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Finding the argument $\theta$ of a complex number I want to find the Argument of $z = -\sqrt{2 - \sqrt{3}} + i\sqrt{2 + \sqrt{3}}$ where $z$ is a complex number of the form $z = a + bi$.
I find that the modulus is $2$, but am having trouble simplifying $\theta = \arctan\left(\frac{\sqrt{2 + \sqrt{3}}}{-\sqrt{2 - \sqrt{3}}}\right)$. I can put the squareroot sign over the whole fraction, but that still doesn't really help me get an actual number for theta.
ALSO: It is a rule that I add $\pi$ to theta if $a < 0$. Why? Looking at the diagram of the triangle form by the complex vector $z$ and the real axis, the 'triangle' is in the second quadrant. Hence, the theta we find with $\theta = \tan^{-1}(\frac{b}{a})$ is the angle closes to the real axis on the left-side. However, we measure angle going counter-clockwise. Shouldn't we do the following computation to get the angle going clock-wise: $\pi - \tan^{-1}(\frac{b}{a})$?
|
These numbers may look pretty complicated but notice that if we square both sides a lot can be simplified.
$$ z^2 = \big((2 - \sqrt{3}) - (2 + \sqrt{3})\big) - 2\sqrt{(2 - \sqrt{3})(2 + \sqrt{3})}\,i
= -2\sqrt{3} - 2i = 4 \left(\cos \left(\frac{7\pi}{6}\right) + i\sin \left(\frac{7\pi}{6}\right) \right) $$
So now we know the argument of $z^2$ is $\frac{7\pi}{6}$, we can relate this to the argument of $z$ according to de Moivre's theorem
$$ (\cos \theta + i\sin \theta)^2 = \cos (2\theta + 2n\pi) + i\sin (2\theta + 2n\pi) $$
If $\theta$ is the argument of $z$, the argument of $z^2$ is $$2\theta + 2n\pi = \frac{7\pi}{6}$$
Solving that, we get
$$\theta = \frac{7\pi}{12} + n\pi = \frac{7\pi}{12}, \frac{19\pi}{12} $$
We got two answers, because one is the argument of $z$, and the other is the argument of $-z$.
Since $\cos \theta < 0$ and $\sin \theta > 0$, $\theta$ must lie on the second quadrant (between $\pi/2$ and $\pi$), therefore $ \theta = \frac{7\pi}{12}$
|
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"url": "https://math.stackexchange.com/questions/1130834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\begin{align}
& \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} = \\
& \lim_{x\to0}e^{\ln\left(\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\left(\frac{\sin(x)}{x}\right)\right)} = \\
& e^{\lim_{x\to0}\frac{\ln\left(\left(\frac{\sin(x)}{x}\right)\right)}{x^2}} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \\
& e^{\lim_{x\to0}\frac{x}{2x \sin(x)}\cdot\frac{\cos(x)x -1\cdot\sin(x)}{x^2}} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\tan(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\tan(x)}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{1}{\cos^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(1 - \infty\right)} = \\
& e^{-\infty} = 0\\
\end{align}
$$
Edit #1:
Continuing after the mistake of the $\tan(x)$:
$$\begin{align}
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\cot(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\frac{1}{\tan(x)}}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{1}{\tan(x)}}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{\frac{-1}{\cos^2(x)}}{\tan^2(x)}}{1} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{-1}{\sin^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(-\infty - \infty\right)} = 0\\
\end{align}
$$
|
:( I have not enough reputation to even add a comment...
Yes, the cot<-->tan problem.
Comment: it should be equal to $e^{-\frac{1}{6}}$, if the cot<-->tan was not there.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using integrating factor to find a series solution for an ODE Consider the ODE
$$xy''-y=0$$
Find the solution $y_1$ in the form of a power series, and use an integrating factor in $$y_2=y_1\int\frac{exp(-\int P(x)dx)}{y_1^2}dx$$
to determine $y_2$
To find $y_1$, I put the ODE in standard form $y''-\frac{0y'}{x}-\frac{xy}{x^2}$, giving me $p(x)=0$ and $q(x)=x$. From this I found the indical equation to be $r(r-1)=0$, leaving the two roots $r=0$ and $r=1$.
One of the roots gives a power series solution using $y=\sum\limits_{n=0}^\infty c_nx^{n+r}$, leaving the expression $\sum\limits_{n=0}^\infty (n+r)(n+r-1)c_nx^{n+r-1}-\sum\limits_{n=1}^\infty c_{n-1}x^{n+r-1}=0$.
Thus, considering $r=1$, this defines $$c_n=\frac{c_{n-1}}{(n+r)(n+r-1)}=\frac{c_{n-1}}{(n+1)n}$$
From this the first few terms of the power series for $y_1$ can be found to be
$$c_1=\frac{1}{2}c_0 \quad \quad c_2=\frac{c_1}{6}=\frac{1}{12}c_0 \quad \quad c_3=\frac{c_2}{12}=\frac{1}{144}c_0$$
Thus $y_1=c_0x(1+\frac{x}{2}+\frac{x^2}{12}+\cdots)=x\sum\limits_{n=0}^\infty \frac{x^n}{(n+1)!n!}$
This is where I get stuck. From the indical equation I know $P(x)=0$, so $\int P(x)dx=x$, so $exp(-\int P(x)dx)=e^{-x}$. I have also calculated $y_1^2=x^2+x^3+\frac{11x^4}{12}+\cdots$
Thus $$y_2=\left(x+\frac{x^2}{2}+\frac{x^3}{12}+\cdots\right) \int \frac{e^{-x}}{x^2+x^3+\frac{11x^4}{12}+\cdots}dx$$
But I have absolutely no idea how to do this. I know the fraction within the integral can be changed to $\frac{12e^{-x}}{x^2(11x^2+12x+12+\cdots)}$ but I don't know if that is helpful.
How do I integrate this?
|
You can factor out the $x^2$ term from the denominator, as you noticed, to get:
$$Q(x) = \frac{e^{-x}}{x^2 \, (1+P(x)) } \approx \\ \frac{1}{x^2} (1-x+x^2/2 - x^3/6 + x^4/24 + \ldots) \, (1-P+P^2-P^3+ \ldots ), $$
where $P(x) = x+11 x^2/12 + \ldots $ and use has been made of the Taylor expansion of both $e^{-x}$ and $1/(1+P)$. You can add up as many terms as you wish to get a desired precision.
Hope this is useful.
Cheers!
|
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|
Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$2^4 \cdot\prod_{n=0}^3 \sin\left(\frac{(2n+1)\pi}{16}\right)$$
After that I tried using the formula:
$$\sin\left(1 \frac{\pi}n\right)\sin\left(2 \frac{\pi}n\right) \cdots \sin\left((n-1) \frac{\pi}n\right)= \frac n {2^{n-1}}$$
to find the value of the expression in the parenthesis in the first equation, but currently I'm stuck. I don't think this method will lead me to the correct answer.
Any help would be appreciated!
|
Hint: $1+\cos \theta = 2\cos^2\frac{\theta}{2}$, and use:
$\cos \frac{\theta}{2} = \dfrac{\sin \theta}{2\sin \frac{\theta}{2}}$, or
try to create a quartic equation and use Viete's formula for the product of zeroes.
|
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"url": "https://math.stackexchange.com/questions/1133590",
"timestamp": "2023-03-29T00:00:00",
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|
Find Jordan basis of a given matrix $A=\begin{bmatrix}
-1 & 4 & -4 \\
0 & 1 & 0 \\
1 & -2 & 3
\end{bmatrix}$
The eigenvalues of the matrix are all $1$. The dimension of it's eigenspace is 2 so the Jordan normal form of the matrix is \begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
this is all confirmed by WolframAlpha.
Now, an eigenvector for $1$ is $(0,1,1)$ but when I try t solve $AP=PJ$ where
$P=\begin{bmatrix}
0 & a & d \\
1 & b & e \\
1 & c & f
\end{bmatrix}$
I get $1+b=b$ for the middle element. Why is $(0,1,1)$ a bad choice for the first column here, and how do I find the Jordan basis in this case?
|
The characteristic polynomial is given by
$$\begin{align} p (\lambda) &= (-1-\lambda)(1 - \lambda)(3 - \lambda) - (-4)(1- \lambda)(1) \\&= (1 - \lambda)[-(-1-\lambda)(3-\lambda) - 4]\\&= -(\lambda - 1)^3\end{align}$$
Which gives you eingenvalue $\lambda = 1$ of multiplicity $3$.
To find the eigenvectors, solve
$$(A - I)\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix} = 0.$$
We get $u = \begin{pmatrix}2u_2 - 2u_3\\u_2\\u_3\end{pmatrix}$.
Take two linearly independent eigenvectors, e.g. $u = \begin{pmatrix}0\\1\\1\end{pmatrix}$ (your same idea) and $u' = \begin{pmatrix}2\\0\\-1\end{pmatrix}$. Now to find a generalized vector we make
$$(A - I) v = u'$$
Then $v = \begin{pmatrix}-1 + 2v_2 - 2v_3\\v_2\\v_3\end{pmatrix}$, we may take $v = \begin{pmatrix}-1\\0\\0\end{pmatrix}$. A basis would then be
$$\Bigg\{\begin{pmatrix}0\\1\\1\end{pmatrix},\begin{pmatrix}2\\0\\-1\end{pmatrix},\begin{pmatrix}-1\\0\\0\end{pmatrix}\Bigg\}$$
As we have two linearly independent eigenvectors there is going to be two Jordan blocks, a $1 \times 1$ block corresponding to $u$ and $2\times 2$ block corresponding to $u'$ and $v$. That gives us
$$J =\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}$$
|
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|
differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+(x^2+2x-8)(6x^2+3)$
$(4x^4+4x^3+6x^2+6x+6x^4+12x^3-48x^2+3x^2+6x-24)$
$=10x^4+16x^3+39x^2+6x-18$
|
Up to $$(2x^3+3x)(2x+2)+(x^2+2x-8)(6x^2+3)$$ your work was fine but, immediately after, you have some mistakes since, developing, you are supposed to get $$(4 x^4+4 x^3+6 x^2+6 x)+(6 x^4+12 x^3-45 x^2+6 x-24)=10 x^4+16 x^3-39 x^2+12 x-24$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me that:
$$\frac{x}{(x-1)^2} = \frac{1}{(x-1)} + \frac{1}{(x-1)^2}$$
How to intuitively think about this partial fraction expansion? I've seen some examples but suddenly these conter intuitive examples opo out and I get confused. I can check that his is true but I couldn't find this expansion by myself
Then:
$$\int \frac{x}{(x-1)^2} dx = \int \frac{1}{(x-1)} + \frac{1}{(x-1)^2}dx = \ln (x-1) + (x-1)^{-1}$$
Then:
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+\frac{x}{(x-1)^2}dx = \frac{x^2}{2} + 2x + 3[\ln(x-1)+(x-1)^{-1}]$$ but wolfram alpha gives another answer. What I did wrong?
|
You can write $\displaystyle\frac{x}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}$; then multiply by $(x-1)^2$ to get $x=A(x-1)+B$.
Letting $x=1$ gives $B=1$, and comparing the coefficients of $x$ gives $A=1$.
(Notice that $\displaystyle\int\frac{1}{(x-1)^2}=-\frac{1}{x-1}+C$.)
You could also find $\displaystyle\int\frac{x}{(x-1)^2}$ by letting $u=x-1, x=u+1, dx=du$ to get
$\hspace{.4 in}\displaystyle\int\frac{u+1}{u^2}du=\int\left(\frac{1}{u}+\frac{1}{u^2}\right) du$,
or use integration by parts with $u=x, du=dx, dv=(x-1)^{-2}dx, v=-(x-1)^{-1}$ to get
$\;\;-x(x-1)^{-1}+\ln|x-1|+C$
|
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|
System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expression and then manipulating, and I also thought of integer solutions(none exist). How do I solve it then?
|
Note that
\begin{align}
f(k)&=(a+k)(b+k)(c+k)(d+k) \\
&=k^4+(a+b+c+d)k^3+(a b+a c+a d+b c+b d+c d) k^2
+(a b c+a b d+a c d+b c d) k+a b cd \\
&=k^4 + k^3 e_1 + k^2 e_2 + k e_3 + e_4.
\end{align}
Now you have a linear system of 4 equations and 4 unknowns ($e_1,e_2,e_3,e_4$).
Solve it and find $f(5)$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int \tan^3\theta d\theta = \int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta$$
$$p = \cos\theta \implies dp = -\sin\theta d\theta$$
$$\int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta = -\int\frac{(1-p^2)(-\sin\theta)}{p^3 }d\theta = -\int \frac{1-p^2}{p^3}dp = -\int \frac{1}{p^3}dp +\int \frac{1}{p}dp = -\frac{p^{-2}}{-2}+\ln|p| = -\frac{(\cos\theta)^{-2}}{-2}+\ln|\cos\theta|$$
$$x = \tan^2\theta \implies \tan\theta= \sqrt{x}\implies \theta = \arctan\sqrt{x}$$
$$= -\frac{(\cos\arctan\sqrt{x})^{-2}}{-2}+\ln|\cos\arctan\sqrt{x}|$$
But the result seems a little bit different than wolfram alpha. I Know there may be easier ways to solve this integral but my question is about this method I choose, specifically.
Is the answer correct? Also, if it is, is there a way to reduce $\cos\arctan$ to something simpler?
|
Another aproach to the integral : substitution $x=t^2$
$$\int\sqrt{\frac{x}{x+1}} \;\mathrm{d}x = \int\frac{2t^2 \mathrm{d}t}{\sqrt{t^2+1}} $$
By per partes :
$$I = \int\frac{t^2 \mathrm{d}t}{\sqrt{t^2+1}} = t\sqrt{t^2+1}-\int\sqrt{t^2+1}\;\mathrm{d}t = t\sqrt{t^2+1}-\int\frac{t^2+1}{\sqrt{t^2+1}}\;\mathrm{d}t $$
Ergo $$I = t\sqrt{t^2+1}-I-\operatorname{arcsinh}{t}$$
So, because original integral was $2I$, solution is therefore :
$$\int\sqrt{\frac{x}{x+1}} \;\mathrm{d}x = \sqrt{x}\sqrt{x+1}-\operatorname{arcsinh}{\sqrt{x}} +C $$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
In which interval is the value of $\sum\limits_{k=1}^\infty \frac{1}{{\left(k + 3\right)}^{2}} $ The question is
But none of these possible answers makes any sense to me.
I know that $\displaystyle\sum_{k=1}^\infty \left( \displaystyle\frac{1}{{\left(k + 3\right)}^{2}} \right)$ is an over estimate of $\displaystyle\int_{1}^\infty \displaystyle\frac{1}{{\left(x + 3\right)}^{2}} \, dx = \displaystyle\frac{1}{4}$
So I think the value should be in the interval of $[1/16,L]$ where $L > 1/4$ but none of these answers reflect this.
|
We can actually find the value of this series:
Note that:
$$\sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{1}{1} + \frac{1}{4} + \frac{1}{9} + \dots + \frac{1}{n^2} = \frac{\pi^2}{6}$$
Now, take a look at the actual sum. Let $S$ be the value of our sum:
$$S = \sum_{k = 1}^{\infty} \frac{1}{(k+3)^2} = \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots$$
Next, if we add to both sides the first original $3$ terms we get:
$$S + \frac{1}{1} + \frac{1}{4} + \frac{1}{9} = \frac{1}{1} + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots = \frac{\pi^2}{6}$$
Now we simply subtract to get the value of $S$:
$$S = \frac{\pi^2}{6} - \left(\frac{1}{1} + \frac{1}{4} + \frac{1}{9}\right) \approx 0.2838$$
Looking at the answers, we see that it is: The second choice
|
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|
From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
|
$$\cos(a+b)=\cos a \cos b -sin a sin b\\cos(x+x)=cos x \cos x -\sin x \sin x\\cos(2x)=cos^2x-sin^2x\\cos(2x)=cos^2x-(1-cos^2x)=2cos^2x-1\\or =(1-\sin^2x)-\sin^2x\\so\\1-cos(2x)=2sin^2x\\$$put there x instead of 2x $$1-cos(x)=2sin^2(\frac{x}{2})\\\frac{1-cos(x)}{sinx}=\frac{2sin^2(\frac{x}{2})}{2sin(\frac{x}{2})cos(\frac{x}{2})}=\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})}=\\tan(\frac{x}{2})$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1144072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving $\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \frac35$ $$\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)$$
Can someone help me to solve it?
result of online calculator: 3/5
|
Hint
$$\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)\left(\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+2}+\sqrt{x-3}}\right)=\frac{\big(\sqrt{x+1}-\sqrt{x-2}\big)\big(\sqrt{x+2}+\sqrt{x-3}\big)}{5}$$
|
{
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|
$\int^{\infty}_{-\infty}u(x,y) \,d y$ independent of x I need to prove that
$$I = \int^{\infty}_{-\infty}u(x,y) \,dy$$
is independent of $x$ and find its value, where
$$u(x,y) = \frac{1}{2\pi}\exp\left(+x^2/2-y^2/2\right)K_0\left(\sqrt{(x-y)^2+(-x^2/2+y^2/2)^2}\right)$$
and $K_0$ is the modified Bessel function of the second kind with order zero.
Evaluating the integral numerically with Mathematica for different values of $x$ gives the result of $2.38$, but I want to know if it is possible to show analytically.
Increasing $x$ results in an increase of the exponential term on the left, but it also then strongly increases the argument of modified Bessel function, thus reducing its value.
To show that integral is independant of $x$, it is sufficient to show that $\int^{\infty}_{-\infty}\frac{\, d}{\, dx}u(x,y) = 0$ but any differentiation looks more and more ugly.
EDIT
Mathematica test:
x = 100
NIntegrate[
(1/(2 Pi))* Exp[x*x/2 - y*y/2] BesselK[0,
Sqrt[(x - y)*(x - y) + (x*x/2 - y*y/2)*(x*x/2 -
y*y/2 )]], {y, -Infinity, x, Infinity}, MaxRecursion -> 22]
This gives an answer of $0.378936$ independent of the choice of $x$. In the earlier calculation I missed the factor $\frac{1}{2\pi}$.
|
Although I'm about 7 years late, here is an answer anyway for anyone interested:
Claim
$$I = \frac{e}{2} \sqrt{\pi} \, \text{erfc} (1)$$
and is thus independent of $x$.
Proof.
By https://dlmf.nist.gov/10.32#E10
$$K_0 (z) = \frac{1}{2} \int_{0}^{\infty} \exp \left(-t-\frac{z^2}{4t}\right) \, \frac{dt}{t}$$
This allows us to write
$$\begin{align}I &= \int_{0}^{\infty} u(x,y) \, dy + \int_{-\infty}^{0} u(x,y) \, dy \\&=\frac{1}{2\pi}\int_0^\infty e^{-v^2/2}\int_0^\infty\frac{e^{xv}e^{-t-v^2[1+(x-v/2)^2]/(4t)}+e^{-xv}e^{-t-v^2[1+(x+v/2)^2]/(4t)}}t\,dt\,dv\\ &=\frac{1}{2\pi}\int_0^\infty\int_0^\infty \frac{\cosh\left(xv\left(\frac{v^2}{4t}+1\right)\right)}{t}\exp\left(-t-\frac{1+2t+x^2}{4t}v^2-\frac{v^4}{16t}\right)\,dv\,dt\end{align}$$
We now enforce the substitution $t = s v^2 \implies dt = v^2 \,ds$:
$$\begin{align}I&=\frac{1}{2\pi} \int_{0}^{\infty} \int_{0}^{\infty}\frac{\cosh\left(x v \left(\frac{1}{4s}+1\right)\right)}{s} \exp\left(-sv^2-\frac{1+2sv^2+x^2}{4s}-\frac{v^2}{16s}\right) \, dv \, ds\\&=\frac{1}{2\pi}\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cosh\left(xv\left(\frac{1}{4s}+1\right)\right)}{s}\exp\left(-\frac{1+x^2}{4s}\right)\exp\left(-v^2 \frac{(1+4s)^2}{16s}\right)\,dv\,ds\end{align}$$
Since
$$\int_{0}^{\infty}\cosh(a v)\exp\left(-v^2 b^2\right) \, dv=\frac{\sqrt{\pi}}{2b}\exp\left(\frac{a^2}{4b^2}\right)$$
$$\implies I = \frac{1}{2\pi} \int_{0}^{\infty} \frac{\exp \left(-\frac{1+x^2}{4s}\right)}{s} \cdot \frac{\sqrt{\pi}}{2\left(\frac{1+4s}{4\sqrt{s}}\right)}\exp \left(\frac{4s x^2 \left(\frac{1}{4s}+1\right)^2}{(1+4s)^2}\right)\, ds$$
This results in the integral being independent as we wanted since the $x^2$ terms cancel.
$$\implies I = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\exp\left(-\frac{1}{4s}\right)}{\sqrt{s} (1+4s)} \, ds\stackrel{s\,\mapsto\frac{1}{s}}{=}\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\exp\left(-\frac{s}{4}\right)}{\sqrt{s} (4+s)} \, ds=\frac{e}{2} \sqrt{\pi} \, \text{erfc} (1)=0.378\cdots$$
$\square$
|
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|
Identity with Harmonic and Catalan numbers Can anyone help me with this.
Prove that
$$2\log \left(\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}\right)=\sum_{n=1}^{\infty}\binom{2n}{n}\left(H_{2n-1}-H_n\right)\frac{x^n}{n}$$
Where $H_n=\sum_{k=1}^{n}\frac{1}{k}$.
The left side is equal to $$2\log(C(x))=2\log\left(\frac{2}{1+\sqrt{1-4x}}\right)$$ where $C(x)=\sum_{n=0}^{\infty}C_n x^n$ is the generating function of the Catalan numbers.
|
Note: Please note that according to the calculation below the factor $(H_{2n-1}-H_n)$ in OPs RHS should be removed.
The following identity is valid
\begin{align*}
\sum_{n=1}^{\infty}\binom{2n}{n}\frac{x^n}{n}=2\log\left(\frac{2}{1+\sqrt{1-4x}}\right)
\end{align*}
The initial point for all calculations is the generating function of the central binomial coefficient
\begin{align*}
\sum_{n= 0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}\tag{1}
\end{align*}
convergent for $|x|<\frac{1}{4}$.
