Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:
$$|z-2i| = 2|z+3| \Leftrightarrow \\
|x+yi-2i|=2|x+yi+2|\Leftrightarrow \\
\sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\
\sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\
y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\
-3y^2-4y=5x^2+24x+26 \Leftrightarrow \\
???$$
What do I do next?
| I assume this means locus and not plane..
$$ \begin{align*}
|z-2i|&= 2|z+3|\\
|x+iy-2i|&= 2|x+iy+3| \\
|x+i(y-2)|&= 2|(x+3)+i(y)| \\
\sqrt{x^2+(y-2)^2}&=2\sqrt{(x+3)^2+y^2}\\
x^2+(y-2)^2&=4\left(x^2+6x+9+y^2\right)\\
x^2+y^2-4y+4&=4x^2+24x+36+4y^2 \\
3x^2+24x+32+3y^2+4y&=0\\
x^2+8x+\frac{32}{3}+y^2+\frac{4}{3}y&=0 \\
x^2+8x+y^2+\frac{4}{3}y &= -\frac{32}{3}\\
x^2+8x+16 +y^2+\frac{4}{3}y +\frac{4}{9}&= -\frac{32}{3}+16+\frac{4}{9} \\
(x+4)^2+\left(y+\frac{2}{3}\right)^2&=\frac{52}{9}
\end{align*}$$
Which means our locus is a circle with centre $\left(-4,\frac{-2}{3}\right)$ and radius $\frac{2\sqrt{13}}{3}$.
| {
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"url": "https://math.stackexchange.com/questions/2279971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof without words: $1+8\times\text{triangular number}$ is an odd perfect square A recent question asked how to show that $8T_n+1$ is a perfect square if $T_n$ is a triangular number. This follows immediately from $T_n=\frac12 n(n+1)\implies 8T_n+1=4n^2+4n+1=(2n+1)^2$.
Can this be proven without words?
| $$
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$$
| {
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An example of evaluating an integral using partial fraction decomposition
How do you evaluate the integral
\begin{equation*}
\int \frac{3x^3+ 7x^2+ 5x+ 3}{(x+1)^2(x^2+1)}\,\mathrm{d}x
\,?
\end{equation*}
This is just an exercise that I wanted to have typed up for my class. Maybe it'll help other students to see an example worked out.
| Notice that if we were to expand the denominator of the integrand, the highest power of $x$ would be $4$, which is less than the highest power in the numerator. So this expression is ripe to be rewritten with partial fraction decomposition. Since $x^2+1$ cannot be factored, the decomposition must look like
\begin{equation*}
\frac{3x^3+ 7x^2+ 5x+ 3}{(x+1)^2(x^2+1)}
= \frac{A}{(x+1)} + \frac{B}{(x+1)^2}+ \frac{Cx+D}{(x^2+1)}
\,.
\end{equation*}
Multiplying out the right-hand side, we get that
\begin{align*}
&\frac{A}{(x+1)} + \frac{B}{(x+1)^2}+ \frac{Cx+D}{(x^2+1)}
\\\\=\quad &\frac{A(x+1)(x^2+1) + B(x^2+1) + (Cx+D)(x+1)^2}{(x+1)^2(x^2+1)}
\\\\=\quad &\frac{(A+C)x^3+(A+B+2C+D)x^2+(A+C+2D)x+(A+B+D)}{(x+1)^2(x^2+1)}
\,.
\end{align*}
Equating the coefficients in this expression with the coefficients in the original expression that we want this one to be equal to, we get the following system of equations:
\begin{cases}
A+C=3
\\ A+B+2C+D=7
\\ A+C+2D = 5
\\ A+B+D=3
\end{cases}
Now we've just got to play around with these equations to solve for our unknowns. This can be done many ways. One way to do it would be to notice that the second and fourth lines are quite similar. If we take the second line and subtract the fourth line, we get that $2C=4$, so $C=2$. Now looking at line one, $A=1$. Looking at the third line, $D=1$. And since $A=D=1$, looking at the fourth line, $B=1$ too. These values for $A$,$B$,$C$, and $D$ are very nice. It's like someone planned this problem to work out nicely.
Anyways, our decomposition must be
\begin{equation*}
\frac{3x^3+ 7x^2+ 5x+ 3}{(x+1)^2(x^2+1)}
= \frac{1}{(x+1)} + \frac{1}{(x+1)^2}+ \frac{2x+1}{(x^2+1)}
\,,
\end{equation*}
and we can now easily evaluate the integral:
\begin{align*}
&\int \frac{3x^3+ 7x^2+ 5x+ 3}{(x+1)^2(x^2+1)}\,\mathrm{d}x
\\\\=\quad &\int\left(\frac{1}{(x+1)} + \frac{1}{(x+1)^2}+ \frac{2x+1}{(x^2+1)}\right)\,\mathrm{d}x
\\\\ =\quad &\int\frac{1}{(x+1)}\,\mathrm{d}x + \int\frac{1}{(x+1)^2}\,\mathrm{d}x+ \int\frac{2x+1}{(x^2+1)}\,\mathrm{d}x
\\\\ = \quad &\int\frac{1}{(x+1)}\,\mathrm{d}x + \int\frac{1}{(x+1)^2}\,\mathrm{d}x+ \int\frac{2x}{(x^2+1)}\,\mathrm{d}x + \int\frac{1}{(x^2+1)}\,\mathrm{d}x
\\\\ =\quad &\ln|x+1| - \frac{1}{x+1}+ \ln|x^2+1| + \arctan(x) + C
\end{align*}
And of course if we wanted to, we could go crazy with logarithms and write this as
\begin{align*}
&\ln|x+1| - \frac{1}{x+1}+ \ln|x^2+1| + \arctan(x) + C
\\\\=\quad &\ln\left|C(x^3+x^2+x+1)e^{\arctan(x) - \frac{1}{x+1}}\right|
\,.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Multivariable limit using polar coordinates?
$$\lim_{(x,y)\to(0,0)} \frac {2\sin (x^2 + y^2) + y^3}{3x^2+3y^2}$$
I am actually very new to polar coordinates so I'm a little bit confused about this part. I substitute this into this:
$\frac {2sin (r^2) + (r cos \theta)^3}{3r^2}$
But I am not even sure if it's right and what should I do after this.
Does the limit even exist? I need an explanation. Thank you!
| \begin{align}
\lim_{(x,y)\rightarrow (0,0)} \frac{2\sin(x^2+y^2)+y^3}{3x^2+3y^2} &= \lim_{r\rightarrow 0} \frac{2\sin(r^2)+r^3\sin^3\theta}{3r^2}\\
&=\lim_{r\rightarrow 0} \frac{2\sin(r^2)}{3r^2} + \lim_{r\rightarrow 0}\frac{r^3\sin^3\theta}{3r^2}\\
&= \frac{2}{3} + \lim_{r\rightarrow 0}\frac{r\sin^3\theta}{3}\\
&= \frac{2}{3} + 0\\
&=\frac{2}{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the sum $\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}$ I want to calculate the sum : $$\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1} $$
I differentiated it, so $$\sum_{n=1}^{\infty}x^{3n}$$
by putting $ x = \sqrt[3]{t}$ it becomes :
$$\sum_{n=0}^{\infty}t^n$$
so $$f'(x) = \frac{1}{1-x^3}$$
by integrating it I should get the sum right ?
Thanks in advance!
| \begin{eqnarray*}
S=\sum_{n=0}^{\infty} \frac{x^{3n+1}}{3n+1} = \sum_{n=0}^{\infty} \int_{0}^x x^{3n} dx = \int_{0}^x \frac{dx}{1-x^3}
\end{eqnarray*}
Now do partial fractions
\begin{eqnarray*}
\frac{1}{1-x^3}= \frac{x+2}{3(1+x+x^2)}-\frac{1}{3(x-1)}
\end{eqnarray*}
Integrating this gives
\begin{eqnarray*}
S=\frac{1}{6} \ln(1+x+x^2)+\frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{2x+1}{\sqrt{3}} \right)-\frac{\pi}{6 \sqrt{3}} +\frac{1}{3} \ln(1-x)
\end{eqnarray*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the volume of the solid inside both $x^2+y^2+z^2=a^2$ and $(x-a/2)^2+y^2=(a/2)^2$ using a triple integral on cylindrical coordinates I am asked to find the volume of the solid inside both $x^2+y^2+z^2=a^2$ and $(x-a/2)^2+y^2=(a/2)^2$ using a triple integral on cylindrical coordinates.
I have this triple integral setup, which makes so much sense to me. Nonetheless, the result of the integral is not $\frac{2a^3}{9} \left( 3 \pi - 4 \right)$, as it should be according to the book's answer (Larson, 10th ed.)
$$\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{a \cos(\theta)} \int_{- \sqrt{a^2 - r^2}}^{\sqrt{a^2 - r^2}} \ r \ dzdrd\theta$$
Am I missing something?
Thank you.
| In order to better understand the solid region $B$ enclosed by both surfaces in order to better evaluate our limits of integration, we can visualize the surfaces as follows:
In cylindrical coordinates, we can express the cylinder as:
$$ r = a \cos \theta \qquad (0 \leq \theta \leq \pi)$$
and the sphere as:
$$ r^2 + z^2 = a^2$$
We see that, as we traverse all the $(r,\theta)$ or $(x,y)$ points within the cylinder's projection on the $x$-$y$ plane (a circle), the values of $z$ we want are between $\pm\sqrt{a^2-r^2}$
Finally, we recall that:
$$ dV = dx \, dy \, dz = r \, dr \, d\theta \, dz $$
So:
\begin{align*}
\iiint_B \, dV &= \int_{0}^{\pi} \int_0^{a \cos \theta}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} r \, dz \, dr \, d\theta \\
&= \int_{0}^{\pi} \int_0^{a \cos \theta} 2\int_{0}^{\sqrt{a^2-r^2}} r \, dz \, dr \, d\theta \\
&= \int_{0}^{\pi} \int_0^{a \cos \theta} 2r\sqrt{a^2-r^2} \, dr \, d\theta \\
&= \int_{0}^{\pi} \left[ -\frac{2}{3}(a^2-r^2)^{3/2} \right]_0^{a \cos \theta} \, d\theta \\
&= \frac{2}{3}\int_{0}^{\pi} (a^3 - a^3 \sin^3 \theta) \, d\theta \\
&= \frac{2a^3}{3}\int_{0}^{\pi} (1 - \sin^3 \theta) \, d\theta \\
&= \frac{2a^3}{12}\int_{0}^{\pi} (4 - 3\sin \theta + \sin 3\theta) \, d\theta \\
&= \frac{a^3}{6} \left[4\theta + 3 \cos \theta - \frac{1}{3}\cos 3\theta \right]_0^{\pi} \\
&= \frac{a^3}{6} \left(4\pi - 3 + \frac{1}{3} - 3 + \frac{1}{3} \right) \\
&= \frac{a^3}{6} \left(4\pi - \frac{16}{3} \right) \\
&= \frac{2a^3}{9} \left(3\pi - 4 \right)
\end{align*}
The problem that you probably ran into is parametrizing from $\theta = -\pi/2$ to $\theta = \pi/2$. This is a valid parametrization, but when you reach this point:
$$ L = \frac{2}{3}\int_{-\pi/2}^{\pi/2} \left[a^3 - (\sin^2 \theta)^{3/2} \right] \, d\theta $$
We have that:
$$ (\sin^2 \theta)^{3/2} \neq \sin^3 \theta $$
for $\sin \theta < 0$, which occurs for $-\pi/2 \leq \theta < 0$.
So we choose a parametrization where we keep $\sin \theta \geq 0$, which is why we use $0 \leq \theta \leq \pi$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polynomial sum relation Given the following polynomial: $$P(x)=(x^2+x+1)^{100}$$ How do I find : $$\sum_{k=1}^{200} \frac{1}{1+x_k} $$ Is there a general solution for this type of problem cause I saw they tend to ask the same thing for $\sum_{k=1}^{200} \frac{1}{x_k}$? Also how do I find the coefficient of $a_1$ and the remainder for $$P(x)/(x^2+x)$$ (/=divided) (I found the coef of a1 is 100 adn the remainder is 1 but im not sure)
|
Is there a general solution for this type of problem
Consider for example $P(x)=(x^2+x+1)^n\,$.
$\sum_{k=1}^{2n} \frac{1}{x_k}$
Let $y_k = \frac{1}{x_k} \ne 0\,$ then $P(\frac{1}{y_k})=P(x_k)=0\,$, so $y_k$ are the roots of: $$Q(y)=y^{2n} P\left(\frac{1}{y}\right)=y^{2n}\left(\frac{1}{y^2}+\frac{1}{y}+1\right)^n=(1+y+y^2)^n=y^{2n}+\binom{n}{0,1,n-1}y^{2n-1}+\cdots\,$$
Then by Vieta's relations $\sum_{k=1}^{2n} y_k=-\binom{n}{0,1,n-1}=-n\,$.
$\sum_{k=1}^{2n} \frac{1}{1+x_k}$
Let $z_k=x_k+1\,$, then $P(z_k-1)=P(x_k)=0\,$, so $z_k$ are the roots of:
$$\require{cancel}
R(z) = P(z-1)=\left((z-1)^2+(z-\bcancel{1})+\bcancel{1}\right)^n=(z^2-z+1)^n
$$
Then, by an argument entirely similar to the above, it follows that $\sum_{k=1}^{2n} \frac{1}{z_k}=\binom{n}{0,1,n-1}=n\,$.
find the coefficient of $a_1$ and remainder for $P(x)/(x^2+x)$
$$
P(x)=\big((x^2+x)+1\big)^n=(x^2+x)^n+\binom{n}{1}(x^2+x)^{n-1}+\cdots+\binom{n}{n-1}(x^2+x) + 1
$$
All terms are multiples of $x^2+x$ except the very last one, so the remainder of the division is $1\,$ which is a constant polynomial so $a_1=0\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taylor expansion of composite functions Suppose one has a function of the form: $$f(x)=\sqrt{1+a\cos^{2}(x)}$$ where $a$ is some parameter.
Firstly, what is the correct method for Taylor expanding such a composite function? Should one just repeatedly use the chain rule $(f\circ g)'(x)=f'(g(x))g'(x)$ to determine the coefficients of the Taylor expansion, such that it has the form: $$(f\circ g)(x)=(f\circ g)(x_{0})+(f\circ g)'(x_{0})(x-x_{0})+\frac{1}{2}(f\circ g)''(x_{0})(x-x_{0})^{2}+\cdots$$ or is there a different method?
Secondly, although not technically a Taylor expansion, if the parameter $a<<1$ is it reasonable to approximate the function as $$\sqrt{1+a\cos^{2}(x)}\approx 1+\frac{a}{2}\cos^{2}(x)$$ i.e. naively plugging in $a\cos^{2}(x)$ to the Taylor expansion for $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\mathcal{O}(x^{3})$?
| Your method would work, but it'd involve computing a lot of unpleasant derivatives. A better method would be to expand $cos^2(x)$, then expand the square root:
$$\sqrt{1+a \cos^2(x)} = \sqrt{1 + a \left(1 - \frac{1}{2} x^2 + \mathcal{O}(x^4)\right)^2} \\
= \sqrt{1 + a \left(1 - x^2 + \mathcal{O}(x^4)\right)} \\
= \sqrt{1 + a - a x^2 + \mathcal{O}(x^4)} \\
= \sqrt{1+a} \sqrt{1 - \frac{a}{1+a} x^2 + \mathcal{O}(x^4)} \\
= \sqrt{1+a} \left(1 - \frac{a}{2(1+a)} x^2 + \mathcal{O}(x^4)\right) \\
= \sqrt{1+a} - \frac{a}{2\sqrt{1+a}} x^2 + \mathcal{O}(x^4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Iterated sequence - general formula I have a question regarding writing a general formula for an iterated sequence of variables.
Say we have
$$a_0=1$$
$$a_1=\frac{3}{2}a_0$$
$$a_2=\frac{3}{2}a_1-\frac{1}{2}a_0$$
$$a_3=\frac{3}{2}a_2-\frac{1}{2}a_1$$
So in general we have $a_i=\frac{3}{2}a_{i-1}-\frac{1}{2}a_{i-2}$. The solution this this iterated sequence is $a_i=2-2^{-i}$. What method should one use to come up with this solution? ?
Thanks in advance.
| Starting from your general term
$$a_n=\frac{3}{2}a_{n-1}-\frac{1}{2}a_{n-2}$$
substituting value of $a_{n-1}$ in this , it will transform into $$a_{n}=\frac{7}{4}a_{n-2}-\frac{3}{4}a_{n-3}$$
which is also equals to $$a_n=\frac{2^{2+1}-1}{2^2}a_{n-2}-\frac{2^2-1}{2^2}a_{n-3}$$
again substituting the value of $a_{n-2}$ in it , it will transform into (I hope you will be understanding from where I am substituting values, of course from your general terms)$$a_n=\frac{15}{8}a_{n-3}-\frac{7}{8}a_{n-4}$$
which is also equals to$$a_n=\frac{2^4-1}{2^3}a_{n-3}-\frac{2^3-1}{2^3}a_{n-4}$$
conclusion goes to$$a_n=\frac{2^{x+1}-1}{2^x}a_{n-x}-\frac{2^x-1}{2^x}a_{n-(x+1)}$$
put $x=n-1$ and take it from here, you will get your original answer that is $$a_n=2-2^{-n}$$
| {
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Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$ I am new to this topic so I appreciate any help on this.
a)$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{x^2y}{x^4 +y^2}$$
For $x=0 \lor y=0 : f(x,y) \longrightarrow 0$ but for $y = x^2$ the limit is $\frac{1}{2}$ so the limit does not exist.
b)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{e^{xy^3}-1}{x^2 +y^4}$$
I don't have any ideas here
c)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = (x^2 +y^2)^{xy}$$
Let $x \geq y$ then $|x^2 +y^2|^{xy} \leq |2y^2|^{y^2} \longrightarrow 1 \quad $for $y \longrightarrow 0$ but for $x = \frac{1}{y} \Longrightarrow f(x,y) = (\frac{1}{y^2} + y^2)^1 \longrightarrow \infty \quad$ for $y \longrightarrow 0 $ so the limit does not exist.
| Another way to process is to set $y=mx^\alpha$ with some well chosen $\alpha>0$ so that the various sums $x^a+y^b$ become homogeneous.
Of course $m=m(x,y)$ is not a constant, just the ratio of the rate of convergence of $x$ and $y$.
Part a:
With $y=mx^2$
$\displaystyle f(x,y)=\frac{x^2y}{x^4+y^2}=\frac{mx^4}{x^4+m^2x^4}=\frac{m}{1+m^2}$
This is bounded but can take multiple values (e.g $m=1,\ f(x,x^2)=\frac 12$ and $m=0,\ f(x,0)=0$), so there is no limit in $(0,0)$.
Part b:
With $x=my^2$
$\displaystyle f(x,y)=\frac{e^{x^3y}-1}{x^2+y^4}\sim\frac {x^3y}{x^2+y^4}=\frac{m^3y^7}{m^2y^4+y^4}=\frac{m^3y^3}{1+m^2}=\underbrace{\bigg(\frac{m^2}{1+m^2}\bigg)}_\text{bounded}xy\to 0$
So there is a limit and it is $0$.
Part c:
The previous method do not work well here, but polar coordinates do
$f(x,y)=(x^2+y^2)^{xy}=\exp(xy\ln(x^2+y^2))=\exp(\underbrace{\cos(\theta)\sin(\theta)}_\text{bounded}\underbrace{r^2\ln(r^2)}_{\to 0})\to e^0=1$
| {
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Confusion about proving $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$ by induction I know what the answer to this question is, but I am not sure how the answer was reached and I would really like to understand it! I am omitting the base case because it is not relevant for my question.
Inductive hypothesis:
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$$ is true when $n = k$ and $k > 1$
Therefore: $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1}$$
Inductive step:
Prove that $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+1+1} = \frac{k+1}{k+2}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)}\right] + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+2}$$
What I am confused about is where the $\frac{1}{(k+1)(k+2)}$ comes from in the first line of the inductive step. Can someone please explain this in a little more detail? The source of the answer explains it as "break last term from sum", but I am unclear on what that means.
| The inductive hypothesis is
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{k(k+1)}
=\frac{k}{k+1}.\tag1$$
You need to prove that (1) implies the statement got from (1) by replacing
$k$ by $k+1$. This is
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{(k+1)(k+2)}
=\frac{k+1}{k+2}\tag2$$
but instead of (2) you have
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{k(k+1)}
=\frac{k+1}{k+2}$$
which is wrong, and this is causing your confusion. The inductive
proof starts by recognising that
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{(k+1)(k+2)}
=\left[\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{k(k+1)}
\right]+\frac1{(k+1)(k+2)}.$$
| {
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How to take $\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$? The integral:
$$\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$$
If num were greater than denum I would just devide it normally with long division, but it is not, how should I handle it then?
| $$\int_{0}^{1}\frac{x^2+1}{x^4+1}\,dx + \int_{1}^{+\infty}\frac{x^2+1}{x^4+1}\,dx = 2\int_{0}^{1}\frac{1+x^2}{1+x^4}\,dx = 2\int_{0}^{1}\frac{1+x^2-x^4-x^6}{1-x^8}\,dx$$
hence the value of the integral is provided by the (Dirichlet) series
$$ 2\sum_{k\geq 0}\left(\frac{1}{8k+1}+\frac{1}{8k+3}-\frac{1}{8k+5}-\frac{1}{8k+7}\right) $$
that can be computed through the digamma reflection formula. Its value is just
$$ \boxed{I = \color{red}{\frac{\pi}{\sqrt{2}}}} .$$
The given integral is straightforward to compute through the residue theorem, too.
| {
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Surface integral of a pyramid in a sphere Calculate surface integral: $$I = \iint \limits_{S} \frac{xy}{x^2 + y^2} d{S}$$ where S is surface determined by sides of pyramid $x + y + z = 1 \phantom\ (x, y, z \ge 0 ),$ inside sphere $(x - \frac{1}{2})^2 + (y - \frac{1}{3})^2 + z^2 \le \frac{1}{4}.$
How should I parameterize the surface $S$? I am also having a question regarding the case of the flux integral: How should I find the unit normal vector of the surface $S$?
| The parts of the surface that give a non-zero contribution to the integral are
1) the one over the face $x + y + z = 1$
2) the one over the face $z=0$.
(on the other two faces $x\cdot y=0$ and therefore $f(x,y)=0$).
As regards $I_1$, we have that $dS=\sqrt{3}dxdy$ and
$$I_1=\sqrt{3}\iint_{D_1}\frac{xy}{x^2 + y^2}\,dxdy$$
where $D_1=\{(x - \frac{1}{2})^2 + (y - \frac{1}{3})^2 +(1-x-y)^2\le \frac{1}{4},y\geq 0,x+y\leq 1\}$.
For $I_2$, $dS=dxdy$ and
$$I_2=\iint_{D_2}\frac{xy}{x^2 + y^2}\,dxdy.$$
where $D_2=\{(x - \frac{1}{2})^2 + (y - \frac{1}{3})^2 \le \frac{1}{4},y\geq 0,x+y\leq 1\}$.
| {
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Prove $\int_{-\frac{1}{4}}^{\frac{1}{4}} \left(\left(H_{x+\frac{1}{4}}-H_{x-\frac{1}{4}}\right) \cos (2 \pi x)+\pi \right) \, dx=\text{Si}(\pi )$ I recently stumbled upon a curious relationship between harmonic numbers and the Wilbraham-Gibbs constant:
$$\int_{-\frac{1}{4}}^{\frac{1}{4}} \left(\left(H_{x+\frac{1}{4}}-H_{x-\frac{1}{4}}\right) \cos (2 \pi x)+\pi \right) \, dx=\text{Si}(\pi )$$
Where $H_n$ is the $nth$ harmonic number, and $\mathrm{Si}(\pi)$ is the sine integral of $\pi$ (Wilbraham-Gibbs constant).
