How do you find the Taylor polynomial of degree 4 for #f(x) = cosh(x)# about #x = 0 # by using the given Taylor polynomial for #e^x#?

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Answer 1 · 2017-05-09T22:51:03

#cosh x = 1+x^2/2+x^4/24+o(x^4)#

The MacLaurin series for #e^x# is:

#e^x = sum_(n=0)^oo x^n/(n!)#

Truncating the series at #n=4# we obtain the Taylor polynomial of degree 4:

#e^x = 1 + x +x^2/2 +x^3/6+x^4/24 +o(x^4)#

and substituting #-x# we have:

#e^(-x) = 1 - x +x^2/2 -x^3/6+x^4/24 +o(x^4)#

Using the exponential form of #cosh x# then we have:

#cosh x = (e^x+e^(-x))/2 = 1/2( 1 + x +x^2/2 +x^3/6+x^4/24 + 1 - x +x^2/2 -x^3/6+x^4/24)+o(x^4)#

and we can see that the terms of even order cancel each other, so that:

#cosh x = 1+x^2/2+x^4/24+o(x^4)#