Catalan Numbers:
We can integrate (1) from $0$ to $x$ and obtain
\begin{align*}
\int_{0}^x\sum_{n= 0}^{\infty}\binom{2n}{n}t^ndt&=\int_{0}^x\frac{dt}{\sqrt{1-4t}}\\
&=\frac{1}{4}\int_{1-4x}^{1}\frac{du}{\sqrt{u}}\tag{2}\\
&=\left.\frac{1}{2}\sqrt{u}\right|_{1-4x}^{1}\\
&=\frac{1}{2}\left(1-\sqrt{1-4x}\right)
\end{align*}
Comment:
*
*In (2) we substitute $u=1-4t$ and $du=-4dt$
Now dividing LHS and RHS of (2) by $x$, we get the generating function of the Catalan numbers
\begin{align*}
\sum_{n= 0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}&=\frac{1}{2x}\left(1-\sqrt{1-4x}\right)\\
&=\frac{2}{1+\sqrt{1-4x}}
\end{align*}
$$$$
OP's identity: Just another variation based upon (1)
We again perform integration but we put the first term of the series to the right side, divide both sides by $t$ and we observe:
\begin{align*}
\int_{0}^x\sum_{n=1}^{\infty}\binom{2n}{n}t^{n-1}dt
&=\int_{0}^x\left(\frac{1}{t\sqrt{1-4t}}-\frac{1}{t}\right)dt\\
&=\int_{0}^{x}\frac{1-\sqrt{1-4t}}{t\sqrt{1-4t}}dt\tag{3}\\
&=\int_{1-4x}^{1}\frac{1-\sqrt{u}}{(1-u)\sqrt{u}}du\\
&=\int_{1-4x}^{1}\frac{du}{(1+\sqrt{u})\sqrt{u}}\\
&=2\int_{\sqrt{1-4x}}^{1}\frac{ds}{1+s}\tag{4}\\
&=2\left.\log(1+s)\right|_{\sqrt{1-4x}}^{1}\\
&=2\left(\log 2 - \log \left(1+\sqrt{1-4x}\right)\right)\\
&=2\log\frac{2}{1+\sqrt{1-4x}}
\end{align*}
Comment:
*
*In (3) we substitute $u=1-4t$ and $du=-4dt$
*In (4) we substitute $s=\sqrt{u}$ and $ds=\frac{1}{2\sqrt{u}}du$
We obtain from LHS and RHS of (4) the identity
\begin{align*}
\sum_{n=1}^{\infty}\binom{2n}{n}\frac{x^n}{n}=2\log\left(\frac{2}{1+\sqrt{1-4x}}\right)
\end{align*}
and the claim follows.
Note: OPs identity can be found in Interesting Series involving the Central Binomial Coefficient by D.H. Lehmer. It is also stated in Series with Central Binomial Coefficients, Catalan Numbers, and Harmonic Numbers by K.N. Boyadzhiev with a reference to Lehmer's paper. Both papers contain interesting related identities.
|
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|
Continuity of $f(x,y)=x^3\sin\left(\frac 1x\right)+y^2$ at $(0,0)$ Consider $$f(x,y)= \begin{cases} x^3\sin(1/x) + y^2\quad & x\ne 0 \\ y^2 &\text{otherwise}\end{cases}$$
Prove that $f$ is continuous at $(0,0)$.
I do not have an experience in proving continuity for functions of several variables so any hint or outline of the proof is much appreciated.
|
Here's what we know: For all $x$, $-1 \leq \sin{(\frac{1}{x})} \leq 1$, which means for all $x, y$ with $x > 0$:
$$-x^{3} + y^{2} \leq x^{3} \sin{(\frac{1}{x})} + y^{2} \leq x^{3} + y^{2}$$
Now both of the functions on the left and right are continuous at $(0,0)$, and taking the limit as $(x,y) \to (0,0)$ for each of them gives $0$. But we have:
$$\lim \limits_{\substack{(x,y) \to (0,0) \\ x > 0 }} -x^{3} + y^{2} \leq \lim \limits_{\substack{(x,y) \to (0,0) \\ x > 0 }} x^{3} \sin{(\frac{1}{x})} \leq \lim \limits_{\substack{(x,y) \to (0,0) \\ x > 0 }} x^{3} + y^{2}$$ which implies $\lim \limits_{\substack{(x,y) \to (0,0) \\ x > 0 }} x^{3} \sin{(\frac{1}{x})} + y^{2} = 0$. This limit holds for all possible paths where $x > 0$.
Now, if $x < 0$, then that just means $$-x^{3} + y^{2} \geq x^{3} \sin{(\frac{1}{x})} + y^{2} \geq x^{3} + y^{2}$$ but by the same argument as above, this still proves that under the paths with $x < 0$, we still have $\lim \limits_{\substack{(x,y) \to (0,0) \\ x < 0 }} x^{3} \sin{(\frac{1}{x})} + y^{2} = 0$
|
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|
Let $A$ be a $3$X$3$ matrix whose eigenvalues are $1$, $2$, $3$. Find $\det(B)$ where $B = A^2 + A^T$. Let $A$ be a $3$X$3$ matrix whose eigenvalues are $1$, $2$, $3$. Find $\det(B)$ where $B = A^2 + A^T$.
I know that $\det(A) = 6$, but I cannot proceed after $|A^2 + A^T|$. Any hints as to how to approach the problem?
|
The answer, it seems, is that it depends. If we have $$ A = \left[ \begin{array}{ccc}
1 & 0 & 1 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}
\right]$$
Then we have $\det(A^2 + A^T) = 120$.
On the other hand, if we have $$ A = \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}
\right]$$
Then we have $\det(A^2 + A^T) = 144$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
List the elements of $\mathbb Z_2 \times \mathbb Z_3$ and write its operation table (the notation is additive). $\mathbb Z_2 \times \mathbb Z_3 = \{(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)\}.$
$$
\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2) \\
(0, 1) & (0, 1) & (0, 2) & (0, 3) & (1, 1) & (1, 2) & (1, 3) \\
(0, 2) & (0, 2) & (0, 3) & (0, 4) & (1, 2) & (1, 3) & (1, 4) \\
(1, 0) & (1, 0) & (1, 1) & (1, 2) & (2, 0) & (2, 1) & (2, 3) \\
(1, 1) & (1, 1) & (1, 2) & (1, 3) & (2, 1) & (2, 2) & (2, 3) \\
(1, 2) & (1, 2) & (1, 3) & (1, 4) & (2, 2) & (2, 3) & (2, 4) \\
\end{array}
$$
Please, check my work.
|
As Omne Bonum mentions, adding two members of a group must yield a member of the group. Using the notation given in the question, in $\mathbb Z_2 \times \mathbb Z_3$,
$$(a, b) + (c, d) = ((a+c)\mod 2, \ (b+d)\mod 3)$$
Here is the corrected addition table.
$$\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2) \\
(0, 1) & (0, 1) & (0, 2) & (0, 0) & (1, 1) & (1, 2) & (1, 0) \\
(0, 2) & (0, 2) & (0, 0) & (0, 1) & (1, 2) & (1, 0) & (1, 1) \\
(1, 0) & (1, 0) & (1, 1) & (1, 2) & (0, 0) & (0, 1) & (0, 2) \\
(1, 1) & (1, 1) & (1, 2) & (1, 0) & (0, 1) & (0, 2) & (0, 0) \\
(1, 2) & (1, 2) & (1, 0) & (1, 1) & (0, 2) & (0, 0) & (0, 1) \\
\end{array}$$
|
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|
Minimum value of $\displaystyle \left(x_{1}-x_{2}\right)^2+\left(12-\sqrt{1-x^2_{1}}-\sqrt{4x_{2}}\right)^2.$ The Minimum value of $\displaystyle \left(x_{1}-x_{2}\right)^2+\left(12-\sqrt{1-x^2_{1}}-\sqrt{4x_{2}}\right)^2\;,$ Where $x_{1}\;,x_{2}\in \mathbb{R}$.
$\bf{My\; Trial}::$ Here we have to find Distance between the point $\displaystyle A\left(x_{1}\;,12-\sqrt{1-x^2_{1}}\right)$ and
point $\displaystyle B\left(x_{2}\;,2\sqrt{x_{2}}\right).$
But i did not understand in which the curve these $2$ points lies.
Help me Thanks
|
One point is in the circle with radius 1 and with center at (0,12) and the other point is in a parabola $x= y^2/4$. I used EXCEL and found that the shortest distance and the minimum value is 7.944277 and the points are (0.45,11.06971) on the circle and (4.0,4.0). As the other responder said, you could use calculus to find the minimum.
The shortest distance for any point outside of a circle with coordinates (x,y) is the normal line to the center of the circle.
Thus $D = \sqrt{x^2+(y-12)^2}$ where $y = 2\sqrt{x}$
$\frac{dD}{dx} = \frac{2x + 2(2\sqrt{x} - 12).\frac{1}{\sqrt{x}}}{2(D)} = 0$
$x + 2\sqrt{x} -12 = 0$
$(\sqrt{x}(x+2))^2 = 144$
$x^3+4x^2+4x-144 = 0$
A casual observation will let you kmow x=4 is a root for this cubic function.
The other roots are non-real.
Thus the point is (4,4) in the parabola.
Now the normal from that point will have a slope $=>(m_1m_2) = -1 => \frac{1}{2}m_2 = -1=> m_2 = -2$ where $m_1 => 2y(\frac{dy}{dx}) = 4=> \frac{dy}{dx} = m_1 =\frac{1}{2}$
The normal equation =>$ y-4 = -2(x-4)$
Now substitite the y would get from the circle where the normal would intersect.
$12-\sqrt(1-x^2) -4 = -2x+8 => 8-\sqrt{(1-x^2)} = -2x+8 => 1-x^2 =4x^2=>5x^2 = 1=> x^2 =\frac{1}{5} => x = \sqrt{\frac{1}{5}}$
Thus the other point is $(\sqrt{\frac{1}{5}},11.105572)$
Minimum$ D = \sqrt{(\sqrt{\frac{1}{5}}-4)^2+(11.105572-4)^2}$
$ = \sqrt{(12.622289+50.489153} = 7.944271$ precisely.
|
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|
What is the coefficient of $x^{10}$ in $\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2$? I did (parcially) the following exercise:
There are $10$ identical gift boxes. Each one must be colored with a unique color and there are the colors red, blue, green and yellow. It's possible to color a maximum of $2$ boxes with red, and a maximum of $3$ colors with blue. Write the ordinary generating function associated with the problem and find the number of ways to color 10 boxes.
I've managed to find the generating function:
$$(1+x+x^2)(1+x+x^2+x^3)(1+x+x^2+\dots)^2=\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2\\ \frac{(1-x^3)(1-x^4)}{(1-x)^4}$$
But for calculating the number of colored boxes, I'd have:
$$\frac{(1-x^3)(1-x^4)}{(1-x)^4}=[1-x^3][1-x^4] \left[ \frac{1}{(1-x)} \right]^4$$
The expansion of $\left[\frac{1}{(1-x)}\right]^4$ is:
$$\left[\frac{1}{(1-x)}\right]^4=\sum_{j=0}^\infty {4+j-1 \choose j} y^k$$
But this doesn't seem too revealing. I don't know how to proceeed the counting in this exercise.
|
You're almost there. Expand $(1-x^3)(1-x^4)$ as $1-x^3-x^4+x^7$; when you take the $x^{10}$ coefficient of the product of this with $1/(1-x)^4$, you'll get the four terms with $j=10,10-3,10-4,10-7$ (with appropriate signs):
$$ \binom{13}{10}-\binom{10}{7}-\binom{9}{6}+\binom{6}{3} = 102.$$
|
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|
If $\text{lcm}(a,b,c,d)=a+b+c+d$, then $\gcd(abcd,15)>1$ $\text{LCM}(a,b,c,d)=a+b+c+d$, where $a,b,c,d\in\mathbb Z^+$.
Prove that $abcd$ is divisible by at least one of $3$ and $5$.
I am not sure how to create this proof. Does it involve divisibility tests? How should I set it out?
|
We suppose that $a\ge b\ge c\ge d\ge 1$ and $s=a+b+c+d$ is the lowest common multiple of $a$, $b$, $c$ and $d$.
Now, by contradiction, we suppose that $s$ is not a multiple of $3$ nor $5$.
Note that if $a=b=c=d$ then $4a=s$ (because $s$ is the sum) and $s=a$ (because $s$ is the lcm). This is not possible.
*
*As $a<s<4a$ and $s$ is a multiple of $a$, then $s=2a$ (can't be $3a$) and $b+c+d=a$, so $b<a$
*$a=b+c+d\le 3b$. Hence $2b<s=2a\le 6b$. So $s=4b$ (can't be $3b$, $5b$ or $6b$). $a=2b$ and $c+d=b$, so $c<b$
*$b=c+d\le 2c$. Hence $4c<s=4b\le 8c$. So either $s=7c$ or $s=8c$ (can't be $5c$ or $6c$).
Now two possibilities :
*
*$s=8c=4b=2a$ Hence $d=c$, and $s=lcm(a,b,c,d)=a$. A contradiction
*$s=7c=4b=2a$ Hence, there is a $k$ such that $a=14k$, $b=7k$, $c=4k$ and $d=3k$ (because $s=a+b+c+d=2a=28k$), so $d$ is a multiple of $3$, contradiction.
Hence, $s$ is a multiple of $3$ or $5$ and as $s$ divides $abcd$, so is $abcd$.
Note that it is obviously always a multiple of $2$ also, because if $a$, $b$, $c$, $d$ are all odd, then $s=a+b+c+d$ would be even, a contradiction.
Some examples :
*
*only multiple of $3$ : 4, 3, 3, 2
*only multiple of $5$ : 5, 2, 2, 1
*multiple of $15$ : 15, 10, 3, 2
After that, you can prove with a similar method that there only 9 different (with $a\ge b\ge c\ge d$) solutions with $gcd(a,b,c,d)=1$, and that all other solutions are multiples of one of this nine primitive solution.
$$
\begin{array}{cccc|c}
a&b&c&d&s\\\hline
4& 3& 3& 2 & 12\\
4& 4& 3& 1 & 12\\
5& 2& 2& 1 & 10\\
6& 4& 1& 1 & 12\\
9& 6& 2& 1 & 18\\
10& 5& 4& 1 & 20\\
12& 8& 3& 1 & 24\\
15& 10& 3& 2 & 30\\
21& 14& 6& 1 & 42\\
\end{array}
$$
|
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|
Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0<r< \infty\}$ under the mapping $w=\frac{z}{z-1}$.
Here is what I got so far. First I got the reverse function
$$z= \frac{w}{w-1}=\frac{2u+2iv}{u-1+iv}$$
I did some algebra and I got $x=r \cos t=\frac{u^2 -u+v^2}{(u-1)^2 +v^2}$ and $y=r \sin t = \frac{v}{(u-1)^2 +v^2}$.
For here, I'm not sure how to use the given boundary to find the boundary of the image. I wonder if anyone would give me a hint please.
|
For the sector $\{z=re^{it}\colon 0<r<\infty\quad\text{and}\quad 0\leq t\leq \pi/4\}$, we have in Cartesian coordinates that $y > 0$ and $0<y\leq x$. Then let $z=x+iy$ so
$$
w = \frac{z}{z-1} = \frac{x+iy}{x-1+iy}.
$$
Since $x,y>0$ and $y\leq x$,
$$
(x-1)^2+y^2\leq x^2+y^2
$$
Therefore, we now know $x\geq 1/2$ and $y>0$. or we have that $\Im\{z\} > 0$ and $\Re\{z\}\geq 1/2$. Now you have already determined the real and imaginary parts of $z$ to be
\begin{align}
f(u,v) &= \frac{u^2-u+v^2}{(u-1)^2+v^2}\tag{1}\\
g(u,v) &= \frac{-v}{(u-1)^2+v^2}\tag{2}
\end{align}
From $(2)$, we are able to determine $\Im\{w\} < 0$ since
$$
\frac{-v}{(u-1)^2+v^2} > 0\Rightarrow -v > 0\Rightarrow v < 0
$$
Now, let's look at $\lvert z\rvert^2\geq 1/4$ since $\Re\{z\}\geq 1/2$
$$
0<\frac{1}{4}\leq\frac{(u^2-u+v^2)^2+v^2}{\bigl[(u-1)^2+v^2\bigr]^2}
$$
so $0 < 1/4\leq (u^2-u+v^2)^2+v^2$. Recall that $\Im\{z\} < 0$ so $-v^2<0$ and we get $-v^2 < 1/4\leq (u^2-u+v^2)^2$. Additionally, since $v<0$, $\sqrt{v^2} = -v$ so we get
$$
\frac{1}{2}\leq u^2-u+v^2+iv\Rightarrow \frac{1}{2}\leq \Bigl(u-\frac{1}{2}\Bigr)^2 + \Bigl(v+\frac{i}{2}\Bigr)^2 -\frac{1}{4}+\frac{1}{4}
$$
Therefore, $\frac{1}{\sqrt{2}}\leq\lvert w - 1/2+i/2\rvert$. The domain of $w$ must be
$$
\bigl\{w\in\mathbb{C}\mid\Im\{w\} < 0\quad\text{and}\quad \lvert w - 1/2+i/2\rvert\geq 1/\sqrt{2}\bigr\}
$$
|
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"url": "https://math.stackexchange.com/questions/1159067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Summation of sine by considering the imaginary part of exp(ik*seta)
prove that
$$
\sum\limits_{k=o}^{n} {\sin k\theta} = \frac{\cos{\frac {1} {2}}\theta - \cos(n + {\frac 1 2}\theta)} {2\sin \frac{1} {2}\theta}
$$
I would like to solve this by considering the imaginary part of $\sum {\exp(ik\theta)}$.
This is an alternative method suggested by the paper.
$$
S_n = \sum\limits_{k=0}^n {e^{ik\theta}} = 1+e^{i\theta} + e^i\theta + \cdots +e^{in\theta}
$$
$$
= \frac{1 - e^{in\theta}} {1-e^{i\theta}}
$$
$$
=\frac{1-(\cos n\theta + i\sin n\theta)}{1-(\cos\theta + i\sin \theta)}
$$
Considering only the imaginary parts:
$$
\sum\limits_{k=o}^{n} {\sin k\theta} = \frac {\sin n\theta}{\sin \theta} = \frac {\sin ((n+{\frac 1 2})-\frac {1} {2})\theta}{{2\sin {\frac{1} {2}}\theta cos {\frac{1} {2}}\theta}}
$$
$$
= \frac {\sin (n + {\frac {1} {2}}) \theta \cos {\frac{1} {2}}\theta - \sin {\frac {1} {2}}\theta \cos (n+{\frac{1} {2}})\theta} {2\sin {\frac{1} {2}}\theta cos {\frac{1} {2}}\theta}
$$
Now I have all the parts required + some extra terms that I need to get rid of.
How do i do that?
Many thanks in advance.
|
HINT:
Try using
$$\begin{align}
S_n&=\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\\\\
&=\frac{(1-e^{i(n+1)\theta})(1-e^{-i\theta})}{2(1-\cos(\theta))}
\end{align}$$
The imaginary part is
$$\frac{(\sin(n\theta)+\sin(\theta)-\sin((n+1)\theta)}{2(1-\cos(\theta))}$$
Now, using $1-\cos(\theta)=2\sin^2(\frac{\theta}{2})$, $\sin(\theta)=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$, and the addition angle formulae for the sine and cosine functions show
$$\begin{align}
\text{Im}(S_n)&=\frac{\sin(n\theta)(1-\cos(\theta))+\sin(\theta)(1-\cos(n\theta))}{4\sin^2(\theta/2)}\\\\
&=\frac{\sin(n\theta)\sin(\theta/2)+\cos(\theta/2)(1-\cos(n\theta))}{2\sin(\theta/2)}\\\\
&=\frac{\cos(\theta/2)-\cos((n+1/2)\theta)}{2\sin(\theta/2)}
\end{align}$$
|
{
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"url": "https://math.stackexchange.com/questions/1168691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Geometric proof of $\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$ It is well-known that $$\sin{20^\circ}\sin{40^\circ}\sin{80^\circ}=\frac{\sqrt{3}}{8}$$
It follows that $$\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$$
But how to prove this by geometry?
Thank you.
|
By using Briggs formulas,
$$\sin A \sin B = \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right),$$
$$\sin C \cos D = \frac{1}{2}\left(\sin(C+D)+\sin(C-D)\right),$$
we have:
$$\begin{eqnarray*} \sin A \sin B \sin C &=& \frac{1}{2}\left(\cos(A-B)\sin C-\cos(A+B)\sin C\right)\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)+\sin(A+B-C)-\sin(A+B+C)\right)\end{eqnarray*} $$
and we just need to prove:
$$ -\sin 20^\circ + \sin 100^\circ + \sin 60^\circ -\sin 140^\circ = \sin 60^\circ $$
or:
$$ \sin 40^\circ = \sin 80^\circ- \sin 20^\circ = 2\sin 30^\circ \cos 50^\circ$$
that is trivial since $2\sin 30^\circ=1$ and $\sin 40^\circ=\cos 50^\circ$.
Another proof. The given identity is equivalent to:
$$\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{3\pi}{9}\sin\frac{4\pi}{9}=\frac{3}{16}$$
or to:
$$ \prod_{k=1}^{8}\sin\frac{k\pi}{9} = \frac{9}{256} $$
that follows from the well-known identity:
$$ \prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1171625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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|
Singular value decomposition for matrices that are not square? I understand that the Singular Value Decomposition is defined as SVD = $U\Sigma V^T$ , but I am slightly confused about the calculations when the matrix is not square.
For example, I have the matrix:
$$
\begin{bmatrix}
1 & -1 \\
-2 & 2 \\
2 & -2
\end{bmatrix}
$$
When I am solving for $V$, however, I am missing the last component. Have I done something wrong when calculating for matrices that are not square matrices?
$$\det(A^T A - \lambda I) =
\begin{bmatrix}
2 - \lambda & -4 & 4 \\
-4 & 8 - \lambda & -8\\
4 & -8 & 8 - \lambda
\end{bmatrix}
$$
$\lambda = 0, 2, 16$
Eigenvectors respectively are:
\begin{bmatrix}
1 \\
1/2 \\
0
\end{bmatrix}
\begin{bmatrix}
1 \\
2/7 \\
-2/7
\end{bmatrix}
\begin{bmatrix}
0 \\
1 \\
-1
\end{bmatrix}
Therefore $$\Sigma =
\begin{bmatrix}
\sqrt 2 & 0 & 0 \\
0 & \sqrt 16 & 0 \\
0 & 0 & 0
\end{bmatrix}$$
Also, $$V =
\begin{bmatrix}
7/\sqrt 57 & 0 & 2/\sqrt 5 \\
2/\sqrt 57 & 1/\sqrt 2 & 1/\sqrt 5 \\
2/\sqrt 57 & -1/\sqrt 2 & 0
\end{bmatrix}$$
This is the portion I am confused about.
Is $U = AV / \sqrt\lambda $ ? What if I have am missing a vector so that I can only get the first two columns of $U$?
|
First, correct a simple error
$$
\mathbf{A}^{*}\mathbf{A} =
\left[
\begin{array}{rr}
9 & -9 \\
-9 & 9 \\
\end{array}
\right].
$$
This system matrix $\mathbf{A}$ is a rank one matrix (column 2 = $-$column 1) with singular value decomposition
$$
\begin{align}
\mathbf{A}
&=
\mathbf{U}\, \Sigma \, \mathbf{V}^{*} \\
%
\left[
\begin{array}{rr}
1 & -1 \\
-2 & 2 \\
2 & -2 \\
\end{array}
\right]
&=
% U
\left[
\begin{array}{ccc}
\frac{1}{3}
\color{blue}{\left[
\begin{array}{r}
-1 \\
2 \\
-2
\end{array}
\right]}
&
\frac{1}{\sqrt{5}}
\color{red}{\left[
\begin{array}{r}
-2 \\
0 \\
1
\end{array}
\right]}
&
\frac{1}{3\sqrt{5}}
\color{red}{\left[
\begin{array}{r}
2 \\
5 \\
4
\end{array}
\right]}
%
\end{array}
\right]
% sigma
\left[
\begin{array}{cc}
3 \sqrt{2} & 0 \\
0 & 0 \\
0 & 0 \\
\end{array}
\right]
% V
\frac{1}{\sqrt{2}}
\left[
\begin{array}{rc}
\color{blue}{-1} & \color{blue}{1} \\
\color{red}{1} & \color{red}{1} \\
\end{array}
\right].