Is this an expected result, and if not, is it possible to prove? I have confirmed equivalency up to 40,000 digits.
| Using the relation between the Harmonic numbers and digamma function, namely $H_{x} = \gamma + \psi(x +1)$, the relation between the digamma function and the gamma function,
$$\psi(x) = \frac{d}{dx} \, \ln(\Gamma(x)),$$
and the integrals:
\begin{align}
\int_{0}^{1} \sin\left(\frac{\pi \, t}{2}\right) \, \ln\left(\frac{1+t}{1-t}\right) \, dt &= - \frac{2}{\pi} \, \left[ Si\left(\frac{\pi(1-t)}{2}\right) -
Si\left(\frac{\pi(1+t)}{2}\right) + \cos\left(\frac{\pi t}{2}\right) \, \ln\left(\frac{1 + t}{1-t}\right) \right]_{0}^{1} \\
&= \frac{2}{\pi} \, Si(\pi)
\end{align}
\begin{align}
\int_{0}^{1} \sin\left(\frac{\pi \, t}{2}\right) \, \ln\left(cot\left(\frac{\pi(1+t)}{4}\right)\right) \, dt &= (-1) \, \left[ t + \frac{2}{\pi} \, \cos\left(\frac{\pi t}{2}\right) \, \ln\left(
\frac{\cos\left(\frac{\pi t}{4}\right) - \sin\left(\frac{\pi t}{4}\right)}{\cos\left(\frac{\pi t}{4}\right) + \sin\left(\frac{\pi t}{4}\right)} \right)\right]_{0}^{1} \\
&= -1
\end{align}
the following is obtained.
\begin{align}
I &= \int_{-1/4}^{1/4} \left[ \pi + \left(H_{x+1/4} - H_{x-1/4}\right) \, \cos(2\pi x) \right] \, dx \\
&= \frac{\pi}{2} + \int_{-1/4}^{1/4} \left[ \psi\left(x + \frac{5}{4}\right) - \psi\left(x + \frac{3}{4}\right) \right] \, \cos(2 \pi x) \, dx
\end{align}
Using integration by parts yields
\begin{align}
I &= \frac{\pi}{2} + 2\pi \, \int_{-1/4}^{1/4} \sin(2\pi t) \, \ln\left( \frac{\Gamma\left(t + \frac{5}{4}\right)}{\Gamma\left(t + \frac{3}{4}\right)}\right) \, dt \\
&= \frac{\pi}{2} + \frac{\pi}{2} \, \int_{-1}^{1} \sin\left(\frac{\pi t}{2}\right) \, \ln\left( \frac{\Gamma\left(\frac{5+t}{4}\right)}{\Gamma\left( \frac{3+t}{4}\right)}\right) \, dt \text{ (obtained by setting $t \to t/4$) } \\
&= \frac{\pi}{2} \, \left[ 1 + \int_{0}^{1} \sin\left(\frac{\pi t}{2}\right) \,
\ln\left[\frac{\Gamma\left(\frac{5+t}{4}\right) \, \Gamma\left(\frac{3-t}{4}\right)}{\Gamma\left( \frac{3+t}{4}\right) \, \Gamma\left(\frac{5-t}{4}\right)} \right] \, dt \right] \\
&= \frac{\pi}{2} \, \left[ 1 + \int_{0}^{1} \sin\left(\frac{\pi t}{2}\right) \,
\ln\left[\frac{1+t}{1-t} \, cot\left(\frac{\pi (1+t)}{4}\right) \right] \, dt \right] \\
&= \frac{\pi}{2} \, \left[ 1 + \frac{2}{\pi} \, Si(\pi) -1 \right] \\
&= Si(\pi).
\end{align}
Hence,
$$\int_{-1/4}^{1/4} \left[ \pi + \left(H_{x+1/4} - H_{x-1/4}\right) \, \cos(2\pi x) \right] \, dx = Si(\pi).$$
| {
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Continuing "Pascal's triangle" for negative binomial exponents Does there exist a pattern for the coefficients in a negative binomial expansion? This question is not about the explicit formula, but rather if there exist a continuation for Pascal's triangle.
$$\begin{array}l
(a+b)^{-2} &=&&&& \color{red}?\\
(a+b)^{-1} &=&&&& \color{red}?\\
(a+b)^{0} &=&&&& 1\\
(a+b)^{1} &=&&& 1a &+& 1b\\
(a+b)^{2} &=&& 1a^2 &+& 2ab &+&1b^2\\
(a+b)^{3} &=& 1a^3 &+& 3a^2b &+& 3ab^2 &+& 1b^3 &
\end{array}$$
It would obviously not be a triangle given that it's an infinite sum, but it seems reasonable that there should be a similar interpretation.
| There is a continuation respecting the addition law
\begin{align*}
\binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1}
\end{align*}
This way we can write
\begin{array}{l|rrrrrrrrrr}
(1+x)^{-3}&\color{grey}{0}&\color{grey}{0}&1&-3x&6x^2&-10x^3&15x^4&-21x^5&38x^6&\ldots\\
(1+x)^{-2}&\color{grey}{0}&\color{grey}{0}&1&-2x&3x^2&-4x^3&5x^4&-6x^5&7x^6&\ldots\\
(1+x)^{-1}&\color{grey}{0}&\color{grey}{0}&1&-x&x^2&-x^3&x^4&-x^5&x^6&\ldots\\
(1+x)^{0}&\color{grey}{0}&\color{grey}{0}&1&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}\\
(1+x)^{1}&\color{grey}{0}&\color{grey}{0}&1&x&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}\\
(1+x)^{2}&\color{grey}{0}&\color{grey}{0}&1&2x&x^2&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}\\
(1+x)^{3}&\color{grey}{0}&\color{grey}{0}&1&3x&3x^2&x^3&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}\\
\end{array}
For negative exponents $-n$ with $n>0$ we have
\begin{align*}
(1+x)^{-n}&=\sum_{j=0}^\infty\binom{-n}{j}x^j
=\sum_{j=0}^\infty\binom{n+j-1}{j}(-1)^jx^j\\
\end{align*}
Hint: See table 164 in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.
| {
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Bound for the equation $a^2 = b^2 + 4n$ For a fixed natural number $n$, I want to study the integer solutions of $$a^2 = b^2 + 4n.$$ There is always at least one solution, namely $$(n+1)^2 = (n-1)^2 + 4n.$$ There can be others; for example, the solutions of $a^2 = b^2 + 24$ are $$(a,b)
= (\pm 7, \pm 5), \; \; (\pm 5, \pm 1).$$
Conjecture: there are no solutions $(a,b)$ with $|a| > n+1$.
I thought of factoring $4n = a^2 - b^2 = (a-b)(a+b)$. So there should definitely not be solutions $|a| > 4n$ since at least one of the factors would be greater than $4n$. Can we get all the way down to $n+1$?
| If $|a|=n+2$ then $b^2=(n+2)^2-4n$, which is strictly greater than $n^2$ but less than $(n+2)^2$, hence it must be $(n+1)^2$, so $n^2+4=n^2+2n+1$ so $n=3/2$(not possible)
Now if $|a|=n+3$ then $(n+3)^2-4n$ is a perfect square, which is strictly greater than $(n+1)^2$ but less than $(n+3)^2$, hence it must be $(n+2)^2$, so $n^2+2n+9=n^2+4n+4$ so $n=5/2$(not possible)
Now if $|a|>n+3$
$a^2>a^2-4n>(a-1)^2$
Hence it is never a perfect square as it's lying between two consecutive perfect squares, hence your conjecture is true.
| {
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Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
The only method in doing this that I know was to just plug it in into the calculator, but I really had no clue on how to do this without the aid of a calculator,
I was thinking more along the lines of using the Taylor polynomial series but that really didn't work out for me though.
| $$\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}=\sqrt{\frac{1-\cos\frac{5\pi}{6}}{2}}+\sqrt{\frac{1-\cos\frac{\pi}{6}}{2}}=$$
$$=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}+\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\frac{\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}}{2\sqrt2}=$$
$$=\frac{\sqrt3+1+\sqrt3-1}{2\sqrt2}=\sqrt{\frac{3}{2}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit Relation Involving Second Moment Saw this limit relation in a paper, but I can't seem to prove it:
\begin{equation}
\mathbb{E}(X^2) = \lim_{t \rightarrow 0} \frac{2}{t^2} [\mathbb{E}_X(e^{tX})-1]
\end{equation}
I might be missing something basic, but does anyone know how? Not sure if we need to assume $\mathbb{E}(X)=0$ here too.
| $$
e^{tX} = 1 + tX + \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots
$$
$$
e^{tX} - 1 = tX + \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots
$$
\begin{align}
\operatorname{E}(e^{tX} - 1) & = \operatorname{E}\left( tX + \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots \right) \\[10pt]
& = \operatorname{E} \left( \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots \right) \text{ if } \operatorname{E}(X)=0 \\[10pt]
\frac 2 {t^2} \operatorname{E}(e^{tX} - 1) & = \operatorname{E} \left( X^2 + \frac{t X^3} 3 + \frac {t^2 X^4}{12} + \frac{t^3 X^5}{60} + \cdots \right) \text{ if } \operatorname{E}(X) = 0.
\end{align}
I think L'Hopital's rule could also be used.
| {
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Representation of $e$ as a descending series I saw on Wikipedia: List of representations of e that
$$e=3+\sum_{k=2}^{\infty}\frac{-1}{k!(k-1)k}$$
It was also mentioned that this identity come from consideration on ways to put upper bound of $e$.
Can anyone give me a hint how we can derive this identity? I have tried to look at its Taylor expansion but that approach seems to fail miserably.
Thanks in advance.
| You can derive the following telescoping sum for $k\ge2$:
\begin{align*}
\frac{1}{k!}+\frac{1}{k!(k-1)k}
&=\frac{1}{k!}\left(1+\frac{1}{(k-1)k}\right)\\
&=\frac{1}{k!}\left(1+\frac{1}{k-1}-\frac{1}{k}\right)\\
&=\frac{k-1}{kk!}+\frac{1}{(k-1)k!}\\
&=\frac{k-1}{kk!}+\frac{1}{(k-1)k(k-1)!}\\
&=\frac{k-1}{kk!}+\frac{1}{(k-1)(k-1)!}-\frac{1}{k(k-1)!}\\
&=\frac{1}{k!}-\frac{1}{kk!}+\frac{1}{(k-1)(k-1)!}-\frac{1}{k!}\\
&=\frac{1}{(k-1)(k-1)!}-\frac{1}{kk!}
\end{align*}
Hence, if you go back to the original series representation of $e$ as you did:
\begin{align*}
e+\sum_{k=2}^{\infty}\frac{1}{k!k(k-1)}
&=2+\sum_{k=2}^{\infty}\frac{1}{k!}+\frac{1}{k!k(k-1)}\\
&=2+\sum_{k=2}^{\infty}\frac{1}{(k-1)(k-1)!}-\frac{1}{kk!}\\
&=2+1-\lim_{n\rightarrow\infty}\frac{1}{nn!}=3
\end{align*}
| {
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Finding the square root $\sqrt{4-2\sqrt{3}}$ How do you solve this:
$$X^2=4-2\sqrt{3}$$
I don't really know how to solve for $X$. I initially got $X=\sqrt{4-2\sqrt{3}}$ but I don't know how to simplify it.
| Let $\,a=\sqrt{4-2\sqrt{3}}\,$, $\,b=\sqrt{4+2\sqrt{3}}\,$ and note that $b \gt a \gt 0\,$, then:
$$\require{cancel}
\begin{cases}
a^2+b^2 = 4-\cancel{2\sqrt{3}} + 4+\cancel{2\sqrt{3}} = 8 \\
ab = \sqrt{(4-2\sqrt{3})(4+2\sqrt{3})} = \sqrt{16 - 2 \cdot 3} = 2
\end{cases}
$$
It follows that:
$$
\begin{cases}
\begin{align}
a+b &= \sqrt{a^2+b^2+2ab}=\sqrt{12}=2 \sqrt{3} \\
b-a &= \sqrt{a^2+b^2-2ab}=\sqrt{4} = 2
\end{align}
\end{cases}
$$
Therefore $a = \frac{1}{2}\big((a+b)-(b-a)\big)=\frac{1}{2}\left(2 \sqrt{3} - 2\right)=\sqrt{3}-1\,$.
| {
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$f(x^2)=(f(x))^2=f(f(x))$, $\forall x\in\mathbb R$. Prove that, there are infinitely many polynomial $f$ of degree $\geq1$, such that
$f(x^2)=(f(x))^2=f(f(x)), \forall x\in\mathbb R$. can only find three of them namely, $f(x)=0,1,x^2$.
Please help.
| Clearly the degree of the polynomial must be $2$, and the leading coefficient must be $1$.
Let the polynomial be $x^2+bx+c$
we need:
$x^4+bx^2+c=(x^2+bx+c)^2=((x^2+bx+c)^2+b(x^2+bx+c)+c$.
Lets separate into the five coefficients to get this system:
$1=1^2=1^3$ (coefficients of $x^4$)
$0=2b=2b$ (coefficients of $x^3$)
$b=b^2+2c=b^2+2c+b$ (coefficients of $x^2$) ( so $b=0$)
$0=0=0$ (coefficients of $x^2$)
$c=c^2=c^2+bc+c=c^2+c$ ( so $c=0$).
Hence the only polynomial of positive degree is $x^2$
| {
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Find all complex roots for $z^6 + (1 + i)z^3 + i = 0$ Find all complex roots of $z^6 + (1 + i)z^3 + i = 0$.
I've tried by first settings $w = z^3$, so then I get $$w^2 + (1 + i)w + i = 0$$
Now I rewrite it as $$(w+(\frac{1+i}{2}))^2 = -i + (\frac{1+i}{2})^2$$
Set $w+(\frac{1+i}{2}) = a+bi$
$$(a+bi)^2 = -i + 2i=i$$
Which implies that
$$a^2-b^2=0$$
$$a^2+b^2=\sqrt{0^2+1^2} = 1$$
$$2ab=1$$
So I get $w=\frac{1+i}{2}\pm(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)$. Not sure where to go from here. I get answers for $w$, but how do I get the answers for $z$?
| The first step is good. The roots of $w^2+(1+i)w+i=0$ can be computed with the usual formula:
$$
\frac{-(1+i)\pm\sqrt{(1+i)^2-4i}}{2}
$$
but it's easier finding two numbers having sum $-1-i$ and product $i$: they're $-1$ and $-i$.
Now just find the cube roots of $-1=e^{\pi i}$ and $-i=e^{3\pi i/2}$.
If you want to do it the hard way, with $w=a+bi$ you get
$$
a^2-b^2+2abi+a+bi+ai-b+i=0
$$
so
\begin{cases}
a^2-b^2+a-b=0\\
2ab+a+b+1=0
\end{cases}
The first equation can be rewritten as $(a-b)(a+b+1)=0$. For $a=b$, the second equation is $2a^2+2a+1=0$, which has no solution. With $b=-1-a$ you get
$$
-2a-2a^2=0
$$
so $a=0$ or $a=-1$.
You seem to have done wrongly the completion of the square.
Note that
$$
(1+i)^2=2i
$$
so
$$
-i+\frac{(1+i)^2}{4}=-i+\frac{i}{2}=-\frac{i}{2}
$$
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Make a list the elements of the subgroup in S7 Consider the permutation $$a=\begin{pmatrix}1234567\\3276514\end{pmatrix}$$ in S7 .List the elements of the subgroup in S7.
Hello, I just graduated from high school and in my summer time I'm reading about abstract algebra I want to learn more about the subject, but I do not understand how the book explains the part of permutation maybe I need a course, but I'll take it on University. Someone could explain or help me with this problem so I can understand better.
| The notation
$$a = \begin{pmatrix} b_1 & b_2 & \cdots & b_n \\ c_1 & c_2 & \cdots & c_n \end{pmatrix}$$
denotes a function with $a(b_1) = c_1, a(b_2) = c_2,\dots,a(b_n) = c_n$. This is usually given with $a_1 = 1, a_2 = 2, \dots, a_n = n$ but if you think about it, the order that the columns are given in doesn't matter. That is,
$$\begin{pmatrix}1&2&3&4&5&6&7\\3&2&7&6&5&1&4\end{pmatrix} \text{ and } \begin{pmatrix} 3&2&7&6&5&1&4\\7&2&4&1&5&3&6 \end{pmatrix}$$
denote the same function. This is helpful because $a^2(i) = a(a(i)) = i$ and we can easily compute this by stacking the two matrices on top of each other:
$$ \begin{pmatrix} \begin{pmatrix}1&2&3&4&5&6&7\\3&2&7&6&5&1&4\end{pmatrix} \\ \begin{pmatrix} 3&2&7&6&5&1&4\\7&2&4&1&5&3&6 \end{pmatrix} \end{pmatrix} \to \begin{pmatrix}1&2&3&4&5&6&7 \\7&2&4&1&5&3&6 \end{pmatrix} = a^2. $$
Notice that if you apply $a$ to $1$ you get $a(1) = 3$ and then if you apply $a$ again you get $a(a(1)) = a(3) = 7$ and the same works for $a(a(2)) = 2$ and $a(a(3)) = 4$ and so on.
Thus you can compute products with this stacking technique. Since the subgroup $\langle a \rangle$ generated by $a$ is nothing but the set of powers of $a$, this lets you compute the subgroup.
Note that this description of the subgroup generated by $a$ is only valid for a finite group. For an infinite group you need to add in $a^{-1}$ and its powers. The reason you can avoid this for finite groups is because in that case, $a^{-1}$ is always equal to a positive power of $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove inequality $a^3b^3+2b^3c^3+3a^3c^3\le 0$ For $a^3+b^3+c^3=0$. Prove that $a^3b^3+2b^3c^3+3a^3c^3\le 0$
i think from $a^3+b^3+c^3=0$ we have one of the three numbers is zero, the other two are opposites.Suppose $a;b$ opposites and $c=0$
| For simplicity, let $a^3=x,b^3=y,c^3=z$. Then: $x+y+z=0\Rightarrow x=-(y+z).$
$$xy+2yz+3xz=$$
$$-y(y+z)+2yz-3z(y+z)=$$
$$-y^2-2yz-3z^2=$$
$$-(y+z)^2-2z^2\le0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Use Parseval's equation and the table of Fourier series to evaluate $\sum\frac{1}{(1+n^2)^2}$
Use Parseval's equation and the table of Fourier series to evaluate $$\sum\frac{1}{(1+n^2)^2}$$
So I have used this method before to show $$\sum\frac{1}{n^4} = \frac{\pi^4}{32}$$ however, for this question I am struggling.
Basically, off the sheet I have from university I have used $$F(x) = \left|\sin\left(\pi\times\frac xL\right)\right|
, -L<x<L, \to \frac 2\pi - \frac 4\pi \sum \frac{1}{n^2-1}\times\cos\left(\frac{n\pi x}{L}\right)$$
I find $a_0$ and $a_n$ then
I apply Parseval's theorem then it becomes clear I have gone wrong...
| It is easy to show that the complex Fourier Series for $e^{bx}$, $x\in[0,2\pi]$ is given by
$$e^{bx}=\frac{e^{2\pi b}-1}{2\pi}\sum_{n=-\infty}^\infty \frac{e^{inx}}{b-in}$$
whereupon isolating the series becomes
$$\sum_{n=-\infty}^\infty \frac{e^{inx}}{b-in}=\frac{2\pi e^{bx}}{e^{2\pi b}-1} \tag1$$
Differentiating $(1)$ with respect to $b$ and multiplying by $-1$ reveals
$$\begin{align}
\sum_{n=-\infty}^\infty \frac{e^{inx}}{(b-in)^2}&=-\frac{d}{db}\left(\frac{2\pi e^{bx}}{e^{2\pi b}-1} \right)\\\\
&=\left(\frac{2\pi e^{bx}}{(e^{2\pi b}-1)^2}\right)\left(2\pi e^{2\pi b}-(e^{2\pi b}-1)x \right)\tag 2
\end{align}$$
Next, setting $b=1$ in $(2)$ yields
$$\begin{align}
\sum_{n=-\infty}^\infty \frac{e^{inx}}{(1-in)^2}&=\left(\frac{2\pi e^{x}}{(e^{2\pi }-1)^2}\right)\left(2\pi e^{2\pi }-(e^{2\pi }-1)x \right)\tag 3
\end{align}$$
Finally, applying Parseval's Theorem to $(3)$ we find that
$$\begin{align}
\sum_{n=-\infty}^\infty \frac{1}{(n^2+1)^2}&=\left(\frac{2\pi}{(e^{2\pi}-1)^2}\right)^2\,\frac{1}{2\pi}\int_0^{2\pi }e^{2x}(2\pi e^{2\pi}-(e^{2\pi}-1)x)^2\,dx\\\\
&=\frac{\pi}2 \left(\pi \text{csch}^2(\pi)+\text{coth}(\pi)\right)
\end{align}$$
whereupon solving for the series of interest yields
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{(n^2+1)^2}=-\frac12 +\frac{\pi}4 \left(\pi \text{csch}^2(\pi)+\text{coth}(\pi)\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put
*
*$z=y+1$
*$x=n^2+3$
*$y=3n^2+4+(n^4-n)/2$
what does it mean that $6$ actually requires all three terms?
| Notice all odd numbers are:
$1 = 1^2 - 0^2$
$3 = 2^2 - 1^2$
$5 = 3^2 - 2^2$ etc..
so if odd numbers are $\{o_1,o_2,...\}$ then $o_n=n^2-(n-1)^2$.
Now even numbers $\{e_1,e_2,...\}$ are simply $e_n=o_n+1^2$.
regarding the number 6, if any shorter representation exists, it must be one of the following:
1) $x^2 + y^2$ - you can easily see no such exists.
2) $x^2 - z^2$ - if this is it, $x,y\in\{1,2,3,4\}$ otherwise their difference will be greater than 6, but again, no such combination exists...
Hence all 3 must be non-zero
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Can the nested radical expression $\sqrt{8 + \sqrt{48}}$ be simplified? Solving a pythagorean problem I came up with the following radical:
$$C = \sqrt{1^2+(\sqrt{3} + 2)^2}$$
I can simplify this to:
$$\sqrt{8+4\times\sqrt{3}} = \sqrt{8+\sqrt{48}}$$
Is there a way to simplify this further? I'm not satisfied with the double radical, but the sum is an obstacle to further simplification, am I correct?
| A nested radical $\sqrt{u + \sqrt{v}}$ can be simplified if $u^2 - v$ is a perfect square. Since $$8^2 - 48 = 64 - 48 = 16 = 4^2$$ this nested radical can be simplified.
