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Product of repeated cosec.
$$P = \prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$$
I realize that there must be some sort of trick in this.
$$P = \csc^2(1)\csc^2(3).....\csc^2(89) = \frac{1}{\sin^2(1)\sin^2(3)....\sin^2(89)}$$
I noticed that: $\sin(90 + x) = \cos(x)$ hence,
$$\sin(89) = \cos(-1) = \cos(359)$$
$$\sin(1) = \cos(-89) = \cos(271)$$
$$\cdots$$
$$P \cdot P = \frac{\cos^2(-1)\cos^2(-3)....}{\sin^2(1)\sin^2(3)....\sin^2(89)}$$
But that doesnt help?
|
$$\sin(2n+1)x=(2n+1)\sin x+\cdots+(-1)^n2^{2n}\sin^{2n+1}x$$
If we set $2n+1=45,2^{44}\sin^{45}x-\cdots+45\sin x-\sin45x=0$
If we set $\sin45x=\sin45^\circ,45x=360^\circ m+45^\circ$ where $m$ is any integer
$\implies x=8^\circ m+1^\circ$ where $0\le m\le44$
$\implies Q=\prod_{m=0}^{44}\sin(8^\circ m+1^\circ)=\dfrac1{\sqrt2}\cdot\dfrac1{2^{44}}$
$Q^2=\dfrac1{2\cdot2^{88}}$
Clearly, $\{\sin^2(8^\circ m+1^\circ);0\le m\le44\}=\{\sin^2(2r-1)^\circ,1\le r\le45\}$
as $x=1^\circ,9^\circ,17^\circ,25^\circ,33^\circ,41^\circ,49^\circ,57^\circ,65^\circ,73^\circ,81^\circ,89^\circ,$
$97^\circ[\sin97^\circ=\sin(180-97)^\circ=\sin83^\circ],$
$105^\circ[\sin75^\circ]$ and so on
$\cdots$
$m=22\implies8m+1=177\implies\sin177^\circ=\sin(180-3)^\circ=\sin3^\circ$
$\cdots$
$m=44\implies8m+1=353\implies\sin353^\circ=-\sin7^\circ$
|
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|
Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].
The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$.
My (wrong) solution I thought that I would get the solution by calculating the orthogonal projection $w=a+bx+cx^2$ of $v=\sin x$ on subspace $W=\langle u_1,u_2,u_3 \rangle=\langle 1,x,x^2\rangle$ using Gramm matrix. Then I have
$$\begin{pmatrix}\langle u_1, u_1\rangle&\langle u_1, u_2\rangle&\langle u_1, u_3\rangle\\
\langle u_2, u_1\rangle&\langle u_2, u_2\rangle&\langle u_2, u_3\rangle\\
\langle u_3, u_1\rangle&\langle u_3, u_2\rangle& \langle u_3, u_3\rangle\end{pmatrix}
\begin{pmatrix} a\\b\\c\end{pmatrix}=
\begin{pmatrix} \langle u_1, v\rangle\\
\langle u_2,v\rangle\\
\langle u_3, v\rangle \end{pmatrix}$$
So then
$$\begin{pmatrix} \int_0^\pi 1&\int_0^\pi x&\int_0^\pi x^2\\
\int_0^\pi x& \int_0^\pi x^2& \int_0^\pi x^3 \\
\int_0^\pi x^2&\int_0^\pi x^3&\int_0^\pi x^4\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} \int_0^\pi \sin x\\\int_0^\pi x\sin x\\\int_0^\pi x^2 \sin x \end{pmatrix}$$
But solving these equations didn't give me any good answer. So my question is - is my way of solving it completely wrong (if so, can you give me hints how to do it otherwise)?
Thank you
Edit
Then$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
\frac{\pi^2}{2}&\frac{\pi^3}{3}&\frac{\pi^4}{4}\\
\frac{\pi^3}{3}&\frac{\pi^4}{4}&\frac{\pi^5}{5}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\\pi\\\pi^2-4 \end{pmatrix}$$
$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&\frac{\pi^4}{12}&\frac{4\pi^5}{45}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&0&\frac{\pi^5}{180}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
Edit II My way (I have found the mistake I did) will get the result
$$f(x)=\dfrac{60(\pi^2-12)}{\pi^5}x^2-\dfrac{60(\pi^2-12)}{\pi^4}x+\dfrac{12(\pi^2-10)}{\pi^3}$$
which I hope is the right answer with error approx. $0,000936$
|
Is not the direct modern equivalent of Bhaskara's parabola approximation for Sine curve good enough as a start?
$$ y(x) = \dfrac{x(\pi-x)}{2},$$
$$ y(\pi/2)= \pi^2/8;y^{'}(\pi/2) =0; y^{''}(\pi/2) =-1. $$
And cannot the amplitude alone be improved/adjusted from here further to minimize error by least squares?
|
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|
Finf $f(x)$ which is a second degree polynomial, such that $f(1)=0$ and $f(x) = f(x-1)$ I must find a function $f(x) = ax^2+bx+c$ such that:
$$f(1) = a+b+c=0\\f(x)=f(x-1)\implies ax^2+bx+c = a(x-1)^2+b(x-1)+c\implies\\ax^2+bx+c = ax^2+(-2a+b)x+a-b+c\implies\\a = a, b = -2a+b, c = a-b+c$$
but this results fo $a=b=c=0$. What am I doing wrong?
|
If $f(x) = f(x-c)$
for all $x$,
where $c$ is a constant,
and $f$ is a polynomial,
then
$f$ is constant.
Proof:
Let
$g(x) = f(x)-f(0)$.
Then $g(0) = 0$.
Also,
$g(-nc) = g(0) = 0$
for all positive integers $n$.
Since a polynomial of
degree $d$ has at most $d$
distinct zeros,
and $g$ has an infinite
number of zeros,
$g$ must be zero.
Therefore
$f(x) = f(0)$
for all $x$.
(If there is anything
original here,
I apologize.)
|
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|
Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational
How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
|
Let $\alpha=\sqrt{3}+ \sqrt{5}+ \sqrt{7}$.
Then $\alpha$ is a root of $x^8-60 x^6+782 x^4-3180 x^2+3481=0$.
Since this is a monic polynomial with integer coefficients, the rational root theorem tells you that $\alpha$ is either irrational or an integer.
Now
$\quad 1.7 < \sqrt 3 < 1.8 $
$\quad 2.2 < \sqrt 5 < 2.3 $
$\quad 2.6 < \sqrt 7 < 2.7 $
gives
$\quad 6.5 < \alpha < 6.8 $
which proves that $\alpha$ is not an integer.
We can avoid even these fine estimates: if $\alpha$ is an integer then it must divide $3481=59^2$, but clearly $1 < \alpha < 3\sqrt 7 < 9 < 59 $.
|
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|
How to show that $zw=1\implies w = z^{-1}, z = w^{-1}$? I need to show that
$$zw=1\implies w = z^{-1}, z = w^{-1}$$
But I don't know how to start. I've tried to consider $z=a+bi, w = c+di$ then:
$$zw = ac+adi+bci-bd\implies zw = ac-bd + (ad+bc)i = 1\implies \\ac-bd = 1\\ad+bc = 0$$
$$ac = 1+bd\\ad = -bc$$
$$c = \frac{1+bd}{a}\\d = \frac{-bc}{a}$$
$$c = \frac{1+b(\frac{-bc}{a})}{a}=\frac{1-\frac{b^2c}{a}}{a} = \frac{a-b^2c}{a^2}\implies ca^2 = a-b^2c\implies ca^2+b^2c=a\implies c[a^2+b^2] = a\implies c = \frac{a}{a^2+b^2}$$
Using $$d=\frac{-bc}{a}\implies d = \frac{-b\frac{a}{a^2+b^2}}{a} = \frac{-b}{a^2+b^2}$$
LoL I did it :D
|
Hint Multiply both $zw=1$ by $z^{-1}$. Do the same for $w$.
Also, how do you define $z^{-1}$?
Since complex multiplication is commutative, the definition of $z^{-1}=w$ is $zw=1$. So there is absolutely nothing to prove....
|
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|
Prove $x^n + x < (x^n)x$ using induction. I need to prove $x^n$ + x < $x^n\cdot$ x, n $\in$ N, x $\in$ R>2 using induction. I started by
$x^n$ + x + (x^(n+1)+x) < ($x^n\cdot$ x) + (x^(n+1)+x)
I simplified to this:
< 2x^(n+1) + x
I am stuck here. All help is appreciated.
|
Let $P(n): x^n+x< x^{n+1}, n \in \mathbb{N}, x > 2$.
Check $P(1)$ is true. $x^1+x = 2x < x^2 \iff x(x-2) > 0$ is true since $x > 2$.
Assume $P(n)$ is true, i.e. $x^n + x < x^{n+1}$, prove $P(n+1)$ is true.
$x^{n+1} + x = x\cdot x^n + x < x(x^{n+1}-x) + x = x^{n+2} - x^2 + x < x^{n+2} - x^2 + x^2 = x^{n+2}$ since $x^2 > x$. Thus $P(n+1)$ is true, and by MPI the statement is true.
|
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|
Integration of $\int\log(\sqrt{1-x}+\sqrt{1+x}) \, dx$ Integration of
$$\int\log\left(\sqrt{1-x}+\sqrt{1+x}\right) \, dx$$
Please help to go through this problem as i have started with putting $x$= $cos2y$.
|
It is shorter to integrate by parts, setting $ u = \ln\bigl(\sqrt{1-x} + \sqrt{1 + x}\bigr) $, hence:
\begin{align*}
\mathrm d\mkern1.5mu u & = \frac{\frac12\Bigl(\dfrac{-1}{\sqrt{1-x}} + \dfrac{1}{\sqrt{1 + x}}\Bigr)}{\sqrt{1-x} + \sqrt{1 + x}}\,\mathrm d\mkern1.5mu x= \frac12\frac{\sqrt{1-x}-\sqrt{1 + x}}{\sqrt{(1-x²)}\Bigl(\sqrt{1-x} + \sqrt{1 + x}\Bigr)}\,\mathrm d\mkern1.5mu x\\& = -\frac{1}{4x}\frac{\bigl(\sqrt{1-x}-\sqrt{1 + x}\bigr)² }{\sqrt{(1-x²)}}\,\mathrm d\mkern1.5mu x =\frac{1}{2x}\frac{\sqrt{1-x²}-1}{\sqrt{1-x²}}\,\mathrm d\mkern1.5mu x
\end{align*}
Setting $I=\displaystyle\int\log(\sqrt{1-x}+\sqrt{1+x})\,\mathrm d\mkern1.5mu x$, you get:
\begin{align*}
I & =x\ln\bigl(\sqrt{1-x}+\sqrt{1+x}\bigr)-\frac{1}{2}\int\Bigl(1-\frac{1}{\sqrt{1-x²}}\Bigr)\,\mathrm d\mkern1.5mu x\\
& = x \ln\bigl(\sqrt{1-x}+\sqrt{1+x}\bigr)+\frac12(\arcsin x-x).
\end{align*}
|
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|
When does the curve $y = {x^p}\cos \frac{\pi }{x}(0 < x \leqslant 1)$ have finite length? When does the curve $y = {x^p}\cos \frac{\pi }{x}(0 < x \leqslant 1)$ have finite length?
I cannot figure out how to solve this problem. Thanks for your help.
|
Thanks for everyone! With your hints I have figured out a solution.
We need to check if the integral
$\begin{gathered}
\int_0^1 {\sqrt {1 + y'{{(x)}^2}} dx} = \int_1^{ + \infty } {\frac{1}{{{t^2}}}\sqrt {1 + y'{{(\frac{1}{t})}^2}} dt} \hfill \\
= \int_1^{ + \infty } {\frac{1}{{{t^2}}}\sqrt {1 + \frac{1}{{{t^{2(p - 1)}}}}{{(p\cos \pi t + \pi t\sin \pi t)}^2}} dt} \hfill \\
\end{gathered} $
is convergent.
If $p > 1$, we have an estimate
$\begin{gathered}
\int_n^{n + 1} {\frac{1}{{{t^2}}}\sqrt {1 + \frac{1}{{{t^{2(p - 1)}}}}{{(p\cos \pi t + \pi t\sin \pi t)}^2}} dt} \hfill \\
\leqslant \int_n^{n + 1} {\frac{1}{{{t^2}}}(1 + \frac{{p + \pi t}}{{{t^{p - 1}}}})dt} \leqslant \frac{1}{{{n^2}}}(1 + \frac{{p + 2\pi n}}{{{n^{p - 1}}}}) \hfill \\
\end{gathered} $
from which we know the integral is convergent.
If $p \leqslant 1$, we have an estimate
$\begin{gathered}
\int_{2n + \frac{1}{6}}^{2n + \frac{1}{3}} {\frac{1}{{{t^2}}}\sqrt {1 + \frac{1}{{{t^{2(p - 1)}}}}{{(p\cos \pi t + \pi t\sin \pi t)}^2}} dt} \hfill \\
\geqslant \int_{2n + \frac{1}{6}}^{2n + \frac{1}{3}} {\frac{1}{{{t^2}}}(\frac{{p + \pi t}}{{2{t^{p - 1}}}} - 1)dt} \geqslant \frac{1}{6}\frac{1}{{{{(3n)}^2}}}(\frac{{p + 2\pi n}}{{2{{(2n)}^{p - 1}}}} - 1) \hfill \\
\end{gathered} $
from which we know the integral is not convergent.
|
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permutation & combinations How many odd three digit numbers are there when tens digit is greater than units digit and hundreds digit is greater than tens digit?
*
*$225$
*$ 45$
*$ 50$
*$230$
My attempt:
The units digit can be $1$ or $3$ or $5$ so $3$ ways ($9$ cannot be taken)
units digit when = 1 _ _ _1 ten's digit = 2,3,4,5,6,7,8
in this if ten's digit = 2 hundred's digit = 3,4,5,6,7,8,9.. -- 7 ways
in this if ten's digit = 3 hundred's digit = 4,5,6,7,8,9... 6 ways
so on upto ten's = 8 hundreds = 9 --1 way i.e, nothing but sum of first 7 terms i.e, 28
similarly for unit's= 3 ten's digit = 4,5,6,7,8 hundred's digit = ,5,6,7,8,9.. -- 5 ways
so on upto ten's = 8 hundreds = 9 --1 way i.e, nothing but sum of first 5 terms i.e, 15
similarly for unit's = 5 the ways can be 6 total 28+15+6 = 49
one number remained is 987 with this it is 50 ways.
calculating this take more time can anyone reduce this or is there any formula for this type of ques...
|
For a number to be odd, it must end with $1$ or $3$ or $5$ or $7$
Ending with 1: Let us fix $1$ at one's place now we have to fill ten's and hundred's place such that tens digit is greater than units digit and hundreds digit is greater than tens digit.
Placing $2$ at ten's place leaves us with $[3,4,5,6,7,8,9]$ i.e, $7$ choices.
Placing $3$ at ten's place leaves us with $[4,5,6,7,8,9]$ -- $6$ choices.
Placing $4$ at ten's places leaves us with $[5,6,7,8,9]$ i.e, 5 choices... so on.
No. of odd three digit numbers with 1 at one's place = $7+6+5+4+3+2+1 = 28$
Ending with 3: Fixing $3$ at one's place and filling ten's place with $4,5,6,7,8,9$ respectively leaves us with $5,4,3,2,1$ ways that sum up to $15$
Ending with 5: Fixing $5$ at one's place and filling ten's place with $6,7,8,9$ respectively leaves us with $3,2,1,0$ ways that sum up to $6$
Ending with 7: Fixing $7$ at one's place and filling ten's place with $8$ and hundred's place with $9$ gives us number $987$ with counts $1$ to total sum.
Total numbers : $28+15+6+1 = 50$
|
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|
Number of solution of Frobenius equation Oke I am trying to find all presentable of a number $n$ as sum $ax+(a+1)y$ where $a=0,1,\ldots$ and $x,y\geq0$ are integers.
I find that
$5=1+1+1+1+1=1+1+1+2=1+2+2=2+3=5$ so we have $5$ ways.
$7=1+1+1+1+1+1+1=1+1+1+1+1+2=1+1+1+2+2=1+2+2+2=2+2+3=3+4$ so we have $7$ ways.
I suppose that if $p\geq3$ is a prime number then there are exactly $p$ ways. However, I found that
$11=1+1+1+1+1+1+1+1+1+1+1=1+1+1+1+1+1+1+1+1+2=1+1+1+1+1+1+1+2+2=1+1+1+1+1+2+2+2=1+1+1+2+2+2+2=1+2+2+2+2+2$
and
$=2+2+2+2+3=2+3+3+3=3+4+4 =11$ so there are only $10$ ways.
My question are that: What is the formula for the number ways to preprent a prime $p$ to sum of $x$ numbers $a$ and $y$ numbers $a+1$ where $a,x,y$ are non-negative integers?
|
This should hold for any $n$, not just $n$ prime. For the induction step from $n-1$ to $n$, note that a generic sum is
$$n-1 = ax + (a+1)y,$$
or
$$n = ax + (a+1)y +1 = a(x-1) + (a+1)(y+1),$$
assuming $x$ is not zero. So in this case, a sum for $n-1$ corresponds to one for $n$. If x is zero, then use
$$n-1= ay,$$
or
$$n = ay+1= a(y-1) + (a+1),$$
and we again relate a sum in $n-1$ to one in $n$.
Additionally, you have a new sum of $n$ as $n$ 1s, so the induction step adds 1 breakdown over that for $n-1$.
This needs fleshing out (inversing the step, etc), of course.
|
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|
Modular Arithmetic with large exponents! Decide whether each of the following is true or false without using a calculator:
The problem is:
$$11^{99}\equiv 1\pmod{5}$$
Now I know I can break the $11$ into $(10+1)^{99}$ and maybe rewrite it as $(10+1)^{98}\times (10+1)$
and then realize that $10+1\equiv 1\pmod{5}$ and then just deal with $(10+1)^{98}$ but that means I have to do this $97$ times until at the end I have $10+1\equiv 1\pmod 5$ but I feel like this is a very stupid way to look at it. Is there a little trick of sorts that I can use on problems like this when the exponents are huge?
|
There are many ways to approach this.
One way is to note that for all $n$: if $\gcd(a,n) = 1$:
$a^{\phi(n)} = 1$ (mod $n$). This is known as Euler's Theorem.
Here, we have $a = 11, n = 5$, and $\phi(5) = 4$, so we have:
$11^{99} = (11^{96})(11^3) = (11^{4})^{24}(11^3) = (1^{24})(11^3) = 11^3$ (mod $5$).
Then it is easy to see:
$11^3 = (5\cdot 20 + 1)(5 \cdot 2 + 1) = 5\cdot (200 + 22) + 1$ so that:
$11^3 = 1$ (mod $5$).
It is also easy to prove by induction on $k$, that $11 = 1$ (mod $5$) $\implies 11^k = 1$ (mod $5$).
Certainly this is self-evident for $k = 1$, since $11 = 2\cdot 5 + 1$.
Suppose now that $11^{k-1} = 1$ (mod $5$). This means there is some integer $t$, so that:
$11^{k-1} = 5t + 1$.
Then $11^k = (11^{k-1})(11) = (5t + 1)(10 + 1) = 5\cdot(10t + t + 2) + 1$, so taking:
$t' = 10t + t + 2$, we see $11^k = 5t' + 1$, that is: $11^k = 1$ (mod $5$).
Now this holds for ANY natural number $k$, so in particular, $k = 99$.
|
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|
Riemann integral problem 4 How can I evaluate the following limit with Riemann integral not with telescoping series:
$$ u_n=\frac{1}{n}\sum_{k=1}^n n\cdot \arctan\left(\frac{1}{k^2+k+1}\right),$$
$$\lim _{n\to \infty }u_n=\lim _{n\to \infty }\frac{1}{n}\sum_{k=1}^n n\cdot \arctan\left(\frac{1}{k^2+k+1}\right)=\lim_{n\to \infty\ }\frac{1}{n}\sum _{k=1}^nf\left(\frac{k}{n}\right)=\int_0^1\:f\left(x\right)dx$$
|
Integration by parts gives:
$$\int_{0}^{1}\arctan\left(\frac{1}{1+x+x^2}\right)\,dx = \arctan\frac{1}{3}+\int_{0}^{1}\frac{x(1+2x)}{(1+x^2)(1+(x+1)^2)}\,dx$$
and the last integral equals:
$$ \int_{0}^{1}\left(\frac{x}{1+x^2}-\frac{x+1}{1+(x+1)^2}+\frac{1}{1+(x+1)^2}\right)\,dx =\frac{2\log 2-\log 5}{2}+\arctan 2-\frac{\pi}{4}$$
so, in the most compact form:
$$\int_{0}^{1}\arctan\left(\frac{1}{1+x+x^2}\right)\,dx = \arctan\frac{3}{4}+\log\frac{2}{\sqrt{5}}.$$
Don't you think something went wrong when converting your partial sums into Riemann sums?
|
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|
Simultaneusly solving $2x \equiv 11 \pmod{15}$ and $3x \equiv 6 \pmod 8$ Find the smallest positive integer $x$ that solves the following simultaneously.
Note: I haven't been taught the Chinese Remainder Theorem, and have had trouble trying to apply it.
$$
\begin{cases}
2x \equiv 11 \pmod{15}\\
3x \equiv 6 \pmod{8}
\end{cases}
$$
I tried solving each congruence individually.
The inverse for the first is 8: $x \equiv 8\cdot11 \equiv 88 \equiv 13 \pmod{15}$.
The inverse for the second is 3: $x \equiv 3\cdot6 \equiv 18 \equiv 2 \pmod{8}$.
But I can't figure out how where to go from here.
|
Let's proceed completely naively and see where it takes us.
The first congruence is equivalent to $x \equiv 13 \pmod{15}$. This is the same as $x = 13 + 15n$ for any integer $n$. Let's use this in the second congruence.
$$\begin{align}
3x &\equiv 6 \pmod 8 \\
3(13 + 15n) &\equiv 6 \pmod 8 \\
39 + 45n &\equiv 6 \pmod 8 \\
7 + 5n &\equiv 6 \pmod 8 \\
5n &\equiv 7 \pmod 8.
\end{align}$$
This lets you determine a solution for $n$. In particular, you'll find that $n \equiv 3 \pmod 8$, or rather $n = 3 + 8l$ for any integer $l$.
