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First derivative meaning in this case If we have a function:
$$f(x)=\frac{x}{2}+\arcsin{\frac{2x}{1+x^2}}$$
And it's first derivative is calculated as:
$$f'(x)=\frac{1}{2}+\frac{1}{\sqrt{1-\big(\frac{2x}{1+x^2}\big)^2}}\frac{2+2x^2-4x^2}{(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}\cdot(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}=$$
$$\frac{1}{2}+\frac{2\cdot sgn(1-x^2)}{1+x^2}$$
Why did my teacher say the critical points are at $x=\pm1$? Then she wrote:
$$f_+'(1)=-\frac{1}{2}$$
$$f_-'(1)=\frac{3}{2}$$
and I'm not sure what she meant by that eather, nor by the following:
$$f'(x)=
\begin{cases}
\frac{x^2-3}{2(x^2+1)}, & |x|>1 \\
\frac{x^2+5}{2(x^2+1)}, & |x|<1
\end{cases}$$
Thank you for your time.
|
If $\arctan x=y,x=\tan y$ and $-\dfrac\pi2\le y\le\dfrac\pi2$
(See definition of the principal values of inverse trigonometric functions)
$$\dfrac{2x}{1+x^2}=\cdots=\sin2y$$
$$\implies\arcsin\dfrac{2x}{1+x^2}=\arcsin(\sin2y)=\begin{cases}2y&\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\iff-1\le x\le1\\
\pi-2y & \mbox{if }2y>\dfrac\pi2\\
-\pi-2y & \mbox{if }2y<-\dfrac\pi2 \end{cases} $$
From this relation, we can easily discern the change of sign of $$\dfrac{d\left(\arcsin\dfrac{2x}{1+x^2}\right)}{dx}$$ for the different ranges of values of $x$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $|\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$ I need to show that $ \forall x,y \in \mathbb R, |\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$
I tried using concavity of log function: $\log(1 + x^2) - \log(1 + y^2)=\log(\frac{1 + x^2}{1 + y^2})=\log(\frac{x^2y^2}{(1 + y^2)y^2}+\frac{1}{1 + y^2}) \ge \frac{2(\log(x)-\log(y))}{1+y^2}$
Also the middle value theorem: $0<\frac{\log(1 + x^2) - \log(1 + y^2)}{x^2-y^2} <1$
But both attempts has not led too far.
|
We can write the term of interest as
$$\left|\log(1+x^2)=\log(1+y^2)\right|=\left|\int_y^x \frac{2t}{1+t^2}\,dt\right|$$
Now, the integrand $f(t)=\frac{2t}{1+t^2}$ attains its maximum value of $1$ when $t=1$. Therefore, the inequality follows immediately from the mean value theorem.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the two other sides in a 15-30-135 triangle A triangle has angle measures of 15, 30, and 135 degrees. The side opposite the 15 angle is x feet, the side opposite the 30 angle is y feet, and the side opposite the 135 angle is 2 feet.
Find x and y without the law of sine.
|
Apply sine rule in given triangle as follows
$$\frac{x}{\sin 15^\circ}=\frac{y}{\sin 30^\circ}=\frac{2}{\sin 135^\circ}$$
considering first & third,
$$\frac{x}{\sin 15^\circ}=\frac{2}{\sin 135^\circ}$$
$$x=\frac{2\sin 15^\circ}{\sin 135^\circ}=\frac{2\sin (45^\circ-30^\circ)}{\sin 45^\circ}=\frac{2\left(\frac{1}{\sqrt 2}\frac{\sqrt 3}{2}-\frac{1}{\sqrt 2}\frac{1}{2}\right)}{\frac1{\sqrt 2}}=\color{red}{\sqrt 3-1}$$
similarly, considering second & third,
$$\frac{y}{\sin 30^\circ}=\frac{2}{\sin 135^\circ}$$
$$y=\frac{2\sin 30^\circ}{\sin 135^\circ}=\frac{2\cdot \frac{1}{2}}{\frac{1}{\sqrt 2}}=\color{red}{\sqrt 2}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $\sin x+\sin^{2} x=1$ , Find $\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
If $\sin x+\sin^{2} x=1$, then the value of
$\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
is equal to
$a.)\ 0 \\
b.)\ 1 \\
c.)\ 2 \\
\color{green}{d.)\ \sin^{2} x} $
$\boxed{\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2\\
=\cos^{6} x\left(\cos^{6} x+3\cos^{4} x+3\cos^{2} x+1\right)+2\sin^{2} x+\sin x-2\\
=\sin^{3} x\left(\cos^{2} x+1\right)^{3}+2\sin^{2} x+\sin x-2\\
=\sin^{3} x\left(\sin x+1\right)^{3}+2\sin^{2} x+\sin x-2\\
=\left(\sin x[\sin x+1]\right)^{3}+2\sin^{2} x+\sin x-2\\
=\left(\sin^{2} x+\sin x\right)^{3}+2\sin^{2} x+\sin x-2\\
=1+\sin^{2} x+\sin^{2} x+\sin x-2\\
=\sin^{2} x} $
I found this solution but considering the time as $1-2$ min to solve this question,
I am looking for a short and simple way.
I have studied maths up to $12$th grade
|
HINT:
$$\sin x=1-\sin^2x\implies(\cos^2x)^2=\sin^2x=1-\cos^2x$$
$$\implies\cos^4x+\cos^2x-1=0$$
Divide $$\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $$ by $\cos^4x+\cos^2x-1$ to get a reminder of $O(\cos^2x)$
|
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|
Value of $\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$ To find the value of:
$\displaystyle\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$
The answer says: $1/2$
My attempt:
$=(\displaystyle\sum_{k=1}^\infty 1/{k^2})-1-(\sum_{k=1}^\infty 1/{k^3})+1+(\sum_{k=1}^\infty 1/{k^4})-1-\cdots$
$=(1+1/2^2+1/3^2+1/4^2+\cdots)-(1+1/2^3+1/3^3+1/4^3+\cdots)+(1+1/2^4+1/3^4+1/4^4+\cdots)-\cdots$
$=(1/2^2-1/2^3+1/2^4-\cdots)+(1/3^2-1/3^3+1/3^4-\cdots)+(1/4^2-1/4^3+1/4^4-...)+\cdots$
$=S_1+S_2+S_3+\cdots$
With:
$S_2=(1/2^2-1/2^3+1/2^4-\cdots)$
$S_3=(1/3^2-1/3^3+1/3^4-\cdots)$
$S_4=(1/4^2-1/4^3+1/4^4-\cdots)$
...
It looks like geometry series with alternating signs.
If I apply the geometric series value: $S=\frac{1}{1-z}$
Gives:
$S_2=2-1-1/2=1/2$
$S_3=3/2-1-1/3=1/2-1/3$
$S_4=4/3-1-1/4=1/3-1/4$
...
Original equation:
$=1/2+1/2-1/3+1/3-1/4+1/4-1/5+\cdots=1$
This is different than the answer (1/2) given.
Could any one help? Thanks,
|
You've calculated $S_2$ wrong. The sum $\sum_{n=2}^\infty(−1/2)^n=\frac1{1−(−1/2))}−1+1/2=1/6$
|
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|
$\iint (y^2-x^2) e^{xy} dxdy$ $\iint_G(y^2-x^2)e^{xy}dxdy$ on G = $\{ (x,y) | x \ge 0, 1 \le xy \le 4, 0 \le y-x \le 3 \}$
I have tried a substitution of
$u = xy$
$v = y-x$
which got me to $\bar G = \{ (u,v) | 1 \le u \le 4, 0 \le v \le 3 \}$
and
$\iint_\bar G \left(\frac{v}{\sqrt{v^2+4u}}\right)^{\frac12} e^u dudv$
(note that $x = -\frac12v+\frac12\sqrt{v^2+4u}$
and that $y = \frac12v+\frac12\sqrt{v^2+4u}$ )
But from here on i'm stuck.
|
A slightly different change of variables which simplifies a lot the boundary conditions :
$\begin{cases}
s=y-x\\
p=xy\\
\end{cases}$
The Jacobian is : $-\begin{vmatrix}
-1 & 1 \\
y & x \\
\end{vmatrix}^{-1}=\frac{1}{y+x}$
$(y^2-x^2)\frac{1}{y+x}=y-x=s$
$\iint_{1 \le xy \le 4, 0 \le y-x \le 3}(y^2-x^2)e^{xy}dxdy =
\int_1^4\int_0^3 s e^p ds\;dp=
\int_1^4 e^p dp\int_0^3 s ds=
(e^4-e)\frac{3^2}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$ Exercise from Ahlfor's Complex:
Given $a,b \in \mathbb C$, with $|a| <1$, $|b|<1$, prove:
$$\left|\frac{a-b}{1-\overline{a}b}\right| <1.$$
My argument:
Lemma:
If $\alpha, \beta \in \mathbb R$ and $\alpha, \beta <1$, then $\alpha^2+\beta^2 < 1 + \alpha^2\beta^2$.
To prove the lemma, just write $\alpha = (1-\epsilon_1)$, $\beta = (1-\epsilon_2)$ with $0 < \epsilon_i < 1$. Then, expanding out both sides, we get
$$ 1+\alpha^2\beta^2 = \alpha^2 + \beta^2 + 4\epsilon_1\epsilon_2 - 2\epsilon_1^2\epsilon_2 - 2\epsilon_1\epsilon_2^2 + \epsilon_1^2\epsilon_2^2,$$ but $\epsilon_i^2 < \epsilon_i$, so $$4\epsilon_1\epsilon_2 - 2\epsilon_1^2\epsilon_2 - 2\epsilon_1\epsilon_2^2 > 0.$$
The result follows.
Now, $$\begin{eqnarray*}
\left|\frac{a-b}{1-\overline{a}b}\right| &=& \frac{|a-b|}{|1-\overline{a}b|} \\
&\leq& \frac{|a-b|}{|1-|a||b| \, |} \\
\end{eqnarray*}
$$
$$
\begin{eqnarray*}
\frac{|a-b|^2}{|1-|a||b| \, |^2}
&\leq& \frac{|a|^2+|b|^2-2|a||b|}{1+|a|^2|b|^2-2|a||b|} \\
&<& 1
\end{eqnarray*}$$
by the lemma, and since $x^2<1$ implies $x<1$ for $x \in \mathbb R$, the result follows.
Where I've used the reverse triangle inequality and the fact that $|z| \geq \Re(z).$
Any slicker ways to do this? Any issues with this?
|
We have
$$
\begin{align*}
|1 - \overline{a}b|^2 - |a - b|^2 &= \left(1 - \overline{a}b \right)\left(1 - a\overline{b}\right) - (a - b)\left( \overline{a} - \overline{b}\right) \\
&= 1 - a\overline{a} - b\overline{b} + a\overline{a}b\overline{b} \\
&= \left( 1 - |a|^2\right)\left( 1 - |b|^2\right)\\
&> 0.
\end{align*}
$$
It follows that $|1 - \overline{a}b| > |a - b|$, hence
$$\left| \frac{a-b}{1 - \overline{a}b}\right| = \frac{|a-b|}{\left| 1 - \overline{a}b\right|} < 1.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Solution of Differential equation $\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$
Solution of Differential equation $$\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$$
$\bf{My\; Try::}$ Let $x=r\sec \theta$ and $y=r\tan \theta\;,$ Then $x^2-y^2=r^2$ and $xdx-ydy=rdr$
and $$\frac{y}{x} = \frac{\tan \theta }{\sec \theta} = \sin \theta\Rightarrow \frac{xdy-ydx}{x^2} = \cos \theta d\theta$$
So $$xdy-ydx=x^2\cos \theta = r^2\sec^2 \theta\cdot \cos \theta d\theta = r^2\sec \theta d\theta$$
So $$\frac{rdr}{r^2\sec \theta d\theta} = \sqrt{\frac{1+r^2}{r^2}}\Rightarrow \int \frac{dr}{\sqrt{1+r^2}} = \int \sec \theta d\theta$$
So $$\ln\left|1+\sqrt{1+r^2}\right| = \ln \left|\sec \theta +\tan \theta\right|+\ln \mathcal{C}$$
So $$\left|1+\sqrt{1+x^2-y^2}\right|= \left|\mathcal{C}\cdot \left(\frac{x+y}{\sqrt{x^2-y^2}}\right)\right|$$
My Question is can we solve it any other shorter way, If yes then plz explain here
Thanks
|
A better change of variables (in hyperbolic system of coordinates) is :
$\begin{cases}
x=r \cosh(t)\\
y=r \sinh(t)\\
\end{cases}$
$\begin{cases}
dx=dr \cosh(t)+r \sinh(t) dt\\
dy=dr \sinh(t)+r \cosh(t) dt\\
\end{cases}$
Bringing them into :
$$\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$$
leads to :
$$\frac{1}{r}\frac{dr}{dt}=\sqrt{\frac{1+r^2}{r^2}}$$
$$t=\pm\int \frac{dr}{\sqrt{1+r^2}}=\pm \sinh^{-1}(r)+c$$
$$r=\pm\sinh(t-c)$$
The solution on parametric form is :
$\begin{cases}
x=\pm\sinh(t-c) \cosh(t)\\
y=\pm\sinh(t-c) \sinh(t)\\
\end{cases}$
The elimination of $t$ from the parametric system leads to the implicit equation :
$$x^2-y^2-y\:\cosh(c)+x\:\sinh(c)=0$$
|
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|
How do you find the probability of A winning if the probability of getting a favourable outcome in the $r^{th}$ turn is a function of $r$? Problem:
Two players A and B are playing snake and ladder. A is at 99 and he needs 1 to win in rolling of a dice. However, he is always allowed to re-throw the dice if 6 appears.
What is the probability that A will win eventually before B gets a chance, if the probability of getting 1 in the $r^{th}$ throw is $\frac{1}{3^r}$ and that of getting 6 in the $r^{th}$ throw is $\frac{2r-1}{4r}$?
My attempt:
We know that A can win before B gets a chance only if he rolls {$1$},{$6$,$1$},{$6$,$6$,$1$} and so on.
In the $r^{th}$ turn, we have the probability: $$\frac{1}{3^r}\cdot\frac{1}{4}\cdot\frac{3}{8}\cdot\cdot\cdot\frac{2r-3}{4(r-1)}$$
$$=\frac{1}{3^r}\cdot\frac{1}{4^{r-1}}\left(\frac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot(2r-3)}{1\cdot2\cdot3\cdot4\cdot\cdot\cdot(r-1)}\right)$$
$$=\frac{1}{3^r}\cdot\frac{1}{4^{r-1}}\left(\frac{(2r-2)!}{(r-1)!\cdot(r-1)!\cdot2^{r-1}}\right)$$
$$=\frac{1}{3}\cdot\frac{1}{24^{r-1}}\left(\frac{(2r-2)!}{(r-1)!\cdot(r-1)!}\right)$$
Therefore, we have the probability as
$$\sum_{r=1}^{\infty} \frac{1}{3}\cdot\binom{2r-2}{r-1} \frac{1}{24^{r-1}}$$
Taking $r-1$=$n$
$$\frac{1}{3}\sum_{n=0}^{\infty} \binom{2n}{n} \frac{1}{24^{n}}$$
I got stuck at the last step because I do not know how to evaluate that summation. Any help with the summation/providing an alternate way to solve this question will be appreciated.
|
The generating function for the central binomial coefficients is
\begin{align*}
\sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad |z|<\frac{1}{4}
\end{align*}
This is an application of the binomial series
\begin{align*}
(1+z)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}z^n\qquad |z|<1, \alpha\in\mathbb{C}
\end{align*}
and the relation
\begin{align*}
\binom{-\frac{1}{2}}{n}=\frac{(-1)^n}{4^n}\binom{2n}{n}
\end{align*}
Can you proceed?
|
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|
Prove that $\frac{(2n)!}{(n!)^2}-1$ is divisible by $(2n+1)$
Prove that $$\frac{(2n)!}{(n!)^2}-1$$ is divisible by $(2n+1)\;,$ Where $n\in \mathbb{N}$ and $n>1$
$\bf{My\; Try::}$ Let $$S = \frac{(2n)!}{n!^2}-1 = \frac{2^n(2n-1)(2n-3)\cdot \cdot ........\cdot 3 \cdot 2 \cdot 1}{n!}-1$$
Now How can we prove that $2^n(2n-1)(2n-3)\cdot \cdot \cdot .... 3\cdot 2 \cdot 1$ is divisible by $n!$ and odd number
Help Required, Thanks
|
$\dfrac{(2n)!}{(n!)^2} - 1 = \dbinom{2n}{n}-1$
This is the count of ways to arrange $n$ discrete objects into two equal piles, excluding a particular arrangement.
$\begin{array}{l:l:l} n & \binom{2n}n-1 & 2n+1
\\ 1 & 1 & 3
\\ 2 & 5 & 5
\\ 3 & 19 & 7
\\ 4 & 69 & 9
\\ \vdots & \vdots & \vdots
\end{array}$
Only occasionally is $\binom{2n}n -1$ divisible by $2n+1$
|
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|
Laurent series expansion of $f$ Find the Laurent series expansion of $f(z)=\dfrac{1}{2z^2-13z+15}$ about the annulus $\dfrac{3}{2}<|z|<5$.
I did like this :
$f(z)=\dfrac{2}{7}(\dfrac{3}{3-2z}-\dfrac{1}{z-5})$
Then I took $\dfrac{3}{3-2z}=\dfrac{1}{1-\dfrac{2z}{3}}=(1-\dfrac{2z}{3})^{-1}=1+\dfrac{2z}{3}+(\dfrac{2z}{3})^2+...$
I took $\dfrac{1}{z-5}=\dfrac{-1}{5}(1-\dfrac{z}{5})^{-1}=\dfrac{-1}{5}(1+\dfrac{z}{5}+\dfrac{z^2}{25}+...)$
I calculated the co-efficients of $z$ and $z^2$ i.e $a_1,a_2$; I found $a_1=\dfrac{47}{75}$ and $a_2=\dfrac{4}{9}-\dfrac{1}{125}$ but the answer given is $\dfrac{a_1}{a_2}=5$
Why am I wrong?Please give some details
|
Here is a somewhat detailed description of the Laurent expansion of $f(z)$ around $z=0$.
The function
\begin{align*}
f(z)&=\frac{1}{2z^2-13z+15}
=-\frac{1}{7}\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\frac{1}{z-5}\\
\end{align*}
has two simple poles at $\frac{3}{2}$ and $5$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $\frac{3}{2}$ and $5$ and see they determine three regions.
\begin{align*}
|z|<\frac{3}{2},\qquad\quad
\frac{3}{2}<|z|<5,\qquad\quad
5<|z|
\end{align*}
*
*The first region $ |z|<\frac{3}{2}$ is a disc with center $0$, radius $\frac{3}{2}$ and the pole $\frac{3}{2}$ at the boundary of the disc. In the interior of this disc all two fractions with poles $\frac{3}{2}$ and $5$ admit a representation as power series at $z=0$.
*The second region $\frac{3}{2}<|z|<5$ is the annulus with center $0$, inner radius $\frac{3}{2}$ and outer radius $5$. Here we have a representation of the fraction with pole $\frac{3}{2}$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $5$ admits a representation as power series.
*The third region $|z|>5$ containing all points outside the disc with center $0$ and radius $\frac{3}{2}$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}
We can now obtain the Laurent expansion of $f(z)$ at $z=0$ for all three regions. Here we consider the second region
*
*Region 2: $\frac{3}{2}<|z|<5$
\begin{align*}
f(z)&=-\frac{1}{7}\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\frac{1}{z-5}\\
&=-\frac{1}{7z}\sum_{n=0}^{\infty}\left(-\frac{3}{2}\right)^{n}\frac{1}{(-z)^n}
+\frac{1}{7}\sum_{n=0}^{\infty}\frac{1}{(-5)^{n+1}}(-z)^n\\
&=-\frac{1}{7}\sum_{n=1}^{\infty}\left(\frac{3}{2}\right)^{n-1}\frac{1}{z^n}
-\frac{1}{7}\sum_{n=0}^{\infty}\frac{1}{5^{n+1}}z^n\\
\end{align*}
We observe $\frac{a_1}{a_2}=\frac{[z^1]f(z)}{[z^2]f(z)}=\frac{-\frac{1}{7\cdot 5^2}}{-\frac{1}{7\cdot 5^3}}=5$.
|
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|
$f(x) - f'(x) = x^3 + 3x^2 + 3x +1; f(9) =?$ Given the following $f(x) - f'(x) = x^3 + 3x^2 + 3x +1$
Calculate $f(9) = ?$
I have tried to play with different number of derivatives. Also tried to solve it by equations.
Maybe there is some geometric meaning that could shade the light ?
I feel it is no complex problem at all.
Thanks for your solution
|
\begin{equation}
f(x) = ax^4 + bx^3 + cx^2 + dx + e\\f'(x) = 4ax^3 + 3bx^2 + 2cx + d\\f(x) - f'(x) = ax^4 + (b-4a)x^3 + (c-3b)x^2 + (d-2c)x + e - d = x^3 + 3x^2 + 3x +1\\a=0, b=1, c=6, d=15, e=16
\end{equation}
So, the function is $f(x)=x^3 + 6x^2 + 15x + 16$ and $f(9) = 1366$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Legendre symbol, what is it? I am reading wiki article about Legendre symbol and I don't understand the power meaning. Can you please explain the next expression.
$$\left(\frac ap\right)\equiv a^{\frac{p-1}{2}}\pmod p$$
|
Since you say you have read about Legendre symbol on Wikipedia, you should already know that $\left(\frac{a}{p}\right)$ is defined as $0$ if $p\mid a$ and as $\pm1$ for $p\nmid a$, depending on whether $a$ is a quadratic residue modulo $p$ or not.
According to Euler's criterion, the congruence
$$\left(\frac{a}{p}\right) \equiv a^{(p-1)/2}$$
holds for any odd prime $p$.
There is nothing mysterious about the expression $a^{(p-1)/2}$, it is just a usual exponentiation. (It is useful to notice that $(p-1)/2$ is a positive integer if $p$ is an odd prime.)
Let us have a look on specific examples. For $p=5$ we have $(p-1)/2=2$ and we get
$$\begin{array}{|c|c|c|}
\hline
a & a^2 & \left(\frac{a}{5}\right) \\\hline
0 & 0 & 0 \\\hline
1 & 1 & 1 \\\hline
2 & 4 &-1 \\\hline
3 & 9 &-1 \\\hline
4 &16 & 1 \\\hline
\end{array}
$$
If you look at the second and third column, they are indeed congruent modulo $5$.