Suppose that
$$\sqrt{8 + \sqrt{48}} = \sqrt{a} + \sqrt{b}$$
where $a$ and $b$ are rational numbers. Squaring both sides of the equation yields
$$8 + \sqrt{48} = a + b + 2\sqrt{ab}$$
Matching rational and irrational parts yields the system of equations
\begin{align*}
a + b & = 8 \tag{1}\\
2\sqrt{ab} & = \sqrt{48} \tag{2}
\end{align*}
Since $\sqrt{48} = 4\sqrt{3}$, we obtain
\begin{align*}
2\sqrt{ab} & = 4\sqrt{3}\\
\sqrt{ab} & = 2\sqrt{3}\\
ab & = 12\\
b & = \frac{12}{a}
\end{align*}
Substituting $12/a$ for $b$ in equation 1 yields
\begin{align*}
a + \frac{12}{a} & = 8\\
a^2 + 12 & = 8a\\
a^2 - 8a + 12 & = 0\\
(a - 2)(a - 6) & = 0
\end{align*}
Hence, $a = 2$ or $a = 6$. If $a = 2$, then $b = 6$, so we obtain
$$\sqrt{8 + \sqrt{48}} = \sqrt{2} + \sqrt{6}$$
If $a = 6$, then $b = 2$, which yields the same solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compare Fourier series $$g(x) = \sum_{n = 1}^{\infty} a_n \sin(nx)$$
$$h(x) = -\frac{a_1}{2 \pi } x $$
How to prove that $g(x) \ge h(x)$ on $ (0, \pi) $, where $a_n$ decreases to $0$ monotonically?
I tried to make a Fourier series from $R(x) = x$, I got $ \displaystyle R(x) = \sum_{n = 1}^{\infty} \frac{(-1)^n}{n} \sin(nx)$, so $\displaystyle h(x) = -\frac{a_1}{2 \pi } \sum_{n = 1}^{\infty} \frac{(-1)^n}{n} \sin(nx) $ but I don't know how to continue.
| I don't see a way to use the Fourier series of $h$ (which you have slightly miscalculated, a factor of $-2$ is missing).
A summation by parts works, however. Note that
$$2\sin \alpha \sin \beta = \cos (\alpha - \beta) - \cos (\alpha + \beta),$$
so
\begin{align}
2 g(x) \sin (x/2)
&= \sum_{n = 1}^{\infty} a_n \bigl(2\sin (nx) \sin(x/2)\bigr) \\
&= \sum_{n = 1}^{\infty} a_n\bigl(\cos \bigl((n-1/2)x\bigr) - \cos \bigl((n+1/2)x\bigr)\bigr) \\
&= \sum_{n = 1}^{\infty} a_n \cos \bigl((n-1/2)x\bigr) - \sum_{m = 2}^{\infty} a_{m-1}\cos \bigl((m-1/2)x\bigr) \\
&= a_1 \cos (x/2) - \sum_{m = 2}^{\infty} \bigl(a_{m-1} - a_m\bigr)\cos\bigl((m-1/2)x\bigr) \\
&\geqslant a_1 \cos (x/2) - \sum_{m = 2}^{\infty} (a_{m-1} - a_m) \\
&= a_1 \bigl(\cos (x/2) - 1\bigr).
\end{align}
Since $\sin (x/2) > 0$ for $x \in (0,\pi)$, the inequality $g(x) \geqslant h(x)$ for $x\in (0,\pi)$ is equivalent to
$$ 2g(x)\sin (x/2) + \frac{a_1}{\pi} x\sin (x/2) \geqslant 0\tag{1}$$
there. From the above computation, it suffices to show
$$\frac{x}{\pi}\sin (x/2) + \cos (x/2) \geqslant 1\tag{2}$$
for $x \in (0,\pi)$, or, writing $x = 2y$,
$$f(y) := \frac{2}{\pi} y \sin y + \cos y \geqslant 1\tag{$2'$}$$
for $0 < y < \frac{\pi}{2}$. We see that $f(0) = f(\pi/2) = 1$, and compute
$$f'(y) = \frac{2}{\pi}y\cos y - \biggl(1 - \frac{2}{\pi}\biggr)\sin y.$$
For $0 < y < \frac{\pi}{2}$, $f'(y) = 0$ is equivalent to
$$\frac{2}{\pi-2}\cdot y = \tan{y},\tag{3}$$
and $(3)$ has a unique solution $y_0$ in that interval.
Since $f'(y) = \bigl(\frac{4}{\pi} - 1\bigr) y + O(y^2)$ near $0$, we see that $f'(y) > 0$ for $0 < y < y_0$ and $f'(y) < 0$ for $y_0 < y < \frac{\pi}{2}$. Thus $(2')$, equivalently $(2)$, and consequently $(1)$, follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dice roll: Probability of getting a given sum + a particular set of numbers The problem goes as: "An unbiased usual six-sided die is thrown three times. The sum of the numbers coming up is $10$. What is the probability that $2$ has appeared at least once?"
Options:
$A. 1/36$
$B. 5/36$
$C. 91/216$
$D. 1/18$
My approach: I understood this as finding the probability of getting at least one $2$ (say event $A$) in each triplet subject to the condition that the sum of the numbers in the triplet $= 10$ (say event $B$). Using the rules of conditional probability,
$P(A|B)=P(A \cap B)/P(B)$
Manually listing all triplets that sum to $10$, I found $P(B)=27/216$ and counting the triplets that had at least one $2$ I found $P(A \cap B)=12/216$. Thus getting the answer $4/9$ which is definitely wrong looking at the options, I've no idea how to go ahead of this.
| The ways to get a sum of $10$ are as follows:
$$1 + 3 + 6$$
$$1 + 4 + 5$$
$$1 + 5 + 4$$
$$1 + 6 + 3$$
$$\color{green}{2 + 2 + 6}$$
$$\color{green}{2 + 3 + 5}$$
$$\color{green}{2 + 4 + 4}$$
$$\color{green}{2 + 5 + 3}$$
$$\color{green}{2 + 6 + 2}$$
$$3 + 1 + 6$$
$$\color{green}{3 + 2 + 5}$$
$$3 + 3 + 4$$
$$3 + 4 + 3$$
$$\color{green}{3 + 5 + 2}$$
$$3 + 6 + 1$$
$$4 + 1 + 5$$
$$\color{green}{4 + 2 + 4}$$
$$4 + 3 + 3$$
$$\color{green}{4 + 4 + 2}$$
$$4 + 5 + 1$$
$$5 + 1 + 4$$
$$\color{green}{5 + 2 + 3}$$
$$\color{green}{5 + 3 + 2}$$
$$5 + 4 + 1$$
$$6 + 1 + 3$$
$$\color{green}{6 + 2 + 2}$$
$$6 + 3 + 1$$
There are $4 + 5 + 6 + 5 + 4 + 3 = 27$ total possibilities, of which $12$ involve a two. This gives $$\frac{12}{27}=\boxed{\frac{4}{9}\,}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \cot 25^\circ$ So I'm trying to come up with an answer to this question for hours now. I don't know what I'm doing wrong and none of the calculators on the internet couldn't help so I figured I should ask people.
What have I done so far:
$\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \frac{(\frac{2\sin(45^\circ+20^\circ)}{\sqrt{2}})^2}{\cos 40^\circ} = \frac{\sin^2 65^\circ}{\cos 40^\circ} = \frac{1 - \cos 130^\circ}{\cos 40^\circ} = \frac{1 + \cos 50^\circ}{\cos 40^\circ} = \frac{2\cos^2 25^\circ}{\cos 40^\circ}$ ... etc.
I can't seem to figure out where to go from here so I'm just stuck.
I also tried the classic approach:
$\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \frac{1 + \sin 40^\circ}{\cos 40^\circ} = \frac{1}{\cos 40^\circ} + {\tan 40^\circ} = \sec 40^\circ + \tan 40^\circ$
But how can I prove that
$\sec 40^\circ + \tan 40^\circ = \cot 25^\circ$ ?
What am I doing wrong? Any hints or solutions would be great. Thanks in advance.
| Expanding the square, using relationship $2 \sin a \cos a= \sin 2a$ and the definition of $\cot$, we have to show that
$$\frac{[(\sin20^\circ)^2 + (\cos20^\circ)^2] + \sin 40^\circ}{\cos40^\circ} = \dfrac{\cos25^\circ}{\sin25^\circ}$$
As the content of the square brackets is 1, it is equivalent to show that :
$$\sin25^\circ + \sin25^\circ \sin 40^\circ=\cos40^\circ\cos25^\circ$$
or
$$\sin25^\circ =\cos40^\circ\cos25^\circ - \sin25^\circ \sin 40^\circ$$
or
$$\sin25^\circ =\cos(40^\circ+25^\circ)$$
or
$$\sin25^\circ =\cos(65^\circ)$$
which is evidently true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Manipulating an algebraic equation This came in a competition, I went back home and tried it but was unsuccessful.
If $$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3}$$ then -
$$1) ~ax =by
~~2)~ xy = ab
~~3)~ ay = bx
~~4) ~ax = b$$
It is clear that if the following given condition holds for the variables then there would be a case in which $a = x$ and $b = y$. Hence then options should also satisfy this and hence we can eliminate options a and d. But I was not able to solve the question further. I tried adding and subtracting 1 from both sides but found nothing helpful.
Moreover we are required to show if the correct option is sufficient to show that the first condition is true for the given variables.
| $$0=\frac{a^3+3ab^2}{3a^2b+b^3} -\frac{x^3+3xy^2}{3x^2y+y^3}=$$
$$=\frac{(ay-bx)(3a^2x^2+3b^2y^2+a^2y^2+b^2x^2-8abxy)}{(3a^2b+b^3)(3x^2y+y^3)}=$$
$$\frac{(ay-bx)(3(ax-by)^2+(ay-bx)^2)}{(3a^2b+b^3)(3x^2y+y^3)},$$
which says that $ay=bx$ or
$$(ax-by)^2+(ay-bx)^2=0.$$
In the last case we obtain $ay=bx$ again.
Thus, the answer is $ay=bx$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the locus of centres of circles I have to find the locus of centres of circles that pass through the point $(8,0)$ and tangent to circle $x^2+y^2=100$
$(x_0, y_0)$ - centres we are looking for
I decided to try something with distances, so I made a the equation $(x_0^2-8)^2+y_0^2=r^2$ It is the distance from center to point
Also I found the equation for tangent for our big circle $x_1x+y_1y-100=0$, where $x_1, y_1$ a point on the circle
Equatation for distance from center to tangent is $\frac {(x_1x_0+y_1y_0-100)^2}
{x_1^2+y_1^2} =r^2$
But I dont have any results
| Note that $P = (8,0)$ is located inside the circle $O: x^2 + y^2 = 10^2$, which is centered at the origin and has radius $10$. A circle tangent to $O$ and passing through $P$ consequently will be internally tangent to $O$. Suppose such a circle, which we will label $Q$, has center $(x_Q, y_Q)$; then $(x_Q, y_Q)$ is equidistant from $P$ and the point of tangency of $Q$ to $O$, and this distance is the radius of $Q$, which we will denote $r_Q$. That is to say, $$r_Q^2 = (x_Q - 8)^2 + (y_Q - 0)^2 = \left(10 - \sqrt{x_Q^2 + y_Q^2}\right)^2.$$ The rightmost expression arises from the fact that the point of tangency of two circles lies on the line joining their centers; thus the distance of $(x_Q, y_Q)$ to the point of tangency of $Q$ to $O$ is equal to the radius of $O$ minus the distance of $(x_Q, y_Q)$ to the origin.
Simplifying the RHS equality then gives $$(x_Q - 8)^2 + y_Q^2 = 10^2 - 20 \sqrt{x_Q^2 + y_Q^2} + x_Q^2 + y_Q^2,$$ or $$16x_Q + 36 = 20 \sqrt{x_Q^2 + y_Q^2},$$ or $$(4x_Q + 9)^2 = 25(x_Q^2 + y_Q^2),$$ or $$9x_Q^2 - 72x_Q + 25y_Q^2 = 81,$$ Completing the square gives $$9(x_Q - 4)^2 + 25y_Q^2 = 225,$$ or in standard form, $$\frac{(x_Q - 4)^2}{5^2} + \frac{y_Q^2}{3^2} = 1.$$ This is an ellipse with center $(4,0)$, major semiaxis $5$, and minor semiaxis $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\sum\limits_n(-1)^{n-1}\frac{2^{n+1}}{2^{2n}-1}$ in terms of $\sum\limits_n\frac1{2^n-1}$ and $\sum\limits_n\frac1{2^n+1}$ Question:
let $$\sum_{n=1}^{+\infty}\dfrac{1}{2^n-1}=E,\sum_{n=1}^{+\infty}\dfrac{1}{2^n+1}=F$$ where $E,F$ are constant,(in fact,$E$ is Erdős-Borwein Constant ),Find the sum
$$f=\sum_{n=1}^{+\infty}(-1)^{n-1}\dfrac{2^{n+1}}{2^{2n}-1}$$
I tried this which I found in my texbook which seems relative, but Im not sure how to apply it to the problem:
$$f=\sum_{n=1}^{+\infty}(-1)^{n-1}\left(\dfrac{1}{2^n+1}+\dfrac{1}{2^n-1}\right)=\sum_{n=1}^{+\infty}\dfrac{(-1)^{n-1}}{2^n-1}+\sum_{n=1}^{+\infty}\dfrac{(-1)^{n-1}}{2^n+1}$$ following can't try
| It's not possible to express $f$ only with $E$ and $F$ because of
$\displaystyle\sum_{n=1}^\infty (-1)^{n-1}\frac{2^{n+1}}{2^{2n}-1}=$
$\displaystyle -\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n-1} +\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n+1}+\frac{i}{2}\sum_{n=1}^\infty\frac{1}{2^n+i} -\frac{i}{2}\sum_{n=1}^\infty\frac{1}{2^n-i}$
$\displaystyle -\frac{1}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n+\sqrt{2}} +\frac{1}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n-\sqrt{2}}+\frac{i}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n+i\sqrt{2}} -\frac{i}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n-i\sqrt{2}}$ .
You can define $\enspace\displaystyle g(x):=\prod_{n=1}^\infty (1+\frac{x}{2^n})\enspace$ (with the hope, that it exists a closed form for this product) and express $\enspace f\enspace$ by $\enspace\displaystyle \frac{d}{dx}\ln g(x)\enspace$ with $\enspace x\in\{-1;1;-i;i;-\sqrt{2};\sqrt{2};-i\sqrt{2};i\sqrt{2}\}$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$
Find the closed formula for the sequence and hence find the limit.
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}
\implies a_{n+1} = \frac{4a_n} {2a_n - 3}
\implies a_{n+1} = \frac{\frac{4a_n} {a_n} } {\frac{2a_n}{a_n} - \frac{3} {a_n} }
\implies a_{n+1} = \frac{4} {2 - \frac{3} {a_n} }
\implies \frac{1 } {a_{n+1}} = \frac{2 - \frac{3} {a_n} } {4}
\implies \frac{1 } {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}$$
Then go to where????
| So you got $\frac{1} {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}, a_1=1.$
Denote: $b_n=\frac{1}{a_n}$, then the equation becomes $b_{n+1}=-\frac{3}{4}b_n+\frac12$ with $b_1=\frac{1}{a_1}=1$.
Now apply the general formula: $$\bbox[lightgreen]
{
If \ \ y_{n+1}=ay_n+b,y_1=c, \ then: y_n=\left(c+\frac{b}{a-1}\right)a^{n-1}-\frac{b}{a-1}.
\qquad
} $$
Thus:
$$b_n=\left(1+\frac{\frac12}{-\frac34-1}\right)\left(-\frac34\right)^{n-1}-\frac{\frac12}{-\frac34-1}=\frac{5}{7}\left(-\frac34\right)^{n-1}+\frac27$$ and $$a_n=\frac{1}{b_n}=\frac{7}{2+5\left(-\frac34\right)^n}$$
And the limit: $$\lim_\limits{n\to+\infty} a_n=\frac72.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve the equation $x^4+(x-1)(x^2-2x+2)=0$ With $x \in \mathbb{R}$. Solve the equation $$x^4+(x-1)(x^2-2x+2)=0$$
Idea $1$ : $(x^2-x+1)(x^2+2x-2)=0$
Idea $2$ :$(2x^2+x-1)^2=(3x-3)^2$
Idea $3$ :$<=>x^4(x-1)x^2-2(x-1)^2=0$
Let $y=x-1 \dots$
I need another way because I am collecting methods of solving with quartic equations
|
Solve $x^4+(x-1)(x^2-2x+2)=0$
You can solve this using Newton-Raphson :
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}$$
Where \begin{align}&f(x)=x^4+x^3-3x^2+4x-2\\
&f'(x)=4x^3+3x^2-6x+4\\\end{align}
To guess $x_0$ :
\begin{align}&f(1)>0
\\&f(0)<0
\\\end{align}
So $x_0=1$ :
\begin{align}&x_1=0.5\\
&x_2=0.8\\
&x_3\approx0.6765\\
& \dots \\
&\dots \\
&\dots \\
&x_{22}\approx0.73205 \\\end{align}
The first Solution is $\boxed{x\approx0.73205}$ now Apply long division :
$$\frac{x^4+x^3-3x^2+4x-2}{x-0.73205 }=x^3+1.73205 x^2-1.73205 x+2.73205$$
Now solve this equation with the same method you should get the $2$nd Solution if there is one
$$x^3+1.73205x^2-1.73205 x+2.73205 \approx 0$$
If there is a Solution apply long division again to find another one until you come to a function that doesn't have Solutions where $x \in\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:
This then gave me the following equation to solve:
\begin{align}y^2&=2+3+4+x\\
&\Downarrow\\
x&=y^2-9\end{align}
And the following equations to do so:
\begin{align}\frac {ya}2 &=4\\
\frac {bc}2 &=3\\
\frac {yd}2 &=2\\
a+b&=y\\
c+d&=y\end{align}
I solved these to obtain:
$$a=2,
b=2,
c=3,
d=1,
y=4$$
And thus $$x=4^2-9=7$$
My question:
Is there another way I could have solved this, using the hint to think outside the box?
| There is a wonderful generalization of the solution for this problem:
$$A = \sqrt{(a + b + c)^2 – 4ac}.$$
Where $a$, $b$ and $c$ are the known areas, and $A$ is the unknown area.
Caveat: $a$ and $c$ must be the triangles whose long legs are the side of the square, in this case $7$ and $9$.
$$A = \sqrt{729 – 4 \times 9 \times 7} = \sqrt{729- 252} = 21.84.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 12,
"answer_id": 8
} |
Let $f(x) = 6\operatorname{arcsec}(2x)$. Find $f'(x)$. Let $f(x) = 6\operatorname{arcsec}(2x)$. Find $f'(x)$.
$$6 \cdot \frac 1{2x\sqrt{(2x)^2-1}} \cdot 2$$
$$\frac{12}{2x\sqrt{4x^2-1}}$$
$$\frac 6 {x\sqrt{4x^2-1}}$$
Why is that wrong?
| You see to say that $\displaystyle \sqrt{x^2-1} = \frac 1 2 \sqrt{(2x)^2 - 1}.$ That is incorrect. In fact
$$
\sqrt{x^2-1} = \frac 1 2 \cdot 2\sqrt{x^2-1} = \frac 1 2 \cdot \sqrt{4}\cdot\sqrt{x^2-1} = \frac 1 2 \sqrt{4x^2-4} = \frac 1 2 \sqrt{(2x)^2-4}.
$$
Now let $y = 6\operatorname{arcsec}(2x).$ Then
\begin{align}
\frac y 6 & = \operatorname{arcsec}(2x) \\[10pt]
\sec\frac y 6 & = 2x \\[10pt]
\frac 1 2 \sec\frac y 6 & = x \\[12pt]
\frac 1 2 \left(\sec\frac y 6\right) \cdot \left(\tan\frac y 6 \right) \cdot \frac 1 6 & = \frac{dx}{dy} \\[10pt]
x \cdot \left( \tan \frac y 6 \right) \cdot \frac 1 6 & = \frac{dx}{dy} \\[10pt]
\pm x \cdot \sqrt{\left( \sec\frac y 6 \right)^2 -1} \cdot \frac 1 6 & = \frac{dx}{dy} \\[10pt]
\pm \frac x 6 \cdot \sqrt{(2x)^2 - 1} & = \frac{dx}{dy} \\[10pt]
\frac{dy}{dx} & = \frac{\pm6}{x\sqrt{(2x)^2 - 1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Obtaining an explicit expression of $U_n$ given by $U_{n+1}=\frac{2}{3}U_n -1$ and $U_0=2$
Let $(U_n)_n\geq 0$ be the sequence defined by: $$U_0=2\qquad, U_{n+1}=\frac{2}{3}U_n -1\quad\text{for all } n \geq 0$$ .
My question Here is : How do I write $U_n$ with a function of $n$ ?
| You can use the method of generating functions. Let $U(x)=\sum_{n=0}^\infty U_n x^n$. We wish to solve the recurrence
$$
U_{n+1}=\frac{2}{3}U_n -1;\quad (n\geq0)\tag{1}
$$
where $U_0=2$. Multiply both sides of (1) by $x^n$ and sum on $n$ to get that
$$
\frac{U(x)-2}{x}=\frac{2}{3}U(x)-\frac{1}{1-x}.\tag{2}
$$
Solve for $U(x)$ and use partial fractions to get that
$$
U(x)
=\frac{2-3x}{(1-x)(1-(2/3)x)}
=\frac{-3}{1-x}+\frac{5}{1-(2/3)x}
=\sum_{n=0}^\infty(-3+5(2/3)^n)x^n
\tag{3}
$$
Hence
$$
U_{n}=-3+5\left(\frac{2}{3}\right)^n;\quad (n\geq0).
\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Find $(x + 1)^{−3}$ in $\mathbb{Z}_2[x]/(x^5 + x^4 + 1)$ In a previous question, I established that $x^5 + x^4 + 1$ is irreducible in $\mathbb{Z}_2[x]/(x^5 + x^4 + 1)$, so it will not be divisible by $x+1$. Hence $x^5+x^4+1$ will be coprime to $(x+1)^3 = x^3+x^2+x+1$
So far I have used the Euclidean algorithm to get:
$x^5+x^4+1 = (x^3+x^2+x+1)(x^2+1) + x$
$x^3+x^2+x+1 = (x^2+1)(x+1) + 0$
However, I know I have done something wrong - because the Euclidean algorithm should show that the gcd of $x^5+x^4+1$ and $x^3+x^2+x+1$ is $1$, not $(x^2+1)$.
| Here's my approach to this one, which does not really depend too much on the standard theorems on polynomial rings--except for the Euclidean Algorithm--but does rely on a few simple observations about the polynomial
$p(x) = x^5 + x^4 + 1 = 0 \in \Bbb Z_2[x]/(x^5 + x^4 + 1) \tag{1}$
and a lot of grinding away at synthetic division. Very brutesy-forcey. I have adopted the notation
$p(x) = x^5 + x^4 + 1 \tag{2}$
because it allows to save typing and $\LaTeX$ all over the place. I will also,for similar reasons, use the shorthand
$R = \Bbb Z_2[x]/(p(x)). \tag{2}$
These things being said:.
We have
$p(x) = x^5 + x^4 + 1 = 0 \in R; \tag{3}$
this much is, I think, obvious. Now (3) implies
$x^5 + x^4 = 1 \in R, \tag{4}$
which follows from adding $1$ to each side of (3) and using the fact that $R$ is of characteristic $2$.