Going back, we see that $x = 13 + 15n = 13 + 15(3 + 8l) = 13 + 45 + 120l = 58 + 120l$, or rather that $x \equiv 58 \pmod{120}$. $\diamondsuit$
|
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|
How to evaluate $\left(\cos{\frac{5\pi}{9}}\right)^{11}+\left(\cos{\frac{7\pi}{9}}\right)^{11}+\left(\cos{\frac{11\pi}{9}}\right)^{11}$ How to evaluate $$\left(\cos{\frac{5\pi}{9}}\right)^{11}+\left(\cos{\frac{7\pi}{9}}\right)^{11}+\left(\cos{\frac{11\pi}{9}}\right)^{11}?$$
I found the problem on this page.
|
Starting from Frieder's final expression, there is a terrible closed form since $$\cos \left(\frac{\pi }{9}\right)=\frac{\sqrt[3]{1-i \sqrt{3}}+\sqrt[3]{1+i \sqrt{3}}}{2 \sqrt[3]{2}}$$ $$\cos \left(\frac{2\pi }{9}\right)=2\cos^2 \left(\frac{\pi }{9}\right)-1$$ $$\sin \left(\frac{\pi }{18}\right)=\sqrt{\frac{1-\cos \left(\frac{\pi }{9}\right)}{2}}$$ After Lucian's comment, this is $$\frac{231}{512}-\frac{\sqrt[3]{\frac{3}{2}} \left(1-i \sqrt{3}\right)
\sqrt[3]{A}}{4096}-\frac{132463\ 3^{2/3} \left(1+i \sqrt{3}\right)}{2048\ 2^{2/3}
\sqrt[3]{A}}$$ with $A=-42951219+93177701 i \sqrt{3}$.
|
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|
Residue Integration I am attempting to calculate the integral of $\frac{(1+sin(\theta))}{(3+cos(\theta))}$ from $0$ to $2\pi$. I have already changed $sin$ and $cos$ into $\frac{1}{2i(z-z^{-1})}$ and $\frac{1}{2(z+z^{-1})}$. I am really stuck now. Can anyone please guide me?
|
If direct way can be considered, just as Dr.MW already answered, tangent half-angle substitution $t=\tan \frac \theta 2$ makes the problem simple since $$I=\int\frac{(1+sin(\theta))}{(3+cos(\theta))}d\theta=\int\frac{(t+1)^2}{t^4+3 t^2+2}dt=\int \frac{2 t}{t^2+1} dt+\int\frac{1-2 t}{t^2+2}dt$$ $$I=\log \left(1+t^2\right)+\frac{\tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)}{\sqrt{2}}-\log \left(2+t^2\right)=\frac 1{\sqrt{2}}\tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+\log\Big(\frac{1+t^2}{2+t^2} \Big)$$ Back to $\theta$ (if required), $$I=\frac{1}{\sqrt 2}\tan ^{-1}\left(\frac{\tan \left(\frac{\theta }{2}\right)}{\sqrt{2}}\right)-\log (3+\cos (\theta ))$$ So, since, as explained by Dr.MW,$$
\int_0^{2\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta=\int_{-\pi}^{\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta
$$ the bounds for $t$ are $-\infty$ and $+\infty$, so the logarithmic terms does not contribute and the result is just $\frac{\pi}{\sqrt 2}$.
More generally, assuming $a \leq \pi$,$$\int_{-a}^{a} \frac{1+\sin \theta}{3 + \cos \theta} d\theta=\sqrt{2} \tan ^{-1}\left(\frac{a}{\sqrt{2}}\right)$$
|
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|
how to solve $2(2y-1)^{\frac{1}{3}}=y^3+1$ how can I solve $2(2y-1)^{\frac{1}{3}}=y^3+1$
I got by $x=\frac{y^3+1}{2}$ that $y=\frac{x^3+1}{2}$ but I was told earlier that can't say that $x=y$.
So I can I solve this equation?
Thanks.
|
You correctly set $x^3=2y-1$, so the equation becomes
$$
\begin{cases}
2y=x^3+1\\
2x=y^3+1
\end{cases}
$$
Subtract them:
$$
2(y-x)=x^3-y^3
$$
or
$$
(x-y)(x^2+xy+y^2)+2(x-y)=0
$$
This factors as
$$
(x-y)(x^2+xy+y^2+2)=0
$$
but the second factor is positive for all $x$ and $y$, because $x^2+xy+y^2\ge0$.
Therefore you conclude $x=y$, so plugging in the equation $2y=x^3+1$ you get
$$
x^3-2x+1=0
$$
that's easy to solve because it factors as $(x-1)(x^2+x-1)=0$.
There is a different way to solve the particular problem, but using more general principles (thanks to ivancho for having suggested it).
Your equation is in the form
$$
f^{-1}(y)=f(y)
$$
where
$$
f(y)=\frac{y^3+1}{2}
$$
is an increasing function (hence invertible) taking on every real value. Decreasing would be the same, of course.
If $y$ is a solution of your equation and we set $x=f(y)$, then from $f^{-1}(y)=f(y)$ we get $f^{-1}(y)=x$ and so $y=f(x)$.
Suppose $(x,y)$ is a solution of $x=f(y)$ and $y=f(x)$. If we had $x<y$, then, by the fact that $f$ is increasing, we would have
$$
f(x)<f(y)
$$
so $y<x$. Similarly, from $x>y$ we get $f(x)>f(y)$ or $y>x$. In both cases we get a contradiction.
|
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|
Prove that $1-\frac{1}{1+\frac{\alpha}{nm}}\leq \sqrt{1-\frac{1}{1+\frac{\alpha}{n}}}\sqrt{1-\frac{1}{1+\frac{\alpha}{m}}}$. For what values of the real parameter $\alpha$ the following inequality is true?
$$1-\frac{1}{1+\frac{\alpha}{nm}}\leq \sqrt{1-\frac{1}{1+\frac{\alpha}{n}}}\sqrt{1-\frac{1}{1+\frac{\alpha}{m}}}$$
for each $n,m\in\mathbb N$.
|
$$1-\frac{1}{1+\frac{\alpha}{nm}}\leq \sqrt{1-\frac{1}{1+\frac{\alpha}{n}}}\sqrt{1-\frac{1}{1+\frac{\alpha}{m}}}$$
Coverts to:
$$\frac{\alpha}{\alpha+mn}\le\sqrt{\frac{\alpha}{\alpha +m}}\sqrt{\frac \alpha{\alpha+n}}$$
Or if $\alpha\ne0$ we can cancel it and get:
$$(\alpha+m)(\alpha+n)\le(\alpha+mn)^2\\
\alpha^2+(m+n)\alpha+mn\le \alpha^2+m^2n^2+2\alpha mn\\
\alpha\ge\frac{mn(mn-1)}{m+n-2mn}$$
Anyways $\alpha=0$ is a solution too. And $\alpha\ge0$ since we have it in squareroots.
|
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|
Proving $ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction Question:
Let $P(n)$ be the statement that $1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2} <2- \dfrac{1}{n}$. Prove by mathematical induction. Use $P(2)$ for base case.
Attempt at solution:
So I plugged in $P(2)$ for the base case, providing me with $\dfrac{1}{4} < \dfrac{3}{2}$ , which is true.
I assume $P(n)$ is true, so I need to prove $P(k) \implies P(k+1)$.
So $\dfrac{1}{(k+1)^2} < 2 - \dfrac{1}{k+1}$.
I don't know where to go from here, do I assume that by the Inductive hypothesis that it's true?
|
For $n\geq 2$, let $S(n)$ denote the statement
$$
S(n) : 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}.
$$
Base step ($n=2$): $S(2)$ says that $1+\frac{1}{4}=\frac{5}{4}\leq\frac{3}{2}= 2-\frac{1}{2}$, and this is true.
Inductive step: Fix some $k\geq 2$ and suppose that $S(k)$ is true. It remains to show that
$$
S(k+1) : 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}\leq 2-\frac{1}{k+1}
$$
holds. Starting with the left-hand side of $S(k+1)$,
\begin{align}
1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}
&\leq 2-\frac{1}{k}+\frac{1}{(k+1)^2}\quad\text{(by $S(k)$)}\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k^2+k+1}{k(k+1)}\right)\tag{simplify}\\[1em]
&< 2-\frac{1}{k+1}.\tag{$\dagger$}
\end{align}
we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k+1)$ is true, thereby completing the inductive step.
By mathematical induction, for any $n\geq 2$, the statement $S(n)$ is true.
Addendum: How did I get from the "simplify" step to the $(\dagger)$ step? Well, the numerator is $k^2+k+1$ and the denominator is $k^2+k$. We note that, $k^2+k+1>k^2+k$ (this boils down to accepting that $1>0$). Since $\frac{1}{k+1}$ is being multiplied by something greater than $1$, this means that what is being subtracted from $2$ in the "simplify" step is larger than what is being subtracting from $2$ in the $(\dagger)$ step.
Note: It really was unnecessary to start your base case at $n=2$. Starting at $n=1$ would have been perfectly fine. Also, note that this exercise shows that the sum of the reciprocals of the squares converges to something at most $2$; in fact, the series converges to $\frac{\pi^2}{6}$.
|
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|
diophantine equation $x^3+x^2-16=2^y$ Solve in integers: $x^3+x^2-16=2^y$.
my attempt:
of course $y\ge 0$, then $2^y\ge 1$, so $x\ge 1$.
for $y=0,1,2,3$ there is no good $x$.
so $y\ge 4$ and we have equation $x^2(x+1)=16(2^z+1)$, where $z=y-4\ge 0$.
what now?
|
Not my answer. Copied.
$x^2(x+1) = 2^y + 16$
Since LHS is an integer, then we must have $y \ge 0$.
Since RHS is a positive integer, then we must have $x \ge 1$.
$x^2(x+1)$ is strictly increasing for $x \ge 0$.
$2^y + 16$ is strictly increasing for $y \ge 0$.
$$\begin{array}{n|c|c|}
\hline
n & n^2(n+1) & 2^n + 16 \\ \hline
0 & 0 & 17 \\
1 & 2 & 18 \\
2 & 12 & 20 \\
3 & 36 & 24 \\
4 & 80 & 32 \\
5 & 150 & 48 \\
6 & 252 & 80\\ \hline
\end{array}$$
Note that the table indicates that $(x,y) = (4,6)$ is a solution.
So any solution involving $y \ge 7$ will require $x \ge 5$.
We will show that there is no solution for $y \ge 7$.
So we can assume now that $x \ge 5$ and $y \ge 7$.
\begin{align}
x^2(x+1) &= 2^y + 16\\
x^2(x+1) &= 16(2^{y-4} + 1)\\
\end{align}
Note that $2^{y-4}+1$ must be an odd integer.
So if $x$ is an odd integer, $\gcd(x+1,2^{y-1}+1) = 1$.
$\quad$ Hence $x+1 | 16$
$\quad$ Remembering that $x \ge 5$, we must have $x = 7$ or $x = 15$.
$\quad$Case $1: x = 7$
\begin{align}
16(2^{y-4}+1) &= 392\\
2(2^{y-4}+1) &= 49 & \text{Has no solution.}\\
\end{align}
$\quad$Case $2: x = 15$
\begin{align}
16(2^{y-4}+1) &= 3600\\
2^{y-4}+1 &= 225 & \\
2^{y-4} &= 224 \\
2^{y-4} &= 32 \times 7 & \text{Has no solution.}\\
\end{align}
So if $x$ is an even integer, $\gcd(x^2,2^{y-1}+1) = 1$.
$\quad$ So $x^2 | 16$.
This can't happen since $x \ge 5$.
|
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|
How to calculate sum of combinations with different n and k Input: $[X,Y]$ and $L$
Output : no of increasing sequence of length L and all elements should be $X\le i \le Y$ e.g: for $[6,7]$ and $2$ sequences are $6,66,67,7,77.$
For the above question my answer is the following sequence
$a+b+c+d+..$ where
$a=N$
$b=\sum_{i=1}^{N} i$
$c = \sum_{i=1}^{N}\sum_{j=1}^{i} j $
$d = \sum_{i=1}^{N}\sum_{j=1}^{i}\sum_{k=1}^{j} k$
$...$ and the no of elements in the sequence is $D = Y-X +1$
is there any formula to the above sequence?
Edit :
For the above I found that multiset is answer. Then again I encountered other issues.
how to efficiently calculate
$\sum_{k=1}^{D} {n+k-1 \choose k}$ mod 10^6+3, $1\le n,D\le 10^9$?
|
Notice that $${n \choose m}={n-1 \choose m-1}+{n-1 \choose m}$$
Hence
\begin{align*}
\sum_{k=1}^{D} {n+k-1 \choose k} &= \sum_{k=1}^{D} {n+k-1 \choose n-1} \\
&= {n \choose n} + \sum_{k=1}^{D} {n+k-1 \choose n-1} - {n \choose n} \\
&= {n \choose n} + {n+1-1 \choose n-1}+ \cdots +{n+D-1 \choose n-1} - {n \choose n} \\
&= {n+1 \choose n} + {n+2-1 \choose n-1} +\cdots +{n+D-1 \choose n-1} - {n \choose n} \\
&= {n+2 \choose n} + {n+3-1 \choose n-1} +\cdots +{n+D-1 \choose n-1} - {n \choose n} \\
&= \cdots\\
&= {n+D \choose n}-1
\end{align*}
Use Lucas' theorem to compute it for $1\le n,D\le 10^9$ (notice that $10^6+3$ is a prime number).
|
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|
Find $n$ such that $(m-1)(m+3)(m-4)(m-8)+n$ is a perfect square for all $m$ Find $n$ such that $(m-1)(m+3)(m-4)(m-8)+n$ is a perfect square for all $m$
I am thinking of starting like this
$(m-1)(m+3)(m-4)(m-8)+n = k^2 \implies (m-1)(m+3)(m-4)(m-8) = k^2-n$
Honestly somewhat scared of expanding the products on left hand side. Any hints/help on the next step ? Thanks!
|
The trick is to do $(m-1)(m+3)(m-4)(m-8)=\underbrace{(m-1)(m-4)}\underbrace{(m+3)(m-8)}$ since $-1-4=-5=3-8$ giving same coefficient of m:
$$(m-1)(m+3)(m-4)(m-8)+n=(m^2-5m+4)(m^2-5m-24)+n\stackrel{y=m^2-5m+4}=y(y-28)+n=y^2-28y+n=(y-14)^2+n-14^2\implies n=14^2=196$$
|
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|
How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +...+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ I know how to reason $$1\cdot2 + 2\cdot3 + 3\cdot4 + n(n-1) = \frac{1}{3}n(n-1)(n+1)$$
However, I'm stuck on proving $$1\cdot1 + (2\cdot1 + 2\cdot2) + (3\cdot1+3\cdot2+3\cdot3) + \cdots +(n\cdot 1+...+n\cdot n) = \frac{1}{24}n(n+1)(n+2)(3n+1).$$
Could somebody shed some light on me?
|
A similar but slightly different variation of Jack's proof above.
$$\begin{align}
&1\cdot (1)\\
+&2\cdot(1+2)\\
+&3\cdot(1+2+3)\\
+&4\cdot(1+2+3+4)\\
+&\vdots\qquad\vdots \; \quad\ddots\quad \ddots\\
+&n\cdot(1+2+3+\cdots+n)\\
&=\sum_{i=1}^ni\sum_{j=1}^i j=\sum_{i=1}^ni\sum_{j=1}^i {j\choose 1}\\
&=\sum_{i=1}^ni{i+1\choose 2}\\
&=\sum_{i=1}^n\left[(i+2){i+1\choose 2}-2{i+1\choose 2}\right]\\
&=\sum_{i=1}^n\left[3{i+2\choose 3}-2{i+1\choose 2}\right]\\
&=3{n+3\choose 4}-2{n+2\choose 3}\\
&={n+2\choose 3}\left[3\left(\frac{n+3}4\right)-2\right]
\qquad\qquad\text{using }{n+3\choose 4}=\frac{n+3}4{n+2\choose 3}\\
&=\frac{(n+2)(n+1)n}{1\cdot 2\cdot 3}\cdot \left[ \frac{3n+1}4\right]\\
&=\frac1{24}n(n+1)(n+2)(3n+1)\qquad\blacksquare
\end{align}$$
NB - the above makes use of the identity
$$\sum_{i=1}^n{i+m\choose m+1}={n+1+m\choose m+2}$$
|
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|
Find the last three digits of $17^{256}$ Find the last three digits of
$ 17 ^{256} $
We have to check mod $1000$
I tried to check some patterns but in vain.!
|
As $1000=8\times 125$ comptute first $17^{256}$ modulo $8$ and modulo $256$, then use the Chinese Remainder Theorem to recover $17^{256}\mod 1000$.
Modulo $8$: $\enspace 17\equiv 1\mod 8$, hence $17^{256}\equiv 1 \mod 8$.
Modulo $125$:
By Euler's theorem, $n^{\varphi(125)}\equiv 1\mod125$ for all $n$. As $\varphi(125)=100$, we have $17^{256}=17^{56} \mod 125$. More over we can check $17^{50}\equiv -1 \mod 125$, hence:
$$17^{56}\equiv -17^6\equiv -69\equiv 56\mod 125.$$
$17^{256}\bmod 1000\,$ is the solution of the system of congruences:
$$\begin{cases}x\equiv 1\mod 8\\x\equiv 56\mod 125\end{cases}$$
The extended euclidean algorithm yields Bézout's identity:
\begin{array}[t]{c@{\qquad}r@{\qquad}r@{\qquad}c}
r_i & u_i & v_i & q_i\\
\hline
125 & 1 & 0 & \\
8 & 0 & 1 & 15\\
\hline
5 & 1 & -15 & 1 \\
3 & -1 & 16 & 1\\
2 & 2 & -31 & 1 \\
1 & -3 & 47 \\
\hline
\end{array}
\begin{align*}&-3\times 125+47\times 8=1,\\
\text{whence}\quad x\equiv-3\times 125&\times\color{red}{1}+47\times 8 \times\color{red}{56}=20681\equiv \color{red}{681} \mod 1000.\end{align*}
|
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|
How can i calculate Total no. of digit in $2^{100}\cdot 5^{75}$
How can i calculate Total no. of digit in $2^{100}\cdot 5^{75}$
$\bf{My\; Try::}$ I have used $$\log_{10}(2) = 0.3010$$.
Now Total no. of digit in $$x^y = \lfloor \log_{10}x^y\rfloor +1$$
Now $$\log_{10}(2^{100}\cdot 5^{75}) = 100\cdot \log_{10}(2)+75\log_{10}(5) = 100\cdot \log_{10}(2)+75-75\log_{10}(2)$$
So we get $$\log_{10}(2^{100}\cdot 5^{75})=30.10+75=105.10$$
So We get no. of Digit in $$2^{100}\cdot 5^{75} = \lfloor \log_{10}(2^{100}\cdot 5^{75})\rfloor +1 = 105+1 = 106$$
Can we solve it without using $\log\;,$ If yes then plz explain me, Thanks
|
I will note that your answer went from $100 \log_{10}(2) + 75 - 75\log_{10}(20)$ to $100 \log_{10}(2) + 75$. The general form of argument is correct, but taking this mistake into account the answer is $\lfloor 25\log_{10}(2) + 75\rfloor+1 = \lfloor 75 + 7.525...\rfloor+1 = 83$.
Doing this without logarithms is fairly straightforward. Factor out the terms $2^{75}5^{75} = 10^{75}$, reducing the problem to adding 75 to the number of digits of $2^{25}$. For this I will note $2^{10} = 1024$, so from the crude estimate $1000 < 2^{10} < 1500$, we can obtain:
$$
10^7 < 1000^2(32) < (2^{10})(2^{10})(2^5) < 1500^2 (32) = 1000^2(1.5)^2(32) = 1000^2(72)< 10^8
$$
Thus, the number of digits of $2^{25}$ is 8, and the answer is 83.
|
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|
Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$
Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|x^3-c^3|=|x-c||x^2+xc+c^2|<ε.$$
Does this make sense or are the steps done in the right way?
|
Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-c|<\delta$, then $|x^3-c^3|<\epsilon$.
Now,
$
|x^3-c^3| = |x-c||x^2+cx+c^2|
$
If $|x-c|<1$, then
$||x|-|c|| \leq |x-c|<1$
i.e. $||x|-|c||<1 \implies -1<|x|-|c|<1 \implies |x|<|c|+1$
so that
$$
|x^2+cx+c^2|\leq |x|^2+|c||x|+|c|^2 < (|c|+1)^2+|c|(|c|+1)+|c|^2=3|c|^2+3|c|+1,
$$
and so
$$
|x^3-c^3| = |x-c||x^2+cx+c^2|<(3|c|^2+3|c|+1)|x-c|.
$$
Therefore, for every $\epsilon >0$, there exists a $\delta=\inf\left\{1,\frac{\epsilon}{3|c|^2+3|c|+1}\right\}>0$, such that if $ 0<|x-c|<\delta$ then
$\begin{align} |x^3-c^3|&=|x-c||x^2+cx+c^2|\\
&<\frac{\epsilon}{3|c|^2+3|c|+1}\cdot(3|c|^2+3|c|+1)\\
&=\epsilon \end{align}
$
Thus, by the definition of limit of a function
$$
\lim_{x\to c}x^3=c^3.
$$
|
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|
evaluate $\int \frac{\tan x}{x^2+1}\:dx$ $$\int \frac{\tan x}{x^2+1}\, \mathrm dx$$
I used By-parts method setting $u=\tan x$ and $\, \mathrm dv=\frac{1}{x^2+1}\, \mathrm dx$, but then I got an integral that's more complicated
I also thought of trigonometric substitution, setting $x=\tan\theta$, but how am going to substitute that for the $\tan x$ in numerator?