For $p=7$ we get the following table:
$$\begin{array}{|c|c|c|}
\hline
a & a^3 & \left(\frac{a}{7}\right) \\\hline
0 & 0 & 0 \\\hline
1 & 1 & 1 \\\hline
2 & 8 & 1 \\\hline
3 &27 &-1 \\\hline
4 &64 & 1 \\\hline
5 &125&-1 \\\hline
6 &216&-1 \\\hline
\end{array}
$$
(We know that quadratic residues modulo $7$ are $(\pm1)^2\equiv1\pmod 7$, $(\pm2)^2\equiv4\pmod 7$ and $(\pm3)^2\equiv2\pmod 7$.)
You may notice that instead of calculating $a^3$ we might calculate $a^3\bmod 7$, which would is a bit easier. For example $5^3=5^2\cdot 5 \equiv 25 \cdot 5 \equiv 4\cdot 5 \equiv 20 \equiv -1 \pmod 7$.
This is more about the the basic properties of congruences rather than about the Legendre symbol, but I will try to explain the previous computation, per request in a comment by the OP.
The only thing we are using here is the fact that if $a_1\equiv b_1\pmod n$ and $a_2\equiv b_2\pmod n$, then also $a_1b_1\equiv a_2b_2\pmod n$.
Using this rule we know that
$$25\cdot 5\equiv 4\cdot 5\pmod 7$$
since $25\equiv 4\pmod 7$.
In fact, we have even easier way to calculate $5^3 \bmod 7$. If we notice that $5\equiv -2 \pmod 7$ then
$$5^3 \equiv (-2)^3 \equiv -8 \equiv -1 \pmod 7.$$
For more about similar computations, have a look at How do I compute $a^b\,\bmod c$ by hand? and other related posts.
|
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|
$ \int\frac{\sin(nx) \sin x}{1-\cos x} \,dx$ by elementary methods What is an elementary way to show that for positive integer $n$
$$
\int\frac{\sin(nx) \sin x}{1-\cos x} \,dx= x + \frac{\sin (nx)}{n} + 2 \sum_{k=1}^{n-1}\frac{\sin(kx)}{k}
$$
This cropped up when trying to answer Proving $\int_0^{\pi } f(x) \, \mathrm{d}x = n\pi$ at a beginning calculus level, that is, without using contour integration or Fourier integrals or Parseval's theorem (but with clever tricks allowed). There, it would suffice to know that the integral from zero to $\pi$ is $\pi$.
I've tried to do this by induction but got nowhere (except that the base for $n=1$ is easy).
I've tried expanding as a series and matching terms but it got pretty messy.
|
Work backwards:
Let $$F(x)=x + \frac{\sin (nx)}{n} + 2 \sum_{k-1}^{n-1}\frac{\sin(kx)}{k}$$
Then
$$F'(x)=1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx)$$
Therefore, you need to prove that
$$\frac{\sin(nx) \sin x}{1-\cos x} = 1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx)$$
or equivalently that
$$\sin(nx) \sin(x)= (1-\cos(x)) \left(1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx) \right)(**)$$
If there is no typo in your problem, this should be a relatively easy induction question.
Or better, if your students know complex numbers, just do the standard
$$z =\cos(x)+i\sin(x)$$
and calculate
$$1+2z+2z^2+..+2z^{n-1}+z^n=2(1+z+z^2+..+z^{n-1}+z^n)-1-z^n=2\frac{1-z^{n+1}}{1-z}-1-z^n \\
=\frac{2-z^{n+1}-z^n}{1-z}-1=\frac{(2-z^{n+1}-z^n)(1-\bar{z})}{(1-\cos(x))^2+\sin(x)^2}-1\\
=\frac{2-z^{n+1}-z^n-2\bar{z}+z^n+z^{n-1}}{2-2\cos(x)}-1\\
=\frac{2-z^{n+1}-2\bar{z}+z^{n-1}}{2-2\cos(x)}-1\\$$
By taking the real parts we get:
$$
\left(1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx) \right)=\frac{2-\cos((n+1)x)-2\cos(x)+\cos((n-1)x)-2+2\cos(x)}{2-2\cos(x)} \\
=\frac{-\cos((n+1)x)+\cos((n-1)x)}{2-2\cos(x)}=\frac{2 \sin(nx)\sin(x)}{2-2\cos(x)}
$$
which is exactly what you need to prove.
P.S. You can also try to prove $(**)$ by writing
$$(1-\cos(x)) \left( \sum_{k-1}^{n-1}\cos(kx) \right)=2\sin^2(\frac{x}{2})\left( \sum_{k-1}^{n-1}\cos(kx) \right)$$ and use the fact that
$$\sum_{k-1}^{n-1}\cos(kx) \sin(\frac{x}{2})$$
is telescopic.
|
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|
Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
Find the polynomial $P$ of smallest degree with rational coefficients
and leading coefficient $1$ such that $$ P(49^{1/3}+7^{1/3})=4 $$
(Source:NYSML)
My attempt
Let $$ 49^{1/3}+7^{1/3}=x$$
Then,
\begin{array}
((49^{1/3})^3 &=(x-7^{1/3})^{3} \\
49 &=x^3-7-3x^2\cdot 7^{1/3}+3x \cdot 7^{2/3} \\
\end{array}
So I have $$x^3+x^2 (-3\cdot 7^{1/3})+x(3\cdot 7^{2/3})-56 =0 $$
Define $$Q(x)=x^3+x^2 (-3\cdot 7^{1/3})+x(3\cdot 7^{2/3})-56 $$
making the substitution $x=7^{1/3} \cdot t $,
I have $$F(t)=\cfrac{Q(7^{1/3} \cdot t)}{7}=t^3+t^2(-21)+21t-56$$
Now I am quite stuck on how I should continue...
|
Your first attempt (cubing $49^{1/3}+7^{1/3}$) came close to finishing. Let $a$ be our number. Cubing, we find that
$$a^3=56+3(7^{5/3}+7^{4/3})=56+21a.$$
So if $Q(x)$ is the polynomial $x^3-21x-56$, then $Q(a)=0$. Dealing with the fact that we want $P(a)=4$ is now easy.
We sketch a proof that this polynomial $P(x)$ has minimal degree. First show that $a$ is not a root of a polynomial of degree $1$ with rational coefficients. This can be handled by the Rational Root Theorem. To show that $a$ is not a root of a polynomial of degree $2$ with rational coefficients, suppose that $W(x)$ is such a polynomial. Then the remainder $R(x)$ when we divide $P(x)$ by $W(x)$ has degree $\le 1$, and $R(a)=0$, which is impossible.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving Trigonometric Equation.
Solve for $\theta$ $[0°<\theta<180°]$
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
My solution is here:
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
After using the transformation formula, I got
$$\sin3\theta\cos\theta=\cos3\theta\cos\theta.$$
I could not proceed from here.
|
Using Prosthaphaeresis Formulas as suggested,
$$ \sin 2\theta+\sin 4\theta = \cos \theta+\cos 3\theta \\
2\sin 3\theta \cos \theta = 2\cos 2\theta \cos \theta \\
\cos \theta \, (\sin 3\theta-\cos 2\theta) = 0 $$
If $\cos \theta=0,\theta=(2n+1)90^\circ$ where $n$ is any integer
We need $0<(2n+1)90^\circ<180^\circ\implies-1<n<1\implies n=0$
Else $\cos 2\theta=\sin 3\theta=\cos(90^\circ-3\theta)$
$2\theta=360^\circ m\pm(90^\circ-3\theta)$
$+\implies\theta=72^\circ m+18^\circ$ and we need $0<72^\circ m+18^\circ<180^\circ\implies m=?$
$-\implies\theta=-360^\circ m+90^\circ$
Can you take it from here?
|
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"timestamp": "2023-03-29T00:00:00",
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|
finding real roots by way of complex I was given
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)( x - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)$$
$$( x + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)( x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)$$
He then mltiplied both pairs,resulting in:
$$(x^2 - \sqrt{2}x + 1) and (x^2 + \sqrt{2}x + 1)$$
How do I get the first four imaginary points, or the two distinct because I realize that each has a conjugate.
|
change equation to
$$x^4 = -1 = e^{\pi i}$$
$$x = e^{(\frac{\pi}{4} + k\frac{\pi}{2})i}$$
where $k = 0, 1, 2, 3$. All four roots are evenly disitributed on the unit circle in complex plane.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1646603",
"timestamp": "2023-03-29T00:00:00",
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|
The sum of the following infinite series $\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$
The sum of the following infinite series $\displaystyle \frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$
$\bf{My\; Try::}$ We can write the given series as $$\left(1+\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots\right)-1$$
Now camparing with $$(1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\cdots$$
So we get $\displaystyle nx=\frac{4}{20}$ and $\displaystyle \frac{n(n-1)x^2}{2}=\frac{4\cdot 7}{20\cdot 30}$
So we get $$\frac{nx\cdot (nx-x)}{2}=\frac{4\cdot 7}{20\cdot 30}\Rightarrow \frac{4}{20}\cdot \left(\frac{4-20}{20}\right)\cdot \frac{1}{2}x^2=\frac{4}{20}\cdot \frac{7}{30}$$
But here $x^2=\text{Negative.}$
I did not understand how can I solve it
Help me, Thanks
|
Since the question asks about $X=\frac4{20}(1+\frac7{30}(1+...)))$, consider
$$1+\frac1{10}(1+\frac4{20}(1+\frac7{30}(...)))$$
The $10,20,30$ have a factor $1,2,3$ which will become $n!$ in the denominator.
Then $\frac1{10},\frac4{10},\frac7{10}$ increase by $3/10$ each time. Take the factor $3/10$ out, and we have $\frac13,\frac43,\frac73$ which increase by 1 each time. Let $x=3/10,n=1/3$.
$$1+xn+\frac{x^2}{2!}n(n+1)+\frac{x^3}{3!}n(n+1)(n+2)+...\\=(1-x)^{-n}=0.7^{-1/3}$$
This equals $1+\frac1{10}(1+X)$,
so your sum is $X=10(0.7)^{-1/3}-11$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+x^2+x^3)+3*2/2((x+x^2+x^3)^2+3*2*1/6(x+x^2+x^3)^3…..$
The coefficient of $x^3$ will be $10$, it is multiplied by $x^9$ outside, so coefficient of $x^{12}$ is $10$.
Can you please explain?
|
From the OP, the coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + \cdots)^3$ is equal to that of $x^3$ in $(1+x+x^2+x^3)^3$. This is equivalent to asking the number of ways to pick one $x^i$ below in each row so that the product of the $x^i$ picked in each row is of the form $kx^3$ for some number $k$.
$$ \require{enclose} \bbox[border:2px solid red]
{
\begin{array}{c|c|c|c}
x^0 & x^1 & x^2 & x^3 \\ \hline
x^0 & x^1 & x^2 & x^3 \\ \hline
x^0 & x^1 & x^2 & x^3
\end{array}
}
$$
If we focus on indices, we will find out that this is equivalent to asking the number of ways of choosing one number from each row so that the sum of three chosen numbers adds up to three.
$$ \bbox[border:2px solid red]
{
\begin{array}{c|c|c|c}
0 & 1 & 2 & 3 \\ \hline
0 & 1 & 2 & 3 \\ \hline
0 & 1 & 2 & 3
\end{array}
}
$$
Therefore, the problem is asking for $$\#\{x,y,z\in\Bbb Z_0^+ \mid x \color{blue}{\fbox+} y \color{red}{\fbox+} z = 3\}.$$
Hence, the answer is very simple: ${5 \choose 2} = 10$. First, imagine that we have a $5\times 1$ grid.
\begin{array}{|l|l|l|l|l|}
\hline \\
&&&& \\ \hline
\end{array}
Then you choose two grids to put $\color{blue}{\fbox+}$ and $\color{red}{\fbox+}$. These two plus signs symbolises $x\color{blue}{\fbox+}y\color{red}{\fbox+}z=3$. Therefore, $\color{blue}{\fbox+}$ should be at the left of $\color{red}{\fbox+}$. The picture below serves as an example.
\begin{array}{|l|l|l|l|l|}
\hline \\
&\color{blue}{\fbox+}&\color{red}{\fbox+}&& \\ \hline \tag{*} \label{*}
\end{array}
Fill the remaining grids with three $\enclose{circle}{1}$ to see what happens.
\begin{array}{|l|l|l|l|l|}
\hline \\
\enclose{circle}{1}&\color{blue}{\fbox+}&\color{red}{\fbox+}&\enclose{circle}{1}&\enclose{circle}{1} \\ \hline
\end{array}
Therefore, this example shows one possibility $x=1,y=0,z=2$. You may make up others by choosing other combinations in \eqref{*}.
|
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|
Evaluation of $\int_{0}^{1}\frac{\arctan x}{1+x}dx$
Evaluation of $$\int_{0}^{1}\frac{\tan^{-1}(x)}{1+x}dx = \int_{0}^{1}\frac{\arctan x}{1+x}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}dx$$
Then $$\frac{dI}{da} = \frac{d}{da}\left[\int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}\right]dx = \int_{0}^{1}\frac{x}{(1+a^2x^2)(1+x)}dx$$
So we get $$I = \frac{1}{a^2}\int_{0}^{1}\frac{x}{(x+1)(x^2+k^2)}dx\;,$$ Where $\displaystyle k=\frac{1}{a^2}$
Now Using Partial fraction for $$\frac{x}{(x^2+k^2)(x+1)} = \frac{Ax+B}{x^2+k^2}+\frac{C}{x+1} = \frac{(A+C)x^2+(A+B)x+(B+Ck^2)}{(x^2+k^2)(x+1)}$$
So we get $A+C=0$ and $A+B=1$ and $B+Ck^2=0$
So we get $\displaystyle A=\frac{1}{1+k^2}$ and $\displaystyle B = \frac{k^2}{1+k^2}$ and $\displaystyle C=-\frac{1}{1+k^2}$
So $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\int_{0}^{1}\left[\frac{x+k^2}{x^2+k^2}-\frac{1}{x+1}\right]dx$$
So $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\left[\frac{1}{2}\ln(x^2+k^2)-k\tan^{-1}\left(\frac{x}{k}\right)-\ln(x+1)\right]_{0}^{1}$$
So we get $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\left[\frac{1}{2}\ln(1+k^2)-\ln(k)-k\tan^{-1}\left(\frac{1}{k}\right)-\ln 2\right]$$
So So we get $$\frac{dI}{da}=\frac{1}{1+a^2}\left[\frac{1}{2}\ln(1+a^2)-\frac{1}{a}\tan^{-1}(a)-\ln2\right]$$
Now integration of above expression is very lengthy, Is there is any other method
If yes Then plz explain here, Thanks
|
I have another way, although the method in the other answer is pretty sweet and mine a bit longer, I decided to post it anyway.
Substitute $\tan^{-1}x=t$, The integral changes as;-
$$\int_{0}^{\pi/4}\frac{t\sec^2 tdt}{1+\tan t}$$
Apply by parts in this to get,
$$=t\log_e|1+\tan t|-\int_{0}^{\pi/4}\log_e|1+\tan t|dt$$
Let this integral be = I,
Then $$I=\int_{0}^{\pi/4}\log_e|1+\tan t|dt=\int_{0}^{\pi/4}\log_e|1+\frac{1-tan t}{1+\tan t}|dt$$ (putting $t=\pi/4-t$ in the integral)
$$\implies I=\int_{0}^{\pi/4}\log_e|\frac{2}{1+\tan t}|dt=\frac{\pi}{4}\log_e{2}-I$$
$$\implies I= \frac{\pi}{8}\log_e{2}$$
Now you can put this above and see that the integral evaluates to $$\frac{\pi}{8}\log_e{2}$$
|
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|
the maximum value of $(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$
If $a, b, c, d \in [\frac12, 2]$ and $abcd =1$, find the maximum value of
$$(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$$
Thought: $$\begin{split}(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a) &= \frac{(ab+1)(bc+1)(cd+1)(da+1)}{abcd} \\ & \stackrel{\text{C-S}}{\le} \sqrt{(a^2+1)(b^2+1)}\sqrt{(b^2+1)(c^2+1)}\sqrt{(c^2+1)(d^2+1)}\sqrt{(d^2+1)(a^2+1)} \\ & = (a^2+1)(b^2+1)(c^2+1)(d^2+1)\end{split}$$
|
This problem is not new: I have found it together with a solution (written in latex + chinese) in :
http://kuing.orzweb.net/viewthread.php?action=printable&tid=3286
Here is one of the interesting solutions presented there, with a slight personal adaptation.
We begin by a lemma that will be used in the sequel:
The convex function defined by $f(x)=x+\frac{1}{x}$ on interval $[\frac{1}{2},2]$ reaches its maximum values at the limit points of its domain, i.e. for $x=1/2$ or $x= 2$, with $f(1/2)=f(2)=2.5$.
Now for the proof. Let us remind the constraint $abcd=1 \ \ \ (1)$
We start from $E=\frac{(ab+1)(bc+1)(cd+1)(da+1)}{abcd} $.
Taking into account (1) and changing the previous factor order into:
$E=[(ab+1)(cd+1)][(bc+1)(da+1)]$
We expand inside each bracketed expression ; using (1), we obtain:
$E=(\sqrt{ab}+\sqrt{cd})^2 (\sqrt{bc}+\sqrt{ad})^2 \ \ \ (2)$
Let $m=\sqrt{ab}$ and $n=\sqrt{bc}$. Using (1), we have:
$E=\left(\left(m+\frac{1}{m}\right)\left(n+\frac{1}{n}\right)\right)^2=\left(mn+\frac{1}{mn}+\frac{m}{n}+\frac{n}{m}\right)^2 = (f(mn)+f(\frac{m}{n}))^2$
where $f$ is a notation introduced in the lemma ; we are in position to apply it (see Appendix explaining why $mn$ and $\frac{m}{n}$ are in $[0.5,2]$) ; it gives the final majorization:
$E \leq (2.5 + 2.5)^2=25 \ \ \ (5)$
Besides, taking
$a=2, b=2, c=\dfrac{1}{2}, d=\dfrac{1}{2}$
in expression $E$ yields this value $25$, which is thus the maximal value of expression $E$.
Remark : this solution uses neither the AGM, nor Cauchy-Schwarz inequality.
APPENDIX: The domains of $\frac{m}{n}$ and $mn$.
Using (1), $\frac{m}{n}=\sqrt{\frac{a}{c}}$. But $\frac{1}{2} \leq a \leq 2$ and $\frac{1}{2} \leq c \leq 2$. Thus $\frac{m}{n} \in [\frac{1}{2},2]$.
$mn=\sqrt{abc}\sqrt{b}=\sqrt{\frac{1}{d}}\sqrt{b}=\sqrt{\frac{b}{d}}$. Thus by a similar reasoning as before, $mn \in [\frac{1}{2}, 2]$.
|
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|
Among the following, which is closest to $\sqrt{0.016}$? Among the following, which is closest in value to $\sqrt{0.016}$?
A. $0.4$
B. $0.04$
C. $0.2$
D. $0.02$
E. $0.13$
My Approach:
$(\frac{16}{1000})^\frac{1}{2} = (\frac{4}{250})^\frac{1}{2} = \frac{2}{5\cdot\sqrt{10}} = \frac{\sqrt{2}}{5\cdot\sqrt{5}}$
But I don't know how to reach to answer.
Thank You!
|
Though squaring each and checking works equally as well, here's another way to figure it out:
$$
\sqrt{0.016}=\frac{\sqrt{1000}}{\sqrt{1000}}\sqrt{0.016}=\frac{\sqrt{16}}{\sqrt{1000}}=\frac{4}{\sqrt{1000}}\approx\frac{4}{\sqrt{1024}}=\frac{4}{\sqrt{2^{10}}}=\frac{4}{2^{5}}=\frac{4}{32}=\frac{1}{8}=0.125
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The area between two curves - is my answer correct?
Find the area between a parabola $y=x^2-2$ and an asymptote to $y=\sqrt{x^2+4x}+2x $ for $x \to -\infty$. As an answer, write the area multiplied by 6.
This is a very easy problem, but I made some careless mistake at first and my answer was incorrect. However, the answer listed for this problem is just 1.
First, we find the asymptote:
$$\lim_{x \to -\infty} \sqrt{x^2+4x}+2x=|x|+2x=x$$
Then we find the intersection points:
$$x^2-2=x$$
$$x_1=-1,~~~~~x_2=2$$
Now the area is:
$$S=\int^{2}_{-1}(x-x^2+2)dx=2-\frac{1}{2}-\frac{8}{3}-\frac{1}{3}+4+2=4.5$$
Now the final answer is $6S=4.5 \cdot 6=27$
Where is my mistake? How could anyone get 1 as a 'correct' answer?
|
We have $\sqrt{x^2+4x}+2x\sim x-2+\frac{2}{x}+ O(\frac{1}{x^2})$ as $x\rightarrow -\infty$. Thus the asymptote is $y=x-2$. The points of intersection are the solutions of $x^2-2=x-2,$ i.e. $x=0\,$ and $x=1$. And therefore the answer is
$$ 6 \Big|\int_0^1 (x^2-2 - (x-2)) dx\Big|= 6 \Big|\int_0^1 (x^2- x) dx\Big| =
6\Big|[\frac{x^3}{3}-\frac{x^2}{2}]_0^1\Big|= 6\Big|-\frac{1}{2}\Big| = 1 $$
|
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|
Distributing identical objects to different people. What is the number of ways in which we can distribute 12 identical oranges among 4 children such that every child gets at least one and no child gets more than 4 ?
Till now ,My attempts have focused on first finding out all such ways in each person gets at least one orange by ${ 13 \choose 3}$ and then trying to find out number of those ways in which at least one person gets more than 4 oranges.
( Basically trying to apply to the inclusion- exclusion principle )
However, I have not been able to figure out the former ( i.e. number of ways in which each person gets more than 4 oranges ).
Is this approach the most suitable one , or should I look at something different ?
|
The number of ways $12$ oranges can be distributed to four children if each child gets at least one orange is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 12 \tag{1}$$
in the positive integers. A particular solution of equation 1 in the positive integers corresponds to the placement of three addition signs in the eleven spaces between successive ones in a row of $12$ ones. For instance,
$$1 + 1 1 + 1 1 1 + 1 1 1 1 1 1$$
corresponds to the solution $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, and $x_4 = 6$. Hence, the number of solutions of equation 1 in the positive integers is
$$\binom{11}{3}$$
Since each child receives at most four oranges, we must exclude those solutions in which one or more of the variables exceeds $4$. Since $3 \cdot 5 = 15 > 12$, at most two of the variables can exceed $4$ simultaneously.
Suppose $x_1 > 4$. Let $y_1 = x_1 - 4$. Then $y_1$ is a positive integer. Substituting $y_1 + 4$ for $x_1$ in equation 1 yields
\begin{align*}
y_1 + 4 + x_2 + x_3 + x_4 & = 12\\
y_1 + x_2 + x_3 + x_4 & = 8 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers with $\binom{7}{3}$ solutions. By symmetry, there are $\binom{7}{3}$ solutions in which one of the four variables exceeds $4$. Hence, there are
$$\binom{4}{1}\binom{7}{3}$$
solutions in which one variable exceeds $4$.