\end{align}
$$
Blue vectors are in range spaces, red vectors in null spaces.
The thin SVD uses the range space components only:
$$
\mathbf{A} =
% U
\frac{1}{3}
\color{blue}{\left[
\begin{array}{r}
-1 \\
2 \\
-2
\end{array}
\right]}
% S
\left( 3\sqrt{2} \right)
% V
\frac{1}{\sqrt{2}}
\left[
\begin{array}{rc}
\color{blue}{-1} & \color{blue}{1}
\end{array}
\right].
$$
You may benefit from this example:
SVD and the columns — I did this wrong but it seems that it still works, why?
|
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"question_score": "4",
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|
Is $53$ expressible in this form? It seems as if prime numbers may always be expressed in the form $a\cdot 2^b+c \cdot 3^d$ for some nonnegative integers $b,d$ and $a,c\in \{-1,0,1\}$.
Examples:
$2=1\cdot 2^1+0\cdot 3^d$
$3=0\cdot 2^b+1\cdot 3^1$
$5=1\cdot 2^1+1\cdot 3^1$
$7=1\cdot 2^2+1\cdot 3^1$
$11=1\cdot 2^3+1\cdot 3^1$
$13=1\cdot 2^2+1\cdot 3^2$
$17=1\cdot 2^3+1\cdot 3^2$
$19=1\cdot 2^4+1\cdot 3^1$
$23=1\cdot 2^5+(-1)\cdot 3^2$
$29=1\cdot 2^5+(-1)\cdot 3^1$
$31=1\cdot 2^5+(-1)\cdot 3^0$
$37=1\cdot 2^6+(-1)\cdot 3^3$
$41=1\cdot 2^5+1\cdot 3^2$
$43=1\cdot 2^4+1\cdot 3^3$
$47=1\cdot 2^7+(-1)\cdot 3^4$
We hit a brick wall at $53$. Can anyone confirm if $53$ is/isn't expressible?
What about $n\in \mathbb{N}$ in general? Is it possible to always express $n$ in this form?
Thanks.
EDIT: I invented this question on my own, there are no sources.
Note: The reason I started with prime numbers is because I found it hard to find an expression for $6$, whereas prime numbers continued to be easy to find expressions for (easy until $53$, that is).
|
Let's count the positive integers up to $X$ of the shape $2^b+3^d$. It is easy to see that there are at most a constant times $\log^2 X$ of them. That the same statement is true for integers of the shape $|2^b-3^d|$ is much less obvious, but follows from what are called lower bounds for linear forms in logarithms (for this exact problem, they were first used by Ellison in the 1970s). In particular, the number of positive integers in $[1,X]$ that can be written as $a \cdot 2^b + c \cdot 3^d$ is at most a constant times $\log^2 X$. It follows that, for any dense enough set of integers, most of them will not be of the form $a \cdot 2^b + c \cdot 3^d$. In the case of the primes, there are of order $X/\log X$ of them up to $X$, essentially 100% of which are not of the desired form. A like conclusion is reached if you ask the same question about, for example, squares, cubes, or Carmichael numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1173460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Finding matrix $A$ knowing that $A^2 = B$ Let $B$ be the $3\times3$ matrix
$$
\begin{pmatrix}
1&8&5\\
0&9&5\\
0&0&4
\end{pmatrix}.
$$
How can I find a triangular matrix $A$ with positive diagonal entries such that $A^2 = B$ ?
|
As per my comments, we have
$$\begin{bmatrix}1&x&y\\0&3&z\\0&0&2\end{bmatrix}^2 = \begin{bmatrix}1&8&5\\0&9&5\\0&0&4\end{bmatrix}$$
and wish to find $x, y, z$. This gives
$$
\cases{4x = 8\\3y + xz = 5\\5z = 5}
$$
where we immediately get the solutions $x = 2, z = 1$, which again gives $y = 1$. So we have
$$
A = \begin{bmatrix}1&2&1\\0&3&1\\0&0&2\end{bmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1175327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
What are some rigorous definitions for sine and cosine? Here are some of my ideas:
1. Addition Formula: $\sin{x}$ and $\cos{x}$ are the unique functions satisfying:
*
*$\sin(x + y) = \sin x \cos y + \cos x \sin y $
*$\cos(x + y) = \cos x \cos y - \sin x \sin y$
*$\sin 0 = 0\quad$ and $\quad\displaystyle{\lim_{x \rightarrow 0} \frac{\sin x }{x} = 1}$
*$\cos 0 = 1\quad$ and $\quad\displaystyle{\lim_{x \rightarrow 0} \frac{1-\cos x}{x} = 0}$
2. Taylor Series:
*
*$\displaystyle{\sin x = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!}\;x^{2n+1}}$
*$\displaystyle{\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\;x^{2n}}$
3. Differential Equations: $\sin(x)$ and $\cos(x)$ are the unique solutions to $y'' = -y$, where $\sin(0) = \cos^\prime(0) = 0$ and $\sin^\prime(0) = \cos(0) = 1$.
4. Inverse Formula: We have:
$$\begin{align}
\arcsin x &= \phantom{\frac{\pi}{2} + } \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt \\[6pt]
\arccos x &= \frac{\pi}{2} - \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt
\end{align}$$
Then $\sin x$ is the inverse of $\arcsin x$, extended appropriately to the real line, and $\cos x$ is similar.
Question: Are there any others that you like? In particular, are there any good rigorous ones coming from the original geometric definition?
|
My preferred definition is:
$\cos x$ and $\sin x$ are the real and imaginary parts of the
exponential function $\exp(ix)$.
Since we have:
$$
\begin{split}
e^{ix}= \sum_{k=0}^\infty\dfrac{(ix)^k}{k!}&=1+ix+\dfrac{(ix)^2}{2!}+\dfrac{(ix)^3}{3!}+\cdots+\dfrac{(ix)^n}{n!}+\cdots\\
&=1-\dfrac{(x)^2}{2!}+\dfrac{(x)^4}{4!}-\dfrac{(x)^6}{6!}+ \cdots +i\left[ x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+\cdots \right]\\
&=
\sum_{k=0}^\infty\dfrac{(-1)^k\,x^{2k}}{(2k)!}+i \sum_{k=0}^\infty\dfrac{(-1)^k\,x^{2k+1}}{(2k+1)!}\end{split}
$$
we find the usual series definitions.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1176098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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|
Tough inequality in positive reals numbers. Let $a, b, c$ be positive real. prove that
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
Thanks
|
This inequality seems to be somehow too weak, so I'll prove the much stronger inequality:
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq 2\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
$$1 + \frac ab + \frac ac \ge 3\sqrt[3]\frac{a^2}{bc} = \frac{3a}{\sqrt[3]{abc}}$$
$$1 + \frac ba + \frac bc \ge 3\sqrt[3]\frac{b^2}{ac} = \frac{3b}{\sqrt[3]{abc}}$$
$$1 + \frac ca + \frac cb \ge 3\sqrt[3]\frac{c^2}{ab} = \frac{3c}{\sqrt[3]{abc}}$$
Sum all the inequalities and we have:
$$3 + \frac ac + \frac ab + \frac ba + \frac bc + \frac ca + \frac cb \ge 3\frac{a+b+c}{\sqrt[3]{abc}}$$
$$1 + \frac ac + \frac ab + \frac ba + \frac bc + \frac ca + \frac cb + 1 \ge 2\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right) + \left(\frac{a+b+c}{\sqrt[3]{abc}} - 3\right)$$
The last term is obviously positive, since $a+b+c \ge 3\sqrt[3]{abc}$ from AM-GM, hence the proof.
We have equality when $a=b=c$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give me some hints? Thanks in advance!
|
First, let us cube the number at hand:
$$\begin{align*}
x^3
&=7+\sqrt{50}+7-5\sqrt{2}+3(7+\sqrt{50})^\frac{2}{3}(7-5\sqrt{2})^\frac{1}{3}+3(7+\sqrt{50})^\frac{1}{3}(7-5\sqrt{2})^\frac{2}{3}\\
&=14+3((49-50)(7+\sqrt{50}))^\frac{1}{3}+3((49-50)(7-5\sqrt{2}))^\frac{1}{3}\\
&=14-3x
\end{align*}$$
Then, $x^3=14-3x \implies x^3+3x-14=0$.
The only real solution of this equation is $x=2$, which is a rational number.
QED.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integral fraction of polynomials I have this problem:
$$\int \frac{-2x^2+6x+8}{x^4-4x+3}dx$$
I have tried using partial fractions, but I can't get solution. Thank you for any advice.
|
$x=1$ is a double root of $x^4-4x+3$ (it is also a root of the derivative), hence it is divisible by $(x-1)^2$, hence you'll get $x^4-4x+3=(x-1)^2(x^2+ax+b)$, and partial fractions decomposition has the form:
$$\frac{-2x^2+6x+8}{x^4-4x+3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac{Cx+D}{x^2+ax+b}.$$
You'll obtain $A\ln\lvert x-1\rvert-\dfrac B{x-1} +C'\ln \lvert x^2+ax+b\rvert+D'\arctan(\text{something})$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1182910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Erroneously Finding the Lagrange Error Bound Consider $f(x) = \sin(5x + \pi/4)$ and let $P(x)$ be the third-degree Taylor polynomial for $f$ about $0$. I am asked to find the Lagrange error bound to show that $|(f(1/10) - P(1/10))| < 1/100$. Because $P(x)$ is a third-degree polynomial, I know the difference is in the fourth degree term. So I found the fourth derivative to be $f(x) = 625 \sin(\pi/4 + 5x)$. Then I substituted $1/10$ into the fourth derivative to find $M$. I substituted the $M$ I found and $x$ as $1/10$ into the formula for a Lagrange remainder: $(M(1/10)^4) / 4!$ This calculates to be $0.002498$. The answer is that the Lagrange error is $1 / 384$. Could someone please help me understand where I am going wrong?
|
We have:
$$f(x) = \sin\left(5x + \dfrac{\pi}{4}\right)$$
The third degree Taylor Polynomial is given by:
$$T_3(x) = -\frac{125 x^3}{6 \sqrt{2}}-\frac{25 x^2}{2 \sqrt{2}}+\frac{5 x}{\sqrt{2}}+\frac{1}{\sqrt{2}}$$
The error term is given by:
$$R_{n+1} = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \le \dfrac{M}{(n+1)!}(x-a)^{n+1}$$
We have $n= 3$, thus:
$$\dfrac{d^4}{dx^4} \left( \sin\left(5x + \dfrac{\pi}{4}\right)\right) = 625 \sin \left(5 x+\frac{\pi }{4}\right)$$
We have $a = 0, n= 3, x = \dfrac{1}{10}$ and the max of sine is $1$, so this yields:
$$R_{n+1} \le \dfrac{M}{(n+1)!}(x-a)^{n+1}= \dfrac{625}{4!} \left(\dfrac{1}{10}\right)^4 = \dfrac{1}{384}$$
|
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|
Solve $\arg(-1/z)=-2\pi/3$ and $|1-\frac2z|=1$ $\arg(\frac{-1}z)=\frac{-2\pi}3$
what does this mean? how to i get the $\arg(z)$ from this.
I'm thinking of reciprocals.
$\left|1-\frac2z\right|=1$ how do i solve for this as well.. i'm confused when i negative sign appears
|
Set
$z = r(\cos \theta + i \sin \theta); \tag{1}$
then $\arg(z) = \theta$, and
$z^{-1} = r^{-1}(\cos \theta - i \sin \theta), \tag{2}$
as may easily be checked by a simple complex multiplication:
$(r(\cos \theta + i \sin \theta))(r^{-1}(\cos \theta - i \sin \theta)) = rr^{-1}(\cos^2 \theta + \sin^2 \theta) = rr^{-1}(1) 1. \tag{3}$
It follows that
$-z^{-1} = -r^{-1}\cos \theta + r^{-1} i\sin \theta, \tag{4}$
whence
$\tan (-\dfrac{2\pi}{3}) = \tan (\arg(-z^{-1})) = \dfrac{\sin \theta}{-\cos \theta} = \dfrac{\sin (-\theta)}{\cos (-\theta)} = \tan (-\theta) \tag{5}$
or
$\theta = \dfrac{2\pi}{3}. \tag{6}$
The modulus or magnitude $r = \vert z \vert$ of $z$ varies independently of $\arg(z) = \theta$; thus we may take any
$z = r(\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}) \tag{7}$
with $0 < r < \infty$ as a solution in this case. And not altogether incidentally, we have
$\cos \dfrac{2\pi}{3} = -\dfrac{1}{2}; \;\; \sin \theta = \dfrac{\sqrt{3}}{2}, \tag{8}$
whence we may write
$z = r(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}); \tag{9}$
the set of all $z \in \Bbb C$ such that
$\arg(z^{-1}) = -\dfrac{2\pi}{3} \tag{10}$
is precisely the ray emanating from, but not including, the origin in the direction $e^{2\pi i/3}$.
As for the other problem,
$\vert 1 - \dfrac{2}{z} \vert = 1, \tag{11}$
both my approach and answer differ from those of incognito. Looking at equation (11) geometrically, we ask what it tells us, and we see that it says that $2/z$ differs from $1$ by a complex number of unit modulus; that is, we must have
$\dfrac{2}{z} = 1 - e^{i\theta}; \tag{12}$
it is easily seen that $z$ satisfying (12) also obeys (11), provided we forbid the value $\theta = 0$; (12) yields
$z = \dfrac{2}{1 - e^{i\theta}}, \tag{13}$
in which we allow $\theta \in (0, 2\pi)$. We can in fact write such $z$ in "$x + iy$" form:
$z = \dfrac{2(1 - e^{-i\theta})}{(1 - e^{-i\theta})(1 - e^{i\theta})} = \dfrac{1 - e^{-i\theta}}{1 - \cos \theta} = \dfrac{1 - \cos \theta + i \sin \theta}{1 - \cos \theta} = 1 + i\dfrac{\sin \theta}{1 - \cos \theta}; \tag {14}$
we note from (14) that the real part of $z$ is the constant $1$; as for the imaginary part, $\sin \theta / (1 - \cos \theta)$ takes on every real value precisely once as $0 \to \theta \to 2\pi$; to see this, obverve that the derivative of
$y(\theta) = \dfrac{\sin \theta}{1 - \cos \theta} \tag{15}$
is
$y'(\theta) = \dfrac{\cos \theta (1 - \cos \theta) - \sin^2 \theta}{(1 - \cos \theta)^2} = \dfrac{\cos \theta - 1}{(1 - \cos \theta)^2} = \dfrac{1}{\cos \theta - 1 }< 0 \tag{16}$
since $\cos \theta < 1$ for $\theta \in (0, 2\pi)$; $y(\theta)$ is monotonically decreasing, hence takes on any value at most once; to see that $y(\theta)$ takes on every real value at least once, we note that
$\lim_{\theta \to 0^+} y(\theta) = \lim_{\theta \to 0^+} \dfrac{\cos \theta}{\sin \theta} = \infty \tag{17}$
by a simple application of L'Hopital's rule, which may similarly be used to show that
$\lim_{\theta \to 2\pi^{-}} y(\theta) = -\infty; \tag{18}$
thus the range of $y(\theta)$ is indeed all of $\Bbb R$. As the differential topologists might put it, $y(\theta)$ is a diffeomorphism 'twixt $(0, 2\pi)$ and $\Bbb R$. But the real point, for the present purposes, is that the solutions of (11) are in fact precisely the set of complex numbers $z$ with $\Re(z) = 1$.
We check that $z = 1 + iy$ solves (11); we have:
$z^{-1} = \dfrac{1 - iy}{1 + y^2}; \tag{19}$
$\dfrac{2}{z} = \dfrac{2 - 2iy}{1 + y^2}; \tag{20}$
$1 - \dfrac{2}{z} = 1 - \dfrac{2 - 2iy}{1 + y^2} = \dfrac{1 + y^2 - 2 + 2iy}{1 + y^2} = \dfrac{(y + i)^2}{(y + i)(y - i)} = \dfrac{y + i}{y - i}; \tag{21}$
finally,
$\vert 1 - \dfrac{2}{z} \vert = \dfrac{\vert y + i \vert}{\vert y - i \vert} = 1 \tag{22}$
since $y - i = \overline{y + i}$. Things check out.
This answer differs from incognito's solution to (11). Apparently, he (and here I use the masculine form of the pronoun since the relevance of gender distinction seems to disappear when one anonymously adopts a name like incognito!) assumed both (11) and (10) were meant to be solved concurrently, wheras I treated them as separate issues. This makes each of us, in our own way, correct. But it is true that there are more solutions to (11) than $x = 1$, $y = -\sqrt{3}$. Indeed, $y$ is not determined by the relation $\vert z - 2 \vert = \vert z \vert$, as his own calculations reveal. Indeed, his work on this question may be seen as an elegant indication that $1 + iy$ gives all $z$ satisfying (11).
|
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|
A conjecture about quadratic residues given $p \equiv 5 \pmod 8$ Original Problem
$p$ is a prime that is congruent to $5$ modulo $8$ and $a$ is a quadratic residue modulo $p$.
Prove that excactly one of $x_1=a^{\frac{p+3}{8}},x_2=(2a)(4a)^{\frac{p-5}{8}}$ is the solution to the congruence $x^2 \equiv a \pmod p$.
What I have already proved
$x_1^2 \equiv a \pmod p$ if $a^{\frac{p-1}{4}} \equiv 1 \pmod p$.
$x_2^2 \equiv a \pmod p$ if $a^{\frac{p-1}{4}} \equiv -1 \pmod p$.
Corollary
Assume that $a \equiv r \pmod p,\quad (1\leq r\leq p-1)$. Then $a^{\frac{p-1}{4}} \equiv r^{\frac{p-1}{4}} \pmod p$, and hence
(a). $x_1^2 \equiv a \pmod p$ if $r^{\frac{p-1}{4}} \equiv 1 \pmod p$.
(b). $x_2^2 \equiv a \pmod p$ if $r^{\frac{p-1}{4}} \equiv -1 \pmod p$.
Examining the cases where $p=5$ and $p=13$, I came up with a conjecture.
Conjecture
(I). $r^{\frac{p-1}{4}} \equiv 1 \pmod p$ if $r$ is odd.
(II). $r^{\frac{p-1}{4}} \equiv -1 \pmod p$ if $r$ is even.
Remark
Notice that $b^2 \equiv a \equiv r \pmod p$ for some nonzero integer $b$.
By Euler's Criterion, we have that $r^{\frac{p-1}{4}} \equiv (b^2)^{\frac{p-1}{4}} \equiv b^{\frac{p-1}{2}} \equiv \genfrac(){}{0}{b}{p} \pmod p$.
Equivalent form of the conjecture
(III). $b$ is a quadratic residue modulo $p$ if $r$ is odd.
(IV). $b$ is a quadratic nonresidue modulo $p$ if $r$ is even.
Can you prove or disprove (III) and (IV) using only the law of quadratic reciprocity?
|
Let $p=8k+5$. By Euler's Criterion, $a^{4k+2}\equiv 1\pmod{p}$ and therefore $a^{2k+1}\equiv \pm 1\pmod{p}$.
Suppose that $a^{2k+1}\equiv 1\pmod{p}$. Then $a^{2k+2}\equiv a\pmod{p}$, and $a$ is congruent modulo $p$ to the square of $a^{k+1}$.
Now suppose that $a^{2k+1}\equiv -1\pmod{p}$. Since $p$ is of the form $8k+5$, it follows that $2$ is a quadratic non-residue of $p$, and therefore $2^{4k+2}\equiv -1\pmod{p}$. It follows that $2^{4k+2}a^{2k+1}\equiv 1\pmod{p}$, and therefore
$$2^{4k+2}a^{2k+2}\equiv a\pmod{p}.$$
Thus the square of $2^{2k+1}a^{k+1}$, that is, of $(2a)(4a)^k$, is congruent to $a$ modulo $p$.
|
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algorithm for positive integer solutions of equation $a^3+b^3=22c^3$ This is a look-a-like to Fermat's last theorem for $n=3$, but it has solutions! I believe that its solution requires knowledge of the techniques of algebraic or analytic number theory which I don't have.
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Might have found something to generate more from an initial solution.
Suppose $a^3+b^3=22c^3$. If want the smallest such $a$ and $b$, they have to be relatively prime to each other and to $c$. $a^3+b^3=(a+b)(a^2+b^2-ab)$. Via trial and error, $a^2+b^2-ab$ cannot be divisible by 11 unless both $a$ and $b$ are, which is not allowed. Combined, this tells us $a+b$ is divisible by $22$.
After substitution: $a^3+(22k-a)^3=22c^3$. After expanding, cancelling the term cubic in $a$ and finding the discriminant of the resulting quadratic, one finds $c^3=11^2k^3+3n^2k=k(11^2k^2+3n^2)$ where $a=11k+n$ and $b=11k-n$. So $k$ and $n$ must be relatively prime in order for $a$ and $b$ to be relatively prime. This leads us to consider two scenarios.
If $k$ is divisible by 3, it can be shown that it must also be divisible by 9. Then $c$ must also be divisible by 3. If $c=3p$ and $k=9q$, then $p^3=11^2\cdot27\cdot q^3+qn^2=q[3\cdot(33)^2q^2+n^2]$. Since $q$ and $n$ are relatively prime then q and the term in square brackets must also be relatively prime. If the product of two relatively prime numbers is a perfect cube, then the factors themselves must be perfect cubes.
This suggests there is a $x$ so that $x^3=3\cdot(33)^2q^2+n^2$.
The form of the equation remains largely unchanged upon multiplication by 27.
$27x^3=81\cdot(33)^2q^2+27n^2\implies (3x)^3=(33)^2(9q)^2+3(3\cdot n)^2$ Again we have an expression of the form a perfect cube is the sum of three times a perfect square plus some other perfect square. A similar situation occurs if $k$ is not taken to be divisible by 3.
By arguments above we also have that a perfect cube has that form if we don't take that 3 divides $k$.
So possible solutions have the form $x^3 = 3\cdot(33)^2r^6+n^2$ where $q=r^3$ or $x^3=11^2\cdot r^6+3n^2$ where $k=r^3$.
We have an example of the first where $r=6$ and $n=4085$. This leads to $a=17299, b=25579, c=9954$, a solution mentioned elsewhere.
|
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|
Can anyone tell me if this is correct? Suppose that the temperature of a metal plate is given by $T(x; y) = x^2 +2x+y^2$, for points $(x, y)$ on the elliptical plate defined by $x^2 + 4y^2 <= 24$.
Find the maximum and minimum temperatures on the plate.
This is what i have done so far.
Finding critical point:
$T(x)=2x+2$, $T(y)=2y$. Equating to $0$, $x=-1$, $y=0$.
Critical point is $(-1,0)$ and is a minimum.