I tried to use websites like symbolab & wolfram to evaluate the integral but I got no result.
|
The Laurent series of tan(x) is
$$\sum_{n=1,3,5..}^{\infty }\frac{8x}{(n\pi )^2-4x^2}$$
so
$$\frac{\tan(x)}{1+x^2}=\sum_{n=1,3,5..}^{\infty }\frac{8x}{\left [(n\pi )^2-4x^2 \right ](1+x^2)}$$
use the partial fraction to get
$$\sum_{n=1,3,5,..}^{\infty }\frac{8x}{((n\pi)^2+4 )(1+x^2)}+\frac{8}{((n\pi)^2+4 )(n\pi -2x)}-\frac{8}{((n\pi)^2+4 )(n\pi +2x)}$$
$$\int \frac{\tan x}{1+x^2}dx=C+\sum_{n=1,3,5,..}^{\infty }\frac{4}{(n\pi )^2+4}\left [ \log(1+x^2)-\log(n\pi -2x)-\log(n\pi +2x) \right ]$$
hence
$$\int \frac{\tan x}{1+x^2}dx=C+\sum_{n=1,3,5,..}^{\infty }\frac{4}{(n\pi )^2+4}\left [ \log(\frac{1+x^2}{(n\pi )^2-4x^2}) \right ]$$
|
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|
Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
|
If $64= (3y^2)^2 + 4x^4$, then $3y^2 =\sqrt{ 64 - 4x^4}$. Plugging that to $x^2+y^2$ we get:
$$x^2+\frac{\sqrt{64 - 4x^4}}{3}.$$
Now you have to maximalize a function of one variable. Remember that if $64 = 4x^4+9y^4$, then (setting $y = 0$) we obtain $16 \ge x^4$, therefore $|x| \le 2$.
|
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|
Prove ${4n \choose 2n} = {\frac{1\cdot3\cdot5\cdots(4n-1)}{(1\cdot3\cdot5\cdots(2n-1))^{2}}}{2n \choose n}$ Prove that prove $\dbinom{4n}{2n} = \dfrac{1\cdot3\cdot5\cdots(4n-1)}{(1\cdot3\cdot5\cdots(2n-1))^2} \dbinom{2n}{n}$ using mathematical induction. I have looked all over the internet, been able to prove a similar problem, but this one has me stumped. Quick help would be appreciated! Thanks!
|
Hints:
*
*$\dfrac{1\times 3\times 5\times \cdots \times (4n-1)}{(1\times 3\times 5\cdots \times (2n-1))^{2}} $ $= \dfrac{1\times 2\times 3\times 4\times 5\times \cdots \times (4n-1)\times 4n}{(1\times 2\times 3\times 4\times 5\times \cdots \times (2n-1)\times 2n)^{2}} \times \dfrac{(2\times 4\times \cdots \times 2n)^{2}}{2\times 4\times \cdots \times 4n}$
*$\dfrac{1\times 2\times 3\times 4\times 5\times \cdots \times (4n-1)\times 4n}{(1\times 2\times 3\times 4\times 5\times \cdots \times (2n-1)\times 2n)^{2}} = \dfrac{(4n)!}{(2n!)^2}$
*$\dfrac{(2\times 4\times \cdots \times 2n)^{2}}{2\times 4\times \cdots \times 4n}=\dfrac{(2^n\times 1\times 2\times \cdots \times n)^{2}}{2^{2n}\times 1\times 2\times \cdots \times 2n}=\dfrac{(n!)^2}{(2n)!}$
|
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|
Find the function that equals to $1-x^3+x^6-x^9+ \cdots$
Find the function that equals to $1-x^3+x^6-x^9+ \cdots$ for all $|x| < 1$
I know that $\frac{1}{1+x} = 1-x+x^2-x^3+...$ But I couldn't find the pattern here
|
Substitute $y=x^{3}$
$$S = 1-y+y^{2}-y^{3}+\cdots$$
$$S = \frac{1}{1+y} = \frac{1}{1+x^{3}}$$
|
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|
How do I solve these questions using Diophantine equations? I have been told that it is easier to solve the below 2 questions using Diophantine equations instead of simply trial and error.
1) Find the smallest positive integer which, when divided by 6, gives a remainder of 1 and when divided by 11, gives a remainder
of 6.
2) What is the least positive integer which has remainders of 1, 1 and 5 when divided by 3, 5 and 7 respectively?
In terms of my progress, I have attempted to generalize these questions by for example the first question:
Since the integer has a remainder of 1 when divided by 6, it must be of the form 6m + 1
By the way, I have no prior knowledge of Diophantine equations and would appreciate both a brief introduction as well as an explanation of how to apply this knowledge to the above questions.
Thank you :)
|
One should maybe say that using this approach replaces the guesswork by some modeling and applying a solution algorithm, so it is more systematic.
Problem 1):
From the equations for the divisions
$$
u = 6x+1 =11y+6
$$
we get a Diophantine linear equation
$$
6x-11y=5
$$
which is the name for a linear equation with integer coefficients
$$
a x + b y = c
$$
where one looks for integer solutions $x$, $y$. It has either no solution or infinite many solutions, depending on if
$$
\gcd(a,b) \,\rvert\, c
$$
where $\gcd(a,b)$ is the greatest common denominator of $a$ and $b$.
Here $\gcd(6,-11)=1$ and $1$ divides $5$, so we have solutions.
A particular solution can be obtained by applying the
extended Euclidean algorithm.
$$
\begin{array}{|c|c|c|c|c|}
\hline
i & q_{i-1} & r_i & s_i & t_i \\
\hline
0 & & 6 & 1 & 0 \\
\hline
1 & & -11 & 0 & 1 \\
\hline
2 & 6:-11 = 0 & 6- 0\cdot (-11) = 6 & 1-0\cdot 0=1 & 0-0\cdot 1=0 \\
\hline
3 & -11:6=-1 & -11 - (-1)\cdot 6 = -5 & 0 -(-1)\cdot 1=1 & 1-(-1)\cdot 0=1\\
\hline
4 & 6:-5=-1 & 6 - (-1) \cdot (-5) = 1 & 1-(-1)\cdot 1=2 & 0-(-1)\cdot 1=1\\
\hline
5 & -5:1=-5 & -5 - 1 \cdot (-5) = 0 & 1-(-5)\cdot 2=11 & 1-(-5)\cdot 1=6\\
\hline
\end{array}
$$
The row for $i=4$ gives $\gcd(6,-11) = 1$, $s = 2$, $t = 1$.
Checking: $6\cdot 2 + (-11) \cdot 1 = 1$, OK.
This solves
$$
a s + b t = \gcd(a,b) = 1 \Rightarrow a (5s) + b (5t) = 5
$$
Thus $(x,y)=(5\cdot 2, 5\cdot 1) = (10,5)$ is a particular solution.
The general solution is all solutions of the homogenous equation
$$
a x + b y = 0 = a x - t ab + t ab + b y = a(x-tb) + b (y + at)
$$
for $t \in \mathbb{Z}$ plus one particular solution.
Here this gives $(10+11t,5+6t)$ for $t \in \mathbb{Z}$ and
$u = 6(10+11t)+1=61-66t$, the smallest positive one is $61$.
Problem 2): The second example gives
$$
u = 3x+1 = 5 y + 1 = 7 z + 5
$$
So we could try
$$
3x-5y=0 \wedge 3x+1=7z+5
$$
which gives
$$
y=(3/5)x \wedge 3x-7z=4
$$
This leads to $(x,z)=(6+7t,2+3t)$ and $y = (3/5)(6+7t) = (18+21t)/5$ which is integer for $t=2$, giving $(x,y,z)=(20,12,8)$ and $u=61$ again.
WA seems to think so as well (link).
|
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|
Prove $\lim\limits_{n \to \infty} \frac{n^2+1}{5n^2+n+1}=\frac{1}{5}$ directly from the definition of limit.
Prove $\lim\limits_{n \to \infty} \frac{n^2+1}{5n^2+n+1}=\frac{1}{5}$ directly
from the definition of limit.
So far ive done this:
Proof:
It must be shown that for any $\epsilon>0$, there exists a positive integer $N$ such that for $n\in \mathbb{N}$ with $n\ge N$,
one has
$$\left|\frac{n^2+1}{5n^2+n+1}-\frac{1}{5}\right| <\epsilon \text{ (or equivalently,}\frac{1}{5}-\epsilon \lt \frac{n^2+1}{5n^2+n+1} \lt \frac{1}{5} + \epsilon \text{).}$$
$$\begin{align}
\left|\frac{n^2+1}{5n^2+n+1}-\frac{1}{5}\right| & = \left|\frac{(5(n^2+1))-(1(5n^2+n+1))}{5(5n^2+n+1)}\right| \\
& = \left|\frac{5n^2+5-5n^2-n-1}{25n^2+5n+5}\right| \\
& = \left|\frac{4-n}{25n^2+5n+5}\right| \\
\end{align}$$
How do i go about choosing the specific terms or factors of $n$ to compare against the numerator and denominator in this particular problem and i guess these type of problems in general?
Like in the answer: https://math.stackexchange.com/a/553466/227134
How do you know to setup the inequality $4n+7\le5n$ for the numerator and so forth?
I cant quite figure out what to put on the right side of $4-n\le $ ?
|
$$|\frac{4-n}{25n^2+5n+5}|< ϵ\\\frac{|4-n|}{|25n^2+5n+5|}< ϵ\\\frac{|n-4|}{|25n^2+5n+5|}< ϵ\\$$ now $n∈N ,n \rightarrow \infty $so $n-4>0 ,|n-4|=n-4 $ also $25n^2+5n+5>0 \rightarrow |25n^2+5n+5|=25n^2+5n+5$ $$\frac{n-4}{25n^2+5n+5} <\frac{n-4}{25n^2+5n}<\frac{n}{25n^2+5n}=\frac{1}{5n+1} < ϵ\\5n+1 > \frac{1}{ϵ}\\5n>\frac{1}{ϵ}-1 \\n>\frac{1}{5}(\frac{1}{ϵ}-1)\\ $$
|
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Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove that given inequality holds for $n+1$ case by using the given $n$ case. Ok, so this is what I've got by now:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{(n+1)^2}\overset{?}{>}1$$
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1$$
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
From the $n$ case we know that:
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}$$
So we basically have to prove that:
$$1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}\frac{1}{n}$$
Since
$$n^2+1<n^2+2<\dots<n^2+2n+1<2n^2+n$$
for $n\geq2$, then also:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$
.
.
$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$
so then we have:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$
|
Your reasoning is a little overbearing on the sums, so here's a simplification although it looks like it's basically the same thing. Define $F(n)=\sum_{k=n}^{n^2}\frac{1}{k}$. Then
$$F(n+1)=F(n)-\frac{1}{n}+\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}.$$
By the induction hypothesis, $F(n)>1$. So it suffices to prove:
$$\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}-\frac{1}{n}>0,$$
which is equivalent to:
$$\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}>\frac{1}{n}.$$
There are $2n+1$ terms on the left side, so
$$\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}\geq \frac{2n+1}{n^2+2n+1}=\frac{2n+1}{(n+1)^2}.$$
Now just show $(2n+1)/(n+1)^2 > 1/n$ by cross multiplying.
|
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|
Inverse function. A function $h$ is defined by $h:x\rightarrow 2-\frac{a}{x}$, where $x\neq 0$ and $a$ is a constant. Given $\frac{1}{2}h^2(2)+h^{-1}(-1)=-1$, find the possible values of $a$.
Can someone give me some hints? Thanks
|
Thank you for this problem. I was not familiar with the notation $h^2(x)=h(h(x))$, but this is clearly what must be meant in this problem since interpreting $h^2(x)=(h(x))^2$, yields no real solutions.
If $h(x)=2-\frac{a}{x}$, then we can solve for $h^{-1}(x)$ in the following manner.
$x=2-\frac{a}{h^{-1}(x)} \Rightarrow x-2=-\frac{a}{h^{-1}(x)} \Rightarrow \frac{1}{x-2}=-\frac{h^{-1}(x)}{a} \Rightarrow -\frac{a}{x-2}=h^{-1}(x)=\frac{a}{2-x}$
Therefore $h^{-1}(-1)=\frac{a}{3}$
Now if we interpret $h^2(x)$ to mean $h(h(x))$, then $$h^2(x)=h(2-\frac{a}{x})=2-\frac{a}{2-\frac{a}{x}}=2-\frac{a}{\frac{2x-a}{x}}=2-\frac{ax}{2x-a}=\frac{4x-2a-ax}{2x-a}$$
So that $\frac{1}{2}h^2(2)=\frac{1}{2}\frac{8-2a-2a}{4-a}=\frac{1}{2}\frac{8-4a}{4-a}=\frac{4-2a}{4-a}$
Plugging these values into your equation yields
$$\frac{4-2a}{4-a}+\frac{a}{3}=-1 \Rightarrow \frac{12-6a+4a-a^2}{12-3a}=-1 \Rightarrow 12-2a-a^2=3a-12$$
Bringing everything to one side gives us the following quadratic equation
$$a^2+5a-24=0$$
This equation is factorable and gives us $(a+8)(a-3)=0 \Rightarrow a=3$ and $a=-8$ are solutions.
|
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|
Prove equation $(ad-bc)(a-c)^2 = (b-d)^3$, if polynomials has common root
$$\begin{split}
W(x) &= x^3 + ax + b \wedge a,b \in \mathbb{R}
&\wedge \mathbb{D}_W &= \mathbb{R}\\
G(x) &= x^3 + cx + d \wedge c,d \in \mathbb{R}
&\wedge \mathbb{D}_G &= \mathbb{R}
\end{split}
$$ Prove:
$$ \left(\exists p \in \mathbb{R}\right)\left(W(p) = 0 = G(p)\right)
\Longrightarrow \left((ad-bc)(a-c)^2 = (b-d)^3\right)$$
I cannot prove it. It's obvious if $p = 0$ cause $d=b=0$. But otherwise I don't see solution. How can I prove it?
I noticed that $pa+b =cp+d$, so $(b-d) = p(c-a)$. Therefore
$(b-d)^3 =
%p^2(c-a)^2(b-d)=
(a-c)^2 \cdot p^2(b-d)$.
Now I just have to show, that $p^2(b-d) = (ad-bc)$.
I tried ($b \neq d)$ show that:
$$ \frac{ad}{b-d} - \frac{bc}{b-d} = p^2$$
From first observation, if $c \neq a$, $p^2 = \left( \frac{b-d}{c-a} \right)^2$, but it's now worthless.I don't see next step. Can you show me, how I should finish that proof?
|
Let $y$ be the common root. We then have
$$y^3 + ay+b = 0 \text{ and }y^3+cy+d = 0$$
This means we have
$$ay+b = cy+d \implies y = \frac{d-b}{a-c}$$
Hence,
$$\left(\frac{d-b}{a-c}\right)^3 + a \left(\frac{d-b}{a-c}\right) + b = 0$$
This gives us
$$(d-b)^3 + a(d-b)(a-c)^2 + b(a-c)^3 = 0 \implies (b-d)^3 = (a-c)^2(ad-ab+ba-bc)$$
which simplifies to what you are after.
|
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|
Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$
Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .
$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfrac{\sqrt{3}}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
&=\dfrac{1.73}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
\end{align}$
$\begin{align}\sin 18^{\circ}&=\sin (30-12)^{\circ}\\~\\
&=\sin (30^{\circ})\cos (12^{\circ})-\cos (30^{\circ})\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{\sqrt3}{2}\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{1.73}{2}\sin (12^{\circ})\\~\\
\end{align}$
$\begin{align}\tan 23^{\circ}&=\dfrac{\sin (30-7)^{\circ}}{\cos (30-7)^{\circ}}\\~\\
&=\dfrac{\sin (30)^{\circ}\cos 7^{\circ}-\cos (30)^{\circ}\sin 7^{\circ}}{\cos (30)^{\circ}\cos 7^{\circ}+\sin (30)^{\circ}\sin 7^{\circ}}\\~\\
\end{align}$
is their any simple way,do i have to rote all values of of $\sin,\cos $ from $1,2,3\cdots15$
I have studied maths upto $12$th grade.
|
You may exploit:
$$ \sin(60^\circ)=\frac{1}{2}\sqrt{3},\quad \sin(18^\circ)=\frac{1}{4}\left(\sqrt{5}-1\right),\quad \tan(22^\circ 30')=\sqrt{2}-1$$
then use some form of interpolation. For instance, in a neighbourhood of $x=\frac{\pi}{3}$:
$$ \sin(x)=\frac{1}{2}\sqrt{3}+\frac{1}{2}\left(x-\frac{\pi}{3}\right)-\frac{\sqrt{3}}{4}\left(x-\frac{\pi}{3}\right)^2-\frac{1}{12}\left(x-\frac{\pi}{3}\right)^3+\ldots $$
so by taking $x=\frac{\pi}{3}+\frac{9\pi}{180}$ we have:
$$ \sin(69^\circ) \approx \frac{\sqrt{3}}{2}+\frac{\pi}{40}-\frac{\pi^2\sqrt{3}}{1600}=0.933\ldots\tag{1}$$
and in a neighbourhood of $x=\frac{\pi}{8}$ we have:
$$ \tan(x)=(\sqrt{2}-1)+(4-2\sqrt{2})\left(x-\frac{\pi}{8}\right)+(6\sqrt{2}-8)\left(x-\frac{\pi}{8}\right)^2+\ldots $$
so by taking $x=\frac{\pi}{8}+\frac{\pi}{360}$ we have:
$$ \tan(23^\circ) \approx \sqrt{2}-1+(4-2\sqrt{2})\frac{\pi}{360}=0.424\ldots \tag{2}$$
At last, we have:
$$ \sin(18^\circ)=\frac{\sqrt{5}-1}{4}\approx 0.309\ldots\tag{3}$$
|
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|
How can I find the sum of the series $\sum_{n=0}^\infty {(-1)^n \over 4^n}$ or show that it diverges using the geometric series test? First, I reindexed it:
$$\sum_{n=0}^\infty {(-1)^n \over 4^n} = \sum_{n=1}^\infty {(-1)^{n-1} \over 4^{n-1}} = \sum_{n=1}^\infty {\left(-1 \over 4\right)}^{n-1} $$
So now I'm pretty sure it's in the form $ar^{n-1}$, but using the formula $\left(a \over {1-r}\right)$, I'm not sure I understand what $a$ would be, I'm guessing that $r$ = -$1 \over 4$
|
If we break the summations down we get:
$$S = \sum_{i=0}^{n}\frac{1}{4^{2(n-1)}} - \sum_{i=0}^{n}\frac{1}{4^{2n-1}}$$
That is the sum of the reciprocals of 4 to the power of the even positive numbers, then you subtract the reciprocals of 4 to the power of the odd positive numbers. $2(n-1)$ is the $n_{th}$ even number, and likewise $2n-1$ is the $n_{th}$ odd number. If take the summation with the even numbers:
$$S_{even} = \frac{1}{4^0} + \frac{1}{4^2} + ... \frac{1}{4^{2(n-1)}}$$
Multiply by the common factor $\frac{1}{4^2}$:
$$\frac{S_{even}}{4^2} = \frac{1}{4^2} + \frac{1}{4^4} + ... \frac{1}{4^{2n}}$$
Now subtract the multiplied sum:
$$S_{even} - \frac{S_{even}}{4^2} = \frac{1}{4^0} -\frac{1}{4^{2n}} $$
We can see that all of the terms cancel each other out, except for the first term and the last term in the second summation. We can also see that: $$\lim\limits_{n \to \infty} \frac{1}{4^{2n}} = 0$$
So:
$$S_{even} - \frac{S_{even}}{4^2} = \frac{1}{4^0} - 0$$
Rearrange:
$$S_{even} = \frac{16}{15}$$
Now we do the same for the odd summation:
$$S_{odd} = \frac{1}{4^1} + \frac{1}{4^3} + ... \frac{1}{4^{2n-1}}$$
Multiply through by the common factor ,then subtract that from the original:
$$S_{odd} - \frac{S_{odd}}{4^2} = \frac{1}{4^1} -\frac{1}{4^{2n+1}} $$
But:
$$\lim\limits_{n \to \infty} \frac{1}{4^{2n+1}} = 0$$
So:
$$S_{odd} - \frac{S_{odd}}{4^2} = \frac{1}{4^1} - 0 $$
Rearrange:
$$S_{odd} = \frac{4}{15}$$
Go back to the original series and substitute in $S_{even}$ and $S_{odd}$:
$$ \begin{align*}S&= \frac{16}{15} - \frac{4}{15}\\ S&= \frac{12}{15}\\ S&=\frac{4}{5}\end{align*}$$
|
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|
Volume of a hole through cylinder (from the side) I need to calculate the volume of a circular hole in a cylinder and I've come across a problem. The problem is finding the "cap-volume", which is needed to complete the volume of the hole. I created a quick example of the problem.
Cylinder which will be drilled
Hole. The cap is shown with the magenta coloured curves
|
Let $R$ and $r$ be the radius of the cylinder and the hole. We will assume $k = \frac{r}{R} < 1$.
We will further assume the hole is drilled toward the center and perpendicular to the axis of the cylinder. Under these assumptions, the volume of the cap is given by
$$\begin{align}
\verb/Vol/_{cap}
&= 4
\int_0^r \int_0^{\sqrt{r^2-x^2}} \left( \sqrt{R^2-x^2} - \sqrt{R^2 - r^2} \right) dy dx\\
&= 4 \int_0^r \sqrt{r^2 - x^2}\left(\sqrt{R^2 - x^2} - \sqrt{R^2-r^2}\right) dx\\
&= 4Rr^2 \left[ \int_0^1 \sqrt{(1-x^2)(1-k^2x^2)} dx - \frac{\pi}{4}\sqrt{1-k^2}\right]\\
&= 4Rr^2 \left[\frac{(1+k^2)E(k) -(1-k^2)K(k)}{3k^2} - \frac{\pi}{4} \sqrt{1-k^2}\right]
\end{align}
$$
where
$$\begin{align}
K(k)
&= \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-k^2\sin^2\theta}} d\theta
= \int_0^1 \frac{1}{\sqrt{(1-k^2x^2)(1-x^2)}} dx
\\
E(k)
&= \int_0^{\frac{\pi}{2}} \sqrt{1-k^2\sin^2\theta} d\theta
= \int_0^1 \sqrt{\frac{1-k^2x^2}{1-x^2}} dx
\end{align}$$
are the complete elliptic integrals of the first and second kind respectively.
Similarly, the volume of the hole is given by
$$\verb/Vol/_{hole} =
8Rr^2 \left[\frac{(1+k^2)E(k) -(1-k^2)K(k)}{3k^2} \right]$$
|
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|
How to evaluate volume of oblique frustum of right circular cone with elliptical section?
Let there be an oblique frustum, with an elliptical section, of a right circular cone with apex point O & cone angle $2\alpha=60^{o}$. It is obtained by cutting the cone by a plane at a normal distance $OM=h=20 cm$ & making an angle $\theta=60^{o}$ with the axis OO' of the frustum (cone) (as shown in the above diagram). How to evaluate the volume of this oblique frustum with elliptical section?
Any help is greatly appreciated!
|
We will not only find the volume but prove that the cross-section is indeed an ellipse.
We look for the intersection (see figure) of the cone:
$$
x^2+y^2=z^2\,\tan^2\alpha\tag1
$$
with the plane:
$$
z=\frac d{\sin\theta}+\frac x{\tan\theta}\tag2
$$
Substituting $(2)$ into $(1)$ results in:
$$\begin{align}
&x^2+y^2=\left(\frac d{\sin\theta}+\frac x{\tan\theta}\right)^2\tan^2\alpha\\
&\iff
\lambda^2\left(x-\frac {d\,\tan^2\alpha}{\lambda^2\sin\theta\tan\theta}\right)^2+y^2=
\frac{d^2\tan^2\alpha}{\sin^2\theta}\left(1+\frac{\tan^2\alpha}{\lambda^2\tan^2\theta}\right)\tag3
\end{align}
$$
with $\displaystyle\lambda=\sqrt{1-\frac{\tan^2\alpha}{\tan^2\theta}}$.
The expression $(3)$ is the equation of an ellipse with
$$
b^2=\frac{d^2\tan^2\alpha}{\sin^2\theta}\left(1+\frac{\tan^2\alpha}{\lambda^2\tan^2\theta}\right)=\frac{d^2\tan^2\alpha}{\lambda^2\sin^2\theta},\quad
a=\frac b\lambda
$$
and the corresponding area
$$
A=\pi ab=\pi\frac{b^2}\lambda
=\pi\frac{d^2\tan^2\alpha}{\lambda^3\sin^2\theta}\tag4.
