An identity of hyperbolic functions

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  • #1
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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
 

Answers and Replies

  • #2
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You cannot "ignore the coefficients". They are important.

Can you use complex numbers?

Did you try induction?
 
  • #3
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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
 
  • #4
PeroK
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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
What is ##cosh(x) + sinh(x)##? Maybe it's something quite simple?
 
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  • #5
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$$\cosh(x)+\sinh(x)=2e^x~~\rightarrow~~(\cosh(x)+\sinh(x))^n=(2e^x)^n=2^ne^{nx}=2^{n-1}(\cosh(nx)+\sinh(nx))$$
The prefix ##2^{n-1}## is redundant
 
  • #6
PeroK
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$$\cosh(x)+\sinh(x)=e^x$$
 
  • #7
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$$\cosh(x)+\sinh(x)=\frac{1}{2}(e^x+e^{-x})+\frac{1}{2}(e^x-e^{-x})=e^x$$
Thank you PeroK and mfb
 
  • #8
Ssnow
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The prefix ##2^{n−1}## is redundant
not redundant, simply there isn't ...
 

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