On the boundary, $ x^2 + 4y^2 = 24$
$g(x,y)=x^2 + 4y^2$
$g(x)=2x, g(y)=8y$
$2x+2=A2x$---------(1)
$2y =A8y$---------(2)
$x^2+4y^2=24$------(3)
When solving from equation 1 and 3 im getting $ x=-1,y=|(23/4)^{0.5}|,$ and $x=|24^{0.5}|, y=0$ and when from eqn 2 and 3 im getting $x=|24^{0.5}|,y=0, A= 0.25 , x=-4/3, y=|(50/9)^{0.5}|$.
Is this correct? am getting different values when using equation $1$ and $2$.
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hint:Equation $(2)$ gives $A = 1/4$, or $y = 0$. If $A = 1/4 \to 2x+2 = \dfrac{x}{2}\to x = -4/3$.
If $A = 2, (2) \to y = 0, (1) \to x = 1 \to x^2+4y^2 = 1^2 + 4\cdot 0^2 = 1 \neq 24 \to A \neq 2$.
|
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|
Evaluating $\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right)$ $$\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) = \;?$$
I have been trying to see if it can be written as sum of two telescope terms but it looks tricky. Any help ?
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Hint: The sum is a rearrangement of the terms of the alternating harmonic series.
Can you find a series to compare term-wise to? Note that $-\frac{1}{2n} = - \frac{1}{4n} - \frac{1}{4n}$.
Since everyone is very hectic in the comments, let me be clear here:
The sum converges. The rearrangement itself is not enough to prove this, of course, since the harmonic series converges conditionally. However, my hint that it is a rearrangement was to hint at the possibility that we can use a comparison to prove it.
Full solution
$
\sum \left( \frac{1}{4n-3} + \frac{1}{4n-1} - \frac{1}{4n} - \frac{1}{4n} \right) = \sum \left(\frac{1}{4n-3} - \frac{1}{4n-2} + \frac{1}{4n-1} - \frac{1}{4n}\right) + \sum \left( \frac{1}{4n-2} - \frac{1}{4n}\right).$
$= \ln 2 + \frac{1}{2} \sum \left( \frac{1}{2n-1} - \frac{1}{2n} \right) = \frac{3}{2} \ln 2.$
|
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critical numbers of a complex function $ f(x) = 2 x - 5 x^{\frac{2}{5}} $ Critical numbers of $ f(x) = 2 x - 5 x^{2/5} $ are 1,$0$
How come $0$ is also a critical number?
I found the critical number 1 this way:
$f'(x) = 2 - 5 \ \frac {2}{5} \ x^{-\frac{3}{5}}$
$ x^{\frac{3}{5}} = 1$
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Critical values of a function occur where the derivative is zero or where the derivative is undefined.
Since $f'(x)=2 - 2x^{\frac{-3}{5}} = 2 - \frac{2}{x^{\frac{3}{5}}} = 2 - \frac{2}{\sqrt[5]{x^3}}$
Hence, $f'(0)$ is undefined since we are dividing by zero.
Also, $f'(1) =0,$ therefore the critical values are 0 and 1.
|
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Why is quadratic integer ring defined in that way? Quadratic integer ring $\mathcal{O}$ is defined by
\begin{equation}
\mathcal{O}=\begin{cases}
\mathbb{Z}[\sqrt{D}] & \text{if}\ D\equiv2,3\ \pmod 4\\
\mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]\ & \text{if}\ D\equiv1\pmod 4
\end{cases}
\end{equation}
where $D$ is square-free.
I understand $\mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]$ is not closed under multiplication if $D\equiv2,3\pmod 4$. But still, isn't it more natural to define $\mathcal{O}=\mathbb{Z}[\sqrt{D}]$ for all square-free $D$? (In that case, it really seems like 'quadratic integer'.) I wonder what is the motivation of this definition.
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Have you studied norms yet? If $x$ is an algebraic number in a given quadratic domain such that the leading coefficient of its minimal polynomial is 1, then its norm in that domain is an integer from $\mathbb{Z}$ and $x$ is called an algebraic integer.
Consider these two numbers: $$\frac{1}{2} + \frac{\sqrt{-7}}{2}$$ and $$\frac{1}{2} + \frac{\sqrt{7}}{2}.$$ The former is an algebraic number in $\mathbb{Q}(\sqrt{-7})$, the latter is an algebraic number in $\mathbb{Q}(\sqrt{7})$. Both are algebraic numbers, but only one of them is an algebraic integer. Observe that $$N\left(\frac{1}{2} + \frac{\sqrt{-7}}{2}\right) = \left(\frac{1}{2} - \frac{\sqrt{-7}}{2}\right)\left( \frac{1}{2} + \frac{\sqrt{-7}}{2}\right) = 2$$ but $$N\left(\frac{1}{2} + \frac{\sqrt{7}}{2}\right) = \left(\frac{1}{2} - \frac{\sqrt{7}}{2}\right)\left( \frac{1}{2} + \frac{\sqrt{7}}{2}\right) = -\frac{3}{2}.$$
We should have seen that coming because $-7 \equiv 1 \pmod 4$ but $7 \equiv 3 \pmod 4$. If $D \not\equiv 1 \pmod 4$, then only numbers of the form $a + b\sqrt{D}$ with $a, b \in \mathbb{Z}$ have $N(a + b\sqrt{D}) \in \mathbb{Z}$. But if $D \equiv 1 \pmod 4$, then it's also possible for numbers of the form $\frac{a}{2} + \frac{b\sqrt{D}}{2}$ to have integer norm, provided $a$ and $b$ have the same parity.
No number of another form in $\mathbb{Q}(\sqrt{D})$ can have integer norm. For example, $$N\left(2 + \frac{\sqrt{-7}}{2}\right) = \left(2 - \frac{\sqrt{-7}}{2}\right)\left( 2 + \frac{\sqrt{-7}}{2}\right) = \frac{23}{4}.$$ I've seen short but dense proofs of this in a couple of different books.
|
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Prove that the series $\sum_{1}^{\infty}\frac{k}{(k+1)(k+2)(k+3)}$ converges and find its limit I try to split the summand into differences, but that seems to be a futile way in our case right here, because the numerator is $k$, instead of a given number.
A closely-related series, say $\sum_{1}^{\infty}\frac{2}{(k+1)(k+2)(k+3)}$, can however be directly tackled by writing its partial sum as
$$\sum_{1}^{n}\frac{1}{(k+1)(k+2)} - \frac{1}{(k+2)(k+3)}.$$
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$$\begin{align}\sum \limits_{k=1}^{\infty}\frac k{(k+1)(k+2)(k+3)}&= \sum \limits_{k=1}^{\infty}\dfrac{2}{k+2} - \dfrac{3}{2(k+3)} - \dfrac{1}{2(k+1)} \\~\\&=\frac{3}{2}\left(\sum \limits_{k=1}^{\infty}\dfrac{1}{k+2} -\dfrac{1}{k+3}\right) + \frac{1}{2}\left(\sum \limits_{k=1}^{\infty}\dfrac{1}{k+2}-\frac{1}{k+1}\right) \\~\\ &= \frac{3}{2}\left(\frac{1}{1+2}\right) + \frac{1}{2}\left(-\frac{1}{1+1}\right) \\~\\ &=\frac{1}{4}\end{align}$$
|
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Unexplicit sum evaluation (Putnam)
For positive integers $n,$ let the numbers $c(n)$ be determined by the rules $c(1)=1,c(2n)=c(n),$ and $c(2n+1)=(-1)^nc(n).$ Find the value of $$S = \sum_{n=1}^{2013}c(n)c(n+2).$$
Let $S_k$ represent the $k$th partial sum.
$$S_1 = c(1)c(3)$$
Since, $c(2n+1) = (-1)^nc(n)$ it follows: $c(3) = (-1)^1c(1) = -1$
$a_2 = c(2)c(4)$. $c(2*1) = 1$ and $c(2*2) = 1$ hence, $a_2 = 1$
$$\implies S_2 = a_1 + a_2 = 0$$
$$a_3 = c(3)c(5) = (-1)(1) = -1$$
$$\implies S_3 = -1$$
But the $1, -1, 1, -1, ....$ pattern doesnt work because of $S_2 = 0$.
$$S = -1 + \sum_{n=2}^{2013} c(n)c(n+2) = -1 + c(2013)c(2015) + \sum_{n=2}^{2012} c(n)c(n+2)$$
But apart from that I don't really have an idea.
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Split into even and odd terms.
$$\begin{align} S &= \sum_{n=1}^{1006} c(2 n) c(2 n+2) + \sum_{n=1}^{1006} c(2 n+1) c(2 n+3) + c(1) c(3) \\ &= \sum_{n=1}^{1006} c(n) c(n+1) + \sum_{n=1}^{1006} (-1)^n c(n) (-1)^{n+1} c(n+1) - 1 \\ &= \sum_{n=1}^{1006} c(n) c(n+1) - \sum_{n=1}^{1006} c(n) c(n+1) - 1 \\ &= -1\end{align} $$
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$\frac{1}{z^2}$ is holomorphic
I have to show that $z\mapsto\frac1{z^2}$ is holomorpic on $\mathbb C\setminus\{0\}$ and compute its $n$-th derivative
I know that $\frac{1}{z^2}=\sum\limits_{n\ge0}(-1)^n(n+1)(z-1)^n$, so it has a power series representation. Is that sufficient to conclude that the function is analytic, which is stronger than holomorphic. I thought analytic means inifinitely differentiable and holomorphic only continuously differentiable, how can one then compute $n$-th derivative of a holomorphic function, if $n$ is greater than $1$? However I computed it;
$\left(\frac{1}{z^2}\right)^{(n)}=\sum\limits_{k\ge n}(-1)^k(n+1)n!(z-1)^{k-n}$ is that correct ?
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Given$$f(z) = {z^{ - 2}}$$
then, for$$z = x + iy$$
we may write$$f(x + iy) = \frac{{{x^2} - {y^2}}}{{{{({x^2} + {y^2})}^2}}} + i\frac{{ - 2xy}}{{{{({x^2} + {y^2})}^2}}}$$
For$$\begin{gathered}
u(x,y) = \frac{{{x^2} - {y^2}}}{{{{({x^2} + {y^2})}^2}}} \hfill \\
v(x,y) = \frac{{ - 2xy}}{{{{({x^2} + {y^2})}^2}}} \hfill \\
\end{gathered}$$
Cauchy-Riemann Equations are satisfied.
For n-th order derivative, we get for $n \in \mathbb{N}$:
$$\frac{{{d^n}f}}{{d{z^n}}}(z) = \left\{ {\begin{array}{*{20}{c}}
{{{( - 1)}^{2n - 1}}(2n)! \cdot z \cdot f{{(z)}^{2n - 1}}} \\
{{{( - 1)}^{2n}}(2n + 1)!f{{(z)}^{2n - 1}}}
\end{array}} \right.$$
One gets this by carefully calculating:
$$\begin{gathered}
\frac{{df}}{{d{z^1}}}(z) = {( - 1)^1} \cdot 2! \cdot z \cdot f{(z)^2} \hfill \\
\frac{{{d^3}f}}{{d{z^3}}}(z) = {( - 1)^3} \cdot 4! \cdot z \cdot f{(z)^3} \hfill \\
\frac{{{d^5}f}}{{d{z^5}}}(z) = {( - 1)^5} \cdot 6! \cdot z \cdot f{(z)^4} \hfill \\
\vdots \hfill \\
\hfill \\
\frac{{{d^2}f}}{{d{z^2}}}(z) = 3! \cdot f{(z)^2} \hfill \\
\frac{{{d^4}f}}{{d{z^4}}}(z) = 5! \cdot f{(z)^3} \hfill \\
\frac{{{d^6}f}}{{d{z^6}}}(z) = 7! \cdot f{(z)^4} \hfill \\
\vdots \hfill \\
\end{gathered}$$
|
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Number theory: $a\in A\iff \frac{1}{2}-a\in A$. Let $A$ be the set of all $a\in \mathbb{Q}$ for which there exist $x,y,z\in \mathbb{Z}$ not all $=0$ such that $$a=\frac{xy+yz+zx}{x^2+y^2+z^2}.$$ Prove that $$a\in A\iff \frac{1}{2}-a\in A.$$
Note: This is a special case of a more general result I discovered while investigating Diophantines of the form $a(x^2+y^2+z^2)=b(xy+yz+zx)$ and already have a solution using methods I invented for that purpose, but this is indirect/unmotivated. I am looking for a more direct solution.
It might help to know the "more general result:"
Proposition. If $b\in A$ then $a\in A\iff \frac{b-a}{1+a-2ab}\in A.$
However, none of these look as nice as the case where $b=\frac{1}{2}=\frac{1\cdot 1+1\cdot 4+4\cdot 1}{1^2+1^2+4^2}$, which gives rise to the above problem: $a\in A\iff \frac{\frac{1}{2}-a}{1+a-a}=\frac{1}{2}-a\in A.$
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Let
$$
a = \frac{xy+yz+zx}{x^2+y^2+z^2}
$$
as above, note that this is equivalent to
$$
1+2a = \frac{(x+y+z)^2}{x^2+y^2+z^2}
$$ and let $a'=\frac12-a$.
If $x=y=z$ then $a=1$ and
$$
a' = -\frac12 = \frac{-1\cdot 2 -1\cdot 2 + 1}{(-1)^2+(-1)^2+2^2} \in A
$$
Otherwise, if $x,y,z$ are not all equal then
$$
1+2a' = 2-2a = \frac{u+v+w}{x^2+y^2+z^2} = \frac{(u+v+w)^2}{u^2+v^2+w^2}
$$
where
$$
u = x^2+y^2-z(x+y) \\
v = y^2+z^2-x(y+z) \\
w = z^2+x^2-y(z+x) \\
u+v+w = (x-y)^2+(y-z)^2+(z-x)^2>0
$$
clearly $u,v,w$ are not all zero and hence
$$
a' = \frac{uv+vw+wu}{u^2+v^2+w^2} \in A
$$
Thus $a\in A \implies \frac12-a \in A$, and the other direction follows immediately because $a'\in A \implies \frac12-a' = a \in A$.
|
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Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pick an odd from the 1st sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 2nd sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 3rd sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 4th sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 5th sample space is $\dfrac{1}{2}$
The final result is:
$\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ = $\dfrac{3}{5}$
Is this reasoning correct? Are there any simpler ways to solve this problem?
|
There's another (exhaustive) way of seeing it, and it's counting.
Let us represent the act of picking a number $a$ and then a number $b$ by $(a,b)$, we then have $5\times 4=20$ of this possible "actions". Now, the cases that favors your situation are:
$$(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3)\ .$$
Since there are $12$ favor cases, the probablity is $12/20=3/5$. It's not easier, it's just another way of seeing it.
|
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|
Differentiation of Power Series Let $$f(x)= \sum_{n=0}^{\infty} \frac{x^n}{n!} $$
for $x\in \mathbb R$.
Show $f′ =f$.
Note: Do not use the fact that $f(x) = e^x$. This is true but has not been established at this point in the text.
|
Here is a proof which avoids uniform convergence.
Let $$f(x) = \sum_{n = 0}^{\infty}\frac{x^{n}}{n!}\tag{1}$$ Using multiplication of infinite series it is easy to show that $$f(x)f(y) = f(x + y)\tag{2}$$ for all real $x, y$ (the proof of this also requires binomial theorem for positive integer index).
Next we can see that if $0 < x < 2$ then we have
\begin{align}
\frac{f(x) - 1}{x} &= 1 + \frac{x}{2!} + \frac{x^{2}}{3!} + \cdots\notag\\
&\leq 1 + \frac{x}{2} + \frac{x^{2}}{2^{2}} + \frac{x^{3}}{2^{3}} + \cdots\notag\\
&= 1 + \frac{x/2}{1 - (x/2)} = 1 + \frac{x}{2 - x}\notag
\end{align}
Thus for $0 < x < 2$ we have $$1 \leq \frac{f(x) - 1}{x}\leq 1 + \frac{x}{2 - x}\tag{3}$$ and taking limits when $x \to 0^{+}$ and using Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{f(x) - 1}{x} = 1\tag{4}$$ This also shows that $$\lim_{x \to 0^{+}}f(x) = \lim_{x \to 0^{+}}1 + x\cdot\frac{f(x) - 1}{x} = 1 + 0\cdot 1 = 1\tag{5}$$ Again using equation $(2)$ we have $$f(x)f(-x) = f(x + (-x)) = f(0) = 1$$ so that $f(-x) = 1/f(x)$ and therefore
\begin{align}
\lim_{x \to 0^{-}}\frac{f(x) - 1}{x}&= \lim_{y \to 0^{+}}\frac{1 - f(-y)}{y}\text{ (putting }x = -y)\notag\\
&= \lim_{y \to 0^{+}}\frac{f(y) - 1}{y}\cdot\frac{1}{f(y)}\notag\\
&= 1\cdot 1 = 1\text{ (using equations (4) and (5))}\notag
\end{align}
It now follows that $$\lim_{x \to 0}\frac{f(x) - 1}{x} = 1\tag{6}$$ Finally we have
\begin{align}
f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\
&= \lim_{h \to 0}\frac{f(x)f(h) - f(x)}{h}\notag\\
&= \lim_{h \to 0}f(x)\cdot\frac{f(h) - 1}{h}\notag\\
&= f(x)\cdot 1 = f(x)\text{ (using equation (6))}\notag
\end{align}
|
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|
Integral of rational function with trigonometric functions $$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}
$$
I saw this problem online and it looked like an interesting/difficult problem to try and tackle. My attempt so far is to use tangent half-angle substitution.
Let $t= \tan^2 (\frac{x}{2})$, then $dt= \frac{1}{2} \sec^2 (\frac{x}{2})\;dx.$ With algebra we get that $dx= \frac{2}{1+t^2}dt,$ $\sin x = \frac{2t}{1+t^2},$ and $\cos x = \frac{1-t^2}{1+t^2}.$ Therefore we get the following:
$$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4} = \int \frac{\frac{2}{1+t^2}}{(\sqrt{\frac{1-t^2}{1+t^2}}+\sqrt{\frac{2t}{1+t^2}})^4}dt = 2\int \frac{1}{1+t^2}\cdot \frac{(1+t^2)^2}{(\sqrt{1-t^2}+ \sqrt{2t})^4}dt
$$
$$
= 2\int \frac{1+t^2}{(\sqrt{1-t^2}+ \sqrt{2t})^4}dt
$$
I expanded the denominator, but only made things worse. According to wolfram alpha, the solution is in terms of elementary functions. Can you give me any clues or hints on how to evaluate this beast. Thank you!
|
You may write
$$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}=\int \frac{1}{(1+ \sqrt{\tan x})^4}\frac{dx}{\cos^2 x}
$$ then make the change of variable $$\displaystyle u=\sqrt{\tan x},\quad \frac{1}{\cos^2 x}=1+u^4, \quad dx=\frac{2u}{1+u^4}du,$$ to just obtain
$$
\begin{align}
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}&=2\int \frac{u}{(1+u)^4}du=-\frac 1 3 \frac{3u+1}{(1+u)^3}+C\\\\&=-\frac1{(1+\sqrt{\tan x})^2}+\frac{2}{3(1+\sqrt{\tan x})^3}+C
\end{align}
$$
|
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|
Proving that a sequence is increasing Problem: A sequence $(a_n)$ is defined recursively as follows, where $0<\alpha\leqslant 2$:
$$
a_1=\alpha,\quad a_{n+1}=\frac{6(1+a_n)}{7+a_n}.
$$
Prove that this sequence is increasing and bounded above by $2$. What is its limit?
Ideas: How should I go about starting this proof? If the value of $a_1$ were given, I could show numerically and by induction that the sequence is increasing. But no exact value of $\alpha$ is given.
|
First, $0\leq a_1\leq2$.
Second, $a_{n+1}=\frac{6(1+a_n)}{7+a_n}=6-\frac{36}{7+a_n}$.
So, if $0\leq a_n\leq2$,
$\qquad$then $7\leq 7+a_n\leq 9$,
$\qquad$then $\frac{1}{7}\geq\frac{1}{7+a_n}\geq\frac{1}{9}$,
$\qquad$then $\frac{36}{7}\geq\frac{36}{7+a_n}\geq\frac{36}{9}$,
$\qquad$then $-\frac{36}{7}\leq-\frac{36}{7+a_n}\leq-\frac{36}{9}$,
$\qquad$then $0\leq6-\frac{36}{7}\leq6-\frac{36}{7+a_n}\leq6-\frac{36}{9}=2$,
then $0\leq a_{n+1}\leq2$.
By induction $0\leq a_n\leq2$ for all $n$.
Also $a_{n+1}-a_n=\frac{6(1+a_n)}{7+a_n}-a_n=\frac{6-a_n-a_n^2}{7+a_n}=\frac{(a_n+3)(2-a_n)}{7+a_n}>0$, as long as $0\leq a_n\leq2$.
Therefore $a_{n+1}\geq a_n$.
|
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|
General solution of $(1-x^2)y''-2y=0$ about $x_0=0$? I've expanded this differential equation as a series to obtain the recurrence relation $$a_{n+2}=\frac{a_n(n^2-n+2)}{n^2+3n+2}.$$ I don't know how to find $a_n$ in terms of $a_0$ and $a_1$ so that I can get the general solution. How do I do this?
|
This is not an answer but it is too long for a comment.
As commented by Winther, finding something nicer than the general recurence seem to be almost impossible.
Playing with a CAS and starting from what Winther wrote, what I obtained is the following monster for $a_{2n}$ $$\frac{ 4^{n-1} \cosh \left(\frac{\sqrt{7} \pi }{2}\right) \Gamma \left(\frac{1}{4}
\left(1-i \sqrt{7}\right)\right) \Gamma \left(\frac{1}{4} \left(1+i
\sqrt{7}\right)\right) \Gamma \left(n-\frac{i \sqrt{7}}{4}-\frac{1}{4}\right)
\Gamma \left(n+\frac{i \sqrt{7}}{4}-\frac{1}{4}\right)}{\pi ^2 \Gamma (2 n+1)}a_0$$ $a_{2n+1}$ is as nice.