%=\frac{\pi d^2\tan^2\alpha\sin\theta}{\left(\sin^2\theta-\cos^2\theta\tan^2\alpha\right)^\frac32},\tag4
$$
Observe that this is the area of the ellipse in the plane $z=0$ (marked red in the figure). The area of the corresponding ellipse (marked blue in the figure) in the plane $(2)$ is
$$
A'=\frac A{\sin\theta}\tag5,
$$
so that finally
$$
V=\frac13A'd=\frac{\pi d^3\tan^2\alpha}{3\lambda^3\sin^3\theta}
=\frac{\pi d^3\sin^2\alpha\cos\alpha}{3\left[\sin(\theta+\alpha)\sin(\theta-\alpha)\right]^\frac32}.\tag6
$$
Substituting in $(6)$ the given values of $d,\theta,\alpha$ one obtains:
$$
V=2000\sqrt\frac23\pi~\text{cm}^3.
$$
PS. $d=h$ from the question.
|
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|
Integration: $\int\frac{1}{(x^2+x+1)^{1/2}} dx$ Find the value of $$\int\frac{1}{(x^2+x+1)^{1/2}} dx$$
Anyone can provide hint on how to integrate this, and how you know what method to use? (I mean, is there any general guideline to follow for solving?)
Thank you!
|
If you make the Euler substitution $t=x+\sqrt{x^2+x+1}$, then $\displaystyle x=\frac{t^2-1}{2t+1}$ and $\displaystyle dx=\frac{2(t^2+t+1)}{(2t+1)^2}dt$;
so $\displaystyle\int\frac{1}{\sqrt{x^2+x+1}}dx=\int\frac{1}{t-\frac{t^2-1}{2t+1}}\cdot\frac{2(t^2+t+1)}{(2t+1)^2}dt=\int\frac{2}{2t+1}dt=\ln\lvert2t+1\lvert+C$
$\displaystyle=\ln\left(2x+2\sqrt{x^2+x+1}+1\right)+C$
|
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|
Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful.
I need to find the angle "x" which is:
a)$\frac{\pi}{8}\quad$ b)$\frac{\pi}{6}\quad$ c)$\frac{\pi}{4}\quad$ d)$\frac{5\pi}{6}\quad$ e)$\frac{3\pi}{4}\quad$ f)$\frac{3\pi}{8}$
|
If $\sin^2y=\sin^2A\iff\cos2y=1-2\sin^2y=\cdots=\cos2A$
$\iff2y=2m\pi\pm2A\iff y=m\pi\pm A$ where $m$ is any integer
We have $\sin^2(2x)=\dfrac12=\left(\sin\dfrac\pi4\right)^2$
$\implies2x=r\pi\pm\dfrac\pi4=\dfrac\pi4\left(4r\pm1\right)$ where $r$ is any integer
|
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|
Write a function as $\sum _{n=0} ^{\infty} a_n x^n$ We have $f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4$.
Now I want to write this as $\sum _{n=0} ^{\infty} a_n x^n$.
What I got:
$f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4 = x^4 (1+ x + x^2 + x^3 + x^4 + x^5)^4$
We know: $\frac{1}{1-x} = \sum _{n=0} ^{\infty} x^n$.
Note that $\displaystyle \frac{d^3}{dx^3} (\frac{1}{1-x})= \frac{6}{(1-x)^4}$.
Then $\frac{1}{(1-x)^4} = \frac{1}{6} \sum n (n-1) (n-2) x^{n-3}$, what gives
$x^4 \frac{1}{(1-x)^4} = \frac{1}{6} \sum n (n-1) (n-2) x^{n+1} = \frac{1}{6} \sum (n-1) (n-2) (n-3) x^{n}$
If I look now on Wolfram Alpha at the expanded form, this expression is only correct for $4 \leq n \leq 9$. For $n \geq 10$ it gives other values of the coefficients. What goes wrong?
|
(can't leave a comment)
You'll want the multinomial series formula for this purpose:
$$\left(\sum_{j=1}^k a_j\right)^n=\sum_{\stackrel{n_j \geq 0}{\sum_{j=1}^k n_j=n}}\frac{n!}{\prod_{j=1}^k n_j!}\prod_{j=1}^k a_j^{n_j}$$
where the condition on the sum on the right means to sum all over the integer partitions of the power $n$.
|
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|
$a^2 = 2b^3 = 3c^5$ Find the smallest value of $abc$. We have following equation:
$a^2 = 2b^3 = 3c^5$
Where $a, b, c$ are natural numbers.
Find the smallest possible value of product $abc$.
|
Introducing any prime factors apart from $2$ and $3$ makes things worse. Therefore we may assume $a=2^\alpha3^\beta$, which leads to
$$b^3=2^{2\alpha-1}3^{2\beta},\qquad c^5=2^{2\alpha}3^{2\beta-1}\ .$$
therefore we have to find the smallest $\alpha\geq0$, $\beta\geq0$ such that
$$2\alpha-1=0\quad(3),\qquad 2\alpha=0\quad(5)\ ,$$
and
$$2\beta=0\quad(3),\qquad2\beta-1=0\quad(5)\ .$$
The obvious solutions are $\alpha=5$, $\>\beta=3$, and lead to the triple $(a,b,c)=(864,72,12)$, so that the minimal product is $abc=746\,496$.
|
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|
Number of times $2^k$ appears in factorial
For what $n$ does: $2^n | 19!18!...1!$?
I checked how many times $2^1$ appears:
It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$
I checked how many times $2^2 = 4$ appears:
It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$
I checked how many times $2^3 = 8$ appears:
It appears in, $8!, 9!, ..., 19!$ meaning, $8^{12} = 2^{36}$
I checked how many times $2^{4} = 16$ appears:
It appears in, $16!, 17!, 18!, 19!$ meaning, $16^{4} = 2^{16}$
In all,
$$2^{18} \cdot 2^{32} \cdot 2^{36} \cdot 2^{16} = 2^{102}$$
But that is the wrong answer, its supposed to be $2^{150}$?
|
A simple trick to compute $k$ such that $2^k|n!$ is to compute $\sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor$, this is because $n$ has $[n/2]$ numbers divided by $2$, if we pick out these numbers and find out that there're $[n/4]$ numbers divided by $4$.. If we continue this procedure, we see that $$k=1\cdot\left(\left\lfloor\frac{n}{2}\right\rfloor-\left\lfloor\frac{n}{4}\right\rfloor\right)+2\left(\left\lfloor\frac{n}{4}\right\rfloor-\left\lfloor\frac{n}{8}\right\rfloor\right)+\ldots=\sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor$$. In this case, we have to sum
$$0+1+1+3+3+4+4+7+7+8+8+10+10+11+11+15+15+16+16=150.$$
Your fault is that your did not count the contribution of those which is not the power of $2$. For instance, there's $14$ in $14!$..
|
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|
Am I misinterpreting this matrix determinant property? I was reading matrix determinant properties from wikipedia.
The property reads
$\det(cA) = c^n \det(A)$ for $n \times n$ matrix.
However I am not able to realize it. What I find is $\det(cA) = c\det(A)$
For example, multiplying matrix by 2 and then taking the determinant of the resultant matrix:
$
2\begin{bmatrix}
4 & 5 & 6 \\
6 & 5 & 4 \\
4 & 6 & 5 \\
\end{bmatrix}=
\begin{bmatrix}
8 & 10 & 12 \\
6 & 5 & 4 \\
4 & 6 & 5
\end{bmatrix}
$
and
$
\begin{vmatrix}
8 & 10 & 12 \\
6 & 5 & 4 \\
4 & 6 & 5
\end{vmatrix}=60
$
Now first taking the determinant and then multiplying by 2 yields the same result:
$$
2\begin{vmatrix}
4 & 5 & 6 \\
6 & 5 & 4 \\
4 & 6 & 5 \\
\end{vmatrix}
= 2 \cdot 30 = 60
$$
Where I am mistaking?
|
$$2 \begin{bmatrix} 4 & 5 & 6 \\ 6 & 5 & 4 \\ 4 & 6 & 5 \end{bmatrix} =
\begin{bmatrix} 8 & 10 & 12 \\ 12 & 10 & 8 \\ 8 & 12 & 10 \end{bmatrix}.$$
You only multiplied the first row by 2.
|
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|
Factorize Trigonometric Equation: $ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $ I have a problem with the following trigonometric equation:
$$ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $$
It's from the book Engineering Mathematics 7th edition by Stroud.
The book is giving the answer, but I can't seem to be able to find out how to factorize it. I can't figure it out.
Consider the following solution:
This equation can be written as $3\sin ^2x-\cos ^2x = 2\sin x \cos x.$
That is: $3\sin ^2x-2\sin x \cos x-\cos ^2 x = 0.$
That is: $(3\sin x + \cos x)(\sin x - \cos x) = 0.$
So that $3 \sin x \cos x = 0$ or $\sin x - \cos x = 0.$
If $3 \sin x + \cos x = 0,$ then $\tan x = \frac{-1}{3},$ and so $x = -0.3218 ± n \pi,$ and if $\sin x - \cos x = 0,$ then $\tan x = 1,$ and so $x = \frac{\pi}{4}.$
If anyone could help me understand how to factorize this equation to get the one shown in the image it would help me very much.
Thank you in advance.
|
Let $u = \sin x$; let $v = \cos x$. Then the equation
$$3\sin^2x - 2\sin x\cos x - \cos^2x = 0$$
becomes
$$3u^2 - 2uv - v^2 = 0$$
To split the linear term, we must find two numbers with product $3 \cdot -1 = -3$ and sum $-2$. They are $-3$ and $1$. Hence,
\begin{align*}
3u^2 - 2uv - v^2 & = 0\\
3u^2 - 3uv + uv - v^2 & = 0 && \text{split the linear term}\\
3u(u - v) + v(u - v) & = 0 && \text{factor by grouping}\\
(3u + v)(u - v) & = 0 && \text{extract the common factor}
\end{align*}
Now, substitute $\sin x$ for $u$ and $\cos x$ for $v$.
|
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|
Sum limit. Please tell if correct. I just solved this limit, and the result provided by the book is different.
$$ \lim_{x\to 1} \frac{x+x^2+x^3 + ... + x^n - n}{x-1} $$
I turned this into:
$$ \lim_{x\to 1} \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \frac{x^3-1}{x-1} + ... + \frac{x^n-1}{x-1} $$
$$ \lim_{x\to 1} 1 + x + (x + x^2) + (x + x^2 + x^3) +...+(x + x^2 + ... + x^{n-1})$$
$$ 1 + 1 + 2 +3 +...+ n-1 $$
My result: $ \frac{n(n-1)}{2}+1 $
Books result: $ \frac{n(n-1)}{2} $
Have i missed something?
|
Your method of solving is correct and nice, but there are some mistakes. First
$$\frac{x^k-1}{x-1}=x^{k-1}+x^{k-2}+\ldots+x+1$$
and thus
$$\frac{x+x^2+\ldots+x^n-n}{x-1}=\sum_{k=1}^n\frac{x^k-1}{x-1}=$$
$$=\sum_{k=1}^n\left(1+x+\ldots+x^{k-1}\right)=1+2+\ldots+n=\frac{n(n+1)}2$$
One can also do the following, though perhaps you haven't yet studied this: define
$$f(x)=x+x^2+\ldots+x^n\;,\;\;n\in\Bbb N\;\;\text{a constant}$$
Then your limit is this functions' derivative at $\;x=1\;$ :
$$\lim_{x\to1}\frac{x+x^2+\ldots+x^n-n}{x-1}=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=f'(1)$$
and since $\;f'(x)=1+2x+3x^2+\ldots+nx^{n-1}\;$ , we get that
$$f'(1)=1+2+3+\ldots+n=\frac{n(n+1)}2$$
|
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|
How can a binomial coefficient can be approximated by using Stirling's formula? I've met some difficulties with such question:
How can we approximate a binomial coefficient by using a Stirling's factorial approximation.
I've evaluate a little bit and got this
How can I transform the right side of this equation for getting estimation like (1 + ?/n + O(1/n^2))
|
Thank you all for your answers, but I mentioned that I have to use slightly different factorial approximation:
$$
n!=\sqrt{2\pi n}\cdot(\frac{n}{e})^n\cdot\left(1+\frac{1}{12n}+\frac{1}{288n^2}+O(\frac{1}{n^3})\right)
$$
So, by using this approximation it's necessary to obtain approximation for the binomial coefficient $\binom{an}{n}$. Sorry that I defined the task conditions rudely:
$a$ - is a constant and $n\rightarrow \infty$.
So, after doing some conversions I've got this one:
$$\sqrt{\frac{a}{2\pi(a-1)n}}\cdot\left(\frac{a^a}{(a-1)^{a-1}}\right)^n\cdot\frac{1+\frac{1}{12an}+\frac{1}{288(an)^2}}{\left({1+\frac{1}{12n}+\frac{1}{288n^2}}\right)\cdot\left(1+\frac{1}{12(a-1)n}+\frac{1}{288((a-1)n)^2}\right)}$$
Then let's transform the right side of the expression:
$$\frac{1+\frac{1}{12an}+\frac{1}{288(an)^2}}{\left({1+\frac{1}{12n}+\frac{1}{288n^2}}\right)\cdot\left(1+\frac{1}{12(a-1)n}+\frac{1}{288((a-1)n)^2}\right)}=\\=\displaystyle{\frac{1+\frac{1}{12an}+O(\frac{1}{n^2})}{1+\frac{a}{12(a-1)n}+O(\frac{1}{n^2})}}$$
Finally, we get:
$$\displaystyle{1+X=\frac{1+\frac{1}{12an}}{1+\frac{a}{12(a-1)n}}}\rightarrow X=\frac{a-1-a^2}{a(12(a-1)n+a)}$$So, the right side of our approximation can be submitted as:
$$1+\frac{a-1-a^2}{12a(a-1)}\cdot\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$
And the whole answer is:
$$\sqrt{\frac{a}{2\pi(a-1)n}}\cdot\left(\frac{a^a}{(a-1)^{a-1}}\right)^n\cdot\left(1+\frac{a-1-a^2}{12a(a-1)}\cdot\frac{1}{n}+O\left(\frac{1}{n^2}\right)\right)$$
|
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|
Show that if $p$ is an odd prime, then the congruence $x^2\equiv1\pmod{p^{\alpha}}$ has only two solutions, $x\equiv1,x\equiv-1\pmod{p^{\alpha}}$.
Show that if $p$ is an odd prime, then the congruence $x^2 \equiv 1 \pmod{p^{\alpha}}$ has only two solutions, which are $x \equiv 1, x \equiv -1 \pmod{p^{\alpha}}$.
Clearly $x \equiv 1, x \equiv - 1 \pmod{p^{\alpha}}$ are solutions. We'll show that there are no other solutions. Suppose that $x^2 \equiv 1 \pmod{p^{\alpha}}$. Then
$$
\begin{align}
x^2-1 \equiv 0 \pmod{p^{\alpha}} &\Longleftrightarrow (x+1)(x-1) \equiv 0 \pmod{p^{\alpha}} \\
&\Longrightarrow p^{\alpha} \mid (x+1)(x-1) \\
&\Longrightarrow p \mid (x+1)(x-1)
\end{align}
$$
By Euclid's Lemma, if a prime divides a product $ab$, then it must divide either $a$ or $b$. So $p \mid (x+1)$ or $p \mid (x-1)$. Suppose $p \mid x + 1$. Since $x +1$ and $x-1$ differ by a factor of $2$ and $p > 2$ ($p$ odd), it follows that $\gcd(p^{\alpha},x-1) = 1$. Thus $p^{\alpha} \mid x + 1 \Longrightarrow x \equiv -1 \pmod{p^{\alpha}}$. Similarly, if $p \mid (x-1)$, then $\gcd(p^{\alpha},x+1) = 1 \Longrightarrow p^{\alpha} \mid (x-1) \Longrightarrow x \equiv 1 \pmod{p^{\alpha}} \text{. } \Box$
Is my proof correct? Criticism appreciated.
|
IMHO there is something missing. Your last implication says, in effect, that if $p\mid x-1$ then $p^\alpha\mid x-1$, which is certainly not true.
The point you need to make (earlier in the proof) is this: since $x+1$ and $x-1$ differ by $2$, they cannot both be multiples of $p$ (because $p$ is greater than $2$). Therefore, the $p^\alpha$ which is a factor of $(x-1)(x+1)$ must be entirely a factor of $x-1$, or entirely a factor of $x+1$.
|
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|
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.
I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find $f(x\pm c)$ which is irreducible:
$f(x+1)=x^4+4x^3+2x^2-4x+13$, but 13 has the divisors: 1 and 13, so don't exist a prime number p such that to apply the first condition: $p|a_i, i\ne n$; the same problem for $f(x-1)=x^4+...+13$
For $f(x+2)=x^4+8x^3+20x^2+16x+16$ is the same problem from where we go, if we set p=2, that means $2|8, 2|20, 2|16$, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for x-2
Now I'll verify for $f(x\pm3)$, but I think it will be fall... I think if I verify all constant $f(x\pm c)$ it doesn't work with this method... so have any idea how we can prove that $f$ is irreducible?
|
You've seen that $f(x)$ has no roots, so you want to exclude factorizations of the form
$$f(x) = (x^2 + ax + b)(x^2 + cx + d)$$
Since $f(x) = f(-x)$, the above implies
$$f(x) = (x^2 - ax + b)(x^2 - cx + d)$$
Here $a,b,c$, and $d$ are integers by Gauss's Lemma.
So a given root $r$ of $x^2 - ax + b$ is a root of $x^2 + ax + b$ or $x^2 + cx + d$.
If $r$ is a root of $x^2 + ax + b$, it is the root of the difference $x^2 + ax + b - (x^2 - ax + b) = 2ax$, which implies $a = 0$ since zero is not a root of $f(x)$.
If $r$ is a root of $x^2 + cx + d$ it is similarly a root of the difference $(c + a)x + (d - b)$, and since $f(x)$ has no rational roots $c = -a$ and $d = b$.
So either $a = 0$ or $c = -a$ and $d = b$. Since the argument is entirely symmetric in the two factors, we also either have $c = 0$ or $c = -a$ and $d = b$. Hence we have two possibilities: Either $a = c = 0$ or $c = -a$ and $d = b$.
In the first case we have
$$x^4 - 4x^2 + 16 = (x^2 + b)(x^2 + d)$$
But the roots of $y^2 - 4y + 16$ are irrational (they're not even real) so this can't happen.
In the second case we have
$$x^4 - 4x^2 + 16 = (x^2 + ax + b)(x^2 - ax + b) = x^4 + (2b - a^2) x^2 + b^2$$
Hence $b = \pm 4$ and therefore either $8 - a^2 = -4$ or $-8 - a^2 = -4$, neither of which has rational solutions.
Hence $f(x)$ is irreducible.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is Vieta the only way out? Let $a,b,c$ are the three roots of the equation $x^3-x-1=0$. Then find the equation whose roots are $\frac{1+a}{1-a}$,$\frac{1+b}{1-b}$,$\frac{1+c}{1-c}$.
The only solution I could think of is by using Vieta's formula repeatedly which is no doubt a very messy solution. Is there any easier and more slick way of doing this ?
|
Transform the equation. Since all the roots are symmetric, say $$y=\frac {1+x}{1-x}\implies x(y+1)=y-1\implies x=\frac {y-1}{y+1}$$ Substitute this expression in place of $x$ in the original equation and simplify. $$\require{cancel}\begin{align}f(y)&=\biggl(\frac {y-1}{y+1}\biggr)^3-\frac {y-1}{y+1}-1=0\\&\implies(y-1)^3-(y+1)^3-(y-1)(y+1)^2=0\\&\implies (y+1)^3-(y-1)^3+(y-1)(y+1)^2=0\\&\implies \cancel{y^3}+3y^2+\bcancel{3y}+1-\cancel{y^3}+3y^2-\bcancel{3y}+\xcancel{1}+y^3+y^2-y-\xcancel{1}=0\\&\implies y^3+7y^2-y+1=0\end{align}$$This is the required answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing the complex integral? I am dealing with the following:
$$\int_{0}^{\infty}\frac{x\sin(x)}{(x^2+a^2)(x^2+b^2)}dx$$
Furthermore, I know $a,b>0$ and I know $a\neq b$. I believe this is using Jordan's Lemma? I see that the singularities in the upper half of the plane are $ai$ and $bi$ where $R>a$. I'm stuck writing out my function of z. I think it should be: $$f(z)=\frac{1}{(z^2+a^2)(z^2+b^2)}.$$
But I do not know how exactly the $x\sin(x)$ numerator comes back into play.
|
$x\sin(x)$ is an even function, so your integral is equal to
$$
\int_0^\infty\frac{x\sin(x)}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x
=\frac12\int_{-\infty}^\infty\frac{x\sin(x)}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x
\tag{1}
$$
We can break up $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$ and compute
$$
\begin{align}
\frac12\int_{-\infty}^{\infty}\frac{z\sin(z)}{(z^2+a^2)(z^2+b^2)}\,\mathrm{d}x
&=\frac1{4i}\int_{\gamma^+}\frac{ze^{iz}}{(z^2+a^2)(z^2+b^2)}\,\mathrm{d}z\tag{2}\\
&-\frac1{4i}\int_{\gamma^-}\frac{ze^{-iz}}{(z^2+a^2)(z^2+b^2)}\,\mathrm{d}x\tag{3}
\end{align}
$$
where
$$
\gamma^\pm=[-R,R]\cup Re^{\pm i[0,\pi]}\tag{4}
$$
as $R\to\infty$, since the integrals along $Re^{\pm i[0,\pi]}$ vanishes.
Now, we can use
$$
\frac1{z^2+a^2}=\frac1{2ia}\left(\frac1{z-ia}-\frac1{z+ia}\right)\tag{5}
$$
To get that the sum of residues inside $\gamma^+$ is
$$
\overbrace{\frac1{2ia}\frac{ia\,e^{-a}}{b^2-a^2}}^{\large\text{at $z=ia$}}\overbrace{+\frac1{2ib}\frac{ib\,e^{-b}}{a^2-b^2}}^{\large\text{at $z=ib$}}\tag{6}
$$
and the sum of residues inside $\gamma^-$ is
$$
\overbrace{-\frac1{2ia}\frac{-ia\,e^{-a}}{b^2-a^2}}^{\large\text{at $z=-ia$}}\overbrace{-\frac1{2ib}\frac{-ib\,e^{-b}}{a^2-b^2}}^{\large\text{at $z=-ib$}}\tag{7}
$$
Combining $(2)$ with $(6)$ and $(3)$ with $(7)$, remembering $(1)$ and the direction of the contours, we get
$$
\begin{align}
\int_0^\infty\frac{x\sin(x)}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x
&=\frac{2\pi i}{4i}\left(\frac{e^{-a}}{b^2-a^2}+\frac{e^{-b}}{a^2-b^2}\right)\\
&=\frac\pi2\frac{e^{-a}-e^{-b}}{b^2-a^2}\tag{8}
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the center
$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
This means that $\frac{\sqrt{b^2 - 4ac}}{2a}$ is the (horizontal) distance from the vertex to the roots. If I squint, the two squared-ish quantities being subtracted under a square root sign reminds me of the Euclidean distance formula
$$\sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2}$$
Is there a connection? If not, is there any intuitive or geometric reason why $\frac{\sqrt{b^2 - 4ac}}{2a}$ should be the horizontal distance from the vertex to the roots?
|
Suppose that $a\ne 0$ and $b^2-4ac\geq 0$. Let $f(x)=ax^2+bx+c$ and
$$
x_0= -\frac{b}{2a},\; x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, \;x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}.
$$
It follows that
$$
y_0=f(x_0)=\frac{-b^2+4ac}{4a},\;y_1=f(x_1)=0, \; y_2=f(x_2)=0.
$$
The distances from vertex to roots:
$$
d_1=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}=\sqrt{\left(-\frac{b}{2a}-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b^2+4ac}{4a}\right)^2}=\frac{\sqrt{(b^2-4ac)^2+4(b^2-4ac)}}{4|a|}>\frac{\sqrt{b^2-4ac}}{2|a|},
$$
$$
d_2=\sqrt{(x_2-x_0)^2+(y_2-y_0)^2}=\sqrt{\left(-\frac{b}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b^2+4ac}{4a}\right)^2}=\frac{\sqrt{(b^2-4ac)^2+4(b^2-4ac)}}{4|a|}>\frac{\sqrt{b^2-4ac}}{2|a|}.
$$
|
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|
How to evaluate the $\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series? How to evaluate the $\displaystyle\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series?
