationships (2). Gather them into 3 groups,
*
*the first one, with $I_3$ alone.
*the second group with terms $\alpha_kB$.
*the third group with terms $\beta_kB^2$.
It is easy to see that $\sum_{k=0}^{\infty} \alpha_k=\sin \theta$ and $\sum_{k=0}^{\infty} \beta_k=1-\cos \theta$, proving (0).
Remark: A similar, simpler, computation exists in 2D. Here it is. Let $J=\begin{bmatrix}0&-1\\1&0 \end{bmatrix}$ (representing a $\pi/2$ rotation, with eigenvalues $i$ and $-i$). A basic relationship for $J$ is $J^2=-I$, thus $J^3=-J$, etc. Proceeding in the same way as we have done upwards, one obtains
$$Exp(\theta J)=(\cos\theta)I_2+(\sin\theta)J$$
which is nothing else than a matrix version of the classical $e^{i\theta}=(\cos\theta)+(\sin\theta)i$ but can also be seen and proved as the particular case of (0) with $a=0, b=0, c=1.$
|
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|
Limit of tan function This is a question from an old tutorial for a basic mathematical analysis module.
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning?
Since $\frac{\pi}{4}+\frac{1}{n} \to \frac{\pi}{4}$ as $n \to \infty$,
it seems to me that $\tan(\frac{\pi}{4}+\frac{1}{n}) \to \tan(\frac{\pi}{4}) = {1}$,
and hence $\tan^n(\frac{\pi}{4}+\frac{1}{n}) \to 1^n = 1$.
Additionally, my tutor has given a hint, to use Squeeze theorem along with the definition $e = \lim\limits_{n \to \infty }(1 + {1\over n})^n$, but I can't see how these are to be used.
Edit: Here's another attempt I've made.
After using addition formula for tangent, we get
$$\frac{1 + \tan{\frac{1}{n}}}{1- \tan{\frac{1}{n}}} = 1 - \frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = 1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}} = [(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}}$$
so given that $\lim\limits_{n \to \infty}{\frac{\tan{\frac{1}{n}}}{\frac{1}{n}}}=1$, we have
$$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = \lim\limits_{n \to \infty}[(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2n\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = e^\frac{2*1}{1-0} = e^2$$
I'm really hoping that this method works as well! So sorry for the ugly formatting, I couldn't figure out some parts.
|
Using the addition angle formula for the tangent function we can write
$$\tan^n\left(\frac{\pi}{4}+\frac1n\right)=\left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n$$
Note that $\tan(x)=x+O(x^3)$. Then, we have
$$\begin{align}
\left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n&=\left(\frac{1+\frac1n +O\left(\frac1{n^3}\right)}{1-\frac1n +O\left(\frac1{n^3}\right)}\right)^n\\\\
&=\left(\frac{1+\frac1n }{1-\frac1n }\right)^n\left(\frac{1+\frac{O\left(\frac1{n^3}\right)}{1+\frac1n}}{1 +\frac{O\left(\frac1{n^3}\right)}{1-\frac1n }}\right)^n\\\\
&\to \frac {e}{e^{-1}}\,\frac{1}{1}\,\,\text{as}\,\,n\to \infty\\\\
&=e^2
\end{align}$$
And we are done!
|
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|
Pade approximant for the function $\sqrt{1+x}$ I'm doing the followiwng exercise:
The objective is to obtain an approximation for the square root of any given number using the expression
$$\sqrt{1+x}=f(x)\cdot\sqrt{1+g(x)}$$
where g(x) is an infinitesimal. If we choose $f(x)$ as an approximation of $\sqrt{1+x}$, then we can calculate $g(x)$:
$$g(x)=\frac{1+x}{f^2(x)}-1$$
$f(x)$ can be chosen as a rational function $p(x)/q(x)$, such that $p$ and $q$ have the same degree and it's Mclaurin series is equal to the Mclaurin series of the function $\sqrt{1+x}$ until some degree. Find a rational function $f(x):=p(x)/q(x)$, quotient of two linear polynomials, such that the McLaurin series of $p(x)-\sqrt{1+x}\cdot q(x)$ have the three first terms equal to $0$.
How can I do this? Has something to be with the Pade approximant?
Any hint would be really appreciated. Thanks for your time.
|
Using the idea of continued fractions, observe that
$$
\sqrt{1+x}=1+(\sqrt{1+x}-1)=1+\frac{x}{2+(\sqrt{1+x}-1)}\\=1+\cfrac{x}{2+\cfrac{x}{2+(\sqrt{1+x}-1)}}=…
$$
continuing in an obvious repetitive pattern.
The computation of the (degree-balanced) partial fractions of this can be implemented (in Magma CAS) as
F<x>:=RationalFunctionField(Rationals());
p := 0;
for k in [1..6] do
p := x/(2+x/(2+p));
print 1+p;
end for;
with the results (for instance using http://magma.maths.usyd.edu.au/calc/)
(3*x + 4)/(x + 4)
(5*x^2 + 20*x + 16)/(x^2 + 12*x + 16)
(7*x^3 + 56*x^2 + 112*x + 64)/(x^3 + 24*x^2 + 80*x + 64)
(9*x^4 + 120*x^3 + 432*x^2 + 576*x + 256)/(x^4 + 40*x^3 + 240*x^2 + 448*x + 256)
(11*x^5 + 220*x^4 + 1232*x^3 + 2816*x^2 + 2816*x + 1024)/(x^5 + 60*x^4 + 560*x^3 + 1792*x^2 + 2304*x + 1024)
(13*x^6 + 364*x^5 + 2912*x^4 + 9984*x^3 + 16640*x^2 + 13312*x + 4096)/(x^6 + 84*x^5 + 1120*x^4 + 5376*x^3 + 11520*x^2 + 11264*x + 4096)
as the first six balanced Pade approximants.
|
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|
How to evaluate this limit using Taylor expansions? I am trying to evaluate this limit:
$\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4}$
I know that I need to use Taylor expansions for $\sin x -x$, $\cos x -1$ and $e^x-1$. I also realise that all of these are just their regular Taylor expansions with their first term removed so the series starts at $n=1$ rather than at 0.
However, when I actually try to evaluate I get stuck at:
$\lim_{x \to 0} \dfrac{(-\frac{1}{3!}+\frac{x^2}{5!}...)(-\frac{9x^2}{2!}+\frac{81x^4}{4!}...)}{(\frac{1}{x}+\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}...)^4}$
I don't know how to proceed from here.
|
Using basic limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\left( \frac{\cos x-1}{x^{2}}\right) &=&-\frac{1}{2} \\
\lim_{x\rightarrow 0}\left( \frac{x}{e^{x}-1}\right) &=&1
\end{eqnarray*}
which can be evaluated using L'HR, it follows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{(\sin x-x)(\cos x-1)}{x(e^{x}-1)^{4}}
&=&\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{%
\cos x-1}{x^{2}}\right) \left( \frac{x}{e^{x}-1}\right) ^{4} \\
&=&\left( \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}\right) \left(
\lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}}\right) \left( \lim_{x\rightarrow
0}\frac{x}{e^{x}-1}\right) ^{4} \\
&=&\left( -\frac{1}{6}\right) \left( -\frac{1}{2}\right) \left( 1\right) ^{4}
\\
&=&\frac{1}{12}.
\end{eqnarray*}
|
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Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.
Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: Why one should never divide by an expression that contains a variable.
So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?
|
The equation says$$ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}}=\frac{x^2 + 9}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}.$$
So either $x^2+9=0$ and $x=\pm3i$, or $\sqrt{x^2+9}=\sqrt{x^2+1}$, which is impossible (by squaring).
|
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Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's rule, that is,
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x} = \lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 }$$
since $\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \frac {1 - 1}{ 1- 1} = \frac00$. Thus, we use the L'Hopital's rule again. that is,
$$\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \lim_{x\rightarrow 0}\frac{-\sin x}{2 \sec^2 x \sec x \tan x}$$
since $\lim_{x\rightarrow 0} \frac{-\sin x}{2 \sec^2 x \sec x \tan x}= \frac{0}{2(1)(0)} = \frac00$. Thus, it always goes to zero by zero. But the answer for this question is $\frac{-1}2$. How is that?!! there must be something missing that either I forget or misunderstand.
|
$$
\lim_{x\to0}\frac{\sin x-x}{\tan x-x}\overset{L'H}{=}\lim_{x\to0}\frac{\cos x-1}{\frac{1}{\cos^2x}-1}=\lim_{x\to0}\frac{\frac{\cos x-1}{x^2}}{\frac{1-\cos^2 x}{x^2\cos^2x}}
$$
now,
$$
\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac{1}{2}
$$
and
$$
\lim_{x\to0}\frac{1-\cos^2 x}{x^2\cos^2x}=\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\frac{1}{\cos^2x}=1
$$
so, the desired limit is $-\frac{1}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence of a series implies convergence of another series Let $a_1,a_2,\cdots$ be a sequence of real numbers with $a_i\geq 0$. If $\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty$ then show that $\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}<\infty$ for each real sequence $x_i$ with $x_i\geq 0$ and $\lim \inf x_n >0$.
Suppose $x_n\geq 1$ for all $n$ then $1+x_na_n\geq 1+a_n$ for all $n$. So, $\frac{1}{1+x_na_n}\leq \frac{1}{1+a_n}$ and in particular, we have
$$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}<\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty.$$
So, we are done in this case.
We can assume that $\lim \inf x_n$ is less than $1$.