Then
$x^4(x + 1) = x^5 +x^4 = 1, \tag{5}$
which proves that $1 + x$ is in fact a unit in $R$ and that its inverse is $x^4$. We may now take the third power of each side of (5); we find
$x^{12}(1 + x)^3 = 1; \tag{6}$
from this we see that
$(1 + x)^{-3} = x^{12}. \tag{7}$
In a sense, we are now done. However, for the sake of good form if nothing else, the effort should be nade to reduce $(1 + x)^{-3}$ to a polynomial of degree less than $5$, since in principle $R$ need contain no powers of $x$ greater than $4$ by virtue of (3) and (4). We accomplish this reduction by computing the renainder of $x^{12}$ when divided by $p(x)$, via synthetic division. And here is where brute force comes into play. In what follows I shall recite the steps of the computation, presenting the quotient and remainder at each one, denoting the quotient by $q(x)$ and the remainder by $r(x)$:
1.) we divide $p(x)$ into $x^{12}$, and find
$q(x) = x^7, r(x) = x^{11} + x^7; \tag{13}$
2.) $p(x)$ into $r(x)$:
$q(x) = x^6, r(x) = x^{10} + x^7 + x^6; \tag{14}$
3.) $p(x)$ into $r(x)$:
$q(x) = x^5, r(x) = x^9 + x^7 + x^6 +x^5; \tag{15}$
4.) $p(x)$ into $r(x)$:
$q(x) = x^4, r(x) = x^8 + x^7 + x^6 +x^5 +x^4\tag{16}$
5.) $p(x)$ into $r(x)$:
$q(x) = x^3, r(x) = x^6 +x^5 +x^4 +x^3; \tag{17}$
6.) $p(x)$ into $r(x)$:
$q(x) = x, r(x) = x^4 +x^3 +x; \tag{18}$
since $\deg r(x) < 5$, we may stop at this point, in accord with the Euclidean Algorithm. According to these calculations, we then have
$(1 + x)^{-3} = x^4 +x^3 +x. \tag{19}$
We can in fact check (20) as follows:
$(x^4 + x^3 + x)(1 + x)^3 = (x^4 + x^3 + x)(1 + x)^2(1 + x); \tag{20}$
since
$(1 + x)^2 = 1 + x^2 \tag{21}$
in $R$, we have
$(x^4 + x^3 + x)(1 + x)^2 = (x^4 + x^3 + x)(1 + x^2) = x^6 + x^5 + x^4 + x; \tag{22}$
at last, we note that via (3)
$x^6 + x^5 + x^4 + x = x(x^5 + x^4 + 1) + x^4 = x^4, \tag{23}$
and thus
$(x^4 + x^3 + x)(1 + x)^3 = x^4(1 + x) = x^5 + x^4 = 1, \tag{24}$
validating our calculation of
$(1 + x)^{-3} = x^4 + x^3 + x \in R. \tag{25}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$
Solve for $x:0\leq x \leq \frac{\pi}{2}$
$$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$$
My attempt,
I changed it into $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$
| HINT: use the substituion
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
where $$\tan\left(\frac{x}{2}\right)=t$$
then you will have
$$\frac{16t(1-t^2)}{(1+t^2)^2}-\frac{2\sqrt{6}t}{1+t^2}-\sqrt{6}\frac{(1-t^2)}{1+t^2}+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find when $\frac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}$ is an integer
Find all natural numbers $n$ such that $$\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}$$ is an integer.
For $n = 1$ we have $\dfrac{18}{3} = 6$.
For $n = 2$ we have $\dfrac{126}{18} = 7$.
For $n = 3$ we have $\dfrac{972}{126} = \dfrac{54}{7}$
For $n = 4$ we have $\dfrac{7938}{972} = \dfrac{49}{6}$.
I don't seem to find any other integers for the fraction for $n > 2$. How can we find all of the $n$ such that it is an integer?
| Hint: It looks like considering the top and bottom $\mod 4$ narrows it down quite a bit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof by induction, factorials and exponents I've spent many hours trying to find proof by induction to this problem:
$$(n!)^2<\left(\frac{(n+1)(2n+1)}{6}\right)^n$$
for n > 1.
Does anyone have any idea how to solve it? Any help would be appreciated.
| Using AM-GM inequality (which can be proved by induction on the number of terms),
$$\sqrt[n]{1^2\cdot 2^2\cdot3^2 \cdots n^2} \le \frac{1^2 + 2^2+3^2+\cdots + n^2}{n}$$
Equality holds iff $1^2 = 2^2 = \ldots = n^2$, which means equality does not hold for $n>1$.
On the left hand side,
$$(n!)^2 = 1^2 \cdot2^2 \cdot 3^2 \cdots n^2$$
which can be proved by induction on $n$.
On the right hand side,
$$1^2 + 2^2+3^2+\cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$
which can also be proved by induction on $n$.
Joining the three links together,
$$\sqrt[n]{(n!)^2} < \frac{(n+1)(2n+1)}{6}$$
Taking the $n$th power on both sides (which preserves order as both sides are positive) gives the required inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Infinitely many positive integers $n$ such that $n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1)$
Prove that there exist infinitely many positive integers $n$ such that $$n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1).$$
We need to prove that there are infinitely many $n$ such that $\prod_{m=0}^{n-1}(m^2+1)$ is divisible by $n^2+1$. Some examples I found for when it is divisible are $n = 3,7,8,13,17,18,21$, but I didn't see how to find infinitely many.
| Since $(m-i)(m-1+i)=\left(m^2-m+1\right)+i$, multiplying both sides by their conjugates gives
$$
(m^2-m+1)^2+1=\left(m^2+1\right)\left((m-1)^2+1\right)\tag{1}
$$
For $m\ge2$, we have that $n=m^2-m+1\gt m$. Therefore, $(1)$ shows that
$$
\left.n^2+1\,\,\middle|\,\,\prod_{k=0}^{n-1}\left(k^2+1\right)\right.\tag{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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The dimension of the class of matrices that suffice a diagonalization property. Let $A$ be a $2\times 2$ real matrix with distinct real eigenvalues. For a $2\times 2$ real matrix $B$, let $C(B)$ denote the $4\times 4$ matrix whose block form is
\begin{equation*}
\begin{bmatrix}
A & B\\
0 & A
\end{bmatrix}_.
\end{equation*}
I need to prove that the subset of $M_{2\times 2}(\mathbb{R})$ consisting of all $B$ for which the corresponding $C(B)$ is diagonalizable forms a two dimensional subspace of $M_{2\times 2}(\mathbb{R})$.
Any suggestions?
|
Disclaimer: I tried to keep this as low level as possible and it took a fair bit of work. There are probably more slick ways to do this with more advanced theory.
Let $x$ and $y$ denote the eigenvalues of $A$. Then the eigenvalues of $C(B)$ are also $x$ and $y$ (its characteristic polynomial is $(\lambda - x)^2(\lambda-y)^2$). $C(B)$ will be diagonalizable if and only if its minimal polynomial splits into distinct linear factors. In this case, that means that $C(B)$ is diagonalizable if and only if its minimal polynomial is $(\lambda - x)(\lambda - y)$. Therefore we need to have:
\begin{align*}
0 &= \begin{bmatrix}
A - xI & B \\ 0 & A - xI
\end{bmatrix} \begin{bmatrix} A-yI & B \\ 0 & A-yI\end{bmatrix} \\
&= \begin{bmatrix} (A - xI)(A-yI) & (A - xI)B + B(A-yI) \\ 0 & (A - xI)(A-yI) \end{bmatrix} \\
&= \begin{bmatrix} 0 & (A - xI)B + B(A-yI) \\ 0 & 0 \end{bmatrix}_.
\end{align*}
Note that $(A-xI)(A-yI) = 0$ because $(\lambda - x)(\lambda - y)$ is the minimal polynomial of $A$. Anyway, cleaning up the top right entry and setting it to zero yields:
$$ AB + BA = (x + y)B.$$
Define $T: M_{2\times 2}(\mathbb{R}) \to M_{2\times 2}(\mathbb{R})$ by $T(X) = AX + XA$. What we have discovered is that $C(B)$ is diagonalizable if and only if $B$ is an eigenvector of $T$ corresponding to the eigenvalue $x+y$. We now will show that the eigenspace of $T$ corresponding to $x+y$ is $2$-dimensional.
Let $P$ be the matrix such that $D := P^{-1}AP = \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix}_.$ Define $U(X) = DX + XD$. Then
\begin{align*}
P^{-1}T(X) P &= P^{-1}(AX + XA) P \\
&= P^{-1}AXP + P^{-1}XAP \\
&= (P^{-1}AP)(P^{-1}XP) + (P^{-1}XP)(P^{-1}AP) \\
&= D(P^{-1}XP) + (P^{-1}XP)D \\
&= U(P^{-1}XP).
\end{align*}
So if $T(B) = (x+y)B$, then:
\begin{align*}
(x+y)P^{-1}BP &= P^{-1}T(B)P \\
&= U(P^{-1}BP).
\end{align*}
Similarly if $U(X) = (x+y)X$, then you can verify that:
\begin{align*}
(x+y)PXP^{-1} &= PU(X)P^{-1} \\
&= T(PXP^{-1}).
\end{align*}
Therefore we have shown that the eigenspaces corresponding to $x+y$ for $T$ and $U$ are isomorphic (conjugation is an isomorphism and we showed that the map to each other under conjugation).
Finally we get to the easy part. We can now work with the much simpler $U$.
$$U\left( \begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = \begin{bmatrix} 2ax & b(x+y) \\ c(x+y) & 2dy \end{bmatrix}_.$$
Now to have $$ (x+y) \begin{bmatrix} a & b \\ c & d \end{bmatrix} = U\left( \begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = \begin{bmatrix} 2ax & b(x+y) \\ c(x+y) & 2dy \end{bmatrix} $$
we must have $a(x+y) = 2ax$ and $d(x + y) = 2dy$. Thus $a(x-y) = ax - ay = 0$. So $a = 0$ or $x-y=0$. But $x \neq y$, so $a = 0$. Similarly $d=0$. Note that $b$ and $c$ have not constraints on them. Therefore the eigenspace of $U$ corresponding to $x+y$ is $2$-dimensional, which implies that the eigenspace of $T$ corresponding to $x+y$ is $2$-dimensional. Finally we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Planes and Augmented Matrices I seem to be having trouble with solving the following system of equations so that an infinite number of solutions would arise. The question is "Find the values of $p$ and $q$ for which the following system of equations has an infinite number of solutions and clearly explain my reasoning:"
$$2x+y+z=5, x-y+z=3, -2x+py+2z=q$$
What I've managed to do is convert it as an augmented matrix and tried to solve from there:
$$\left[\begin{array}{rrr|r} 2 & 1 & 1 & 5\\ 1 & -1 & 1 & 3 \\ -2 & p & 2 & q\end{array}\right]$$
However, when I tried to solve the augmented matrix, I ended up with a solution that has $p$ on both sides in the third row. The answer in my textbook says that: "$t(p+10)=q+2$ has infinitely many solutions for $t$ when $p+10=0$ and $q+2=0, {\therefore}{p=-10, q=-2}$."
I can't seem to find that solution, no matter what I tried, so any help with the matrix to reach that solution would be greatly appreciated!!
| Begin with the augmented matrix
$$
\left(
\begin{array}{ccc|c} 2 & 1 & 1 & 5\\ 1 & -1 & 1 & 3 \\ -2 & p & 2 & q\end{array}
\right)
$$
and row reduce this by
*
*multiplying the first row by $-1/2$ and adding it to the second row, and
*adding the first row to the last row
to get:
$$
\left(
\begin{array}{ccc|c} 2 & 1 & 1 & 5\\ 0 & -\frac{3}{2} & \frac{1}{2} & \frac{1}{2} \\ 0 & p+1 & 3 & 5+q\end{array}
\right).
$$
Then multiply the second row by $-2/3$ to get:
$$
\left(
\begin{array}{ccc|c} 2 & 1 & 1 & 5\\ 0 & 1 & -\frac{1}{3} & -\frac{1}{3} \\ 0 & p+1 & 3 & 5+q\end{array}
\right).
$$
Next,
*
*multiply the second row by $-1$ and add it to get first row, and
*multiply the second row by $-(p+1)$ and add it to the third row:
$$
\left(
\begin{array}{ccc|c} 2 & 0 & \frac{4}{3} & \frac{16}{3}\\ 0 & 1 & -\frac{1}{3} & -\frac{1}{3} \\ 0 & 0 & 3+\frac{p+1}{3} & 5+q+\frac{p+1}{3} \end{array}
\right).
$$
The equation corresponding to the last row in the augmented matrix is
$$\left(3+\frac{p+1}{3}\right)z= 5+q+\frac{p+1}{3}.$$
Multiply both sides by three to obtain:
$$
(p+10)z = 15+3q+p+1,
$$
which is
$$
(p+10)z = p+3q+16.
$$
So what we want to do is find $p$ and $q$ so that $p+10=0$ and $p+3q+16=0$. This would mean that the system of equations will have an infinite number of solutions.
So $p=-10$ and $q=-2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Orthonormalbasis of a isometry $2 \times 2$ matrix I am a bit stuck in this easy question.
Let $A\in \mathbb R^{2 \times2}$ be a isometry so $A^{-1} = \bar{A}^T$. Show that $\left\{\begin{pmatrix} a \\ c \end{pmatrix},\begin{pmatrix} b \\ d \end{pmatrix}\right\}$ is a orthonormal basis.
$A = \begin{pmatrix} a&b \\ c&d \end{pmatrix} \Longrightarrow \bar{A}^T = \begin{pmatrix} a&c \\ b&d \end{pmatrix}$
$A \cdot \bar{A}^T = \begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd&c^2+d^2 \end{pmatrix} = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} $
I tried to use Gram-Schmidt but too many variables were left so I couldn't prove it.
Now i also know that is has to be
$\left\langle\begin{pmatrix} a \\ c \end{pmatrix}, \begin{pmatrix} b \\ d \end{pmatrix}\right\rangle =0 \Longrightarrow ab+cd = 0 $ and $\left\langle\begin{pmatrix} a \\ c \end{pmatrix}, \begin{pmatrix} a \\ c \end{pmatrix}\right\rangle =1 \Longrightarrow a^2+c^2 = 1$ but I also can't prove that because I only have $a^2+b^2 = 1 = c^2 + d^2$ and $ac+bd = 0$.
| Hint: Write
$$
v_1 = \pmatrix{a\\c} \qquad v_2 = \pmatrix{b\\d}
$$
Verify that for $A = \pmatrix{a&b\\c&d}$, we have
$$
A^TA = \pmatrix{
\langle v_1,v_1 \rangle & \langle v_2,v_1 \rangle\\
\langle v_1,v_2 \rangle & \langle v_2,v_2 \rangle}
$$
Since $A^{-1} = {A^T}$, we have ${A^T}A = I$.
| {
"language": "en",
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Equivalent quadratic form I want to find an integral quadratic form $a^2+2bxy+z^2$ with $b\equiv 0 \mod 81$ which is properly integrally equivalent to $3x^2+2xy+12y^2$.
In matrix worm that means that there exists a $2\times 2$-matrix $T$ over $\mathbb{Z}$ with $\det(T)=1$ and with $$T^T \begin{pmatrix} 3 & 1 \\ 1 & 12 \end{pmatrix} T = \begin{pmatrix} a & b \\ b & c \end{pmatrix}.$$
I tried as Ansatz matrices $T$ of the form $$\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \text { and } \begin{pmatrix} 1 & 1 \\ n-1 & n \end{pmatrix}.$$ With the first two I got congruences wit no solutions ($3n+1 \equiv 0 \mod 81$ and $12n+1 \equiv 0 \mod 81$), whereas with the third one I got and $12n^2-10n+2 \equiv 0 \mod 81$, which has the solution 41 (thank you WolframAlpha), but I don't know how to find it, so it doesn't make much sense to simply accept it.
Is there an easy way to find the equivalent form?
| There is an easy step-by-step solution:
we want $81|12n^2-10n+2$, which is equivalent to $81|6n^2-5n+1$.
$81|6n^2-5n+1$ implies $3|-5n+1$, so $n\equiv 2\pmod{3}$, so $n=3k+2$ for some $k$ ($k\in\{0,1,2,\dots,26\}$ if you're looking for a solution $n\in\mathbb{Z}_{81}$).
put $n=3k+2$ into $81|6n^2-5n+1$.
you will get $81|54k^2 + 57k + 15$, which is equivalent to $27| 18k^2+ 19k + 5$.
now we see that it has to be $9|19k + 5$, or $9|k+5$, so $k=9l+4$ for some $l$ ($l\in\{0,1,2\}$ if $n\le 81$).
put it into the divisibility and the end is near (you will have $3|163l+41$, equivalent to $3|l+2$, so for $l\in\{0,1,2\}$ we see that $l$ turns out to be $1$, what finally gives $n=3(9\cdot 1+4)+2=41$).
| {
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"timestamp": "2023-03-29T00:00:00",
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sum of integer parts of roots
If $\alpha,\beta,\gamma$ are real roots of the equation $\displaystyle \frac{1}{x}+\frac{1}{x-1}+\frac{1}{x-2} = \frac{11}{54}x^2.$ Then $\lfloor \alpha \rfloor + \lfloor \beta \rfloor +\lfloor \gamma \rfloor $
$\bf{Attempt}$ From equation $$\frac{x^2-3x+2+x^2-2x+x^2-x}{x(x-1)(x-2)} = \frac{11}{54}x^2$$
So $$\frac{3x^2-6x+2}{x^3-3x^2+2x} = \frac{11}{54}x^2$$
$$162x^2-324x+108 = 11x^5-33x^4+22x^3$$
$$11x^5-33x^4+22x^3-162x^2+324x-108 =0$$
Could some help me how to solve it, thanks
| HINT: use that $$\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x-2}-\frac{11}{54}x^2=-\frac{(x-3) \left(11 x^4+22 x^2-96 x+36\right)}{54 (x-2) (x-1) x}$$
| {
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"question_score": "1",
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How to find basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T$?
Find the basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T:M^{\mathbb R}_{2 \times 2} \to M^{\mathbb R}_{2 \times 2}$ defined as $T(A)=A-A^t$ for all $A \in M^{\mathbb R}_{2 \times 2}$.
Let $A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$. Then:
$$
T(A)=\begin{pmatrix} a&b\\c&d \end{pmatrix}-\begin{pmatrix} a&c\\b&d \end{pmatrix}=\begin{pmatrix} 0&b-c\\c-b&0 \end{pmatrix}\stackrel{R_2 \gets-1\cdot R_2}{=}\begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix}
$$
In order to find $\ker T$ we'll evalute $T(A)=0$:
$$
\begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix}=\begin{pmatrix} 0&0\\0&0 \end{pmatrix}\Rightarrow b=c \Rightarrow \ker T=span \Biggl\{ \begin{pmatrix} 1&0\\0&0 \end{pmatrix}, \begin{pmatrix} 0&1\\1&0 \end{pmatrix},\begin{pmatrix} 0&0\\0&1 \end{pmatrix} \Biggr\}
$$
Thus $\dim(\ker T)=3$.
Regarding $\mathrm{Im} T$:
$$
ImT=\Biggl\{ \begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix} \text{such that}\quad b,c, \in \mathbb R \Biggr\}=span\Biggl\{ \begin{pmatrix} 0&1\\1&0 \end{pmatrix} \Biggr\}
$$
Thus $\dim(\mathrm{Im} T)=1$.
Are my calculations correct?
| You can use the fact that every square matrix is, in a unique way, the sum of a symmetric and an antisymmetric matrices.
Any symmetric matrix belongs to the kernel of $T$, whereas the antisymmetric matrix
$$
A=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
$$
does not belong to $\ker T$. Since the space of symmetric matrices has dimension $3$, with basis
$$
\left\{
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
\right\}
$$
this is also a basis for $\ker T$.
A basis of the image consists of
$$
T(A)=\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}
$$
| {
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For which $c$ there exists $t\in\mathbb{R}^+$ such that $x_1(t)=0$ and $x_2(t)>0$ for the system $x_1'(t)=-200kx_2(t)-c\ ,\ x_2'(t)=-100kx_1(t)$? For which $c\in\mathbb{R}$ there exists $t\in\mathbb{R}^+$ such that $x_1(t)=0$ and $x_2(t)>0$ for the system $x_1'(t)=-200kx_2(t)-c\ ,\ x_2'(t)=-100kx_1(t)\ ,\ x_1(0)=10^4\ ,\ x_2(0)=5\cdot 10^3\ ,\ k>0$.
What I did: I found this form of $X(t)$:
$$\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=(\frac{c}{400\sqrt2k}+2500(1+\sqrt2))e^{-100\sqrt2kt}\begin{pmatrix}\sqrt2\\1\end{pmatrix}+(\frac{c}{400\sqrt2k}+2500(1-\sqrt2))e^{100\sqrt2kt}\begin{pmatrix}-\sqrt2\\1\end{pmatrix}
-\begin{pmatrix}0\\\frac{c}{200k}\end{pmatrix}$$
If $x_1=0$ we get:
$$t=\frac{10^2\cdot ln(\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)})}{2\sqrt2k}$$
Which means that for $x_1$ to be able to reach $0$ we must have:
$$c>10^6\cdot\sqrt2k(\sqrt2-1)$$
I wasn't able to simplify the expression for $x_2=0$ and arrive at a final answer.
| You already expressed $t$ (such that $x_1(t)=0$ in terms of $c$). By the way, there should be
$$t=\frac{ln\left(\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)}\right)}{200\sqrt2k}.$$
So it remains to make the respective substitution in the formula for $x_2(t)$ and then to find when $x_2(t(c))>0$.
That is
$$\left(\frac{c}{400\sqrt2k}+2500(1+\sqrt2)\right)\sqrt{\frac{c+10^6\cdot\sqrt2k(1-\sqrt2)}{c+10^6\cdot\sqrt2k(1+\sqrt2)}}+$$
$$\left(\frac{c}{400\sqrt2k}+2500(1-\sqrt2)\right)\sqrt{\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)}}>\frac{c}{200k}$$
$$\left(c+10^6\sqrt2k(1+\sqrt2)\right)\sqrt{\frac{c+10^6\cdot\sqrt2k(1-\sqrt2)}{c+10^6\cdot\sqrt2k(1+\sqrt2)}}+$$
$$\left(c+10^6\sqrt2k(1-\sqrt2)\right)\sqrt{\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)}}>2\sqrt2c$$
$$2\sqrt{(c+10^6\cdot\sqrt2k(1-\sqrt2))(c+10^6\cdot\sqrt2k(1+\sqrt2)})>2\sqrt2c$$
$$(c+10^6\cdot\sqrt2k(1-\sqrt2))(c+10^6\cdot\sqrt2k(1+\sqrt2))>2c^2$$
$$10^6\cdot2\sqrt2kc-10^{12}\cdot 2k^2>c^2$$
$$c^2-10^6\cdot2\sqrt2kc+10^{12}\cdot 2k^2<0$$
$$(c-10^6\sqrt2k)^2<0,$$
which is impossible.
PS. This simple answer suggests that there may be a more simple solution.
| {
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How to find inverse Laplace transform of $\frac{\sqrt{s} \sqrt{a+s}}{b+s}$ How to find inverse Laplace transform of
$$\frac{\sqrt{s} \sqrt{a+s}}{b+s}$$
where $a$ and $b$ constants, ${a,b}>0$
I tried to solve it, but I couldn't.
EDITED.
Numerical solution for $a=1,b=1$.