It made sense to first try and build the numerator using power series that are commonly used:
$\displaystyle2\sin(x)=\sum_{k=0}^\infty \dfrac{2(-1)^kx^{2k+1}}{2k+1!} = 2x -\frac{x^3}{3}+\frac{x^6}{60} + \dotsb$
$\displaystyle-\arctan(x)=\sum_{k=0}^\infty \dfrac{(-1)^{k+1}x^{2k+1}}{2k+1} = -x +\frac{x^3}{3}-\frac{x^6}{6} + \dotsb$
$\displaystyle-x\cos(x^2)=\sum_{k=0}^\infty \dfrac{(-1)^{k+1}x^{4k+1}}{2k!} = -x +\frac{x^5}{2}+ \dotsb$
Hence,
$\displaystyle\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5} =
\lim\limits_{x\to 0} \dfrac{[2x -\frac{x^3}{3}+\frac{x^6}{60} + \dotsb] + [-x +\frac{x^3}{3}-\frac{x^6}{6} + \dotsb] + [x +\frac{x^5}{2}+ \dotsb]} {x^5}$
In similar problems, the there is an easy way to take out a common factor that would cancel out with the denominator, resulting in an easy-to-calculate limit. Here, however, if we were to take a common factor from the numerator, say, $x^6$, then we would end up with an extra $x$
What possible strategies are there to solve this question?
|
if you add up the expansions you have given us, you get 1/2 +0. You take a common factor of $x^5$ from the numerator and divide through you get 1/2+ x(...) --> 1/2
|
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|
Computing $\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2 + x + 2} \, dx$ I wish to compute
$$\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2 + x + 2} \, dx$$
The singularities are $\pm \frac{\sqrt{7}}{2}i - \frac{1}{2}$
I then make a half-circle contour $C = \{Re^{it}, t \in [0,\pi], R \in [0,\infty) \}$
$$ \int_{-\infty}^{\infty} \frac{\cos(x)}{((x - (-\frac{\sqrt{7}i}{2} - \frac{1}{2}))(x - (\frac{\sqrt{7}i}{2} - \frac{1}{2}))} dx = \int_{C} \frac{\cos(x)}{((x - (-\frac{\sqrt{7}i}{2} - \frac{1}{2}))(x - (\frac{\sqrt{7}i}{2} - \frac{1}{2}))} dx$$
Is this correct so far?
Then we can apply Cauchy Integral Formula yes?
|
We have $x^2+x+2 = (x+a)(x+b)$, where $a+b=1$ and $ab=2$. Hence, we have
$$\dfrac{\cos(x)}{x^2+x+2} = \dfrac{\cos(x)}{(x+a)(x+b)} = \dfrac1{b-a}\left(\dfrac{\cos(x)}{x+a} - \dfrac{\cos(x)}{x+b}\right)$$
Our integral is of the form
$$I = \dfrac1{b-a}\int_{-\infty}^{\infty}\left(\dfrac{\cos(x)}{x+a} - \dfrac{\cos(x)}{x+b}\right)dx = \dfrac{F(a) - F(b)}{b-a}$$
where $F(y) = \displaystyle\int_{-\infty}^{\infty} \dfrac{\cos(x)}{x+y}dx$. We have
$$F(y) = \int_{-\infty}^{\infty} \dfrac{\cos(t-y)}t dt = \cos(y)\int_{-\infty}^{\infty} \dfrac{\cos(t)}tdt + \sin(y)\int_{-\infty}^{\infty} \dfrac{\sin(t)}tdt$$
Note that the principal value of $\displaystyle \int_{-\infty}^{\infty} \dfrac{\cos(t)}tdt$ is zero, since the integrand is odd. This gives us
$$F(y) = \sin(y) \int_{-\infty}^{\infty} \dfrac{\sin(t)}tdt = \pi \sin(y)$$
Hence, we have
$$I = \pi \cdot \dfrac{\sin(a)-\sin(b)}{b-a}$$
|
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|
Solve the following recurrence relation: $S(1) = 2$; $S(n) = 2S(n-1)+n2^n, n \ge 2$ Solve the following recurrence relation:
$$\begin{align}
S(1) &= 2 \\
S(n) &= 2S(n-1) + n 2^n, n \ge 2
\end{align}$$
I tried expanding the relation, but could not figure out what the closed relation is:
+-------+------------------------------------------------+
| n | S(n) |
+-------+------------------------------------------------+
| 1 | 2 |
| 2 | 2S(1)+2*2^2 = 2*2 + 2*2^2 = 2^2 + 2*2^2 = |
| | = 4 + 8 = 12 |
| 3 | 2S(2)+3*2^3 = 2*(2^2 + 2*2^2) + 3*2^3 = |
| | = 8 + 16 + 24 = 48 |
| 4 | 2S(3)+4*2^4 = 2*(2*2^2 + 2*2*2^2 + 3*2^3) + |
| | + 4*2^4 = |
| | = 16 + 32 + 48 + 64 = 160 |
| 5 | 2S(4)+5*2^5 = 2*(2*2*2^2 + 2*2*2*2^2 + |
| | + 2*3*2^3 + 4*2^4) + 5*2^5 = |
| | = 32 + 64 + 96 + 128 + 160 |
| | = 384 + 96 = 480 |
+-------+------------------------------------------------+
Edit: is it $S(n)=2^n\frac{n(n+1)}{2}$?
|
We can write $S(n)=2+2(2^2)+3(2^3)+\cdots+n(2^n)$. However, we can simplify the problem slightly by generalizing: replace $2$ with $x$ to get
$$f_n(x)=x+2x^2+3x^3+\cdots+ nx^n = x(1+2x+3x^2+\cdots+nx^{n-1})=x \frac{d}{dx}(1+x+\cdots+x^n)$$
Now, using the formula for a geometric sum, $1+x+\cdots+x^n=\frac{1-x^{n+1}}{1-x}$, and differentiating we have
$$f_n(x)=x \frac{d}{dx}\left(\frac{1-x^{n+1}}{1-x}\right)=x\frac{(1-x)(-(n+1)x^{n})+(1-x^{n+1})}{(1-x)^2}$$
Plugging back in with $x=2$, we have $$S(n)=f_n(2)=2((n+1)2^n+1-2^{n+1})=2((n-1)2^n+1)=(n-1)2^{n+1}+2.$$
I believe there was a typo in the problem, I solved $S(1)=2;S(n)=S(n-1)+n2^n$, which was the formula in the problem but not in the title. What follows is a solution the the recurrence in the title.
Let $T(n)=S(n)/2^n$. Then dividing the formula $S(n)=2S(n-1)+n2^n$ by $2^n$, we get
$$T(n)=T(n-1)+n,\qquad T(1)=1.$$
Plugging in and using the standard formula for triangular number, $T(n)=1+2+3+\cdots+n=\frac{n(n+1)}{2}$. Thus, $S(n)=n(n+1)2^{n-1}$
|
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|
Convert Riemann sum to definite integral $ \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} $ The limit
$\quad\quad \displaystyle \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} $
is the limit of a Riemann sum for a certain definite integral
$\quad\quad \displaystyle \int_a^b f(x)\, dx $
a = ?
b = ?
f(x) = ?
Attempt at solution:
So I know:
$ \displaystyle \sum_{i=1}^{n} f(a + i dx) dx $ = $ \int_a^b f(x)\, dx $
and that $x_i = a + i \cdot \frac {b-a}{n}$
So I will try to rewrite my riemann sum as convenient:
$\frac {3}{n} (7 \cdot \frac {3i}{n} + 6) $
so my $dx$ is $\frac {3}{n}$
since +6 is supposed to be my a, b is therefore 9 because $dx = \frac {b-a}{n}$
but that 7 inside the parenthesis spoils everything... so I can't say what's inside the parenthesis is equal to $x_i$
So what's next?
|
You were so close with $\frac {3}{n} (7 \cdot \frac {3i}{n} + 6)$. You can simply factor the $7$ outside the sum as follows:
Given the Riemann sum definition of the definite integral,
$$\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(a + i \, \mathrm{d}x) \, \mathrm{d}x = \int_a^b f(x)\, \mathrm{d}x$$
we manipulate to find $f(x)$:
\begin{align}
\lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} &= \lim_{n\to\infty}\sum_{i=1}^{n} \left(6 + 7 i \cdot\frac{3}{n} \right)\frac{3}{n} \\
&= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} 7\left(\frac{6}{7} + i \cdot\frac{3}{n} \right)\frac{3}{n} \\
&= 7\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \left(\frac{6}{7} + i \cdot\frac{3}{n} \right)\frac{3}{n}.
\end{align}
So, we have $\mathrm{d}x=\frac{3}{n}$, $a=\frac{6}{7}$, $b=3+a=\frac{27}{7}$ and $f(x)=x$.
Hence,
\begin{align}
\lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} &=7\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \left(\frac{6}{7} + i \cdot\frac{3}{n} \right)\frac{3}{n} \\
&= 7 \int_{\frac{6}{7}}^{\frac{27}{7}} x \,\mathrm{d}x \\
&= 7 \left[\frac{1}{2} x^2 \right]_{\frac{6}{7}}^{\frac{27}{7}} \\
&= 7 \left[\frac{729}{98} - \frac{36}{98} \right] \\
&= 49.5
\end{align}
|
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|
Integral-derivative issue The derivative of $sin^2(x)$ is $2sin(x)cos(x)$. You can also write it as $sin(2x)$.
If we integrate $\sin(2x)$ we get $-0.5\cos(2x)$ and according to calculator does not equal $\sin^2(x)$. Help?
|
Both functions ($\sin^2$ and $-\frac 12 \cos(2\cdot)$) differ by a constant, as any two function $\mathbf R \to \mathbf R$ having the same derivative do. Note that
\begin{align*}
\cos(2x) &= \cos^2 x - \sin^2 x\\
&= 1 - \sin^2 x - \sin^2 x\\
&= 1 - 2\sin^2 x\\
\iff \sin^2 x &= \frac 12 - \frac 12\cos(2x)
\end{align*}
|
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|
Evaluate $ \lim_{x\to 0} \frac{\tan(4x)}{\sin(7x)}$ Evaluate $$ \lim_{x\to 0} \ \frac{\tan(4x)}{\sin(7x)}$$
I am stuck after I convert tan(4x) into sin (4x) / cosine(4x)
|
Recall that $\lim_{x\to0} \sin(x)/x = 1$ and $\lim_{x\to 0} \tan(x)/x = 1$.
Then $$\lim_{x\to 0} \frac{\tan(4x)}{\sin(7x)} = \lim_{x\to 0} \frac{7}{7} \cdot \frac{4x}{4x} \cdot \frac{\tan(4x)}{\sin(7x)} = \lim_{x\to 0} \frac{4}{7} \frac{\tan(4x)}{4x} \cdot \frac{7x}{\sin(7x)} = \frac47 \cdot 1 \cdot 1 = \frac47.$$
Once we accept that $\lim_{x\to0} \frac{\sin(x)}{x} =1$ it immediately follows that $\lim_{x\to0} \frac{\tan(x)}{x} = 1$, since $$\lim_{x\to0} \frac{\tan(x)}{x} = \lim_{x\to0} \frac{\sin(x)}{x} \cdot \frac{1}{\cos(x)} = 1 \cdot 1 = 1$$
|
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|
Maclaurin series for a function Provided I have the function
\begin{equation*}
f(x)=(1+x)^{1/x},
\end{equation*}
and I want to calculate a 3rd order Maclaurin series, how can that be done without taking direct derivatives (as this seems hard..). I know that
\begin{equation*}
(1+x)^{1/x}=e^{ln(1+x)/x},
\end{equation*}
and the Maclaurin series for $e^x$ is easy to prove, so I think it's a good direction..
|
$$\frac{\ln(1+x)}x=1-\frac x2+\frac{x^2}3-\frac{x^3}4+o(x^3)$$
Substitute $-\frac x2+\frac{x^2}3-\frac{x^3}4$ to $u$ in the development of $\mathrm e^u$, first computing the succesive powers of $u$:
\begin{align*}
u^2&=\frac{x^2}4-\frac{x^3}3+o(x^3),\\
u^3&-\frac{x^3}8+o(x^3),
\end{align*}
so that $$(1+x)^{\tfrac 1x}=\mathrm e\Bigl(1+u+\frac{u^2}2+\frac{u^3}6+o(u)\Bigr)=\mathrm e\Bigl(1-\frac x2+\frac{11}{48}x^2-\frac 7{16}x^3+o(x^3)\Bigr).$$
|
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|
Let $a,b$ be relative integers such that $2a+3b$ is divisible by $11$. Prove that $a^2-5b^2$ is also divisible by $11$. The divisibility for $11$ of $a^2 - 5b^2$ can be easily verified; in fact: $$a \equiv \frac {-3}{2}b \pmod {11}$$ therefore $$\frac {9}{4}\cdot b^2 - 5b^2 = 11(-\frac{b^2}{4}) \equiv 0 \pmod {11}.$$
The solution doesn't need the use of the rules of modular-arithmetic. How can I demonstrate it?
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My first thought was to think about a difference of squares: $(2a+3b)(2a-3b) = 4a^2 - 9b^2$. So if $2a+3b$ is divisible by 11, then $4a^2-9b^2$ as well. And therefore $4a^2-9b^2+11b^2 = 4a^2+2b^2$ is divisible by 11.
That's not what you asked about. But is there some number $k$ such that if $k(4a^2+2b^2)$ is divisible by 11, then $a^2-5b^2$ is as well? (Hint: yes.)
|
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|
How to solve 3 variable in 2 equation? This paper is abstracted from 2007 British Mathematics Olympiad Round 1 Question 2.
I am currently practicing grade 8 (Singapore Secondary 2) for the upcoming Singapore Mathematics Olympiad(SMO). Even before solving the question,is it even possible to solve 3 variable of 2 equation?(As from the title of the question).
Question:
Find all solutions in positive integers $x,y,z$ to the simultaneous equations.
$x+y-z=12$
$x^2+y^2-z^2=12$
Of course the first thing that I almost done is $x+y-z=x^2+y^2-z^2$ which is definitely not the way.
I cannot use elimination method as in the end there will be 2 variable in an equation.
I tried using substitution method like $x=12-y+z$,so,$(12-y+z)^2+y^2-z^2=12$ and substituting $x^2=12-y^2+z^2$ where $x=\sqrt {12-y^2+z^2}$,so,$\sqrt {12-y+z}+y-z=12$
Therefore,I get these equations.
*
*$(12-y+z)^2+y^2-z^2=12$
*$\sqrt {12-y+z}+y-z=12$
If I expand,these equation ended up
*
*$z^2-yz+y\sqrt {12-y^2+z^2}-z\sqrt {12-y^2+z^2}=78$ (Squared version) OR $\sqrt {12-y^2+z^2}+y-z=12$ (Non-Squared Version)
*$y^2-12y+12z-yz=-66$
From here I don't see how to eliminate one variable away.I am sure what I am doing is wrong or expanded wrongly.Any other ways to do this question?
|
$$x+y-12=z \Rightarrow x^2+y^2-(x+y-12)^2=12$$
This leads to
$$xy-12x-12y+78=0$$
or
$$(x-12)(y-12)=66$$
Now check all the possible ways of writing $66$ as a product of 2 integers.
|
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|
How many $n-$digit number that contain only digits $ 1,2,3,4,5,6$ How many $n- $digit numbers can be formed from the digits $1,2,3,4,5$ and 6, which contains the numbers $1$ and $2$ as neighbours.
Let $p_n$ be the number of n-digit numbers which consist only of the digits 1,2,3,4,5,6, which contains the numbers $1$ and $2$ as neighbours.
$$p_{n+1}=p_{n}+(?)$$
Now How to find $(?)$
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Start by making $n$-symbol words from the five symbols $[12],3,4,5,6$; there are $\binom{n+4}{4}$ ways to do this (use "stars and bars".) For each such word, you have two possible orderings of the $1$ and $2$. So
$$p_n=2\binom{n+4}{4}.$$
And $p_{n+1}-p_n=2\binom{n+5}{4}-2\binom{n+4}{4}=2\binom{n+4}{3}$.
|
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|
Trouble constructing $\mathbb Z_3[x]/(x^2+1)$ If I have $\mathbb Z_3[x]/(x^2+2x+2)$, I can construct a field by letting $x^2=x+1$. The reps are:
$0$
$1$
$x$
$x^2=x+1$
$x^3=x^2+x=x+1+x=2x+1$
$x^4=2x^2+x=2x+2+x=2$
$x^5=2x$
$x^6=2x^2=2x+2$
$x^7=2x^2+2x=2x+2+2x=x+2$
However, when I try this for $\mathbb Z_3[x]/(x^2+1)$, letting $x^2=-1=2$, it does not work. $x^2+1$ is irreducible over $\mathbb Z_3$ and thus we know $\mathbb Z_3[x]/(x^2+1)$ is a field.
Here's what I get
$0$
$1$
$x$
$x^2=2$
$x^3=2x$
$x^4=2x^2=4=1$
Why doesn't this work with $x^2+1$?
|
If I understand you correctly, the issue is that there's no guarantee that $x$ will be a multiplicative generator for the group of units for the resulting field. You might to choose another element. One choice might be $x+2$, which squares to $x^2+4x+4 \equiv x$.
|
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|
Extracting real and imaginary numbers from a complex number How can I get the real number and the imaginary number from: $$\frac{3+i}{5-12i}$$
|
$$\frac{3+i}{5-12i}=$$
$$\left|\frac{3+i}{5-12i}\right|e^{\arg\left(\frac{3+i}{5-12i}\right)i}=$$
$$\frac{\left|3+i\right|}{\left|5-12i\right|}e^{\arg\left(\frac{3}{169}+\frac{41}{169}i\right)i}=$$
$$\frac{\sqrt{3^2+1^2}}{\sqrt{5^2+12^2}}e^{\tan^{-1}\left(\frac{41}{3}\right)i}=$$
$$\frac{\sqrt{10}}{\sqrt{169}}e^{\tan^{-1}\left(\frac{41}{3}\right)i}=$$
$$\frac{\sqrt{10}}{13}e^{\tan^{-1}\left(\frac{41}{3}\right)i}=$$
$$\frac{\sqrt{10}}{13}\left(\cos\left(\tan^{-1}\left(\frac{41}{3}\right)\right)+\sin\left(\tan^{-1}\left(\frac{41}{3}\right)\right)i\right)=$$
$$\frac{\sqrt{10}}{13}\cos\left(\tan^{-1}\left(\frac{41}{3}\right)\right)+\frac{\sqrt{10}}{13}\sin\left(\tan^{-1}\left(\frac{41}{3}\right)\right)i=$$
$$\frac{\sqrt{10}}{13}*\frac{3}{13\sqrt{10}}+\frac{\sqrt{10}}{13}*\frac{41}{13\sqrt{10}}i$$
$$\frac{3}{169}+\frac{41}{169}i$$
|
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|
Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that:
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove the general case and then say that it is true for the specific case, but for the specific case I decided to write out the term and show that they were identical since they were so few.
For the LHS
$$\sum_{k = 0}^{4} (1+x)^k = (1+x)^0 + (1+x)^1 + (1+x)^2 + (1+x)^3 + (1+x)^4$$
$$ = (1) + (1 + x) + (x^2 +2x + 1) + (x^3 + 3x^2 + 3x + 1) + (x^4+4 x^3+6 x^2+4 x+1)$$
$$=5 + 10x + 10x^2 + 5x^3 + x^4$$
For the RHS:
$$\sum_{k=1}^5 \binom{5}{k}x^{k-1} = \binom{5}{1}x^{1-1} + \binom{5}{2}x^{2-1} + \binom{5}{3}x^{3-1} + \binom{5}{4}x^{4-1} + \binom{5}{5}x^{5-1}$$
$$= \frac{5!}{1!4!} x^0 + \frac{5!}{2!3!} x^1 + \frac{5!}{3!2!} x^2 + \frac{5!}{4!1!} x^3 + \frac{5!}{5!0!} x^4$$
$$ = 5 + 10x + 10x^2 + 5x^3 + x^4$$
That completes the first step of the question. So far so good.
For the general case, I started by using the binomial theorem for the binom and then writing out the inner-most sum:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k = 0}^{n-1} \sum_{i = 0}^{k} \binom{k}{i}x^i = \sum_{k = 0}^{n-1} \left( \binom{k}{0}x^0 + \binom{k}{1}x^1 + ... + \binom{k}{k}x^{k}\right)$$
I can imagine that each step in the sum will decide the coefficients for the various powers of x and thus be identical to the RHS of the general case.
However, I run of out steam here and do not at the moment see any obvious way forward. What are some productive approaches for the general case? Am I doing things too complicated?
|
Try induction in $n$, and then use $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} $.
|
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|
For which values does the Matrix system have a unique solution, infinitely many solutions and no solution? Given the system:
$$\begin{align}
& x+3y-3z=4 \\
& y+2z=a \\
& 2x+5y+(a^2-9)z=9
\end{align}$$
For which values of a (if any) does the system have a unique solution, infinitely many solutions, and no solution?
So I am getting that it has:
infinitely many solutions at: (-1)
No solution at (1)
Unique solution at (-1,1)
AM I RIGHT??
|
Note that your system is equivalent to the matrix equation
$$
\begin{bmatrix}
1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9
\end{bmatrix}
\begin{bmatrix}
x\\ y\\ z
\end{bmatrix}
=
\begin{bmatrix}
4\\ a\\9
\end{bmatrix}
$$
Since
$$
\det\begin{bmatrix}
1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9
\end{bmatrix}=a^2-1
$$
this system is guaranteed a unique solution for $a\neq\pm 1$ (do you know why?).
