Question

How do you find the power series for `f(x)=sinhx+xcoshx` and determine its radius of convergence?

Answer 1

`f(x) = 2x + (2x^3)/3 + x^5/(20) + x^7/(630) + ... \ \ \ \ x in RR`

Explanation:

There are several ways to develop a solution, we could start from first principles and form the Taylor series for the hyperbolic functions using their derivatives; we could also use the definitions of the hyperbolic functions in terms of `e^x`; or we could use the fact that the Taylor Series for these functions are commonly used and very often available in formula books for exams.

I'll initially start with the assumption that we know what the TS for the hyperbolic functions are. That is:

`sinhx = x + x^3/(3!) + x^5/(5!) + x^7/(7!) + ... \ \ \ \ x in RR`
`coshx = 1 + x^2/(2!) + x^4/(4!) + x^6/(6!) + ... \ \ \ \ x in RR`

And so the power series for `f(x)=sinhx+xcoshx` is given by:

`f(x) = x + x^3/(3!) + x^5/(5!) + x^7/(7!) + ...`
`\ \ \ \ \ \ \ \ \ \ \ +x{1 + x^2/(2!) + x^4/(4!) + x^6/(6!) + ...}`
`\ \ \ \ \ \ \ = x + x^3/(3!) + x^5/(5!) + x^7/(7!) + ...`
`\ \ \ \ \ \ \ \ \ \ \ +x + x^3/(2!) + x^5/(4!) + x^7/(6!) + ...`
`\ \ \ \ \ \ \ = 2x + (2x^3)/3 + x^5/(20) + x^7/(630) + ...`

If you need the derivation of the hyperbolic power series then this can be done using the power series for `e^x`

`e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... \ \ \ \ x in RR`

Along with the definitions:

`sinhx = (e^x-e^-x)/2`
`coshx = (e^x+e^-x)/2`

Giving:

`sinhx = 1/2{ 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... }`
`\ \ \ \ \ \ \ \ \ \ \ -1/2{ 1 - x + x^2/(2!) - x^3/(3!) + x^4/(4!) + ... }`
`\ \ \ \ \ \ \ \ \ = x + x^3/(3!) + x^5/(5!) + x^7/(7!) + ...`

`coshx = 1/2{ 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... }`
`\ \ \ \ \ \ \ \ \ \ \ +1/2{ 1 - x + x^2/(2!) - x^3/(3!) + x^4/(4!) + ... }`
`\ \ \ \ \ \ \ \ \ = 1 + x^2/(2!) + x^4/(4!) + x^6/(6!) + ...`