Plot:
| One method of consideration may be applied to the following. By using the series expansion
$$\sqrt{1-x} = 1 - \sum_{k=1}^{\infty} \frac{(2k-2)! \, x^{k}}{2^{2k-1} \, k! \, (k-1)!}.$$
Now the fraction in question becomes:
\begin{align}
\frac{\sqrt{s \, (s+a)}}{s+b} &= \frac{\sqrt{\left(s+ \frac{a}{2}\right)^{2} - \frac{a^{2}}{4}}}{s+b} = \frac{s + \frac{a}{2}}{s+b} \cdot \sqrt{1 - \frac{a^{2}}{4 \, \left(s + \frac{a}{2}\right)^{2}}} \\
&= \frac{s + \frac{a}{2}}{s+b} - \sum_{k=1}^{\infty} \frac{(2k-2)! \, \left(\frac{a}{2}\right)^{2k}}{2^{2k-1} \, k! \, (k-1)!} \cdot \frac{1}{(s+b) \, \left(s + \frac{a}{2}\right)^{2k-1}}.
\end{align}
By using the convolution theorem it is developed into:
\begin{align}
\frac{(2k-2)!}{(s+b) \, \left(s + \frac{a}{2}\right)^{2k-1}} &\Doteq \int_{0}^{t} e^{-b (t-u)} \, u^{2k-2} \, e^{- \frac{a u}{2}} \, du \\
&\Doteq \frac{e^{-bt}}{\left(\frac{a}{2} - b\right)^{2k-1}} \, \int_{0}^{\left(\frac{a}{2} - b\right) \, t} e^{-x} \, x^{2k-2} \, dx \\
&\Doteq \frac{e^{-bt}}{\left(\frac{a}{2} -b\right)^{2k-1}} \, \gamma\left(2k-1, \left(\frac{a}{2} - b\right) \, t \right),
\end{align}
where $\gamma(a,x)$ is the incomplete Gamma function. Putting all the components together yields
$$\frac{\sqrt{s \, (s+a)}}{s+b} \Doteq \delta(t) + \left(\frac{a}{2} - b\right) \, e^{-b t} \, \left[1 - 2 \, \sum_{k=1}^{\infty} \frac{\left(\frac{a}{2 \, (a-2b)}\right)^{2k} \, \gamma\left(2k-1, \left(\frac{a}{2} -b\right) \, t \right) }{ k! \, (k-1)!} \right]. $$
Notice that if $b = \frac{a}{2}$ then this reduces to
$$\frac{\sqrt{s \, (s+a)}}{s+\frac{a}{2}} \Doteq \delta(t) - \frac{a^{2} \, t}{8} \, e^{- \frac{a \, t}{2}} \, {}_{1}F_{2}\left(\frac{1}{2}; \frac{3}{2} , 2; \frac{a^{2} \, t^{2}}{16} \right).$$
| {
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"source": "stackexchange",
"question_score": "2",
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How to solve $\lim _{x\to \pi/2}\left(\tan x-1\right)\left(1-\tan\left(\frac{x}{2}\right)\right) =\infty\cdot 0$? $\lim _{x\to \pi/2}\left(\tan x-1\right)\left(1-\tan\left(\frac{x}{2}\right)\right) $ Is there a way to solve this limit without any fancy trigonometric identity?
| Assuming that you can use Taylor series.
First, let $x=y+\frac \pi 2$ to make $$(\tan (x)-1)\left(1-\tan \left(\frac{x}{2}\right)\right) =(\cot (y)+1)\left(\tan \left(\frac{y}{2}+\frac \pi 4 \right)-1\right) $$ Now, using $$\cot(y)=\frac{1}{y}-\frac{y}{3}+O\left(y^3\right)$$ $$\tan \left(\frac{y}{2}+\frac \pi 4 \right)=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ then
$$A=(\cot (y)+1)\left(\tan \left(\frac{y}{2}+\frac \pi 4 \right)-1\right)=\left(1+\frac{1}{y}-\frac{y}{3}+O\left(y^3\right) \right)\left(y+\frac{y^2}{2}+O\left(y^3\right)\right)$$ Expanding and simplifying $$A=1+\frac{3 y}{2}+O\left(y^2\right)$$ which shows the limit and how it is approached.
| {
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Find the measure of angle $MAN$. We construct two points $M,N$ such that perimeter of triangle $MCN$ is equal to half of the perimeter of the square.Find the measure of angle $MAN$.
The case $CM=CN$ leads to a simple equation so we can find the measure of $MAN$ using trigonometry.But in general case I don't no How to use equality of perimeter maybe we have to construct the same triangle in another place and use the sum of their perimeter.But I didn't get anything from this method.Any hints?
| I think I understood about which square you say!
I think your problem is the following.
Let $ABCD$ is a square with $AB=a$. $M\in BC$ and $N\in CD$ such that $CN+CM+MN=2a$. Find $\measuredangle MAN$.
Let $MC=x$ and $CN=y$.
Hence, $$x+y+\sqrt{x^2+y^2}=2a,$$
$$AM=\sqrt{(a-x)^2+a^2}=\sqrt{2a^2-2ax+x^2}=\sqrt{\frac{1}{2}\left(x+y+\sqrt{x^2+y^2}\right)^2-\left(x+y+\sqrt{x^2+y^2}\right)x+x^2}=$$
$$=\sqrt{\sqrt{x^2+y^2}\left(y+\sqrt{x^2+y^2}\right)}.$$
By the same way we obtain:
$$AN=\sqrt{\sqrt{x^2+y^2}\left(x+\sqrt{x^2+y^2}\right)}.$$
Thus,
$$\measuredangle MAN=\arccos\frac{AM^2+AN^2-MN^2}{2AM\cdot AN}=$$
$$=\arccos\frac{4a^2-2a(x+y)+x^2+y^2-x^2-y^2}{2\sqrt{x^2+y^2}\sqrt{\left(y+\sqrt{x^2+y^2}\right)\left(x+\sqrt{x^2+y^2}\right)}}=$$
$$=\arccos\frac{2a(2a-x-y)}{2\sqrt{x^2+y^2}\sqrt{x^2+xy+y^2+(x+y)\sqrt{x^2+y^2}}}=$$
$$=\arccos\frac{a}{\sqrt{2a^2}}=\arccos\frac{1}{\sqrt2}=45^{\circ}.$$
Done!
| {
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if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:
| $$1+\dfrac a{x-a}=\dfrac x{x-a}$$
$$1+\dfrac a{x-a}+\dfrac{bx}{(x-a)(x-b)} =x\left[\dfrac1{x-a}+\dfrac b{(x-a)(x-b)}\right]=\dfrac{x^2}{(x-a)(x-b)}$$
$$1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}=\cdots=\dfrac{x^3}{(x-a)(x-b)(x-c)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to use mittag-leffer expansion to compute $\texttt{sech} z$? A question concerning a paper from Kesten (1986) In Kesten's 1986 paper (Limit distribution of Sinai's Random Walk) we read:
The reference is E.T. Whittaker and G.N. Watson, A Course of Modern Analysis, 4th ed. (Cambridge Univ. Press, Cambridge, 1952).
We see that:
Using this result we can begin to do some manipulations:
First write
$$\frac{1}{\cosh(\sqrt{2\theta})} = \texttt{sech} \sqrt{2\theta} = \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 + 8\theta} $$
Then we come back to the expression we are interested in:
$$\Bbb{E}\{ e^{-\theta L}; L>0\} =\frac{1}{2\theta} \bigg[1 - \texttt{sech}(\sqrt{2\theta})\bigg]\\
= \frac{1}{2\theta} \bigg[1 - \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 + 8\theta}\bigg] $$
But how do we obtain the second line
$$ \frac{16}{\pi} \sum_{k=0}^\infty \frac{(-1)^k}{(2k + 1)}\frac{1}{(2k+1)^2\pi^2 + 8\theta} ?$$
| Note first that $\texttt{cosh} (0) = 1$ so
$$ 1 = \frac{1}{\cosh(0)} = \texttt{sech} (0) = \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 }$$
Therefore
$$ \Bbb{E}\{ e^{-\theta L}; L>0\} =\frac{1}{2\theta} \bigg[1 - \texttt{sech}(\sqrt{2\theta})\bigg]\\
= \frac{1}{2\theta} \bigg[ \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 } - \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 + 8\theta}\bigg]\\
=\frac{4\pi}{2\theta}\bigg[\sum_{k=0}^\infty \frac{8\theta (2k + 1) (-1)^k }{(2k+1)^2\pi^2} \frac{1}{(2k+1)^2\pi^2 + 8\theta}\bigg]\\
=\frac{16}{\pi}\bigg[\sum_{k=0}^\infty \frac{ (-1)^k }{(2k+1)} \frac{1}{(2k+1)^2\pi^2 + 8\theta}\bigg]$$
which is the desired result
| {
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How to calculate multiplicative inverses in $GF(2^3)$ without the Euclidean algorithm The problem I have concerns finite field arithmetic in $GF(p^k)$.
I know how to find multiplicative inverses using the extended Euclidean algorithm, but for my exams I need to calculate multiplicative inverses in $GF(2^3)$ without it.
What's the best way to do this? Is there even a convenient way?
The irreducible polynomial I have is $x^3 + x + 1$.
| As an alternative, build the table of the so called discrete logarithm. Let $a$ be a root of $f = x^{3} + x + 1$. Then
\begin{align}
&a^{0} = 1\\
&a^{1} = a\\
&a^{2} = a^{2}\\
&a^{3} = a + 1\\
&a^{4} = a^{2} + a\\
&a^{5} = a^{2} + a + 1\\
&a^{6} = a^{2} + 1\\
&a^{7} = 1\\
\end{align}
So to compute the inverse of $a^{2} + a$, say, you note that $a^{2} + a = a^{4}$. Since $a^{7} = 1$, the inverse of $a^{4}$ is $a^{3} = a + 1$.
Note that in building the table you are doing Euclidean divisions in a simplified form. For instance
$$
a^{5} = a^{4} a = (a^{2} + a) a = a^{3} + a^{2} = a + 1 + a^{2}
$$
since $a$ is a root of $x^{3} + x + 1$, and thus $a^{3} = a + 1$.
| {
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Find the minimum of the value $3x^2-2xy$ if $\frac{x^2}{4}-y^2=1$ Let $x,y\in R$,such
$$\dfrac{x^2}{4}-y^2=1$$
find the minium of the
$$3x^2-2xy$$
I think $x=2\sec{t},y=\tan{t}$,then
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$
| Let $3x^2-2xy=k$.
Since $x^2=4y^2+4$, we obtain:
$$k=3x^2-2xy\geq3x^2-(x^2+y^2)=2(4+4y^2)-y^2=8+7y^2\geq8.$$
Hence,
$$3x^2-2xy=k\left(\frac{x^2}{4}-1\right)$$ or
$$\left(3-\frac{k}{4}\right)x^2-2xy+ky^2=0,$$
which gives
$$1-k\left(3-\frac{k}{4}\right)\geq0$$ or
$$\frac{k^2}{4}-3k+1\geq0,$$ which is
$k\geq2(3+2\sqrt2)$ or $k\leq2(3-2\sqrt2)$ and since $k\geq8$, we obtain:
$$k\geq2(3+2\sqrt2).$$
The equality occurs for $k=2(3+2\sqrt{2})$, $\frac{x^4}{4}-y^2=1$ and $\frac{x}{y}=\frac{1}{3-\frac{2(3+2\sqrt2)}{4}}=2(3+2\sqrt2),$
which says that the equality occurs.
Thus, we got the answer:
$$\min_{\frac{x^2}{4}-y^2=1}(3x^2-2xy)=2(3+2\sqrt2).$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$
Let $\{a,b,c\}\subset\mathbb{R^+}$ such that $a+b+c =1$. Prove that :
$$a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$$
$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$
$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$
Now what ?
| the trick is here to write
$$3(a^2+b^2+c^2)\geq (a^2+b^2+c^2)(a+b+c)$$ this is equivalent to
$$(a-b)(a^2-b^2)+(b-c)(b^2-c^2)+(c-a)(c^2-a^2)\geq 0$$
which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ I have the following problem:
How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$
My attempt:
\begin{align*}
\exp(ix)=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}
\end{align*}
Now split the sum in 4 sums because $i^n$ has period 4, more precisely $i^{4n}=1$, $i^{4n+1}=i$, $i^{4n+2}=-1$ and $i^{4n+3}=-i$ for $n\geq 0$.
\begin{align*}
\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}&=\sum_{n=0}^{\infty}\frac{(ix)^{4n}}{(4n)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+1}}{(4n+1)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+2}}{(4n+2)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+3}}{(4n+3)!}\\
&=\sum_{n=0}^{\infty}\left(\frac{x^{4n}}{(4n)!}-\frac{x^{4n+2}}{(4n+2)!}\right)+i\sum_{n=0}^{\infty}\left(\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+3}}{(4n+3)!}\right)
\end{align*}
Then $4n$ and $4n+2$ are nonnegative and even numbers and $4n+1$ and $4n+3$ are odd. How can I use this to convert the two last expressions into
\begin{align*}
\cos x&=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m}}{(2m)!}\\
\sin x&=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m+1}}{(2m+1)!}
\end{align*}
I don't want to see proofs like
\begin{align*}
\exp(ix)&=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}\ldots\\
&=\Big(1-\frac{x^2}{2!}+\frac{x^4}{4!}\mp\ldots\Big)+i\Big(x-\frac{x^3}{3!}+\frac{x^5}{5!}\mp\ldots\Big)\\
&=\cos x+i\sin x
\end{align*}
|
We obtain
\begin{align*}
\cos(x)&=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}\\
&=\sum_{{n=0}\atop{n \text{ even}}}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}+\sum_{{n=0}\atop{n \text{ odd}}}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}\\
&=\sum_{{n=0}}^{\infty} \frac{x^{4n}}{(4n)!}-\sum_{{n=0}}^{\infty} \frac{x^{4n+2}}{(4n+2)!}\\
&=\sum_{{n=0}}^{\infty}\left( \frac{x^{4n}}{(4n)!}- \frac{x^{4n+2}}{(4n+2)!}\right)
\end{align*}
and similarly for $\sin(x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving for $x$ in the absolute value equation $$|x-1|+|2x-3|=|3x-4|$$
I tried to solve the cases separately for the range of values of $x$, but it did not work. Can anyone please show me the way.
| Twisted/Confusing way to solve this problem:
$|x-1|+|2x-3|=|3x-4|$...eq(I). Let us call the three expressions as $A, B, C$
, so we have $A+B=C$
$\begin{matrix} \text{when} & \text{expression}\\ x\geq1&A=x-1 & i\\x<1& A=-(x-1) & ii\\ x\geq\frac{3}{2} & B=2x-3 & iii\\ x<\frac{3}{2} & B=-(2x-3) & iv\\ x\geq \frac{4}{3} & C=3x-4 & v\\ x<\frac{4}{3} & C=-(3x-4) & vi
\end{matrix}$
*
*When we say that $x\geq \frac{3}{2}$ we also say that $x\geq1$ and $x\geq \frac{4}{3}$; $(\frac{3}{2}-\frac{4}{3}=\frac{1}{6}\Rightarrow \frac{3}{2}>\frac{4}{3})$
LHS of eq (I) $= i + iii = v$ RHS of eq (I)
*
*When we say that $x\leq 1$ we also say that $x\leq \frac{4}{3}$ and $x\leq \frac{3}{2}$
LHS of eq (I) $= ii + iv = vi $ RHS of eq (I)
for all the values of $x \leq 1 $ and $ x \geq \frac{3}{2}$, eq (I) holds
*
*When $ 1 < x < \frac{4}{3} $, LHS of eq (I) $i + iv \neq vi $ RHS of eq (I), eq (i) does not hold
*When $ \frac{4}{3} < x < \frac{3}{2}$, LHS of eq (I) $ i + iv \neq v$ RHS of eq (I) ,eq (I) does not hold
So the solution is $x\leq 1$ and $x\geq \frac{3}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate per unit cost? I need help understanding this math problem from my TEAS Mometrix Practice exam. It gives an explanation but I still don't understand. The question is:
Mandy can buy $4$ containers of yogurt and $3$ boxes of crackers for $\$9.55$. She can buy $2$ containers of yogurt and $2$ boxes of crackers for $\$5.90$. How much does one box of crackers cost?
The answer is $ \$2.25$.
I understand that I have to set up the problem like this:
\begin{align}
4x+3y &= 9.55 \\
2x+2y &= 5.90
\end{align}
but the explanation says that I have to multiply the bottom equation by $-2$; therefore, it becomes $-4x+ -4y= -11.80$
and then I have to add the equations
\begin{align}
4x+3y &= 9.55\\
-4x+-4y &= -11.80
\end{align}
by top to bottom getting either $\pm 2.25$
Can you explain why this is the answer? And also why does one have multiply by -2?
| There are several ways to solve this or similar problems we have
$$4 \cdot x + 3 \cdot y = 9.55 \\
2 \cdot x + 2 \cdot y = 5.90$$
One approch as given in your answer is to multiply the second equation by 2 or -2 you then have $4 \cdot x$ in both equations so can cancel for $x$ by adding the two equations or subtracting the equations as appropriate. The reason this works is that by multiplying all terms by the same number you have created another true statement.
This is not the only approach however, Another approach would be for example to solve one equation for say $y$ then substituting it in the other.
$$2 \cdot x + 2 \cdot y = 5.90 \Rightarrow y = \frac{5.90 - 2 \cdot x}{2}$$
Putting that in the other equation we have
$$4 \cdot x + 3 \cdot y = 4 \cdot x + 3 \cdot \frac{5.90 - 2 \cdot x}{2} = 4 \cdot x + 8.85 - 3 \cdot x= 9.55 \Rightarrow x = 9.55 - 8.85 = 0.7$$
Putting this back in the other equation
$$ 2 \cdot x + 2 \cdot y = 2 \cdot 0.7 +2 \cdot y = 5.90 \Rightarrow y = \frac{5.9 - 1.4}{2} = 2.25$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1$ When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:
$$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$
This might be preliminary knowledge, I'm not good at mathematics, any body could give me a hint?
| Since $1= 2-1$, you can multiply by $2-1$ and it won't change the value.
$$(2-1)(2^0+2^1+2^2+\cdots + 2^n) $$
$$=2(2^0+2^1+2^2+\cdots + 2^n) -1(2^0+2^1+2^2+\cdots + 2^n)$$
$$= (2^1+2^2+2^3+\cdots + 2^{n+1}) - (2^0+2^1+2^2+\cdots +2^n)$$
$$= 2^{n+1}-2^0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Possibilities for the side and angle of the triangle We have the triangle ABC with $c=18$, $\alpha=\frac{\pi}{6}$ and the median through $C$ has the length $5$. I want to determine all the possibilities for $a$ and $\gamma$.
I have done the following:
From the median we have the following:
$$m_c=\sqrt{\frac{2a^2+2b^2-c^2}{4}}\Rightarrow m_c^2=\frac{2a^2+2b^2-c^2}{4} \Rightarrow 4m_c^2=2a^2+2b^2-c^2 \\ \Rightarrow 4\cdot 25=2a^2+2b^2-18^2 \Rightarrow 2a^2+2b^2=100+324 \Rightarrow 2a^2+2b^2=424 \\ \Rightarrow a^2+b^2=212 \Rightarrow b^2=212-a^2$$
From the cosine law we have that $$a^2=b^2+c^2-2\cdot b\cdot c\cdot \cos \alpha \Rightarrow a^2=212-a^2+c^2-2\sqrt{212-a^2}\cdot c\cdot \frac{\sqrt{3}}{2} \\ \Rightarrow 2a^2=212+324-18\sqrt{3}\sqrt{212-a^2}\Rightarrow 2a^2=536-18\sqrt{3}\sqrt{212-a^2} \\ \Rightarrow a^2=268-9\sqrt{3}\sqrt{212-a^2} \ \ \
\ (1)$$
From the sine law we have that $$\frac{\sin\alpha}{a}=\frac{\sin \gamma}{c} \Rightarrow c\sin \alpha=a\sin \gamma \Rightarrow 18\cdot \frac{1}{2}=a\sin \gamma \Rightarrow a\sin \gamma=9 \ \ \ \ (2)$$
So from the equation we solve for $a$ and from the equation we solve for $\gamma$, right?
Are they then all the posiibilities for $a$ and $\gamma$ ?
| I think it is easier to find $C$ by straightedge and compass, then study $\gamma$.
Let $M$ be the midpoint of $AB$: $C$ must lie on a circle centered at $M$ with radius $5$, and on a fixed line from $A$. It follows that there are two solutions, namely $C_1$ and $C_2$ in the diagram below.
By the secant-tangent theorem we have $AC_1\cdot AC_2=4\cdot 14=56$ and by depicting the midpoint of $C_1 C_2$ (that is the foot of an altitude in the isosceles triangle $MC_1 C_2$) we also have
$$ \frac{AC_1+AC_2}{2}=9\cos 30^{\circ} = \frac{9}{2}\sqrt{3}$$
hence $AC_1$ and $AC_2$ are given by $\frac{9\sqrt{3}\pm\sqrt{19}}{2}$. In both cases $BC$ can be computed from the cosine theorem, then $\sin\gamma$ can be derived from the sine theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Application Of De Moivre & Euler's Formulae Can anyone show me how I can prove that $\sin x \cos (3x) = \frac{1}{4} \sin (7x) - \frac{1}{4}\sin (5x) + \frac{1}{2}\sin x$?
I tried using Euler's formulae
$$\sin x= \frac{e^{ix}-e^{-ix}}{2i}$$
and
$$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$
but the simplification didn't help at all.
PS: Simplify starting from the left.
| It should be $\sin { x } \cos ^{ 2 }{ 3x } $
$$\frac { 1 }{ 4 } \sin { 7x } -\frac { 1 }{ 4 } \sin { 5x } +\frac { 1 }{ 2 } \sin { x } =\frac { 1 }{ 4 } \left( \sin { 7x } -\sin { 5x } +2\sin { x } \right) =\\ =\frac { 1 }{ 4 } \left( 2\sin { \frac { 7x-5x }{ 2 } \cos { \frac { 7x+5x }{ 2 } } } +2\sin { x } \right) =\frac { 1 }{ 2 } \left( \sin { x } \left( \cos { 6x } +1 \right) \right) =\\ =\frac { \sin { x } }{ 2 } \left( 2\cos ^{ 2 }{ 3x } \right) =\sin { x } \cos ^{ 2 }{ 3x } \\ $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the inequality $\frac{x}{x+1} - \frac{1}{x-3} - 2 \leq 0$ I tried to solve
$$ \dfrac{x}{x+1} - \dfrac{1}{x-3} - 2 \leq 0$$
and I got
$$S = \{x \in \mathbb{R} \mid x \leq -\sqrt{5} \vee x \geq \sqrt{5}\}$$
but I have plotted and here is the function graph. It doesn't seem to be a quadratic inequality.