Now the augmented systems for $a=1$ is
$$
\begin{bmatrix}
1&3&-3&4\\ 0&1&2&1\\ 2&5&-8&9
\end{bmatrix}
$$
Row-reducing this matrix gives
$$
\DeclareMathOperator{rref}{rref}\rref\begin{bmatrix}
1&3&-3&4\\ 0&1&2&1\\ 2&5&-8&9
\end{bmatrix}
=
\begin{bmatrix}
1&0&-9&0\\ 0&1&2&0\\ 0&0&0&1
\end{bmatrix}
$$
This system is not consistent (why?) so the original system has no solution for $a=1$.
Can you repeat the process for $a=-1$?
Addendum. You mention in your question that you're having trouble taking determinants. To find the determinant computed above we can expand about the first column:
\begin{align*}
\det\begin{bmatrix}
\color{blue}1 & \color{red}3 & \color{red}{-3}\\ \color{blue}0 & \color{green}1 & \color{green}2 \\ \color{blue}2 & \color{purple}5 & \color{purple}{a^2-9}
\end{bmatrix}
&= (\color{blue}{1})\cdot\det
\begin{bmatrix}\color{green}1&\color{green}2\\\color{purple}5 & \color{purple}{a^2-9} \end{bmatrix}-(\color{blue}0)\det\begin{bmatrix}\color{red}3 & \color{red}{-3}\\ \color{purple}5 & \color{purple}{a^2-9}\end{bmatrix}+(\color{blue}{2})\det\begin{bmatrix}\color{red}3 & \color{red}{-3}\\ \color{green}1 & \color{green}2 \end{bmatrix} \\
&= \left(a^2-9-10\right)-(0)+2\,\left(6+3\right) \\
&= a^2-19+18 \\
&= a^2-1
\end{align*}
|
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|
Inequality for sides and height of right angle triangle
Someone recently posed the question to me for the above, is c+h or a+b greater, without originally the x and y lengths. I used this method: (mainly pythagorus)
$a^2+b^2=c^2=(x+y)^2=x^2+y^2+2xy$
$a^2=x^2+h^2$ and $b^2=y^2+h^2$
therefore $x^2+h^2+y^2+h^2=x^2+y^2+2xy$
$x^2+y^2+2h^2=x^2+y^2+2xy$
so $2h^2=2xy$ and $$xy=h^2$$
also $Area={ab\over 2}={ch\over 2}$ so $$ab=ch$$
$(a+b)^2=a^2+b^2+2ab$
$(c+h)^2=c^2+h^2+2ch=a^2+b^2+xy+2ab$
therefore $$(a+b)^2+xy=(c+h)^2$$
so $$c+h>a+b$$
I feel like I have made a mistake somewhere, is this incorrectly generalised? And is there an easier way to show the inequality. Also, if this is correct can it be expanded to non right angle triangle, I tried to do this using trig but was pretty much going in circles, thanks!
|
Perhaps easier, in the right triangle, once you note $ab=ch$, the smaller sum is when the terms are closer to each other, and clearly $c>a,b>h$, so $c+h>a+b$.
|
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|
Compute power of a matrix $A$ as $n\rightarrow \infty$ We are given $A^p=A ...A$(p times)
And we are given matrix A:
$A=\begin{vmatrix}0.6&-0.4&0\\-0.4&0.6&0\\0&0&0.5\end{vmatrix}$
I need to compute $A^p$ as p approach Infinity.
By some calculations results that the values of matrix A get always close to $0$ as n grows to infinity, but i actually need a proof(something concrete).
In the web i found a person saying that:
The usual technique for doing this (in cases where the limit exists) is to diagonalise the matrix. If $A=P^{-1}DP$ , where $P$ is invertible and $D$ is the diagonal matrix whose entries are the eigenvalues of $A$, then $A^n=p^{-1}D^nP$. The matrix $D^n$ is easy to compute: it is diagonal, and its entries are those of $D$ raised to the power $n$. The limit matrix will only exist if all the eigenvalues are either $1$ or have absolute value less than $1$. In that case, $D^n$ will converge to a limit matrix $D_0$, and $A^n$ will converge to $P^{-1}D_0P$ .
Now i will compute the Eigenvalues of my matrix:
$(0.6-\lambda)^3-0.08+0.16\lambda=(\lambda^2-1.2λ+0.02)(0.5-\lambda)=(0.02λ-0.2)(5\lambda-1)(0.5-\lambda)$
Therefore:
$\lambda_1= 1.$
$\lambda_2 = 0.5 $
$\lambda_3 = 0.2 $
But from now on I'm stuck.
|
Everything is in the passage you quoted.
THe matrix $A$ has three distinct real eigenvalues, hence it is diagonalisable:
$A=PDP^{-1}$, where $D$ is a digonal matrix with eigenvalues of $A$ and $P$ is a matrix whose columns are eigenvectors of $A$.
In your case it easy to find these eigenvectors: $(1,-1,0)^T$ corresponds to $\lambda_1=1$, $(1,1,0)^T$ corresponds to $\lambda_2=0.2$ and $(0,0,1)^T$ corresponds to $\lambda_3=0.5$. In other words, $D=\begin{pmatrix}1&0&0\\
0&0.2&0\\
0&0&0.5\end{pmatrix}$ and $P=\begin{pmatrix}1&1&0\\
-1&1&0\\
0&0&1\end{pmatrix}$.
Now we can say that $$A^N=PD^NP^{-1} = P\begin{pmatrix}1^N&0&0\\
0&0.2^N&0\\
0&0&0.5^N\end{pmatrix}P^{-1}\to P\begin{pmatrix}1&0&0\\
0&0&0\\
0&0&0\end{pmatrix}P^{-1}$$
$$=\frac 12\begin{pmatrix}1&1&0\\
-1&1&0\\
0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\
0&0&0\\
0&0&0\end{pmatrix}\begin{pmatrix}1&-1&0\\
1&1&0\\
0&0&1\end{pmatrix}=\frac 12 \begin{pmatrix}1&-1&0\\
-1&1&0\\
0&0&0\end{pmatrix}$$
|
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|
Integral giving wrong values Given following function:
$x^3 + 1$
(source: gyazo.com)
Find the area that is selected in red lines.
So I solved the root $\sqrt[3]{-1} = -1$ so $x = -1$
So now I have to create the 3 integrals:
$S_1 = \int_{-2}^{-1}\begin{pmatrix}0 - (x^3 + 1)\end{pmatrix}dx = \begin{bmatrix} \dfrac{-x^4}{4} - x\end{bmatrix}_{-2}^{-1}\textrm{}$
$\begin{bmatrix}
\dfrac{-(-2)^4}{4} + 2
\end{bmatrix} = -2$
$\begin{bmatrix}
\dfrac{-(-1)^4}{4} + 1
\end{bmatrix} = \dfrac{3}{4}$
$-2 + \dfrac{3}{4} = -\dfrac{3}{4}$
$ => S_1 = -1\dfrac{1}{4}$
$S_2 = \int_{-1}^{0}\begin{pmatrix}x^3 + 1\end{pmatrix}dx = \begin{bmatrix} \dfrac{x^4}{4} + x\end{bmatrix}_{-1}^{0}\textrm{}$
$\begin{bmatrix}
\dfrac{(-1)^4}{4} - 1
\end{bmatrix} = -1\dfrac{3}{4}$
$=> S_2 = -\dfrac{3}{4}$
$S_3 = \int_{0}^{3}\begin{pmatrix}x^3 + 1\end{pmatrix}dx = \begin{bmatrix} \dfrac{x^4}{4} + x\end{bmatrix}_{0}^{3}\textrm{}$
$\begin{bmatrix}
\dfrac{3^4}{4} + 3
\end{bmatrix} = 23\dfrac{1}{4}$
$=> S_3 = 23\dfrac{1}{4}$$
$S_1 + S_2 + S_3 = -1\dfrac{1}{4} - \dfrac{3}{4} + 23\dfrac{1}{4} = 21\dfrac{1}{4}$
And the correct answer is $26\dfrac{3}{4}$
What did I do wrong? I've checked myself many times, but I am failing to see my mistake... thanks!
|
Hint: If you have $\begin{bmatrix} -\frac{x^4}{4}-x \end{bmatrix}_a^b$, then you have to calculate $-\frac{b^4}{4}-b-\left(-\frac{a^4}{4}-a \right)=-\frac{b^4}{4}-b+\frac{a^4}{4}+a$
You have to substract the whole term, if you insert the lower bound. And both signs change, if you remove the brackets.
Additionally you have to take the absolute value of the subareas.
|
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|
Finding Limit Using Taylor Polynomial
Find the limit $$\displaystyle{\lim_{x\to4}}\frac{\left(1-\cos \left( x-4 \right)\right)^4\,\ln \left( x-3 \right)}{\left(e^{\left(x-4\right)^2}-1\right)^2\,\sin ^5\left(\pi \,x\right)}$$
Using Taylor Polynomial and Peano reminder
So first thing would be to move $\lim_{x\to 4}$ to $\lim_{x\to 0}$
So we get $$\displaystyle{\lim_{x\to0}}\frac{\left(1-\cos \left( x \right)\right)^4\,\ln \left( x+1 \right)}{\left(e^{\left(x\right)^2}-1\right)^2\,\sin ^5\left(\pi \,x\right)}$$
using Talyor expansion we get:
$$\displaystyle{\lim_{x\to0}}\frac{1-(1-\frac{x^2}{2}+\frac{x^4}{4}+o(x^6))^4\cdot x-\frac{x^2}{2}-\frac{x^3}{3}+o(x^6)}{x^2+\frac{x^4}{2}+o(x^6)\cdot (\pi^5 x^5-\frac{5\pi^7 x^7}{6}+\frac{23\pi^9x^9}{72}+o(x^{11})}$$
how should I processed from here? and why does the taylor expansion of $\sin ^5\left(\pi \,x\right)= $ this?
|
Using Taylor for this simple limit is not a good idea. However if one does wish to use the Taylor series it can be done by a little simplification at first (similar to what I present below) and then using first term (i.e. upto $x^{1}$) of the Taylor series for $\sin x, \log(1 + x), e^{x}$ and for $\cos x$ you need to go upto $x^{2}$.
On the other hand without Taylor, we can put $x = 4 + h$ so that $h \to 0$ and then we have
\begin{align}
L & = \lim_{x \to 4}\frac{(1 - \cos(x - 4))^{4}\log (x - 3)}{\left(e^{(x - 4)^{2}} - 1\right)^2\sin ^{5}(\pi x)}\notag\\
&= \lim_{h \to 0}\frac{(1 - \cos h)^{4}\log (1 + h)}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi(4 + h))}\notag\\
&= \lim_{h \to 0}\frac{(1 - \cos h)^{4}\log (1 + h)}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi h)}\text{ (may use Taylor after this step)}\tag{1}\\
&= \lim_{h \to 0}\frac{(1 - \cos h)^{4}}{h^{8}}\cdot\frac{\log (1 + h)}{h}\cdot\frac{h^{9}}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi h)}\notag\\
&= \frac{1}{2^{4}}\lim_{h \to 0}\frac{h^{9}}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi h)}\notag\\
&= \frac{1}{2^{4}}\lim_{h \to 0}\frac{h^{4}}{\left(e^{h^{2}} - 1\right)^2}\cdot\frac{(\pi h)^{5}}{\sin ^{5}(\pi h)}\frac{1}{\pi^{5}}\notag\\
&= \frac{1}{2^{4}}\cdot\frac{1}{\pi^{5}} = \frac{1}{16\pi^{5}}
\end{align}
Here we have used standard limits
$$\lim_{x \to 0}\frac{\sin x }{x} = 1, \lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}, \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1, \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$
If we wish to use Taylor after step $(1)$ above then we can note that
\begin{align}
\cos h &= 1 - \frac{h^{2}}{2} + o(h^{2})\notag\\
e^{h^{2}} &= 1 + h^{2} + o(h^{2})\notag\\
\sin(\pi h) &= \pi h + o(h)\notag\\
\log(1 + h) &= h + o(h)\notag
\end{align}
It follows from the above Taylor expansions that
\begin{align}
\lim_{h \to 0}\frac{1 - \cos h}{h^{2}} &= \frac{1}{2}\notag\\
\lim_{h \to 0}\frac{\log(1 + h)}{h} &= 1\notag\\
\lim_{h \to 0}\frac{h^{2}}{e^{h^{2}} - 1} &= 1\notag\\
\lim_{h \to 0}\frac{\pi h}{\sin(\pi h)} &= 1\notag
\end{align}
Thus continuing from equation $(1)$ we have
\begin{align}
L &= \lim_{h \to 0}\frac{(1 - \cos h)^{4}\log (1 + h)}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi h)}\notag\\
&= \lim_{h \to 0}\left(\frac{(1 - \cos h)}{h^{2}}\right)^{4}\cdot\frac{\log (1 + h)}{h}\left(\frac{h^{2}}{e^{h^{2}} - 1}\right)^{2}\left(\frac{\pi h}{\sin(\pi h)}\right)^{5}\frac{1}{\pi^{5}}\notag\\
&= \frac{1}{2^{4}}\cdot 1\cdot 1^{2}\cdot 1^{5}\cdot\frac{1}{\pi^{5}} = \frac{1}{16\pi^{5}}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Triangular Array's Recursive Formula Breakdown I have the following polynomials:
$$1$$
$$z-1$$
$$z^2-2z+3$$
$$z^3-3z^2+9z-15$$
$$z^4-4z^3+18z^2-60z+93$$
$$z^5-5z^4+30z^3-150z^2+465z-725$$
$$...$$
They are generated both recursively and explicitly. The recursive formula is
$$p_n(z)=z^n-\sum_{k=0}^{n-1}\binom{n}{k}\frac{1+(-1)^{n-k}+2(n-k)(-1)^{n-k-1}}{2}p_k(z)$$
Placing the unsigned coefficients in a lower triangular array,
$$
\begin{matrix}
1\\
1&1\\
1&2&3\\
1&3&9&15\\
1&4&18&60&93\\
1&5&30&150&465&725
\end{matrix}
$$
If we call the first row and column the 0-th row and column, then the columns 0 through 3 satisfy the following recurrence relation: If $T(n,k)$ is the entry in the nth row and kth column, then
$$T(n,k)=(2k-1)T(n-1,k-1)+T(n-1,k)$$
where $T(n,k)=0$ if $n<k$, and the explicit formula
$$T(n,k)=\binom{n}{k}(2k-1)!!$$
What would make an obvious pattern in the triangle break down after the 3rd column? If it were to continue, then $T(4,4)=7T(3,3)=105$, but instead we have 93. I know there are many sequences that mimic others very early and diverge from the desired results, but I'm trying to identify in the recursive formula the terms that would be responsible for this. Any thoughts?
|
$$p_n(z)=z^n-\sum_{k=0}^{n-1}\binom{n}{k}\frac{1+(-1)^{n-k}+2(n-k)(-1)^{n-k-1}}{2}p_k(z)$$
$T(n, k) = [z^{n-k}]p_n(z)$ and you're particularly interested in $T(n, n) = [z^0]p_n(z)$
$$T(n, n) = [n = 0] - \sum_{k=0}^{n-1}\binom{n}{k}\frac{1+(-1)^{n-k}+2(n-k)(-1)^{n-k-1}}{2}T(k, k)$$
*
*$T(0, 0) = 1$ is a special case.
*$T(1, 1) = -\binom{1}{0}T(0, 0) = -1$
*$T(2, 2) = -\left(-\binom{2}{0}T(0,0) + \binom{2}{1}T(1,1)\right) = -(-1-2) = 3$
*$T(3, 3) = -\left( 3\binom{3}{0}T(0, 0) - \binom{3}{1}T(1, 1) + \binom{3}{2}T(2, 2) \right) = -(3 + 3 + 9) = -15$
*$T(4, 4) = -\left( -3\binom{4}{0}T(0, 0) + 3\binom{4}{1}T(1, 1) - \binom{4}{2}T(2, 2) + \binom{4}{3}T(3, 3) \right) = -(-3 - 12 - 18 - 60) = 93$
So basically the coincidence with the odd factorial seems to be nothing more than an coincidence. $T(2, 2) = \binom{2}{0} + \binom{2}{1}$ just happens to be $3!!$. In $T(3, 3)$, the factors of 3 in $3\binom{3}{0}$ and $\binom{3}{1}$ come from different places. In general the coefficient of $-T(n-1, n-1)$ in the sum that makes up $T(n, n)$ is $\binom{n}{n-1} = n$, but there's no reason for the remaining terms to sum to $-(n-1)T(n-1, n-1)$.
|
{
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|
Perpendicular vectors in 3d Suppose a vector $v$ in $\mathbb{R}^3 $
How can I find two arbitrary unit vectors $u$ and $u^*$, that are perpendicular to each other and $v$ ?
There are infinitely many solutions, but I cannot hand pick them. I need some function $Q(v)$ = $(u, u^*)$ that deterministically finds a solution, for all non-zero vectors in $\mathbb{R}^3$.
This is probably a quaternion question, but I would prefer notation in terms of linear algebra.
|
$$u^\star = a \times u$$
Pick an arbitrary vector $a$ which is not parallel to $u$ and do a cross product. The result is perpendicular to both vectors. You can use a fixed vector such as $a=\hat{x}$, $a=\hat{y}$ or $a=\hat{z}$ by selecting the least parallel (lowest $a\cdot u$ value).
Alternatively pick any point is space with coordinates $(a,b,c)$ and construct a 3×3 rotation matrix where each column is a unit mutually perpendicular vector
$$ \begin{align}
E(a,b,c) & = \begin{bmatrix}
\frac{\sqrt{b^2+c^2}}{\sqrt{a^2+b^2+c^2}} & 0 & \frac{a}{\sqrt{a^2+b^2+c^2}} \\
\frac{-a b}{\sqrt{a^2+b^2+c^2}\sqrt{b^2+c^2}} & \frac{c}{\sqrt{b^2+c^2}} & \frac{b}{\sqrt{a^2+b^2+c^2}} \\
\frac{-a c}{\sqrt{a^2+b^2+c^2}\sqrt{b^2+c^2}} & \frac{-b}{\sqrt{b^2+c^2}} & \frac{c}{\sqrt{a^2+b^2+c^2}}
\end{bmatrix} \end{align} $$
|
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|
Find the length of chord $BC$.
On a semicircle with diameter $AD$. Chord $BC$ is parallel to the diameter.Further each of the chords $AB$ and $CD$ has length of $2$ cm while $AD$ has the length $8$ cm.Find the length of $BC$.
$a.)7.5\quad cm\\
\color{green}{b.)7\quad cm}\\
c.)7.75\quad cm\\
d.)\text{cannot be determined}$
I constructed $BO$ and with the help of cosine rule i found $\angle AOB$.then i found $\angle BOC$
and then again applying cosine rule in $\triangle BOC$ i found $BC$ ,
but i m looking for a more simple short way.
I have studied maths up to $12th$ grade.
|
Drop altitudes from $B$ and $C$ to $AD$, and call them $X, Y$ respectively.
Then, by symmetry, $AX = DY = 4 - \frac{BC}{2}$.
Furthermore $OX = \frac{BC}{2}$.
Also, $OB = 4$ since $AD = 8$.
Now use pythagorean theorem twice:
$AX^2 + BX^2 = AB^2$
$OX^2 + BX^2 = OB^2$
$OX^2 - AX^2 = OB^2 - AB^2$
$\frac{1}{4}BC^2 - (4 - \frac{BC}{2})^2 = 12$
Solving gives $BC = 7$.
|
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|
Proving $\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}$ I've been going through some of my notes when I found the following inequality for $a,b,c>0$ and $abc=1$:
$$
\begin{equation*}
\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}
\end{equation*}
$$
This was what I attempted, but that yielded no result whatsoever:
$$
\begin{align}
\sqrt{2}(a+b+c) &\geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}\\
&\geq \sqrt{abc+a²} + \sqrt{abc+b²} + \sqrt{abc+c²}\\
&\geq \sqrt{a(bc+a)}+\sqrt{b(ac+b)}+\sqrt{c(ab+c)}\\
&\geq \sqrt{a\left(\frac{1}{a}+a\right)}+\sqrt{b\left(\frac{1}{b}+b\right)}+\sqrt{c\left(\frac{1}{c}+c\right)}
\end{align}
$$
With this we then have
$$
\begin{align}
\sqrt{1+a²} &= \sqrt{a\left(\frac{1}{a}+a\right)}\\
&=\sqrt{a\left(\frac{1}{a}+\frac{a²}{a}\right)}\\
&=\sqrt{a\left(\frac{1+a²}{a}\right)}\\
&=\sqrt{1+a²}
\end{align}
$$
This yields nothing but frustration and gets me back to step 0
|
Hint
$$\sqrt{2}x-\sqrt{x^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{x},x>0$$
it is easy to prove by derivative.
so
$$\sqrt{2}a-\sqrt{a^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{a}\tag{1}$$
$$\sqrt{2}b-\sqrt{b^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{b}\tag{2}$$
$$\sqrt{2}c-\sqrt{c^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{c}\tag{3}$$
$(1)+(2)+(3)$
$$\sqrt{2}(a+b+c)\ge\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}$$
|
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|
A set of numbers Problem: Let $E(x)$ be the number defined by the following expression
\begin{equation*}
E(x)=\sqrt[3]\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}+\sqrt[3]\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}
\end{equation*}
where $x$ is a real number and
$\sqrt[3]{Z}$ denotes the real cubic root of the real number $Z$.
Determine the set
$ \{E(n); n\in \mathbb{N}$ and $n > 1\}$
|
Let: $t=E(n)$ for some $n\gt 1$. So: $$t=\sqrt[3]{\frac {n^3-3n+(n^2-1)\sqrt{n^2-4}}{2}}+\sqrt[3]{\frac {n^3-3n-(n^2-1)\sqrt{n^2-4}}{2}}$$ From here: $$t^3=\frac {n^3-3n+(n^2-1)\sqrt{n^2-4}}{2}+\frac {n^3-3n-(n^2-1)\sqrt{n^2-4}}{2}+3t\sqrt[3]{\frac {(n^3-3n+(n^2-1)\sqrt{n^2-4})(n^3-3n-(n^2-1)\sqrt{n^2-4})}{4}}$$ But: $$(n^3-3n+(n^2-1)\sqrt{n^2-4})(n^3-3n-(n^2-1)\sqrt{n^2-4})=(n^3-3n)^2-(n^2-1)^2(n^2-4)=n^6-6n^4+9n^2-(n^6-6n^4+9n^2-4)=4$$ Thus: $$t^3-3t-n^3+3n=0$$ Which means: $$(t-n)(t^2+n^2+tn-3)=0$$ But the equation: $t^2+n^2+tn-3=0$ hasn't real root for $n\gt 2$. And when $n=2$, we have: $$E(2)=2$$ And finally: $E(n)=n$ for every $n\gt 1$
|
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|
What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}}$ and so: $$\lim \limits _ {n \to \infty} n - n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}} = 0$$
Because both $\frac{a}{n}$ and $\frac{b}{n}$ tend to $0$.