Just to get some idea, considering some simple sequences. Let $x_n=\frac{1}{2}+\frac{1}{n}$. Then $$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}=\sum_{n=1}^{\infty}\frac{2n}{2n+(n+2)a_n}=2\sum_{n=1}^{\infty}\frac{n}{2n+(n+2)a_n}=2\sum_{n=1}^{\infty}\frac{1}{2+(1+2/n)a_n}$$
Now, clealy $2+(1+2/n)a_n\geq 1+a_n$ which implies $$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}=2\sum_{n=1}^{\infty}\frac{1}{2+(1+2/n)a_n}\leq 2\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty.$$
So, we are done. There is nothing special about $2$.. Any $k$ would guarantee the convergence.
For $x_n=\frac{1}{k}+\frac{1}{n}$ then $$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}=k\sum_{n=1}^{\infty}\frac{1}{k+(1+k/n)a_n}\leq k\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty.$$
For this sequence, $x_n=\frac{1}{k}+\frac{1}{n}$, the $\lim \inf x_n $ is $\frac{1}{k}>0$. I want to do the same for general $1>\epsilon >0$. Given $1>\epsilon >0$ there exists $k$ such that $\epsilon>\frac{1}{k}$.
So, $$\lim \inf x_n=\epsilon >\frac{1}{k}=\lim \inf \frac{1}{k}+\frac{1}{n}$$
Then i want to say $x_n\geq \frac{1}{k}+\frac{1}{n}$. So, $$\frac{1}{1+x_na_n}\leq \frac{1}{1+(1/k+1/n)a_n}$$ and in particular,
$$\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}\leq \sum_{n=1}^{\infty} \frac{1}{1+(1/k+1/n)a_n}\leq k\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty.$$
I just want to know if there are any gaps or false statements.
|
Convergence only depends on the tail of the series.
If $\liminf x_n \ge b > 0$, then there is some $N$ such that for $n > N$, $x_n > b/2$. Let $B = \max(1, 2/b)$.
Then for $n > N$,
$$\dfrac{1}{1+x_n a_n} \le \dfrac{1}{1 + (b/2) a_n} \le \dfrac{B}{1+a_n}$$
By a Comparison test, $\sum_n 1/(1+x_n a_n)$ converges.
|
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|
How do I solve for $m$ and $n$ While reading about nested radicals, I came across a theorem that said $\sqrt{m\sqrt[3]{4m-8n}+n\sqrt[3]{4m+n}}=\pm\frac {1}{3}\left(\sqrt[3]{(4m+n)^{2}}+\sqrt[3]{4(4m+n)(n-2n)}+\sqrt[3]{(n-2n)^{2}}\right)$
So I tried an easy example ($\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}$) and got this System of Equation: $$\begin{cases}m^3(4m-8n)=-4\\n^3(4m+n)=5\end{cases}\text{or} \begin{cases}m^3(4m-8n)=5\\n^3(4m+n)=-4\end{cases}$$
My question: How do I solve for $m$ and $n$ without too much tedious work, and how do I know which System to discard?
|
Eliminate $m$ from the first system with
$$m=\frac{5-n^4}{4n^3},$$
giving
$$(5-n^4)^3(20-36n^4)=1024n^{12},$$
a quartic equation in $n^4$
$$9(n^4)^4-396(n^4)^3+750(n^4)^2-1500(n^4)+625=0.$$
Even Wolfram cannot find a closed-form solution !
With $-4$ in the RHS of the first equation, the equation is
$$9(n^4)^4+116(n^4)^3+750(n^4)^2-1500(n^4)+625=0,$$
which has the root $n^4=1$.
|
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|
Multiplying binomials to come up with $ y^8 - 256 $ $$\ { (y^4 + 16) }{ (y^2 + 4) }{ (y + 2) }{ (y - 2) }$$
How do I multiply these to come up with $\ {y^8 - 256}$
|
$$(a+b)(a-b)=a^2-b^2$$
$$\ { (y^4 + 16) }{ (y^2 + 4) }\color{red}{{ (y + 2) }{ (y - 2)} }=(y^4+16)(y^2+4)\color{red}{(y^2-4)}$$
$$(y^4+16)\color{red}{(y^2+4){(y^2-4)}}=(y^4+16)\color{red}{(y^4-16)}$$
$$\color{red}{(y^4+16)(y^4-16)}={(y^8-256)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers $$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$
My attempt
$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$
$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$
$x+938^2 + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$
$2x + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} = 1116^2 - 938^2 - 140^2$
$x + \sqrt{x^2 + 2(938^2 + 140^2)x+(938*140)^2} = 1116^2 - 938^2 - 140^2$
At this point trying to solve for x inside the sqrt in the quadratic gives me an imaginary number. How is it possible to solve this?
|
For this sort of problem, sometimes it will be useful to replace
explicit numbers by variables to avoid distractions.
Let $a = 938$, $b = 140$, $c = 938 + 140 + 38 = 1116$.
Let $u = \sqrt{x+a^2}$ and $v = \sqrt{x+b^2}$, the equation at hand becomes
$$u + v = c\tag{*1}$$
Notice $u^2 - v^2 = (x+a^2) - (x+b^2) = a^2-b^2$, we have
$$u - v = \frac{u^2 - v^2}{u+v} = \frac{a^2-b^2}{c}\tag{*2}$$
Combine $(*1)$ and $(*2)$, we have
$\displaystyle\;u = \frac{c^2 + a^2-b^2}{2c}$ and hence
$$\begin{align}
x = u^2 - a^2
& = \left(\frac{c^2 + a^2-b^2}{2c}\right)^2 - a^2
= \left(\frac{(c+a)^2 - b^2}{2c}\right)\left(\frac{(c-a)^2-b^2}{2c}\right)\\
& = \frac{(c+a+b)(c+a-b)(c-a+b)(c-a-b)}{4c^2}\\
& = \frac{2194 \cdot 1914 \cdot 318 \cdot 38}{4\cdot 1116^2}
= \frac{352392601}{34596}
\approx 10185.93481905423
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
So many logs with different bases
$ \large { 6 }^{ \log _{ 5 }{ x } }\log _{ 3 }( { x }^{ 5 } ) -{ 5 }^{ \log _{ 6 }{ 6x } }\log _{ 3 }{ \frac { x }{ 3 } } ={ 6 }^{ \log _{ 5 }{ 5x } }-{ 5 }^{ \log _{ 6 }{ x } }$
The sum of the solutions to the equation above can be expressed as $a^{b/c} + d$ find $abc+d$.
I can't find a way out... But if $x = 1$ then the equation is true. I think $d = 1$. I guess first I've to simplify the equation. But can't find any way... The bases of the logs are also different. any hint will be helpful....
P.S: Problem Collected from Brilliant.org
|
$$ \large { 6 }^{ \log _{ 5 }{ x } }\log _{ 3 }( { x }^{ 5 } ) -{ 5 }^{ \log _{ 6 }{ 6x } }\log _{ 3 }{ \frac { x }{ 3 } } ={ 6 }^{ \log _{ 5 }{ 5x } }-{ 5 }^{ \log _{ 6 }{ x } }$$
$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-{5}^{\log _{6}{6}+\log_6x}(\log_{3}x-\log_3 3)={6}^{\log _{ 5 }{5}+\log_5x}-{5}^{\log _{6}{x}}$$
$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-{5}^{1+\log_6x}(\log_{3}x-1)={6}^{1+\log_5x}-{5}^{\log _{6}{x}}$$
$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-5\cdot{5}^{\log_6x}\cdot\log_{3}x+5\cdot{5}^{\log_6x}=6\cdot{6}^{\log_5x}-{5}^{\log _{6}{x}}$$
$$\large 5\cdot{6}^{\log _{5}{x} }\log _{3}{x}-5\cdot{5}^{\log_6x}\cdot\log_{3}x+5\cdot{5}^{\log_6x}=6\cdot{6}^{\log_5x}-{5}^{\log _{6}{x}}$$
Let $\log_5x=a; \log_3x=b; \log_6x=c.$
$$\large 5\cdot{6}^{a}\cdot b-5\cdot{5}^{c}\cdot b+5\cdot{5}^{c}=6\cdot{6}^{a}-{5}^{c}$$
$$5b(6^a-5^c)=6(6^a-5^c)$$
$$(6^a-5^c)(5b-6)=0$$
Then $6^a-5^c=0$ or $5b-6=0$
Case 1)$6^a-5^c=0 \Leftrightarrow 6^{\log_5x}=5^{\log_6x}$
$$6^{\log_5x}=5^{\log_6x}\Leftrightarrow x=1$$
Case 2) $5\log_3x=6$
$$\log_3x=\frac65 \Rightarrow x=\sqrt[5]{3^6}$$
|
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|
$(x-1)(y-2)=5$ and $(x-1)^2+(y+2)^2=r^2$ intersect at four points $A,B,C,D$. Centroid of $\Delta ABC$ lies on $y=3x-4$, then the locus of $D$ $(x-1)(y-2)=5$ and $(x-1)^2+(y+2)^2=r^2$ intersect at four points $A,B,C,D$. If centroid of $\Delta ABC$ lies on $y=3x-4$, then what is the locus of $D$?
I did try a couple of things, but I honestly have no idea how to approach this. I checked if $(x-1)(y-2)=5$ represented a pair of straight lines, but it doesn't.
