Question

68. If \(f(x) = {(1 + {x^3})^{30}}\), what is \({f^{58}}(0)\)

Short answer

\({f^{58}}(0) = 0\)

Step-by-step solution

Step 1: The Maclaurin series

To solve such problems you need to remember that : The Maclaurin series of a polynomial is the polynomial itself!

Maclaurin series of f(x) is

\(\sum\limits_{n = 0}^\infty {\frac{{{f^n}(0){x^n}}}{{n!}}} \)

Note the term : \(\frac{{{f^{58}}(0){x^{58}}}}{{58!}}\)

If we can find the coefficient of \({x^{58}}\)in the expansion of \({(1 + {x^3})^{30}}\)we can use the following relation

Coefficient of \({x^{58}} = \frac{{{f^{58}}(0)}}{{58!}}\)

Step 2: Further simplification

Note that powers of x in the expansion of \({(1 + {x^3})^{30}}\)must be multiples of 3, but 58 is not a multiple of 3.

Therefore there is no term in the expansion of \({(1 + {x^3})^{30}}\)with power 58. Therefore the coefficient of \({x^{58}}\)is 0.

Therefore,

\(0 = \frac{{{f^{58}}(0)}}{{58!}}\)\({f^{58}}(0) = 0\)