What would give a correct answer:
Plotting the function $$f(x) = x - \sqrt{x+a} \sqrt{x+b}$$
Clearly indicates that it has an asymptote in $- \frac{a+b}{2}$. This result can be obtained multiplying the numerator and the denominator by $n + \sqrt{n+a} \sqrt{n+b}$:
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = $$
$$-\lim \limits _{n \to \infty} \frac {n(a+b)}{n + \sqrt{n+a} \sqrt{n+b}} - \lim \limits _{n \to \infty} \frac {ab}{n + \sqrt{n+a} \sqrt{n+b}}$$
The second limit is clearly $0$ and the first one gives the correct answer (dividing the numerator and denominator by $n$).
Why the first way I tried is wrong? I might have done something silly but I cannot find it.
|
Another approach:
$$\sqrt{a+n}\sqrt{b+n}=\sqrt{ab+(a+b)n+n^2}$$ and
$$\begin{align}n-\sqrt{n+a}\sqrt{n+b}&=\frac{n^2-(n+a)(n+b)}{n+\sqrt{n^2+(a+b)n+ab}}\\
&=\frac{-(a+b)n-ab}{n+\sqrt{n^2+(a+b)n+ab}}\\
&=\frac{-(a+b)-\frac{ab}{n}}{1 + \sqrt{1+\frac{a+b}{n} + \frac{ab}{n^2}}}
\end{align}$$
The numerator converges to $-(a+b)$ and the denominator converges to $2$.
|
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|
What can the various ways of integrating $\int \frac { x ^2 }{ (x \sin (x) +\cos(x))^2} \, dx$ $$
\int \frac {x^2}{(x\sin(x) + \cos(x))^2} \, \mathrm{d}x
$$
Well I found a method for solving this sum in a book saying that : We can multiply and divide the expression by $x\cos(x)$ and then apply integration by parts.
*But that method/trick is quite difficult to spot when one gets the integral for the first time.Is there any alternative "easier to spot"/"more general" technique to solve this integral?
|
$\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)^2}dx$$
We can write
$$\displaystyle (x\sin x+\cos x) = \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \sin x+\frac{1}{\sqrt{1+x^2}}\cdot \cos x\right\} = \sqrt{1+x^2}\cdot \cos \left(x-\phi \right)$$
So Here $$\displaystyle \cos x\ \phi = \frac{1}{\sqrt{1+x^2}}$$ and $\displaystyle \sin \phi = \frac{x}{\sqrt{1+x^2}}$ and $\tan \phi = x\Rightarrow \phi = \tan^{-1}(x)$
So Integral
$$\displaystyle I = \int\frac{x^2}{(1+x^2)\cdot \cos^2(x-\phi)}dx = \int \sec^2 (x-\phi)\cdot \frac{x^2}{1+x^2} dx = \int \sec^2 (x-\tan^{-1}(x))\cdot \frac{x^2}{(1+x^2)}dx$$
Now Let $$\left(x-\tan^{-1}x\right) = t\;,$$ Then $\displaystyle \left(1-\frac{1}{1+x^2}\right)dx = dt\Rightarrow \frac{x^2}{1+x^2}dx = dt$
So $$\displaystyle I = \int \sec^2 t dt = \tan t +\mathcal {C} = \tan \left(x-\tan^{-1} x\right)+\mathcal{C} = \left(\frac{\tan x-x}{1+x\cdot \tan x}\right)+\mathcal{C}$$
So $$\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)^2}dx = \left(\frac{\sin x- x\cos x}{\cos x+x\sin x}\right)+\mathcal{C}$$
|
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|
Determining the limit of this series $$ \sum_{n=0}^\infty \frac{(-2)^n + 2^{3n}}{3^n4^n} = $$
$$ \sum_{n=0}^\infty \frac{(-1)^n2^n}{3^n4^n} + \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n = $$
$$ \sum_{n=0}^\infty (-1)^n\frac{1}{6^n} + \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n = $$
The second sum is a geometric series, and its limit is
$$ \frac{1}{1 - 2/3} = 3 $$
However, I am having trouble finding the limit of the first Sum. How else can I approach this question?
|
Both are geometric series:
$$
\sum_{n=0}^\infty \frac{(-2)^n}{3^n \cdot 4^n} = \sum_{n=0}^\infty \left(\frac{-2}{3\cdot 4}\right)^n = \sum_{n=0}^\infty \left(\frac{-1}{6}\right)^n
$$
and
$$
\sum_{n=0}^\infty \frac{2^{3n}}{3^n\cdot 4^n} =
\sum_{n=0}^\infty \frac{8^n}{3^n\cdot 4^n} =
\sum_{n=0}^\infty \left(\frac{8}{3\cdot 4}\right)^n =
\sum_{n=0}^\infty \left(\frac{2}{3}\right)^n
$$
|
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|
Determinant of matrices without expanding Show that $$\begin{array}{|ccc|}
-2a & a + b & c + a \\
a + b & -2b & b + c \\
c + a & c + b & -2c
\end{array} = 4(a+b)(b+c)(c+a)\text{.}$$
I added the all rows but couldn't get it.
|
We can deduce from the structure of the matrix that (a) the determinant will be a symmetric polynomial in $a,b,c$ with every monomial with non-zero coefficient having degree $3$, and (b) that the coefficient of $a^3$, $b^3$, and $c^3$ will be $0$. So
$$
\det= k\,(a^2 b+ a^2 c+ b^2 a+ b^2 c+ c^2 a+ c^2 b)+m\,abc
$$
for some constants $k,m$. To find $k,m$, we substitute in values of $(a,b,c)$ into the matrix.
*
*When $a=0$, $b=1$, $c=1$ we get $8=\begin{vmatrix} 0 & 1 & 1 \\ 1 & -2 & 2 \\ 1 & 2 & -2 \\ \end{vmatrix}=\det=2k$, so $k=4$.
*When $a=1$, $b=-1$, $c=1$ we get $0=\begin{vmatrix} -2 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & -2 \\ \end{vmatrix}=\det=4 \times 2-m$, so $m=8$.
Thus
\begin{align*}
\det &= 4(a^2 b+ a^2 c+ b^2 a+ b^2 c+ c^2 a+ c^2 b)+8abc \\
&= 4(a+b)(b+c)(c+a)
\end{align*}
as desired.
|
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|
Simple trigonometry equation The previous class we were doing trigonometry exercises. Before the class finished, our teacher wrote exercises on the table. I am stuck with the following one:
$$
\cos(2x) + 1 + 3\sin x = 0
$$
I have come up with this:
$$
1= \sin^2 x + \cos^2 x
$$
$$
\cos(2x) = \cos^2 x - \sin^2 x
$$
When we substitute we get
$$
2\cos^2 x + 3\sin x = 0
$$
Need to find $x\ldots$
|
Although the procedure is almost same as given but I will explain it to the answer $$\cos(2x)+1+3\sin x=0$$ $$2-2\sin^2 x+3\sin x=0$$
$$2\sin^2 x-3\sin x-2=0$$ $$\implies (\sin x-2)(2\sin x+1)=0$$ $$ \implies \sin x-2=0 \implies\sin x=2\quad \text{but}\quad -1\leq\sin x\leq 1 $$ Hence we have $$ 2\sin x+1=0 \implies\sin x=-\frac{1}{2}=\sin\left(\frac{-\pi}{6}\right)$$ Writing the general solution, we get $$ \color {#0b4}{x = n\pi+ \frac{\pi}{6}} \quad \text{&}\quad \color{#0b4} {x=2n\pi- \frac{\pi}{6}} \quad (\forall \quad n\in I) $$ Where, $n$ is an integer.
|
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|
How can I show this inequality: $-2 \le \cos \theta (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$
Show that $$-2 \le \cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$$ for all value of $\theta$.
Trial: I know that $0\le \sin^2 \theta \le1 $. So, I have $\sqrt3 \le \sqrt{\sin ^2 \theta +3} \le 2 $. After that I am unable to solve the problem.
|
Use this well known inequality
$$-\dfrac{a^2+b^2}{2}\le ab\le\dfrac{a^2+b^2}{2},a,b\in R$$
so
$$-\dfrac{\cos^2{\theta}+4\sin^2{\theta}}{2}\le\cos{\theta}\cdot 2\sin{\theta}\le\dfrac{\cos^2{\theta}+4\sin^2{\theta}}{2}\tag{1}$$
$$-\dfrac{4\cos^2{\theta}+\sin^2{\theta}+3}{2}\le2\cos{\theta}\cdot\sqrt{\sin^2{\theta}+3}\le \dfrac{4\cos^2{\theta}+\sin^2{\theta}+3}{2}\tag{2}$$
then (1)+(2)
$$-4\le2\cos{\theta}(\sin{\theta}+\sqrt{\sin^2{\theta}+3})\le 4$$
so
$$-2\le\cos{\theta}(\sin{\theta}+\sqrt{\sin^2{\theta}+3})\le 2$$
|
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|
Sum of binomial coefficients from $\binom{2m+1}{0}$ to $ \binom{2m+1}{m} $ Prove that: $\displaystyle 4^m = \binom{2m+1}{0}+\binom{2m+1}{1}+\binom{2m+1}{2}+\ldots + \binom{2m+1}{m} $
From the Binomial Theorem:
$\displaystyle (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$
If $\displaystyle a = b = 2$, then we have
$\displaystyle 4^n = \sum_{k=0}^{n}\binom{n}{k}2^{n-k}2^k = 2^n\sum_{k=0}^{n}\binom{n}{k} = 2^n\cdot 2^n = 4^n$
Which is the same as $\displaystyle \sum_{k=0}^{2m}\binom{2m}{k}$ if $a = b = 1$ and $n =2m$ (with $4^m$ though).
What I'm having trouble with is the expansion. I've tried two things:
1) $\displaystyle \binom{2m}{0} = \binom{2m+1}{0}$ and applying the Pascal Identity to the remaining binomial coefficients:
\begin{align*}\sum_{k=0}^{2m}\binom{2m}{k} &= \binom{2m}{0}+\binom{2m}{1}+\binom{2m}{2}+\binom{2m}{3}+ \ldots \\
&= \binom{2m+1}{0} + \binom{2m+1}{2} + \binom{2m+1}{4} + \ldots
\end{align*}
2) Applying the Pascal Identity from the 1st binomial coefficient:
\begin{align*}\sum_{k=0}^{2m}\binom{2m}{k} &= \binom{2m}{0}+\binom{2m}{1}+\binom{2m}{2}+\binom{2m}{3}+ \ldots \\
&= \binom{2m+1}{1} + \binom{2m+1}{3} + \binom{2m+1}{5} + \ldots\end{align*}
I don't know how to get the coefficients I'm asked to have.
Thanks!!
|
Your sum is the first half of the binomial expansion of $(1+1)^{2m+1}$ -- and since $\binom{2m+1}n = \binom{2m+1}{2m+1-n}$, the sum is equal to the other half, so the sum is
$$\frac{(1+1)^{2m+1}}2 = \frac{2^{2m+1}}2 = 2^{2m} = 4^m $$
|
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|
How to prove that this matrix is positive definite? Let $\mathbf{A}=\begin{pmatrix}a^2+b^2 & b^2 & b^2 & ... & b^2 \\ b^2 & a^2+b^2 & b^2 & ... & b^2\\ \vdots & b^2 & \ddots & & b^2 \\ b^2 & \dots & & & a^2+b^2 \end{pmatrix}$, where $a,b\ne 0$. How can I be sure that this matrix is positive definite?
Any help would be appreciated.
|
Since $\mathbf{A}=a^2 \mathbf{I} + b^2 \mathbf{J}$ where $\mathbf{I}$ is an identity matrix and $\mathbf{J}$ has all ones, $$z^T \mathbf{A} z = z^T(a^2 \mathbf{I} + b^2 \mathbf{J})z = a^2 z^T \mathbf{I} z + b^2 z^T \mathbf{J} z.$$ If $z = (z_1, \ldots, z_n) \neq (0, \ldots, 0)$ then $$z^T I z = z_1^2 + \ldots z_n^2 > 0$$ and $$z^T J z = (z_1 + \ldots + z_n)^2 \ge 0.$$ Therefore $$a^2 z^T \mathbf{I} z + b^2 z^T \mathbf{J} z > 0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1305477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Roots of the equation $x^2 + px + q = 0$ If $\tan A$ and $\tan B$ are the roots of the equation $x^2 + px + q = 0 $ , show that $$\sin^2(A+B) + p \sin(A+B)\cos(A+B) + q \cos^2(A+B) = q$$
I tried using the result that $\tan A + \tan B = -p $ and $(\tan A)(\tan B) = q$ and tried substituting in the original equation but was unable to make any headway.
|
As $\tan(A+B)=\cdots=\dfrac p{q-1}$
Method $\#1:$
$\implies(1-q)\sin(A+B)+p\cos(A+B)=0$
Multiplying both sides by $\sin(A+B),$
$(1-q)\sin^2(A+B)+p\cos(A+B)\sin(A+B)=0$
$\iff\sin^2(A+B)+p\cos(A+B)\sin(A+B)+q(\cos^2(A+B)-1)=0$
$\iff\sin^2(A+B)+p\cos(A+B)\sin(A+B)+q\cos^2(A+B)=q$
Method $\#2:$
Dividing the numerator & the denominator by $\cos^2(A+B)=\dfrac1{\sec^2(A+B)}=\dfrac1{1+\tan^2(A+B)},$
$$\sin^2(A+B) + p \sin(A+B)\cos(A+B) + q \cos^2(A+B) =\dfrac{\tan^2(A+B)+p\tan(A+B)+q}{\tan^2(A+B)+1}$$
$$=\dfrac{p^2+p^2(q-1)+q(q-1)^2}{p^2+(q-1)^2}$$
Now, $p^2+p^2(q-1)+q(q-1)^2=q[p^2+(q-1)^2]$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
What is the principal cubic root of $-8$? According to my book it should be a real number, and according to WolframAlpha it should be $1+1.73i$
What is the correct answer?
|
Consider $x^3 = - 8 $. It "seems" we can just say $x = -2$ because $(- 2)^3 = -8$ but $- 2$ is not the official Primary Solution!
Using De Moivre's theorem again:
$$x^3 = - 8$$
$$( r \operatorname{cis}(\theta) )^3 = -8$$
$$r^3 \operatorname{cis}(3\theta) = 8\operatorname{cis}( 180^\circ + 360^\circ n)$$
$$r^3 = 8 \text{ and } 3\theta = 180^\circ + 360^\circ n $$
$$r = 2 \text{ and } \theta = 60^\circ + 120^\circ n = 60^\circ, 180^\circ, 300^\circ $$
$$x_1 = 2\operatorname{cis}( 60^\circ ) = 1 + i\sqrt3$$
$$x_2 = 2\operatorname{cis}(180^\circ) = - 2$$
$$x_3 = 2\operatorname{cis}(240^\circ) = 1 - i\sqrt3$$
The Primary Solution is $x_1 = 1 + i\sqrt3 ≈ 1 + 1.732i$
So the cube root of $-8$ or $(-8)^{\frac13}=1+1.732i$ and NOT $-2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
integral $\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$ I want to compute this integral
$$\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$$
where $0<b \leq a$.
I have this results
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2sin^2(t)+b^2cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
But I don't know how to prove this equality.
Which can help me,
Thanks for all.
|
One trick to facilitate analysis is to write
$$\sin^2x=\frac{1-\cos 2x}{2}$$
and
$$\cos^2x=\frac{1+\cos 2x}{2}$$
Thus,
$$a^2\sin^2(t)+b^2\cos^2(t)=\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos 2t$$
For the integral of interest, we can write
$$\begin{align}
I_1&=\int_0^{\pi} \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\
&=\frac{1}{2}\int_0^{2\pi}\frac{\cos t}{\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos t}dt\\\\
&=\frac{1}{(b^2-a^2)}\int_0^{2\pi}\frac{\cos t}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt\\\\
&=\frac{2\pi}{(b^2-a^2)}-\frac{b^2+a^2}{(b^2-a^2)^2}\int_0^{2\pi}\frac{1}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt
\end{align}$$
Can you finish now?
SPOLIER ALERT
SCROLL OVER SHADED AREA TO SEE ANSWER
Starting with the last term $I=\frac{2\pi}{(b^2-a^2)}-\frac{b^2+a^2}{(b^2-a^2)^2}\int_0^{2\pi}\frac{1}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt$, we move to the complex plane by letting $z=e^{it}$. Then, $I$ becomes $$\begin{align}I&=\frac{2\pi}{(b^2-a^2)}-\frac{b^2+a^2}{(b^2-a^2)^2}\oint_C \frac{-2i}{z^2+2\frac{b^2+a^2}{b^2-a^2}\,z+1}dz\\\\&=\frac{2\pi}{(b^2-a^2)}-\frac{b^2+a^2}{(b^2-a^2)^2}\left(2\pi i \frac{-2i}{-2\frac{2ab}{a^2-b^2}}\right)\\\\&=\frac{2\pi}{(b^2-a^2)}+\frac{2\pi}{ab}\frac{a^2+b^2}{2(a^2-b^2)}\\\\&=\frac{-2\pi}{ab}\left(\frac12 -\frac{a}{a+b}\right)\end{align}$$
NOTE:
For $I_2=\int_0^{\pi} \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$, enforce the substitution $t= \pi -x$. Then, we see
$$\begin{align}
I_2&=\int_0^{\pi} \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\
&=\int_{\pi}^{0} \frac{\sin 2(\pi-x)}{a^2\sin^2(\pi-x)+b^2\cos^2(\pi-x)}(-1)dx\\\\
&=\int_{0}^{\pi} \frac{-\sin (2x)}{a^2\sin^2(x)+b^2\cos^2(x)}dx\\\\
&=-I_2
\end{align}$$
Thus, we have $I_2=-I_2$ which, of course, implies $I_2=0$!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing $ \int_0^{2 \pi } \frac{dt}{a^2 \cos^2 t + b^2 \sin^2 t} = \frac{2 \pi}{ab}$ The question:
Let $\gamma$ be a contour such that $0 \in I(\gamma),$ where $I$ is the interior of the contour. Show that
$$\int_\gamma z^n \, \text{d}z = \begin{cases} 2\pi i & \text{if } n = -1 \\ 0 & \text{otherwise} \end{cases}$$
By taking $\gamma$ as the ellipse
$$\{ (x,y) : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \},$$
show that $$ \int_0^{2 \pi } \frac{dt}{a^2 \cos^2 t + b^2 \sin^2 t} = \frac{2 \pi}{ab}.$$
The question's answer uses the Deformation theorem and the fact that the first integral has the given value if the contour is a unit circle. However, the two contours must not overlap, so it seems like this should only be true for a contour that either always has magnitude less than one or greater than one. In Mathematica, the final integral was true for the case that $a = 1.5$ and $b=0.4$. What am I missing?
Edit: The radius of the circle cancels in the integral if $n = -1$, so then it does hold irrespective of the radius.
|
Let us consider the ellipsis in the complex plane $z=a\cos t+ib\sin t$, $t \in [0,2\pi]$, from which $dz=(-a\sin t+ib\cos t)dt$, then
\begin{align}
&2\pi i=\int_\gamma z^{-1}dz=\int_0^{2\pi}\frac{-a\sin t+ib\cos t}{a\cos t+ib\sin t}dt=\\
&\qquad=\int_0^{2\pi}\frac{(b^2-a^2)\sin t \cos t+iab}{a^2\cos^2 t+b^2\sin^2 t}dt=\\
&\qquad(b^2-a^2)\int_0^{2\pi}\frac{\sin t \cos t}{a^2\cos^2 t+b^2\sin^2 t}dt+iab\int_0^{2\pi}\frac{1}{a^2\cos^2 t+b^2\sin^2 t}dt
\end{align}
from which the result (the first integral vanishes for symmetry reasons, or also because the first member is purely imaginary).
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
|
Suppose $\sqrt2$ is arational, then there exist $p,q$ two natural numbers such that $$\color{Red}{p^2=q^2+q^2.}$$ Then by the parametric solution of Pythagoras Equation there exist two natural numbers $a,b$ such that $a\gt b\ge 1$ and $p=a^2+b^2,$ $$\color{Green}{q=a^2-b^2=2ab.}$$
Now, if $r=a+b$ and $s=a,$ then $$\color{Red}{r^2=s^2+s^2}$$ with $r\lt p,\ s\lt q.$ Hence, by the Infinite Descent there are no such $p$ and $q$ natural numbers.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $f=(x+i)^{10}+(x-i)^{10}$ have all real roots We have $f=(x+i)^{10}+(x-i)^{10}$ and we need to prove that $f$ have all the roots in $\mathbb{R}$.
Here is all my steps:
*
*Suppose that $z\in\mathbb{R}$ is a root of $f\Rightarrow (z+i)^{10}+(z-i)^{10}=0$
Therefore: $f(z)=\sum_{k=0}^{10}\left[\left(\dbinom{10}{k}\cdot z^{10-k}\cdot i^k\right)\left(1+(-1)^k\right)\right]=0$
I don't have ideea how can I prove that $f$ have all real roots
Suppose that $z=a+bi$ and we need to prove that $b=0$:
$$\Rightarrow f(z)=\sum_{k=0}^{10}\left(\dbinom{10}{k}\cdot a^{10-k}\cdot i^k\right) \left[(b+1)^k+(b-1)^k \right]=0$$
$$\Rightarrow \left[(b+1)^k+(b-1)^k \right]=0$$
$$\Rightarrow b+1=-b+1$$
$$\Rightarrow b=0$$
Therefore $f$ have all the roots in $\mathbb{R}$
*
*Here is a photo with the proof of real axis, I don't know if is correct:
|
$x=a+bi$
$$(x+i)^{10}+(x-i)^{10}=0 \Rightarrow (x+i)^{10}=-(x-i)^{10} \Rightarrow (\frac{x+i}{x-i})^{10}=-1 \\ \Rightarrow \left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=i \\ \Rightarrow \frac{(a+(b+1)i)(a-(b-1)i)}{(a+(b-1)i)(a-(b-1)i)}=i \Rightarrow \frac{a^2-a(b-1)i+a(b+1)i+(b^2-1)}{a^2+(b-1)^2}=i \Rightarrow \frac{(a^2+(b^2-1))+2ai}{a^2+(b-1)^2}=i$$
That means that $$\frac{a^2+(b^2-1)}{a^2+(b-1)^2}=0 \text{ and } \frac{2a}{a^2+(b-1)^2}=1$$
$$\frac{a^2+(b^2-1)}{a^2+(b-1)^2}=0 \Rightarrow a^2+b^2-1=0 \Rightarrow b^2=1-a^2$$
$$\frac{2a}{a^2+(b-1)^2}=1 \Rightarrow 2a=a^2+(b-1)^2 \Rightarrow 2a=a^2+b^2-2b+1 $$
Solving for $a$ and $b$ we get the following solutions:
$a=0, b=1$ and $a=1, b=0$
We reject $(a, b)=(0, 1)$ because then the denominator would be $0$.
