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Mimi_x3

  • one year ago

im not able to figure this out

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  1. Mimi_x3
    • one year ago
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    https://gyazo.com/883588ab3191a7dbb78de630c9b1549c

  2. Mimi_x3
    • one year ago
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    how does sin (...) sin(..) = cos(...)-cos(...)

  3. DanJS
    • one year ago
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    link is broken...

  4. Mimi_x3
    • one year ago
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    apparently its accordance to this https://gyazo.com/0df138c31be5b67523ca6f3ea60ca281

  5. Mimi_x3
    • one year ago
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    itrs not borken

  6. inkyvoyd
    • one year ago
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    FOURIER SERIES

  7. Mimi_x3
    • one year ago
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    NOT YET

  8. inkyvoyd
    • one year ago
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    YES IT IS ALMOST THERE

  9. DanJS
    • one year ago
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    oh it was my browser tha tbroke, sorry hah

  10. Mimi_x3
    • one year ago
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    well ya IM LIKE 5 WEEKS BEHIND LOL

  11. anonymous
    • one year ago
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    It's just product to sum formulas.

  12. inkyvoyd
    • one year ago
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    http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3

  13. Mimi_x3
    • one year ago
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    i need to know this https://gyazo.com/883588ab3191a7dbb78de630c9b1549c

  14. inkyvoyd
    • one year ago
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    miwi review http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3

  15. inkyvoyd
    • one year ago
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    actually the top of the link at http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx

  16. DanJS
    • one year ago
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    hell yeah, pauls notebook is awesome

  17. Mimi_x3
    • one year ago
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    oh FML

  18. Mimi_x3
    • one year ago
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    I WANNA DROP THIS COURSE

  19. DanJS
    • one year ago
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    Arthur Matuk has awesome video series for DE on MIT OCW, got me an A using that series ...

  20. inkyvoyd
    • one year ago
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    thanks for info dan - gonna take ODEs next semester :o

  21. Mimi_x3
    • one year ago
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    it still doesnt tell me why sin(..)sin(...) = cos(..) - cos(..)

  22. inkyvoyd
    • one year ago
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    http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex1 doesn't help?

  23. zepdrix
    • one year ago
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    @Mimi_x3 as Concentrationalizing mentioned, it's the trig product to sum formula:\[\large\rm \sin(a)\sin(b)=\frac{1}{2}\left[\cos(a-b)-\cos(a+b)\right]\]

  24. Mimi_x3
    • one year ago
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    why does my prof say this https://gyazo.com/55c13f04ae511455aa39b45ddc333030

  25. Mimi_x3
    • one year ago
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    (havent done maths for 3 years)

  26. inkyvoyd
    • one year ago
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    that's a trig fromula from precalc

  27. zepdrix
    • one year ago
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    the angle addition formula? Hmm not sure why he included that... thinking

  28. inkyvoyd
    • one year ago
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    the minus on top of the plus means the signs are opposite

  29. inkyvoyd
    • one year ago
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    zepdrix do you think he was using it to derive the formulas you linked too?

  30. zepdrix
    • one year ago
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    Hmm :\ I don't think so.\[\large\rm \cos(A-B)=\cos(A) \cos(B)+\sin(A) \sin(B)\]Subtracting cos(A)cos(B) from each side,\[\large\rm \cos(A-B)-\cos(A) \cos(B)=\sin(A) \sin(B)\]Which doesn't quite get us there :d Hmm

  31. zepdrix
    • one year ago
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    Oh wait wait wait.. maybe this.\[\large\rm (1)\qquad \cos(A-B)=\cos(A) \cos(B)+\sin(A) \sin(B)\]\[\large\rm (2)\qquad \cos(A+B)=\cos(A) \cos(B)-\sin(A) \sin(B)\]I'm going to combine these equations like this: \(\large\rm (1)-(2)\), giving us:\[\large\rm \cos(A-B)-\cos(A+B)=\begin{align}&~~~~\cos(A) \cos(B)+\sin(A) \sin(B)\\&-\cos(A) \cos(B)+\sin(A) \sin(B)\end{align}\]Ah yes, I guess it does lead to the identity I mentioned :)\[\large\rm \cos(A-B)-\cos(A+B)=2\sin(A) \sin(B)\]

  32. dan815
    • one year ago
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    what do u wanna know aobut inner product

  33. dan815
    • one year ago
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    you know how to take inner / dot produt of 2 vecotrs now think of it like taking infinite dimensional dot product where sin(x) is telling you like the value at eacch dimension in your vector

  34. dan815
    • one year ago
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    |dw:1443255318087:dw|

  35. Empty
    • one year ago
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    \[e^{ix} = \cos x + i \sin x\]\[e^{-ix}=\cos x - i \sin x\]

  36. UnkleRhaukus
    • one year ago
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    \[\begin{align} \int\limits_{-\pi}^\pi\sin(mx)\sin(nx)\text dx &=\frac12\int\limits_{-\pi}^\pi\cos((m-n)x)-\cos((m+n)x)\text dx\\ &=\int\limits_{0}^\pi\cos((m-n)x)-\cos((m+n)x)\text dx\qquad (\text{even integrand})\\\\ &=\frac{\sin((m-n)x)}{m-n}-\frac{\sin((m+n)x)}{m+n}\Big|_0^\pi\\ &=\frac{\sin(\pi(m-n))}{m-n}-\frac{\sin(\pi(m+n))}{m+n}\\ &\hspace{16em} p=m-n\\ &\hspace{16em} q=m+n\\ \text{I for }m\neq n &\qquad=\frac{\sin(\pi p)}{p}-\frac{\sin(\pi q)}{q}\\ &\qquad=\frac 0p-\frac0q\\ &\qquad=0\\ \\ \text{II for }m=n=0 &\qquad=\lim\limits_{p\rightarrow 0}\frac{\sin(\pi p)}{p}-\frac{\sin(2\pi m)}{2m}\\ &\qquad\stackrel{\text{l'H}}=\pi\frac{\cos(\pi p)}{1}-2\pi\frac{\cos(2\pi m)}{2}\\ &\qquad=\pi-\frac{2\pi}2\\ &\qquad=0\\ \\ \text{III for }m=n\neq0&\qquad=\lim\limits_{p\rightarrow 0}\frac{\sin(\pi p)}{p}-\frac{\sin(\pi q)}{q}\\ &\qquad\stackrel{\text{l'H}}=\pi\lim\limits_{p\rightarrow 0}\frac{\cos(\pi p)}{1}-\frac 0q\\ &\qquad=\pi-0\\ &\qquad=\pi\\ \\ &=\pi\delta_{m,n}(1-\delta_{m,0}) \end{align}\]

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