I also tried substituting the value of $(x-1)$ from the first equation in the equation of the circle, but the only useful result I got was that the sum of y-coordinates is zero since the equation after substitution does not have a $y^3$ term, only $y^4$, $y^2$, $y$ and constant terms.
Apart from that, I have no other ideas.
|
Point D is the intersection of line $y=3x$ (parallel to $y=3x-4$ and through origin) and curves: $(x-1)(3x-2)=5$ and $(x-1)^2+(y)^2=r^2$ .
$\begin{cases} y=3x \\(x-1)(y-2)=5 \\ (x-1)^2+(y+2)^2=r^2 \end{cases}$
$(x-1)(3x-2)=5$
$3x^{2}-5x-3=0$
$\begin{array}{} x_{D}=\frac{5+\sqrt{61}}{6} & y_{D}=\frac{5+\sqrt{61}}{2} \end{array}$
solving numerically
$(x_{D}-1)^2+(y_{D}+2)^2=r^2$
$r^2=\frac{335+\sqrt{97600}}{9}≈71.93444300402957$
$\left\{ \begin{array}{} (x-1)(y-2)=5 & ⇒ y=\frac{2x+3}{x-1} \\ (x-1)^2+(y+2)^2=r^2 \end{array} \right\}$
$sol=\left( \begin{array}{left } x_{A}=-6.788336312 & y_{A}=1.358014369 \\ x_{B}=.599099974 & y_{B}=-10.471937333 \\ x_{C}=8.054194725 & y_{C}=2.708798126 \\ x_{D}=2.135041613 & y_{D}=6.405124838 \end{array} \right)$
checking if the centroid of triangle ABC belongs to the line $y = 3x-4$
$\begin{array}{} \text{Centroid (G)} & x_{G}=\frac{x_{A}+x_{B}+x_{C}}{3} & y_{G}=\frac{y_{A}+y_{B}+y_{C}}{3} \end{array}$
$G=(0.621652796,-2.135041613)$
$3·x_{G}-4=y_{G}$
|
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|
solve $a_n=5a_{n-1}-4a_{n-2}+3\cdot2^n$ with initial conditions $a_0=1, a_1=10$ so i am pretty sure that i have solve the homogeneous solution correctly. $a_n^h = B\cdot 4^n+C\cdot1^n$ however I am not so confident on the particular solution. Here was my attempt.
Since $f(n)=3\cdot2^n$ I knew that the particular solution would be in the form of $D*2^n$ so i plugged that in and got $D 2^n=5 D 2^{n-1}-4 D 2^{n-2} +3\cdot2^n $ I then did some work to find D=-6. The way I did that was divide everything by $2^{n-2}$ and got the equation $4D=10D-4D+12$.
So my question am i doing this correctly or was I wrong to starting out by saying the equation was of the form $D\cdot2^n$. In class we always did it by finding $s$, $t$, and $m$ where $s =2$, $t$ would be the degree so t=0 and m=multiplicity which i always get confused about. Thanks for the help
|
As a way of checking, you could always fall back on the generating function route:
$$\begin{align*} f(z) &= \sum_{n=0}^\infty a_n z^n \\
&= a_0 + a_1 z + \sum_{n=2}^\infty (5a_{n-1} - 4a_{n-2} + 3(2^n))z^n \\
&= 1 + 10z + 5z(-1 + f(z)) - 4z^2 f(z) + 3\sum_{n=2}^\infty (2z)^n \\
&= 1 + 5z + (5z - 4z^2) f(z) + \frac{12z^2}{1-2z}. \end{align*}$$
Thus, $$\begin{align*} f(z) &= \frac{(1+z)(1+2z)}{(1-2z)(1-z)(1-4z)} \\
&= \frac{2}{1-z} - \frac{6}{1-2z} + \frac{5}{1-4z} \\
&= \sum_{n=0}^\infty 2z^n - 6(2z)^n + 5(4z)^n \\
&= \sum_{n=0}^\infty (2-6(2^n)+5(4^n))z^n, \end{align*}$$ giving $$a_n = 2-6(2^n)+5(4^n).$$
|
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|
For positive $a$, $b$, $c$ with $a^4+b^4+c^4=3$, then $\sum_{cyc}\frac{a^2}{b^3+1}\geq \frac32$
If $a$, $b$, $c$ $\in (0, \infty)$, $a^4+b^4+c^4=3$, then: $$\sum_{cyc}\frac{a^2}{b^3+1}\geq \frac32$$
original problem image
I have been into inequalities lately and I am stuck with this. I used a famous inequality at first
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3 \left(\frac{a^4+b^4+c^4}{3}\right)^{1/4}$$
From this, I just had to prove that $x^3+1-2x \le 0$ for all $x \in (0,1)$ but infact this is not true.
Can anyone help me out?
|
Note that
$$\frac{1}{b^3+1} - \Big(1 - \frac{b^2+b}{4}\Big) = \frac{b(b^2+3b+1)(b-1)^2}{4(b+1)(b^2-b+1)}\ge 0.$$
It suffices to prove that
$$\sum_{\mathrm{cyc}} a^2\Big(1 - \frac{b^2+b}{4}\Big) \ge \frac{3}{2}$$
or
$$a^2+b^2+c^2 - \frac{1}{4}(a^2b^2+b^2c^2+c^2a^2) - \frac{1}{4}(a^2b+b^2c+c^2a)\ge \frac{3}{2}.$$
Note that
$$a^2b+b^2c+c^2a \le \sqrt{(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)}.$$
It suffices to prove that
$$a^2+b^2+c^2 - \frac{1}{4}(a^2b^2+b^2c^2+c^2a^2) - \frac{1}{4} \sqrt{(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)} \ge \frac{3}{2}.$$
Let $p = a^2+b^2+c^2, \ q = a^2b^2 + b^2c^2 + c^2a^2$.
It follows from $a^4+b^4+c^4 = 3$ that $p^2 - 2q = 3$ or $q = \frac{p^2 - 3}{2}.$
It is easy to obtain $\sqrt{3}\le p \le 3.$
We need to prove that
$$p - \frac{1}{4}q - \frac{1}{4}\sqrt{pq} \ge \frac{3}{2}$$
or
$$8p-p^2-9 \ge \sqrt{2p(p^2-3)}.$$
Clearly, $8p-p^2-9 > 0$. It suffices to prove that $$(8p-p^2-9)^2 - 2p(p^2-3)\ge 0$$
or $$(p-3)(p^3-15p^2+37p-27)\ge 0.$$ It is obvious. We are done.
|
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|
Solving a radical equation for real $x$
Solve for $x \in \mathbb{R}$
$$\dfrac{\sqrt{x^2-x+2}}{1+\sqrt{-x^2+x+2}} - \dfrac{\sqrt{x^2+x}}{1+\sqrt{-x^2-x+4}} = x^2-1$$
I tried squaring the equation but it became a sixteen degree equation. I also tried substitutions, but that didn't help. There must be some elegant solution to it in its current form.
Any help will be appreciated.
Thank you.
|
Firstly, we need to find what values of $x$ are acceptable. Solving the inequalities:
$$\begin{array}[t]{rl}
x^2-x+2 &\ge 0\\
x^2+x &\ge 0 \\
-x^2+x+2 &\ge 0\\
-x^2-x+4 &\ge 0
\end{array}
$$
yields
$$\left.\begin{array}{l}
x \in \mathbb R\\
x \in (-\infty,-1] \cup [0,+\infty)\\
x \in [-1,2] \\
x\in \left[ \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}\right]
\end{array}\right\}\implies x\in \{-1\} \cup \left[0,\frac{-1+\sqrt{17}}{2}\right].
$$
We can see that $x = -1$ is not a solution, so we are looking for non - negative solutions.
For $ x \in [0,1)$ the RHS is negative. But in that case we have that the first numerator is greater than the second one (this can be derived easily by solving the inequality $\sqrt{x^2-x+2} \gt \sqrt{x^2+x}$) and the first denominator is less than the second denominator, thus the first fraction is always greater than the second one, so their difference is positive. Thus there are no solutions in $[0,1)$.
Similarly, for $x\in \left(1,\frac{-1+\sqrt{17}}{2}\right]$ we have that the RHS is positive and the LHS is negative.
This implies that the only real solution is $x = 1$.
|
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|
Solve $ 1 + \frac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \frac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $
Solve for $x \in \mathbb{R}$
$$ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $$
I tried some substitutions and squaring but that didn't help. I also tried to use inequalities as done in my previous problem, but that too didn't help.
|
I hope that my solution is simpler:
\begin{eqnarray}
&&1+\frac{\sqrt{x+3}}{1+\sqrt{1-x}}=x+\frac{\sqrt{2x+2}}{1+\sqrt{2-2x}}\\
&\Longleftrightarrow& x-1+\frac{\sqrt{2x+2}}{1+\sqrt{2-2x}}-\frac{\sqrt{x+3}}
{1+\sqrt{1-x}}=0\\
&\Longleftrightarrow& x-1+\frac{\sqrt{2x+2}+\sqrt{2x+2}\sqrt{1-x}-\sqrt{x+3}-\sqrt{x+3}\sqrt{2-2x}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}=0\\
&\Longleftrightarrow& x-1+\frac{\frac{x-1}{\sqrt{2x+2}+\sqrt{x+3}}+\frac{4(x-1)}{\sqrt{2x+2}\sqrt{1-x}+\sqrt{x+3}\sqrt{2-2x}}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}=0\\
&\Longleftrightarrow& (x-1)\left(1+\frac{\frac{1}{\sqrt{2x+2}+\sqrt{x+3}}+\frac{4}{\sqrt{2x+2}\sqrt{1-x}+\sqrt{x+3}\sqrt{2-2x}}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}\right)=0\\
&\Longleftrightarrow& x=1.