However, subtracting $\binom{4}{1}\binom{7}{3}$ from $\binom{11}{3}$ removes those solutions in which two of the variables exceed $4$ twice. Since we only want to remove such solutions once, we must add the number of solutions in which two of the variables exceed $4$.
Suppose $x_1$ and $x_2$ exceed $4$. Let $y_1 = x_1 - 4$; let $y_2 = x_2 - 4$. Then $y_1$ and $y_2$ are positive integers. Substituting $y_1 + 4$ for $x_1$ and $y_2 + 4$ for $x_2$ in equation 1 yields
\begin{align*}
y_1 + 4 + y_2 + 4 + x_3 + x_4 & = 12\\
y_1 + y_2 + x_3 + x_4 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the positive integers with one solution (each variable is equal to $1$). By symmetry, there is one solution for each of the $\binom{4}{2}$ ways in which two of the variables exceed $4$, so there are
$$\binom{4}{2}\binom{3}{3}$$
solutions in which two of the variables exceed $4$.
By the Inclusion-Exclusion Principle, the number of ways the twelve oranges can be distributed to four children so that each child receives at least one and at most four oranges is
$$\binom{11}{3} - \binom{4}{1}\binom{7}{3} + \binom{4}{2}\binom{3}{3}$$
|
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|
Constant length of segment of tangent
Prove that the segment of the tangent to the curve $y=\frac{a}{2}\ln\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\sqrt{a^2-x^2}$ contained between the $y$-axis and the point of tangency has a constant length.
What I have done: First of all I found out slope $m$ of the tangent to the curve at any general point $(h,k)$, which after simplifying comes out to be
$m=\frac{1}{\sqrt{a^2-h^2}}[2h-\frac{a^2}{h}]$
After that I found the equation of the tangent at the general point which comes out to be
$y-k=\left(\frac{1}{\sqrt{a^2-h^2}}(2h-\frac{a^2}{h})\right)(x-h)$
Then I found the $y$-coordinate at $x=0$ which comes out to be $k+\frac{a^2-2h^2}{\sqrt{a^2-h^2}}$
Then I found distance between points $(h,k)$ and $(0,k+\frac{a^2-2h^2}{\sqrt{a^2-h^2}})$ which comes out to be $\sqrt{\frac{3h^4-3a^2h^2+a^4}{a^2-h^2}}$ which shows length of segment depends on $h$ but this was not we had to prove.
Please tell me where I have made the mistake.
|
Here are some steps:
\begin{align*}
y & =\frac{a}{2}\ln\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\sqrt{a^2-x^2}\\
\frac{dy}{dx} & = \frac{a}{2}\left(\frac{a-\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\right) \, \frac{d}{dx}\left(\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}\right)+\frac{x}{\sqrt{a^2-x^2}}\\
& = \frac{a}{2}\left(\frac{a-\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\right) \, \left(\frac{(a-\sqrt{a^2-x^2})\color{red}{\frac{-x}{\sqrt{a^2-x^2}}}-(a+\sqrt{a^2-x^2})\color{blue}{\frac{x}{\sqrt{a^2-x^2}}}}{(a-\sqrt{a^2-x^2})^2}\right)+\frac{x}{\sqrt{a^2-x^2}}\\
& =\frac{a}{2}\left(\frac{1}{a+\sqrt{a^2-x^2}}\right)\, \left(\frac{\color{red}{\frac{-2ax}{\sqrt{a^2-x^2}}}}{(a-\sqrt{a^2-x^2})}\right)+\frac{x}{\sqrt{a^2-x^2}}\\
& = \frac{-a^2x}{x^2\sqrt{a^2-x^2}}+\frac{x}{\sqrt{a^2-x^2}}\\
\end{align*}
Can you simplify now?
|
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|
Find the unknown vectors of a parallelogram and equilateral triangle I am not looking for the answers, could someone help break the questions down into simpler terms. I can find out what the answers are so that is not the goal.
Vectors $p$, $q$ and $r$ are represented on the diagram as shown:
*
*$BCDE$ is a parallelogram
*$ABE$ is an equilateral triangle
*$\left|p\right| = 3$
*Angle $ABC = 90^\circ$
(a) Evaluate $p\cdot(q + r)$
I am not sure what this means. Is it saying evaluate Multiplying vector $p$ by the sum of $q$ plus $r$?
(b) Express ${\vec {AC}}$ in terms of $p$, $q$ and $r$
For this question, is it saying the path to $EC$ using the vectors.
Would it be $q + p + r$?
(c) Given that ${\vec{AE}\cdot\vec{EC} = 9{\sqrt{3}-{9\over2}}}$, find $\left|r\right|$
I don't understand this question, I am not sure how I find the length of $r$ from the values given.
I am specifically looking for the explanations of $a$ and $c$. I don't understand how I can deduce $\left|r\right|$ from the values.
|
(a) First distribute the dot product over addition as in
$$p\cdot(q+r) = p\cdot q + p\cdot r$$
and notice that $p$ and $r$ are perpendicular and so $p\cdot r = 0$. And so we are left with $p\cdot q$. We know that the angle between $p$ and $q$ is $\frac{\pi}{3}$ and that $\|p\|=\|q\|=3$ because we are told $ABE$ is an equilateral triangle. But we also know that for any vectors $a,b$ that $a\cdot b=\|a\|\|b\|\cos{\theta}$ where $\theta$ is the angle between $a$ and $b$. Thus
\begin{align}
p\cdot q &= \|p\|\|q\|\cos{\frac{\pi}{3}} \\
&= (3)(3)\frac{1}{2} \\
&= \frac{9}{2}
\end{align}
and so
$$p\cdot(q+r) = p\cdot q + p\cdot r = \frac{9}{2} + 0 = \frac{9}{2}$$
(b) Notice two things. First that, because $BCDE$ is a parallelogram, that $\vec{EB}=\vec{DC}$. And, second, that $\vec{EB}=p-q$. Now to get $\vec{AC}$ just add vectors head to tail starting at $A$ until we get to $C$. Thus,
\begin{align}
\vec{AC} &= p + r + \vec{DC} \\
&= p + r + \vec{EB} \\
&= p + r + (p - q) \\
&= 2p + r - q
\end{align}
(c) Notice that $\vec{AE}=q$ and $\vec{EC} = r + \vec{DC}$. From (b) we know that $\vec{DC}=p-q$ and so $\vec{EC} = r + p - q$. From (a) we know that $q\cdot p = \frac{9}{2}$. Thus
\begin{align}
\vec{AE}\cdot\vec{EC} &= q\cdot(r + p - q) \\
&= q\cdot r + q\cdot p - q\cdot q \\
&= q\cdot r + \frac{9}{2} - \|q\|^2 \\
&= q\cdot r - \frac{9}{2}
\end{align}
But we are told this equals $9\sqrt{3} - \frac{9}{2}$ and so we have that $q\cdot r = 9\sqrt{3}$. We know that $q\cdot r=\|q\|\|r\|\cos{\theta}$ and from simple geometry we see that $\theta = \frac{\pi}{6}$ and so
\begin{align}
q\cdot r = 9\sqrt{3} &= \|q\|\|r\|\cos{\frac{\pi}{6}} \\
&= 3\|r\|\frac{\sqrt{3}}{2} \\
&\implies \\
\|r\| &= 6.
\end{align}
|
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|
Probability that $7^m+7^n$ is divisible by $5$
If $m,n$ are chosen from the first hundred natural numbers with replacement, the probability that $7^m+7^n$ is divisible by $5$ is?
$$7^m+7^n=7^m(1+7^{n-m}), n\ge m$$
The above expression is divisible by $5$ only if $n-m=4k+2$. The max value of $k$ is $24$.
So is the number of possibilities $48$? ($24$ for each $n>m$ and $m>n$)
|
The possibilities of ending digits of $7^m$ or $7^n$ : $9,3,1,7$
Favourable cases : $(9,1),(3,7),(1,9),(7,3)$
Probability : $\frac{4}{4\times 4} =\frac{1}{4}$
|
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|
How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$? How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$?
My Attempt
Let $f(x,y)=x^2 + 4y^2 − 2xy − 2x − 4y − 8$ . So $f(x,0)=x^2 − 2x − 8$ . $f(x,0)$ has two roots $x=4 , -2$ . So (4,0), (-2,0) are solution of the given equation. Same way solving for $f(0,y)=0$ we get $ y=2 , -1$ are roots and hence (0,2), (0,-1) are solutions. I have tried to factorize $f(x,y)$ or writing it as sum of squares but could not succeed. Is there any other solutions? How do I find them?
|
HINT:
$$x^2-2x(1+y)+4y^2-4y-8=0$$
$$x=\dfrac{2(1+y)\pm\sqrt{\{2(1+y)\}^2-4(4y^2-4y-8)}}2=1+y\pm\sqrt{9+5y-3y^2}$$
We need
$$0\le9+5y-3y^2\iff3y^2-5y-9\le0$$
Now roots of $3y^2-5y-9=0$ are $\dfrac{5\pm\sqrt{25+108}}2$
Now $(x-a)(x-b)\le0, a\le b\implies a\le x\le b$
|
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|
Find $m$ so that $(m+4\cdot41)(m^2+4^2\cdot41^2)$ is a square number
Question: Find the minimum positive odd interger $m$ so that $(m+4\cdot41)(m^2+4^2\cdot41^2)$ is a square number.
Any suggestion will be appreciated. Thanks.
|
HINT:
Since $m$ is an odd positive integer, so we must have
$$(m+4\cdot 41)(m^2+4^2\cdot 41^2)\equiv 1 \pmod 8$$
$$m^3+4^3\cdot 41^3 + m\cdot 4\cdot 41(m+4\cdot 41) \equiv 1 \pmod 8$$
|
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|
Prove by induction that $n^2 < 2^n$ for all $n \geq 5$? So far I have this:
First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.
Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.
So I started with $(n+1)^2 = n^2 + 2n + 1$. Because $n^2 < 2^n$ by the hypothesis, $n^2 + 2n + 1$ < $2^n + 2n + 1$. As far as I know, the only way I can get $2^{n+1}$ on the right side is to multiply it by $2$, but then I get $2^{n+1} + 4n + 2$ on the right side and don't know how to get rid of the $4n + 2$. Am I on the right track, or should I have gone a different route?
|
Remark that
$$2^{n+1} = 2 \cdot 2^n > 2 n^2 = n^2 +n^2> n^2+2n+1 = (n+1)^2.$$
Indeed $n^2-2n-1=(n-1)^2-2>0$ for $n \geq 5$.
|
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|
What is the maximum of $\ 2(a+b)-ab$ if we have: $\ a^2+b^2=8\ (a,b\ real)$ What is the maximum of $\quad 2(a+b)-ab\ $ if we have: $a^2+b^2=8 \ (a,b\ real) $
My work is as follows:
According to AM-GM inequality:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2} \Rightarrow $$
$$ab\le4 $$
On the other hand , from $\ a^2+b^2=8\ ,\ $we have: $\ 2(a+b)-ab=2\sqrt{8+2ab}-ab$
We can get this: $$2\sqrt{8+2ab}\le8$$
Is it possible to somehow combine these two inequalities to find the maximum value of $\ 2(a+b)-ab$ ?
I think in this solution technique we must find the minimum value of $ab\ $ , but how???
|
Another, easier method(without trigonometry and differentiation):
$$a^2+b^2 = (a + b)^2 - 2ab = 8 \rightarrow ab=\frac{(a + b)^2}{2} - 4$$
Substitute in your expression. Now you have to find the maximum of the following:
$$2(a + b) - \frac{(a + b)^2}{2} + 4$$Let $t = a + b$. Since $a, b$ are arbitrary reals, then $t$ is also arbitrary real.
$$f(t) = -\frac{1}{2}t^2 + 2t + 4$$
$f(t)$ is a quadratic function, since the coefficient before $t^2$ is negative, the parabola is turned downwards. Its vertex is $t_0 = -\frac{b}{2a} = 2$, hence f(t) attains maximum value when $t = 2$ and that value is $$f(2) = 6$$
|
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|
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requires going from:
Expression 1: $$-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2$$
to:
Expression 2: $$(a+b-c)(a-b+c)(a+b+c)(-a+b+c)$$
I can see that these two expressions are equal by working in reverse and multiplying out the second expression to get the first but how does one go from Exp.1 to Exp.2?
|
$4a^2b^2-(a^2+b^2)^2+2(a^2+b^2)c^2-c^4=(2ab)^2-((a^2+b^2)^2-2(a^2+b^2)c^2+c^4))=(2ab)^2-(a^2+b^2-c^2)^2$
now you go to last step.
|
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What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I suppose that I can't do it in this way. Why?
EDIT:
This is my solution:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)
\Longleftrightarrow
\sin(2x) - \sin(x) = 0
\Longleftrightarrow
2\cos(\frac{3x}{2})\sin(\frac{x}{2}) = 0
\Longleftrightarrow
\cos(\frac{3x}{2}) = 0 \vee \sin(\frac{x}{2}) = 0$$
Proper solution is $\cos(x) = \frac{1}{2} \vee \sin(x) = 0$
|
Equations $\frac{\sin x}{\cos x} = 2\sin x$ and $\sin x =2\sin x\cos x$ are indeed equivalent since for $\cos x = 0$ you have $\sin x =\pm 1$, which doesn't give solution for the second equation.
So, to solve it, we have that either $\sin x = 0$ or $\cos x = \frac 1 2$, thus solutions are given by $x = 2k\pi$, $x = \pi + 2k\pi$, $x = \pm\frac\pi 3 + 2k\pi$, $k\in\mathbb Z$.
To summarize, there is no error in your manipulations.
|
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How to solve the non-homogeneous linear recurrence $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$? The problem: $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$
First I solved the associated homogeneous recurrence and got $a_n = A(1)^n = A$, where A is a constant, but I got stuck solving the rest. My final answer was $a_n=2n^2+n+1$ while my textbook has $a_n=n^2+2n+1$
Here is my work:
$ a_{n+1} - a_n = 2n+3 $
$pn=nd_1+d_2$ for all n
plug in and solve for $d_1$:
$(n+1)d_1+d_2-nd_1-d_2=2n+3$
$d_1=2n+3$
then solve for $d_2$:
$a_0=1=a_{n+1}-2n-3=(n+1)d_1+d_2-2n-3$
$d_2=1-n(2n+3)$
therefore
$p_n=nd_1+d_2 = 2n^2+n+4$ and $a_n=A+2n^2+n+4$
then solve for A:
$a_0=1=A +0+0+4$ so $A=-3$
therefore
$a_n=2n^2+n+1$
|
If $ a_{i+1} - a_i = 2i+3 $ then $\sum_{i=0}^{n-1}(a_{i+1} - a_i)=\sum_{i=0}^{n-1}(2i+3)$.
Observe that $\sum_{i=0}^{n-1}(a_{i+1} - a_i)=a_n-a_0$ and $\sum_{i=0}^{n-1}(2i+3)=n(n-1)+3n$.
Hence $a_n=1+n(n-1)+3n=(n+1)^2$.
|
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Suppose $n$ divides $3^n + 4^n$. Show that $7$ divides $n$. Suppose $n \geq 2$ and $n$ is a divisor of $3^n + 4^n$. Prove that $7$ is a divisor of $n$.
My work so far:
I had a hypothesis that if $n| 3^n + 4^n$, then $n = 7^k$ for some $k\in\mathbb{N}$. But this is not necessarily so. Take $n = 7⋅379$, where $3^7 + 4^7 = 7^2⋅379$. Then, $3^7+4^7$ divides $3^n + 4^n$, and since $n$ divides $3^7+4^7$, we must have $n|3^n+4^n$.
|
First, we note that $3 \nmid 3^n + 4^n$ and $4 \nmid 3^n + 4^n$. So $3 \nmid n$ and $2 \nmid n$.
Now, reworking $3^n + 4^n \equiv 0 \pmod n$. Since n is odd, we find
\begin{equation}
3^n \equiv (-4)^n \pmod n.
\end{equation}
Since $\gcd(3, n) = 1$, we can take the inverse of $3 \mod n$, i. e. there exists $3^{-1}$ so that $3 \cdot 3^{-1} \equiv 1 \pmod n$. Multiplying both sides by $(3^{-1})^n$ gives
\begin{equation}
(-4 \cdot 3^{-1})^n \equiv 1 \pmod n.
\end{equation}
Now, if $O_m(k)$ denotes the order of $k \mod m$, we know from algebra that
\begin{equation}
O(-4 \cdot 3^{-1}) \mid n.
\end{equation}
Now, to show that $7 \nmid n$ for any solution, let's assume $7 \nmid n$ for some $n \in \mathbb{N}$. Let's assume $n > 1$ is the smallest solution so that $7 \nmid n$ and $n \mid 3^n + 4^n$.
If we can show that either $n = 1$ or $7 \mid n$ or there exists an $m < n$ satisfying these properties, we're done.
Let's split 2 cases:
(Case 1): $O_n(-4 \cdot 3^{-1}) = 1$ Then we're done, since that implies (using uniqueness of inverses):
\begin{equation}
(-4 \cdot 3^{-1}) \equiv 1 \pmod n \implies -4 \cdot 3^{-1} \equiv 3 \cdot 3^{-1} \pmod n \\ \implies -4 \equiv 3 \pmod n \implies n \mid 7 \implies n = 1 \vee n = 7.
\end{equation}
(Case 2): $O_n(-4 \cdot 3^{-1}) > 1$. Let $d = O_n(-4 \cdot 3^{-1})$.
Then $(-4 \cdot 3^{-1})^d \equiv 1 \pmod n$, and since $d \mid n$ we have $(-4 \cdot 3^{-1})^d \equiv 1 \pmod d$. $d$ is also odd, and
\begin{equation}
-4^d (3^{-1})^d \equiv 1 \pmod d \implies -4^d \equiv 3^d \pmod d
\end{equation}
If $O_d(-4 \cdot 3^{-1}) = 1$, then by the reasoning above $d \mid 7$. Since $d > 1$, we have $d = 7$. Since $d \mid n$, $7 \mid n$.
If $O_d(-4 \cdot 3^{-1}) > 1$, let m = $O_d(-4 \cdot 3^{-1})$. Then $m > 1$, $7 \nmid m$ and $m \mid 3^m + 4^m$. $m < n$, since $m \mid \varphi(n)$. So we found an $m < n$ satisfying the above conditions, implying $n$ is not the smallest.
|
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|
Methods for Integrating $\int \frac{\cos(x)}{\sin^2(x) +\sin(x)}dx$ So I've found that there's the Weierstrass Substitution that can be used on this problem but I just want to check I can use a normal substitution method to solve the equation:
$$\int \frac{\cos(x)}{\sin^2(x) +\sin (x)}dx$$
Let $u = \sin(x)$
$du = \cos(x)\, dx$
$dx = \frac {1}{\cos(x)\,} du$
Which becomes:
$$\int \frac{\cos(x)}{u^2 + u} \frac{1}{\cos(x)}du$$
$$\int \frac{1}{u^2 + u}du$$
Factor out u from denominator:
$$\int \frac{1}{u(u + 1)}du$$
Integrate as a partial fraction:
$$\int \frac{1}{u} - \frac{1}{(u + 1)}du$$
Which integrates as:
$$\ln|u| - \ln|(u + 1)| + C$$
Subtitute $u = \sin(x)$ back in and simplifies to:
$$\ln \left|\frac{\sin(x)}{\sin(x)+1} \right| + C$$
Is this correct? From the Weierstrass Substitution, one gets:
$$\ln \left|\tan \left(\frac{x}{2}\right)\right|-2\ln \left|\tan \left(\frac{x}{2}\right)+1\right| +C $$
|
By using Weierstrass Substitution,
let $t = \tan{\frac{x}{2}}, \,dt = \frac{1}{2}\sec^{2}{\frac{x}{2}}\,dx$
$=> 2\cos^{2}{\frac{x}{2}}\,dt = \frac{2}{1+t^{2}}\,dt = dx$
$\because \sin{x} = \frac{2t}{1+t^{2}}, \cos{x} = \frac{1-t^{2}}{1+t^{2}}$
$\therefore \int \frac{\cos{x}}{\sin^{2}{x}+\sin{x}}\,dx = \int \frac{(\frac{1-t^{2}}{1+t^{2}})(\frac{2}{1+t^{2}})}{(\frac{2t}{1+t^{2}})^2+(\frac{2t}{1+t^{2}})}\,dt = \int \frac{2(1-t^{2})}{4t^{2}+2t(1+t^{2})}\,dt = \int \frac{2(1-t)(1+t)}{2t(1+t^{2}+2t)}\,dt = \int \frac{1-t}{t(1+t)}\,dt$
$= \int \frac{1+t-2t}{t(1+t)} = \int \frac{1}{t}\,dt -2 \int \frac{1}{1+t}\,dt = \ln|t| - 2\ln|1+t| + C$
$ = \ln|\tan\frac{x}{2}|-2\ln|1+\tan\frac{x}{2}|+C$
Therefore your statement is correct!
|
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|
check workings for polynomial function from graph
The diagram shows a curve with equation of the form $y = kx(x + a)^2$,
which passes through $(-2, 0)$, $(0, 0)$ and $(1, 3)$.
What are the values of $a$ and $k$.
To find $k$, I would set $f(x) = 1$.
$f(1) = k(x + 2)(x + 0)(x - 1)$
$3 = k(1 + 2)(x + 1)(1 - 1)$
$3 = k(3)(1)(0)$
This would make $k = 0$ which does not appear to be right.
|
$$y=kx(x+a)^2\quad \quad \quad (-2, 0),(0, 0) , (1, 3)$$
$$\text{for} (0,0):\quad \quad \quad f(0)=0=0\\
\text{for}(1,3):\quad \quad \quad f(1)=k(1+a)^2=3\\
\text{for}(-2,0):\quad \quad \quad f(-2)=-2k(-2+a)^2=0$$
so you have two equations with two variables:
$$\begin{cases}k(1+2a+a^2)=3\\
-2k(4-4a+a^2)=0\end{cases}$$
$$\begin{cases}k+2ak+ka^2=3\\
-8k+8ka-2ka^2=0\end{cases}$$
Solve, you should get $\color{red}{a=2,k=\frac 1 3}$
|
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|
How to find $a+b$ in square $ABCD,$? In square $ABCD,$ $AB=1.$ $BEFA$ and $MNOP$ are congruent. $BE= a - \sqrt b.$ Where $a$ and $b$ are both primes. How to find $a+b$? I have no idea how to do this, can this be proved with simple geometry?
|
Let the side of the square be $p$ (so we can better track dimensions) and let $|\overline{BE}| = q$; and let $\theta = \angle EMP$. We have two equations:
$$\begin{align}
q + p \sin\theta + q \cos\theta &= p \\
p \cos\theta + q \sin\theta &= p
\end{align}$$
Solving for $\cos\theta$ and $\sin\theta$ gives
$$\cos\theta = \frac{p^2 - p q + q^2}{p^2 - q^2}\qquad \sin\theta = \frac{p\, (p - 2 q)}{p^2 - q^2}$$
Since $\sin^2\theta + \cos^2\theta = 1$, we deduce
$$\frac{p\,(p - 2 q)\,(p^2 - 4 p q + q^2)}{(p^2 - q^2)^2} = 0$$
so that $q = p/2$ (extraneous), or $q = p\,( 2 + \sqrt{ 3 } )$ (extraneous), or $q = p\,(2 - \sqrt{3})$ (bingo!).
