Question

Maclaurin series for trig function

Answer 1

lol ... which one?

Answer 2

Find the maclaurin series for:
\[f(x) = x*\cos(x ^{3})\]
Given that:
\[\cos(x) = 1-\frac{ x ^{2} }{ 2 } + \frac{ x ^{4} }{ 4! } - \frac{ x ^{6} }{ 6! } + ... = \sum_{n=0}^{\infty} \frac{ (-1)^{n}x ^{2n} }{ (2n)! }\]
Show working.

Answer 3

well, determine cos(u) and let u=x^3 .... which boils down to replaceing x by x^3 .... then multiply thru by x

Answer 4

\[\cos(x) = \sum_{n=0}^{\infty} \frac{ (-1)^{n}x ^{2n} }{ (2n)! }\]
\[\cos(u) = \sum_{n=0}^{\infty} \frac{ (-1)^{n}u ^{2n} }{ (2n)! }\]
\[\cos(x^3) = \sum_{n=0}^{\infty} \frac{ (-1)^{n}(x^3) ^{2n} }{ (2n)! }\]
\[x*\cos(x^3) = \sum_{n=0}^{\infty} \frac{ (-1)^{n}~ x*(x^3) ^{2n} }{ (2n)! }\]

Answer 5

so \[xcos(u)=1-x\frac{ u ^{2} }{ 2 } + x\frac{ u ^{4} }{ 4! }...\]
then \[xcos(x^{3})=1-x \frac{ (x ^{3})^{2} }{ 2 }+x \frac{ (x ^{3})^{4} }{ 4! }...?\]

Answer 6

thats a fair assessment yes ... it doesnt use the maclaurin perse, but fair in my eyes

Answer 7

Opps, xcos(x^3) = x - x(x3)^2.... right?

Answer 8

Thanks!

Answer 9

yes:) x(1) for the first term