Use taylor series to evaluate #f^11 (0)# for #f(x) = x^3 cos(x^2)# ?

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Answer 1 · 2018-06-10T01:06:24

#(11!)/(4!) = 1663200#

We know that the Taylor expansion if #cos x# around #x = 0# is given by

#cos x = 1-x^2/(2!)+x^4/(4!)-... implies#

#cos (x^2) = 1-x^4/(2!)+x^8/(4!)-... #

This yields the Taylor expansion of the function #x^3cos(x^2)# around #x=0# as

#x^3cos( x^2) = x^3-x^7/(2!)+x^11/(4!)-... #

Since the coefficient of #x^n# in the Taylor expansion is given by

#c_n = f^{(n)}/(n!)#

we have

#f^{(11)}(0)/(11!) = 1/(4!)#

and thus

#f^{(11)}(0) = (11!)/(4!) = 1663200#