As Winther said, this kind of beauty just reflects the fact that the solution of the differential equation involves ugly hypergeomatric functions $$y=c_1 \, _2F_1\left(-\frac{1}{4}-\frac{i \sqrt{7}}{4},-\frac{1}{4}+\frac{i
\sqrt{7}}{4};\frac{1}{2};x^2\right)+i c_2 x \, _2F_1\left(\frac{1}{4}-\frac{i
\sqrt{7}}{4},\frac{1}{4}+\frac{i \sqrt{7}}{4};\frac{3}{2};x^2\right)$$
|
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|
Integral of $\big((1+\cos(x))\sin(x)\big)^2$ What is
$$\int \big((1+\cos(x))\sin(x)\big)^2dx$$
?
|
use $$\begin{align}(1 + \cos t)^2\sin^2 t &= \sin^2 t + 2\sin^2 t \cos t + \sin^2 t \cos^2 t\\
&=\frac 12 - \frac 12 \cos 2t + 2\sin^2 t \cos t + \frac 18 - \frac 18 \cos 4t \\
&=\frac 58 - \frac 12 \cos 2t + 2\sin^2 t \cos t - \frac 18 \cos 4t \end{align}$$
now we can integrate $$\int (1 + \cos t)^2\sin^2 t\, dt = \frac 58 t - \frac 14 \sin 2t + \frac 23 \sin^3 t - \frac 1{32} \sin 4t + C $$
|
{
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|
Does $a \leq b + c $ imply $a^2 \leq (b+c)^2 + (b-c)^2$? Givens
$$
a \leq b + c
$$
or
$$
a^2 \leq b^2 + c^2+2bc
$$
Can we prove that?:
$$
a^2 \leq (b+c)^2 + (b-c)^2 = 2b^2 + 2c^2
$$
|
If
$$a^2 \leq b^2+c^2 + 2bc= (b+c)^2,$$
then
$$a^2 \leq (b+c)^2 + (b-c)^2$$
as $(b-c)^2 \geq 0$.
|
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|
Proving $\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$ for all $n\geq 1$ by induction How prove the following equality:
$a_n$:=$\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$
$1$.presumption: $(-1)^1 \cdot 1^2+(-1)^2\cdot2^2=(2 \cdot 1+1) \cdot 1=3$ that seems legit
$2$.precondition:
$a_{n-1}$= $(2(n-1)+1)(n-1)$=$2n^2-3n+1$
for $k=n$
$a_n$= $\sum\limits_{k=1}^{2n} {(-1)^n \cdot n^2}$+$a_{n-1}$
$a_n$=${(-1)^n \cdot n^2}$+$2n^2-3n+1$
But that last equation seems somehow wrong to get $(2n+1)\cdot n$
|
Note: Here is a clear, clean way of going about proving your result going from $k$ to $k+1$:
For $n\geq 1$ let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^{2n}(-1)^i\cdot i^2 = (2n+1)\cdot n.
$$
To prove this, we argue by induction on $n$.
Base case ($n=1$): For $S(1)$, we have that
$$
\sum_{i=1}^{2}(-1)^i\cdot i^2=[(-1)^1\cdot 1^2]+[(-1)^2\cdot 2^2]=-1+4=3=(2 \cdot 1+1) \cdot 1. \quad\large\checkmark
$$
Inductive step ($S(k)\to S(k+1)$): Fix some $k\geq 1$ and suppose that
$$
S(k) : \sum_{i=1}^{2k}(-1)^i\cdot i^2 = (2k+1)\cdot k
$$
holds. To be shown is that
$$
S(k+1) : \sum_{i=1}^{2k+2}(-1)^i\cdot i^2 = (2k+3)(k+1)
$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{2k+2}(-1)^i\cdot i^2 &= \sum_{i=1}^{2k}(-1)^i\cdot i^2 + [(-1)^{2k+2}\cdot(2k+2)^2] + [(-1)^{2k+1}\cdot(2k+1)^2]\quad\text{(by defn. of $\Sigma$)}\\[1em]
&= [(2k+1)\cdot k] + [4k^2+8k+4] - [4k^2+4k+1]\qquad\text{(by $S(k)$; simplify)}\\[1em]
&= [2k^2+k]+[4k+3]\qquad\text{(simplify)}\\[1em]
&= 2k^2+5k+3\qquad\text{(simplify)}\\[1em]
&= (2k+3)(k+1),\qquad\text{(factor)}
\end{align}
we see that the right-hand side of $S(k+1)$ follows, completing the inductive step.
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
|
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|
$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$, sum, area of convergence & center I want to find the
$$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$$
now this is how I thought about doing it but I get stuck.
$$\sum _{n=0}^{\infty } x^n=\frac{1}{1-x}$$ given that absolute value of x is less then zero.
$$\int x^n \, dx=\frac{x^{n+1}}{n+1}$$
I keep integrating the result until I get
$$\frac{x^{n+3}}{(n+3) (n+1) (n+2)}=\frac{3 x^2}{4}-\frac{1}{2} x^2 \log (1-x)-\frac{x}{2}+x \log (1-x)-\frac{1}{2} \log (1-x)$$
I thought about multiplying by $$\frac{(n+1) (n+2)}{x^3}$$
However, the answer should be this: $$\sum _{n=0}^{\infty } \frac{x^n}{n+3}=-\frac{\log (1-x)}{x^3}-\frac{1}{x^2}-\frac{1}{2 x}$$
On the interval [-1,1), and $\frac{1}{3}$ if x=0.
Could someone please show me what I did wrong and how the steps should be to get the correct answer?
|
I approached it in a slightly different way using Taylor expansions.
The sum for $x=0$ is trivial.
For $x \in (-1,1)$
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\dots$$
$$-\log(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\dots$$
$$-\log(1-x)-x-\frac{x^2}{2}=\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\dots$$
$$-\frac{\log(1-x)}{x^3}-\frac{1}{x^2}-\frac{1}{2x}=\frac{1}{3}+\frac{x}{4}+\frac{x^2}{5}+\dots = \sum _{n=0}^{\infty } \frac{x^n}{n+3}$$
|
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|
Show that the substitution $t=\tan\theta$ transforms the integral ${\int}\frac{d\theta}{9\cos^2\theta+\sin^2\theta}$, into ${\int}\frac{dt}{9+t^2}$ To begin with the $d\theta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:
$${\int}\frac{1}{9\cos^2\theta+\sin^2\theta}\,d\theta$$
I tried working backwards
$$\frac{d}{d\theta}\tan\theta=\sec^2\theta\,\,\,\,{\Rightarrow}\,\,\,\,d\,\tan\theta=\sec^2\theta\,d\theta$$
$${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{9+\tan^2\theta}$$
$$\tan\theta=\frac{\sin\theta}{\cos\theta}\,\,\,\,{\Rightarrow}\,\,\,\,\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$
$${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{\left(9+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}$$
$$9=\frac{9\cos^2\theta}{\cos^2\theta}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta}{\cos^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta+\sin^2\theta}{\cos^2\theta}\right)}$$
$${\Rightarrow}\,\,\,\,{\int}\frac{\color{red}{\cos^2\theta\,\sec^2\theta}\,d\theta}{9\cos^2\theta+\sin^2\theta}$$
Now I have to prove that $$\cos^2\theta\,\sec^2\theta=1$$
but I don't think it is... What have I done wrong? Regards Tom
|
$\sec^2\theta = \frac{1}{\cos^2\theta} \implies \cos^2\theta\sec^2\theta = \frac{\cos^2\theta}{\cos^2\theta} = 1$
|
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|
Odd divisibility induction proof Prove that for odd $n>3$
$$64\ | \ n^4-18n^2+17$$
I checked that for $n=5$ it works. I think I need to assume that for $2n+1$ it holds and show that $2n+3$ also holds. Any ideas?
|
we have:
$$n^4-18n^2+17+64=(n^2-9)^2 $$
and because $(n^2-9)=(n-3)(n+3)$ is divisible by $8$ for odd numbers we can conclude.
By induction:
Assume that $n^4-18n^2+17=(n^2-9)^2-64$ is divisible by $64$ for an odd number $n$ we want to prove that this still true for $n+2$, we have
$$\begin{align}(n+2)^4-18(n+2)^2+17+64&= ((n+2)^2-9)^2\\
&=(n^2-9+4(1+n))^2 \\
&=(n^2-9)^2+2.4(1+n)(n^2-9)+2.16.(1+n)^2\end{align}$$
and because $n$ is odd we have $2$ divides $(1+n)$, and by induction hypothesis $8$ divides $(n^2-9)$ and $64$ divides $(n^2-9)^2$ which implies the result.
|
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|
Is the sequence defined by the recurrence $ a _ { n + 2 } = \frac 1 { a _ { n + 1 } } + \frac 1 { a _ n } $ convergent?
Let $ a _ 0 = 1 $, $ a _ 1 = 1 $ and $ a _ { n + 2 } = \frac 1 { a _ { n + 1 } } + \frac 1 { a _ n } $ for every natural number $ n $. How can I prove that this sequence is convergent?
I know that if it's convergent, it converges to $ \sqrt 2 $ since if $ \lim \limits _ { n \to \infty } a _ n = a $ then:
$$ \lim _ { n \to \infty } \left ( a _ { n + 2 } - \frac 1 { a _ { n + 1 } } - \frac 1 { a _ n } \right) = a - \frac 2 a = 0 \text ; $$
$$ \therefore \quad a ^ 2 = 2 \text . $$
Now it's easy to see that every $ a _ n $ is positive, so $ a \ge 0 $ and thus $ a = \sqrt 2 $.
Assuming the sequence is convergent, I can calculate an estimation of the rate of convergence too. Let $ \epsilon _ n := a _ n - \sqrt 2 $. We have:
$$ \epsilon _ { n + 2 } = \frac 1 { a _ { n + 1 } } - \frac 1 { \sqrt 2 } + \frac 1 { a _ n } - \frac 1 { \sqrt 2 } = - \frac { a _ { n + 1 } - \sqrt 2 } { \sqrt 2 a _ { n + 1 } } - \frac { a _ n - \sqrt 2 } { \sqrt 2 a _ n } = - \frac { \epsilon _ { n + 1 } } { \sqrt 2 a _ { n + 1 } } - \frac { \epsilon _ n } { \sqrt 2 a _ n } \text . $$
Now because $ a _ n \sim \sqrt 2 + \epsilon _ n $ and $ \lim \limits _ { n \to \infty } \epsilon _ n = 0 $, therefore from the above equation:
$$ \epsilon _ { n + 2 } \lesssim - \frac { \epsilon _ { n + 1 } + \epsilon _ n } 2 \text , $$
which yields $ \epsilon _ n \lesssim \alpha \left ( \frac { - 1 - \sqrt 7 i } 4 \right) ^ n + \beta \left( \frac { - 1 + \sqrt 7 i } 4 \right) ^ n $ for some complex constants $ \alpha $ and $ \beta $, using induction on $ n $. Equivalently, we have $ \epsilon _ n \lesssim \left( \frac 1 { \sqrt 2 } \right) ^ n \bigl( A \cos ( n \theta ) + B \sin ( n \theta ) \bigr) $ for $ \theta = \arctan \frac { \sqrt 7 } 4 $ and some real constants $ A $ and $ B $, since $ \left| \frac { - 1 \pm \sqrt 7 i } 4 \right| = \frac 1 { \sqrt 2 } $ and $ \arg \frac { - 1 \pm \sqrt 7 i } 4 = \pi \mp \theta $. Hence we get the rough estimation $ | \epsilon _ n | \lesssim C 2 ^ { - \frac n 2 } $ for some real constant $ C $, and $ \frac 1 { \sqrt 2 } $ is a good guess for the rate of convergence.
(Edit: Thanks to Alex Ravsky for the confirming graphs in his answer.)
Edit (some more of my thoughts):
Let $ b _ n := \min \left\{ a _ n , a _ { n + 1 } , \frac 2 { a _ n } , \frac 2 { a _ { n + 1 } } \right\} $. It's easy to see that $ b _ n \le a _ n \le \frac 2 { b _ n } $ and $ b _ n \le a _ { n + 1 } \le \frac 2 { b _ n } $. Now using induction we can prove that $ b _ n \le a _ { n + m } \le \frac 2 { b _ n } $. Especially, $ a _ { n + 2 } \ge b _ n $ and $ \frac 2 { a _ { n + 2 } } \ge b _ n $ which yields $ b _ { n + 1 } \ge b _ n $. The problem can be solved if I show that the sequence $ ( b _ n ) _ { n = 0 } ^ \infty $ increases to $ \sqrt 2 $.
|
Graphs, illustrating asymptotic behavior of the sequence $\{a_n\}$. The graphs suggest that $$(a_n-\sqrt{2})\sqrt{2}^n=O(1).$$
Added:
|
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|
calculate the serie $\sum \frac{6^n}{(3^{n+1}-2^{n+1})(3^n - 2^n)}$ The following serie is obviously convergent, but I can not figure out how to calculate its sum :
$$\sum \frac{6^n}{(3^{n+1}-2^{n+1})(3^n - 2^n)}$$
|
You can use one of the following identities to telescope the series: $$\dfrac{3^{n+1}}{3^{n+1}-2^{n+1}}-\dfrac{3^n}{3^n-2^n}=\dfrac{2^n}{3^n-2^n}-\dfrac{2^{n+1}}{3^{n+1}-2^{n+1}}=\dfrac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$$
|
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|
When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is even $x^2\equiv (1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
Thus, when $p\equiv 1\pmod{4}$, the congruence $x^2\equiv -1\pmod{p}$ is soluble.
Point of contention: I understand the general argument
I understand the relation
$(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
But I cant work out how $x^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
How is $((\frac{p-1}{2})!)^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
|
You can see it the following way:
$$\begin{align}(1\cdot 2\cdot...\cdot r)\underbrace{((p-1)\cdot(p-2)\cdot...\cdot(p-r))}_{\equiv (-1)(-2)(-3)\cdots(-r)}\pmod{p}
\end{align}$$
so every $i$ appears two times in this product so this product is equivalent to $r!(-1)^r r!$ and this gives you $r!^2$
|
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|
Solve the following integral: $ \int \frac{x^2}{x^2+x-2} dx $ Solve the integral: $ \int \frac{x^2}{x^2+x-2} dx $
I was hoping that writing it in the form $ \int 1 - \frac{x-2}{x^2+x-2} dx $ would help but I'm still not getting anywhere.
In the example it was re-written as $ \int 1 - \frac{4}{3x+6} - \frac{1}{3x-3} dx $
but I am not sure how this was accomplished. Any ideas? I am more interested in the method than the answer.
|
While partial fraction expansion is certainly the natural choice, here is a slightly different approach that continues from the OP's attempt.
$$\begin{align}
\frac{x^2}{x^2+x-2}&=1-\frac{x-2}{x^2+x-2}\\\\
&=1-\frac12 \frac{2x-4}{x^2+x-2}\\\\
&=1-\frac12 \frac{2x+1-5}{x^2+x-2}\\\\
&1-\frac12 \frac{2x+1}{x^2+x-2}+ \frac{5/2}{x^2+x-2}\\\\
&=1-\frac12 \frac{2x+1}{x^2+x-2}+ \frac{5/2}{(x+1/2)^2-(3/2)^2}
\end{align}$$
Now, the first term is trivial to integrate. The second term is a perfect differential since the numerator is the derivative of the denominator. And the last term is set up for a hyperbolic trigonometric substitution.
|
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|
Can $\oint_{|z|=2}z^3 \bar {z} e^\frac{1}{(z-1)} dz$ be solved? How we can calculate the result of following Integral?
$$\oint_{|z|=2}z^3 \bar {z} e^\frac{1}{z-1} \mathrm{d}z$$
|
$$\begin{align}
\oint_{|z=2|} z^3\bar z e^{\frac{1}{(z-1)}}dz&=\oint_{|z=2|} 4z^2 e^{\frac{1}{(z-1)}}dz\\
&=8\pi i \text{Res}_{z=1}\left(z^2 e^{\frac{1}{(z-1)}}\right)\\
\end{align}$$
The residue $\text{Res}_{z=1} \left(z^2 e^{\frac{1}{(z-1)}}\right)$ can be found as follows
Note that $z^2=(z-1)^2+2(z-1)+1$ and the Laurent expansion of $e^{\frac{1}{z-1}}$ is
$$e^{\frac{1}{(z-1)}}=1+\frac{1}{z-1}+\frac{1}{2!}\frac{1}{(z-1)^2}+\frac{1}{3!}\frac{1}{(z-1)^3}+ \cdots$$
wherein we observe that the residue at $z=1$ of $z^2e^{\frac{1}{(z-1)}}$ comes from the 3 terms
$$\begin{align}
&(1)\,\,(z-1)^2 \times \frac{1}{3!}\frac{1}{(z-1)^3}=\frac16 (z-1)^{-1}\\
&(2)\,\,2(z-1) \times \frac{1}{2!}\frac{1}{(z-1)^2}=1 (z-1)^{-1}\\
&(3)\,\, 1 \times \frac{1}{z-1}=1 (z-1)^{-1}
\end{align}$$
Thus the residue is $\frac16+1+1=\frac{13}{6}$. Therefore,
$$\begin{align}
\oint_{|z=2|} z^3\bar z e^{\frac{1}{(z-1)}}dz&=\oint_{|z=2|} 4z^2 e^{\frac{1}{(z-1)}}dz\\
&=8\pi i \text{Res}_{z=1}\left(z^2 e^{\frac{1}{(z-1)}}\right)\\
&=8\pi i \frac{13}{6}\\
&=\frac{52\pi i}{3}
\end{align}$$
|
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|
Question about matrix multiplication notation I have the following matrices:
$A=\begin{pmatrix}
-\frac{2}{3} & \frac{1}{3} & 0 \\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{2} \\
\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\
\end{pmatrix}$
; $B=\begin{pmatrix}
2 & 3\\
2 & 0 \\
0 & 3\\
\end{pmatrix}$
; $C=\begin{pmatrix}
6 & 3& 4 \\
6 & 6 & 0 \\
\end{pmatrix}$
;$u=\begin{pmatrix}2 \\ -5 \\ 4 \end{pmatrix}$
I want to calculate the following:
$\sum_j a_{ij}u_j$
$\sum_{jk} c_{ij}a_{jk}b_{kl}$
but I am not used to this kind of notation. Does the first one mean to multiply all $i$-rows of A with the $j$-column of u?
$\implies AB=\begin{pmatrix}
-\frac{2}{3} & \frac{1}{3} & 0 \\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{2} \\
\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\
\end{pmatrix} \cdot \begin{pmatrix}2 \\ -5 \\ 4 \end{pmatrix}=\begin{pmatrix}-3 \\ 4 \\ 0 \end{pmatrix}$?
I am not sure what the second one is trying to say? Maybe $C\cdot A\cdot B$?
|
$a_{ij}$ denotes the $i^{th}$ row and $j^{th}$ column of $A$. The first summation means keeping $i$ fixed, increment the $j$ parameter as below for each $i$.
$$\sum_j a_{ij}u_j = a_{i1}u_1+a_{i2}u_2+a_{i3}u_3$$
This translates to the product $A.U$.
Similarly for the second sum.
|
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|
Find the coefficient of $x^{30}$. Find the coefficient of $x^{30}$ in the given polynomial
$$
\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}\right)^5
$$
I don't know how to solve problems with such high degree.
|
$$[x^{30}]\left(\frac{1-x^{13}}{1-x}\right)^5 = [x^{30}]\sum_{k=0}^{5}\binom{5}{k}(-1)^k x^{13k}\sum_{n\geq 0}\binom{n+4}{4}x^n \tag{1}$$
hence the LHS of $(1)$ equals:
$$\binom{5}{0}\binom{34}{4}-\binom{5}{1}\binom{21}{4}+\binom{5}{2}\binom{8}{4}=\color{red}{17151.}\tag{2}$$
|
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|
Evaluate $\int_{-\pi}^\pi \! \cos(kx)\cos^n(x) \, \mathrm{d}x$ My question is:
Evaluate $$\int_{-\pi}^\pi \! \cos(kx)\cos^n(x) \, \mathrm{d}x$$ for $k=0,1,...,(n-1)$ and $n \in \mathbb{N}$.
I've tried integration by parts but without much success. Any help/trick for this integral? Thanks!
|
For every $n\in \mathbb{N}$ we have:
\begin{eqnarray}
\cos(kx)\cos^n(x)&=&\frac{e^{ikx}+e^{-ikx}}{2}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^n\\
&=&\frac{e^{ikx}+e^{-ikx}}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{i(n-p)x}e^{-ipx}\\
&=&\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}\left[e^{i(n+k-2p)x}+e^{i(n-k-2p)x}\right]\\
&=&\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{i(n+k-2p)x}+\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{i(n-k-2p)x}\\
&=&\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{i(n+k-2p)x}+\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose n-p}e^{i[n-k-2(n-p)]x}\\
&=&\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{i(n+k-2p)x}+\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{i(-n-k+2p)x}\\
&=&\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{i(n+k-2p)x}+\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{-i(n+k-2p)x}\\
&=&\frac{1}{2^n}\sum_{p=0}^n{n\choose p}\frac{e^{i(n+k-2p)x}+e^{-i(n+k-2p)x}}{2}\\
&=&\frac{1}{2^n}\sum_{p=0}^n{n\choose p}\cos(n+k-2p)x
\end{eqnarray}
Case 1: $n+k$ is even. Let
$$
A(k,n)=\{0\le p\le n:\, 2p\ne n+k\}.
$$
We have
\begin{eqnarray}
\int_{-\pi}^\pi\cos(kx)\cos^n(x)\,d&=&2\int_0^\pi\cos(kx)\cos^n(x)\,dx=\frac{1}{2^{n-1}}\sum_{p=0}^n{n\choose p}\int_0^\pi\cos(n+k-2p)x\,dx\\
&=&\frac{\pi}{2^{n-1}}{n\choose \frac{n+k}{2}}+\frac{1}{2^{n-1}}\sum_{p \in A(k,n)}{n\choose p}\int_0^\pi \cos(n+k-2p)x\,dx\\
&=&\frac{\pi}{2^{n-1}}{n\choose \frac{n+k}{2}}+\frac{1}{2^{n-1}}\sum_{p\in A(k,n)}{n\choose p}\left[\frac{\sin(n+k-2p)x}{n+k-2p}\right]_0^\pi\\
&=&\frac{\pi}{2^{n-1}}{n\choose \frac{n+k}{2}}.
\end{eqnarray}
Case 2: $n+k$ is odd. We have:
\begin{eqnarray}
\int_{-\pi}^\pi\cos(kx)\cos^n(x)\,d&=&2\int_0^\pi\cos(kx)\cos^n(x)\,dx=\frac{1}{2^{n-1}}\sum_{p=0}^n{n\choose p}\int_0^\pi\cos(n+k-2p)x\,dx\\
&=&\frac{1}{2^{n-1}}\sum_{p=0}^n{n\choose p}\left[\frac{\sin(n+k-2p)x}{n+k-2p}\right]_0^\pi\\
&=&0.
\end{eqnarray}
|
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|
Maya Lists the Positive Divisors
Maya lists all the positive divisors of $2010^2$. She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
$2010^2 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 67^2$
It has, $(3)(3)(3)(3) = 3^4 = 81$ total divisors.
Now I need to find how many are perfect squares. It certainly has two, $1^2, 2010^2$. But this is complicated, any hints?
|
The perfect squares must have an even power of each prime factor. Therefore, you could count the divisors of $2010$; the square of each of these is a square divisor of $2010^2$.
More precisely, since $2010^2$ has 4 distinct pairs of divisors, if $a$ is a square divisor of $2010^2$ and $2$ is a factor of $a$, then $2^2$ is also a factor of $a$. Extending this idea, there are $2^4=16$ total square divisors of $2010^2$.
Edit (from comments):
Claim: Let $a^2\mid b^2$, then $a\mid b$.
Proof: We can prime factorize $a$ and $b$ as $a=p_1^{a_1}\cdots p_n^{a_n}$ and $b=p_1^{b_1}\cdots p_n^{b_n}$. In this prime factorization, some of the powers $a_i$ may be zero. Since $a^2\mid b^2$, it follows that $2a_i\leq 2b_i$. Then, $a_i\leq b_i$, so it follows that $a\mid b$.
From this, it follows that the square divisors of $b^2$ are precisely the divisors of $b$, squared.
|
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|
What is the tip for this exact differential equation? $$ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} $$
I have multiplied the left part $x^2+y^2$ for $x dx + y dy$ getting
$$(x^3+xy^2+y)dx+(x^2y+y^3-x)dy=0$$
And the derivative test give me:
$\frac{dM}{dy}= 0+2xy+1$ and $\frac{dN}{dx} = 2xy+0-1$.