\end{eqnarray}
|
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|
Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$. Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$.
My method:
If $n = 10$, $2^n \gt n^3$ where $2^{10} \gt 10^3$ which is equivalent to $1024 \gt 1000$, which holds for $n = 10$. $2^k \gt k^3$.
$2^{k + 1} \gt (k + 1)^3$
$2^{k + 1} \gt (k + 1)(k + 1)(k + 1)$
$2^{k + 1} \gt (k^2 + 2k + 1)(k + 1)$
$2^{k + 1} \gt (k^3 + 3k^2 + 3k + 1)$
$(k + 1) \cdot 2^k \gt k^3 + 3k^2 + 3k + 1$
Since $2^{10} \gt 10^3$, the inequality holds for $n = 10$. Assume $2^k \gt k^3$ for $k \ge 10$. We show that $2^{k + 1} \gt (k + 1)^3$. Hence, $2^{k + 1} = 2 \cdot 2^k \gt 2k^3$
$= k^3 + k^3 \ge k^3 + 10k^2$
$= k^3 + 10k^2 = k^3 + 4k^2 + 6k^2 \ge k^3 + 4k^2 + 6 \cdot 10 = k^3 + 4k^2 +60$
$k^3 + 4k^2 + 60 \gt k^3 + 3k^2 + 3k + 1$
Therefore, $2^{k + 1} \gt (k + 1)^3$. Hence, $2^n \gt n^3$ for every integer $n \ge 10$.
I was trying to fix this but I am not sure how to go about doing so.
|
Hint: if $n\geq 10$, $$
\frac{(n+1)^3}{n^3}=\frac{n^3+3n^2+3n+1}{n^3}=1+\frac 3n+\frac 3{n^2}+\frac 1{n^3}\leq 1+\frac 3{10}+\frac 3{100}+\frac 1{1000}<2.
$$
|
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|
How many phone number with $8$ digits exist s.t divide $2,3,5$ and there is no repetitive digit in it? Here is my approach:
The last digit should be $0$ and the first digit does not $0$. Hence there are $9$ choices for the first digit, $8$ for second,...,$3$ for seventh.
So there are $9.8.7.6.5.4.3$ phone numbers that they don't divide $2,5$.
Now I divide $9.8.7.6.5.4.3$ by $3$ and get approximation $9.8.7.6.5.4$
Because from each of three number (statistically!) one them is divisible by $3$.
But what is the exact answer?
|
Recall that a number is divisible by $3$ if and only if its digit sum is divisible by $3$. We will be choosing $7$ digits from $1,2,\dots,9$ whose digit sum is divisible by $3$.
The sum of the digits $1$ to $9$ is divisible by $3$, so the two digits we do not choose must have sum divisible by $3$. There are $12$ ways to do this, and then $7!$ ways to arrange them in a row, for a total of $12\cdot 7!$.
To see there are $12$ ways to decide which two digits we do not use, note that we could throw away two of $3,6,9$ or throw away one of $1,4,7$ and one of $2,5,8$.
|
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|
Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$
I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from Mathematica
|
The condition can be presented in form
$$a\ln a +b\ln b + c\ln c = 0.$$
Let
$$a\leq b\leq c,$$
then
$$\ln a \leq \ln b \leq \ln c,$$
and we can use Chebyshev sum inequality:
$$3(a\ln a +b\ln b + c\ln c) \geq (a+b+c)\ln abc,$$
$$abc\leq 1.$$
Evidently, the expression $a^2b+b^2c+c^2a$ achieves maximum when $abc=1.$
The standard way of solving the problem on a conditional extremum is the method of Lagrange multipliers, which reduces it to a system of equations.
The maximum value of function
$$f(a,b,c,λ)=a^2b+b^2c+c^2a+λ(abc-1)$$
for $a,b,c>0$ on the interval
is reached or at its edges, or in one of the points with zero partial derivatives
$$f'_a=0,\quad f'_b=0,\quad f'_c=0,\quad f'_λ=0,$$
or
$$\begin{cases}
2ab+c^2+\lambda bc = 0\\
2bc+a^2+\lambda ca = 0\\
2ca+b^2+\lambda ab = 0\\
abc-1 = 0,
\end{cases}$$
then
$$\begin{cases}
a(2ab+c^2) = b(2bc+a^2)=c(2ca+b^2),\\
abc=1,
\end{cases}$$
$$\begin{cases}
2b^2c -c^2a = a^2b\\
b^2c -2c^2a = -a^2b\\
abc=1,
\end{cases}$$
$$c^2a = b^2c = a^2b,\quad abc=1,$$
$$\dfrac cb =1, \dfrac ba =1, \quad abc=1,$$
$$a=b=c=1,$$
$$f(a,b,c,\lambda)=3.$$
Note that
$$\lim_{a\to 0} a\ln a = 0,$$
so on the edge $a=0$ we have to minimize
$f(b,c,\lambda) = b^2c + \lambda(bc-1),$
and then we get the system
$$\begin{cases}
2bc+\lambda c = 0\\
b^2+\lambda b = 0\\
bc-1=0
\end{cases}$$
with solution
$$b=2,\quad c=\dfrac12,\quad f = 2 < 3.$$
That means that
$$\boxed{a^2b+b^2c+c^2a \leq 3}$$
|
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|
How to prove this trigonometric integral? $$ \displaystyle \int_{-\pi/4}^{\pi/4} {{\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)}^{\cos(2t)} \ dx} = \frac{\pi}{2 \sin(\pi \cos^2 t)}$$
I could simplify it to
$\displaystyle \int_0^1 {\left(t^n + \frac{1}{t^n}\right) \ \frac{dt}{1+t^2}}, \ n = \cos 2t $
From here, I can think of expanding into sums but that doesn't seem a good option. Also, getting back to trigonometric form is also an option but it would get us to reduction formula which will be messy.
What is a straight, neat and easy approach to solve it?
|
\begin{align*}
\frac{\cos x-\sin x}{\cos x+\sin x} &=
\tan \left( \frac{\pi}{4}-x \right) \\
\int_{-\pi/4}^{\pi/4}
\left( \frac{\cos x-\sin x}{\cos x+\sin x} \right)^{n} dx &=
\int_{-\pi/4}^{\pi/4} \tan^{n} \left( \frac{\pi}{4}-x \right) dx \\
&= \int_{0}^{\pi/2} \tan^{n} u \, du \\
&= \frac{\pi}{2} \sec \frac{n\pi}{2} \, , \quad -1<n<1 \\
&= \frac{\pi}{2} \sec \frac{\pi \cos 2t}{2} \\
&= \frac{\pi}{2} \sec \frac{\pi (2\cos^2 t-1)}{2} \\
&= \frac{\pi}{2} \csc (\pi \cos^2 t) \\
\end{align*}
|
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|
$a,b,c >0$, and $ab+bc+ca=3$, prove $(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$ $a,b,c >0$, and $ab+bc+ca=3$, prove
$$(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$
I think the equality is only achieve when $a=b=c=1$. The condition $ab+bc+ca=3$ is necessary. I used the estimation $x^x \geqslant \frac12 (x^2+1)$ but cannot proceed further.
|
It is not an answer, but rather some results which can be helpful for the further investigations.
*
*$abc \leq 1$
$\rhd$ $$\frac{ab +bc + ac}{3} \geq \sqrt[3]{a^2b^2c^2}\Rightarrow\sqrt[3]{a^2b^2c^2}\leq 1$$
$\lhd$
*One can prove that
$$\forall a_i > 0,\forall p_i>0: \sum\limits_{1}^{n}p_i = 1$$
the following inequality is fulfilled
$$\sum\limits_{i =1}^{n}p_i a_i \geq \prod\limits_{i = 1}^{n}a^{p_i}_{i}.$$
Let us denote
\begin{align}
\frac{a}{a+b+c} &= \alpha\\
\frac{b}{a+b+c} &= \beta,\\
\frac{c}{a+b+c} &= \gamma,
\end{align}
then $\alpha+\beta+\gamma = 1$ and, according to the previous formula ($a_i = \frac1a, \frac1b, \frac1c$),
$$(a^ab^bc^c)^{\frac{3}{a+b+c}} =(a^\alpha b^\beta c^\gamma)^3 \geq \Big(\frac{a+b+c}{3}\Big)^3 = \Big(\frac{a^3 + b^3 + c^3 + 9(a+b+c) - 3abc}{27}\Big)\geq \Big(\frac{a^3 + b^3 + c^3}{27} + \frac{a+b+c}{3} - \frac19\Big),$$
where the last inequality uses 1.
|
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|
Given two lines in Cartesian form, find the vector equation of a line which passes through the intersection of two lines. Given two lines in Cartesian form, find the vector equation of a line which passes through the intersection of two lines (and is perpendicular to both). No points given just the two equations. What are the steps to solve this?
For example:Line 1: $x+1=\frac{y}{3}=-z$, Line 2: $2x+1=2y+1=z-\frac{3}{2}$.
|
$x+1 = 2x+1 \Leftrightarrow x = 0$
$\frac{y}{3} = 2y + 1 \Leftrightarrow y = 6y + 3 \Leftrightarrow y = - \frac{3}{5} $
$ z - \frac{3}{2} = -z \Leftrightarrow 2z = \frac{3}{2} \Leftrightarrow z = \frac{3}{4} $
So the point $P$ of the intersection is : $P=(0,-\frac{3}{5},\frac{3}{4})$
The parallel vectors to the lines are : $n_1 = (1,3,-1) $ and $n_2 = (\frac{1}{2}, \frac{1}{2}, 1)$.
