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Math Help - proof by induction ...

  1. #1
    Junior Member
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    proof by induction ...

    I am asked to prove the following by induction...

    y=(e^x)sinx => d(^n)y/dx = [2^(n/2)] (e^x) [sin(x+npi/4)]

    unfortunately I m trying to establish that this is true for n=1 and I am stuck already.

    I get dy/dx = (e^x)(sinx-cosx) and not much passed that...

    I have tried to square both sides and it gets me as far as...

    dy/dx = [2(1/2)](e^x)[1/2 - sinxcosx]

    ....?
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  2. #2
    MHF Contributor red_dog's Avatar
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    First of all we have

    \sin a+\cos a=\sin a+\cos\left(\frac{\pi}{2}-a\right)=2\sin\frac{\pi}{4}\cos\left(\frac{\pi}{4}-a\right)=

    =\sqrt{2}\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+a\right)=\sqrt{2}\sin\left(a+\frac{\  pi}{4}\right)

    Now, for n=1:

    (e^x\sin x)'=e^x(\sin x+\cos x)=\sqrt{2}e^x\sin\left(x+\frac{\pi}{4}\right)

    Suppose that (e^x\sin x)^{(n)}=2^{\frac{n}{2}}e^x\sin\left(x+\frac{n\pi}  {4}\right)

    We have to prove that (e^x\sin x)^{(n+1)}=2^{\frac{n+1}{2}}e^x\sin\left(x+\frac{(  n+1)\pi}{4}\right)

    (e^x\sin x)^{(n+1)}=\left[(e^x\sin x)^{(n)}\right]'=\left[2^{\frac{n}{2}}e^x\sin\left(x+\frac{n\pi}{4}\right  )\right]'=

    =2^{\frac{n}{2}}e^x\left[\sin\left(x+\frac{n\pi}{4}\right)+\cos\left(x+\fra  c{n\pi}{4}\right)\right]=

    =2^{\frac{n}{2}}e^x\sqrt{2}\sin\left(x+\frac{n\pi}  {4}+\frac{\pi}{4}\right)=2^{\frac{n+1}{2}}e^x\sin\  left(x+\frac{(n+1)\pi}{4}\right)
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  3. #3
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    really thanks! i will study it very carefully
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