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Astrophysics

  • one year ago

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?

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  1. Astrophysics
    • one year ago
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    @ganeshie8 @Empty

  2. Astrophysics
    • one year ago
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    |dw:1437076681063:dw|

  3. Empty
    • one year ago
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    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.

  4. Astrophysics
    • one year ago
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    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?

  5. Astrophysics
    • one year ago
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    \[\left(\begin{matrix}n \\ k\end{matrix}\right) = \frac{ n! }{ k!(n-k)! }\] (Using this to remind myself)

  6. Astrophysics
    • one year ago
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    \[(1+e^{i t})^8 = \sum_{k=0}^{8} \left(\begin{matrix}8 \\ k\end{matrix}\right)1^{8-k}e^{i t k}\]

  7. Astrophysics
    • one year ago
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    Any idea what t represents here?

  8. SolomonZelman
    • one year ago
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    I am comfortable with nCr and nPr formulas that use factorial....

  9. Empty
    • one year ago
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    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

  10. Astrophysics
    • one year ago
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    Alright thanks, I'll try them out :P

  11. Empty
    • one year ago
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    I'll give you a vague hint, it's related to vectors in equilibrium. :O

  12. Empty
    • one year ago
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    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.

  13. Astrophysics
    • one year ago
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    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

  14. Astrophysics
    • one year ago
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    Mhm vectors in equilibrium

  15. Empty
    • one year ago
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    Ahhh almost but not quite right, the integral will not have a \(2 \pi\) out front.

  16. Astrophysics
    • one year ago
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    Oh oops! Hhahaha

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

  18. Astrophysics
    • one year ago
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    \[\frac{ 1 }{ i } = - i\]

  19. Empty
    • one year ago
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    Perfect! Now there are some simplifications to make, and to help, use: \[e^{i \theta} = \cos \theta + i \sin \theta\]

  20. Astrophysics
    • one year ago
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    right right eulers formula

  21. anonymous
    • one year ago
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    it's just using a contour integral to pick out a certain point using the residue theorem or the Cauchy integral formula

  22. anonymous
    • one year ago
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    $$\frac{f^{(n)}(0)}{n!}=\frac1{2\pi i}\oint\frac{f(z)}{z^{n+1}}\,dz$$ 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 $$f(z)=(1+z)^8=\sum_{k=0}^8\binom8k z^k=\sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}z^k$$

  23. anonymous
    • one year ago
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    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 $$\frac1{2\pi i}\oint\frac{(1+z)^8}{z^4}\, dz=\binom83$$

  24. anonymous
    • one year ago
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    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 \(z:[a,b]\to\mathbb{C}\)

  25. anonymous
    • one year ago
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    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 $$\binom83=\frac1{2\pi}\int_{-\pi}^\pi\frac{(1+e^{it})^8}{e^{3it}}\, dt$$

  26. anonymous
    • one year ago
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    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

  27. Astrophysics
    • one year ago
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    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.

  28. anonymous
    • one year ago
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    it's the same thing as a line integral along a curve, only the scalar field and path/contour are complex

  29. anonymous
    • one year ago
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    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

  30. anonymous
    • one year ago
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    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

  31. Astrophysics
    • one year ago
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    Ah, interesting, thank you!

  32. Astrophysics
    • one year ago
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    @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.

  33. Astrophysics
    • one year ago
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    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 \[-i(-1)+i(-1) = i-i = 0 \]

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

  35. Empty
    • one year ago
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    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.

  36. Astrophysics
    • one year ago
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    |dw:1437104350913:dw|

  37. Empty
    • one year ago
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    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

  38. Astrophysics
    • one year ago
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    Yeah haha, it is -pi to pi, fixed! |dw:1437110426405:dw|

  39. Empty
    • one year ago
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    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!

  40. Astrophysics
    • one year ago
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    Yup, I realized that haha, that was kind of neat, so this is just showing how to do complex integrals?

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