So, the solution is $(a, b)=(1, 0)$.
That means that $x=1 \in \mathbb{R}$.
$$$$
Do the same for the case $\left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=-i$
|
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|
integer solution of $(x-y)(x+y)xy=z^2$ By wolfram alpha, integer solution of $(x-y)(x+y)xy=z^2$ is $x=y=z=0$.
How to show that there are not another solutions with $z \neq 0$.
Thanks.
|
If $(x,y)=k$ then $k^4(x_1-y_1)(x_1+y_1)x_1y_1= z^2$ which implies an equation (we use the same notation) $(x-y)(x+y)xy=z^2$ with $(x,y)=1$; it follows that $(x+y,x-y)=1$ ($x+y=mX$ and $x-y=mY$ give clearly a contradiction). Hence $x-y, x+y, x, y$ are four coprime integers and each of them must be a square, $x, y$ being of different parity (if not then $2|(x-y,x+y)=1$).
We have three pythagorean triples to which we apply the well known parametrization
$(a,b,c) =(p^2-q^2, 2pq, p^2+q^2)$:
(1)………………$x^2=y^2+u^2$
(2)……………..$x^2+y^2=v^2$
(3)…………….$x^4+y^4=(uv)^2$
If $y$ were odd then $x$ would be odd in (1) and even in (2), absurde, so $y$ must be even.
Now from (3)
$x^2=t^2+s^2, y^2=2ts$ implies $ {x^4-y^4=(t^2-s^2)^2}$ , $t$ and $s$ of distinct parity.
Since $t^2+s^2=x^2$, we write $t=m^2-n^2, s=2mn$ and $x=m^2+n^2$ with $(m,n)=1$.
It follows $y^2=2(m^2-n^2)(2mn)=4(m^2-n^2)mn$
Hence $m=p^2, n=q^2$ and $y^2=4(p^2-q^2)p^2q^2$ and since y is even,$(p^2-q^2)(p^2+q^2)p^2q^2= w^2$.
Thus, from $(x^2-y^2)(x^2+y^2)x^2y^2=z^2$ we have got another equation
$(p^2-q^2)(p^2+q^2)p^2q^2= w^2$ similar to the first one with the values of $p,q$ less than the values of $x, y$. This process can be iterated indefinitely therefore, by infinite descent, we finish the proof.
|
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|
Tangent numbers are divisible by $2^{n}$ Let us consider a $$\tan(z) = \sum_{n=1}^{\infty}{T_{2n-1} \cdot \frac{z^{2n-1}}{(2n-1)!}}$$.
So, it can be shown that $$T_{2n+1}=\frac{(-1)^{n} 4^{n+1}(4^{n+1}-1) B_{2n+2}}{2n+2} $$ where $B_{2n+2}$ is the $2n+2$ th Bernoulli number.
How to prove that $T_{2n+1}$ is divisible by $2^{n}$?
The first idea is to derive it using induction, but the induction step seem to be too much complicated. Are there another possible approaches?
Any help would be much appreciated.
|
Take a look at the first several derivatives of $\tan(z)$:
$$
\begin{align}
\tan(z)&=\tan(z)\\
\sec^2(z)&=\tan(z)^2+1\\
2\sec^2(z)\tan(z)&=2\tan^3(z)+2\tan(z)\\
6\tan^2(z)\sec^2(z)+2\sec^2(z)&=6\tan^4(z)+8\tan^2(z)+2\\
24\tan^3(z)\sec^2(z)+16\tan(z)\sec^2(z)&=24\tan^5(z)+40\tan^3(z)+16\tan(z)\\
\end{align}
$$
This suggests that the $n$th derivative of tangent is a certain polynomial in $\tan(z)$. That is, $$\frac{d^n}{dz^n}\tan(z)=P_n(\tan(z))$$ These polynomials $P$ satisfy a recursion that is easy to prove: $$P_n(X)=\frac{d}{dX}P_{n-1}(X)\cdot\left(X^2+1\right)$$ and in fact proving this recursion proves the existence of the polynomials.
We can give names to the coefficients of these polynomials: $$P_n(X)=\sum_{k=0}^{n+1}c_{n,k}X^k$$ and be sure to define $c_{n,k}$ to be $0$ for negative $k$ or $k$ larger than $n+1$. The recursion tells us $$c_{n,j}=(j-1)c_{n-1,j-1}+(j+1)c_{n-1,j+1}$$
Inductively, assume that $2^n$ divides $c_{2n+1,k}$ for all $k$. (This is true when $n=1$, as is apparent from the fourth row [third derivative] above.) Then $$\begin{align}
c_{2n+3,k}
&=(k-1)c_{2n+2,k-1}+(k+1)c_{2n+2,k+1}\\
&= (k-1)(k-2)c_{2n+1,k-2}+(k-1)(k)c_{2n+1,k}+(k+1)(k)c_{2n+1,k}+(k+1)(k+2)c_{2n+1,k+2}
\end{align}
$$
Since each of the $c_{2n+1,*}$ are divisible by $2^n$ by the induction assumption, and since each of $(k-1)(k-2)$, $(k-1)(k)$, $k(k+1)$, and $(k+1)(k+2)$ are even, it follows that $c_{2n+3,k}$ is divisible by the next higher power of $2$, namely $2^{n+1}$. So by induction, each $c_{2n+1,k}$ is divisible by $2^n$.
You are specifically asking about $c_{2n+1,0}$, the values of $\left.P_{2n+1}(X)\right|_{X=0}=\left.\frac{d^{2n+1}}{dz^{2n+1}}\tan(z)\right|_{z=0}$, which are of course the Maclaurin series coefficients for $\tan(z)$ (prior to division by a factorial).
|
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|
Finding streamlines from complex potential. I'm currently studying a Fluid Dynamics module and mock exam question has me completely stumped, I have been given the complex potential and shown it to be in the form given, however when trying to separate the complex potential into velocity potential and stream function I cannot seem to do the manipulation. The question is as follows:
$w(z)=\frac{m}{2\pi}log[z^2+a^2]$
Show that the streamlines are given by $x^2-y^2+a^2=kxy$
where k is a constant.
Any help would be appreciated! Thanks in advance.
|
You'll have to accept or justify the definition that
$$(1) \quad \ln(a+b \cdot i)={{\ln(a^2+b^2)} \over 2}+\left({{\pi \cdot sign(b)} \over 2}-\tan^{-1}{ a \over b} \right) \cdot i$$
$$W=C \cdot \ln(z^2+a^2)$$
Where $C={m \over {2 \cdot \pi}}$
The stream function is given by the imaginary part of the complex potential. Assuming $z=x+y \cdot i$
$$z^2=x^2+2 \cdot x\cdot y \cdot i-y^2$$
Comparing with (1), we can see that the real and imaginary parts inside the logarithm are
$$a=x^2-y^2+a^2$$
and
$$b=2 \cdot x \cdot y \cdot i$$
The imaginary part of (1) is the stream function
$$\Psi=\left({{\pi \cdot sign(b)} \over 2}-\tan^{-1}{ a \over b} \right)$$
Substituting in and simplifying we get...
$$(2) \quad \Psi={C \over 2} \cdot \left(2 \cdot \tan^{-1} \left({{x^2-y^2+a^2} \over {2 \cdot x \cdot y}} \right)-\pi \cdot sign(x \cdot y) \right)$$
You'll want to justify this next part in more detail. The streamlines are where the stream function is constant. We'll get rid of the right hand side of (2) under justification that it's constant. Then since everything added to the left should be constant, substitute the value for a new constant. We'll get...
$$(3) \quad {{2 \cdot \Psi} \over C}+B=D=2 \cdot \tan^{-1} \left({{x^2-y^2+a^2} \over {2 \cdot x \cdot y}} \right)$$
$$(4) \quad tan^{-1}(D)=2 \cdot L=k={{x^2-y^2+a^2} \over {x \cdot y}}$$
finally,
$$(5) \quad x^2-y^2+a^2=k \cdot x \cdot y$$
|
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Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent
First i subbed numbers in
$$\lim_{n \to \infty} \frac{(-1)^n}{1+\sqrt{n}} = \frac{-1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} - \frac{-1}{1+\sqrt{3}}$$
So it's divergent
1 LIMIT
$$\lim_{n \to \infty} \frac{1}{1+\sqrt{n}}=0 \quad \text{hence divergent} $$
2 $a_{n}$ and $a_{n+1}$
$$ \frac{1}{1+\sqrt{n}}>\frac{1}{1+\sqrt{n+1}} $$
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}}> 0 $$
hence increasing
or
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}} = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n+1}} $$
$$ \therefore \sqrt{n+1}-\sqrt{n}> 0$$ hence decreasing
$ \sum_{n=1}^{\infty} \left\lvert \frac{-1^n}{1+\sqrt{n}} \right\rvert = \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} $
$$ \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} = \frac{1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} + \frac{1}{1+\sqrt{3}} $$
if this is convergent then the whole thing is absolutely convergent but i don't know how to prove this
UPDATE
I got something out for the second part
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There are so many mistakes (and so severe) in your "solution" that it is a superhuman task to discuss them. Your series is convergent, Leibniz's test solves the problem in a second. The series is not absolutely convergent by an application of the limit comparison test (compare your series with $\sum \frac 1 {\sqrt n}$ which is known to be divergent).
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$x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers.
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
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It is possible to find out $x^2+y^2$, as $196$, which will be $(x-y)(x+y)$, can be factorised as five possible integer products, $14\times 14$, $7\times28$, $4\times 49$, $2\times 98$, $1\times 196$.
Of these only one product leads to both positive and integer solutions for $x$ and $y$. Can you work out which of the products this will be?
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How to solve this inequality, with the hypothesis more complicated than the conclusion? Given $x,y,z \in \mathbb{R}$ and $x,y,z>2,$ I want to show that if,
$$\frac{1}{x^2-4}+\frac{1}{y^2-4}+\frac{1}{z^2-4} = \frac{1}{7}$$
then,
$$\frac{1}{x+2} + \frac{1}{y+2} + \frac{1}{z+2} \leq \frac{3}{7}.$$
I follow the solution here: http://artofproblemsolving.com/community/c6h514107_inequality_by_poru_loh
but I don't know how to alter it to fit this problem?!
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Here is an adaptation of hyperbolictangent's answer in your given link.
Let $S:= \frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}$. Note that
\begin{align*}
3-4S &= \frac{x^2-4}{x^2-4} + \frac{y^2-4}{y^2-4} + \frac{z^2-4}{z^2-4} - 4 \left(\frac{x-2}{x^2-4}+\frac{y-2}{y^2-4}+\frac{z-2}{z^2-4}\right)\\
&= \frac{(x-2)^2}{x^2-4}+\frac{(y-2)^2}{y^2-4}+\frac{(z-2)^2}{z^2-4}\\
&= \frac{x-2}{x+2} + \frac{y-2}{y+2} + \frac{z-2}{z+2}.
\end{align*}
Thus,
\begin{align*}
\frac{1}{7} \cdot (3-4S)
&= \left(\frac{1}{x^2-4}+\frac{1}{y^2-4}+\frac{1}{z^2-4}\right)
\left(\frac{x-2}{x+2} + \frac{y-2}{y+2} + \frac{z-2}{z+2}\right)\\
&\ge S^2 & \text{Cauchy-Schwarz}
\end{align*}
So we have
$$0 \ge S^2+\frac{4}{7}S-\frac{3}{7} = \left(S+\frac{2}{7}\right)^2 -\frac{25}{49}$$
$$\frac{5}{7} \ge S+\frac{2}{7}$$
$$\frac{3}{7} \ge S$$
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Why is $\sqrt{-i} \neq i\sqrt{i}$? I wanted to figure out the square root of $-i$.
Since $\sqrt{-x} = i\sqrt{x}$,
$\sqrt{-i}$ should equal $i\sqrt {i}$, however, WolframAlpha said it was false.
However, if I do say that $\sqrt{-i} = i\sqrt{i}$, I can replace $\sqrt{i}$ with $\dfrac {1+i} {\sqrt{2}}$, leaving me with $\sqrt{-i} = \dfrac {i(1+i)} {\sqrt{2}}$
Now I distribute, giving me $\sqrt{-i} = \dfrac {i-1} {\sqrt{2}}$
Then I square both sides, giving me $-i = \dfrac {(i-1)(i-1)} {2}$
Then I FOIL, giving me $-i = \dfrac {-2i} {2}$
Then the $2$'s cancel out, leaving $-i = -i$, making $\sqrt {-i} = i\sqrt{i}$ a true statement.
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You are running up against a complication that arises with roots of complex numbers: they are not single-valued, but multi-valued. If you express the question "what is $ \ \sqrt{-i} \ $ ?" as "what are the roots of the equation $ \ x^2 \ - \ (-i) \ = \ 0 \ ? $ " , there are two solutions. DeMoivre's Theorem tells us that the square-roots of $ \ -i \ = \ \cos(\frac{3 \pi}{2}) \ + \ i \ \sin(\frac{3 \pi}{2}) \ $ are
$$ \cos(\frac{3 \pi}{4}) \ + \ i \ \sin(\frac{3 \pi}{4}) \ = \ \frac{-1 + i}{\sqrt{2}} \ \ \text{and} \ \ \cos(\frac{7 \pi}{4}) \ + \ i \ \sin(\frac{7 \pi}{4}) \ = \ \frac{1 - i}{\sqrt{2}} \ \ . $$
The square-roots of $ \ i \ = \ \cos(\frac{ \pi}{2}) \ + \ i \ \sin(\frac{ \pi}{2}) \ $ are
$$ \cos(\frac{ \pi}{4}) \ + \ i \ \sin(\frac{ \pi}{4}) \ = \ \frac{1 + i}{\sqrt{2}} \ \ \text{and} \ \ \cos(\frac{5 \pi}{4}) \ + \ i \ \sin(\frac{5 \pi}{4}) \ = \ \frac{-1 - i}{\sqrt{2}} \ \ . $$
We can see that if we multiply each of the square-roots of $ \ -i \ $ by $ \ i \ $ , we will get one of the square-roots of $ \ i \ $ , but the correspondence is not as simple as is suggested by the putative equation $ \ \sqrt{-i} \ = \ i \ \sqrt{i} \ $ .
What WolframAlpha appears to do is take as the principal values for these square-roots the ones where the angle (argument) in the complex plane is in the domain of the arctangent function, $ \ -\frac{\pi}{2} \ < \ \theta \ < \ \frac{\pi}{2} \ $ . When you just look at that, the equation doesn't work. But further down the pages, when you look at all the square-roots, you see that the locations of the two square-roots of $ \ i \ $ get rotated by $ \ \frac{\pi}{2} \ $ (upon multiplication by $ \ i \ $ ) to give the locations of the two square-roots of $ \ -i \ $ . So your equation is "wrong" in the sense that the complex square-root function has to be described more carefully than the one defined only for real numbers.
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Please help to find term's coefficient in the following example I trying find the number of all solutions in the following:
$ x_1 + x_2 + x_3 + x_4 + x_5 = 24 $
where:
2 of variables are natural odd numbers
3 of variables are natural even numbers
none of variables are equal to $0$ or $1$
(all the variables are $>= 2$)
So I've made generating functions according to the limitations as follow:
$ (x^3 + x^5 + x^7 + ... + x^{21} + x^{23})^2 $
$ (x^2 + x^4 + x^6 + ... + x^{22} + x^{24})^3 $
Now I need to find the coefficient of $ x^{24} $
I've tried to solve it but get stuck, here is what I've done:
$ x^6(1 + x^2 + x^4 + ... + x^{18} + x^{20})^2 $
$ x^6(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
$ x^{12}(1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
Now we actually looking for coefficient of $ x^{24-12}=x^{12} $ in:
$ (1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
$ (\frac{1 - x^{21}}{1 - x})^2(\frac{1 - x^{23}}{1 - x})^3 $
Here I'm stuck. How to proceed in order to find coefficient of $ x^{12} $?
Regards.
PS: Solution was given but just out of curiosity, how to solve it using generating-functions?
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Let the numbers be $2a + 1, 2b + 1, 2c, 2d, 2e$, where $a, b, c, d, e$ are natural. Note that this makes the odd numbers greater than or equal to $3$, and the even numbers greater than or equal to $2$.
Then you have $2a + 1 + 2b + 1 + 2c + 2d + 2e = 24 \rightarrow a + b + c + d + e = 11$. You can solve this with a standard stars-and-bars approach.
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Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would be thankful.
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is equal to?
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There is some quadratic $x^2 + cx + d$ such that:
$$(x^2 + 4x + 6)(x^2 + cx + d) = x^4 + ax^2 + b$$
Multiply it out:
$$x^4 + (c + 4)x^3 + (d + 4c + 6)x^2 + (6c + 4d)x + 6d = x^4 + ax^2 + b$$
Equate coefficients:
$$\begin{cases}
c + 4 = 0 \\
d + 4c + 6 = a \\
4d + 6c = 0 \\
6d = b
\end{cases}$$
Use the first equation to find $c$. Use the third to find $d$. Use those to find $a$ and $b$.
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Alternative way to count the number of solutions to the equation $x^2 + y^2 = -1$ over $\Bbb Z /p$
$x^2 + y^2 = -1$ is a weird equation because it has no solutions over $\Bbb R$. I want to count the number of solutions it has over $\Bbb Z / p$ where $p$ is prime.
If $p = 2$ then it has $p$ solutions. This is to do with the fact that squaring is a field automorphism.
If $p \equiv 1 \pmod{4}$ then there is an $i$ such that $i^2 = -1$ so $$x^2 + y^2 = -1 \implies \left({x \over i}\right)^2 + \left({y \over i}\right)^2 = 1 \implies \left({x \over i} + y \right)\left({x \over i} - y \right) = 1$$ which has $p - 1$ solutions.
If $p \equiv 3 \pmod{4}$ then the situation is more complicated. The thing I noticed is that $A = \{x \mid x^2 + y^2 = -1\}$ and $B = \{x \mid y^2 - x^2 = 1\}$ form a partition of $\Bbb Z/p$. Reason being that $x \not\in A \implies (-1 - x^2 \mid p) = -1 \implies (1 + x^2 \mid p) = 1 \implies (\exists y)\,1+x^2 = y^2 \implies x \in B$, and vice versa. Also notice that $A \cap B = \emptyset$. So we get $|A| = |\Bbb Z / p| - |B| = p - |B|$. To determine $|B|$, use the fact that for every $(x,y)$ for which $y^2 - x^2 = 1$, $(x,-y)$ also satisfies the equation, so $|B|$ is the number of solutions to $y^2 - x^2 = 1$ divided by $2$, which is $\frac{p-1}{2} \therefore \,|A| = {p + 1 \over 2}$. Now it's easy to see that the number of solutions to $x^2 + y^2 = -1$ is $2|A|$ which is $p+1$.
Any quicker method?
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Given a finite field $F$, let $C = F[i]/(i^2+1)$. Define the mapping $N:C \to F$ by $N(x+iy) = x^2 + y^2$ and note that it's a multiplicative homomorphism from $C^*$ to $F^*$.
Now let $X = \{x \in C \mid N(x) = 1\}$ and $Y = \{x \in C \mid N(x) = -1\}$. The aim is to determine $|Y|$.
Note that there exists an $e \in Y$. I will omit the proof.
Define the mapping $f: X \to Y$ by $f(x) = ey$. First, the codomain is correct because given $x \in X$, $N(ex) = N(e)N(x) = (-1) \times 1 = -1$. Second, the mapping is injective because $f(x) = f(x') \implies ex = ex' \implies x = x'$. Finally, the mapping is surjective because given $y \in Y$, $N\left({y \over e}\right) = 1$ so ${y \over e} \in X$ and $f\left(y \over e\right) = e{y \over e} = y$.
So $f$ is a bijection. This implies that we only need to count $X$ to find $|Y|$. Here are some ways to do that.
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Michael Spivak's Calculus - Chapter 1, Problem 19 Problem 19. The fact that ${a^2}\ge{0}$ for all the numbers a, elementary as it may seem, is nevertheless the fundamental idea upon which most important inequalities are ultimately based. The great-grandaddy of all inequalities is the Schwarz inequality:
${x_{1}y_{1}+x_{2}y_{2}}\le{\sqrt{x_{1}^2+x_{2}^2}\sqrt{y_{1}^2+y_{2}^2}}$
The three proofs of the Schwarz inequality outlineed below have only one thing in common - their reliance on the fact that $a^2\ge{0}$ for all $a$.
(a) Using Problem 18, complete the proof of the Schwarz inequality.
Solution.
This is the proof I've written.
\begin{array}{ll}
(x_{1}y_{2}-x_{2}y_{1})^{2} & \ge0\\
x_{1}^{2}y_{2}^{2}-2x_{1}x_{2}y_{1}y_{2}+x_{2}^{2}y_{1}^{2} & \ge0\\
x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2} & \ge2x_{1}x_{2}y_{1}y_{2}\\
2x_{1}x_{2}y_{1}y_{2} & \le x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2}\\
x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}+2x_{1}x_{2}y_{1}y_{2} & \le x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2}\\
(x_{1}y_{1}+x_{2}y_{2})^{2} & \le(x_{1}^{2}+x_{2}^{2})(y_{1}^{2}+y_{2}^{2})\\
x_{1}y_{1}+x_{2}y_{2} & \le\sqrt{(x_{1}^{2}+x_{2}^{2})}\sqrt{(y_{1}^{2}+y_{2}^{2})}
\end{array}
(b)Prove the Schwarz inequality by using $2xy\le{x^2+y^2}$ with
$x=\displaystyle{\frac{x_{i}}{\sqrt{x_{1}^2+x_{2}^2}}}$, $y=\displaystyle{\frac{y_{i}}{\sqrt{x_{1}^2+x_{2}^2}}}$
first for $i=1$ and then for $i=2$.
I am not able use the inequality to come up with another proof. Could someone tell me the way, how to approach this.
Also, what are some of the applications of Cauchy-Schwarz inequality? I did google it, but would love to hear it from a mathematician!
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We have
\begin{align}2x_iy_i &\le \frac{x_i^2 \sqrt{y_1^2 + y_2^2}}{\sqrt{x_1^2 + x_2^2}} + \frac{y_i^2 \sqrt{x_1^2 + x_2^2}}{\sqrt{y_1^2 + y_2^2}}\\
&\le \frac{x_i^2 (y_1^2 + y_2^2) + y_i^2 (x_1^2 + x_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
&\le \frac{x_i^2 (y_1^2 + y_2^2) + y_i^2 (x_1^2 + x_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
\to 2 \sum_{i\in\{1,2\}} x_iy_i &\le \frac{(x_1^2 + x_2^2) (y_1^2 + y_2^2) + (y_1^2 + y_2^2) (x_1^2 + x_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
\to \sum_{i\in\{1,2\}} x_iy_i &\le \frac{(x_1^2 + x_2^2) (y_1^2 + y_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
\to \sum_{i\in\{1,2\}} x_iy_i &\le \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}\end{align}
QED
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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.