A vector perpendicular to these 2 is : $n_p = n_1 \otimes n_2 = (\frac{7}{2},-\frac{3}{2},-1) $
$n_p =(\frac{7}{2},-\frac{3}{2},-1) $ will be the parallel vector of your line and your line passes through $P=(0,-\frac{3}{5},\frac{3}{4})$.
So the equation of the line is : $\frac{2x-}{7} = -\frac{2y+ \frac{3}{5}}{3}=-z +\frac{3}{4}$
|
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|
Convert from Nested Square Roots to Sum of Square Roots I am looking for a way to easily discover how to go from a nested root to a sum of roots. For example,
$$\sqrt{10-2\sqrt{21}}=\sqrt{3}-\sqrt{7}$$
I know that if I set $\alpha=\sqrt{10-2\sqrt{21}}$, square both sides, I get
$$\alpha^2=10-2\sqrt{21}$$
Now I recognize that we have a situation where $10=3+7$ and $21=7\cdot 3$, so I can immediately see that we have
$$\alpha^2=10-2\sqrt{21}=3-2\sqrt{21}+7=\sqrt{3}^2-2\sqrt{3}\sqrt{7}+\sqrt{7}^2=(\sqrt{3}-\sqrt{7})^2$$
My question is, it this the only way to approach this problem? This approach mirrors basic algebra 1 methods of factoring quadratics, but I was curious to know if thre are other techniques that can be used to quickly deduce that a nested radical can be simplified to the sum of two radicals. Mathematically, suppose $a,b,c,m,n,r,s\in\mathbb{N}$. Is there a way to quickly determine $m,n,r,s$ in the equation
$$\sqrt{a\pm b\sqrt{c}}=m\sqrt{r}\pm n\sqrt{s}$$
|
Let $a$ and $b$ be non-negative rational numbers such that
$$\sqrt{10 - 2\sqrt{21}} = \sqrt{a} - \sqrt{b}$$
Squaring both sides of the equation yields
$$10 - 2\sqrt{21} = a - 2\sqrt{ab} + b$$
Matching rational and irrational parts yields the system of equations
\begin{align*}
a + b & = 10 \tag{1}\\
-2\sqrt{ab} & = -2\sqrt{21} \tag{2}
\end{align*}
Dividing both sides of equation 2 by $-2$ and squaring yields
$$ab = 21$$
Solving for $b$ yields
$$b = \frac{21}{a}$$
Substituting this expression in equation 1 yields
$$a + \frac{21}{a} = 10$$
Multiplying both sides of the equation by $a$ and solving the resulting quadratic equation yields
\begin{align*}
a^2 + 21 & = 10a\\
a^2 - 10a + 21 & = 0\\
(a - 3)(a - 7) & = 0
\end{align*}
The roots are $a = 3$ and $a = 7$. If $a = 3$, then $b = 21/a = 21/3 = 7$. However, $$\sqrt{3} - \sqrt{7} < 0 \implies \sqrt{10 - 2\sqrt{21}} \neq \sqrt{3} - \sqrt{7}$$
since the principal square root of a number cannot be negative.
If $a = 7$, then $b = 21/a = 21/7 = 3$. Hence,
$$\sqrt{10 - 2\sqrt{21}} = \sqrt{7} - \sqrt{3}$$
If you instead had to evaluate $\sqrt{7 + 4\sqrt{3}}$, set
$$\sqrt{7 + 4\sqrt{3}} = \sqrt{a} + \sqrt{b}$$
|
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|
If $x \geq C$, where $C > 0$ is a constant, then what is the least upper bound for $\dfrac{2x}{x + 1}$? The title says it all.
Since
$$f(x) = \dfrac{2x}{x + 1} = 2\left(1 - \dfrac{1}{x + 1}\right),$$
then because $x \geq C$ where $C > 0$, an upper bound is given by
$$\dfrac{2x}{x + 1} < 2.$$
Note that the lower bound is equivalent to
$$x \geq C \iff x + 1 \geq C + 1 \iff \dfrac{1}{x + 1} \leq \dfrac{1}{C + 1}$$ $$\iff 2\left(1 - \dfrac{1}{x + 1}\right) \geq 2\left(1 - \dfrac{1}{C + 1}\right),$$
so that if $2 - \varepsilon$ is an upper bound for $f(x)$ ($\varepsilon > 0$), then $\varepsilon$ satisfies the inequality
$$2 - \varepsilon \geq f(x) \geq 2\left(1 - \dfrac{1}{C + 1}\right) \implies 0 < \varepsilon < \dfrac{2}{C + 1}.$$
My question is essentially whether we can do better than this.
|
rewrite $f(x)$ in the form $$\frac{2}{1+\frac{1}{x}}$$ and consider the cases $$x>0$$ or $$x<0$$
|
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|
How to prove that $(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$, where $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$ $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$, prove
$$(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$$
I try several trig substitutions but feel hopeless with the cyclic term here. The condition $x^2+y^2+z^2+xyz=4$ made it too difficult to homogenize the inequality. I don't even know how to do brutal force either.
|
Let $x=\frac{2a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{2b}{\sqrt{(a+b)(b+c)}},$ where $a$, $b$ and $c$ are positives.
Thus, the condition gives $z=\frac{2c}{\sqrt{(a+c)(b+c)}}$ and we need to prove that
$$x^2y^2z^2+3xyz+\sum_{cyc}(x^3y^2+x^3z)\leq10$$ or
$$\frac{32a^2b^2c^2}{\prod\limits_{cyc}(a+b)^2}+\frac{12abc}{\prod\limits_{cyc}(a+b)}+\sum_{cyc}\left(\tfrac{16a^3b^2}{\sqrt{(a+b)^5(a+c)^3(b+c)^2}}+\tfrac{8a^3c}{\sqrt{(a+b)^3(a+c)^4(b+c)}}\right)\leq5$$ or
$$\sum_{cyc}\left(16a^3b^2(b+c)\sqrt{\tfrac{a+c}{a+b}}+8a^3c(b+c)\sqrt{(b+c)(a+b)}\right)\leq$$
$$\leq5\prod_{cyc}(a+b)^2-32a^2b^2c^2-12abc\prod_{cyc}(a+b).$$
Now, by AM-GM $$\sqrt{(b+c)(a+b)}\leq\frac{1}{2}(2b+a+c)$$ and
$$ab\sqrt{\frac{a+c}{a+b}}=\frac{ab\sqrt{(a+c)(a+b)}}{a+b}\leq\frac{\left(\frac{a+b}{2}\right)^2\cdot\frac{1}{2}(2a+b+c)}{a+b}=\frac{1}{8}(a+b)(2a+b+c).$$
Thus, it's enough to prove that:
$$\sum_{cyc}\left(16a^3b^3\sqrt{\tfrac{a+c}{a+b}}+2a^2bc(a+b)(2a+b+c)+4a^3c(b+c)(2b+a+c)\right)\leq$$
$$\leq5\prod_{cyc}(a+b)^2-32a^2b^2c^2-12abc\prod_{cyc}(a+b)$$ or
$$16\sum_{cyc}a^3b^3\sqrt{\frac{a+c}{a+b}}\leq\sum_{cyc}(5a^4b^2+a^4c^2+6a^3b^3+2a^4bc+4a^3b^2c+2a^3c^2b-4a^2b^2c^2).$$
Now, by C-S $$\sum_{cyc}a^3b^3\sqrt{\frac{a+c}{a+b}}\leq\sqrt{\sum_{cyc}\frac{a^3b^3}{a+b}\sum_{cyc}a^3b^3(a+c)}.$$
Id est, it's enough to prove that:
$$256\sum_{cyc}\frac{a^3b^3}{a+b}\sum_{cyc}a^3b^3(a+c)\leq\left(\sum_{cyc}(5a^4b^2+a^4c^2+6a^3b^3+2a^4bc+4a^3b^2c+2a^3c^2b-4a^2b^2c^2)\right)^2,$$
which is obviously true after full expanding.
The last part for you.
|
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Find $\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$ $$\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$$
I have no idea about this.