Where´s my mistake?
|
your differential equation $$Mdx + Ndy = (x^3 + xy^2 + y)\, dx +(y^3 + yx^2 - x)\, dy = 0.$$ then we have $$M_y = 2xy + 1, N_x = 2xy - 1 $$ therefore is not exact. we can make it exact by multiplying by $a$ and demanding that $$(aM)_y = (aN)_x \to a_y M + aM_y = a_x N + aN_x \to a_xN- a_y M = a(M_y - N_x) = 2a $$ that is $$a_x(y^3 + yx^2 -x) -a_y(x^3 + xy^2 +y) = 2a\tag 1$$ now choose $$a = a(r), r^2 = x^2 + y^2$$ so that $$ a_x = \frac x r a', a_y = \frac x y a'$$ subbing this in $(1),$ we get $$\frac x r a'\left(yr^2 - x\right) - \frac y r a'\left(xr^2+y\right) = 2a \to -r\frac{da}{dr} = 2a \to a = \frac1{r^2} = \frac1{x^2 + y^2} $$
now that we have an integration factor, we get $$ \frac1{x^2 + y^2}(x^3 + xy^2 + y)\, dx +\frac 1{x^2 + y^2}(y^3 + yx^2 - x)\, dy = 0 \to x\,dx + y\, dy +\frac{ydx - xdy}{x^2 + y^2} = 0 .$$ we have $$ F_x = M = x + \frac y{x^2 + y^2}, F_y = N = y -\frac x{x^2 + y^2} \to \\
F_{xy} = \frac1{x^2+y^2} - \frac{2y^2}{(x^2+y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2},\\ F_{yx} = -\frac 1{x^2 + y^2} + \frac{2x^2}{(x(2 + y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2}$$
integrating $F_x$ i get $$F = \frac 12 x^2 + \tan^{-1}\left(\frac x y\right) + C(y) \to F_y = \frac 1{1 + (x/y)^2}( -\frac x{y^2}) + \frac{dC}{dy} \to F_y = -\frac x{x^2 + y^2} +\frac{dC}{dy} $$
this gives $$ \frac{dC}{dy} = y \to C = \frac 1{2}y^2 + const$$
this has the integral $$\frac 12(x^2 + y^2) + \tan^{-1}\left(\frac xy\right) = const. $$
|
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|
Binomial Theorem of Differentiation? I noticed that $$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$ and it's had me scratching for a little bit. It's easy to see how the cross terms add up but can anyone draw a direct comparison between the behavior being exhibited here and the binomial expansion? It seems to work for the case equivalent to the multinomial theorem as well ($\frac{d^{m}}{dx^{m}} \prod_{k=1}^{n} f_k(x)$). Can someone produce a proof for this more general case (preferably by induction)? Any and all insights are welcome!
|
Taylor series expansion of $f(x+h)$
$$
f(x+h)=f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....
$$
Taylor series expansion of $g(x+h)$
$$
g(x+h)=g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....
$$
Taylor series expansion of $f(x+h)g(x+h)$
$$
f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....
$$
$$
f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....=(f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....)(g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....)
$$
If you order $h^n$, it will give you binom expansion
$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$
Because it has exactly the same coefficient of
$$
=(1+hx +\frac{h^2 x^2 }{2!}+\frac{h^3x^3 }{3!}+....)(1+hy +\frac{h^2 y^2 }{2!}+\frac{h^3y^3 }{3!}+....)=e^{hx}e^{hy}=e^{h(x+y)}
$$
$$
e^{h(x+y)}=1+h(x+y) +\frac{h^2 (x+y)^2 }{2!}+\frac{h^3(x+y)^3 }{3!}+....
$$
If you order $h^n$ of $e^{hx}e^{hy}$ and equal to $h^n$ of $e^{h(x+y)}$ , it will give you binom expansion proof.
$$\frac{(x+y)^n }{n!}=\sum_{i=0}^n \frac{x^i y^{n-i}}{i! (n-i)!} $$
$$(x+y)^n=\sum_{i=0}^n \frac{n! x^i y^{n-i}}{i! (n-i)!} $$
$$(x+y)^n=\sum_{i=0}^n {n \choose i} x^i y^{n-i} $$
Thus it is also true
$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$
|
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|
How to solve $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$? I need to compute $$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}\ dx.$$ I tried it on wolfram but it timed out, maybe because I am on a mobile device. Any hint is appreciated.
|
$\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{(x-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx = \int\frac{(x^2-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x^2+2x+1)}dx$$
Above we multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $(x+1).$
$$\displaystyle = \int\frac{\left(1-\frac {1}{x^2}\right)\cdot \sqrt{x^2\cdot \left(x^2+2x-1+\frac{2}{x}+\frac{1}{x^2}\right)}}{ \left(x+2+\frac{1}{x}\right)}dx$$
Now Let $ \left(x-\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1-\frac{1}{x^2}\right)dx = dt$
So Integral $$\displaystyle = \int\frac{\sqrt{t^2+2t-3}}{t+2}dt = \frac{t^2+2t-3}{(t+2)\sqrt{t^2+2t-3}}dt = \int\frac{t(t+2)-3}{(t+2)\sqrt{t^2+2t-3}}dt$$
So Integral $$\displaystyle = \underbrace{\int\frac{t}{\sqrt{t^2+2t-3}}dt}_{I} - \underbrace{\int\frac{3}{(t+2)}\cdot \frac{1}{\sqrt{t^2+2t-3}}dt}_{J}..........\color{\red}\checkmark.$$
So $$\displaystyle I = \int\frac{t}{\sqrt{t^2+2t-3}}dt = \int\frac{(t+1)-1}{\sqrt{(t-1)^2-2^2}} = \int\frac{(t-1)}{\sqrt{(t-1)^2-2^2}}-\int\frac{1}{\sqrt{(t-1)^2-2^2}}dt$$
Now Let $(t-1) = z\;\;,$ Then $dt = dz$
So $$\displaystyle I = \int\frac{z}{\sqrt{z^2-2^2}}dz-\int\frac{1}{\sqrt{z^2-2^2}}dz = \sqrt{z^2-4}-\ln \left|(t+1)+\sqrt{t^2+2t-3}\right|$$
Now $$\displaystyle J = 3\int\frac{1}{(t+2)\sqrt{t^2+2t-3}}dt = 3\int\frac{1}{(t+2)\sqrt{(t+2)^2-2(t+2)+1-4}}$$
Now Let $(t+2) = u\;,$ Then $dt = du$ and Integral $$\displaystyle = 3\int\frac{1}{u\sqrt{u^2-2u+1-4}}=3\int\frac{1}{u\sqrt{(u-1)^2-4}}du$$
Now $\displaystyle (u-1) = 2\sec \theta \;, $ Then $du= 2\sec \theta \cdot \tan \theta.$
Now after that You can Solve It.
|
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|
How is $\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$? As the title states, how is: $$\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$$
I can't see the pattern. Can someone please help? Thanks.
|
It all comes from the algebraic identity:
$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\dots+1).$$
|
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|
Proof that $0.33333... = \frac{1}{3}$ using $\epsilon-N$ method This proof is quite prevalent on the web, yet I struggle using this particular method.
Wikipedia (http://en.wikipedia.org/wiki/Limit_of_a_sequence) tells us:
We call $x$ the limit of the sequence $\{ x_n\}$ if:
for each real number $\epsilon> 0$ , there exists a natural number $N$
such that, for every natural number $n > N$, we have $|x_n - x| <\epsilon$.
So I start with the sequence $\{ x_n\} = \{0.3, 0.33, 0.333, 0.3333, \ldots\}$.
Then I want $\left|x_n - \dfrac{1}{3} \right| < \epsilon$, since this would mean that $\dfrac{1}{3}$ is the limit of the sequence $\{x_n\}$, by definition:
$$\left|x_n - \frac{1}{3} \right| = \left|\frac{1}{3} - x_n\right| = \dfrac{1}{3} - x_n = \frac{1}{3}10^{-n}$$
So we want $\dfrac{1}{3}10^{-n}<\epsilon$.
Where do I go from here? Where does the $N$ come in?
And also: can I simply say that $0.33333\ldots$ is the limit of the sequence $\{x_n\}$. By theorem a sequence can have at most one limit, thus this must mean that $\dfrac{1}{3} = 0.33333\ldots$, since both are limits of the sequence.
|
You did no justify your value for $x_n$.
The value of a decimal (base 10) string with infinite number of digits 3 behind the period is:
$$
(0.333\cdots)_{10}
= \sum_{k=1}^{\infty} 3\cdot 10^{-k}
= 3 \sum_{k=1}^{\infty} 10^{-k}
= 3 \sum_{k=1}^{\infty} \left(\frac{1}{10}\right)^k
= 3 \lim_{n\to \infty} S_n
$$
for the partial sums
$$
S_n = \sum_{k=1}^{n} \left(\frac{1}{10}\right)^k
$$
In your notation $x_n = 3 S_n$.
Their value can be evaluated by the geometric sum in $q = 1/10$.
$$
S_n
= \frac{1 - (1/10)^{n+1}}{1 - (1/10)} - 1 \\
= \frac{(10^{n+1} - 1)/10^{n+1}}{9/10} - 1 \\
= \frac{10^{n+1} - 1}{9\cdot 10^n} - \frac{9\cdot 10^n}{9\cdot 10^n} \\
= \frac{10^n - 1}{9\cdot 10^n} \\
= 1/9 - 1/(9 \cdot 10^n)
$$
This means $x_n = 1/3 - 1/(3\cdot 10^n)$ and $1/3 - x_n = 1/(3\cdot 10^n)$, which agrees with your value.
For any challenge $\epsilon > 0$ we have
$$
1/(3\cdot 10^n) < \epsilon \Rightarrow \\
1/(3 \cdot \epsilon) < 10^n \Rightarrow \\
\ln(1/3 \cdot \epsilon) < n \ln(10) \Rightarrow \\
N := \frac{\ln(1/3 \cdot \epsilon)}{\ln(10)} < n
$$
|
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|
Find all integers n such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients. I am unsure of how to approach this problem. I have thought about using the Rational root theorem, but I am unsure if this answers the question being asked.
Using the theorem, I get $\frac{p}{q} = \pm 1, \pm 13, \pm \frac{1}{5}$, and $\pm \frac{13}{5}$ as possible roots. Then I use synthetic division and Horner's method to get a remainder of $-(n+8)$. For this to be a solution, $-(n+8)=0$, so $n = -8$. Then I could do this for $+1, +13, -13,$ etc.
Is this the correct approach to answering the original question? Original question:
Find all integers $n$ such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients.
Why would I need to have a rational root to answer the problem? Couldn't I have complex solutions where I can express $5x^2 + nx - 13$ (where n is an integer) as a product of two linear factors with integer coefficients?
I greatly appreciate any insight you could provide on this. It's been about 2 years since I've done any mathematics (a brief foray into Neuroscience turned into a longer expedition than intended) and I am longing to return to the beautiful realm of mathematics. Thanks for your time in reading through this jumbled mathematical thought!
|
If a quadratic polynomial $ax^2 + bx + c$ with integer coefficients, $a \neq 0$, has factorization
$$ax^2 + bx + c = (rx + s)(tx + u) \tag{1}$$
then
\begin{align*}
a & = rt\\
b & = ru + st\\
c & = su
\end{align*}
which can be demonstrated by substituting $0$, $1$, and $-1$ in equation 1, then solving the resulting system of equations. Note that $ac = (rt)(su) = rstu = (ru)(st)$, that is, $b$ is the sum of two integers whose product is $ac$.
For the equation $5x^2 + nx - 13$, this means that $n$ is the sum of two integers with product $5 \cdot -13 = -65$. Since
\begin{align*}
-65 & = 1 \cdot -65 & -65 & = -1 \cdot 65\\
& = 5 \cdot -13 & & = -5 \cdot 13
\end{align*}
the possible integer values for $n$ are
\begin{align*}
1 + (-65) & = -64 & -1 + 65 & = 64\\
5 + (-13) & = -8 & -5 + 13 & = 8
\end{align*}
|
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|
Solve the recurrence of the alternating sum $R_n=R_{n-1}+(-1)^{n}(n+1)^{2}$ I have been trying to solve this recurrence for a few hours, but I haven't been able to find the solution yet:
$R_0=1$
$R_n=R_{n-1}+(-1)^{n}*(n+1)^{2}$.
I have been trying to substitute $T_n=(-1)^{n}*R_n$ and then solving for $T_n$ and got the sum:
$1+3+...+\frac{n(n+1)}{2}$
but this sum $(-1)^{n}\frac{n(n+1)(n+2)}{6}$ didn't give the generalized form that would give the terms of the recurrence.
Additionally, I have been trying to substitute $n=2*a$:
$\sum_{k=1}^a (-1)^{2k}*(2k+1)^{2}+ \sum_{k=1}^a (-1)^{2k-1}*(2k)^2$
and I got $\frac{n*(n+3)}{2}$, but it doesn't seem to be right either.
Please help me and if you can figure out, please tell me what I did wrong.
Edited: I have added the initial values.
|
$$\begin{align}
R_n-R_{n-1}&=(-1)^n(n+1)^2;\qquad R_0=1\\
R_n-\underbrace{R_0}_{=1}&=\sum_{r=1}^{n}(-1)^r(r+1)^2\qquad\text{by telescoping}\\
R_n&=\sum_{r=0}^{n}(-1)^r(r+1)^2\\
&=\sum_{r=1}^{n+1}(-1)^{r-1}r^2\\
\end{align}$$
Note that $-r^2+(r+1)^2=r+(r+1)$.
Hence, for even $n$,
$$\begin{align}
R_n&=1^2
\underbrace{-2^2+3^2}_{2+3}
\underbrace{-4^2+5^2}_{4+5}+\cdots+\underbrace{-n^2+(n+1)^2}_{n+(n+1)}\\
&=1+2+3+4+\cdots+n+(n+1)\\
&=\frac{(n+2)(n+1)}2=\binom {n+2}2
\end{align}$$
and for odd $n$,
$$\begin{align}
R_n&=
\underbrace{1^2-2^2}_{-1-2}+
\underbrace{3^2-4^2}_{-3-4}+\cdots+\underbrace{n^2-(n+1)^2}_{-n-(n+1)}\\
&=-(1+2+3+4+\cdots+n+(n+1))\\
&=-\frac{(n+2)(n+1)}2=-\binom {n+2}2
\end{align}$$
Hence the general solution is
$$R_n=(-1)^n\frac {(n+2)(n+1)}2=(-1)^n \binom {n+2}2\qquad\blacksquare$$
|
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|
Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} $
I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality
|
For inductive step :
$$\begin{align}\sum_{k=1}^{n+1}\frac{1}{k^2}&=\color{red}{\sum_{k=1}^{n}\frac{1}{k^2}}+\frac{1}{(n+1)^2}\\&\lt\color{red}{2-\frac 1n}+\frac{1}{(n+1)^2}\\&=2-\frac{(n+1)^2-n}{n(n+1)^2}\\&=2-\frac{1}{n+1}\left(\frac{n^2+n+1}{n(n+1)}\right)\\&=2-\frac{1}{n+1}\left(1+\frac{1}{n(n+1)}\right)\\&\lt 2-\frac{1}{n+1}\end{align}$$
|
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|
Proving $n^a-n^b$ is divisible by 10 Let $n$ be positive integer. Prove that there exists positive integers $a$ and $b$, with $a \neq b$, such that $n^a-n^b$ is divisible by $10$.
I have tried using mathematical induction and logs but I am really stuck .
|
Here is way of proving that $n^5-n$ works by induction, just so you know it can be done. It does rely on knowing the answer. The pigeonhole answer is the easiest elementary argument, but doesn't tell you which powers to use, and a more sophisticated approach tells you in general which powers will work for numbers other than $10$.
First note that $1^5-1=0$ is divisible by $10$.
Now for the induction step:
$$(n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1=(n^5-n)+10(n^3+n^2)+5n(n^3+1)$$
Now $n(n^3+1)=n(n+1)(n^2-n+1)$ is divisible by both $n$ and $n+1$, so is even (there are other ways of seeing this) so that $5n(n^3+1)$ is divisible by $10$.
Hence if $n^5-n$ is divisible by $10$ so is $(n+1)^5-(n+1)$, and the induction goes through.
|
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|
How can I integrate $\int {dx \over \sqrt{3^2+x^2}} $ using Trigonometric Substitution? $$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}} $$
$$ x =3\tan\theta$$
$$dx = 3\sec^2\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}} d\theta$$
$$\int {3\sec^2\theta \over 3\sec\theta} d\theta = \int \sec\theta$$
$$\ln|\sec\theta + \tan\theta| + C $$
$$\ln\left({\sqrt{9+x^2} \over 3} + {x \over 3}\right)$$
My book says the answer should just be:
$$\ln\left({\sqrt{9+x^2}} + {x}\right) $$
I'm wondering where I went wrong with this?
|
Hint: $$\log (t\times C) = \log t + \log C$$
|
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|
Why does the equation $x^2\equiv2 \pmod 5$ have no solutions? Why does the equation $x^2\equiv2 \pmod 5$ have no solutions?
I did a remainders table and found that $$x^2\equiv0;1;4\pmod 5$$
But is there any way to justify this besides that?
The original equation was $2x^2\equiv9\pmod 5$ but I got it to the form above.
|
The best way is what you have done.
*
*First note that given any $x$, we have
$x\equiv 0\pmod5 \text{ or }x \equiv 1\pmod5 \text{ or }x \equiv 2\pmod5 \text{ or }x \equiv 3\pmod5 \text{ or }x \equiv 4\pmod5$
*Next note that if $x \equiv y\pmod{n}$, then $x^2 \equiv y^2\pmod{n}$. Hence, we have that
*
*If $x \equiv 0 \pmod5$, then $x^2 \equiv 0^2 \pmod5 \equiv 0 \pmod5$.
*If $x \equiv 1 \pmod5$, then $x^2 \equiv 1^2 \pmod5 \equiv 1 \pmod5$.
*If $x \equiv 2 \pmod5$, then $x^2 \equiv 2^2 \pmod5 \equiv 4 \pmod5$.
*If $x \equiv 3 \pmod5$, then $x^2 \equiv 3^2 \pmod5 \equiv 4 \pmod5$.
*If $x \equiv 4 \pmod5$, then $x^2 \equiv 4^2 \pmod5 \equiv 1 \pmod5$.
Hence, we can only have $x^2 \equiv 0,1,4\pmod5$.
|
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|
How would you convert this particular polar equation to cartesian? How would you go about converting the polar equation $r^2 = 4cos(2\theta)$ into a cartesian equation in terms of y?
I have just started working on polar-cartesian equations, but do not yet have sufficient pattern recognition. My current approach of trying to apply the basic trigonometric pythagorean identity is leading me back to where I started.
Any hints on how I should approach this?
|
if you multiply the equation by $r^2,$ you get $$r^4 = 4r^2\cos 2\theta = 4r^2(\cos^2 \theta - \sin^2 \theta) = 4(x^2 - y^2) $$ that is $$(x^2 + y^2 )^2 = 4x^2 - 4y^2 \to y^4+2y^2(x^2+2) +(x^4-4x^2).$$ use the quadratic formula to get $$y^2 = -x^2 - 2\pm\sqrt{(x^2+2)^2 -x^4 +4x^2}=-2-x^2 \pm 2\sqrt{2x^2+1} $$ finally, $$y = \pm\sqrt{2\sqrt{2x^2 + 1} - x^2 - 2}. $$
|
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|
Equations with Sinus. How to find equation solution? I don't understand how to get a solution for sinus equation.
I have:
$$\sin x = \frac{1}{2}$$
\begin{align*}
x & = (-1)^k \cdot \arcsin\left(\frac{1}{2}\right) + 180^\circ \cdot k, k \in \mathbb{Z}\\
& = (-1)^k \cdot 30^\circ + 180^\circ \cdot k, k \in \mathbb{Z}
\end{align*}
Answer: $x = \ldots$
How to find x?
|
what you have $x = (-1)^k 30^\circ+180^\circ \cdot k$ is correct.
but can also be done the following way.
the solutions of $$\sin x = \frac 12$$ are $$x = \sin^{-1}(1/2) + k\cdot 360^\circ, 180 - \sin^{-1}(1/2) + k\cdot 360^\circ \\
x = 30^\circ + k\cdot 360^\circ, 150^\circ + k\cdot 360^\circ, \quad k \text{ any integer.} $$
|
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|
If $M = \int_{o}^{\pi/2}\frac{\cos x}{x+2}dx$ and $N = \int_{0}^{4}\frac{\sin x\cos x}{(x+1)^2}dx$ then the value of $M - N$ is I don't think we can directly solve both the definite integrals and get the answer because I checked these two with WolframAlpha and integration is too difficult.
What I tried is if I execute integration by parts on $M$ then:
$$\int\frac{\cos x}{x+2}dx = \frac{\sin x}{x+2} + \int \frac{\sin x}{(x+2)^2}dx + c$$
and $$N = \frac{1}{2}\int_{0}^{4}\frac{\sin 2x}{(x+1)^2}dx$$
Now we can see that the 2nd element of 1st equation(right side) is somewhat analogous to $N$. But I don't know how to get past this point because the limits and other things are different.
|
If the question is Like this way...
$$\displaystyle M=\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{x+2}dx$$ and $$\displaystyle N=\int_{0}^{\frac{\pi}{4}}\frac{\sin x\cdot \cos x}{(x+1)^2}dx\;,$$ Then find $M-N$
$\bf{Then\; Solution::}$ Given $$\displaystyle M=\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{x+2}dx$$ and $$\displaystyle N=\int_{0}^{\frac{\pi}{4}}\frac{\sin x\cdot \cos x}{(x+1)^2}dx$$
Now We can write $$\displaystyle N= \frac{1}{2}\int_{0}^{\frac{\pi}{4}}\frac{\sin 2x}{(x+1)^2}dx = \frac{1}{2}\int_{0}^{\frac{\pi}{4}}\sin 2x \cdot \frac{1}{(x+1)^2}dx$$
Now Uding Integration by parts, We get
$$\displaystyle N= -\frac{1}{2}\left[\sin 2x \cdot \frac{1}{(x+1)}\right]_{0}^{\frac{\pi}{4}}+\frac{1}{2}\cdot 2\int_{0}^{\frac{\pi}{4}}\cos 2x\cdot \frac{1}{(x+1)}dx$$
Now Let $2x=t\;,$ Then $\displaystyle dx = \frac{1}{2}dt$ and Changing limits, We get
$$\displaystyle N=-\frac{1}{2}\cdot \frac{4}{\pi+4}+\int_{0}^{\frac{\pi}{2}}\frac{2\cos t}{t+2}\cdot \frac{1}{2}dt=-\frac{2}{\pi+4}+\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{x+2}dx$$
Using $$\displaystyle \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$$
So We get $$\displaystyle M=-\frac{2}{\pi+4}+N\Rightarrow M-N = -\frac{2}{\pi+4}.$$
|
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|
Indefinite Integral of a function $$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$
I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.
|
$\int cx^n=c\int x^n=c\frac{x^{n+1}}{n+1}$ when $n\neq -1$
Thus, $\int \frac{1}{5}x^3=\frac{1}{20}x^4$
Also, by the same rule, $\int -2x=-x^2$
$\int \frac{1}{x}=\ln(x)$
Thus, $\int \frac{3}{x}=3\ln(x)$
$\int e^x=e^x$
Thus, by above, the required integral is
$F(x)=\frac{1}{20}x^4-x^2+3\ln(x)+e^x+c$
Done.
|
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|
How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$ How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$?
