Hey, how can I treat combinations/ permutations with integral representations, for example the integrals represented here http://www.wolframalpha.com/input/?i=8C3
I'm assuming it has to do with power/ infinite series? I'm not exactly sure, how would you get this?

- Astrophysics

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Astrophysics

@ganeshie8 @Empty

- Astrophysics

|dw:1437076681063:dw|

- Empty

The thing I think I am able to kinda think, "Ok that is plausible but I don't quite see it yet" is the first one. Since the binomial coefficients come from expanding a binomial, that \((1+e^{it})^8\) seems to be doing something interesting there and then having the special term \(e^{3it}\) divided out some how and that causes it to remain or something. Hmmm cool. As for the bottom one, I wonder if it's similar or not. I guess we should expand it and check it out and see what we can find.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Astrophysics

Oh so lets use binomial theorem \[(a+b)^n = \sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k}b^k\]
ye?

- Astrophysics

\[\left(\begin{matrix}n \\ k\end{matrix}\right) = \frac{ n! }{ k!(n-k)! }\] (Using this to remind myself)

- Astrophysics

\[(1+e^{i t})^8 = \sum_{k=0}^{8} \left(\begin{matrix}8 \\ k\end{matrix}\right)1^{8-k}e^{i t k}\]

- Astrophysics

Any idea what t represents here?

- SolomonZelman

I am comfortable with nCr and nPr formulas that use factorial....

- Empty

Yeah it represents the variable of integration (which ok is probably not the answer you wanted) but since I know you're still getting used to complex numbers this integral will be a little strange for you.
I think it might be a good idea to calculate these integrals and guess what will happen before hand and see what actually happens. If it doesn't make sense (Which it surely won't) then think about it a bit and see if you can figure out a reason for why the hell that would be.
\[\int_0^{2 \pi} e^{ i t} dt\]
\[\int_{- \pi}^{ \pi} e^{ i t} dt\]
\[\int_0^{\pi} e^{ i t} dt\]
It should be pretty quick to just evaluate these integrals, but when you throw complex numbers into things if you're not used to them it can make you doubt certain rules you know to be true so tell me if you have any problems and then we can talk about why these solutions to this combinatoric thing works out. :P

- Astrophysics

Alright thanks, I'll try them out :P

- Empty

I'll give you a vague hint, it's related to vectors in equilibrium. :O

- Empty

I'd say don't spend longer than 5 minutes on this, this is just a quick little intro so that when you come back from trying to figure this out you'll want to hear the explanation lol.

- Astrophysics

I got the first integral to be \[-2\pi e^{2\pi t}\] but this is not making sense to me, I think I have to consider the complex plane

- Astrophysics

Mhm vectors in equilibrium

- Empty

Ahhh almost but not quite right, the integral will not have a \(2 \pi\) out front.

- Astrophysics

Oh oops! Hhahaha

- Astrophysics

Rookie mistake, let me just write it out here \[\int\limits_{0}^{2 \pi} e^{i t} dt = \frac{ e^{ i t} }{ i } |_{0}^{2 \pi} = -i e^{i t} |_{0}^{2\ \pi} = -ie^{i 2\pi}+i \]

- Astrophysics

\[\frac{ 1 }{ i } = - i\]

- Empty

Perfect! Now there are some simplifications to make, and to help, use:
\[e^{i \theta} = \cos \theta + i \sin \theta\]

- Astrophysics

right right eulers formula

- anonymous

it's just using a contour integral to pick out a certain point using the residue theorem or the Cauchy integral formula

- anonymous

$$\frac{f^{(n)}(0)}{n!}=\frac1{2\pi i}\oint\frac{f(z)}{z^{n+1}}\,dz$$ for holomorphic \(z\) using a simple loop contour of winding number \(1\) -- this follows from the fact that we can write \(f\) as a Taylor series: $$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n\\$$ and the fact that the integral of meromorphic \(1/z\) in a loop around 0 is \(2\pi i\) -- the total change in argument -- so here we take $$f(z)=(1+z)^8=\sum_{k=0}^8\binom8k z^k=\sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}z^k$$

- anonymous

it follows $$\frac1{2\pi i}\oint\frac{(1+z)^8}{z^4}\, dz=\frac{f^{(3)}(0)}{3!}$$by looking at the power expansion, \(f^{(3)}(0)/3!\) corresponds to the coefficient of \(z^3\) in the series expansion, so $$\frac1{2\pi i}\oint\frac{(1+z)^8}{z^4}\, dz=\binom83$$

- anonymous

the other integral just follows from the fact that we have $$\oint_{\gamma} f(z)\, dz=\int_a^b f(z(t)) z'(t)\, dt$$where \(\gamma\) is a curve parameterized by \(z:[a,b]\to\mathbb{C}\)