The equation can be written in its recursive form as:
$$f(n) = g(1,n)$$
Where
$$g(x,n) = [x\impliedby n]\cdot (x+ g(x+1,n))^{\frac 1x}+[x=n]\cdot (n)^{\frac 1n}$$
Of course, [] is the indicator function representing of piece wise notation.
|
Here I show that $1.9<Lim_{n→∞} a_n<2$ if:
$a_n=\sqrt{2+\sqrt [3]{3+\sqrt[4]{4+ . . . +\sqrt[n]{n}}}}$
We use following inequalities:
$\sqrt[k]{k+1}>1$ ; $\sqrt[k]k<\sqrt[k]{k+2}$. . (for $k>2$)
The first one is clear and the second one can be proved by induction. The series of numbers of $a_n $is increasing; using inequalities we may write:
$a_n=\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n]{n}}}<\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n-1]{n-1+2}}}< . . .<\sqrt{2+\sqrt[3]{5}}$
Therefore aeries $a_n $ has a limit equal to $a_0$ such that we have:
$a_0<\sqrt{2+\sqrt[3]{5}}<2$
Using inequalities we have:
$a_n>\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n-1]{n}}}>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$
Therefore:
$a_0>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$
We can easily see that:
$\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}>1.9$
That finally gives:
$1,9<a_n<2$
and for you question:
$2.9<a_1+1<3$
|
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|
Compute $\sum_{k=0}^{180} \cos^22k^\circ $ I'm trying to compute the following:
$$\sum_{k=0}^{180} \cos^22k^\circ $$
|
Remark that:
$cos(x+90°) = sin(x)$
And that:
$cos^2(x) + sin^2(x) = 1$
So you can group your terms:
$$
\begin{align*}
\sum_{x=0}^{180} cos^2(x.2°) &= \sum_{x=0}^{44} cos^2(x.2°) \\
& + \sum_{x=45}^{89} cos^2(x.2°) \\
& + \sum_{x=90}^{134} cos^2(x.2°) \\
& + \sum_{x=135}^{179}{cos^2(x.2°)} \\
& + cos^2(180.2°) \\
\\
&= \sum_{x=0}^{44}{cos^2(x.2°)+cos^2(x.2°+90°)} \\
& + \sum_{x=90}^{179}{cos^2(x.2°)+cos^2(x.2°+90°)} \\
& + cos^2(360°) \\
\\
&= \sum_{x=0}^{44}{cos^2(x.2°)+sin^2(x.2°)} \\
& + \sum_{x=90}^{134}{cos^2(x.2°)+sin^2(x.2°)} \\
& + 1 \\
\\
&= \sum_{x=0}^{44}{1} \\
& + \sum_{x=90}^{134}{1} \\
& + 1 \\
&= 91
\end{align*}
$$
See the approximate results at wolfram alpha
|
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|
How to evaluate $\lim _{n\to \infty }\left(\frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use l'Hospital's rule.)
$$\lim _{n\to \infty }\left(\frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}}\right)$$
|
To provide an alternative (may be a little expeditious but also need a little more background) to rationalization, we may use Taylor's expansion for $(1 + x)^\alpha, \alpha > 0$ at $0$ (which is also known as McLaurin expansion):
$$(1 + x)^\alpha = 1 + \alpha x + o(x). \tag{1}$$
In view of $(1)$, the expression of interest can be written as
\begin{align}
& \frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}} \\
= & \frac{2n^{3/2}\sqrt{1 + \frac{3}{4n^2}} - 2n^{3/2}\sqrt{1 - \frac{2}{n}}}{\sqrt{2n}\sqrt{1 + \frac{2}{n}}} \\
= & \sqrt{2}n\frac{\left(1 + \frac{1}{2}\frac{3}{4n^3} + o(1/n^2)\right) - \left(1 - \frac{1}{2}\frac{2}{n} + o(1/n)\right)}{1 + \frac{1}{2}\frac{2}{n} + o(1/n)} \\
= & \frac{\sqrt{2} + o(1)}{1 + o(1)} \\
\to & \sqrt{2}
\end{align}
as $n \to \infty$.
|
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How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+x^2} = \frac{1}{x^2} + \frac{2}{x} - \frac{1}{x+1}$$
Therefore:
$$\int \frac{x^4+1}{x^3+x^2}\,dx = \int \frac{dx}{x^2} + \int \dfrac{2\,dx}{x} - \int \frac{dx}{x+1} = -\frac{1}{x} +2\log \vert x\vert - \log \vert x+1 \vert + C$$
The problem is I was supposed to find: $$\int \frac{x^4+1}{x^3+x^2}\,dx = \frac{x^2}{2} - x - \frac{1}{2} - \log \vert x \vert + 2 \log \vert x+1 \vert + C$$
Where is my mistake?
|
Your $A,B,C$ are wrong. You can't write the expression in that form because the numerator won't have the correct degree.
If you combine the terms, you get
$$\frac{A(x+1)+Bx(x+1) +Cx^2}{x^3+x^2}$$
Note that the numerator is of degree at most $2$.
Instead, just perform polynomial division to get a quotient and a remainder term. You can then do what you did, working with the fractional remainder.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form of $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$? I am trying to find a closed form for the integral $$I=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$$ So far, my reasoning is thus: write, by symmetry through $x=\pi/2$, $$I=2\sum_{n=1}^{\infty}n\int_{\arctan n}^{\arctan (n+1)}\frac{dx}{|\tan x|}=2\sum_{n=1}^{\infty}n\ln\frac{\sin\arctan(n+1)}{\sin\arctan n}$$
Using $\sin{\arctan {x}}=\frac{x}{\sqrt{1+x^{2}}}$, we get: $$I=2\sum_{n=1}^{\infty}n\ln(\frac{(n+1)\sqrt{1+n^2}}{n\sqrt{1+(n+1)^2}})=\sum_{n=1}^{\infty}n\ln\frac{(n+1)^2(1+n^2)}{n^2(1+(n+1)^2)}=\sum_{n=1}^{\infty}n\ln(1+\frac{2n+1}{n^2(n+1)^2})$$ Expanding the logarithm into an infinite series we get $$I=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}(2n+1)^m}{mn^{2m-1}(n+1)^{2m}}$$ Here I am a bit stuck.. Does anyone have any suggestions to go further?
Thank you.
EDIT:
keeping in mind the nice answer below, applying summation by parts to $$I_N=2\sum_{n=1}^{N}n\ln\frac{\sin\arctan(n+1)}{\sin\arctan n}=2\sum_{n=1}^{N}n(\ln\sin\arctan(n+1)-\ln\sin\arctan n)$$ gives $$I_N=2((N+1)\ln\sin\arctan(N+1)+\frac{\ln 2}{2}-\sum_{n=1}^{N}\ln\sin\arctan(n+1))$$ hence: $$I-\ln2=-\sum_{n=2}^{\infty}\ln\frac{n^2}{1+n^2}=\sum_{n=2}^{\infty}\ln\frac{1+n^2}{n^2}=\sum_{n=2}^\infty\sum_{m=1}^\infty\frac{(-1)^{m+1}}{mn^{2m}}= \sum_{m=1}^\infty\frac{(-1)^{m+1}}{m}\sum_{n=2}^\infty n^{-2m}=\sum_{m=1}^\infty\frac{(-1)^{m+1}(\zeta(2m)-1)}{m}$$
Is this valid and helpful?
EDIT 2: Coming back to $$\sum_{n=2}^{\infty}\ln(1+\frac{1}{n^2})=\ln(\prod_{n=2}^{\infty}(1+\frac{1}{n^2}))=\ln(\prod_{n=2}^{\infty}(1-\frac{i^2}{n^2}))=\ln(\prod_{n=1}^{\infty}(1-\frac{i^2}{n^2}))-\ln2$$
$$=\ln(\frac{\sin(i\pi)}{i\pi})-\ln2=\ln\frac{\sinh\pi}{\pi}-\ln2$$ hence $I=\ln\frac{\sinh\pi}{\pi}$
|
Maybe we are lucky. We may notice that:
$$ 1+\frac{2n+1}{n^2(n+1)^2} = 1+\frac{1}{n^2}-\frac{1}{(n+1)^2} $$
and the roots of the polynomial $x^2(x+1)^2+2x+1$ are given by
$$ \alpha = \frac{1}{2}\left(-1-\sqrt{2}-\sqrt{2\sqrt{2}-1}\right), $$
$$ \beta = \frac{1}{2}\left(-1-\sqrt{2}+\sqrt{2\sqrt{2}-1}\right), $$
$$ \gamma = \frac{1}{2}\left(-1+\sqrt{2}-i\sqrt{2\sqrt{2}+1}\right), $$
$$ \delta = \frac{1}{2}\left(-1+\sqrt{2}+i\sqrt{2\sqrt{2}+1}\right), $$
so:
$$ \sum_{n=1}^{N}\log\left(1+\frac{2n+1}{n^2(n+1)^2}\right)=\log\prod_{n=1}^{N}\frac{(n-\alpha)(n-\beta)(n-\gamma)(n-\delta)}{n^2(n+1)^2}$$
can be written in terms of:
$$ \log\prod_{n=1}^{N}\frac{n-\alpha}{n} = \log\frac{\Gamma(N+1-\alpha)}{\Gamma(N+1)\Gamma(1-\alpha)} $$
and through summation by parts the problem boils down to computing:
$$ \sum_{N\geq 1}\log\frac{\Gamma(N+1-\alpha)\Gamma(N+1-\beta)\Gamma(N+1-\gamma)\Gamma(N+1-\delta)}{(N+1)^2\Gamma(N+1)^4\Gamma(1-\alpha)\Gamma(1-\beta)\Gamma(1-\gamma)\Gamma(1-\delta)}\tag{1}$$
where:
$$\log\Gamma(z+1)=-\gamma z+\sum_{n\geq 1}\left(\frac{z}{n}-\log\left(1+\frac{z}{n}\right)\right) $$
probably leads to a massive simplification of $(1)$, or at least the chance to write $(1)$ as a simple integral by exploiting the identities:
$$ \log(m)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-mx}}{x}\,dx,\qquad \log\left(1-\frac{\nu}{n}\right)=\int_{0}^{+\infty}\frac{1-e^{\nu x}}{x e^{nx}}\,dx.$$
However, by Did's comment we simply have:
$$ \log\prod_{n\geq 1}\left(1+\frac{1}{n^2}\right) = \color{red}{\log\frac{\sinh \pi}{\pi}} $$
through the Weierstrass product for the $\sinh$ function.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How does one solve this recurrence relation? We have the following recursive system:
$$
\begin{cases}
& a_{n+1}=-2a_n -4b_n\\
& b_{n+1}=4a_n +6b_n\\
& a_0=1, b_0=0
\end{cases}
$$
and the 2005 mid-exam wants me to calculate answer of $ \frac{a_{20}}{a_{20}+b_{20}} $.