Recalling $p=1$, we have $a = 2$ and $b = 3$, so that $a+b = 5$. $\square$
Note: In this solution, $\theta = 30^\circ$.
|
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|
Fundamental Theorem of Calculus Confusion regarding atan According to this site,
$$ \int \frac{1}{a^2 \cos^2(x) + b^2 \sin^2(x)} \,dx =\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(x)\right)$$
Thus,
$$ \int_0^{\pi} \frac{1}{a^2 \cos^2(x) + b^2 \sin^2(x)} \, dx =\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(\pi)\right)-\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(0)\right)=0-0=0$$
But in fact value of integral is not zero.
What am I doing wrong?
|
This very integral, with a little manipulation, is referenced in the Jeffrey paper so the nice form in that paper can be attained by observing that $\tan^{-1}(\tan x)$ has the same discontinuities as the original integral, so we may hope to cancel them out by setting
$$\frac1{ab}\tan^{-1}\left(\frac{b}a\tan x\right)=\frac1{ab}\left\{x+\tan^{-1}\left(\frac{b}a\tan x\right)-\tan^{-1}(\tan x)\right\}$$ Now,
$$\begin{align}\tan^{-1}\left(\frac{b}a\tan x\right)-\tan^{-1}(\tan x) & =\tan^{-1}\left(\frac{\frac{b}a\tan x-\tan x}{1+\frac{b}a\tan^2x}\right) \\ & =\tan^{-1}\left(\frac{(b-a)\sin2x}{(a+b)-(b-a)\cos2x}\right) \end{align}$$ after a little algebra, so
$$\int\frac1{a^2\cos^2x+b^2\sin^2x}dx=\frac1{ab}\left\{x+\tan^{-1}\left(\frac{(b-a)\sin2x}{(a+b)-(b-a)\cos2x}\right)\right\}+C$$ removes all discontinuities and even checks by differentiation.
|
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|
show this inequality with $a+b+c+d=1$ Let $a,b,c,d\ge 0$,and such $a+b+c+d=1$, show that
$$3(a^2+b^2+c^2+d^2)+64abcd\ge 1$$
use AM-GM
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
it suffices to
$$4\sqrt{abcd}+64abcd\ge 1$$
|
the inequality equals:
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4\ge 0 $
WLOG, let $d=min(a,b,c,d)$
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4=2d\left(\dfrac{1}{3}\sum_{cyc (a,b,c)}(a-d)^2(2a+d)+\sum_{cyc (a,b,c)} a(a-d)^2+(a+b+c-3d)((a-b)^2+(b-c)^2+(a-c)^2)+3(a^3+b^3+c^3+3abc-\sum ab(a+b))+\dfrac{a^3+b^3+c^3}{3}-abc\right)+(a+b+c-3d)^2((a-b)^2+(b-c)^2+(a-c)^2)\ge0$
|
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|
calculate $\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}$? prove the following equation :
$$\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}={s+2r-1\choose r} - {s+2r-1\choose r-1}$$
|
I use two known properties of binomial coefficients: $${n+k \choose k} = (-1)^k {-n-1 \choose k}\tag{$i$}$$ and $$\sum_{i}{a \choose i} {b\choose c-i} = {a+b \choose c}\tag{$ii$}.$$
First,
$$
\sum_{0\le i\le r}\frac{r-i} {r+1}{2r-i\choose r-i}{s+i-2\choose i} \overset{(i)}{=}
(-1)^r \sum_{0\le i\le r-1}\frac{r-i} {r+1}{-r-1\choose r-i}{-s+1\choose i} \\
= (-1)^{r+1}\sum_{0\le i\le r-1}{-r-2\choose r-i-1}{-s+1\choose i} \\\overset{(ii)}{=}
(-1)^{r+1}{-r-s-1\choose r-1} \overset{(i)}{=} {2r+s-1\choose r-1}.\tag{1}
$$
Similarly,
$$
\sum_{0\le i\le r}{2r-i\choose r-i}{s+i-2\choose i} \overset{(i)}{=}
(-1)^r \sum_{0\le i\le r}{-r-1\choose r-i}{-s+1\choose i} \\
\overset{(ii)}{=} (-1)^r{-r-s\choose r} \overset{(i)}{=} {2r+s-1\choose r} \tag{2}.
$$
Subtracting (1) from (2),
$$
\sum_{0\le i\le r}\frac{i+1} {r+1}{2r-i\choose r-i}{s+i-2\choose i} = {2r+s-1\choose r-1} - {2r+s-1\choose r}.
$$
|
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|
Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$
Let $a$ be a constant. Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$.
After computing a few derivatives, the derivatives seem to have factorials in them sometimes and other times not. For example, $\dfrac{d^5}{dx^5} \left (\dfrac{1}{x^2-a^2} \right) = -\dfrac{240(3a^4x+10a^2x^3+3x^5)}{(x^2-a^2)^6}$ while $\dfrac{d^6}{dx^6} \left (\dfrac{1}{x^2-a^2} \right) = -\dfrac{720(a^6+21a^4x^2+35a^2x^4+7x^6)}{(x^2-a^2)^6}$, so I am not immediately seeing the pattern.
|
Just observe that
$$f(x)=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$$
Then
$$f^{(n)}(x)=\frac{(-1)^nn!}{2a}\left[\frac{1}{(x-a)^{n+1}}-\frac{1}{(x+a)^{n+1}}\right]$$
|
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|
Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2} - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5 = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
|
$$ 6 s^2 + s c -c^2 =5 $$
Divide by c^2 . $ t=s/c,$
$$ 6 t^2 + t - 1 = 5/c^2 = 5 ( 1 +t^2) \rightarrow ;\; t^2 + t -6 =0;\; (t-2)(t+3) =0$$
$$ t= \tan x = 2,-3 $$
|
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|
Cubic Equation Related How do I solve the following cubic equation?
$(x+1)^2(x-2)-2nx-6n=0,n \in \mathbb{N}$
$ \therefore (x+1)^2(x-2)-2n(x+3)=0$
I don't know how to solve further.
|
We have
$$(x+1)^2(x-2)-2nx-6n=0,$$
i.e.
$$x^3+(-3-2n)x-2-6n=0\tag1$$
Let $f(x)=x^3+(-3-2n)x-2-6n$. Then,
$$f'(x)=3x^2-3-2n=0\iff x=\pm\sqrt{\frac{3+2n}{3}}.$$
We have
$$\begin{align}f\left(-\sqrt{\frac{3+2n}{3}}\right)\lt 0&\iff (2n+3)\sqrt{\frac{2n+3}{3}}\lt 9n+3\\&\iff (2n+3)^2\cdot\frac{2n+3}{3}\lt (9n+3)^2\\&\iff 8n^2-207n-108\lt 0\\&\iff 1\le n\le 26\end{align}$$
So, we know that $(1)$ has only one real root for $1\le n\le 26$, and that $(1)$ has three distinct real roots for $n\ge 27$.
Here, let
$$x=y+\frac{A}{y}\quad \text{where $\quad A=\frac{2n+3}{3}$}\tag2$$
Then,
$$\begin{align}(1)&\iff \left(y+\frac{A}{y}\right)^3+(-3-2n)\left(y+\frac{A}{y}\right)-2-6n=0\\&\iff y^3+3y^2\cdot\frac Ay+3y\cdot\frac{A^2}{y^2}+\frac{A^3}{y^3}+(-3-2n)y+\frac{A(-3-2n)}{y}-2-6n=0\\&\iff y^3+\frac{A^3}{y^3}-6n-2=0\end{align}$$
Here, let $$y^3=z\tag3$$ to have
$$\begin{align}z+\frac{A^3}{z}-6n-2=0&\iff z^2-(6n+2)z+A^3=0\\&\iff z=3n+1\pm\sqrt{(3n+1)^2-A^3}\end{align}$$
Therefore, from $(2)(3)$, we have
$$x=y+\frac Ay=z^{1/3}+\frac{A}{z^{1/3}}$$
$$\color{red}{x=\sqrt[3]{3n+1\pm\sqrt{(3n+1)^2-((2n+3)/3)^3}}+\frac{1+(2n/3)}{\sqrt[3]{3n+1\pm\sqrt{(3n+1)^2-((2n+3)/3)^3}}}}$$
|
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|
Prove $ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $
Question: Prove $$ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$
RHS: $$ \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$
$$ ⇔ \frac{\frac{1}{\sin^2(x)} +\frac{\cos(x)}{\sin(x)}-2\cos^2(x)}{1-\frac{\cos^2(x)}{\sin^2(x)}} $$
$$ ⇔ \frac{\frac{1}{\sin^2(x)} +\frac{\cos(x)\sin(x)}{\sin^2(x)}-\frac{2\cos^2(x)\sin^2(x)}{\sin^2(x)}}{\frac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}} $$
$$ ⇔ \frac{\frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)}}{\frac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}} $$
$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)} \times {\frac{\sin^2(x)}{\sin^2(x)-\cos^2(x)}} $$
$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)-\cos^2(x)} $$
$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))} $$
Now I am stuck
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You are correct except one sign :
$$\frac{1\color{red}{-}\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}$$
(The error starts at the very beginning.)
Now using $1=\cos^2(x)+\sin^2(x)$,
$$\begin{align}&1-\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)\\&=(1-2\cos(x)\sin(x))(1+\cos(x)\sin(x))\\&=(\cos^2(x)+\sin^2(x)-2\sin(x)\cos(x))(\cos^2(x)+\sin^2(x)+\cos(x)\sin(x))\\&=(\sin(x)-\cos(x))(\sin(x)-\cos(x))(\sin^2(x)+\sin(x)\cos(x)+\cos^2(x))\\&=(\sin(x)-\cos(x))(\sin^3(x)-\cos^3(x))\end{align}$$
|
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|
Can't solve exponential equation using logs? I can't figure out why my method isn't working.
I know it is possible to solve this using a substitution but I don't know when to use the substitution. In general when are you supposed to substitute for, say, u?
Here is how I did it;
We have the following function:
$3^{x-1}+3^{-x+1}$ and the slope of a point is given: $\psi =\frac{8}{3}\ln(3)$
Calculate the coordinates of the point in which the slope equals $\psi$
My approach (Which is wrong):
$$\ln(3)3^{x-1}-\ln(3)3^{-x+1}= \frac{8}{3}\ln(3)$$
$$3^{x-1} - 3^{-x+1} = \frac{8}{3}$$
$$\log_{3}(3^{x-1})-\log_{3}{3^{-x+1}}=\log_{3}(\frac{8}{3})$$
$$(x-1)\log_{3}(3)-(-x+1)\log_{3}(3) = \log_{3}(\frac{8}{3})$$
$$(x-1)(1)-(-x+1)(1) =\log_{3}(\frac{8}{3})$$
$$x-1+x-1 =\log_{3}(\frac{8}{3})$$
$$2x-2=\log_{3}(\frac{8}{3})$$
$$2x = \log_{3}(\frac{8}{3})+2$$
$$x = \frac{1}{2}\log_{3}(\frac{8}{3})+1$$
Now I have a hard time with problems like these in general so all help is appreciated.
-Bowser
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Here after your first line of taking the derivatives, set $u = x-1$
$3^{x-1} - 3^{-x+1} = \frac{8}{3}$
$3^{u} - 3^{-u} = \frac{8}{3}$
Furthermore set $v = 3^u$
$v - \frac{1}{v} = \frac{8}{3}$
$\frac{v^2 - 1}{v} = \frac{8}{3}$
$v^2 - 1= \frac{8v}{3}$
$v^2 - \frac{8v}{3} - 1=0$
$\Delta = \sqrt{ \frac{64}{9} + 4 }$
$v_{1,2} = \frac{\frac{8}{3} +- \sqrt{ \frac{100}{9} }}{2}$
$v_{1,2} = \frac{\frac{8}{3} +- \frac{10}{3} }{2}$
$v_{1} = \frac{18}{6} = 3$
$v_{2} = \frac{-2}{6} = -0.33333...$
Since $v = e^u$, it can not be negative, so only $v = 3$
Now put back $u = ln(v)$ and $x = 1+u$
|
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Geometric inequality involving the inradius For the $\triangle ABC$, let $T$ be the area of the triangle, $a,b,c$ its sides, $p$ the semiperimeter and $r$ the inradius. Prove the following inequality:
$$p^2\ge 2\sqrt3 T+\frac {abc}{p}+r^2.$$
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Using the known formula $pr=T$, we rewrite the inequality as follows
$$ p^2 - \frac{abc}{p} - \frac{T^2}{p^2} \ge 2 \sqrt{3} T$$
Now, using Heron's formula,
$$ \begin{align*}
\mathrm{LHS} &= \frac{1}{4} (a+b+c)^2 - \frac{2abc}{a+b+c} - \frac{(a+b-c)(a-b+c)(-a+b+c)}{4(a+b+c)} \\
&=\frac{1}{4} (a+b+c)^2 - \frac{8abc+(a+b-c)(a-b+c)(-a+b+c)}{4(a+b+c)}\\
&=\frac{1}{4} (a+b+c)^2 - \frac{(a+b+c)\left(2ab+2bc+2ca-a^2-b^2-c^2\right)}{4(a+b+c)}\\
&=\frac{1}{2} \left(a^2+b^2+c^2 \right)
\end{align*}$$
So, the inequality becomes
$$ a^2+b^2+c^2 \ge 4 \sqrt{3} T $$
Let $\gamma \colon \!= \widehat{ACB}$. Then, by the Law of cosines
$$ c^2 = a^2 + b^2 -2ab \cos \gamma $$
By the trigonometric formula of the area of a triangle
$$ T= \frac{1}{2} ab \sin\gamma $$
The inequality becomes then
$$ 2 \left( a^2+b^2 - ab \left(\cos\gamma + \sqrt{3} \sin\gamma \right) \right)\ge 0$$
And this is true since
$$\begin{align*}
a^2+b^2 - ab \left(\cos\gamma + \sqrt{3} \sin\gamma \right) &= a^2+b^2- 2 ab \sin\left(\gamma+ \frac{\pi}{6} \right) \\
& \ge a^2+b^2- 2 ab = (a-b)^2 \ge 0
\end{align*}$$
And you can also see that the equality holds iff $a=b$ and $\gamma= \frac \pi 3$, that is when the triangle is equilateral.
|
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|
Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively. Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively.Find $(a-b).$
I tried to factorize $(1+x+2x^2+3x^3)$ and $(1+x+2x^2+3x^3+4x^4)$ into product of two binomials,but i could not.And i do not have any other method to solve it.
|
The problem is a way of making the point that changing the terms of degree $k$ and higher in the argument can only change the degree $k$-or-higher terms in the result, when applying polynomial or power-series operations (such as taking the 4th power of the argument). If you are performing a calculation to get the answer, something is wrong.
|
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|
How many solutions in integers to the following equation
What is the number of positive integer solutions $(a, b)$ to $2016 + a^2 = b^2$?
We have,
$2016 = (b-a)(b+a) = 2^5 \cdot 3^2 \cdot 7$
$b - a = 2^{t_1} 3^{t_2} 7^{t_4}$ and
$a - b = 2^{t_5} 3^{t_6} 7^{t_7}$
Thus, we must have $t_1 + t_5 = 5$ and $t_2 + t_6 = 2$ and $t_4 + t_7 = 1$.
Total, $\binom{6}{1} \cdot\binom{3}{1} \cdot\binom{2}{1} = 36$.
But the actual answer is $12$, where am I going wrong?
|
First of all, $a$ and $b$ must be integers. Since
$$
a = \frac{(b+a) - (b-a)}{2},\quad b = \frac{(b+a) + (b-a)}{2}
$$
we see that for $a$ and $b$ to be integers, $(b-a)$ and $(b+a)$ must have the same parity. Since at least one of them must be even (there are five powers of $2$ to distribute between them), then both must be even. Therefore, instead of $0\leq t_1, t_5 \leq 6$, we have $1\leq t_1, t_5\leq 5$. Therefore we get $4\cdot 3 \cdot 2 = 24$ possible combinations.
Next, exactly half of the distributions makes $b-a > b+a$, which is not allowed (since there is an odd number of $2$'s and $7$'s, none of the distributions make $b-a = b+a$). Therefore only half of the $24$ combinations from the previous paragraph are actually valid. Thus we get $12$ combinations in total.
|
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|
Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
|
$$
\begin{align}
1
&=\left(\sin^2(x)+\cos^2(x)\right)^3\\
&=\sin^6(x)+3\sin^4(x)\cos^2(x)+3\sin^2(x)\cos^4(x)+\cos^6(x)\\
&=\sin^6(x)+3\sin^2(x)\cos^2(x)+\cos^6(x)\tag{1}\\\\
1
&=\left(\sin^2(x)+\cos^2(x)\right)^2\\
&=\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x)\tag{2}
\end{align}
$$
Subtracting $3$ times $(2)$ from $2$ times $(1)$ gives $-1$. Therefore,
$$
2\left(\sin^6(x)+\cos^6(x)\right)-3\left(\sin^4(x)+\cos^4(x)\right)+1=0\tag{3}
$$
|
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|
Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$
Assume $f$ has a continuous second derivative with $f(0) = f'(0) = f'(1) = 0$ and $f(1) = 1$. Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$.
We must also have that $f'$ is continuous by differentiability. Therefore, by Rolles theorem there exists a $c$ in $[0,1]$ such that $f''(c) = 0$. I am not sure how to use the fact that $f(1) = 1$ to show that $|f''(x)| > 4$.
|
$$f(\frac{1}{2}) = f(0) + f'(0)(\frac{1}{2}) + \frac{f''(\theta_1)}{2}(\frac{1}{2})^2 = f''(\theta_1)\frac{1}{8}$$
$$f(\frac{1}{2}) = f(1) - f'(1)(\frac{1}{2}) + \frac{f''(\theta_2)}{2}(\frac{1}{2})^2 = 1 + f''(\theta_2)\frac{1}{8}$$
Here, $0\le \theta_1 \le 1/2$, and $1/2 \le \theta_2 \le 1$
Combining both, we have
$$ 1 = f''(\theta_1)\frac{1}{8} - f''(\theta_2)\frac{1}{8}$$
Take absolute value,
$$ 1 \le max \{|f''(\theta_1)|, |f''(\theta_2)|\} \frac{1}{4}$$
So c is one of $\theta_1, \theta_2$.
Let
$$f(x) = \begin{cases} 2x^2\, \text{ for } 0 \le x \le 1/2\\ -2(x-1)^2 +1\, \text{ for } 1/2 < x \le 1 \end{cases}$$
$$f''(x) = 4$$, so the equality holds for this f.
|
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|
differentiation problem If $x^{13}y^{7}=(x+y)^{20}$ , then $\frac{dy}{dx}$
directly doing it makes it very complicated so, I did this $\left(\frac{x}{y}\right)^{13}=\left(1+\frac{x}{y} \right)^{20}$.
following are the options for solution
(a) $\frac{y^2}{x^2}$ (b)$\frac{x^2}{y^2}$ (c)$\frac{x}{y}$ (d)$\frac{y}{x}$
thanks for any hints.
|
I differentiated directly,
$$13x^{12}y^7+7x^{13}y^6y'=20(x+y)^{19}(1+y')$$
$${13x^{13}y^7\over {x}}+{7x^{13}y^7y'\over y}={20(x+y)^{20}(1+y')\over {(x+y)}}$$
$$(x+y)(13y+7xy')=20xy(1+y')$$
Cancelling $x^{13}y^7$ and rearranging we get,
$$y'={13y^2-7xy\over {13xy-7x^2}}$$
Divide the numerator and the dinominator by$x^2$
$$y'={13({y\over x})^2-7({y\over x})\over {13({y\over x})-7}}$$
$$y'={y\over x}$$
|
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|
Compute $e^{-x^2} * e^{-x^2}$ How to compute the convolution of $e^{-x^2}$ with itself?
$$e^{-x^2} * e^{-x^2} = \int_{\mathbb R} e^{-(x-y)^2} e^{-y^2}dy = e^{-x^2}\int_{-\infty}^{\infty} e^{2xy - 2y^2} dy$$
I can't solve it. I tried integration by parts so neglecting $e^{-x^2}$:
$$2\int_{-\infty}^{\infty}(2y^2 - xy) e^{2xy - 2y^2}dy$$
please help. thanks.
|
$$ \left( y - \frac{x}{2} \right)^2 = y^2 - xy + \frac{x^2}{4} $$
$$ \left( y - \frac{x}{2} \right)^2 - \frac{x^2}{4} = y^2 - xy $$
$$ 2 \left( y - \frac{x}{2} \right)^2 - \frac{x^2}{2} = 2 y^2 - 2 xy $$
$$ \frac{x^2}{2}-2 \left( y - \frac{x}{2} \right)^2 = 2 xy- 2 y^2 $$
|
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|
Inequality $(x-1)(y-1)(z-1)\geq 8$ provided that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ How can I prove that if $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1,$$ then $(x-1)(y-1)(z-1) \geq 8$?
Edit:
$x,y,z \in \mathbb R_{>0} $
Thanks
|
We have $x\left(\frac1x+\frac1y+\frac1z\right)=x\implies x-1=\frac xy+\frac xz\overset{AM-GM}{≥}\frac{2x}{\sqrt{yz}}$ and similarly for $y$ and $z$. So we have:
$$
(x-1)(y-1)(z-1)\geq \frac{2x}{\sqrt{yz}}\frac{2y}{\sqrt{zx}}\frac{2z}{\sqrt{xy}}=8
$$
|
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|
Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$ Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$
My attempt:
$\lim_{x\to0}(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}=e^2$
So the required limit is in $\frac{0}{0}$ form.
Then i used L hospital form.
