find the taylor series f(x)1/2(e^x + e^-x) ;x=0

since this is just f(x) = cosh(x), and the derivatives are sinh,cosh,sinh,cosh,...

f(x) = 1 + x^2/2! + x^4/4! + x^6/6! + ...

To find the Taylor series of the function f(x) = (1/2)(e^x + e^(-x)) centered at x = 0, we can use the formula for the Taylor series expansion. The general formula for the nth term of the Taylor series expansion of a function f(x) centered at x = a is given by:

f^(n)(a) * (x - a)^n / n!

where f^(n)(a) represents the nth derivative of f(x) evaluated at x = a.

Let's begin by finding the derivatives of f(x):

f(x) = (1/2)(e^x + e^(-x))

Differentiating once:
f'(x) = (1/2)(e^x - e^(-x))

Differentiating again:
f''(x) = (1/2)(e^x + e^(-x))

It seems that we have reached a pattern. Each derivative yields either e^x or e^(-x) terms with a coefficient of (1/2). Therefore, the nth derivative will be (1/2)(e^x + e^(-x)) for even n, and (1/2)(e^x - e^(-x)) for odd n.

Now, let's evaluate these derivatives at x = 0:

f(0) = (1/2)(e^0 + e^(0)) = 1

f'(0) = (1/2)(e^0 - e^(0)) = 0

f''(0) = (1/2)(e^0 + e^(0)) = 1

We can see that after evaluating the derivatives at x = 0, the function f(x) = (1/2)(e^x + e^(-x)) becomes:

f(x) = 1 + x^2/2! + x^4/4! + x^6/6! + ...

This is the Taylor series expansion of f(x) centered at x = 0.