Does someone have a hint on how to solve it?
| Consider $x-3> 0$
$$\dfrac{x}{x+1} - \dfrac{1}{x-3} - 2 \leq 0$$
$$\Leftrightarrow x-\dfrac{x+1}{x-3}-2(x+1)\leq 0 \space\space ||x+1 > 0$$
$$\Leftrightarrow \dfrac{x(x-3)}{x-3}-\dfrac{x+1}{x-3}-\dfrac{2(x+1)(x-3)}{x-3}\leq 0$$
$$\Leftrightarrow x^2-3x-x-1-2(x^2-2x-3) \leq 0 \space \space ||x -3>0$$
$$\Leftrightarrow -x^2+5 \leq 0$$
$$\Leftrightarrow x > \sqrt5$$
So, when $x > 3$, the above holds. You can do the other two cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2366556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rational integral $$\int\dfrac{x^2}{1+x^4}dx$$
I tried many standard approaches, but I didn't get too far! Here's the most promising of them:
$$\int\dfrac{dx}{\frac{1}{x^2}+x^2}$$
knowing that $\left(1/x+x\right)^2=\frac{1}{x^2}+x^2+2$ we can change variables $1/x+x=t$. Unluckily this doesn't work either.
| Hint: Express
$$\frac{x^2}{1+x^4}=\frac{ax+b}{x^2-\sqrt{2}x+1}+\frac{cx+d}{x^2+\sqrt{2}x+1}$$
and find $a,b,c,d$.
| {
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"timestamp": "2023-03-29T00:00:00",
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For $0I only prove that for $0<x<y\leq 1/2$ the result is right as following:
the inequality is equivalent to
$$x^{2x+1}+2x^{x+1}y^y+x y^{2y}<y x^{2y}+2y^{x+1}x^y+y^{2x+1}.$$
First, $x^{2x+1}<y^{2x+1}$ because of $$\left(\frac{y}{x}\right)^{2x+1}>1;$$
Second, $2x^{x+1}y^y<2y^{x+1}x^y$ because of
$$\frac{2y^{x+1}x^y}{2x^{x+1}y^y}=\left(\frac{y}{x}\right)^{x+1-y}>1;$$
Last, for $0<x<y\leq 1/2$,
$xy^{2y}<yx^{2y}$ because of
$$\frac{yx^{2y}}{xy^{2y}}=\left(\frac{y}{x}\right)^{1-2y}\geq 1.$$
| $$\frac{x^x+y^y}{x^y+y^x}<\frac{x^x+y^y}{2\sqrt{x^yy^x}}\ \ (\because a+b>2\sqrt{ab})$$
$$=\frac{1}{2}\left(\sqrt{\frac{x^{2x}}{x^yy^x}}+\sqrt{\frac{y^{2y}}{x^yy^x}}\right)$$
$$<\frac{1}{2}\sqrt{2}\sqrt{\frac{x^{2x}+y^{2y}}{x^yy^x}}\ \ \left(\because a+b<\sqrt{2(a^2+b^2)}\right)$$
$$<\frac{1}{2}\sqrt{2}\sqrt{\frac{2y^{2x}}{x^yy^x}}\ \ (\because0<x<y<1)$$
$$=\sqrt{\frac{y^x}{x^y}}<\sqrt{\frac{y}{x}}.$$
The last inequality is equivalent to $\frac{y^x}{x^y}<\frac{y}{x},0<x<y<1$
which is easy to prove!
| {
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"timestamp": "2023-03-29T00:00:00",
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Need help showing $\cos(x + y) / \sin(x - y) = -2$ if $\tan(\pi/4 - x) = 3 \tan(\pi/4 - y)$. I've tried manipulating the given relations using basic trig identities (which is all I know) but I wasn't able to prove anything.
| $\tan(\pi/4 - x) = 3 \tan(\pi/4 - y)$
$\iff \frac{\sin(\pi/4 - x)}{\cos(\pi/4-x)}=\frac{3 \sin(\pi/4 - y)}{\cos(\pi/4-y)}$
$\iff \frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}=\frac{3 \cos(y)- 3\sin(y)}{\cos(y)+\sin(y)}$
$\iff \cos(x)\cos(y)+\sin(y)\cos(x)-\sin(x)\cos(y)-\sin(x)\sin(y)= 3\cos(x) \cos(y)+3\sin(x)\cos(y)-3\sin(y)\cos(x)-3\sin(x)\sin(y)$
$\iff 2\cos(x) \cos(y) + 4 \sin(x) \cos(y) - 4 \sin(y) \cos(x) - 2 \sin(x) \sin(y) = 0$
$\iff 2 \cos(x + y) + 4 \sin(x -y) = 0$
$\iff \frac{\cos(x + y)}{\sin(x - y)} = -2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $F = Z_3$. Is $f(x) = x^4 + x^3 + x + 2$ irreducible in $F[x]$? I'm posting to check if I did this right.
First approach was to check roots:
$f(0) \neq 0$,
$f(1) \neq 0$,
$f(2) \neq 0$.
Second approach: Brute Force
My options for degree 1 were:
$x$, $x+1$, $x+2$
My options for degree 2 were:
$x^2$, $x^2 + 1$, $x^2+2$, $x^2+x$, $x^2+x+1$, $x^2+x+2$
I took out $x$, $x^2$, $x^2+x$ because I have a constant at the end, so there is no way these are factors.
After some long divisions, I arrived at the fact that
$\frac{x^4 + x^3 + x + 2}{x^2 +1} = x^2 + x + 2$ with $r = 0$
I concluded that this polynomial is reducible in $F[x]$.
Is this enough?
| All you have to do to check that your factorization is correct is multiply it out.
$\begin{align}
(x^2 + 1)(x^2 + x + 2) &= x^2(x^2 + x + 2) + (x^2 + x + 2) \\
&= x^4 + x^3 + 2x^2 + x^2 + x + 2 \\&= x^4 + x^3 + 3x^2 + x + 2\\& = x^4 + x^3 + x + 2\end{align} $
Which is indeed what you started with.
| {
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Special palindromic pattern of $(\frac{10^n-1}{9})^2$ I noticed a repetitive, palindromic pattern when I've performed exponentiation with $10$. That is: $$(\frac{10^n-1}{9})^2 \text{ has a palindromic pattern, }\forall\:n\in [1;9]\:(n\in \mathbb{N})$$
We define $f(n)=(\frac{10^n-1}{9})^2$.
*
*$f(1) = \color{red}{1}$
*$f(2) = \color{red}{1}\color{blue}{2}\color{red}{1}$
*$f(3) = \color{red}{1}\color{blue}{2}\color{green}{3}\color{blue}{2}\color{red}{1}$
$$...$$
*
*$f(9) = \color{red}{1}\color{blue}{2}\color{green}{3}\color{pink}{4}\color{magenta}{5}6\color{grey}{7}\color{brown}{8}\color{orange}{9}\color{brown}{8}\color{grey}{7}6\color{magenta}{5}\color{pink}{4}\color{green}{3}\color{blue}{2}\color{red}{1}$*
I was wondering why this expression has such results. Not only they are palindromic, they are in fact using ascending consecutive digits until the sequence repeats, in descending order. Can anyone explain why the structure of these numbers is so strange?
Thanks in advance!
| \begin{align*}
11 \times 11 &= \begin{matrix} 1 & 1 & \\ & 1 & 1 & + \\ 1 & 2 & 1 \end{matrix} \\ \\
111 \times 111 &= \begin{matrix} 1 & 1 & 1 & & \\ & 1 & 1 & 1 & \\ & & 1 & 1 & 1 & + \\ 1 & 2 & 3 & 2 & 1 \end{matrix} \\ \\
1111 \times 1111 &= \begin{matrix} 1 & 1 & 1 & 1 & & & \\ & 1 & 1 & 1 & 1 & & \\ & & 1 & 1 & 1 & 1 & \\ & & & 1 & 1 & 1 & 1 & + \\ 1 & 2 & 3 & 4 & 3 & 2 & 1 \end{matrix}
\end{align*}
etc. Due to the rotational symmetry of these parallograms, and the fact that, for $n \le 9$ there will be no carry-over in any column, you're going to get symmetric digits.
| {
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Given 2 roots and a remainder of a cubic polynomial, find the remainder of $f(3)$
$f(x)$ is a cubic polynomial and the coefficient of $x^3$ is $k$. If the polynomial has exactly $2$ roots and these roots are $2$ and $-5$ and $f(1) = -48$, find the possible values of $f(3)$.
I've gotten part of the cubic equation to be $(kx+a)(x^2+3x-10)$, then I let $f(x) = (kx+a)(x^2+3x-10)$, substitute in $f(1)$ to end up with $-6k-a=-48$, and I am stuck from here onward. Any hint will be appreciated.
Thanks!
| We have also $2k+a=0$ or $-5k+a=0$.
In the first case $f(x)=k(x-2)^2(x+5)$, which gives $-48=6k$ or $k=-8$ and
$f(3)=-8\cdot1^2\cdot8=-64$.
In the second case $f(x)=k(x+5)^2(x-2)$, which gives $-48=-36k$ or $k=\frac{4}{3}$ and
$f(3)=\frac{4}{3}\cdot8^2\cdot1=\frac{256}{3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $|\cos z\ |^2 + |\sin z\ |^2 \geq 1$ My attempt:
For $z :=x+iy$,
$$\cos z =\cos x \cos iy - \sin x \sin iy \\ = \cos x \cosh y - i \sin x \sinh y\\ \sin z = \sin x \cosh y + i \cos x \sinh y$$
So
$$|\cos z\ |^2 + |\sin z\ |^2 = \cos ^2 x \cosh ^2 y + \sin ^2 x \sinh ^2 y + \cosh^2 y \sin^2x + \cos ^2x \sinh ^2 y\\ = \cos ^2x ( \cosh ^2 y + \sinh ^2 y) + \sin ^2 x (\cosh^2 y + \sinh^2 y) \\ = \cos 2y \leq 1.$$
I'm not sure where I went wrong, and a numerical check on Wolfram Alpha shows that the inequality should be $\geq$ as suggested.
| One of my favourite trig. identities is
\begin{align}\lvert \sin{(x+iy)} \rvert^2 &= \lvert \sin{x}\cosh{y} + i\cos{x}\sinh{y} \rvert^2 \\
&= \sin^2{x}\cosh^2{y}+\cos^2{x}\sinh^2{y} \\
&= \sin^2{x}(1+\sinh^2{y})+(1-\sin^2{x})\sinh^2{y} \\
&= \sin^2{x}+\sinh^2{y}.\end{align}
To get the equivalent for $\lvert \sin{(x+iy)} \rvert^2$, replace $x$ by $\pi/2-x$ and $y$ by $-y$, which just swaps the sines and cosines in the proof (and the sign of the imaginary term, which is irrelevant). So
$$ \lvert \cos{(x+iy)} \rvert^2 = \cos^2{x}+\sinh^2{y}, $$
and then the sum is
$$ \lvert \sin{(x+iy)} \rvert^2 + \lvert \cos{(x+iy)} \rvert^2 = 1+2\sinh^2{y} \geq 1 $$
since the other term is the square of a real.
| {
"language": "en",
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"question_score": "3",
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Find $(a^2+b^2),$ where $(a+b)=\dfrac{a}{b}+\dfrac{b}{a}$ The question is the same .Find $a^2+b^2$. I think we have to find $a$ and $b$ firstly. Given that $a$ and $b$ are integers.
| I assume your formula is $a+b=\dfrac ab+\dfrac ba,$ then $a^2+b^2=ab(a+b)$ and since $a^2+b^2=(a+b)^2-2ab,$ we have $$x^2-2y=xy,$$ where $x=a+b$ and $y=ab.$ From here you can find $x$ in-terms of $y$ as $$x=\dfrac{y\pm\sqrt{y^2+8y}}{2}.$$ Then $$\color{Green}{a^2+b^2=\left(\dfrac{y\pm\sqrt{y^2+8y}}{2}\right)y},$$ where $y$ is a parameter.
Also, if you only interest on $a,b$ integer solutions, we need to find $y$ so that $y^2+8y$ is a perfect square.
Edit:
To do this, note that $z^2=y^2+8y=(y+4)^2-16$ implies $16=(y-z+4)(y+z+4).$ Now find the all possible values of $y-z+4$ and $y+z+4$ using factors of $16.$ Then you will have to solve few simultaneous equations to find corresponding $y.$
Later:
As @MANMAID pointed out $y=1$ is the only possible value and this gives us $$\color{Red}{a^2+b^2=2.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of seven-letter words that use letters from the set $\{\alpha,\beta,\gamma,\delta, \epsilon\}$... Find the number of seven-letter words that use letters from the set $\{\alpha,\beta,\gamma,\delta, \epsilon\}$ and contain at least one each of $\alpha$, $\beta$, and $\gamma$.
My attempt: using inclusion/exclusion
Let A denotes $\alpha$ is missing, B denotes $\beta$ is missing, and C denotes $\gamma$ is missing. Then, $$\begin{aligned}|A\cup B\cup C|&=|A|+|B|+|C|-(|AB|+|AC|+|BC|)+|ABC|\\ &= 2^7+2^7+2^7-(1+1+1)-0\\ &= 381\end{aligned}$$
$|U|= 3^7 (\text{No restriction on the combination of words})$
So, $$\begin{aligned} |U|-|A\cup B\cup C|&= 3^7-[2^7+2^7+2^7-(1+1+1)-0]\\ &= 1806 \end{aligned}$$
I'm not sure if I'm doing this problem correctly at all, because I think the universal set, i.e $|U|$, should be $5^7$ and I not sure if my calculation needs to include the other elements as well. I'm just second guessing myself. Please help me. Thank you.
| Since there are five letters available and each word has seven positions to fill, there are $5^7$ words. From these, we must exclude those cases in which one or more of the letters $\alpha$, $\beta$, or $\gamma$ is missing.
There are three ways to exclude one of the letters $\alpha$, $\beta$, or $\gamma$ and four ways to fill each of the seven positions with the other four letters. Hence, there are
$$\binom{3}{1}4^7$$
ways to exclude one of the letters $\alpha$, $\beta$, $\gamma$.
There are three ways to exclude two of $\alpha$, $\beta$, or $\gamma$ and three ways to fill each of the seven positions with the other three letters. Hence, there are
$$\binom{3}{2}3^7$$
ways to exclude two of the letters $\alpha$, $\beta$, $\gamma$.
There is one ways to exclude all three of the letters $\alpha$, $\beta$, and $\gamma$ and two ways to fill each of the seven positions with the other two letters. Hence, there are
$$\binom{3}{3}2^7$$
ways to exclude all three of the letters $\alpha$, $\beta$, and $\gamma$.
By the Inclusion-Exclusion Principle, the number of seven-letter words that can be formed from the set $\{\alpha, \beta, \gamma, \delta, \epsilon\}$ that contain at least one $\alpha$, at least one $\beta$, and at least one $\gamma$ is
$$5^7 - \binom{3}{1}4^7 + \binom{3}{2}3^7 - \binom{3}{3}1^7$$
If we let $U$, $A$, $B$, and $C$ be the sets you named, then
\begin{align*}
|U| & = 5^7\\
|A| & = 4^7\\
|B| & = 4^7\\
|C| & = 4^7\\
|A \cap B| & = 3^7\\
|A \cap C| & = 3^7\\
|B \cap C| & = 3^7\\
|A \cap B \cap C| & = 2^7
\end{align*}
Hence,
\begin{align*}
|A \cup B \cup C| & = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|\\
& = 4^7 + 4^7 + 4^7 - 3^7 - 3^7 - 3^7 + 2^7\\
& = 3 \cdot 4^7 - 3 \cdot 3^7 + 2^7\\
& = \binom{3}{1}4^7 - \binom{3}{2} 3^7 + \binom{3}{3}2^7
\end{align*}
Thus, the number of permissible words is
$$|U| - |A \cup B \cup C| = 5^7 - \binom{3}{1}4^7 + \binom{3}{2}3^7 - \binom{3}{3}2^7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382192",
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"source": "stackexchange",
"question_score": "1",
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Given $\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$, what is the value of $a+b$?
Given that
$$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$
What is the value of $a+b$
My try
$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$
$\lim\limits_{x\to 0} (\frac {(1+a×cos(x)}{x^2}-\frac {b}{x^2})=1$
$\lim\limits_{x\to 0} \frac {1+a\cos x-bx}{x^2}=1$
Apply L'Hôpital's rule:
$\lim\limits_{x\to 0} \frac {-a\cos x-b}{2x}=1$
Apply L'Hôpital's rule again:
$\lim\limits_{x\to 0} \frac {-a\sin x}{2}=1$ $\to$ $a=-2$
Is my approach right?
| It is:
$$L=\lim_{x\to 0} \frac {x+ax\cos x-b\sin x}{x^3}\stackrel{LR}=\lim_{x\to 0} \frac {1+a\cos x-ax\sin x-b\cos x}{3x^2}\stackrel{1+a-b=0; LR}=\lim_{x\to 0} \frac {-a\sin x-a\sin x-ax\cos x+b\sin x}{6x}\stackrel{LR}=\lim_{x\to 0} \frac {-a\cos x-a\cos x-a\cos x+ax\sin x+b\cos x}{6}=1 \Rightarrow -3a+b=6.$$
Hence:
$$\begin{cases}1+a-b=0 \\ -3a+b=6\end{cases} \Rightarrow a+b=-\frac52-\frac32=-4.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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} |
Given $a+b+c=0$ find the value of $\big(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\big)\big(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\big)$ I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.
First we just expand and find
$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big)+\frac{c-a}{b}\big(\frac{a}{b-c}+\frac{c}{a-b}\big)+\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big)$
Each of these "not yet a number" terms can be expanded to give
$\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big) = \frac{2(c-b)^2}{(c-a)(a-b)}$
$\frac{c-a}{b}\big(\frac{a}{b-c}\big)= \frac{2(a-c)^2}{(c-a)(a-b)}$
$\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big) = \frac{2(b-a)^2}{(b-c)(c-a)}$
Adding each of these terms together and factoring out the two we find that they equal
$2\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=6$
So our total is $9$.
This is problem 160 of The USSR olympiad problem book by Shklarsky, Chentzov and Yaglom, they give the answer and $9$ is correct. Obviously they don't give the method.
| $$\sum_{cyc}\frac{b-c}{a}=\frac{\sum\limits_{cyc}(a^2b-a^2c)}{abc}=\frac{(a-b)(a-c)(b-c)}{abc}$$ and
$$\sum_{cyc}\frac{a}{b-c}=\frac{\sum\limits_{cyc}a(a-b)(c-a)}{(a-b)(b-c)(c-a)}=$$
$$=\frac{\sum\limits_{cyc}(a^2c-a^3-abc+a^2b)}{-(a-b)(a-c)(b-c)}=\frac{\sum\limits_{cyc}(a^2c+a^2b+abc-a^3+abc-3abc)}{-(a-b)(a-c)(b-c)}=$$
$$=\frac{9abc}{(a-b)(a-c)(b-c)},$$
which gives the answer: $9$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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Inequality $\sum\limits_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$ I recently came across the following exercise:
Prove that $$\sum_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$$
for every natural number $n \ge 1$.
I immediately tried by induction, by I did not succeed as - after some trivial algebraic manipulations - I arrive at $\sum_{k=0}^{n+1} \frac{1}{n+k} \le \frac{3}{4} + \text{something non-negative}$.
Thus the solution I have found goes via comparison with the integral of $1/x$: more precisely,
$$
\sum_{k=1}^n \frac{1}{n+k} \le \int_1^n \frac{dx}{n+x} =\ln(2n) - \ln(n+1) = \ln(2) + \ln\left(\frac{n}{n+1}\right)
$$
and the conclusion follows easily as $\ln 2 \le 3/4$ and the last term is non-positive.
I would like to ask first if my solution is correct; secondly, do you have other solutions? I believe it should be very easy to prove.
| By C-S
$$\sum_{k=1}^n\frac{1}{n+k}=1-\sum_{k=1}^n\left(\frac{1}{n}-\frac{1}{n+k}\right)=1-\frac{1}{n}\sum_{k=1}^n\frac{k}{n+k}=$$
$$=1-\frac{1}{n}\sum_{k=1}^n\frac{k^2}{nk+k^2}\leq1-\frac{\left(\sum\limits_{k=1}^nk\right)^2}{n\sum\limits_{k=1}^n(nk+k^2)}=1-\frac{\frac{n(n+1)^2}{4}}{n\cdot\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}}=$$
$$=\frac{7n-1}{2(5n+1)}<0.7\leq\frac{3}{4}.$$
Done!
I think it's interesting that $\ln2=0.6931...$.
C-S forever!!!
| {
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"answer_id": 0
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Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ I have the following question.
Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ for all positive integer $n$.
My attempt:
Firstly, notice that $$n^{30}-n^{14}-n^{18}+n^2$$
$$=n^2(n-1)^2(n+1)^2(n^2+n+1)(n^2-n+1)(n^2+1)^2(n^4-n^2+1)(n^4+1)(n^8+1).$$
Also, $$46410=2\times3\times5\times7\times13\times17$$
My first thought was to use induction, maybe substitute $n+1$ into $n$, then find terms that was already divisible by $46410$, but then I don't think this is a smart way to do it.
Are there any better ways to do it?
| Hint:
Consider the factors of $n^{30}-n^{18}-n^{14}+n^2$
$$(n-1)^2,n^2,(n+1)^2,\left(n^2+1\right)^2,n^2-n+1,n^2+n+1,n^4+1,n^4-n^2+1,n^8+1$$
for any $n>1$ we have an even number and a multiple of $3$ in the first three factors.
With similar arguments it can be shown that at least one of the factor is a multiple of $5,7,11,13,17$ which are the other factors of $46410$.
| {
"language": "en",
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Analysis of Convergence Properties of a Series Approximation to $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$ I recently found a simple slowly converging infinite series approximation to $\sqrt{x}$ which may be well known, but I am struggling to prove it and fully understand its properties. The formula is
$$\sqrt{x} = \sum_{k=0}^\infty \frac{1}{4^k}\binom{2k}{k}\left(1-\frac{1}{x} \right)^k$$ valid for $x>1$, with x being a rational number.
This series is zero at $x={1}$, but appears to converge to $\sqrt{x}$ for values of $x$ between $\frac{1}{2}$ and $1$.
Additional Note w.r.t. @MarkViola's Answer:
With hindsight I do not need to investigate properties of above formula $x<1$ as the formula for $\frac{1}{\sqrt{x}}$ with $(x>1)$ can be found using a very similar method to MarkViola's Answer with
$$\left(1-t\right)^{1/2}=\sum_{k=0}^\infty (-1)^k\binom{1/2}{k}t^k$$
and
$$\binom{1/2}{k}=\frac{(-1)^{k-1}}{4^k \left( 2k-1\right)}\binom{2k}{k} $$
Resulting in
$$\frac{1}{\sqrt{x}} = -\sum_{k=0}^\infty \frac{1}{4^k \left( 2k-1\right)}\binom{2k}{k}\left(1-\frac{1}{x} \right)^k$$
End of Addition
The series approximation does not converge for complex numbers with real component zero, but does appear to correctly converge to $\sqrt{x}$ when calculating the square root of numbers like $(1+i)$
How do I prove the formula and in doing so best find, specify and justify the complete domain of numbers {rational, real, complex} over which such a series approximation formula to $\sqrt{x}$ will converge to the required value?
Some thoughts with Reference to @SimplyBeautifulArt's comment/question.
Introducing Catalan Number's $C_k=\frac{1}{k+1}\binom{2k}{k}$ allows the formula to be split into two parts, thus
$$\sqrt{x} = \sum_{k=0}^\infty \frac{k}{4^k}C_k\left(1-\frac{1}{x} \right)^k + \sum_{k=0}^\infty \frac{1}{4^k}C_k\left(1-\frac{1}{x} \right)^k$$
For example for $\sqrt{5}$
$$\sqrt{5} = \sum_{k=0}^\infty \frac{k}{4^k}C_k\left(1-\frac{1}{5} \right)^k + \sum_{k=0}^\infty \frac{1}{4^k}C_k\left(1-\frac{1}{5} \right)^k$$
with the ratio of the two terms seemingly giving the Golden Ratio $\phi$.