$\lim_{x\to0}\frac{e^2(\frac{2}{x\cos 2x}-\frac{1}{x^2}\log(\frac{1+\tan x}{1-\tan x}))}{2x}$
I am stuck here.
|
Note that $$\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x}$$ and therefore
\begin{align}
L &= \lim_{x \to 0}\dfrac{\left(\tan\left(\dfrac{\pi}{4} + x\right)\right)^{1/x} - e^{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\left(\dfrac{1 + \tan x}{1 - \tan x}\right)^{1/x} - e^{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right)\right) - e^{2}}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2\right) - 1}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2\right) - 1}{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}\cdot\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\
&= e^{2}\lim_{t \to 0}\dfrac{\exp(t) - 1}{t}\cdot\lim_{x \to 0}\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2x}{x^{3}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x + 2\tan x - 2x}{x^{3}}\notag\\
&= e^{2}\left(\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x}{x^{3}} + 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}}\right)\notag\\
&= e^{2}\left(\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x}{\tan^{3}x}\cdot\frac{\tan^{3}x}{x^{3}} + 2\lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}}\right)\text{ (via LHR)}\notag\\
&= e^{2}\left(\lim_{u \to 0}\dfrac{\log\left(\dfrac{1 + u}{1 - u}\right) - 2u}{u^{3}}\cdot 1 + \frac{2}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\right)\text{ (putting }u = \tan x)\notag\\
&= e^{2}\left(\lim_{u \to 0}\dfrac{2\left(u + \dfrac{u^{3}}{3} + o(u^{3})\right) - 2u}{u^{3}}\cdot 1 + \frac{2}{3}\right)\notag\\
&= e^{2}\left(\frac{2}{3} + \frac{2}{3}\right)\notag\\
&= \frac{4e^{2}}{3}\notag
\end{align}
The simplifications done above are obvious and are aimed to simplify the use of series expansions and L'Hospital Rule. Also note that the variable $$t = \frac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2 = \dfrac{\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right)}{\dfrac{2\tan x}{1 - \tan x}}\cdot\dfrac{\tan x}{x}\cdot\frac{2}{1 - \tan x} - 2$$ tends to $1\cdot1\cdot 2 - 2 = 0$ as $x \to 0$ and this fact has been used in the solution provided.
As a rule I prefer to use series expansions only when they are available via memory (like binomial, exponential, logarithmic and sine/cosine series) and try to use algebraic simplification so as to limit the use of series expansion to these simple series. If I need to do complicated stuff with series (like multiplying/dividing series or finding series of composite functions) I show the evaluation of series coefficients.
|
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|
Does the method of substitution always work for solving linear congruence systems? By substitution, I mean if I had this example:
$$ x \equiv 1 \pmod{5}$$ $$x \equiv 2 \pmod{6}$$
Then I would solve it by solving this equation...
$$ x = 1 + 5k \equiv 2 \pmod{6}$$
$$k \equiv 4 \pmod{6}$$
From this, I can substitute k back in for x...
$$x = 1 + 5(4 + 6m)$$
$$x = 21 + 30m$$
Meaning that...
$$x \equiv 21 \pmod{30}$$ (systems with more congruences would be solved by continuing this same process...)
It appears to work even if the moduli are not coprime, unlike one method of the Chinese Remainder Theorem. However, does it work for all systems? If so, please include a mathematical proof/explanation of why.
If not, please include a counterexample showing that it doesn't work.
|
DEFINITION $1.$ Let's say that the system of linear congruences
\begin{align}
x &\equiv a \pmod A \\
x &\equiv b \pmod B
\end{align}
is consistent if $a \equiv b \mod{\gcd(A,B)}$.
THEOREM $2.$ The system of linear congruences
\begin{align}
x &\equiv a \pmod A \\
x &\equiv b \pmod B
\end{align}
has a solution if and only if they are consistent.
PROOF.
Suppose that the system of congruences has a solution, $x$.
Let $G = \gcd(A,B)$. Then $A = mG$ for some integer $m$. So
\begin{align}
x \equiv a \pmod A
&\implies x = a + kA & \text{for some } k \in \mathbb Z\\
&\implies x = a + kmG \\
&\implies x \equiv a \pmod G
\end{align}
So $x \equiv a \pmod A \implies x \equiv a \pmod G$.
Similarly, $x \equiv b \pmod B \implies x \equiv b \pmod G$.
Hence we must have $a \equiv b \pmod{G}$ and we have shown that the system of linear congruences is consistent.
Now suppose that the system of linear congruences is not consistent.
Then $a \not\equiv b \pmod{G}$ Then we can't have $x \equiv a \pmod G$ and $x \equiv b \pmod G$. so the system on linear congruences cannnot have a solution.
COROLLARY $3.$ The family of linear congruences
$\{ x \equiv a_i \pmod A_i\}_{i=1}^N$ has a solution if and only if every pair of linear congruences is consistent.
LEMMA $4.$ Let $\gcd(A, B) = G$. Then
$\left( \dfrac AG \right)^{-1} \left(\bmod \dfrac BG \right)$ exists.
PROOF.
Note that $\dfrac AG$ and $\dfrac BG$ are integers because $G$ divides both $A$ and $B$.
Since $\gcd(A, B) = G$, there must exists integers $u$ and $v$ such that
\begin{align}
Au + Bv = G
&\implies \dfrac AGu + \dfrac BGv = 1 \\
&\implies \dfrac AGu = 1 - \dfrac BGv \\
&\implies \dfrac AGu \equiv 1 \left(\bmod \dfrac BG \right) \\
&\implies \left( \dfrac AG \right)^{-1}
\equiv u \left(\bmod \dfrac BG \right) \\
\end{align}
THEOREM $5.$ If the system of linear congruences
\begin{align}
x &\equiv a \pmod A \\
x &\equiv b \pmod B
\end{align}
is consistent, then the substitution method works.
PROOF.
Suppose that the system of linear congruences is consistent. Again, let
$G = \gcd(A,B)$. Then we must have $a \equiv b \pmod G$. So $G = k(a-b)$ for some integer $k$. Note that $\dfrac AG$ and $\dfrac BG$ are both integers.
Then if $x \equiv a \pmod A$, then $x = a +mA$ for some integer $m$.
Using the substitution method, we get
\begin{align}
x \equiv b \pmod B
&\implies a + mA \equiv b \pmod B \\
&\implies mA \equiv b - a \pmod B \\
&\implies mA \equiv kG \pmod B \\
&\implies mA = Bu + kG & \text{for some integer } u \\
&\implies m\dfrac AG = \dfrac BGu + k \\
&\implies m\dfrac AG \equiv k \left(\bmod \dfrac BG \right) \\
&\implies m \equiv
k \left( \dfrac AG \right)^{-1}
\left(\bmod \dfrac BG \right) \\
&\implies m = u + \dfrac BGv
&\text{for some } u,v\in \mathbb Z \\
\end{align}
So there must exists integers $u$ and $v$ such that
\begin{align}
x &= a +mA \\
&= a +(u + \dfrac BGv)A \\
&= (a + uA) + \dfrac{AB}Gv \\
&= (a + uA) + \operatorname{lcm}(A,B)v \\
\end{align}
Note that we have just proved that $x \equiv b \pmod B$ and
$ x= (a + uA) + \operatorname{lcm}(A,B)v$ implies that
$x \equiv a \pmod A$. Hence the substitution method works.
Note that we have also proved this.
COROLLARY $6.$ If the system of linear congruences
\begin{align}
x &\equiv a \pmod A \\
x &\equiv b \pmod B
\end{align}
is consistent then the substitution method will produce an answer modulo $\operatorname{lcm}(A,B)$.
Using simple induction, we can finally prove this.
THEOREM $7.$ If the family of linear congruences
$\{ x \equiv a_i \pmod A_i\}_{i=1}^N$ is pairwise consistent, then the substitution method will produce an answer modulo
$\operatorname{lcm}\{A_i\}_{i=1}^N$.
|
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|
Prove that a positive integer is composite Let $ n $ be a positive integer that can be written as a sum of two relatively prime squares in two distinct ways, that is $ n = a^2 + b^2 = c^2 + d^2 $ so that $ gcd(a, b) = gcd(c, d) = 1, $ then $ n $ is composite.
So I have successfully factored $ n $ into $ \displaystyle \frac{ac + bd}{a - d}.\frac{ac - bd}{a + d} $ and have proved that $ \displaystyle (a - d) | (ac + bd) $ and $ \displaystyle (a + d) | (ac - bd). $ So each of the $ \displaystyle \frac{ac + bd}{a - d} $ and $ \displaystyle \frac{ac - bd}{a + d} $ is an integer, but I am currently stuck on proving that $ n $ is composite. My intention is to prove that both factors of $ n $ are greater than $ 1, $ but no success at this point. Any hint or suggestion? I have tried to use the fact that $ gcd(a, b) = gcd(c, d) = 1, $ but still no progress.
Thanks.
|
What you ask is equivalent to showing that if $n$ is not composite then $n$ is the sum of two co-prime squares in just one way, or in no way.
The cases $n=1$ is trivial.
If $n$ is prime, suppose $n=a^2+b^2=c^2+d^2.$ with each of $a,b,c,d$ positive and less than $n.$ We will take all congruences modulo $n.$
We have $b\equiv a i$ and $c \equiv \pm d i$ where $i\in Z$ and $i^2\equiv -1.$
Case (1): $d\equiv i c .$ Then $$n^2=(a^2+b^2)(c^2+d^2)=(a c+b d)^2+(a d-b c)^2$$ and $$a c+b d\equiv a c +i^2 a c\equiv 0\equiv a c i-a ic\equiv a d- bc.$$ So $a c+b d=n x$ and $a d-b c=n y$ with $x,y\in Z.$ We have now $$n^2=(x^2+y^2)n^2$$ while $x=a c +b d>0,$ so $x=1$ and $y=0.$ Therefore $$0=y n=a d-bc.$$ With $z=a/b,$ we have also $z=c/d,$ and $$(1+z^2)b^2 =a^2+b^2=n=c^2+d^2=(1+z^2)d^2$$ so $b=d.$ And then (of course) $a=c.$
Case (2): $d\equiv - i c.$ We have $$p^2=(a^2+b^2)(c^2+d^2)=(a c-b d)^2+(a d+b c)^2,$$ and similarly to the methods of Case (1), we obtain $a=d$ and $b=c$. So if $n$ is not composite then $n$ is the sum of two squares in at most one way.
|
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Divergent sum of factorials Is it possible to get an exact value of the sum (using divergent series summation methods)
$$ \sum_{n=0}^\infty~ \frac{(n+k)!}{n!} \quad?$$
where $k$ is a positive integer.
The only other divergent sum of factorials I have seen is $\sum_{n=0}^\infty(-1)^nn!$.
Does anyone know any useful techniques or references?
|
We have
$$
\sum_{n=0}^\infty \frac{(n+k)!}{n!}x^n=\frac{d^k}{dx^k}\sum_{n=0}^\infty x^{n+k}=\frac{d^k}{dx^k}\frac{x^k}{1-x}=\frac{k!}{(1-x)^{k+1}}
$$
for $|x|<1$ and may use the elementary Ramanujan summation (definition) (or simply linearity) and obtain your result, which corresponds to $k!(\sum_{n=0}^\infty 1)^{k+1}$
as a Cauchy product of series (here and here) (not $k!(-1/2)^{k+1}$).
Example $k=0$. We have
$$
\sum_{n=0}^\infty x^n=\frac{1}{1-x}=\frac{x}{1-x}+1~.
$$
As $x/(1-x)=x+x^2+x^3+\cdots$ for $|x|<1$, corresponding to $1+1+1+\cdots=-1/2=\zeta(0)=\sum_{n=1}^\infty 1$, we obtain $\sum_{n=0}^\infty 1=1-1/2=1/2$.
Example $k=1$. We have
$$
\sum_{n=0}^\infty (n+1)x^n=\sum_{n=0}^\infty nx^n + \sum_{n=0}^\infty x^n=\frac{x}{(1-x)^2}+\frac{1}{1-x}=\frac{1}{(1-x)^2}~.
$$
As $x/(1-x)^2=x+2x^2+3x^3+\cdots$ for $|x|<1$, corresponding to $1+2+3+\cdots=-1/12=\zeta(-1)=\sum_{n=1}^\infty n$, we obtain
$\sum_{n=0}^\infty (n+1)=-1/12+1/2=5/12$.
Example $k=2$. We have
$$
\sum_{n=0}^\infty (n+1)(n+2)x^n=\frac{x+x^2}{(1-x)^3}+\frac{3x}{(1-x)^2}+ \frac{2}{1-x}=\frac{2}{(1-x)^3}~.
$$
As $(x+x^2)/(1-x)^3=x+4x^2+9x^3+\cdots$ for $|x|<1$, corresponding to $1+4+9+\cdots=0=\zeta(-2)=\sum_{n=1}^\infty n^2$, we obtain $\sum_{n=0}^\infty (n+1)(n+2)=0+3\times(-1/12)+2\times(1/2)=3/4$.
The elementary Ramanujan summation of a series is consistent with the analytic continuation of Dirichlet series (here, here and here).
For the example $k=2$, using the venerable method of analytic continuation of Dirichlet series to sum the series $\sum_{n=0}^\infty (n+1)(n+2)=2+\sum_{n=1}^\infty (n^2+3n+2)$, corresponding to
$$
F(s)=2+\sum_{n=1}^\infty(n^2+3n+2)n^{-s}=2+\zeta(s-2)+3\zeta(s-1)+2\zeta(s)~,
$$
we obtain $F(0)=2+\zeta(-2)+3\zeta(-1)+2\zeta(0)=2+0+3\times(-1/12)+2\times(-1/2)=3/4$, in agreement with the elementary Ramanujan summation.
It is well-known that, in general, the sum of a divergent series and the sum of the shifted series are different (here and here).
|
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|
Integration Using Inverse Trig Functions
$$\int \frac{1}{(36-4x^2)^{0.5}} dx$$
You can immediately see that,
$$6^2 = 36$$
And,
$$(2x)^2 = 4x^2$$
Using the integration technique as follows:
$$\int \frac{1}{(a^2-x^2)^{0.5}} dx = \arcsin(\frac{x}{a})+C$$
We get,
$$\int \frac{1}{(36-4x^2)^{0.5}} dx = \int \frac{1}{(6^2-(2x)^2)^{0.5}} dx$$
Where $a = 6$ and $x = 2x$. Substituting this into the above general rule, we get
$$\int \frac{1}{(6^2-(2x)^2)^{0.5}} dx = \arcsin(\frac{2x}{6})+C$$
$$=\arcsin(\frac{x}{3})+C$$
However, using this same procedure when we split the denominator gives us a different result:
$$\int \frac{1}{(36-4x^2)^{0.5}} dx = \int \frac{1}{((4)(9-x^2))^{0.5}} dx$$
$$=\int \frac{1}{(4)^{0.5}(9-x^2)^{0.5}} dx$$
$$=\int \frac{1}{2(9-x^2)^{0.5}} dx$$
$$=\frac{1}{2}\int \frac{1}{(9-x^2)^{0.5}} dx$$
Here, $a = 3$ and $x = x$ to give:
$$ = \frac{1}{2}\arcsin(\frac{x}{3})+C$$
Why are both results different? Through differentiation I know that the second result is correct - why does a direct substitution into the general formula yield an incorrect answer?
|
Here's the issue. The general form is $$\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin\left(\frac{x}{a}\right)+C,$$
but your integral is
$$\int\frac{1}{\sqrt{6^2-(2x)^2}}\,dx.$$
The $2x$ is different than $x$. The difference can be eliminated by making the substitution $2x=u$, so $dx=du/2$, and we get
$$\frac{1}{2}\int\frac{1}{\sqrt{6^2-u^2}}\,du=\frac{1}{2}\arcsin\left(\frac{u}{6}\right)+C=\frac{1}{2}\arcsin\left(\frac{x}{3}\right)+C.$$
|
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|
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors in the numerator and denominator:
$$
\left\{\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}{1\cdot 3\cdot 5\cdot 7\cdot 9 \cdot ...\cdot (2n-1)}\right\}
$$
Second, I tried looking at elements of the sequence with common factors removed:
$$
1, \frac{2}{3}, \frac{2}{5}, \frac{2\cdot 4}{5\cdot 7}, \frac{2\cdot 4}{7\cdot 9}, ...
$$
Third, I tried looking at the elements again as fractions without simplifications:
$$
\frac{1}{1}, \frac{2}{3}, \frac{6}{15}, \frac{24}{105}, \frac{120}{945}, ...
$$
Last, I tried searching for similar questions on Stack Exchange and I found one for $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ but I didn't understand how that might apply to my question. So, any hints would be much appreciated.
|
As far as monotonicity is concerned, write the sequence recursively:
\begin{align}
a_1 =&\ 1 \\
a_{n} =&\ a_1\frac{n}{2n + 1}
\end{align}
Then look at the difference between $a_{n + 1}$ and $a_{n}$:
\begin{align}
a_{n + 1} - a_{n} =&\ a_n\frac{n+1}{2(n + 1) + 1} - a_n\\
=&\ a_n\left(\frac{n+1}{2(n + 1) + 1} - 1\right) \\
=&\ a_n\left(\frac{n+1}{2n + 3} - 1\right) \\
=&\ a_n\frac{n + 1 - 2n - 3}{2n + 3} \\
=&\ a_n\frac{-n - 2}{2n + 3}\\
=&\ -a_n\frac{n + 2}{2n + 3}
\end{align}
Clearly this sequence monotonically decreases since the difference between each consecutive element is negative.
As far as convergence goes, one can certainly see that, in the limit that $n \rightarrow \infty$ that this difference will tend towards $\Delta a = -a_n\frac{1}{2}$. The negative is showing that the next term is one half smaller than the previous. For the sequence to converge, $\Delta a$ must tend towards $0$--the only possible way for this to happen is if $a_n$ eventually tends towards $0$. Thus if the sequence converges, it converges to $0$. And I put that in italics because I have not shown that the sequence converges--it seems to me that this is possible through the fact that it's positive definite and monotonically decreases (since it cannot monotonically decrease without crossing zero and not converge--but that's more of an intuitive argument).
|
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|
$P$ is a point on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ and $S$ and $S'$ are its focii If $P$ is a point on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ and $S$ and $S'$ are its focii.
$\angle PSS'=\alpha$ and $\angle PS'S=\beta$, then prove that:
$$
\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)=\frac{1-e}{1+e}
$$
I am trying to do it by taking $P$ as $(0,b)$ but not able to derive required expression. Any suggestion?
|
Another answer is, let the angle $\theta=\angle POS'$, the following identity is straightforward:
$$
\tan\left(\frac{180-\beta}{2}\right) = \sqrt{\frac{1+e}{1-e}} \tan\left(\frac{\theta}{2}\right)
$$
and
$$
\tan\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-e}{1+e}} \tan\left(\frac{\theta}{2}\right)
$$
Hence,
$$
\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)= \sqrt{\frac{1-e}{1+e}} \tan\left(\frac{\theta}{2}\right) \sqrt{\frac{1-e}{1+e}} \cot\left(\frac{\theta}{2}\right) = \frac{1-e}{1+e}
$$
|
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|
Parametrize a curve, Multivariable Calculus I am stuck on what seems to be an easy exercise.
We have $f(x,y) = x^2 + 4xy + y^2 \mbox{ for all } (x,y)$ in $\mathbb{R}^2.$
Now we are supposed to find a parametrization of the intersection curve between $f(x,y)$ and $z = x + 3y.$
I've been stuck for hours now, anyone got any ideas/tips?
Thank you!
|
Setting 'z= z' gives $x^2+ 4xy+ y^2= x+ 3y$. The first thing I would do is "complete the square" on the left: $x^2+ 4xy+ 4y^2- 3y^2= (x+ 2y)- 3y^2= x+ 3y$. Now, let u= x+ y and v= y then x+ 3y= u- v+ 3v= u+ 2v so the equation becomes $u^2- 3v^2= u+ 2v$ or $u^2- u- 3(v^2+ (2/3)v)= 0$ and complete the square again: $u^2- u+ 1/4- 3(v^2+ (2/3)v+ 1/9)= (u- 1/2)^2- 3(v+ 1/3)^2= 1/4+ 1/3= 7/12$. Let p= u- 1/2 and q= v+ 1/3 so the equation becomes $p^2- 3q^2= 7/12$ or $\frac{p^2}{\frac{12}{7}}- \frac{q^2}{\frac{4}{7}}= 1$
|
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|
Which rules are used to make function like one in Laplace Transformations table? I have function like this:
$$\frac{s^2+3s+3}{(2s^2+7s+7)} $$
It needs to be brought to the level of Laplace Transformations from table, like these two:
$$\frac{a}{(s-b)^2 + a^2} $$
$$\frac{s-b}{(s-b)^2 + a^2} $$
I have the solution for this problem, but I can't figure out the logic behind it. Steps included are:
$$\frac{1}{2} - \frac{0.5s+0.5}{2s^2+7s+7} = \frac{1}{2} - \frac{0.5s+0.5}{(s+1.75)^2 + 0.6614^2} = \frac{1}{2} - 0.5*\frac{s+1.75}{(s+1.75)^2 + 0.6614^2} + (0.5*1.75-0.5)*\frac{1}{0.6614}*\frac{0.6614}{(s+1.75)^2+0.6614^2}$$
|
$$\begin{align}
\frac{s^2+3s+3}{(2s^2+7s+7)}&=\frac12 \frac{2s^2+6s+6}{(2s^2+7s+7)}\\\\
&=\frac12 \frac{(2s^2+7s+7)-(s+1)}{(2s^2+7s+7)}\\\\
&=\frac12 -\frac12\frac{s+1}{2s^2+7s+7}\\\\
&=\frac12-\frac14\frac{s+1}{s^2+(7/2)s+(7/2)}\\\\
&=\frac12-\frac14\frac{s+1}{(s+7/4)^2+(7/2)-(49/16)}\\\\
& =\frac12-\frac14\frac{s+(7/4)-(3/4)}{(s+7/4)^2+7/16}\\\\
&=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+7/16}+\frac{3}{16}\frac{1}{(s+7/4)^2+7/16}\\\\
&=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+7/16}+\frac{3}{16}\frac{1}{(s+7/4)^2+7/16}\\\\
&=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}+\frac{3}{16}\frac{1}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}\\\\
&\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}+\frac{3\sqrt 7}{28}\frac{\frac{\sqrt 7}{4}}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}
\end{align}$$
|
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|
Evaluation of the integral $\int{\frac{x+\sin{x}}{1+\cos{x}}\mathrm{d}x}$ by parts I have to evaluate the following integral by parts: $$\int {\dfrac{x+\sin{x}}{1+\cos{x}}}\mathrm{d}x $$
So I tried to put:
$ u = x + \sin{x}$ $~\qquad\rightarrow \quad$ $\mathrm{d}u=\left(1+\cos{x}\right) \mathrm{d}x$
$\mathrm{d}v = \dfrac{\mathrm{d}x}{1+\cos(x)}$ $\quad \rightarrow \quad$ $v = \int{\dfrac{\mathrm{d}x}{1+\cos{x}}}$
But there is an extra integral to do ( the $v$ function) I evaluated it by sibstitution, and I get $v = \tan{\dfrac{x}{2}}$, Now
$$\int {\dfrac{x+\sin(x)}{1+\cos(x)}}\mathrm{d}x = (x+\sin{x})\, \tan{\dfrac{x}{2}}-x +C$$
My question: is it possible to evaluate this integral entirely by parts (without using any substitution) ?