Do you have any idea how to solve this recursive equation to reach a numerical value?
|
Observe that
$$a_{n+1}+b_{n+1}=2a_n+2b_n=2(a_n+b_n)\;,$$
and $a_0+b_0=1$, so in general $a_n+b_n=2^n$.
Quickly calculating a few values, we see that the numbers $b_n$ are a little nicer than the numbers $a_n$:
$$\begin{array}{rcc}
n:&0&1&2&3&4\\
a_n:&1&-2&-12&-40&-112\\
b_n:&0&4&16&48&128\\
\end{array}$$
Concentrating on the $b_n$, we see that
$$b_{n+1}=4(a_n+b_n)+2b_n=2^{n+2}+2b_n\;,$$
so that
$$\begin{align*}
b_n&=2b_{n-1}+2^{n+1}\\
&=2(2b_{n-2}+2^n)+2^{n+1}\\
&=2^2b_{n-2}+2\cdot2^{n+1}\\
&=2^2(2b_{n-3}+2^{n-1})+2\cdot 2^{n+1}\\
&=2^3b_{n-3}+3\cdot 2^{n+1}\\
&\;\;\vdots\\
&=2^kb_{n-k}+k2^{n+1}\\
&\;\;\vdots\\
&=2^nb_0+n2^{n+1}\\
&=n2^{n+1}\;,
\end{align*}$$
so $a_n=2^n-n2^{n+1}=2^n(1-2n)$, and
$$\frac{a_n}{a_n+b_n}=\frac{2^n(1-2n)}{2^n}=1-2n\;.$$
(There are other ways to solve that first-order recurrence for $b_n$; I just picked the most elementary one.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$ I started like this :
$a^2+c^2=b^2(a^2-1)\\c^2 +1=(a^2-1)(b^2-1)$
but it's leads to nowhere.
can you help please ?
|
This is a Pythagorean quadruple problem.there is a good way to prove that the only solution is (0,0,0)
assuming the equation is
$a^2+b^2+c^2=t^2$ ,∀ t=ab
all integer positive solution given by
$a=(l^2+m^2−n^2)/n , b=2l, c=2m$ ,and $t=(l^2+m^2+n^2)/n$
Then
$2l(l^2+m^2−n^2)=l^2+m^2+n^2$
one can find that $(l^2+m^2−n^2)(2l−1)=0$
suppose $2l−1=0$ refused
then
$l^2+m^2−n^2=0⇒a=0 $ ,and $ab=2n$
but $ab=0⇒n=0⇒l^2+m^2=0⇒b=0,c=0$
therefore the only solution is (0,0,0)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determining axis of rotation from the rotation matrix without using eigenvalues and eigenvectors
Consider the following rotation matrix:
$$R=\begin{pmatrix} -\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\
\sqrt{\frac{3}{8}} &\frac{1}{2}& \sqrt{\frac{3}{8}} \\
-\frac{1}{\sqrt{8}}&\sqrt{\frac{3}{4}}&-\frac{1}{\sqrt{8}}\end{pmatrix}$$
Determine the angle and the axis of rotation by using the following
equations:
$$T=R+R^T-[\text{Tr}R-1]I\\\cos{\varphi}=\frac{1}{2}(\text{Tr}R-1)$$
Where $\text{Tr}R$ is the trace of $R$.
This is what I have done so far:
$$R+R^T-(\text{Tr}R-1)I=\begin{pmatrix} -\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\
\sqrt{\frac{3}{8}} &\frac{1}{2}& \sqrt{\frac{3}{8}} \\
-\frac{1}{\sqrt{8}}&\sqrt{\frac{3}{4}}&-\frac{1}{\sqrt{8}}\end{pmatrix}+\begin{pmatrix}-\frac{1}{\sqrt{2}}&\sqrt{\frac{3}{8}}&-\frac{1}{\sqrt{8}} \\ 0&\frac{1}{2}&\sqrt{\frac{3}{4}} \\ \frac{1}{\sqrt{2}} & \sqrt{\frac{3}{8}}&-\frac{1}{\sqrt{8}}\end{pmatrix}-\left (-\frac{1}{\sqrt{2}}+\frac{1}{2}-\frac{1}{\sqrt{8}}-1\right )I \\ $$
$$=\begin{pmatrix} -\frac{2}{\sqrt{2}}&\sqrt{\frac{3}{8}}& \frac{1}{2\sqrt{2}} \\ \sqrt{\frac{3}{8}}&1& \frac{\sqrt{2}\sqrt{3}+\sqrt{3}}{2\sqrt{2}} \\\frac{1}{2\sqrt{2}}&\frac{\sqrt{2}\sqrt{3}+\sqrt{3}}{2\sqrt{2}}&-\frac{2}{\sqrt{8}}\end{pmatrix}=...?$$
How do I determine the axis of rotation from this matrix?
|
Consider Rodrigues' formula for a rotation matrix:
$$
R=I+\sin\varphi K+(1-\cos\varphi)K^2
$$
with
$$
K=\pmatrix{
0&-k_3&k_2\\
k_3&0&-k_1\\
-k_2&k_1&0
}\;,
$$
where $k$ is the (unit) rotation axis.
To my mind it seems easier to just take the antisymmetric part of $R$, which is $\sin\varphi K$, and normalise to get rid of the $\sin\varphi$, but your lecture notes apparently take the other option and extract $K$ from the symmetric part:
$$
R+R^\top=2\left(I+(1-\cos\varphi)K^2\right)
$$
and thus
\begin{align}
R+R^\top-(\operatorname{Tr}R-1)I
&=2\left(I+(1-\cos\varphi)K^2\right)-(\operatorname{Tr}R-1)I\\
&=2\left(I+(1-\cos\varphi)K^2\right)-2\cos\varphi I\\
&=2(1-\cos\varphi)(K^2+I)\;.
\end{align}
$K^2$ has $-k_2^2-k_3^2$, $-k_3^2-k_1^2$ and $-k_1^2-k_2^2$ on the diagonal, so $K^2+I$ has $k_1^2$, $k_2^2$ and $k_3^2$ on the diagonal.
It follows that you must have made a mistake, as the diagonal elements should all have the same sign. Indeed it seems you forgot to subtract the multiple of the identity in the last line.
This doesn't give you the signs of the $k_i$; you're going to have to extract them from the other components of $K^2$, or use the antisymmetric part after all.
|
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|
Prove : $2(x^3+y^3+z^3)+3xyz \ge 3(x^2y+y^2z+z^2x)$ with $x,y,z \gt 0$ At first I tried to divide both side by $xyz$, the inequality became:
$$2\sum {\frac{x^2}{yz}}+3 \ge 3\sum{\frac xy}$$
Let $$\frac xy = a;\frac yz = b;\frac zx = c;$$
So all we have to prove is
$$2\sum \frac ab +3 \ge 3 \sum a $$ with $a,b,c$ being positive real and $abc=1$ .
And then I stuck at this point. Any help ?
|
By Schur inequality (whenever you get $xyz$ with positive coefficient on the higher side, worth trying Schur):
$$x^3+y^3+z^3 + 3xyz \geqslant (x^2y+y^2z+z^2x) + (xy^2+yz^2+zx^2)$$
So it is enough to show
$$x^3+y^3+z^3+(xy^2+yz^2+zx^2) \geqslant 2(x^2y+y^2z+z^2x)$$
which follows from three AM-GMs like $x^3+xy^2 \geqslant 2x^2y$.
Equality is iff $x=y=z$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Sketch the heart and indicate its orientation with arrows $ r = 1 - \cos(\theta)$. Find the area enclosed by the heart Hi all I am trying to figure out how to sketch the heart. Here is what I have tried so far:
$$r = 1 - \cos(\theta) \\
r(r = 1 - \cos(\theta)) \\
r^2 = r - r\cos(\theta) \\
$$
Use the fact that
$$r =\sqrt{x^2 + y^2} \text{ and } x=r\cos(\theta) \text{ to rewrite the equation as , } \\
x^2 + y^2 = \sqrt{x^2 + y^2} + x \\
x^2 - 2x + y^2 - y = 0 \\
\text{Completing the square gives } \\
(x-1)^2 + (y-\frac{1}{2})^2 = -\frac{5}{4}\\
\sqrt{(x-1)^2 + (y-\frac{1}{2})^2 } = \sqrt{\frac{5}{4}} \\
(x-1) + (y-\frac{1}{2}) = \sqrt{\frac{5}{4}} \\
(y-\frac{1}{2}) = -(x-1) + \sqrt{\frac{5}{4}} \\
y = -(x-1) + \sqrt{\frac{5}{4}} + \frac{1}{2} \\
y = -x + \frac{\sqrt{5} + 3}{2}
$$From here we can get a table of values and sketch the function. For the area the formula is
$$\frac{1}{2} \int f(\theta)^2 \,d\theta$$
So should it be
$$\frac{1}{2} \int y = -x + \frac{\sqrt{5} + 3}{2}$$
|
You can easily draw the shape by substituting convenient values for $\theta$. To find the area, use the formula $\displaystyle A=\frac{1}{2}\int_a^{b}r(\theta)^2d\theta$. The integration proceeds as follows:
$$2\cdot \frac{1}{2} \int_0^{\pi}(1-\cos \theta)^2d\theta$$
$$ \int_0^{\pi}(1-2\cos \theta+\cos^2\theta )d\theta$$
$$ \int_0^{\pi}1-2\cos \theta+\frac{1}{2}+\frac{1}{2}\cos 2 x )d\theta$$
$$ \int_0^{\pi}\frac{3}{2}-2\cos \theta+\frac{1}{2}\cos 2 x )d\theta$$
$$ (\frac{3}{2}\theta-2\sin \theta+\frac{1}{4}\sin 2 x) \Bigg|_0^{\pi}$$
$$\frac{3}{2}\pi-0+0-(0-0+0)=\frac{3}{2}\pi$$
|
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|
Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c ^{\frac34}}{2^{\frac34}}$$
I tried the substitution $x=a^4,\ldots$ but I have no idea how to deal with the left- hand side. I tried some C-S but it goes nowhere. I think Bernoulli's inequality may be the only way to prove this inequality.
