Question

use the fact that the taylor series of g(x) = sin(x^2) is:
x^2 - (x^6/3!) + (x^(10)/5!) - (x^(14)/7!) + . . .

to find g"(0), g"'(0), and g^(10)(0)

Answer 1

what's the trick?

Answer 2

For get about the series and just go with the function

Answer 3

since the taylor series is a polynomial, it just derives using the power rule

Answer 4

you might wanna write it up in its summation notation tho

Answer 5

you are given that
\[g(x)\approx x^2-{x^6\over3!}+{x^{10}\over5!}-{x^{14}\over7!}+\ldots\]

Answer 6

notice that when you take the second derivative, you get
\[g''(x)=2-6\times5{x^4\over3!}+10\times9{x^8\over5!}-14\times13{x^{12}\over7!}+\ldots\]

Answer 7

when you take the third derivative, the first term becomes "zero"
\[g'''(x)=-6\times5\times4{x^3\over3!}+10\times9\times8{x^7\over5!}-14\times13\times12{x^{11}\over7!}+\ldots\]

Answer 8

now, when you take the 10^th derivative, the second term reduces to a constant 10!, the third term coefficient becomes \(14\times13\times12\ldots\times5\) and the exponent becomes \(x^4\).
\[g^{(10)}(x)={10!\over5!}-{14!\over4!}{x^4\over7!}+\ldots\]

Answer 9

now plugin "x=0" in each of those epansions.

Answer 10

thank you!

Answer 11

yw