# find the taylor series about x=0 for the following functions. f(x) = e^(x) + e^(-x)

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You need to remember the equation that gives the Taylor's series of the function f, about a value a such that:

`f(x) = f(a) + (f'(a))/(1!)(x - a) + (f''(a))/(2!)(x - a)^2 + ...`

Hence, evaluating the Taylor's series of the function `f(x) = e^x` , about `x = 0` yields:

`f(x) = f(0) + (f'(0))/(1!)(x - 0) + (f''(0))/(2!)(x - 0)^2 + ...`

`f(0) = e^0 => f(0) = 1`

`f'(x) = e^x => f'(0) = 1`

`f''(x) = e^x => f''(0) = 1`

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`f(x) = 1 + x + x^2/2 + x^3/(3!)+...`

You need to evaluate the Taylor's series of the function `f(x) = e^(-x)` , about `x = 0` yields:

`f(x) = f(0) + (f'(0))/(1!)(x - 0) + (f''(0))/(2!)(x - 0)^2 + ...`

`f(0) = e^0 => f(0) = 1`

`f'(x) = -e^x => f'(0) = -1 `

`f''(x) = e^x => f''(0) = 1 `

`f'''(x) = -e^x => f'''(0) = -1`

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`f(x) = 1- x + x^2/2 -x^3/(3!)+ ...`

**Hence, evaluating the Taylor's series of the function `f(x) = e^x + e^(-x)` , about `x = 0` yields:**

`f(x) = 1 + x + x^2/2 + x^3/(3!)+ ... + 1 - x + x^2/2 - x^3/(3!)+ ...`

`f(x) = 2 + 2(x^2/2) + 2(x^2/(4!)) + .... + 2*((x^(2n))/(2n!))`