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

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

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