Find the Taylor expansion of cosh x?

Question

Find the Taylor expansion for cosh x about x=0 ?

given:- cos x taylor expansion formulae

the relation cos x = cosh(ix) may be used here

b)Assume that x and y are real variables

i) What are the domain and range of the function cosh?

ii) By convention, how is the domain of cosh restricted in construcing

the inverse function cosh^-1?

iii)What are the domain and range of the function cosh^-1?

v) Show that cosh^-1x can be written as cosh^-1 x = ln(x+(x^2-1)^1/2)

vi) hence, or otherwiseshow that d/dx(cosh^-1 x)= 1/(x^2-1)^1/2

Answer 1

From cos x = cosh(ix) we get cos(ix) = cosh(i(ix)) = cosh(i^2x) =

cosh (-x) = cosh x, since cosh is an even function.  We know that

cos x = Sum[k =0,inf]{(-1)^kx^(2k)/(2k)!}.  Therefore, cos(ix) =

Sum[k=0,inf]{(-1)^k(ix)^(2k)/(2k)!} = Sum[k=0,inf]{(-1)^k(i^2)^kx^(2k)/(2k)!} = Sum[k=0,inf]{(-1)^k(-1)^kx^(2k)/(2k)!} =

Sum[k=0,inf]{x^(2k)/(2k)!}.