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I An identity of hyperbolic functions

  1. Oct 18, 2016 #1
    Prove: ##(\cosh(x)+\sinh(x))^n=\cosh(nx)+\sinh(nx)##
    Newton's binomial: ##(a+b)^n=C^0_n a^n+C^1_n a^{n-1}b+...+C^n_n b^n## and: ##(a-b)^n~\rightarrow~(-1)^kC^k_n##
    I ignore the coefficients.
    $$(\cosh(x)+\sinh(x))^n=\cosh^n(x)+\cosh^{n-1}\sinh(x)+...+\sinh^n(x)$$
    $$\cosh^n(x)=(e^x+e^{-x})^n=e^{nx}+e^{(n-2)x}+e^{(n-4)x}+...+e^{-nx}$$
    $$\cosh^{n-1}\sinh(x)=(e^x+e^{-x})^{n-1}(e^x-e^{-x})=...=e^{nx}-e^{-nx}$$
    I use this result in the next derivations:
    $$\cosh^{(n-2)}x\sinh^2(x)=[(e^x+e^{-x})^{n-1}(e^x-e^{-x})](e^x-e^{-x})=...=e^{nx}+e^{-nx}-e^{(n-2)x}-e^{-(n-2)x}$$
    $$\cosh^{(n-3)}x\sinh^3(x)=[(e^x+e^{-x})^{n-1}(e^x-e^{-x})](e^x-e^{-x})^2=...=e^{nx}-e^{-nx}+2e^{-(n-2)x}+e^{(n-4)x}-2e^{(n-2)x}-e^{-(n-4)x}$$
    It doesn't lead anywhere
     
  2. jcsd
  3. Oct 18, 2016 #2

    mfb

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    You cannot "ignore the coefficients". They are important.

    Can you use complex numbers?

    Did you try induction?
     
  4. Oct 18, 2016 #3
    I ignored coefficients to see if my method brings me somewhere, later i would have considered them.
    I don't know complex numbers good, i will try to learn. the same for induction
     
  5. Oct 18, 2016 #4

    PeroK

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    What is ##cosh(x) + sinh(x)##? Maybe it's something quite simple?
     
  6. Oct 18, 2016 #5
    $$\cosh(x)+\sinh(x)=2e^x~~\rightarrow~~(\cosh(x)+\sinh(x))^n=(2e^x)^n=2^ne^{nx}=2^{n-1}(\cosh(nx)+\sinh(nx))$$
    The prefix ##2^{n-1}## is redundant
     
  7. Oct 18, 2016 #6

    PeroK

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    $$\cosh(x)+\sinh(x)=e^x$$
     
  8. Oct 18, 2016 #7
    $$\cosh(x)+\sinh(x)=\frac{1}{2}(e^x+e^{-x})+\frac{1}{2}(e^x-e^{-x})=e^x$$
    Thank you PeroK and mfb
     
  9. Oct 18, 2016 #8

    Ssnow

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    not redundant, simply there isn't ...
     
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