I appreciate any ideas
|
$\int \dfrac{x + \sin x}{1+\cos x}dx\\
\int \dfrac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} dx\\
\int \frac{1}{2}x\sec^2\frac{x}{2} + \tan \frac{x}{2} dx\\
\int \frac{1}{2}x\sec^2\frac{x}{2}dx + \int\tan \frac{x}{2} dx$
Now we do integration by parts on the $1^{st}$ integral. we won't evaluate the $2^{nd}$ one quite yet.
$u = x, dv = \sec^2 \frac{x}{2} dx\\ du = dx, v = 2 \tan \frac{x}{2}$
$x\tan\frac{x}{2} - \int \tan\frac{x}{2}+\int\tan\frac{x}{2} \\x\tan\frac{x}{2} + C$
|
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|
Probability of selecting specific balls from an urn? An urn contains seven red balls, seven white balls, and seven blue balls. A sample of 5 balls is drawn at a random without replacement. What is the probability that the sample contains three balls of one color and two of another?
Here was what I tried doing.
I said that the total number of ways to choose 5 balls without any restrictions is $\binom{21}{5}$. This is because there are $21$ balls and $5$ positions that you need to choose.
Next I thought that the number of ways to select the $3$ balls of one color would be $\binom{7}{3}$ since there are $7$ balls of one color. Since there are 3 different colors I also thought you would multiply $\binom{7}{3}$ by 3.I also thought since the remaining two balls must be of a different color there are $2!$ ways to permute the remaining balls.
So my final answer was
$$Probability = \frac{3*2!*\binom{7}{3}}{\binom{21}{5}}$$
Only thing is I'm not sure If my answer is right. One problem that I had dealing with was if the two remaining balls are of the same color then I'm not sure multiplying by $2!$ is right but I'm not really sure how to deal with these cases.
==================================================================
Comment: I was wondering if instead of having the same number of balls lets say you had 7 red balls, 8 white balls, and 9 blue balls. Then how would the formula for probability be affected.
|
Your denominator is correct. For the numerator, we must select one of the three colors from which three balls will be selected, select three of the seven balls of that color, select one of the two remaining colors from which two balls will be selected, and select two balls of that color. Hence, the probability of selecting three balls of one color and two balls of another color when five balls are selected is
$$\frac{\dbinom{3}{1}\dbinom{7}{3}\dbinom{2}{1}\dbinom{7}{2}}{\dbinom{21}{5}}$$
Addendum: If you instead have seven red balls, eight white balls, and nine blue balls, you have six cases. They are
*
*three red, two white $\binom{7}{3}\binom{8}{2}$
*three red, two blue $\binom{7}{3}\binom{9}{2}$
*two red, three white $\binom{7}{2}\binom{8}{3}$
*two red, three blue $\binom{7}{2}\binom{9}{3}$
*three white, two blue $\binom{8}{3}\binom{9}{2}$
*two white, three blue $\binom{8}{2}\binom{9}{3}$
Hence, the probability of selecting three balls of one color and two balls of a different color when five balls are selected from the $7 + 8 + 9 = 24$ balls in the bag is
$$\frac{\dbinom{7}{3}\dbinom{8}{2} + \dbinom{7}{3}\dbinom{9}{2} + \dbinom{7}{2}\dbinom{8}{3} + \dbinom{7}{2}\dbinom{9}{3} + \dbinom{8}{3}\dbinom{9}{2} + \dbinom{8}{2}\dbinom{9}{3}}{\dbinom{24}{5}}$$
|
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|
$a,b,c$ are the sides and $A,B,C$ are the angles of a triangle. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then,
$a,b,c$ are the sides of a $\triangle ABC$ and $A,B,C$ are the respective angles. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then $\sin^2 \bigl(\frac{A}{2}\bigr), \sin^2 \bigl(\frac{B}{2}\bigr)$ and $\sin^2\bigl(\frac{C}{2}\bigr)$ are in which type of progression?(Arithmetic,geometric,harmonic)
My try,
I immediately noticed that $1$ was a root and since the product of the roots was $1$ I got that $a,b,c$ were in harmonic progression. I replaced the reciprocals of $a,b,c$ by $p,q,r$. Wrote $\sin^2\bigl({\theta\over 2}\bigr)$ as $\frac{1-cos(\theta)}{2}$ and replaced $\cos$ by the sides but did not succeed. I also tried by replacing it with $\frac{(s-b)(s-c)}{bc}$ but still could not get the required answer.
If I assume the result I am getting the answer by simplifying the expression.
Please give an elegant solution.
|
This is not an answer but a hint and comment about your assertion you have already got that a,b,c are in H.P. Here a possible way to prove what you want with the squares of the sinus of middle angles.
►Considering the incenter you have
$$\sin^2 (\frac{A}{2})= \frac {r^2}{d_1^2}\left(= \frac{1}{\frac{d_1^2}{r^2}}\right)\\ \sin^2 \bigl(\frac{B}{2}\bigr)=\frac{r^2}{d_2^2}\left(= \frac{1}{\frac{d_2^2}{r^2}}\right)\\\sin^2(\frac{C}{2})=\frac{r^2}{d_3^2}\left(= \frac{1}{\frac{d_3^2}{r^2}}\right)$$
where $r$ is the inradius and $d_1,d_2,d_3$ the distances from the incenter to the vertices $A,B,C$ respectively.
Assuming convenient magnitudes of angles, we can deduce (supposing true the H.P. relation)
$$\frac{d_2^2-d_1^2}{r^2}=\frac{d_3^2-d_2^2}{r^2}\iff 2d_2^2=d_1^2+d_3^2$$
Hence $$2d_2^2=d_1^2+d_3^2\iff \frac{2}{\sin^2 (\frac{B}{2})}=\frac{1}{\sin^2 (\frac{A}{2})}+\frac{1}{\sin^2 (\frac{C}{2})}\iff \text{the three squares of sinus are in H.P.} $$
It follows an alternative to try to prove your statement.
►REMARK.- In an equilateral triangle you can verify the asked property (since $2a^2=a^2+a^2$ and besides $\frac 12=\frac{2}{2+2}$) but in this case the given quadratic equation does not make sense.
|
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|
$x_{k+1}=(x_k+c/x_k)/2$. Prove that {$x_k$} converges and find its limit. Fix any $c>0$.
Let $x_1$ be any positive number and define $x_{k+1}=(x_k+c/x_k)/2$.
a)Prove that {$x_k$} converges and find its limit.
b)Use this sequence to calculate $\sqrt5$, accurate to six decimal places.
I tried to solve $x_{k+1}=(x_k+c/x_k)/2$ in terms of $x_1$ but it all went wrong. How can I solve it when there are two terms of $x_k$ in an equation?
Plus, any hint to question b)?
|
Notice for all $k \ge 1$,
$$\frac{x_{k+1}-\sqrt{c}}{x_{k+1}+\sqrt{c}}
= \frac{\frac12(x_k + \frac{c}{x_k}) - \sqrt{c}}{\frac12(x_k+\frac{c}{x_k}) + \sqrt{c}}
= \left(\frac{x_k-\sqrt{c}}{x_k+\sqrt{c}}\right)^2
$$
Using induction, it is easy to see for all $n \ge 1$, we have
$$\frac{x_n - \sqrt{c}}{x_n + \sqrt{c}} = \alpha^{2^{n-1}}
\iff x_n = \sqrt{c}\left(\frac{1 + \alpha^{2^{n-1}}}{1 - \alpha^{2^{n-1}}}\right)
\quad\text{ where }\quad \alpha = \frac{x_1-\sqrt{c}}{x_1+\sqrt{c}}$$
When $x_1 > 0$, we have $|\alpha| < 1$ and hence $\alpha^{2^{n-1}}$ converges to $0$ as $n \to \infty$.
As a result, $x_n$ converges to $\sqrt{c}\left(\frac{1+0}{1-0}\right) = \sqrt{c}$.
For the special case $c = 5$, we know $\sqrt{5} \approx 2.236$.
If we start from $x_1 = 2$, we have
$$\alpha = \frac{2-\sqrt{5}}{2 + \sqrt{5}} \approx -0.0557$$
Since $\alpha$ is relatively small, even for $n = 1$, we have
$$|x_n - \sqrt{c} | = 2\sqrt{c}\left|
\frac{
\alpha^{2^{n-1}}
}{
1 - \alpha^{2^{n-1}}}
\right|\approx 2\sqrt{c}|\alpha|^{2^{n-1}}$$
For $n = 4$, the RHS is about $4\times 10^{-10}$. This implies
$x_4 = \frac{51841}{23184} \approx 2.23606797791580$
is an approximation of $\sqrt{5} \approx 2.23606797749979$
accurate to around $9$ decimal places.
|
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|
Laurent expansion Question is to write Laurent expansion of $f(z)=\dfrac{1}{(z-2)(z-1)}$ in the annulus $1<|z|<2$ based at $z=0$
I am aware of the method of partial fractions and writing expansions for both $\dfrac{1}{z-1}$ and $\dfrac{1}{z-2}$..
I am rying to do this using Laurent expansion formula..
We have $$f(z)=\sum_{n=-\infty}^{\infty}a_n z^n$$ where $$a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z^{n+1}}dz$$
for any $1<r<2$.
Easiest case is when $n=-1$ then it is just $$a_{-1}=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-1)(z-2)}dz=\frac{1}{2\pi i}\left(\int_{|z|=r}\frac{1}{z-2}-\int_{|z|=r}\frac{1}{z-1}\right)$$
Now, as $r<2$ we see that $\frac{1}{z-2}$ is analytic inside the circle of radius $r$.. so first integral is zero and we are left with second integral..
$$a_{-1}=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-1)(z-2)}dz=-\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{z-1}$$
I got stuck after this...
|
I think i got it...
For $1<r<2$, we have to compute $$a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-2)(z-1)z^{n+1}}dz$$
For $n>0$ we have $$a_{-n}=\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-2)(z-1)}dz
=\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-2)}-\frac{z^{n-1}}{(z-1)}dz$$
As $r<2$, the function $\frac{z^{n-1}}{(z-2)}$ is analytic in the disk $\{|z|\leq r\}$ so first integral is zero. We then have
$$a_{-n}=\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-2)(z-1)}dz
=-\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-1)}dz$$
The function $z^{n-1}$ is analytic so, by cauchy integral formula, we have
$$a_{-n}
=-\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-1)}dz=-z^{n-1}|_{z=1}=-1$$
So, for all $n>0$ we have $a_{-n}=-1$.
For $n>0$ we compute
$$a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-2)(z-1)z^{n+1}}dz$$
By induction, i could see that
$$\frac{1}{(z-2)(z-1)z^{n+1}}=\frac{1}{2^{n+1}}\frac{1}{z-2}-\frac{1}{2^{n+1}}\frac{1}{z}-\cdots-\frac{1}{2}\frac{1}{z^{n+1}}-\frac{1}{z-1}+\frac{1}{z}+\cdots+\frac{1}{z^{n+1}}$$
As $\frac{1}{z-2}$ is analytic in the disk $\{|z|\leq r\}$ the integral is zero. Then, by all integrals are zero except $\frac{1}{z}$ and $\frac{1}{z-1}$.. By cauchy integral formula, $\frac{1}{z-1}$ integral is $1$.. We then have
$$a_n=
\frac{1}{2\pi i}\int_{|z|=r}\left[\frac{1}{2^{n+1}}\frac{-1}{z}-\frac{1}{z-1}+\frac{1}{z}\right]=\frac{-1}{2^{n+1}}-1+1=\frac{-1}{2^{n+1}}$$
So, we have $a_n=\frac{-1}{2^{n+1}}$ for all $n\geq 0$ and $a_{-n}=-1$ for all $n>0$.
We then have $$f(z)=\cdots-\frac{1}{z^2}-\frac{1}{z}-\frac{1}{2}\left(1+\frac{z}{2}+\frac{z^2}{2^2}+\cdots\right)$$
|
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|
Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$
Find all integer solutions $(x, y)$ of the equation
$$\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$$
What have done is that:
$$\frac{1}{x}= \frac{2y-3}{3y}$$
so,
$$x=\frac{3y}{2y-3}$$
If $2y-3 = +1 \text{ or } {-1}$, $x$ will be an integer, so we choose integer $y$
to make $2y-3=1 \text{ or } {-1}$. $y = 2$ or $y = 1$ is such a solution.
Also, $2y - 3$ can be deleted by numerator $3$, so $2y - 3$ can be $3$ or $-3$ too. This gives $y = 3$ or $y = 0$, but $y$ can not be $0$. So far, we have $y=1,2,3$. Finally, $2y-3$ can be deleted by numerator $y$, but how can we find such a $y$?
|
Your approach can be continued to a full solution. Let's see it in detail:
Given integers $x,y$ such that $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$:
At least one of $x,y$ is positive, and by symmetry we can assume that $y$ is positive.
$\frac{1}{x} = \frac{2y-3}{3y}$.
Thus $x = \frac{3y}{2y-3} = 1 + \frac{y+3}{2y-3}$.
If $y > 6$ then $y+3 < 2y-3$ and so $0 < \frac{y+3}{2y-3} < 1$, which makes $x$ not an integer.
Thus $y \le 6$.
It only remains to check all $y$ from $1$ to $6$.
|
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|
Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-(4n-2)^2)$$
$$=2(4)+2(6)+2(12)+2(14)+2(20)+2(22)+\cdots+2(8n-4)+2(8n-2)$$
$$=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]$$
How should I proceed further?
|
In the last line what you have enclosed into rectangular braces, namely $S$, can be rearranged as follows
\begin{align*}
S&=(1+3)+(2+4)+(5+7)+(6+8)+(9+11)+(10+12)+\cdots+[(4n-3)+(4n-1)]+[(4n-2)+4n]\\
&=\sum_{j=1}^{4n}j\\[3pt]
&=\frac{4n(4n+1)}{2}\\[3pt]
&=\boxed{\color{blue}{2n(4n+1)}}
\end{align*}
|
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|
Checking the boundedness of $A_n = \frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \ldots + \frac{x^{2^n}}{1-x^{2^{n+1}}}$ Sequence is given by $$A_n = \frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \ldots + \frac{x^{2^n}}{1-x^{2^{n+1}}}.$$ Please advise me how to show that this sequence is bounded above by 1. 0
|
\begin{align*}
\frac{x^{2^{n}}}{1-x^{2^{n+1}}}+\color{red}{\frac{1}{1-x^{2^{n+1}}}} &=
\frac{1}{1-x^{2^{n}}} \\
\frac{x^{2^{n-1}}}{1-x^{2^{n}}}+\frac{1}{1-x^{2^{n}}} &=
\frac{1}{1-x^{2^{n-1}}} \\
& \: \; \vdots \\
\frac{x}{1-x^{2}}+\frac{1}{1-x^{2}} &=
\color{blue}{\frac{1}{1-x}} \\
A_{n} &= \color{blue}{\frac{1}{1-x}}-
\color{red}{\frac{1}{1-x^{2^{n+1}}}} \\
\lim_{n\to \infty} A_{n} &=
\left \{
\begin{array}{ccc}
\displaystyle \frac{x}{1-x} & \text{for} & -1<x<1 \\
\displaystyle \frac{1}{1-x} & \text{for} & x^2 > 1 \\
\end{array}
\right.
\end{align*}
|
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|
Normal closure of $\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$ The following is a question from an undergrad course in Galois theory:
Find a normal closure $L$ of $K=\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$
I know that normal extensions are splitting fields
Let: $X=\sqrt{11+3\sqrt{13}} \implies X^2=11+3\sqrt{13} \implies X^2-11=3\sqrt{13}\implies ({\frac{X^2-11}{3}})^2-13=0 $
Is this related to the splitting field?
Would the normal closure look something like: $\mathbb{Q}(\sqrt{11}, \sqrt{13})$ since $\sqrt{11}, \sqrt{13} \notin \mathbb{Q}$? I am guessing not since we have the weird embedded square root
I have not really got my head aruond these questions so would very much appreciate your guidance
|
You need to compute the zeros of $X^4-22 X^2+4=9\cdot ((\frac{X^2-11}3)^2-13)$. There is a general formula for equations of degree four which gives you
\begin{align*}
\sqrt{11+3\sqrt{13}} && -\sqrt{11+3\sqrt{13}} && \sqrt{11-3\sqrt{13}} && -\sqrt{11+3\sqrt{13}}
\end{align*}
So the normal closure of $K$ is $\Bbb Q(\sqrt{11+3\sqrt{13}}, \sqrt{11-3\sqrt{13}})$.
Since
$$
\sqrt{11+3\sqrt{13}}\cdot \sqrt{11-3\sqrt{13}}=\sqrt{11^2-9\cdot 13}=\sqrt{4}=2,
$$
you can see that
$$
\sqrt{11-3\sqrt{13}}=\frac{2}{\sqrt{11+3\sqrt{13}}}\in K
$$
and it follows that $K$ is normal already.
|
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|
Convergence of a compound sequence let $a_n=\frac{1}{2}\sqrt{n}+\sum_{k=1}^n(\sqrt{k}-\sqrt{k+\frac{1}{2}})$ be a sequence.
Is this sequence convergent?
|
Write $\sqrt{n} = \sum_{k=1}^{n} \left(\sqrt{k}-\sqrt{k-1}\right)$. Then you get:
$$a_n = \sum_{k=1}^{n}\left(\sqrt{k}-\sqrt{k+1/2}+\frac{1}{2}\sqrt{k}-\frac{1}{2}\sqrt{k-1}\right)=\sum_{k=1}^n \left(\frac 3 2\sqrt{k}-\sqrt{k+1/2}-\frac{1}{2}\sqrt{k-1}\right)$$
This represents $a_n$ as a pure series, with $a_n=\sum_{k=1}^{n} b_k$ with $b_k=
\frac 3 2\sqrt{k}-\sqrt{k+1/2}-\frac{1}{2}\sqrt{k-1}$.
Now $$\sqrt{k+1/2}=\sqrt{k}{\sqrt{1+1/(2k)}} = \sqrt{k}+\frac{1}{4\sqrt{k}} + O\left(\frac{1}{k^{3/2}}\right)$$ and similarly $$\sqrt{k-1}=\sqrt{k}-\frac{1}{2\sqrt{k}}+O\left(\frac{1}{k^{3/2}}\right)$$
So you get
$$b_k=\frac 3 2\sqrt{k}-\sqrt{k+1/2}-\frac{1}{2}\sqrt{k-1}=O\left(\frac1{k^{3/2}}\right)$$
So $\sum_{k=1}^{\infty} b_k$ converges, and you are done.
Wolfram alpha gives an exact value of:
$$\frac{(\sqrt2-4) \zeta(3/2)+4 \pi\sqrt 2}{8 \pi}$$
|
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|
Solve by Mathematical Induction: $2^n+1 \leq 3^n$ I'm very confused on how to prove this.
By using Mathematical Induction prove that, for all positive integers $n$ the
following inequality holds:
$$2^n + 1 ≤ 3^n$$
|
For $n = 1$,
$$3^1 = 3 \ge 2^1 + 1$$
Assume that $2^k + 1 \le 3^k$ for some $k\in \mathbb N$.
For $n = k+1$,
$$\begin{align*}
3^{k+1} &= 3\cdot 3^k\\
&\ge 3\cdot(2^k+1)\\
&= 3\cdot 2^k + 3\\
&= 2\cdot2^k + 1 + 2^k + 2\\
&\ge 2^{k+1} + 1
\end{align*}$$
|
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|
Calculating exponential limit I've been breaking my mind over this one.
Find the limit.
$\lim\limits_{n \to \infty} (\frac{n^2+3}{n^2+5 n-4})^{2n} $
I know it equals $\frac{1}{e^{10}} $ but can't figure out how to find it.
Help?
|
Remember that $$ \left(\frac{n^2+3}{n^2+5 n-4}\right)^{2n}=\exp\left(2n \ln\left(\frac{n^2+3}{n^2+5 n-4}\right)\right).$$ To compute the limit of the inside, use L'Hospital's rule :
\begin{align*}
\lim_{n \to +\infty} 2n \ln\left(\frac{n^2+3}{n^2+5 n-4}\right)& =\lim_{n\to +\infty}\frac{\ln\left(\frac{n^2+3}{n^2+5 n-4}\right)}{\frac{1}{2n}}\\
& =\lim_{n\to +\infty}-\frac{2n^2(2n(n^2+5n-4)-(n^2+3)(2n+5))}{(n^2+3)(n^2+5n-4)} \\
& = -10
\end{align*}Indeed the numerator and the denominator are both polynomial of degree $4$ so juste compare the coefficients of $n^4$.
|
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|
How many ordered pairs $(x,y)$ are there such that ${1\over\sqrt x}+{1\over\sqrt y}={1\over\sqrt {20}}$, where both $x$ and $y$ are positive integers $${1\over \sqrt x}+{1\over \sqrt y}={1\over \sqrt {20}}$$
I could find one ordered pair that satisfies above equation, that is $(80,80)$. But the answer says that there are $3$ ordered pairs satisfying above equation.
|
We have the following equation:
$$\frac{1}{\sqrt x}+\frac{1}{\sqrt y}=\frac{1}{\sqrt {20}}$$
$\sqrt{20}=2\sqrt{5}$:
$$\frac{1}{\sqrt x}+\frac{1}{\sqrt y}=\frac{1}{2\sqrt 5}$$
Multiply both sides by $\sqrt 5$:
$$\frac 1 {\sqrt{\frac x 5}}+\frac 1 {\sqrt{\frac y 5}}=\frac 1 2$$
Basically, we want answers to $\frac 1 a+\frac 1 b=\frac 1 2$ where $a=\sqrt{\frac x 5}$ (meaning $x=5a^2$) and $b=\sqrt{\frac y 5}$ (meaning $y=5b^2$). In order to show $a$ and $b$ have to be integers, look at the beginning of @MichaelBurr's answer. Thus, we can manipulate this equation to get the following:
$$a=\frac{2b}{b-2}$$
Thus, $(b-2) \mid (2b)$. However, $b \equiv 2 \pmod{b-2}$, so $2b \equiv 4 \pmod{b-2}$. This means $(b-2) \mid 4$. Therefore, we have the following possibilities:
*
*$b-2=1 \implies a=6, b=3 \implies x=180, y=45$
*$b-2=2 \implies a=b=4 \implies x=y=80$
*$b-2=4 \implies a=3, b=6 \implies x=45, y=180$
|
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|
Study the convergence of the following series $\sum\limits_{n=1}^\infty \frac{n}{n^2+3} \sin(\frac{1}{\sqrt{n+2}}) $ I have to study the convergence of the following series:
$\sum\limits_{n=1}^\infty \frac{n}{n^2+3} \sin(\frac{1}{\sqrt{n+2}}) $
Is a positive series, so I should divide for $\frac{1}{\sqrt{n+2}}$
and then use the comparison test ?