I try
$x=2r \cos \phi$,
$y=2r \sin \phi$,
$z=z$, but but probably not the way to go
|
Letting $r=\sqrt{x^2+y^2}$ gives $r^2+4z^2=4$ in the 1st equation and $z=r-2$ in the 2nd equation, so
substituting gives $r^2+4(r-2)^2=4$ and so $5r^2-16r+12=0$,
which gives $(5r-6)(r-2)=0$ so $r=\frac{6}{5}$ or $r=2$.
$\;\;$Using $z=r-2$, we have that
if $0\le r\le\frac{6}{5}, \;\;-\frac{1}{2}\sqrt{4-r^2}\le z\le \frac{1}{2}\sqrt{4-r^2}$ $\;\;\;$ and $\;\;\;$ if $\frac{6}{5}\le r\le2, \;\;r-2\le z\le \frac{1}{2}\sqrt{4-r^2}$.
Therefore $\displaystyle V=\int_0^{2\pi}\int_0^{\frac{6}{5}}\sqrt{4-r^2}r\;drd\theta+\int_0^{2\pi}\int_{\frac{6}{5}}^2\big(\frac{1}{2}\sqrt{4-r^2}-(r-2)\big)r\;drd\theta$.
|
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|
How can I finish integrating $\int {\sqrt{x^2-49} \over x} $ using trig substitution? $$\int {\sqrt{x^2-49} \over x}\,dx $$
$$ x = 7\sec\theta$$
$$ dx = 7\tan\theta \sec\theta \,d\theta$$
$$\int {\sqrt{7^2\sec^2\theta - 7^2} \over 7\sec\theta}\left(7\tan\theta \sec\theta \,d\theta\right) = \int \sqrt{7^2\sec^2\theta - 49} \left(\tan\theta d\theta\right)$$
$$ \int\sqrt{7^2(\sec^2\theta - 1)} (\tan\theta \,d\theta) = 7\int\sqrt{\sec^2\theta - 1} (\tan\theta \,d\theta)$$
$$ 7\int \tan^2\theta \,d\theta = 7\int \sec^2\theta - 1 \,d\theta $$
$$ 7\int \sec^2\theta - 7\int d\theta $$
$$ 7\tan\theta - 7\theta + C = 7(\tan\theta - \theta) + C$$
This makes: $$ \theta = \sec^{-1}\left(x \over 7\right)$$
And plugging back in to the indefinite integral:
$$ 7\left(\left(\sqrt{x^2-49} \over 7 \right) - \sec^{-1}\left(x \over 7 \right)\right) + C $$
My question really is, how can I evaluate $\sec^{-1}\left(x \over 7 \right)$ ?
|
If it matters, $\sec^{-1}(x) = \cos^{-1}\left(\frac 1x\right)$. No getting away from inverse trig, though.
|
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|
Inverse of piecewise function I've the following function:
$$f(x)=
\begin{cases}
12x+3, & \text{if $x\ge0$} \\
x+3, & \text{if $x\lt0$}
\end{cases}
$$
What will be its inverse?
For me is $f(x)^{-1}= \frac{x-3}{12}$ per $x\ge 0$ and $3-x$ for $x\lt 0$. Right?
|
To find the inverse of a function, we set $y=f(x)$ and try to solve for $x$ in terms of $y$. You can do this by direct manipulation of each case:
$$
\begin{align}
&y=
\begin{cases}
12x+3, & \text{if $x\ge0$} \\
x+3, & \text{if $x\lt0$}
\end{cases}\\
&\implies
\begin{cases}
y=12x+3, & \text{if $x\ge0$} \\
y = x+3, & \text{if $x\lt0$}
\end{cases}\\
&\implies
\begin{cases}
y=12x+3, & \text{if $12x+3\ge 3$} \\
y = x+3, & \text{if $x+3\lt 3$}
\end{cases}\\
&\implies
\begin{cases}
y=12x+3, & \text{if $y\ge 3$} \\
y = x+3, & \text{if $y\lt 3$}
\end{cases}\\
&\implies
\begin{cases}
x=\frac{y-3}{12}, & \text{if $y\ge 3$} \\
x=y-3, & \text{if $y\lt 3$}
\end{cases}\\
&\implies
x=\begin{cases}
\frac{y-3}{12}, & \text{if $y\ge 3$} \\
y-3, & \text{if $y\lt 3$}
\end{cases}\\
\end{align}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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Can't find solution for equation $100z=a(z+i)(z-i)^2(z-2)+b(z-i)^2(z-2)+c(z+i)^2(z-i)(z-2)+d(z+i)^2(z-2)+e(z+i)^2(z-i)^2$
I already got $b=5+10i, d=-5+10i $ and $e =8$ by eliminating the factors using $z=i, z=-i, z=-2$ but I cant get $a$ and $c$ with that method. What am I doing wrong?
|
$100z=a(z+i)(z-i)^2(z-2)+b(z-i)^2(z-2)+c(z+i)^2(z-i)(z-2)+d(z+i)^2(z-2)+e(z+i)^2(z-i)^2$
Setting $z=i$ gives
$$100i=d(i+i)^2(i-2)\quad\Rightarrow\quad d=-5+10i$$
Setting$z=-i$ gives
$$100(-i)=b(-i-i)^2(-i-2)\quad\Rightarrow\quad b=\color{red}{-}5\color{red}{-}10i$$
Setting $z=\color{red}{+}2$ gives
$$100\cdot 2=e(2+i)^2(2-i)^2\quad\Rightarrow\quad e=8$$
Setting $z=0$ gives
$$\begin{align}&0=ai(-i)^2(-2)+b(-i)^2(-2)+ci^2(-i)(-2)+di^2(-2)+ei^2(-i)^2\\&\Rightarrow 0=2ai+2b-2ci+2d+e\\&\Rightarrow 0=2ai+2(-5-10i)-2ci+2(-5+10i)+8\\&\Rightarrow c=a+6i\tag1\end{align}$$
Setting $z=1$ gives
$$\begin{align}&100=-a(1+i)(1-i)^2-b(1-i)^2-c(1+i)^2(1-i)-d(1+i)^2+e(1+i)^2(1-i)^2\\&\Rightarrow 50=-a(1-i)-b(-i)-c(1+i)-di+2e\\&\Rightarrow 50=-a(1-i)-(-5-10i)(-i)-c(1+i)-(-5+10i)i+16\\&\Rightarrow a(-1+i)+c(-1-i)=14\tag2\end{align}$$
From $(1)(2)$,
$$a(-1+i)+(a+6i)(-1-i)=14\quad\Rightarrow \quad a=-4-3i,\quad c=-4+3i.$$
|
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|
Solving a recurrence relation using generating functions My recurrence relation is
D(n) = D(n - 1) + D(n - 2) + 5(n - 1);
with the initial conditions D(2),D(3) being 6, 17 respectively.
The generating function G(z) for the sequence D(n) is given
I don't know how i got this. Please can anyone give explanation for this?
Thank you.
|
Note that we can work backwards from $D(2)$ and $D(3)$ to discover what $D(1)$ and then $D(0)$ should be if the recurrence is to hold for $D(2)$ and $D(3)$ as well. We find that if $D(0)=0$ and $D(1)=1$, the recurrence does indeed yield $D(2)=6$ and $D(3)=17$, so we can simplify matters by taking $D(0)=0$ and $D(1)=1$ as initial conditions.
There are a couple of slightly different ways to proceed from this point; I prefer the one used in Graham, Knuth, & Patashnik, Concrete Mathematics. Assume that $D(n)=0$ for all $n<0$. Then the recurrence
$$D(n)=D(n-1)+D(n-2)+5(n-1)+5[n=0]+[n=1]\tag{1}$$
holds for all integers $n$, where the terms in square brackets are Iverson brackets. Multiply $(1)$ by $z^n$ and sum over $n$:
$$\begin{align*}
\sum_nD(n)z^n&=\sum_nD(n-1)z^n+\sum_nD(n-2)z^n+5\sum_n(n-1)z^n+5+z\\\\
&=z\sum_nD(n-1)z^{n-1}+z^2\sum_nD(n-2)z^{n-2}+5\sum_nnz^n-5\sum_nz^n+5+z\\\\
&=z\sum_nD(n)z^n+z^2\sum_nD(n)z^n+\frac{5z}{(1-z)^2}-\frac5{1-z}+5+z\;.
\end{align*}$$
If $G(z)=\sum_nD(n)z^n$, we can now write
$$G(z)=zG(z)+z^2G(z)+\frac{5z}{(1-z)^2}-\frac5{1-z}+5+z$$
and collect terms in $G(z)$ on the lefthand side of the equation to get
$$(1-z-z^2)G(z)=\frac{5z}{(1-z)^2}-\frac5{1-z}+5+z\;.$$
As a quick check, note that
$$\begin{align*}
\frac{5z}{(1-z)^2}&-\frac{5}{1-z}+5+z\\
&=5(z+2z^2+3z^3+\ldots)-5(1+z+z^2+\ldots)+5+z\\
&=5(2z^2+3z^3+4z^4+\ldots)-5(z^2+z^3+z^4+\ldots)+z\\
&=z+5z^2+10z^3+15z^4+\ldots\;,
\end{align*}$$
agreeing nicely with
$$\begin{align*}
(1-z-z^2)G(z)&=(1-z-z^2)(z+6z^2+17z^3+38z^4+\ldots)\\
&=z+5z^2+10z^3+15z^4+\ldots\;.
\end{align*}$$
Thus,
$$G(z)=\frac1{1-z-z^2}\left(5+z-\frac5{1-z}+\frac{5z}{(1-z)^2}\right)\;.$$
This is not equal to your
$$\frac1{1-z-z^2}\left(6+11z+\frac{15z^2}{1-z}+\frac{5z^3}{(1-z)^2}\right)\;,$$
as you can check by noting that the constant term when you expand
$$6+11z+\frac{15z^2}{1-z}+\frac{5z^3}{(1-z)^2}$$
is $6$, not $0$.
If you change your function to
$$\frac1{1-z-z^2}\left(6z^2+11z^3-\frac{5z^2}{1-z}+\frac{5z^3}{(1-z)^2}\right)\;,$$
you get the generating function for the $D$ sequence with $D(0)$ and $D(1)$ arbitrarily set to $0$; if you change it to
$$\frac1{1-z-z^2}\left(6+11z-\frac{5}{1-z}+\frac{5z}{(1-z)^2}\right)\;,$$
you get the generating function for the same recurrence, but with initial conditions $D(0)=6$ and $D(1)=17$.
|
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|
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}$ I'm trying to solve evaluate this limit
$$\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}.$$
I've tried to rationalize the denominator but this is what I've got
$$\lim_{x\to\infty}(\sqrt{x-1} - \sqrt{x-2})({\sqrt{x-2} + \sqrt{x-3}})$$
and I don't know how to remove these indeterminate forms $(\infty - \infty)$.
EDIT: without l'Hospital's rule (if possible).
|
Note that
$$\sqrt{x-1}-\sqrt{x-2} = \dfrac1{\sqrt{x-1}+\sqrt{x-2}}$$
and
$$\dfrac1{\sqrt{x-2}-\sqrt{x-3}} = \sqrt{x-2}+\sqrt{x-3}$$
We hence have
$$\dfrac{\sqrt{x-1}-\sqrt{x-2}}{\sqrt{x-2}-\sqrt{x-3}} = \dfrac{\sqrt{x-2}+\sqrt{x-3}}{\sqrt{x-1}+\sqrt{x-2}}$$
We have
$$\dfrac{\sqrt{x-3}}{\sqrt{x-2}}<\dfrac{\sqrt{x-2}+\sqrt{x-3}}{\sqrt{x-1}+\sqrt{x-2}} < 1$$
Now conclude what you want.
|
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|
Showing $s_n = \left(\frac{1}{2}\right)(s_{n-1} + s_{n-2})$ is Cauchy. Let $(s_n)$ be a sequence defined as $s_1 = 1, s_2 = 2$ , and $s_n = \left(\frac{1}{2}\right)(s_{n-1} + s_{n-2})$. Prove that $(s_n)$ is Cauchy.
I can see how it is convergent and Cauchy but not sure how to put it into a formal proof.
|
We have
\begin{align}
s_n - s_{n-1} &= \dfrac{s_{n-1}+s_{n-2}}{2} - s_{n-1}
\\
&= -\dfrac{1}{2}(s_{n-1}-s_{n-2})
\\
&= \cdots
\\
&=(-\dfrac{1}{2})^{n-2}(s_2-s_1)
\\
\end{align}
Let $M=\dfrac{4}{3}|(s_2-s_1)|$, for any $\epsilon$, set $\space N=\log_2(\dfrac{M}{\epsilon})+1$, then for any $n>m, \space m>N,\space m,n\in \mathbb{N}$, there is
\begin{align}
|s_n - s_m| &= |(s_{n}-s_{n-1})+\cdots+(s_{m+1}-s_{m})|
\\
&=|(-\dfrac{1}{2})^{n-2}+\cdots+(-\dfrac{1}{2})^{m-1}||(s_2-s_1)|
\\
&= \left|\dfrac{(-\dfrac{1}{2})^{m-1}(1-(-\dfrac{1}{2})^{n-m})}{1+\dfrac{1}{2}}\right||(s_2-s_1)|
\\
&=\dfrac{2}{3}\dfrac{1}{2^{m-1}}\left|(1-(-\dfrac{1}{2})^{n-m})\right||(s_2-s_1)|
\\
&<\dfrac{M}{2^{m-1}}
\\
&<\epsilon
\end{align}
So $s_n$ is Cauchy sequence.
|
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|
How do I know when the limit of a function at a certain point doesn't exist? $$\lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8}$$
I have this limit. I can't use L'Hopital. After rationalizing the numerator I get:
$$\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(x-8)(\sqrt{7+\sqrt[3]{x}}+3)}$$
Isn't there anything left to do after that? How to know if the limit just doesn't exist?
EDIT: I typed the limit wrong.
|
\begin{align}
\lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8}
&=\lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8} \cdot \frac{\sqrt{7+\sqrt[3]{x}}+3}{\sqrt{7+\sqrt[3]{x}}+3}\tag{1}\\
&=\lim_{x \to 8} \frac{(7+\sqrt[3]{x})-9}{(x-8)(7+\sqrt[3]{x}+3)}\tag{2}\\
&=\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(x-8)(7+\sqrt[3]{x}+3)}\tag{3}\\
&=\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(\sqrt[3]{x}^3-2^3)(7+\sqrt[3]{x}+3)}\tag{4}\\
&=\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(\sqrt[3]{x}-2)(\sqrt[3]{x}^2+2\sqrt[3]{x}+4)(7+\sqrt[3]{x}+3)}\tag{5}\\
&=\lim_{x \to 8} \frac{1}{(\sqrt[3]{x}^2+2\sqrt[3]{x}+4)(7+\sqrt[3]{x}+3)}\tag{6}\\[6pt]
&=\frac{1}{(4+4+4)(7+4+3)}=\frac{1}{168}\tag{7}
\end{align}
Explanation:
$(2)$: I used $(x-y)(x+y)=(x^2-y^2)$
$(5)$: I used $(x^3-y^3)=(x-y)(x^2+xy+y^2)$
|
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|
What are two easiest numerical method to calculate $\sqrt[23]{123456789}$ by hand? What are two easiest numerical methods to calculate $\sqrt[23]{123456789}$ by hand?
I want to compare that two methods myself
Just name the methods
|
Note that $2^{23}=8388608$ and that $10^{23}=100000000000000000000000$
Divide by $8388608$ until a value less than $8388608$ is achieved. Then multiply by $100000000000000000000000$.
Repeat as long as you want until you achieve a value close to $1$.
I discovered that $123456789 = k \times 2^{23} \times \frac 1 {10^{23}}\times 2^{23}\times 2^{23}\times 2^{23}\times \frac 1 {10^{23}}...$
$... \times 2^{23}\times 2^{23}\times 2^{23}\times \frac 1 {10^{23}}\times 2^{23}\times 2^{23}\times 2^{23}\times 2^{23}$, where $k\approx 8.53$
Since $\sqrt [23] k \approx 1$,
$\sqrt [23] {123456789} \approx 2 \times \frac 1 {10} \times 2 \times 2 \times 2 \times \frac 1 {10} \times 2 \times 2 \times 2 \times \frac 1 {10} \times 2 \times 2 \times 2 \times 2=2.048$
|
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|
Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction
Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$.
Here's where I am right now:
Assume $n= k $ is correct:
$$12^k+2(5^{k-1}) = 7k.$$
Let $n= k+1 $:
$$12^{k+1} + 2(5^k)$$
$$12^k(12) + 2(5^k)$$
Any ideas on how I can proceed from here?
|
For $n=k$, we have
$$12^k+2\cdot 5^{k-1} = 7M$$
Multiplying by $12$ throughout, we obtain
$$12^{k+1}+24\cdot 5^{k-1} = 84M \implies 12^{k+1} + (2\cdot 5 + 14)\cdot 5^{k-1} = 84 M$$
which in-turn implies
$$12^{k+1} + 2\cdot 5^k +14\cdot5^{k-1} = 84M \implies 12^{k+1}+2\cdot 5^k = 14\left(6M-5^{k-1}\right)$$
Hence, the statement is true for $n=k+1$.
We have
$$12^n = (7+5)^n = \sum_{k=0}^{n} \dbinom{n}k 7^k \cdot 5^{n-k} = 5^n + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}$$
This gives us
\begin{align}
12^n + 2 \cdot 5^{n-1} & = 2\cdot 5^{n-1} + 5^n + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\\
& = \left(2+5\right)\cdot 5^{n-1} + 7 \cdot \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\\
& = 7 \cdot \left(5^{n-1} + \sum_{k=1}^{n} \dbinom{n}k 7^{k-1} \cdot 5^{n-k}\right)
\end{align}
|
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|
Limit of square root where x approaches infinity I have to calculate the following limit, and I wondered if my solution to the question was true. Here it is:
$$\lim _{x \to -\infty} (\sqrt{(1+x+x^2)}-\sqrt{1-x+x^2})$$
Now I divide by $x^2$ and get:
$$\lim _{x \to -\infty} (\sqrt{\frac{1}{x^2}+\frac{x}{x^2}+\frac{x^2}{x^2}}-\sqrt{\frac{1}{x^2}-\frac{x}{x^2}+\frac{x^2}{x^2}})$$
I know that $$\lim_{x \to -\infty}\frac{1}{x}=0$$ so I get the following:
$$\lim _{x \to -\infty} (\sqrt{0+0+1}-\sqrt{0-0+1})$$
So we get the following:
$$\lim _{x \to -\infty} (\sqrt{1}-\sqrt{1})=0$$
Is my solution correct?
Thanks.
|
$$\sqrt{1+x+x^2}-\sqrt{1-x+x^2}=\frac{(1+x+x^2)-(1-x+x^2)}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$
$$=\frac{2x}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}\stackrel{(1)}=\frac{-2}{\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}+\sqrt{\frac{1}{x^2}-\frac{1}{x}+1}}$$
$$\stackrel{x\to -\infty}\to \frac{-2}{\sqrt{0+0+1}+\sqrt{0-0+1}}=-1$$
$(1)\!\! :\,$ I divide numerator and denominator by $|x|=\sqrt{x^2}$ and assume $x<0$, which I can do because we only care about what happens when $x\to -\infty$.
|
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|
Progression from indefinite integral to definite integral - $\int_{0}^{2\pi}\frac{1}{5-3\cos x} dx$ I'm trying to evaluate the following integral:
$$\int_{0}^{2\pi}\frac{1}{5-3\cos x} dx$$
We can evaluate indefinite one first - $\int\frac{1}{5-3\cos x}dx = \frac{1}{2}\tan^{-1}(2\tan(\frac{x}{2})) + C$. The problem is that $\frac{1}{2}\tan^{-1}(2\tan(\frac{2\pi}{2}))-\frac{1}{2}\tan^{-1}(2\tan(\frac{0}{2}))=0$ but there's a hint on this excercise that the value of this definite integral is greater than $0$. So, what went wrong? What's the trap I've fallen into during evaluation of this integral?
|
the graph of $\cos x$ is symmetric about $x = \pi,$ therefore $$\int_0^{2\pi}\frac{dx}{5-3\cos x} = 2 \int_0^{\pi}\frac{dx}{5-3\cos x} = \tan^{-1}\left(2\tan(x/2)\right)\Big|_0^{\pi} =\frac{\pi}2.$$
|
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|
Need a more compact formula This is a part of solution of a programming contest problem
$$\sum_{i=0}^{k} {x-i \choose 2} $$ given $x-i \ge 2$ is always true.
for
$k=1$,$(x-1)^2$
$k=2$, $(x-1)^2+((x-2)*(x-3)/2)$
$k=3$, $(x-1)^2 + (x-3)^2$
and so on.