|
Version of 29.06.18
HINT
$$\mathbf{\color{brown}{Task\ transformations}}$$
Let
$$
\begin{cases}
&b^3+c^3= 2x^4\\
&c^3+a^3= 2y^4\\
&a^3+b^3= 2z^4
\end{cases}\Rightarrow
\begin{cases}
a^3=-x^4+y^4+z^4\\
b^3= x^4-y^4+z^4\\
c^3= x^4+y^4-z^4\tag1,
\end{cases}$$
then the issue equation transforms to
$$\rlap\bigcirc\!\sum\dfrac{-x^4+y^4+z^4}{z^3}\ge \rlap\bigcirc\!\sum\sqrt[4]{-x^4+y^4+z^4}.$$
$$\rlap\bigcirc\!\sum z\dfrac{-x^4+y^4+z^4}{z^4}\left(1-\dfrac1{\left(1+\dfrac{y^4-x^4}{z^4}\right)^{3/4}}\right)\ge 0.\tag2$$
Using inequality
$$(1+t)^\alpha\le 1+\alpha t,\quad 1>\alpha>0,\quad t\ge-1,\tag3$$
for $\alpha=\dfrac34,$ can be obtained the stronger inequality than $(2):$
$$\rlap\bigcirc\!\sum z\left(1+\dfrac{y^4-x^4}{z^4}\right)\left(1-\dfrac1{1+\dfrac34\dfrac{y^4-x^4}{z^4}}\right)\ge 0,$$
$$\rlap\bigcirc\!\sum \dfrac{y^4-x^4}{z^3}\dfrac{z^4+y^4-x^4}{4z^4+3y^4-3x^4}\ge 0.\tag4$$
$$\mathbf{\color{brown}{Conditions}}$$
Taking in account $(1),$ the boundary conditions are
$$0<x<\sqrt[4]{y^4+z^4},\quad 0<y<\sqrt[4]{z^4+x^4},\quad 0<z<\sqrt[4]{x^4+y^4}.\tag5$$
Taking in account $(5)$, term of $LSH(4)$ is negative iff $x > y.$ If $0<x\le y\le z,$ then the inequality $(4)$ is satisfied.
The task $(4)-(5)$ has rotational symmetry, so WLOG it is required to check only the case
$$x\ge y\ge z>0.\tag6$$
The obtained task $(4)-(6)\ $ is correct and allows a simple proof.
|
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|
The problem of surds and indices If $$ \frac 12 \times \left(\sqrt[3] a + \sqrt[3] b + \sqrt[3] c\right) = \frac 1{\sqrt[3] 3 - 1}\;\;\;(a>b>c)$$
then what is $a - 2b - 3c $ ?
|
Use the factorization $x^3 - 1 = (x - 1)(x^2 + x + 1)$ with $x = \sqrt[3]{3}$ to get $2 = (\sqrt[3]{3}-1)(\sqrt[3]{9} + \sqrt[3]{3} + 1)$, or
$$\frac{1}{\sqrt[3]{3}-1} = \frac{1}{2}(\sqrt[3]{9}+\sqrt[3]{3} + \sqrt[3]{1})$$
So $a = 9$, $b = 3$, $c = 1$.
|
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|
If $g(x) = xf(x^2)$ and $f(x)=\sum_{n=0}^\infty \sin(\frac{\pi}{n+2})x^n$, what is $f^{(20)}(0)$ and $g^{(35)}(0)$? My task is this:
(i)Let $$f(x)=\sum_{n=0}^\infty \sin\left(\frac{\pi}{n+2}\right)x^n.$$ Find $f^{(20)}(0)$ and $g^{(35)}(0)$ when $g(x) = xf(x^2)$.
(ii)Find $$\lim_{n\to\infty}\frac{f(x)-1-\frac{\sqrt(3)x}{2}}{x^2}.$$
My work so far:
After finding the convergence interval $x\in[\pm1,)$, we notice that the $20$th derivative for $x^{20}$ is $20!x^{20-20}= 20!$. If we try the sum formula for this series with $r=x, a_1 = 20!\sin\left(\frac{\pi}{22}\right)$ one should expect to see that $f^{(20)}(x) = 20!\frac{\sin\left(\frac{\pi}{22}\right)}{1-x}\to f^{(20)}(0)=20!\sin\left(\frac{\pi}{22}\right)$.
Although this is the answer I'm not sure if I used this formula correctly. I also have som serious trouble finding my way to $g^{(35)}(0)$. Because $g(x) = xf(x^2) = x\sum_{n=0}^\infty \sin\left(\frac{\pi}{n+2}\right)x^{2n}=\sum_{n=0}^\infty \sin\left(\frac{\pi}{n+2}\right)x^{2n+1}$. We must have that $2n + 1 = 35 \implies n = 17$. I'm thinking that our $g^{(2n+1)}(x)= xf(x^2)=(2n+1)!\frac{x\sin\left(\frac{\pi}{n+2}\right)}{1-x^2}$. This however can't be the case since the answer in my book is $g^{(35)}(0)=35!\sin(\frac{\pi}{19})$. It seems to me that they are using $g^{(2n +1)}= (2n+1)!\frac{\sin\left(\frac{\pi}{n+2}\right)}{1-x^2}$. For the limit we notice that $f^{(0)}=0!\frac{\sin\left(\frac{\pi}{0 + 2}\right)}{1-x}=\frac{1}{1-x}\to\lim_{x\to0}\frac{f(x) - 1 - \sqrt(3)x/2}{x^2}= \lim_{x\to0}\frac{1/(1-x)^2-\sqrt(3)/2}{2x}$, but this doesn't lead me anywhere near the finite answer. So I need some help in understanding their approach and reasoning in addition to comments on my approach with $f^{(20)}$. I'm getting the wrong answer for the limit as I pointed out, so I need someone to show me that aswell. Thanks in advance!
|
Note that $$f'\left(x\right)=\sum_{n\geq0}n\sin\left(\frac{\pi}{n+2}\right)x^{n-1}=\sum_{n\geq1}n\sin\left(\frac{\pi}{n+2}\right)x^{n-1}
$$ since for $n=0
$ there is no addend. Furthermore $$f''\left(x\right)=\sum_{n\geq1}n\left(n-1\right)\sin\left(\frac{\pi}{n+2}\right)x^{n-2}=\sum_{n\geq2}n\left(n-1\right)\sin\left(\frac{\pi}{n+2}\right)x^{n-2}
$$ since for $n=1
$ again there is no addendum. Iterating we have $$f^{\left(20\right)}\left(x\right)=\sum_{n\geq20}n\left(n-1\right)\cdots\left(n-19\right)\sin\left(\frac{\pi}{n+2}\right)x^{n-20}
$$ and if we consider $f^{\left(20\right)}\left(0\right)
$ the only term which is not zero is when the exponent is $n=20$, since $x^{n-20}=1
$, then $$f^{\left(20\right)}\left(0\right)=20!\sin\left(\frac{\pi}{22}\right).
$$ For the other function we can observe that $$g\left(x\right)=\sum_{n\geq0}\sin\left(\frac{\pi}{n+2}\right)x^{2n+1}
$$ so if we take the derivative $$g'\left(x\right)=\sum_{n\geq0}\left(2n+1\right)\sin\left(\frac{\pi}{n+2}\right)x^{2n}
$$ and in this case the series start from $0$ since there is no cancellation. For the second derivative $$g''\left(x\right)=\sum_{n\geq0}\left(2n+1\right)\left(2n\right)\sin\left(\frac{\pi}{n+2}\right)x^{2n-1}=\sum_{n\geq1}\left(2n+1\right)\left(2n\right)\sin\left(\frac{\pi}{n+2}\right)x^{2n-1}
$$ since for $n=0
$ the sum is $0$. So in this case we have to change the starting number of the series only when we have to differentiate an even exponent. Hence $$g^{\left(35\right)}\left(x\right)=\sum_{n\geq17}\left(2n+1\right)\left(2n\right)\cdots\left(2n-33\right)\sin\left(\frac{\pi}{n+2}\right)x^{2n-34}
$$ and again if we consider $g^{\left(35\right)}\left(0\right)
$ the only non zero terms is when $2n-34=0
$ so $n=17$. Hence $$g^{\left(35\right)}\left(0\right)=35!\sin\left(\frac{\pi}{19}\right).$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1796341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.