Many thanks
|
With equivalents, as it's a series with positive terms:
\begin{align*}\frac n{n^2+3}&\sim_\infty\frac1n,\quad \sin\frac1{\sqrt{n+2}}\sim_\infty\frac1{\sqrt{n+2}}\sim_\infty\frac1{\sqrt n}, \\\text{hence}\quad\frac n{n^2+3}\sin\frac1{\sqrt{n+2}}&\sim_\infty\frac 1{n^{3/2}},\quad\text{which converges.}
\end{align*}
|
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|
Whats the probability I roll a 2 on one of the dice if the sum rolled is 8? I have two dice and I want to find the probability of this situation:
Whats the probability I roll a 2 on one of the dice if the sum rolled is 8 ?
Is the answer $\frac{2}{36}$?
|
Consider it intuitively
We can draw a table to show this:
$$\begin{array}{|cc|cccccc|}\hline
&&&&&D_1\\
&&1&2&3&4&5&6\\
\hline
&1&2&3&4&5&6&7\\
&2&3&4&5&6&7&\color{red}8\\
D_2&3&4&5&6&7&\color{red}8&9\\
&4&5&6&7&\color{red}8&9&10\\
&5&6&7&\color{red}8&9&10&11\\
&6&7&\color{blue}8&9&10&11&12\\
\hline\end{array}$$
The numbers in red show where the sum is equal to $8$ and the blue number shows where one of the dice shows a $2$
Therefore we can see that we have a probability of $\dfrac15=\dfrac{\text{sum}=8\text{ and one dice showing }2}{\text{sum}=8}$
Now think mathematically
We can see that
\begin{align}\frac15&=\frac{\frac 1{36}}{\frac5{36}}\\\\
&=\frac1{36}\div \frac{5}{36}\\
&= P(\text{sum}=8\text{ and one dice showing }2)\div P(\text{sum}=8)
\end{align}
|
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|
Find a functional equation for the generating function whose coefficients satisfy the relation Find a functional equation for the generating function whose coefficients satisfy the relation:
$\qquad{}$ $a_n = 3a_{n-1} -2a_{n-2}+2, a_0=a_1=1$
When I solve this, I get the function $g(x)-1-x=3xg(x)-2x^2g(x)$
However, the correct answer is supposed to be $g(x)-1-x = 3xg(x)-3-2x^2g(x)+\frac{2x^2}{1-x}$
Where does the $-3$ and the $\frac{2x^2}{1-x}$ come from?
|
Mostly you forgot to take into account the $+2$ term in the recurrence, though there is one other problem. For $n\ge2$ you have
$$a_nx^n=3a_{n-1}x^n-2a_{n-2}x^n+2x^n\;,$$
so when you sum over $n\ge 2$ you get
$$\sum_{n\ge 2}a_nx^n=3\sum_{n\ge 2}a_{n-1}x^n-2\sum_{n\ge 2}a_{n-2}x^n+2\sum_{n\ge 2}x^n\;.$$
The lefthand side is $g(x)-1-x$, so
$$\begin{align*}
g(x)-1-x&=3x\sum_{n\ge 2}a_{n-1}x^{n-1}-2x^2\sum_{n\ge 2}a_{n-2}x^{n-2}+\frac{2x^2}{1-x}\\
&=3x\sum_{n\ge 1}a_nx^n-2x^2\sum_{n\ge 0}a_nx^n+\frac{2x^2}{1-x}\\
&=3x\big(g(x)-1\big)-2x^2g(x)+\frac{2x^2}{1-x}\\
&=3xg(x)-3x-2x^2g(x)+\frac{2x^2}{1-x}\;.
\end{align*}$$
Note that it’s $-3x$, not $-3$; either you or your answer key seems to have a typo.
|
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|
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$.
$4\cos x - 3 \sin x = k\sin(x - \alpha)$
=> $k(\sin x \cos \alpha - \cos x \sin \alpha)$
=> $k\cos \alpha \sin x - \sin \alpha \cos x$
Equating coefficients:
$k\cos \alpha = - 3$
$k\sin \alpha = 4$
$k = \sqrt{(-3)^2 + 4^2} = 5$
$\alpha$ is in the fourth quarter because $ \cos \alpha$ is positive and $\sin \alpha$ is negative.
$\alpha = \arctan\frac{4}{-3} = -53.1$
$\alpha = 360 - 53.1 = 306.9$
$4\cos x - 3 \sin x = 5\sin(x - 306.9)$
But the answer is $4\cos x - 3 \sin x = 5\sin(x - 233.1)$
|
HINT:
$k\sin\alpha=-4,k\cos\alpha=-3\implies\alpha$ is in the third quadrant
See this
|
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|
Connection between quadratic residue of a number to its factors' Is it true that, If $N$ is product of two coprime numbers greater than 1. Quadratic residues of these numbers are quadratic residue of $N$ and vice versa? Can someone point me to a proof or show me if this is not the case.
In other words, if we start with $n = pq$, where $p, q$ are coprimes. If and only if $x^2 \equiv a \mod n$, then $a \equiv x^2 \mod p$ and $a \equiv x^2 \mod q$
|
We state a precise version of the result you are after.
Let $m$ and $n$ be relatively prime. Then $a$ is a quadratic residue of $mn$ if and only if $a$ is a quadratic residue of $m$ and $a$ is a quadratic residue of $n$.
One direction is easy. Suppose that $a$ is a quadratic residue of $mn$. Then there exists a integer $x$ such that $x^2\equiv a\pmod{mn}$. It follows that $x^2\equiv a\pmod{m}$ and $x^2\equiv a\pmod{n}$, and therefore $a$ is a quadratic residue of $m$ and a quadratic residue of $n$.
The converse is a little more complicated. Suppose that $a$ is a quadratic residue of $m$ and a quadratic residue of $n$. Then there is a $b$ such that $b^2\equiv a\pmod{n}$, and a $c$ such that $c^2\equiv a\pmod{n}$.
By the Chinese Remainder Theorem, there is an $x$ such that $x\equiv b\pmod{m}$ and $x\equiv c\pmod{n}$.
We have $x^2\equiv b^2\equiv a\pmod{m}$ and $x^2\equiv c^2\equiv a\pmod{n}$. It follows that $x^2\equiv a\pmod{mn}$, and therefore $a$ is a quadratic residue of $mn$.
|
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|
Find out all solutions of the congruence $x^2 \equiv 9 \mod 256$. I need to find all the solutions of the congruence $x^2 \equiv 9 \mod 256$.
I tried (apparently naively) to do this:
$x^2 \equiv 9 \mod 256$
$\Leftrightarrow$ $x^2 -9 \equiv 0 \mod 256$ $\Leftrightarrow$ $256 | (x-3)(x+3)$ $\Leftrightarrow$
$256|(x-3)$ or $256|(x+3)$ $\Leftrightarrow$ $x \equiv 3, -3 \mod 256$
but I found out that it's not fully correct because there are other solutions that I didn't succeed to find such as $x = 125$, so I want to know what is the general way to solve such questions? thank you!
|
If $(x-3)(x+3)\equiv0\pmod{2^{m+2}}$ for integer $m\ge1$
both $(x-3),(x+3)$ must be even
$\implies\dfrac{x-3}2\cdot\dfrac{x+3}2\equiv0\pmod{2^m}$
Now $\dfrac{x-3}2-\dfrac{x+3}2$ is odd $\implies$ they have opposite parity
If $\dfrac{x-3}2$ is even, $\dfrac{x-3}2\equiv0\pmod{2^m},x\equiv3\pmod{2^{m+1}}$
Can you take it from here?
Here $m+2=8$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I go about solving this? I have tried substitution, but it is not working for me.
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+1)}+\sin(x)+n\cos(x)}=\int_0^\pi \frac{n dx}{\sqrt{(n^2+1)}+n\sin(x)+\cos(x)}=2
$$
General form of this integral is
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m}
$$
|
First of all:
$$
(\sqrt{n^2+1}+\sin x+n\cos x)(\sqrt{n^2+1}-\sin x-n\cos x)=$$
$$
=n^2+1-n^2\cos^2x-2n\cos x\sin x-\sin^2 x=
$$
$$
=\cos^2x-2n\cos x\sin x+n^2\sin^2 x=(n\sin x-\cos x)^2
$$
So:
$$
\int\frac{\mathrm{d}x}{\sqrt{n^2+1}+\sin x+n\cos x}=\int\frac{(\sqrt{n^2+1}-\sin x-n\cos x)\mathrm{d}x}{(n\sin x-\cos x)^2}
$$
Now use the formula:
$$
\theta=\arctan\frac{b}{a}\qquad a\cos x+b\sin x=\sqrt{a^2+b^2}{\cos(x+\theta)}
$$
To finish this up.
Result:
$$
\int\frac{\mathrm{d}x}{\sqrt{n^2+1}+\sin x+n\cos x}=\frac{1-\sqrt{n^2+1} \sin (x)}{n \sin (x)-\cos (x)}$$
$$\int\frac{n\mathrm{d}x}{\sqrt{n^2+1}+n\sin x+\cos x}=\frac{\sqrt{n^2+1} \sin (x)-n}{n \cos (x)-\sin (x)}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Sum of the series $\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}+....$ If $|x|<1$, find the sum of infinite terms of following series:
$$\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}+....$$
Could someone give me hint to solve this problem. I wrote $r_{th}$ term of the series but can't find sum of $n$ terms. If somehow I can get sum of $n$ term then I can put $n \to \infty $ to get sum of infinite terms.
|
$\frac{1}{(1-x)(1-x^3)}=\frac{1}{x-x^3}(\frac{1}{1-x}-\frac{1}{1-x^3})$
$\frac{x^2}{(1-x)(1-x^5)}=\frac{1}{x-x^3}(\frac{1}{1-x^3}-\frac{1}{1-x^5})$
...
So, assuming n starts from 1, $\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}...=\frac{1}{x-x^3}(\frac{1}{1-x}-\frac{1}{1-x^{2n+1}})$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Definite integral of a positive continuous function equals zero? Let's calculate $$\int_0^{\frac\pi 2} \frac {dx}{\sin^6x + \cos^6x}$$
We have
$$\int \frac {dx}{\sin^6x + \cos^6x} = \int \frac {dx}{1 - \frac 34 \sin^2{2x}}$$
now we substitute $u = \tan 2x$, and get
$$\int \frac {dx}{1 - \frac 34 \sin^2{2x}} = \frac{1}{2} \int \frac {du}{1 + \frac 14 u^2} = \tan^{-1}\frac u2 + C= \tan^{-1}\left(\frac12 \tan 2x\right) + C = F(x)$$
Now, evaluating the primitive function at $x = 0, x = \frac{\pi}2$, we get
$$\int_0^{\frac\pi 2} \frac {dx}{\sin^6x + \cos^6x} = F\left(\frac {\pi}2\right) - F(0) = 0 - 0 = 0$$
But the integrand is positive and continuous, so the integral should be positive!! What have I done wrong?
|
One thing I can see: in your substitution $\;u=\tan 2x\;$ on $\;[0,\pi/2]\;$ , you get
$$0\le x\le\frac\pi2\implies 0\le 2x\le\pi\implies u=\tan2x$$
is not defined on $\;[0,\pi]\;$ , which renders the substitution incorrect.
|
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|
Is there a solution to this differential equation? I am trying to find a function $y(x)$ that is a solution to
$$
\left(a_3 x^3+a_1 x\right) y''(x)-\left(3 a_3 x^2+2 a_1\right) y'(x)+3 a_3\, x \,y(x)=a_0 x^4+a_2
$$
I tried using mathematica but it ran for hours without giving a solution.
Thank you.
|
Notice the LHS of the ODE can be rewritten in operator form:
$$\begin{align}
LHS
&= \left[x\frac{d}{dx} - 2\right]\left[(a_1 + a_3 x^2)\frac{d}{dx} - 3a_3 x\right]y\\
&= \left[ x^3 \frac{d}{dx} x^{-2}\right]\left[ (a_1 + a_3 x^2)^{5/2} \frac{d}{dx}
\frac{1}{(a_1 + a_3 x^2)^{3/2}} \right] y
\end{align}
$$
Unwinding the leftmost differential operator, we find for suitable chosen integration constant $A$, we have
$$\left[ (a_1 + a_3 x^2)^{5/2} \frac{d}{dx}
\frac{1}{(a_1 + a_3 x^2)^{3/2}} \right] y
= x^2 \int^x \frac{a_0 t^4 + a_2}{t^3} dt
= \frac{a_0 x^4 + A x^2 - a_2}{2}
$$
Unwinding the remaining differential operator, we can express $y(x)$ as an integral
$$y(x) = (a_1+a_3 x^2)^{3/2}\left[\frac{y(0)}{a_1^{3/2}} + \frac12 \int_0^x
\frac{a_0 t^4 + A t^2 - a_2}{(a_1 + a_3 t^2)^{5/2}} dt\right]
$$
With help of an CAS, we can evaluate the integral and get
$$\begin{align}
y(x) &=
(a_1+a_3 x^2)^{3/2}\left[\frac{y(0)}{a_1^{3/2}} + \frac{a_0}{2a_3^{5/2}}\sinh^{-1}\left(\sqrt{\frac{a_3}{a_1}}x\right)\right]\\
& \quad - \frac16\left[
\frac{a_0 x (3a_1 + 4 a_3 x^2)}{a_3^2}
-\frac{A x^3}{a_1}
+\frac{a_2 x (3a_1 + 2 a_3 x^2)}{a_1^2}
\right]
\end{align}
$$
|
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|
How many values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
How many different values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
I was wondering about this problem and didn't think it was immediately obvious. The answer can't be $2^{n-1}$ since the combinations all might not be unique. I was thinking of looking at partitions of $\{1,2,\ldots,n, k\}$ where $1 \pm 2 \pm 3 \pm \cdots \pm n = k$ and thus the sum must be even. Therefore, $\dfrac{n(n+1)}{2}+k = \dfrac{n(n+1)+2k}{2}$ must be even which means $n(n+1) + 2k$ must be a multiple of $4$ so $n(n+1)+2k = 4m$. Thus if $n \equiv 1,2 \bmod 4$, then $k \equiv 1,3 \bmod 4$ and if $n \equiv 3,0 \bmod 4$ then $k \equiv 0,2 \bmod 4$.
|
To start with, you can forget about the leading term, $1$.
All that term does is to shift all the values one place to the right
on the number line.
The expression $\pm 2 \pm 3 \pm \cdots \pm n$
has the same number of distinct values as $1 \pm 2 \pm 3 \pm \cdots \pm n$.
From this point until further notice, I'll consider only sums of
the form $\pm 2 \pm 3 \pm \cdots \pm n$.
All possible sums have the same parity.
The minimum sum is $\sum_{i=2}^n(-i) = -\frac{(n+2)(n-1)}{2}$,
when all the $\pm$ signs
are negative. The sum $-\frac{(n+2)(n-1)}{2} + 2$
is clearly not achievable, but $-\frac{(n+2)(n-1)}{2} + 4$ is,
as long as $n \geq 2$.
Now as long as the sum includes two terms $j - (j+1)$
(positive followed by negative), we can change the two terms to
$-j + (j+1)$ and thereby increase the sum by $2$.
If there are no such terms, that is, all the negative terms occur
before any of the positive terms, the sum has the form
$$-2 - 3 - \ldots - (j-1) - j + (j+1) + \ldots + n = k \tag1$$
where $1 \leq j \leq n$. (In the case $j = 1$, all terms are positive.)
We have already considered the case $j = n$.
In the case $j=1$, the sum has the maximum value,
$\frac{(n+2)(n-1)}{2}$,
and in the case $j=2$, the sum has the second greatest possible value,
$\frac{(n+2)(n-1)}{2} - 4$.
Now suppose instead that $3 \leq j \leq n-1$.
Then we can change the sum on line $(1)$ to
$$2 - 3 - \ldots - (j-1) + j - (j+1) + \ldots + n = k + 2.$$
In other words, if $n > 3$, then except for the values
$-\frac{(n+2)(n-1)}{2}$,
$\frac{(n+2)(n-1)}{2} - 4$, and $\frac{(n+2)(n-1)}{2}$,
every possible value of the sum
(including $-\frac{(n+2)(n-1)}{2} + 4$)
corresponds to another sum that is greater by $2$.
So the possible sums are $-\frac{(n+2)(n-1)}{2}$, $\frac{(n+2)(n-1)}{2}$,
and every integer of the same parity from $-\frac{(n+2)(n-1)}{2} + 4$
to $\frac{(n+2)(n-1)}{2} - 4$, inclusive.
The number of integers in
$\left[-\frac{(n+2)(n-1)}{2}, \frac{(n+2)(n-1)}{2}\right]$
is $(n+2)(n-1) + 1$, and those of the same parity are
$\frac{(n+2)(n-1)}{2} + 1$.
So if $n > 3$, the total number of possible sums (excluding the
values $-\frac{(n+2)(n-1)}{2}+2$ and $\frac{(n+2)(n-1)}{2}-2$,
which are not possible sums) is
$$\left(\frac{(n+2)(n-1)}{2} + 1\right) - 2 = \frac12(n^2 + n - 4).$$
For example, if $n=4$ then there are $8$ possible sums:
$-9,-5,-3,-1,1,3,5$, and $9$.
By inspection, this formula also gives the number of sums for $n=3$
(the sums are $-5,-1,1,$ and $5$),
but the number of sums for $n=2$ is $2$, which does not agree with
the formula; $n=2$ must be treated as a special case.
If you modify the original expression by making the first term $\pm1$,
that is, if you consider $\pm 1 \pm 2 \pm \cdots \pm n$,
the minimum is $-\frac{n(n+1)}{2}$,
the maximum is $\frac{n(n+1)}{2}$,
and all integers of the same parity between those values are
possible sums.
|
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|
Prove an identity in a Combinatorics method It is a combinatorics proof. Anyone has any idea on how to prove
$$\sum \limits_{i=0}^{l} \sum\limits_{j=0}^i (-1)^j {m-i\choose m-l} {n \choose j}{m-n \choose i-j} = 2^l {m-n \choose l}\;$$
We need to prove this equation holds for all $l$.
I know that $\sum {n \choose j}{m-n \choose i-j}$ equals to ${m \choose i}$ but has no idea if there is a $(-1)^j$, it seems like a PIE but actually not....
Could anyone help me move forward in the process?
|
Suppose we seek to verify that
$$\sum_{p=0}^l\sum_{q=0}^p (-1)^q
{m-p\choose m-l} {n\choose q} {m-n\choose p-q}
= 2^l {m-n\choose l}$$
where $m\ge n$ and $m-n\ge l.$
This is
$$\sum_{p=0}^l {m-p\choose m-l} \sum_{q=0}^p (-1)^q
{n\choose q} {m-n\choose p-q}.$$
Now introduce the integral
$${m-n\choose p-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p-q+1}} (1+z)^{m-n} \; dz.$$
Note that this vanishes when $q\gt p$ so we may extend the range of
$q$ to infinity, getting for the sum
$$\sum_{p=0}^l {m-p\choose m-l}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} (1+z)^{m-n}
\sum_{q\ge 0} (-1)^q
{n\choose q} z^q
\; dz
\\ = \sum_{p=0}^l {m-p\choose l-p}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} (1+z)^{m-n}
(1-z)^n
\; dz.$$
Introduce furthermore
$${m-p\choose l-p} =
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{l-p+1}} (1+w)^{m-p} \; dw.$$
This too vanishes when $p\gt l$ so we may extend $p$ to infinity, getting
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{l+1}} (1+w)^{m}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z} (1+z)^{m-n}
(1-z)^n
\sum_{p\ge 0} \frac{w^p}{z^p} \frac{1}{(1+w)^p}
\; dz
\; dw.$$
The geometric series converges when $|w/z/(1+w)|\lt 1.$ We get
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{l+1}} (1+w)^{m}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z} (1+z)^{m-n}
(1-z)^n
\frac{1}{1-w/z/(1+w)}
\; dz
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{l+1}} (1+w)^{m}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{m-n}
(1-z)^n
\frac{1}{z-w/(1+w)}
\; dz
\; dw.$$
Now from the convergence we have $|w/(1+w)|<|z|$ which means the pole
at $z=w/(1+w)$ is inside the contour $|z|=\epsilon.$ Extracting the
residue yields (the pole at zero has disappeared)
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{l+1}} (1+w)^{m}
\left(1+\frac{w}{1+w}\right)^{m-n}
\left(1-\frac{w}{1+w}\right)^n
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{l+1}}
(1+2w)^{m-n}
\; dw
\\ = 2^l {m-n\choose l}.$$
|
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|
Carrying out a substituting to evaluate $\int (x + 1) (x^2 + 2 x)^5dx$ The problem is:
$$ \int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx $$
The next step given by WolframAlpha is
$$\int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx\\ =\quad \frac { 1 }{ 2 } \int { { u }^{ 5 }du } $$
(While I realize I am doing somthing worng)
The steps I am taking are : $ \int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx$.
Let $u={ x }^{ 2 }+2x, \, du = 2x+2\, dx$. Then I see I can rewrite $du$ as $du = 2(x+1)dx$ giving me $\frac { du }{ 2 } = (x+1)\, dx$.
I can now see where the $ \frac { 1 }{ 2 } $ is coming from, but I cannot seem to visualize the next steps to get to
$$ \int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx=\frac { 1 }{ 2 } \int { { u }^{ 5 }du } $$
|
Since $(x^2 + 2 x)^5 = u^5$, we can use the equation $(x + 1) \,dx = \frac{1}{2} du$ to write the integral as
$$\int \underbrace{(x^2 + 2 x)^5}_{u^5} \,\underbrace{(x + 1) \,dx}_{\frac{1}{2} du} = \int u^5 \cdot \frac{1}{2} du = \frac{1}{2} \int u^5 \,du .$$
|
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|
Find the interval of convergence $\sum\limits_{n=1}^\infty \frac{(3n)!(x^n)}{(n!)^3}$ $\displaystyle \sum_{n=1}^\infty \dfrac{(3n)!(x^n)}{(n!)^3}$
I already found the interval of convergence to be $\displaystyle -\frac{1}{27} < x < \frac{1}{27}$. I am having trouble checking the endpoint of $\displaystyle x= \frac{1}{27}$. I need to figure out of $\displaystyle \sum_{n=1}^\infty \dfrac{(3n)!\left(\frac{1}{27}\right)^n}{(n!)^3}$ converges or diverges. Any help?
|
Observe that
\begin{align}
\sum_{n=1}^\infty\frac{(3n)!\left(\frac{1}{27}\right)^n}{(n!)^3}
&=\sum_{n=1}^\infty\frac{(3n)!}{\left[3^n(n!)\right]^3}\\
&=\sum_{n=1}^\infty\frac{(3n)!}{\left[3\cdot6\cdot9\cdots3n\right]^3}\\
&=\sum_{n=1}^\infty\left[\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{6^2}\cdots\frac{(3n-2)(3n-1)}{(3n)^2}\right].