In the case of $\sqrt{2}$ the ratio of the two terms below appears to be double the Silver Ratio.
$$\sqrt{2} = \sum_{k=0}^\infty \frac{k}{4^k}C_k\left(1-\frac{1}{2} \right)^k + \sum_{k=0}^\infty \frac{1}{4^k}C_k\left(1-\frac{1}{2} \right)^k$$
| Let $t=1-\frac1x$. Then, we have $\sqrt x =\left(1-t\right)^{-1/2}$.
Applying the generalized binomial theorem reveals
$$\left(1-t\right)^{-1/2}=\sum_{k=0}^\infty (-1)^k\binom{-1/2}{k}t^k\tag1$$
for $|t|<1$ ($x>1$).
We can write the term $(-1)^k\binom{-1/2}{k}$ as
$$\begin{align}
(-1)^k\binom{-1/2}{k}&=\frac{(1/2)(3/2)(5/2)\cdots ((2k-1)/2)}{k!}\\\\
&=\frac{(2k-1)!!}{2^kk!}\\\\
&=\frac{2^k\,k!\,(2k-1)!!}{2^k\,k!\,2^k\,k!}\\\\
&=\frac{(2k)!}{4^k\,(k!)^2}\\\\
&=\frac{1}{4^k}\binom{2k}{k}\tag2
\end{align}$$
Substituting $(2)$ into $(1)$ yields the coveted result
$$\begin{align}
\left(1-t\right)^{-1/2}&=\sum_{k=0}^\infty \frac{1}{4^k}\binom{2k}{k}t^k\\\\
&=\sum_{k=0}^\infty \frac{1}{4^k}\binom{2k}{k} \left(1-\frac1x\right)^k
\end{align}$$
for $x>1$.
| {
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"url": "https://math.stackexchange.com/questions/2384587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Indices/Exponents: Simplify the following expression I've been asked to simplify this expressing it with positive indices/exponents. Could someone please show me in steps how to do this? The answer is shown next to it. Thanks a lot.
$$\dfrac{{(\large a^\frac{-5}{2})}^2 b^2}{\large {a}^3(b^\frac{1}{4})^6} = \dfrac{b^\frac{1}{2}}{a^8}$$
| $$ \begin{eqnarray*}
\left( a^{-\frac 52} \right)^2 = a^{-5}
\\ \frac 1{a^3}=a^{-3}
\\ \dfrac 1 { \left( b^{\frac 14} \right)^6}=b^{-\frac 32}
\end{eqnarray*} $$
so
$$ \begin{eqnarray*}
\dfrac {\left( a^{-\frac 52} \right)^2b^2 }{a^3\left( b^{\frac 14} \right)^6}
&=a^{-5}a^{-3}b^2 b^{-\frac 32}
\\&=a^{-8}b^\frac 12
\end{eqnarray*} $$
| {
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"source": "stackexchange",
"question_score": "1",
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Convergence or divergence of $ \sum \frac{n(n+1)}{4^n} $
Convergence or divergence of
$$ \sum_{n=1}^{\infty} \frac{n(n+1)}{4^n} $$
Considering
$$ a_n= \frac{n(n+1)}{4^n} \leq \frac{n(n+2)}{4^n} =b_n$$
As the integral test is conditioned on having a function positive and decreasing. $a_n$ is first increasing then decreasing.
This $b_n$ is obviously not appropriate. How would you find an upper bound to as to do a comparison test?
What would be another approach ?
Much appreciated
| The series is convergent (by the Root test: $\lim_\limits{n\to\infty} \frac{\sqrt[n]{n(n+1)}}{4}=\frac14<1)$ and its sum can be calculated. Consider the series (for $|x|<1$):
$$\sum_{n=0}^{\infty} x^{n+1}=x+x^2+x^3+x^4+\cdots=\frac{x}{1-x}.$$
Take its derivative:
$$\sum_{n=0}^{\infty} (n+1)x^n=1+2x+3x^2+4x^3+\cdots=\frac{1}{(1-x)^2}.$$
Take derivative again:
$$\sum_{n=1}^{\infty} n(n+1)x^{n-1}=2+6x+12x^2+\cdots=\frac{2}{(1-x)^3}.$$
Multiply both sides by $x$:
$$\sum_{n=1}^{\infty} n(n+1)x^n=2x+6x^2+12x^3+\cdots=\frac{2x}{(1-x)^3}.$$
Substitute $x=\frac14$:
$$\sum_{n=1}^{\infty} \frac{n(n+1)}{4^n}=\frac{2\cdot \frac14}{(1-\frac14)^3}=\frac{32}{27}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a_n = \sum\limits_{r=0}^n \binom{n}{r} b_r$, prove $(-1)^n b_n = \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s$ Suppose that sequences of real numbers satisfy:
\begin{align*}
a_n &= \sum\limits_{r=0}^n \binom{n}{r} b_r \\
\end{align*}
Prove that:
\begin{align*}
(-1)^n b_n &= \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s \\
\end{align*}
My work:
The $n=0$ case:
\begin{align*}
a_0 &= b_0 \\
b_0 &= a_0 \\
\end{align*}
Then $n=1$ case:
\begin{align*}
a_1 &= \binom{1}{0} b_0 + \binom{1}{1} b_1 = b_0 + b_1 \\
-b_1 &= \binom{1}{0} a_0 - \binom{1}{1} a_1 = a_0 - a_1 \\
b_1 &= a_1 - a_0 = b_0 + b_1 - a_0 = b_1 \\
\end{align*}
The inductive step:
\begin{align*}
(-1)^{n+1} b_{n+1} &= \sum\limits_{s=0}^{n+1} \binom{n+1}{s} (-1)^s a_s \\
\end{align*}
I'm not sure where to go from here. I tried substituting $a_s$ into that last equation and it didn't help. Any ideas?
| See binomial transform for more details. Here's a proof using (exponential) generating series.
Let
$$ B(x) = \sum_{n = 0}^\infty b_n \frac{x^n}{n!} $$
be the exponential generating series for $(b_n)$.
Then
\begin{align}
e^x B(x) &= \left( \sum_{i = 0}^\infty \frac{x^i}{i!} \right)\left( \sum_{j = 0}^\infty b_j \frac{x^j}{j!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{i + j = n} b_j \frac{x^i}{i!}\frac{x^j}{j!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{k = 0}^n b_k \frac{x^{n - k}}{(n - k)!}\frac{x^{k}}{k!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{k = 0}^n b_k \frac{n!}{(n - k)!k!} \right) \frac{x^{n}}{n!} \\
&= \sum_{n = 0}^\infty a_n \frac{x^n}{n!} = A(x).
\end{align}
Where $A(x)$ is the exponential generating series for $(a_n)$. That is,
$$ \color{purple}{A(x) = e^x B(x)} $$
and conversely,
$$ B(x) = e^{-x} A(x). \tag{$*$} $$
The product $e^{-x} B(x)$ can be written as
\begin{align}
e^{-x} A(x) &= \left( \sum_{i = 0}^\infty (-1)^{i}\frac{x^i}{i!} \right)\left( \sum_{j = 0}^\infty a_j \frac{x^j}{j!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{i + j = n} (-1)^i a_j \frac{x^i}{i!}\frac{x^j}{j!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{k = 0}^n (-1)^{n - k}a_k \frac{x^{n - k}}{(n - k)!}\frac{x^{k}}{k!} \right) \\
&= \sum_{n = 0}^\infty \left( (-1)^n \sum_{k = 0}^n (-1)^{k}a_k \frac{n!}{(n - k)!k!} \right) \frac{x^{n}}{n!} \\
&= B(x) = \sum_{n = 0}^\infty b_n \frac{x^n}{n!} \tag{by ($*$)}
\end{align}
Comparing the coefficient of $x^n/n!$ on both sides in $(*)$ gives us
$$ b_n = (-1)^n \sum_{k = 0}^n (-1)^{k}a_k \frac{n!}{(n - k)!k!} $$
or equivalently,
$$ \color{blue}{(-1)^n b_n = \sum_{k = 0}^n \binom{n}{k}(-1)^{k}a_k.} $$
| {
"language": "en",
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If $ab+1$ is perfect square there is a $k$ such that $ak+1$ and $bk+1$ are perfect squares
Show that if $ab+1$ is a perfect square for positive integers $a$ and $b$, then there is a positive integer $k$ such that $ak+1$ and $bk+1$ are both perfect squares.
I tried to prove that there is a $k$ such that $\gcd(ak+1, bk+1)=1$ and $(ak+1)(bk+1)$ is a perfect square. But it gave me nothing. Any ideas?
| Just to give a path to the answer given by Ghartal there...
It's reasonably clear that $k$ is going to be dependent on both $a$ and $b$. So, taking $n$ defined by $n^2=ab+1$, we can try $k=a+b$ and see what happens:
$ak+1 = a(a+b)+1 = a^2+ab+1 = a^2+n^2$
$bk+1 = b(a+b)+1 = b^2+ab+1 = b^2+n^2$
So then it's easy to find a couple of solutions - we just need to insert $\pm 2an$ and $\pm 2bn$ respectively to make squares of the RH terms, so $k=a+b\pm2n$ will work:
$ak+1 = a(a+b\pm2n)+1 = a^2+ab{+}1\pm2an = a^2\pm2an+n^2 = (a\pm n)^2$
$bk+1 = b(a+b\pm2n)+1 = b^2+ab{+}1\pm2bn = b^2\pm2bn+n^2 = (b\pm n)^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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solve $\sin x+\cos x=1$
How do we solve $\sin x+\cos x=1$?
I have solved it easily by making the substitution,
$\sin x+\cos x=1=R\sin(x+a)$ which gives the solutions $x=2n\pi, 2n\pi+\pi/2$. But when I do as follows
$$
(\sin x+\cos x)^2=1\implies 2\sin x\cos x=0\implies \sin 2x=0\implies2x=n\pi
$$
For even and odd
$$
2x=2m\pi\text{ or }2x=(2m+1)\pi\\x=m\pi\text{ or } x=m\pi+\frac{\pi}{2}
$$
What am I missing here ?
| One way is this: if $\sin x + \cos x = 1$, then both $\sin x$ and $\cos x$ are non negative, otherwise the other one would be greater than $1$. If one of them is different from $0$ or $1$, one has
$$1 = \sin^2 x + \cos^2 x \lt \sin x + \cos x = 1$$
This is impossible. The only remaining possibilities are $(\sin x, \cos x) = (0, 1) $ or $(1, 0)$, hence $x = 2 k \pi$ or $\frac{\pi}{2} + 2 k \pi$.
| {
"language": "en",
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"question_score": "1",
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Proving the product expansion of $\sin\theta$. Where did I go wrong? You can prove the expansion$$\frac {\sin\theta}{\theta}=\left\{1-\left(\frac {\theta}{\pi}\right)^2\right\}\left\{1-\left(\frac {\theta}{2\pi}\right)^2\right\}\ldots$$By taking the expansion$$\sin n\phi=2^{n-1}\sin\phi\cos\phi\left(\sin^2\frac {\pi}n-\sin^2\phi\right)\left(\sin^2\frac {2\pi}{n}-\sin^2\phi\right)\ldots$$substituting $\phi=\theta/n$ and dividing by$$n=2^{n-1}\sin\frac {\pi}n\sin\frac {2\pi}n\ldots\sin\frac {\pi(n-1)}{n}$$
However, when I try, I always get zero as the answer. I started off with$$\sin\theta=2^{n-1}\sin\frac {\theta}n\cos\frac {\theta}n\left(\sin^2\frac {\pi}n-\sin^2\frac {\theta}n\right)\left(\sin^2\frac {2\pi}n-\sin^2\frac {\theta}n\right)\ldots$$And divided it to get$$\frac {\sin\theta}n=\frac {\sin\frac {\theta}n\cos\frac {\theta}n\left(\sin^2\frac {\pi}n-\sin^2\frac {\theta}n\right)\left(\sin^2\frac {2\pi}n-\sin^2\frac {\theta}n\right)\ldots}{\sin\frac {\theta}n\cos\frac {\theta}n\ldots\sin\frac {\pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$\begin{align*}\sin\theta & =\theta\left\{\frac 12-\frac 12\left(\frac {\theta}n\right)^2\right\}\left\{\frac 13-\frac 13\left(\frac {\theta}{2\pi}\right)^2\right\}\ldots\\\\ & =\theta\prod\limits_{k=1}^{\infty}\left\{\frac 1{k+1}-\frac 1{k+1}\left(\frac {\theta}{k\pi}\right)^2\right\}\\ & =0\end{align*}$$
Question: I'm trying to prove$$\frac {\sin\theta}{\theta}=\prod\limits_{k\geq1}\left\{1-\left(\frac {\theta}{k\pi}\right)^2\right\}$$So where did I go wrong?
| We have
\begin{align}
\sin\theta&=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\
&=2\sin\frac{\theta}{2}\sin\left(\frac{\pi}{2}+\frac{\theta}{2}\right)..........(1).
\end{align}
Similarly in (1) changing $\theta$ into $\frac{\theta}{2}$ and $\frac{\pi}{2}+\frac{\theta}{2}$ successively, we have
$$\sin\frac{\theta}{2}=2\sin\frac{\theta}{2^2}\sin\left(\frac{\pi}{2}+\frac{\theta}{2^2}\right)=2\sin\frac{\theta}{2^2}\sin\left(\frac{2\pi}{2^2}+\frac{\theta}{2^2}\right),$$
and
\begin{align}
\sin\left(\frac{\pi}{2}+\frac{\theta}{2}\right)&=2\sin\left(\frac{\pi}{2^2}+\frac{\theta}{2^2}\right)\cdot\sin\left(\frac{\pi}{2}+\frac{\pi}{2^2}+\frac{\theta}{2^2}\right)\\
&=2\sin\left(\frac{\pi}{2^2}+\frac{\theta}{2^2}\right)\cdot\sin\left(\frac{3\pi}{2^2}+\frac{\theta}{2^2}\right).
\end{align}
Substituting these values in the right-hand side of (1) we have, after rearranging,
$$\sin\theta=2^3\sin\frac{\theta}{2^2}\sin\frac{\pi+\theta}{2^2}\sin\frac{2\pi+\theta}{2^2}\sin\frac{3\pi+\theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$\sin\theta=2^{p-1}\sin\frac{\theta}{p}\sin\frac{\pi+\theta}{p}\sin\frac{2\pi+\theta}{p}...\sin\frac{(p-1)\pi+\theta}{p} ......(3),$$
where $p$ is a power of 2.
The last factor in (3)
$$=\sin\left[\pi-\frac{\pi-\theta}{p}\right]=\sin\frac{\pi-\theta}{p}.$$
The last but one
$$=\sin\frac{(p-2)\pi+\theta}{p}=\sin\frac{2\pi-\theta}{p},$$
and so on.
Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$\sin\theta=2^{p-1}\sin\frac{\theta}{p}\left\{\sin\frac{\pi+\theta}{p}\sin\frac{\pi-\theta}{p}\right\}\left\{\sin\frac{2\pi+\theta}{p}\sin\frac{2\pi-\theta}{p}\right\}...\ \ ....(4).$$
The last factor is
$$\sin\frac{\frac{p}{2}\pi+\theta}{p}\\=\sin\left(\frac{\pi}{2}+\frac{\theta}{p}\right)=\cos\frac{\theta}{p}.$$
Hence (4) is
$$\sin\theta=2^{p-1}\sin\frac{\theta}{p}\left[\sin^2\frac{\pi}{p}-\sin^2\frac{\theta}{p}\right]\left[\sin^2\frac{2\pi}{p}-\sin^2\frac{\theta}{p}\right]\\...\left[\sin^2\frac{\left(\frac{p}{2}-1\right)\pi}{p}-\sin^2\frac{\theta}{p}\right]\cdot\cos\frac{\theta}{p}.........(5)$$
Divide both sides of (5) by $\sin\frac{\theta}{p}$ and let $\theta\to 0$. Then $\lim_{\theta\to 0}\frac{\sin\theta}{\sin\frac{\theta}{p}}=p,$ and we have
$$p=2^{p-1}\cdot\sin^2\frac{\pi}{p}\cdot\sin^2\frac{2\pi}{p}\cdot\sin^2\frac{3\pi}{p}...\sin^2\frac{\left(\frac{p}{2}-1\right)\pi}{p}............(6)$$
Dividing (5) by (6), we have
$$\sin\theta=p\sin\frac{\theta}{p}\left[1-\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{\pi}{p}}\right]\left[1-\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{2\pi}{p}}\right]\left[1-\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{3\pi}{p}}\right]...\\...\left[1-\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{\left(\frac{p}{2}-1\right)\pi}{p}}\right]\cos\frac{\theta}{p}.......(7)$$
Now let $p\to\infty$.
Since
\begin{align}
\lim_{p\to\infty}\left[p\sin\frac{\theta}{p}\right]&=\theta,\\
\lim_{p\to\infty}\left[\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{\pi}{p}}\right]&=\frac{\theta^2}{\pi^2},
\end{align}
and so on, we have
$$\sin\theta=\theta\left(1-\frac{\theta^2}{\pi^2}\right)\left(1-\frac{\theta^2}{2^2\pi^2}\right)\left(1-\frac{\theta^2}{3^2\pi^2}\right)...$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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Construct $4 \times 4$ magic square with fixed "1" The method I have found to generate $4\times 4$ magic squares gives me a result in which the number "1" is at of the corners of the square. How can we extend this to a method to generate a magic square, for a fixed location of number "$1$"?
The number "$1$" can be in $16$ different locations (cells). If we name the cells, from upper left corner: $1, 2, 3,4, 5, \ldots,$ and the number "$1$" is at the $i^{th}$ cell, then how we can fill the other cells to make a magic square?
You can see variations here.
| At first let's define a more powerful magic square,
which we will call $\color{Red}{\text{super-magic square}}$.
By a $\color{Red}{\text{super-magic square}}$ we mean a magic square such the the sum of any arbitrary row is equal to the sum of any arbitrary column is equal to sum of any arbitrary diagonal.
For example suppose the following:
$$\begin {array} {|c|c|c|c|} \hline 1&14&7&12 \\ \hline 15&4&9&6 \\ \hline 10&5&16&3
\\ \hline 8&11&2&13\\ \hline \end {array}$$
here we have:
$\color{Blue}{\text{Columns}}$:
$$\begin {array} {ccccccccc}
1 & + & 15 & + & 10 & + & 8 & = & 34 \\
14 & + & 4 & + & 5 & + & 11 & = & 34 \\
7 & + & 9 & + & 16 & + & 2 & = & 34 \\
12 & + & 11 & + & 3 & + & 13 & = & 34 \\
\end {array}$$
$\color{Green}{\text{Rows}}$:
$$\begin {array} {ccccccccc}
1 & + & 14 & + & 7 & + & 12 & = & 34 \\
15 & + & 4 & + & 9 & + & 6 & = & 34 \\
10 & + & 5 & + & 16 & + & 3 & = & 34 \\
8 & + & 11 & + & 2 & + & 13 & = & 34 \\
\end {array}$$
$\color{Purple}{\text{Diagonals parallel to the main diagonal}}$:
$$\begin {array} {ccccccccc}
1 & + & 4 & + & 16 & + & 13 & = & 34 \\
14 & + & 9 & + & 3 & + & 8 & = & 34 \\
7 & + & 6 & + & 10 & + & 11 & = & 34 \\
12 & + & 15 & + & 5 & + & 2 & = & 34 \\
\end {array}$$
$\color{Pink}{\text{Diagonals which are not parallel to the main diagonal}}$:
$$\begin {array} {ccccccccc}
12 & + & 9 & + & 5 & + & 8 & = & 34 \\
7 & + & 4 & + & 10 & + & 13 & = & 34 \\
14 & + & 15 & + & 3 & + & 2 & = & 34 \\
1 & + & 6 & + & 16 & + & 11 & = & 34 \\
\end {array}$$
$$ %%
%% 1 + 14 + 7 + 12 = 34 ,
%% \\
%% 15 + 4 + 9 + 6 = 34 ,
%% \\
%% 10 + 5 + 16 + 3 = 34 ,
%% \\
%% 8 + 11 + 2 + 13 = 34 , $$
We will prove that, $1$ could be everywhere in a $\color{Red}{\text{super-magic square}}$.
Remark(I):
Consider that a $\color{Red}{\text{super-magic square}}$
($\color{Brown}{\text{of any arbitrary order}}$) is given.
Then if we $\color{Blue}{\text{replace any two arbitrary columns}}$, then the resulting square, is again a $\color{Red}{\text{super-magic square}}$.
Remark(II):
Consider that a $\color{Red}{\text{super-magic square}}$
($\color{Brown}{\text{of any arbitrary order}}$) is given.
Then if we $\color{Green}{\text{replace any two arbitrary rows}}$, then the resulting square, is again a $\color{Red}{\text{super-magic square}}$.
Now by $\color{Blue}{\text{column operations (I)}}$
and by $\color{Green}{\text{row operations (II)}}$ ,
we are able to change "the cell containing 1" to "any desired cell, so we are done!
| {
"language": "en",
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If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ I have become interested in constrained relations among simple cyclic sums involving three positive variables. By simple, I mean so simple that they are also fully symmetric. The "building blocks" of the constraints and relations I have been looking at are:
$$
\sum_{\mbox{cyc}} 1 \equiv 3 \\
\sum_{\mbox{cyc}} a \\
\sum_{\mbox{cyc}} ab \\
\sum_{\mbox{cyc}} a^2 \\
\sum_{\mbox{cyc}} 1/a \\
\sum_{\mbox{cyc}} abc \equiv 3abc \\
$$
So an easy sample would be that $$\frac{\sum_{\mbox{cyc}} abc}{\sum_{\mbox{cyc}} a^2} \leq 1$$
The first really tough one I have encountered is:
If $a$, $b$ and $c$ are positives and $a+b+c=3$, show that:
$$a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
I got to this while trying to prove that if $a+b+c=3$ then $a^2+b^2+c^2 \leq 1/a+1/b+1/c$; that turns out to be untrue, but only by a little bit ($27-15\sqrt{3}\approx 1.019)$.
You might show this using BW, but I would hope to find something easier to follow.
EDIT
The maximum ratio is $(27-15\sqrt{3})$ and it occurs at
$$
\left(a = \sqrt{3}, b=c= \frac{3-\sqrt{3}}{2}\right)
$$
and at the two other cyclic permutations of that point.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $u=1$ and we need to prove that
$$(27-15\sqrt3)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq a^2+b^2+c^2$$ or
$$\frac{(27-15\sqrt3)v^2}{w^3}\geq3u^2-2v^2$$ or $f(w^3)\geq0,$ where
$$f(w^3)=(27-15\sqrt3)u^3v^2-(3u^2-2v^2)w^3.$$
We see that $f$ decreases, which says that it's enough to prove our inequality
for a maximal value of $w^3$, which happens for equality case of two variables.
Since $f(w^3)\geq0$ is homogeneous, it's enough to assume $b=c=1$, which gives
$$(27-15\sqrt3)(a+2)^3\left(2+\frac{1}{a}\right)\geq27(a^2+2)$$ or
$$(a-1-\sqrt3)^2(2(9-5\sqrt3)a^2+7(12-7\sqrt3)a+4(33-19\sqrt3))\geq0,$$
which is obvious.
Done!
| {
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"source": "stackexchange",
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Permutations of seven numbers such that none of the sums are divisible by 3 How many permutations $(a_1,\dots, a_7)$ of $(2,3,4,5,6,7,8)$ exist such that none of the sums
$a_1$,
$a_1+a_2$,
...
$a_1+a_2+\dots a_7$
are divisible by $3$?