- anonymous

so the simplest closed loop winding once about \(0\) that we know is simply the circle with radius \(1\), parameterized by \(z(t)=e^{it}\) on \([-\pi,\pi]\), so \(z'(t)=ie^{it}\) and we have $$\frac1{2\pi i}\oint\frac{(1+z)^8}{z^4}\,dz=\frac1{2\pi i}\int_{-\pi}^\pi\frac{(1+e^{it})^8}{e^{4it}}\cdot ie^{it}\, dt=\frac1{2\pi}\int_{-\pi}^\pi\frac{(1+e^{it})^8}{e^{3it}}\, dt$$so $$\binom83=\frac1{2\pi}\int_{-\pi}^\pi\frac{(1+e^{it})^8}{e^{3it}}\, dt$$

- anonymous

this sort of stuff is a very simple basic trick in complex analysis and has to do with the fact that a holomorphic function \(f(x+iy)=u(x,y)+iv(x,y)\) can be identified with a vector field \(\tilde f(x,y)=(u(x,y),v(x,y),0)\) on \(\mathbb{R}^3\), where the Cauchy-Riemann equations on \(f\) guarantee \(\nabla\times\tilde f=0\) i.e. \(\tilde f\) is conservative, and thus line integrals on closed curves in \(\tilde f\) vanish--that is, however, if \(\tilde f\) does not has a singularity. take for example the case of \(\tilde f=(-y/(x^2+y^2),x/(x^2+y^2),0)\), where the closed line integrals do not vanish

- Astrophysics

Could you actually please explain what exactly a contour integral is, I know it has to do with integrating in the complex plane, but I don't really follow it.

- anonymous

it's the same thing as a line integral along a curve, only the scalar field and path/contour are complex

- anonymous

recall from multivariable calculus for a function \(f:\mathbb{R^2}\to\mathbb{R}\)$$\int_C f\, ds=\int_a^b f(r(t)))\cdot \|r'(t)\|\,dt$$where \(r:[a,b]\to\mathbb{R^2}\) parameterized a path in the plane

- anonymous

in complex analysis we similarly define for \(f:\mathbb{C}\to\mathbb{C}\) $$\int_\gamma f\, dz=\int_a^b f(z(t))\cdot z'(t)\, dt$$ where \(z:[a,b]\to\mathbb{C}\) parameterizes a path in the complex plane

- Astrophysics

Ah, interesting, thank you!

- Astrophysics

@Empty So now simplifying the result of \[-ie^{i 2 \pi}+i\] using eulers equation \[e^{i 2 \pi} = \cos(2 \pi) +i \sin(2 \pi) = 1+ 0\]
\[-i(1)+i \implies -i+i = 0\] so that will be our result, awesome.

- Astrophysics

For the second integral \[\large \int\limits_{- \pi}^{\pi} e^{i t} dt = - i e^{i t} |_{\pi}^\pi = -ie^{i \pi}+ i e ^{- i \pi}\]
Using Euler's equation \[e^{i \pi} = \cos(\pi)+i \sin(\pi) = -1\]
\[e^{i (-\pi)} = \cos(-\pi) + isin (- \pi) = -1\] noting that \[\cos(-x) = \cos(x) ~~\text{even function}\]
hence we get \[-i(-1)+i(-1) = i-i = 0 \]

- Astrophysics

Ok so the last one is different...\[ \large \int\limits_{0}^{\pi} e^{i t} dt = - i e^{ i t} |_{0}^\pi = - i e ^{i \pi} + i = -i(-1)+i = 2i \]

- Empty

Good! So this last one is weird. Now let's look at the circle, since we have really been integrating around a circle, draw out the angles. So for the first one we have: |dw:1437103238999:dw| Imagine taking a unit vector and rotating it through this angle, and at every point adding them all together. Since every vector has another vector 180 degrees away from it, they will all cancel each other out like vectors in equilibrium.

- Astrophysics

|dw:1437104350913:dw|

- Empty

So the blue line is correct, that represents going from 0 to \(\pi\) for the third problem is good, but the pink line needs some work!
Remember we start at \(-\pi\) and go to \(\pi\) which is weird, since a negative angle means we started from rotating backwards. :P

- Astrophysics

Yeah haha, it is -pi to pi, fixed! |dw:1437110426405:dw|

- Empty

Ok so the first two integrals gave the same value, that's because they were both the same integral. They just started at different spots on the circle and went all the way around while the last integral only went halfway around the circle!

- Astrophysics

Yup, I realized that haha, that was kind of neat, so this is just showing how to do complex integrals?

Looking for something else?

Not the answer you are looking for? Search for more explanations.