\end{align}
By using the limit comparison test with the divergent
series $\displaystyle\sum_{n=1}^\infty\frac{1}{n}$, we see that
\begin{align}
&\lim_{n\to\infty}\frac{\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{6^2}\cdots\frac{(3n-2)(3n-1)}{(3n)^2}}{\frac{1}{n}}\\
&\qquad=\lim_{n\to\infty}\left[\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{6^2}\cdots
\frac{(3n-5)(3n-4)}{(3n-3)^2}\cdot\frac{(3n-2)(3n-1)}{9n}\right]\\
&\qquad=\lim_{n\to\infty}\left[\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{3\cdot6}\cdots
\frac{(3n-5)(3n-4)}{(3n-6)(3n-3)}\cdot\frac{(3n-2)(3n-1)}{(3n-3)3n}\right]\\
&\qquad\ge \frac{1\cdot2}{3^2}=\frac{2}{9}.
\end{align}
Hence $\displaystyle\sum_{n=1}^\infty\frac{(3n)!\left(\frac{1}{27}\right)^n}{(n!)^3}$ diverges.
|
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|
Evaluate the following limit: Find
$$\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left[\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right]$$
MY TRY:
$$
\begin{align}
\lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}}
+ \cdots + \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr] \\
&= \lim_{n \to \infty} \frac{\sqrt{n}}{n} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}} + \cdots
+ \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr]
\end{align}
$$
Now using Cauchy first thm of limits
$$
a_n = \frac{\sqrt{n}}{\sqrt{2n}+\sqrt{2n+1}}
$$
The answer should be $\frac{1}{2\sqrt{2}}$.
But the answer is $1/\sqrt{2}$.
|
This is a classic partial sums of integrals problem. Your expression is:
$$\frac{1}{\sqrt{n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$$
write as
$$\frac{1}{n}\sum\limits_{k=1}^n \frac{1}{\sqrt{2\frac{k}{n}}+\sqrt{2\frac{k}{n}+2}}$$
then the limit is the same as the integral
$$\int_0^1\frac{1}{\sqrt{2x}+\sqrt{2x+2}}dx=
\int_0^1\frac{\sqrt{2x}-\sqrt{2x+2}}{-2}dx
$$
|
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|
Solve the recurrence $a_n=3a_{n/3}+2$ given $a_0=1$ and $n$ is a power of $3$ Solve the recurrence $$a_n=3a_{n/3}+2$$ given $a_0=1$ and $n$ is a power of $3$
I am trying to study for my final using my previous quizzes, of which I got this question wrong. My instructor wants me to use the Divide and Conquer technique. I imagine I would at least start with the base cases where $n$ is a power of $3$:$$3^0\rightarrow a_0=1$$$$3^1\rightarrow a_1=3(0)+2=2$$$$3^2\rightarrow a_3=3(2)+2=8$$$$3^3\rightarrow a_9=3(8)+2=26$$The next step I have is solve for $a_n$. Stop when $\frac{n}{3^k}=1$. That implies $n=3^k$ and we substitute $k=log_3n$ times:$$a_n=3a_{n/3}+2$$$$=3(3a_{n/3^2}+2)+2=3^2a_{n/3^2}+6+2$$$$=3^2(3a_{n/3^3}+2)+6+2=3^3a_{n/3^3}+18+6+2$$$$=\cdots $$$$=3^ka_{n/3^k}+2(3^{k-1}+3^{k-2}+\cdots +3+1)$$Here is where I get lost, however. In the answer key, my instructor has this last term equating to:$$=3^ka_{n/3^k}+2\frac{3^k-1}{3-1}$$But I have NO idea how these two expressions equal each other. Can someone point this out to me? I have the steps after that complete solving the recurrence, this step just loses me, and I need to know what's going on with it for my final.
|
Consider:
$$\begin{aligned}
(3-1)(3^{k-1}+3^{k-2}+\cdots +3+1) & =(3^{k}+3^{k-1}+\cdots +3)-(3^{k-1}+3^{k-2}+\cdots +3+1) \\
&=3^k+(3^{k-1}-3^{k-1})+ ...+(3-3)+1\\
&=3^k-1
\end{aligned}$$
rearranging:
$$
3^{k-1}+3^{k-2}+\cdots +3+1=\frac{3^k-1}{3-1}
$$
Which is just the sum of a geometric series formula
|
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Show that $\forall n\in\mathbb{N}$, $14^n$ can be represented as a sum of three perfect squares. Show that $\forall n\in\mathbb{N}$, $14^n$ can be represented as a sum of three perfect squares.
I checked $(\mod 7)$ and deduced that the three squares can be $1,4,2(\mod 7) $ or all divisible by $7$ and for $n \ge 2$ they must be all divisible by $4$. But I am stuck here.
|
You can use induction.
Step 1:
When $n = 1$, we have $14 = 3^2 + 2^2 + 1^2$
When $n = 2$, we have $14^2 = 12^2 + 6^2 + 4^2$
Step 2: Suppose the statment is right for all $1 \le k \le n$.
In particular, $\exists a, b, c > 0$ s.t. $14^{n-1} = a^2 + b^2 + c^2$
Then for $n + 1$,
$14^{n+1} = 14^{n-1+2} = 14^{n-1} \times 14^2 $
$= 14^2\times a^2 + 14^2\times b^2 + 14^2\times c^2$
$= (14 \times a)^2 + (14 \times b)^2 + (14 \times c)^2$
Finally we can say the statement is true.
|
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|
How to solve this double integral involving trig substitution (using tangent function)? This is a question I came across and I cannot find the answer.
By using a substitution involving the tangent function, show that $$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx=\frac{\pi}{4}$$
My attempt
I use trig substitution, by saying
$$\tan(\theta)=\frac{y}{x}$$ which means
$$x\sec^2(\theta)\,d\theta=dy$$
Also, it should be noted that because of this
$$x\sec(\theta)=\sqrt{x^2+y^2}$$
$$x^4\sec^4(\theta)=(x^2+y^2)^2
$$
Thus, when I substitute this information into the integral, I get
$$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2-(x^2\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)} \, d\theta \,dx$$
Then, this simplifies to
$$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2(1-\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx$$
$$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2\sec^2(\theta)}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx=\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})}{\frac{1}{x}}\,d\theta \,dx$$ which leads to
$$\int_0^1 \left[\frac \theta x \right]_0^{\arctan\left(\frac{1}{x}\right)} \,dx = \int_0^1 \frac{\arctan\left(\frac{1}{x}\right)}{x} \, dx_{(3)} $$
At this point I am stuck. How do I evaluate this integral. Am I on the right path? Wolfram Alpha gives an answer other than $\frac{\pi}{4}$ for (3), so I am not sure where I am wrong.
|
Let us start considering $$I=\int\frac{x^2-y^2}{(x^2+y^2)^2}dy$$ Defining $$y=x\tan(\theta)\implies dy=x \sec ^2(\theta )\implies I=\int \frac{\cos (2 \theta )}{x}\,d\theta=\frac{\sin (2 \theta )}{2 x}$$ Back to $x$ $$I=\frac{y}{x^2+y^2}$$ So, $$\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dy=\frac 1 {1+x^2}$$ So, you are left with $$\int_0^1\frac {dx} {1+x^2}$$ Repeat the same change of variable.
|
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How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$. So I've been struggling with this sum for some time and I just can't figure it out. I tried proving by induction that if the sum above is a $S_n$ then $S_{n+1} = 4S_n$, but I didn't really succeed so here I am. Thanks in advance.
|
I can give a mildly ugly proof by induction. Let
$$S_n=\sum_{k=0}^n2^k\binom{2n-k}n=\sum_{k=0}^n2^k\binom{2n-k}{n-k}=\sum_{k=0}^n2^{n-k}\binom{n+k}n\;,$$
and assume that $S_n=4^n=2^{2n}$. Then
$$\begin{align*}
S_{n+1}&=\sum_{k=0}^{n+1}2^{n+1-k}\binom{n+1+k}{n+1}\\
&=\sum_{k=0}^{n+1}2^{n+1-k}\left(\binom{n+k}n+\binom{n+k}{n+1}\right)\\
&=\sum_{k=0}^{n+1}2^{n+1-k}\binom{n+k}n+\sum_{k=0}^{n+1}2^{n+1-k}\binom{n+k}{n+1}\\
&=\binom{2n+1}n+2\sum_{k=0}^n2^{n-k}\binom{n+k}n+\sum_{k=0}^{n+1}2^{n+1-k}\binom{n+k}{n+1}\\
&=\binom{2n+1}n+2^{2n+1}+\sum_{k=1}^{n+1}2^{n+1-k}\binom{n+k}{n+1}\\
&=2^{2n+1}+\binom{2n+1}n+\binom{2n+1}{n+1}+\sum_{k=1}^n2^{n+1-k}\binom{n+k}{n+1}\\
&=2^{2n+1}+\binom{2n+2}{n+1}+\sum_{k=0}^{n-1}2^{n-k}\binom{n+1+k}{n+1}\\
&=2^{2n+1}+\binom{2n+2}{n+1}+\frac12\left(S_{n+1}-\binom{2n+2}{n+1}-2\binom{2n+1}{n+1}\right)\\
&=2^{2n+1}+\binom{2n+2}{n+1}+\frac12\left(S_{n+1}-2\binom{2n+2}{n+1}\right)\\
&=2^{2n+1}+\frac12S_{n+1}\;,
\end{align*}$$
so $\frac12S_{n+1}=2^{2n+1}$, and $S_{n+1}=2^{2n+2}=4^{n+1}$.
|
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Different ways to find limit in infinity I would like to find a limit of the function $f(x)=\frac{x^3+\sqrt{x^6-1}}{4(x-2)^2}$ as $x\rightarrow \infty$ and as $x\rightarrow -\infty$.
In the first case, both numerator and denominator goes to infinity but nominator has higher exponent so $$\lim_{x\rightarrow \infty}\frac{x^3+\sqrt{x^6-1}}{4(x-2)^2}= \infty.$$
In the second case numerator is indefinite $-\infty+\infty$ so I can do "the trick" where I transform the expression by factoring with $x^3-\sqrt{x^6-1}$ and I get
$$\frac{x^3+\sqrt{x^6-1}}{4(x-2)^2}=\frac{x^6-x^6+1}{4(x-2)^2(x^3-\sqrt{x^6-1})}=\frac{1}{4(x-2)^2(x^3-\sqrt{x^6-1})}$$
which tends to $0$ as $x\rightarrow-\infty$. Is this step correct? It seems wrong since I cant think of the reason why I couldn't do the same in the first case and get limit $0$ as $x\rightarrow \infty$ which would, of course, be wrong.
What am I missing?
|
Something I learned when dealing with limits $x \to -\infty$ is that you can do a change in variable.
$$
\lim_{x \to - \infty} \frac{x^3 + \sqrt{x^6 -1}}{4(x-2)^2} \left/ x = -t, \ x \to -\infty \Rightarrow t \to \infty \right/ = \lim_{t \to \infty} \frac{-t^3 + \sqrt{t^6 -1}}{-4(t+2)^2}
$$
Now for big values of $t$ this will behave like
$$
\frac{-t^3 + \sqrt{t^6 -1}}{-4(t+2)^2} \approx \frac{-t^3 + t^3}{-4(t+2)^2} = 0
$$
I hope this illustrates the difference.
|
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About the solution to "Finding the range of $y= \sqrt x + \sqrt{3-x}"$ I was reading the solution of "Find the range of $y = \sqrt{x} + \sqrt{3 -x}$" and I had some points of confusion about the solution posted in the OP.
I wrote here my interpretation of the solution. Could you please let me know if my understanding contains any redundant, incomplete, missing, or erroneous steps?
My understanding of the solution is as follows:
The domain is clearly $[0, 3]$
$\{(x, y):y=\sqrt x + \sqrt {(3-x)} \}=\{(x, y):y^2-3=2\sqrt{x(3-x)}$ and $y \ge 0\}$
Since the RHS in $y^2-3=2\sqrt{x(3-x)}$ is non-negative $\forall x$ in the domain, it is implied that $y^2-3 \ge 0$ $\forall y$ in the solution set. So we can now say that
$\{(x, y):y=\sqrt x + \sqrt {(3-x)} \}=\{(x, y):y^2-3=2\sqrt{x(3-x)}$ and $y \ge \sqrt3\}$
(We do not have to worry about $y \le -\sqrt 3$ because $y \ge 0 \forall x$ in the domain). Now $\sqrt 3$ is just a "boundary" for $y$; we need to check that the $y=\sqrt 3$ is actually a part of the range. Because $2\sqrt{x(3-x)}$ attains the value of $0$ at $x=0$ and $x=3$, $y$ actually attains the value of $\sqrt 3$.Therefore we can establish a minimum of $\sqrt 3$ for the range.
Now, we know that
$\{(x, y):y^2-3=2\sqrt{x(3-x)}$ and $y \ge \sqrt3\}=\{(x, y):4(3x-x^2)=(y^2-3)^2$ and $y \ge \sqrt3\} = \{(x, y): 4x^2-12x+(y^2-3)^2=0$ and $y \ge \sqrt 3 \}$ (We do not need to worry about explicitly restricting $x$ in the set on the set in the middle and on the right, because the interval over which $4(3x-x^2)$ is positive coincides with the domain of the original problem, so we have not introduced/lost any solutions.)
$4x^2-12x+(y^2-3)^2=0$ has real solutions $\iff$ the discriminant is $\ge 0$.
$(-12)^2-4(4)(y^2-3)^2 \ge 0$
$144-16(y-3)^2 \ge 0$
$9 \ge (y^2-3)^2$
Because we already established $y \ge \sqrt 3$, we only need to worry about the positive root
$3 \ge y^2-3$
$6 \ge y^2$
Again, since we established $y \ge \sqrt 3$, we can take
$\sqrt 6 \ge y$
$\sqrt 6$ is only a "boundary" for $y$; we still need to establish that $y=\sqrt 6$ is part of at least one ordered pair in the solution set:
$(\sqrt 6)^2-3 = 2 \sqrt {x(3-x)}$
$ 3 = 2\sqrt {x(3-x)}$
This gives the solution $\dfrac 32$, implying that $y=\sqrt 6$ is contained in the range.
We established a minimum value of $\sqrt 3$ and a maximum value of $\sqrt 6$, both on the interval $[0, 3]$/ Because the original function is continuous on $[0, 3]$, we can safely conclude that the range is $[\sqrt 3, \sqrt 6].$
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(too long for a comment)
I will have to look at the continuity later; but on the issue of checking that $\sqrt 3$ and $\sqrt 6$; is it necessary?
I think checking that both $\sqrt 3$ and $\sqrt 6$ are in the range is necessary in your "solution" because you use continuity.
My reasoning is that these values are in a sense "necessary but not sufficient", meaning that we know the range must be somewhere between $\sqrt 3$ and $\sqrt 6$, but we don't know it's precisely $[\sqrt 3,\sqrt 6]$
I understand what you mean, and I found nothing wrong in your "solution".
By the way, what I was trying to say in the linked solution is not the same as your understanding, and I think it's worth showing with details my proof without using continuity (hence no need to check "the boundary") :
First of all, to say that the range of $y=\sqrt x+\sqrt{3-x}$ is $\sqrt 3\le y\le\sqrt 6$, we have to prove the following two things :
(1) (necessity) If there exists a pair of real numbers $(x,y)$ such that $y=\sqrt x+\sqrt{3-x}$, then $\sqrt 3\le y\le \sqrt 6$.
(2) (sufficiency) For every $y$ satisfying $\sqrt 3\le y\le\sqrt 6$, there exists at least one real $x$ such that $y=\sqrt x+\sqrt{3-x}$.
To make it easy to understand the proof, let us prove $(1)$ and $(2)$ separately.
Proof for (1) (necessity) :
We consider $$y=\sqrt x+\sqrt{3-x}\tag3$$
First of all, we have to have $$0\le x\le 3\tag4$$
If $y\lt 0$, then the LHS of $(3)$ is negative, and the RHS of $(3)$ is non-negative, so there is no $(x,y)$ satisfying $(3)$. So, we have to have
$$y\ge 0\tag5$$
Under $(4)(5)$, we know that $(3)$ is equivalent to
$$y^2=(\sqrt x+\sqrt{3-x})^2,$$
i.e.
$$y^2-3=2\sqrt{x(3-x)}\tag6$$
Here, if $y^2-3\lt 0$, then the LHS of $(6)$ is negative, and the RHS of $(6)$ is non-negative, so there is no $(x,y)$ satisfying $(6)$. So, we have to have
$$y^2-3\ge 0\tag7$$
Under $(4)(7)$, we know that $(6)$ is equivalent to
$$(y^2-3)^2=\left(2\sqrt{x(3-x)}\right)^2,$$
i.e.
$$4x^2-12x+y^4-6y^2+9=0\tag8$$
We want $(8)$ to have at least one real root $x$, so we have to have that the discriminant is non-negative :
$$(-12)^2-4\cdot 4(y^4-6y^2+9)\ge 0,$$
i.e.
$$-\sqrt 6\le y\le \sqrt 6\tag9$$
As a result, if there exists a pair of $(x,y)$ such that $y=\sqrt x+\sqrt{3-x}$, we have to have $(5)(7)(9)$, i.e. $\sqrt 3\le y\le\sqrt 6.$ $\blacksquare$
Proof for (2) (sufficiency) :
As we've seen in the proof for (1), if $\sqrt 3\le y\le \sqrt 6$, then $(3)$ is equivalent to $(8)$.
So, it is sufficient to prove that if $\sqrt 3\le y\le\sqrt 6$, then there exists at least one real $x$ where $0\le x\le 3$ such that $(8)$.
By the way, if $\sqrt 3\le y\le \sqrt 6$, then we have $(-12)^2-4\cdot 4(y^4-6y^2+9)\ge 0$, and so
$$\begin{align}&0\le \frac{12-\sqrt{(-12)^2-4\cdot 4(y^4-6y^2+9)}}{8}\le 3\tag{10}\\\\&\iff 0\le\frac{3-\sqrt{-y^4+6y^2}}{2}\le 3\\\\&\iff 0\le 3-\sqrt{-y^4+6y^2}\le 6\\\\&\iff -3\le -\sqrt{-y^4+6y^2}\le 3\\\\&\iff -3\le\sqrt{-y^4+6y^2}\le 3\\\\&\iff -y^4+6y^2\le 3^2\\\\&\iff (y^2-3)^2\ge 0\end{align}$$
This holds for every $y$ satisfying $\sqrt 3\le y\le\sqrt 6$. So, $(10)$ holds for every $y$ satisfying $\sqrt 3\le y\le\sqrt 6$. $\blacksquare$
|
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Equation with limit $\lim\limits_{n\to \infty}\sqrt{1+\sqrt{x+\sqrt{x^2+...+\sqrt{x^n}}}}=2$
I had never faced with problems like this. Give me, please, a little hint.
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We are trying to solve for $x$ where $\lim_{n\to\infty}a_n=2$ and $a_n=\sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{\dots x^n}}}}$
The trivial possible solutions are $x=0,1$, which do not work.
For $x$
$$\sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{\dots x^n}}}}=\sqrt{1+\sqrt{x}\sqrt{1+\sqrt{1+\sqrt{\frac1x+\sqrt{\frac1{x^4}+\sqrt{\dots x^n}}}}}}$$
$$<\sqrt{1+\sqrt{x}\sqrt{1+\sqrt{1+\sqrt{\dots+1}}}}\tag{If $x>1$}$$
$$=\sqrt{1+\frac{1+\sqrt{5}}{2}\sqrt{x}}$$
So if we set that equal to $2$, we have
$$x>\left(\frac6{1+\sqrt{5}}\right)^2=\frac{27-9\sqrt5}2$$
|
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|
Rationalizing the denominator of $\frac{\sqrt {2}}{\sqrt {x-3}}$ ok, so im reviewing for a math test and the following question is from the practice final exam.
Rationalize the denominator in the example:
$$\frac{\sqrt {2}}{\sqrt {x-3}}$$
After multiplying both the numeration and denominator by the conjugate of the denominator, I got $$\frac{\sqrt {2x+6}}{x-3}$$
But, in the answer key the answer is $$\frac{\sqrt {2x-6}}{x-3}$$
The problem looks quite simple, but I'm not sure what is the answer.
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$$\frac{\sqrt{2}}{\sqrt{x-3}}=\frac{\sqrt{2}\cdot\sqrt{x-3}}{\sqrt{x-3}\cdot\sqrt{x-3}}=\frac{\sqrt{2\cdot(x-3)}}{\left(\sqrt{x-3}\right)^2}=\frac{\sqrt{2x-6}}{x-3}$$
|
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Proving integration formula I want to prove the integration formula
$$
\int \frac {\sqrt {a+bu}}{u} \ du = 2 \sqrt {a+bu}+a \int { \frac {du}{u \sqrt {a+bu}} }.
$$
I tried trigonometric substitution (as $u= \frac {a \tan^2 \theta}{b}$), but it didn't seem to work. I am stuck on how to prove the integration formula.
|
Let's try another substitution.
Let $x=\sqrt{a+bu}\Rightarrow u=\dfrac{x^2-a}{b} \qquad dx=b/2(a+bu)^{-1/2}du=\dfrac{bdu}{2x}\Rightarrow \dfrac{2xdx}{b}=du$
Now we have
${\Large{\int}} \dfrac{2x^2}{b}\cdot\dfrac{b}{x^2-a}dx=2{\Large{\int}} \dfrac{x^2}{x^2-a}dx=2{\Large{\int}} \dfrac{x^2-(x^2-a)}{x^2-a}+\dfrac{x^2-a}{x^2-a}dx=2{\Large{\int}} \dfrac{a}{x^2-a}+1dx$
$=2x+2{\Large{\int}} \dfrac{a}{a+bu-a}b/2(a+bu)^{-1/2}du=2\sqrt{a+bu}+a{\Large{\int}} \dfrac{du}{u\sqrt{a+bu}}$
|
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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.