Is there a reduced form of this?
|
Since
$$(x-i)(x-i-1)=\frac13\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x-i-1)(x-i)\right)$$
one has
$$\begin{align}\sum_{i=0}^{k}\binom{x-i}{2}&=\sum_{i=0}^{k}\frac{(x-i)(x-i-1)}{2}\\&=\frac 12\sum_{i=0}^{k}\frac 13\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x-i-1)(x-i)\right)\\&=\frac 16\sum_{i=0}^{k}\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x-i-1)(x-i)\right)\\&=\frac 16\left((x-1)x(x+1)-(x-k-2)(x-k-1)(x-k)\right)\\&=\frac{k+1}{2}x^2-\frac{(k+1)^2}{2}x+\frac{k(k+1)(k+2)}{6}\end{align}$$
|
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|
finding determinants using different properties The equation is as follows:
$\operatorname{det}(2A^{-1} + 7\operatorname{adj}(A))$
Here I know that $\operatorname{det}(A^{-1}) = (\operatorname{det}(A))^{-1}$ and $\operatorname{det}(kA) = k^n \operatorname{det}(A)$
using these, we know that $\operatorname{det}((2A)^{-1} )= \frac{1}{4} \operatorname{det}(A)^{-1} =\frac{1}{8}$
I do not know a formula that relates to adjoints in this case besides maybe:
$$A \operatorname{adj}(A) = \operatorname{det}(A)I$$
|
Look at this:
since
$A\operatorname{adj}(A) = \det(A)I, \tag{1}$
we have
$\operatorname{adj}(A) = \det(A) A^{-1} \tag{2}$
provided $A^{-1}$ exists. Well, since we are given that (see comments above)
$\det(A) = 2, \tag{3}$
we know $A^{-1}$ does indeed exist, so using (2) we have
$2A^{-1} + 7\operatorname{adj}(A) = 2A^{-1} + 7\det(A)A^{-1} = (2 + 7\det(A))A^{-1}$
$= (2 + 7(2))A^{-1} = 16A^{-1}, \tag{4}$
whence
$\det(2A^{-1} + 7\operatorname{adj}(A)) = \det(16A^{-1}) = (16)^n (\det(A))^{-1},\tag{5}$
where $n = \operatorname{size}(A)$. Now since
$\operatorname{det}((2A)^{-1} )= \frac{1}{4} \operatorname{det}(A)^{-1} =\frac{1}{8}, \tag{6}$
with
$\det(2A) = 2^n \det(A) \tag{7}$
and
$\det((2A)^{-1}) = (\det(2A))^{-1}, \tag{8}$
we see that
$2^{-n} = \dfrac{1}{4}, \tag{9}$
which shows that $n = 2$; thus (5) yields
$\det(2A^{-1} + 7\operatorname{adj}(A)) = \det(16A^{-1})$
$= (16)^2(\det(A))^{-1} = 256 (\det(A))^{-1} = \dfrac{256}{2} = 128.\tag{10}$
|
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|
Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the solution for x.
a)
$$\log_{2} x + \log_{3} x = \log_{4} x \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{\log_{2} 4} \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{2} \Leftrightarrow$$
$$\frac{2 \log_{2} 3 \log_{2}x + 2\log_{2} x - \log_{2}3 \log_{2}x}{2 \log_{2}3} = 0 \Leftrightarrow $$
Moving the denominator over and solving for $\log_{2} x$
$$\frac{\log_{2} x (2\log_{2}3 + 1 - \log_{2}3)}{2\log_{2} 3} = 0 \Leftrightarrow$$
$$\log_{2} x = 0 \Rightarrow x = 2^{0} = 1$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
$$\log_{2} x \frac{\log_{2} x}{\log_{2} 3} - \frac{\log_{2} x}{\log_{2} 4} = 0 \Leftrightarrow$$
$$\frac{4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x}{4 \log_{2}3} = 0 \Leftrightarrow $$
$$4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x = 0 \Leftrightarrow $$
Substituting $t = \log_{2} x$ gives:
$$4t^2 - \log_{2}3t = 0 \Rightarrow$$
$$t^2 - \frac{\log_{2}3t}{4} = 0 \Rightarrow$$
$$t \left(t- \frac{\log_{2}3}{4}\right) = 0$$
$$t_{1} = 0 \Rightarrow x = 1$$
$$t_{2} = \frac{\log_{2}3}{4} \Rightarrow x = 2^{\frac{\log_{2}3}{4}}$$
However, the second solution here should be $\sqrt{3}$, so I must have made a mistake somewhere. Any suggestions?
|
Your mistake in part (b) is on line three where you replaced $\log_24$ with $4$ when it should be replaced by $2$
|
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"url": "https://math.stackexchange.com/questions/1285148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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|
For which primes is $-2$ a quadratic residue? For which primes is $-2$ a quadratic residue?
We are trying to find primes that have solution for $x^2 \equiv -2 \mod p.$ Using the Lagrange symbol I know that $2$ is a quadratic residue when $p \equiv 1$ or $7 \mod 8,$ but when is $-2$ a quadratic residue?
Work so far:
$\left(\dfrac{-2}{p} \right) = \left(\dfrac{-1}{p} \right)\left(\dfrac{2}{p} \right).$
We have that
$\left(\dfrac{-1}{p}\right) = \begin{cases} 1 & : p \equiv 1 \pmod{8} \text{ or } p \equiv 5 \pmod{8} \\ -1 & : p \equiv 3 \pmod{8} \text{ or } p \equiv 7 \pmod{8} \end{cases}$
and
$\left(\dfrac{2}{p} \right) = \left\{
\begin{array}{lr}
1 & : p \equiv 1 \pmod 8 \text{ or } p \equiv 7 \pmod 8 \\
-1 & : p \equiv 3 \pmod 8 \text{ or } p \equiv 5 \pmod 8
\end{array}
\right.$
If $p \equiv 5 \pmod 8,$ then $\left(\dfrac{-2}{p} \right) = \left(\dfrac{-1}{p} \right)\left(\dfrac{2}{p} \right) = 1 \cdot -1 = -1,$ so $-2$ is not a quadratic residue for this particular case. If $p \equiv 7 \mod 8,$ then $\left(\dfrac{2}{p} \right) = 1$ and $\left(\dfrac{-1}{p} \right) = -1 \Longrightarrow \left(\dfrac{-2}{p} \right) = -1 \cdot 1 = -1.$ Hence $-2$ is not a quadratic residue in this particular case either. But if $p \equiv 1 \mod 8$ or $p \equiv 3 \mod 8,$ then $\left(\dfrac{-2}{p} \right) = 1 \cdot 1 = 1$ or $\left(\dfrac{-2}{p} \right) = -1 \cdot -1 = 1.$ Thus $-2$ is a quadratic residue for primes congruent to $1$ or $3$ modulo $8.$
|
Hint:
$$
\left(\frac{-1}{p}\right) = \begin{cases} 1 & : p \equiv 1 \pmod{8} \text{ or } p \equiv 5 \pmod{8} \\ -1 & : p \equiv 3 \pmod{8} \text{ or } p \equiv 7 \pmod{8} \end{cases}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1287274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
General Solution of ODE (complex eigenversion) I am trying to figure out the general solution to the following matrix:
$ \frac{d\mathbf{Y}}{dt} = \begin{pmatrix} -3 & -5 \\ 3 & 1 \end{pmatrix}\mathbf{Y}$
I got a solution, but it is so complex I am not sure, if it's even remotely right....:
My Solution:
$x = 1/3*k_1*(-\sqrt{11}*cos(\sqrt{11}*t))-2*sin(\sqrt{11}*t)+1/3*k_2*(\sqrt{11}*sin(\sqrt{11}*t)-2*cos(\sqrt{11}*t))$
$y = k_1*sin(\sqrt{11}*t)+k_2*cos(\sqrt{11}*t)$
After that my task is to solve the inital-value problem for the same matrix with:
$Y_0=(4, 0)$
I have an idea where to start, should I find the correct solution to the general solution problem, but it looks so complex, that it overwhelms me...
|
The coefficient matrix $A$ has characteristic polynomial
\begin{align}
\left|\begin{array}{cc}
\lambda+3 & 5 \\
-3 & \lambda-1
\end{array}\right|
& = (\lambda+3)(\lambda-1)+15 \\
& = \lambda^{2}+2\lambda+12 \\
& = (\lambda+1)^{2}+11 \\
& = (\lambda+1+i\sqrt{11})(\lambda+1-i\sqrt{11}).
\end{align}
So the eigenvalues of the coefficient matrix are $-1\pm i\sqrt{11}$. There is one eigenvector correspondending to each eigenvalue. If $A$ is the coefficient matrix in the differential equation, then
$$
A-(-1+i\sqrt{11})I=\left[\begin{array}{cc}
-2-i\sqrt{11} & -5 \\
3 & 2-i\sqrt{11}
\end{array}\right]
$$
gives an eigenvector
$$
X_{+} = \left[\begin{array}{c}
-5 \\ 2+i\sqrt{11}
\end{array}\right]
$$
Similarly,
$$
A-(-1-i\sqrt{11})I = \left[\begin{array}{cc}
-2+i\sqrt{11} & -5 \\
3 & 2+i\sqrt{11}
\end{array}\right]
$$
gives an eigenvector
$$
X_{-}= \left[\begin{array}{c}
-5 \\
2-i\sqrt{11}
\end{array}\right]
$$
So the general solution of the equation is
$$
P(t) = \alpha e^{-t+i\sqrt{11}t}X_{+} + \beta e^{-t-i\sqrt{11}t}X_{-}
$$
The solution for $Y_{0}=(4,0)$ is obtained by solving for $\alpha$, $\beta$ such that
$$
\alpha\left[\begin{array}{c} -5 \\ 2+i\sqrt{11}\end{array}\right]+\beta\left[\begin{array}{c} -5 \\ 2-i\sqrt{11}\end{array}\right]=\left[\begin{array}{c} 4 \\ 0 \end{array}\right] \\
\left[\begin{array}{c} \alpha \\ \beta \end{array}\right]
= \frac{1}{10i\sqrt{11}}\left[\begin{array}{cc}
2-i\sqrt{11} & 5 \\
-2-i\sqrt{11} & -5\end{array}\right]\left[\begin{array}{c}4 \\ 0\end{array}\right]
= \frac{2}{5i\sqrt{11}}\left[\begin{array}{c}2-i\sqrt{11} \\ -2-i\sqrt{11}\end{array}\right]
$$
Therefore,
$$
P(t) = \frac{2(2-i\sqrt{11})}{5i\sqrt{11}}e^{-t+i\sqrt{11}t}X_{+}
-\frac{2(2+i\sqrt{11})}{5i\sqrt{11}}e^{-t-i\sqrt{11}t}X_{-}.
$$
The first coordinate of $P(t)$ is
$$
-e^{-t}\left[\frac{4}{i\sqrt{11}}\{e^{i\sqrt{11}t}-e^{-i\sqrt{11}t}\}\right]+e^{-t}\left[2\{e^{i\sqrt{11}t}+e^{-i\sqrt{11}t}\}\right] \\
= -\frac{8}{\sqrt{11}}e^{-t}\sin(\sqrt{11}t)+4e^{-t}\cos(\sqrt{11}t).
$$
The second coordinate of $P(t)$ is
$$
\frac{2(2-i\sqrt{11})(2+i\sqrt{11})}{5i\sqrt{11}}e^{-t+i\sqrt{11}t}
-\frac{2(2+i\sqrt{11})(2-i\sqrt{11})}{5i\sqrt{11}}e^{-t-i\sqrt{11}t} \\
= \frac{12}{\sqrt{11}}e^{-t}\sin(\sqrt{11}t)
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1288783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$ \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} $ Can someone show me how is possible to prove that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15}
\end{equation*}
but without Taylor series. One can use L'Hospital rule if necessary. I was not able.
|
L'Hospital once the fraction becomes
$$\frac{2\sin^2{x}+\tan^2{x}-3x^2}{7x^6}$$
Still an indeterminate
L'hospital 6 more times and rearranging the fraction becomes
$$\frac{4\cos{2x}}{315}+\sec^8{x}-\frac{4\sec^6{x}}{3}+\frac{2\sec^4{x}}{5}-\frac{4\sec^2{x}}{315}$$
And the limit for $x\to 0$ is $1-\frac{4}{3}+\frac{2}{5}=\frac{1}{15}$
7 times l'Hospital is so much of a pain: Taylor in this case is more convenient
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1289063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Problems with a Simple Differential Equation I am trying to solve the following: $y' = (y-5)(y+5)$ if $y(4) = 0$.
So far, I have tried separating the variables and then use partial fractions and have followed these steps:
(1) $A(y+5) + B (y-5) =1$.
(2) For $y = -5$ we have $B = \dfrac{-1}{10}$.
(3) For $ y= 5$, we have $A = \dfrac{1}{10}$.
(4) so we have $\dfrac{1}{10}[ \ln|y^2 - 25|] = x + C$
(5) After a chain of algebraic operations, we have $y = \pm \sqrt{e^{10x + 10C} + 25}$.
However, if I plug in my initial condition, then I get: $0 = e^{40 + 10C} + 25$ which implies $\ln(-25) = 40 + C$ which gets into complex numbers. Any help anyone can provide on where my reasoning went wrong is appreciated.
|
Let's start by decomposing $\frac{1}{(y-5)(y+5)}$ into partial fractions.
We want to to find $A$ and $B$ such that
$$\frac{1}{(y-5)(y+5)} = \frac{A}{(y-5)} + \frac{B}{(y+5)} = \frac{A(y+5)}{(y-5)(y+5)} + \frac{B(y-5)}{(y-5)(y+5)}.$$
Since we have the same denominator for all terms, the numerators of the fractions must then satisfy
$$\tag{0} 1 = A(y+5) + B(y-5)$$
Now, if we take $y=5$ in $(0)$, we get
$$1 = A (5+5) + B(5-5) \Rightarrow A = 1/10.$$
On the other hand, by taking $y=-5$ in $(0)$, we get
$$1 = A (-5+5) + B(-5-5) \Rightarrow B = -1/10.$$
Therefore,
$$\frac{1}{(y-5)(y+5)} = \frac{A}{(y-5)} + \frac{B}{(y+5)} = \frac{1}{10(y-5)} - \frac{1}{10(y+5)}.$$
We are now able to solve the differential equation, by noting that
$$\frac{dy}{dx} = (y-5)(y+5) \Rightarrow \frac{1}{(y-5)(y+5)}dy = dx \Rightarrow \left(\frac{1}{10(y-5)} - \frac{1}{10(y+5)} \right) dy = dx.$$
Integrating this expression on both sides, we get
$$\tag{1} \frac{1}{10}\left(\ln|y-5| - \ln|y+5|\right) = x + C \Rightarrow \ln{\left|\frac{y-5}{y+5} \right|} = 10x + \hat C.$$
Exponentiating both sides of $(1)$, we arrive at
$$\tag{2} \left|\frac{y-5}{y+5} \right| = e^{10x + \hat C} = e^{10x} \cdot \underbrace{e^{\hat C}}_{=K} = Ke^{10x}.$$
Now, plugging the initial condition, $y(4)=0$, in $(2)$, we finally get
$$\underbrace{\left|\frac{0-5}{0+5} \right|}_{=1} = K \cdot e^{10\times4} \Rightarrow K = e^{-40}.$$
If you really want to solve explicitly for $y$, you need to know the following identity
$$\tag{3} \ln{\left| \frac{1+u}{1-u} \right|} = 2 \, \text{arctanh}(u), \; |u|<1.$$
Then,
$$\frac{1}{10}\left(\ln|y-5| - \ln|y+5|\right) = x + C \Rightarrow \ln{\left( |-5| \left|1-\frac{y}{5} \right| \right)} - \ln{\left( |5| \left|1+\frac{y}{5} \right| \right)} = 10x + \hat C,$$
and, resorting to identity $(3)$, we get
$$\ln{\left| \frac{1-\frac{y}{5}}{1+\frac{y}{5}} \right|} = \ln{\left| \left( \frac{1+\frac{y}{5}}{1-\frac{y}{5}} \right)^{-1} \right|} = -\ln{\left| \frac{1+\frac{y}{5}}{1-\frac{y}{5}} \right|} = - 2 \, \text{arctanh}\left( \frac{y}{5} \right), \; |y|<5.$$
Therefore,
$$\tag{4}\text{arctanh}\left( \frac{y}{5} \right) = \bar C - 5x.$$
Applying $\tanh$ on both sides of $(4)$, we arrive at
$$\frac{y}{5} = \tanh{(\bar C - 5x)} \Rightarrow y = 5\tanh{(\bar C - 5x)}.$$
Now, plugging the initial condition, $y(4)=0$, we finally get
$$0 = \tanh{(\bar C - 5 \cdot 4)} \Rightarrow \bar C - 20 = 0 \Rightarrow \bar C = 20,$$
so
$$y(x) = 5 \tanh{(20 - 5x)}.$$
PS: Another way to solve this problem was to notice that $\frac{d}{du} \text{arctanh}(u) = \frac{1}{1-u^2}$, so all we had to do was to write the inital differential equation in that form.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1290586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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|
Find $Z$ transform of given signal Given the discrete signal $h(n)=r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}$ if $n \geq 0$ and $h(n)=0$ otherwise, find the $Z$ transform of $h(n)$.
What I did:
We know that $H(z)=\sum_{n=0}^{\infty}h(n)z^{-n}=\sum_{n=0}^{\infty}r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}z^{-n}$ which is equal to
\begin{equation*}
\frac{1}{\sin \theta}\sum_{n=0}^{\infty}r^n[\sin{(n\theta)}\cos \theta+\cos{(n\theta)}\sin \theta]z^{-n}=\cot \theta\sum_{n=0}^{\infty}r^n \sin(n\theta)z^{-n}+\sum_{n=0}^{\infty}r^n\cos(n\theta)z^{-n}
\end{equation*}
and now I am stuck and don't know how to continue. We wish it was some sort of geometric series in the sum, so we will know a closed form for it, but sadly the $\sin(n\theta)$ and $\cos(n\theta)$ ruin it for us.
How do we continue?
|
The fastest way to find the Z-transform of $h(n)=r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}$ is to utilize the Chebyshev polynomials of the second kind. These polynomials are given as
\begin{align}
U_{n}(\cos \theta) = \frac{\sin(n+1)\theta}{\sin \theta}
\end{align}
with the generating function
\begin{align}
\sum_{n=0}^{\infty} U_{n}(\cos \theta) \, t^{n} = \frac{1}{1 - 2 \cos\theta \, t + t^{2}}.
\end{align}
From these components the desired Z-transform is given by
\begin{align}
\mathcal{Z}\{h(n)\} &= \sum_{n=0}^{\infty} \frac{\sin(n+1)\theta}{\sin\theta} \, \left(\frac{r}{z}\right)^{n} = \frac{z^{2}}{z^{2} - 2 \cos\theta \, r z + r^{2}}.
\end{align}
As to the question of the problem proposed consider the series
$$S_{1} = \sum_{n=0}^{\infty}r^n\cos(n\theta)z^{-n}.$$
The evaluation is as follows.
\begin{align}
S_{1} &= \sum_{n=0}^{\infty}r^n\cos(n\theta)z^{-n} \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} \left( e^{i n \theta} + e^{- i n \theta} \right) \, \left(\frac{r}{z}\right)^{n} \\\
&= \frac{1}{2} \, \left[ \sum_{n=0}^{\infty} \left(\frac{e^{i\theta} r}{z}\right)^{n} + \sum_{n=0}^{\infty} \left(\frac{e^{-i\theta} r}{z}\right)^{n} \right] \\
&= \frac{1}{2} \left[ \frac{z}{z - e^{i\theta} r} + \frac{z}{z - e^{-i\theta} r} \right] \\
&= \frac{z^{2} - r z \, \cos\theta}{z^{2} - 2 r z \, \cos\theta + r^{2}}.
\end{align}
The sine series follows the same pattern and the resulting Z-transform is identical to the one presented above.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find the length of angle bisector $AD$.
In $\triangle ABC$ , the internal bisector of the angle $\angle A$ meets $BC$ at $D$. If $AB=4$, $AC=3$ and $\angle A=60^{\circ}$, then the length of $AD$ is
$a.)\ 2\sqrt3\\
\color{green}{b.)\ \dfrac{12\sqrt3}{7}}\\
c.)\ \dfrac{15\sqrt3}{8}\\
d.)\ \dfrac{6\sqrt3}{7}\\$
With the help of cosine rule i found $BC$
Then with the relation $\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac43$,
I found $BD$ then again applying cosine rule in $\triangle ABD$ i found $AD$.
But i would like to know it there is another short simple way.
I have studied maths up to $12th$ grade.
|
Let the length of AD be $x$ & $\angle ADC=\theta$
Apply law of sine in $\Delta ABD$ $$\frac{\sin(\pi-\theta)}{AB}=\frac{\sin 30^o}{BD} \implies BD=\frac{4}{2\sin\theta}=\frac{2}{\sin\theta}$$
Apply law of cosine in $\Delta ABD$ $$BD=\sqrt{4^2+x^2-2(4)(x)\cos 30^o}\implies \frac{2}{\sin\theta}=\sqrt{x^2-4x\sqrt{3}+16}\tag 1$$
Similarly, apply law of sine in $\Delta ADC$ $$\frac{\sin\theta}{AC}=\frac{\sin 30^o}{DC} \implies DC=\frac{3}{2\sin\theta}$$
Now, apply law of cosine in $\Delta ADC$ $$DC=\sqrt{3^2+x^2-2(3)(x)\cos 30^o}\implies \frac{3}{2\sin\theta}=\sqrt{x^2-3x\sqrt{3}+9}\tag 2$$ Now, dividing the eq (1) by (2), we get
$$\frac{4}{3}=\frac{\sqrt{x^2-4x\sqrt{3}+16}}{\sqrt{x^2-3x\sqrt{3}+9}}\implies 7x^2-12\sqrt{3}x=0 \implies x=\frac{12\sqrt{3}}{7} \space (x>0)$$ Hence, the length of AD is $\frac{12\sqrt{3}}{7}$ Thus option (b) is correct.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1294749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How to solve $12-\sin(\theta)=\cos(2\theta)$? $$12-\sin(\theta)=\cos(2\theta)$$
What's the correct answer on the $[0,2\pi]$?
I started with $12-\sin(\theta)=1-2\sin^2(\theta)$ and then i cant get anything sensible as i end up with $12=\sin(\theta)+2\sin^2(\theta)$
|
$$12-\sin(x)=\cos(2x)\Longleftrightarrow$$
$$12-\cos(2x)-\sin(x)=0\Longleftrightarrow $$
$$11-\sin(x)+2\sin^2(x)=0\Longleftrightarrow $$
$$\frac{11}{2}-\frac{\sin(x)}{2}+\sin^2(x)=0\Longleftrightarrow $$
$$\sin^2(x)-\frac{\sin(x)}{2}=-\frac{11}{2}\Longleftrightarrow $$
$$\frac{1}{16}-\frac{\sin(x)}{2}+\sin^2(x)=-\frac{87}{16}\Longleftrightarrow $$
$$\left(\sin(x)-\frac{1}{4}\right)^2=-\frac{87}{16}\Longleftrightarrow $$
$$\sin(x)-\frac{1}{4}=\pm \frac{i\sqrt{87}}{4}\Longleftrightarrow $$
$$\sin(x)=\pm \frac{i\sqrt{87}}{4}+\frac{1}{4}\Longleftrightarrow $$
So now the solutions:
$$x_1=2\left(\pi n+\tan^{-1}\left(\frac{1}{22}-\frac{i\sqrt{87}}{22}\pm \frac{1}{2}\sqrt{-\frac{570}{121}-\frac{2i\sqrt{87}}{121}}
\right)\right)$$
$$x_2=2\left(\pi n+\tan^{-1}\left(\frac{1}{22}+\frac{i\sqrt{87}}{22}\pm \frac{1}{2}\sqrt{-\frac{570}{121}-\frac{2i\sqrt{87}}{121}}
\right)\right)$$
With $n$ is the element of Z -> the set of integers
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Compute $\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^3+1}} + \frac{1}{\sqrt{n^3+4}} + \cdots + \frac{1}{\sqrt{n^3+n^2}}\right)$ How do I evaluate the following limit?
I guess I should do a comparison, but I've got no clue about what to do. Could you give me a hand?
$$\lim_{n \to \infty}\left( \frac{1}{\sqrt{n^3+1}} + \frac{1}{\sqrt{n^3+4}}
+ \cdots + \frac{1}{\sqrt{n^3+n^2}}\right)$$
|
Easily we can find that, $$\frac{1}{\sqrt{n^3+n^2}}\le \frac{1}{\sqrt{n^3+k^2}}\le\frac{1}{\sqrt{n^3}}.$$for all $k=1,2,...,n$.
Taking summation, $$\sum_{k=1}^n\frac{1}{\sqrt{n^3+n^2}}\le \sum_{k=1}^n\frac{1}{\sqrt{n^3+k^2}}\le \sum_{k=1}^n\frac{1}{\sqrt{n^3}}$$
$$\implies \frac{n}{\sqrt{n^3+n^2}}\le \sum_{k=1}^n\frac{1}{\sqrt{n^3+k^2}}\le \frac{n}{\sqrt{n^3}}$$
Then , by Sandwich theorem , $$\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{\sqrt{n^3+k^2}}=0.$$
|
{
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"url": "https://math.stackexchange.com/questions/1296042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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.