I tried to study the cases of the possible $3$-remainders for members of such permutations, but I found that there are just too many possibilities. I hope there is some simple idea that I just can't see.
| We can't start with a number that is congruent $0 \mod 3$. so we must start with one that is congruent $\pm 1 \mod 3$. If we start $1 \mod 3$ we can't have a $-1 \mod 0$ before we have the second $1\mod 3$. If we start with a $1 \mod 3$ then have some zeros modulo three, another $1 \mod 3$ we will have no more $1\mod 3$s, so we will have to have, at some point, two $-1 \mod 3$s which will add to $0$.
So the first number must be congruent $-1 \mod 3$, then we must have another $-1 \mod 3$ before we have a $1 \mod 3$. After the two $-1 \mod 3$s we can't have a third $-1\mod 3$ until we have a $1 \mod 3$. Once we have two $-1 \mod 3$s and one $1\mod 3$ we must have the third $-1 \mod 3$ before the second $1\mod 3$.
So the order must be $-1\mod 3, *, -1\mod 3, *, 1 \mod 3, *, -1\mod 3, *, 1 \mod 3, * $, where $*$ are zero or more $0\mod 3$. So of the six remaining places, the two $0\mod 3$ can go anywhere.
So there are ${6 \choose 2}$ ways to arrange $(-1\mod 3,0\mod 3,1\mod 3,-1\mod 3,0\mod 3,1\mod 3,-1\mod 3,0\mod 3)$.
$2,5,8 \equiv -1 \mod 3$ and there are $3!$ ways to order those. $2!$ ways to order $4,7 \equiv 1 \mod 3$ and $2!$ ways to order $3,6\equiv 0 \mod 3$.
So there are ${6 \choose 2}3!*2!*2! = \frac {6!}{4!2!}*6*2*2 = \frac{5*6}26*2*2 =5*6*6*2 =360$ ways to do this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple question - How $x^4 +y^4 -2(x-y)^2$ having a minimum at $(0,0)$ I know this is a silly question, but I need to prove that
$$ f(x,y) = x^4 + y^4 -2 (x-y)^2 $$
is having a minima at $(0,0)$.
I tried doing maxima and minima method to it but the determinant i.e $AC- B^2= 0 $
So, we need to prove it algebraically only.
| If $x=\sqrt2$ and $y=-\sqrt2$ then we get a value $-8$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$x^4+y^4-2(x-y)^2\geq-8$$ or
$$x^4+y^4+8-2(x-y)^2\geq0.$$
Now, by AM-GM $$x^4+y^4+8\geq2\sqrt{8(x^4+y^4)}.$$
Thus, it remains to prove that
$$2\sqrt{8(x^4+y^4)}\geq2(x-y)^2$$ or
$$8(x^4+y^4)\geq(x-y)^4$$ or
$$8((x^2+y^2)^2-2x^2y^2)\geq(x^2+y^2-2xy)^2$$ or
$$7(x^2+y^2)^2+4(x^2+y^2)xy-20x^2y^2\geq0$$ or $$7(x^2+y^2)^2+14(x^2+y^2)xy-10(x^2+y^2)xy-20x^2y^2\geq0$$ or
$$7(x^2+y^2)(x^2+y^2+2xy)-10xy(x^2+y^2+2xy)\geq0$$ or
$$(x+y)^2(7x^2-10xy+7y^2)\geq0,$$
which is obvious.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x \to \infty}{\frac{1}{2^x-5^x+3^x}}$. Why is my solution wrong?
Evaluate $$\lim_{x \to\infty}{\frac{1}{2^x-5^x+3^x}}. $$
My attempt:
$$\lim\limits_{x \rightarrow \infty}{\frac{1}{2^x-5^x+3^x}}
= \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{1-\frac{5^x}{2^x}+\frac{3^x}{2^x}}} \\= \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{\frac{1}{5^x}-\frac{5^x}{10^x}+\frac{3^x}{10^x}}\cdot\frac{1}{5^x}} = 0
\cdot\infty \cdot 0 = 0.$$
Someone told me that it is a wrong way to show the limit. Can anyone explain why?
| Because it is not true that $0\times\infty\times0=0$. In fact, $0\times\infty\times0$ is indeterminate.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\dfrac {\pi x}{\sin \pi x} \times \dfrac {1}{\pi^2 x^2})$$
| $${1+\cos\theta\over\tan^2\theta}=\cos^2\theta{1+\cos\theta\over1-\cos^2\theta}=\cos^2\theta{1\over1-\cos\theta}$$
so
$$\lim_{x\to1}{1+\cos\pi x\over\tan^2\pi x}=\lim_{x\to1}{\cos^2\pi x\over1-\cos\pi x}={1\over1+1}={1\over2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Modular Arithmetic Combination I'm trying to find the remainder of $\frac{2^{2010}}{35}$. The first two parts of the question asked for the remainder when dividing $2^{2010}$ by 5 and then by 7.
To solve for the first two problems, I used Euler's Theorem:
$$2^{2010}=(2^{3})^{670}$$
Let $a=2$ and $n=5$, then
$$2^{3}\equiv1\pmod 5$$
$$2^{2010}=(2^{3})^{670}\equiv1^{670}\equiv1\pmod5$$
And so I found that the remainder was 1 (please correct me if I made a mistake).
I repeated the process for dividing by 7, and got the same result, the only difference where the exponents which did not effect the result.
How would I combine the two results, $2^{2010}\equiv1\pmod5$ and $2^{2010}\equiv1\pmod7$?
| $$2^{2010} \pmod{35}$$
Note that $2^2 \equiv 4 \equiv -1 \pmod 5$
So $2^{2010} \equiv (2^2)^{1005} \equiv (-1)^{1005} \equiv -1 \pmod 5$
Note that $2^3 \equiv 8 \equiv 1 \pmod 7$
So $2^{2010} \equiv (2^3)^{670} \equiv 1 \pmod 7$
The Chinese remainder theorem states that there is an isomorphism
$$f:\mathbb Z_{35} \to \mathbb Z_5 \times \mathbb Z_7$$
Beyond that, it states that
$f(n) = (n,n)$ and,
if $f(A) = (1,0)$ and $f(B) = (0,1)$, then $f^{-1}(m,n) = Am + Bn \pmod{35}$
Let $n \equiv 2^{2010} \pmod{35}$. You have found that $n \equiv -1 \pmod 5$ and $n \equiv 1 \pmod 7$.
You what to find the value of $n$ for which $f(n) = (-1,1)$.
First we need to find some number $x$ such that $f(x) = (1,0)$
That means $x \equiv 1 \pmod 5$ and $x \equiv 0 \pmod 7$.
$x \equiv 0 \pmod 7$ means that $x$ is a multiple of $7$. The first few multiples of $7$ are $7, 14, 21$ and we see that
$$21 \equiv 1 \pmod 5 \ \text{and} \ 21 \equiv 0 \pmod 7$$
So $f(21) = (1, 0)$
Next we need to find $y$ such that $y \equiv 0 \pmod 5 \ \text{and} \ y \equiv 1 \pmod 7$
Thr first few multiples of $5$ are $5, 10, 15$ and we see that
$$15 \equiv 0 \pmod 5 \ \text{and} \ 15 \equiv 1 \pmod 7$$
It follows that $f^{-1}(-1,1) \equiv -1(21) + 1(15) \equiv -6 \equiv 29 \pmod{35}$
So your answer is that the remainder is $29$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$ I have no idea how hard this conjecture is to prove:
Any even number $n\ge 36$ can be written as $n=a+b+c+d$ where $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb Z^+$.
Small exceptions are $n=2, 4, 6, 10, 12, 14, 20, 26, 34.\,$ Tested for $n\le 10,000$.
A counter-example would be as interesting as a proof.
$n$ as above must be even
$d-c,c+d|a^2+b^2$
| $$a^2+b^2+c^2=d^2$$
$$a=2ps$$
$$b=2ks$$
$$c=s^2-p^2-k^2$$
$$d=s^2+p^2+k^2$$
$$2n=2ps+2ks+s^2-p^2-k^2+s^2+p^2+k^2$$
$$qt=n=s(p+k+s)$$
Lay on multipliers and pick the right.
Or other item.
$$a=2s(p-k)$$
$$b=2s(p+k)$$
$$c=p^2+k^2-2s^2$$
$$d=p^2+k^2+2s^2$$
$$2n=2p^2+2k^2+4ps$$
$$p(p+2s)=n-k^2$$
It remains to try all possible $k$ . That expression was greater than $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving an infinite product of consecutive square roots Given $a$ and $b$ calculate $ab$
$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
I simplified the terms and further obtained that $ab$ is equal to:
$$ab=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}\cdot7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...}$$
How can I get a finite value?
| Assuming both nested square roots are well-defined, we have $a=\sqrt{7b}$ and $b=\sqrt{2a}$, from which
$ab=\sqrt{14 ab}$ and $ab=\color{blue}{14}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I show that the following series converges in $\mathbb{C}$? Problem Statement
Does the series
$$S := \sum_{n=0}^{\infty}{\frac{1}{2n^2 - i^n}}$$
converge in $\mathbb{C}$?
Attempt
Since
$$i^n = \begin{cases}
1, n \equiv 0 \pmod 4 \\
i, n \equiv 1 \pmod 4 \\
-1, n \equiv 2 \pmod 4 \\
-i, n \equiv 3 \pmod 4 \\
\end{cases}
$$
we can write $S$ as
$$S = S_0 + S_1 + S_2 + S_3$$
where
$$S_0 := \sum_{n=0 \\ n \equiv 0 \pmod 4}^{\infty}{\frac{1}{2n^2 - 1}}$$
$$S_1 := \sum_{n=1 \\ n \equiv 1 \pmod 4}^{\infty}{\frac{1}{2n^2 - i}}$$
$$S_2 := \sum_{n=2 \\ n \equiv 2 \pmod 4}^{\infty}{\frac{1}{2n^2 + 1}}$$
$$S_3 := \sum_{n=3 \\ n \equiv 3 \pmod 4}^{\infty}{\frac{1}{2n^2 + i}}.$$
Divergence Test does not apply, as each of the following are true:
$$\left|\frac{1}{2n^2 - 1}\right| \to 0, n \to \infty$$
$$\left|\frac{1}{2n^2 - i}\right| \to 0, n \to \infty$$
$$\left|\frac{1}{2n^2 + 1}\right| \to 0, n \to \infty$$
$$\left|\frac{1}{2n^2 + i}\right| \to 0, n \to \infty$$
I think Ratio Test also does not apply, as the resulting
$$\left|\frac{s_{n+1}}{s_n}\right|$$
for each of $S_1, S_2, S_3, S_4$ approaches $L=1$ as a limit.
Finally, I think I need to use the Comparison Test. However, I am not sure how to apply the test in this case. Any helpful hints will be appreciated.
| Your series is absolutely convergent since $\sum_{n\geq 1}\frac{1}{2n^2-1}$ is convergent.
What about computing a closed form for it? We are dealing with
$$ \sum_{m\geq 0}\left(\frac{1}{2(4m+1)^2-i}+\frac{1}{2(4m+2)^2+1}+\frac{1}{2(4m+3)^2+i}+\frac{1}{2(4m+4)^2-1}\right)$$
and in general
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b} $$
for any $a\neq b$ with positive real part, with $\psi(x)=\frac{d}{dx}\log\Gamma(x)$. By exploiting the reflection formula for the $\Gamma$ function, $\Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$, we get that the original series equals
$$ \frac{1}{2}+\frac{\pi}{8\sqrt{2}}\left(\tanh\frac{\pi}{4\sqrt{2}}-\cot\frac{\pi}{4\sqrt{2}}\right)+\frac{1+i}{16}\left[\psi\left(\frac{7-i}{8}\right)-\psi\left(\frac{5+i}{8}\right)\right]+\frac{1-i}{16}\left[\psi\left(\frac{3+i}{8}\right)-\psi\left(\frac{1-i}{8}\right)\right]$$
i.e.
$$ \approx 0.70926022705888734623 + 0.19724548429701715455\,i.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Range of $|x+3|+|x-3|=6$ I can solve this equation by som tedious algebra, I got $x_1=3$ and $x_2=-3$. But according to the book the solutions are given by $x\in[-3,3]$, which means that for example $x=1$ and $x=2$ are solutions as well. How can I algebraically show this? Or can I interpret the absolutes as distances along the x-axis and somehow proceed from there?
| $$|x+3|=\begin{cases}
x+3, & \text{ if } x\geq -3\\
-x-3,& \text{ if } x<-3
\end{cases}$$
and
$$|x-3|=\begin{cases}
x-3, & \text{ if } x> 3\\
-x+3,& \text{ if } x\leq3
\end{cases}$$
Thus,
$$|x+3|+|x-3|=\begin{cases}
2x, & \text{ if } x> 3\\
6, &\text{ if }-3 \leq x \leq3\\
-2x,& \text{ if } x<-3
\end{cases}$$
When $x>3$, then $2x>6.$ Similarly if $x<-3,$ then $-2x>6.$
Thus, the set of solutions is $[-3,3].$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$ We have
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot{2n-1\over 2n+1}=4-\pi\tag1$$
I would like to know if there exist a closed form for
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2 =\,??\tag2$$
I was able to roughly estimate it as $\approx\sqrt{8+2\pi}$ but it is not the closed form.
How can we find the closd form for $(2)?$
| This is not an answer to this question but an attempt to answer a generalized question. Our generalization consists in replacing the power two in the last term in the parentheses by a power three.Now, by starting from the (corrected--the term $(1-x^2)$ in the denominator in front of the first power of the arc sine should be raised to the power $3/2$ and not as it is now to the power one) second formula given in the answer by Martin Gales and then by multiplying the right hand side of that formula by $x^2$ and then integrating over $x$ we have derived the following formula:
\begin{eqnarray}
&&\sum\limits_{n=0}^\infty \frac{2^{2 n} x^{2n+1}}{\binom{2 n}{n}} \cdot \frac{(2n-1)^2}{(2n+1)^3}=\\
&&\left(4 \log(1+\sqrt{1-x^2})-4 \log(x)+ \frac{1}{\sqrt{1-x^2}}\right) \arcsin(x) + \\
&&4 \imath \left( \log\left[\frac{\left(x-\imath(1-\sqrt{1-x^2})\right)^2}{2-2 \sqrt{1-x^2}}\right] \log\left[ \frac{1-\sqrt{1-x^2}}{x}\right]-\right.\\
&&\left.Li_2[(-\imath) \frac{1-\sqrt{1-x^2}}{x} ]+Li_2[(\imath) \frac{1-\sqrt{1-x^2}}{x})]\right)+\\
&&2 \int\limits_0^x \left(\frac{\arcsin(t)}{t}\right)^2 \left(1+\log(\frac{x}{t})\right)dt =\\
&&-4 i \text{Li}_2\left(-\frac{i \left(1-\sqrt{1-x^2}\right)}{x}\right)+4 i \text{Li}_2\left(\frac{i \left(1-\sqrt{1-x^2}\right)}{x}\right)+\\
&&-8 i \text{Li}_3\left(-e^{i \sin ^{-1}(x)}\right)+8 i \text{Li}_3\left(e^{i \sin
^{-1}(x)}\right)+\\
&&\left(\text{Li}_2\left(e^{i \sin ^{-1}(x)}\right)-\text{Li}_2\left(-e^{i \sin ^{-1}(x)}\right)\right) \left(8 \sin ^{-1}(x)-4 i \log (2 i x)\right)+\\
&&\sin ^{-1}(x) \left(\frac{1}{\sqrt{1-x^2}}+4 i \log (2
i x) \cos ^{-1}\left(\frac{1}{x}\right)\right)+\\
&&4 \sin ^{-1}\left(\frac{1}{x}\right) \sin ^{-1}(x)^2+2 i \pi \log (2 i x) \cos ^{-1}(x)-14 i \zeta (3)+\\
&&8 \imath \int\limits_1^{\exp(\imath \arcsin(x))} \frac{\log(z)}{z^2-1} \log(z^2-1) dz
\end{eqnarray}
We will evaluate the result further and simplify it later on.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$? I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$
I'm trying to prove that such pair of numbers does not exist, but I'm not sure my proof and my reasoning are correct. Could anyone please check my proof attempt?
First of all, $\exists\frac{1}{a+b}\in\Bbb{R}\iff a\neq -b$
and $\exists(\frac{1}{a}+\frac{1}{b})\in\Bbb{R}\iff a\neq 0 \text{ and } b \neq 0$
$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b} \iff \frac{1}{a+b}=\frac{a+b}{ab}$
$\iff \frac{ab}{a+b}=a+b$
$\iff ab=(a+b)^2$
$\iff ab= a^2 + 2ab + b^2$
$\iff 0 = a^2+2ab+b^2-ab$
$\iff 0= a^2+b^2+ab$
Since $a \neq 0$ and $b \neq 0$; $\exists (ab)^{-1}\in\Bbb{R}$. This allows me to continue like this:
$ 0= a^2+b^2+ab \iff 0(ab)^{-1}=(a^2+b^2+ab)(ab)^{-1}$
$\iff 0 = \frac{a^2+b^2}{ab} + 1$
$\iff -1 = \frac{a^2+b^2}{ab}$
$a^2+b^2 > 0$ because $a \neq 0$ and $b \neq 0$
If a and b have the same sign, then $ab>0$.
So $\frac{a^2+b^2}{ab}$ could be negative only if:
*
*$a>0 \text{ and } b<0$ or
*$a<0 \text{ and } b>0$
Taking into account that $a \neq -b$, any of these two options would be possible only if either:
*
*$|a| > |b|$
*$|b| > |a|$
Assuming that $|a| > |b|$ we have the following:
$|a| > |b| \implies |a|*|a| > |b|*|a|$
$\implies |a|^2 > |ab|$
$\implies \frac{|a|^2}{|ab|}>1$
$\implies |\frac{a^2}{ab}|>1$
And $b \neq 0$ so we also have $|\frac{b^2}{ab}|>0$. Therefore $|\frac{a^2}{ab}|+|\frac{b^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.
This implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.
Assuming $|b|>|a|$ we arrive at the same conclusion:
$|b| > |a| \implies |b|*|b| > |a|*|b|$
$\implies |b|^2 > |ab|$
$\implies \frac{|b|^2}{|ab|}>1$
$\implies |\frac{b^2}{ab}|>1$
And $a \neq 0$ so we also have $|\frac{a^2}{ab}|>0$. Therefore $|\frac{b^2}{ab}|+|\frac{a^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.
So this also implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.
$\blacksquare$
Is this correct??
| It is always true that
$$0\le (|a|-|b|)^2=a^2+b^2-2|a||b|$$
$$|a||b|\le \frac{a^2+b^2}{2}.$$
$$-ab\le|a||b|\le \frac{a^2+b^2}{2}.$$
As $a\not =0$ and $b\not =0$ in this case, it follows
$$-ab\le|a||b|\le \frac{a^2+b^2}{2}<a^2+b^2,$$
and it holds that
$$-ab<a^2+b^2.$$
This result leads to a contradiction considering your development. Therefore, $\not \exists \{a,b\} \subset \Bbb R$ such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$. The existence of this subset $\{a,b\}$ would imply that $-ab=a^2+b^2$, as you've already shown, and that is not possible if $a\not =0$ and $b\not =0$, a necessary condition to prevent division by zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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$s(n^x)$ is not a perfect square for all $x$
Does there exist an $n$ such that $s(n^x)$ is not a perfect square for all positive integers $x$ where $s(m)$ denotes the sum of the digits of a positive integer $m$?
If $n = 5$, for example, then take $x = 8$ to get $5^8 = 390625$, which has digit sum $25 = 5^2$.
If $n = 6$, take $x = 2$ to get $6^2 = 36$, which has digit sum $9 = 3^2$.
If $n = 7$, take $x = 7$ to get $7^7 = 823543$, which has digit sum $25 = 5^2$.
For $n = 19$, take $x = 15$ to get $19^{15} = 15181127029874798299$, which has digit sum $100 = 10^2$.
How can we find such an $n$ or prove that none exists?
| Here is a very handwavy argument that none exists. There are $9 \cdot 10^{k-1}$ numbers with $k$ digits, of which $\sqrt{10^k}-\sqrt{10^{k-1}}=10^{\frac k2}-10^{\frac{k-1}2}=(\sqrt{10}-1)10^{\frac{k-1}2}$ are squares. We imagine that the numbers of interest are randomly squares or not, so a $k$ digit number has probability $\frac {\sqrt{10}-1}910^{-\frac{k-1}2}$ of being a square.
$n^m$ has $m \log_{10}n$ digits and the average digit is $4.5$, so the digit sum is about $4.5m\log_{10}n$, which has $\log_{10}(4.5m\log_{10}n)=\log_{10}m+c$ digits for a suitable $c$. The chance for the digit sum to be square is then $\frac{\sqrt{10}-1}9\cdot 10^{-\frac {\log_{10}m+c}2}=\frac{\sqrt{10}-1}9\cdot10^{-\frac c2}\cdot m^{-\frac 12}$. The sum of these as $m$ goes from $1$ to $\infty$ diverges, so we expect infinitely many cases where the sum of digits of $n^m$ is a square.
Of course, someone could find a number where there is a recurring pattern to make the sum of digits nonsquare. The squares $\bmod 9$ are $0,1,4,7$ so if there were a number that you could prove the digit sums of the powers were always something else we would have an answer. For example, the digit sums of $10^m$ are always $1$. In this cases $1$ is a square, so this is not a problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Remainder when the polynomial $1+x^2+x^4+\cdots +x^{22}$ is divided by $1+x+x^2\cdots+ x^{11}$
Question : Find the remainder when the polynomial $1+x^2+x^4+\ldots +x^{22}$ is divided by $1+x+x^2+\cdots+ x^{11}$.
I tried using Euclid's division lemma, I.e.
$$P_1(x)=1+x^2+x^4+\cdots+x^{22}$$
$$P_2(x)=1+x+x^2+\cdots+x^{11}$$
Then for some polynomial $Q(x)$ and $R(x)$; we have
$$P_1(x)=Q(x)\cdot P_2(x)+R(x)$$
Now, we put the values of $x$ such that $R(x)=0$ and form equations, but this method is way too long and solving the 11 set of equations for 11 variable (Since $R(x)$ a polynomial of at most 10 degree) is impossible to do for a competitive exam where the average time for solving a question is 3 minutes.
Another method is using the original long division method, and following the pattern, we can predict $Q(x)$ and $R(x)$, but it's also very hard and time taking.
I am searching for a simple solution to this problem since last a week and now I doubt even we have a simple solution to this question.
Can you please give me a hint/solution on how to proceed to solve this problem in time?
Thanks!
| $$P_1(x)=\frac{x^{24}-1}{x^2-1}$$
$$P_2(x)=\frac{x^{12}-1}{x-1}$$
$$\frac{P_1(x)}{P_2(x)}=\frac{x^{24}-1}{x^{12}-1}\cdot\frac{x-1}{x^2-1}=\frac{x^{12}+1}{x+1}$$
Then Ruffini's rule tells us that the remainder of this reduced division is the polynomial $x^{12}+1$ evaluated at $-1$, i.e. 2. When the top and bottom of $\frac2{x+1}$ are multiplied by $\frac{x^{12}-1}{x^2-1}$, the denominator becomes $P_2(x)$ and the numerator gives the final answer of $\frac{2(x^{12}-1)}{x^2-1}=2+2x^2+2x^4+2x^6+2x^8+2x^{10}$.
| {
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"url": "https://math.stackexchange.com/questions/2404083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.