Finding Maclaurin Series of f(x)=sin(3t^2) up to Degree 4

[snoggerT]

Find the MacLaurin polynomial of degree 4 for f(x)
f(x)= (integral from 0 to x) sin(3t^2)



[f^(n)(0)/n!]*x^n

The Attempt at a Solution

- I took the 4th derivative of sin(3t^2) and got:

f''''(x)= -108sin(3t^2)-1296t^2cos(3t^2)+1296t^4sin(3t^2)

Not real sure what to do from there. I plugged 0 in for t to find my f^(n)(0) and got 0,1,1,-1,-1 ...but I'm not sure if that's right. Can somebody please check my derivative and point me in the right direction for finding the series?

 

[Avodyne]

This problem is easier than you're making it out to be.

Since f(x) is given by the integral, to get the degree 4 and lower terms in f(x), we need only the degree 3 and lower terms in the integrand. Then, since sin(z) = z + O(z^3), we have sin(3t^2) = 3t^2 + O(t^6), and so we need only keep the first term.

 

[snoggerT]

This problem is easier than you're making it out to be.

Since f(x) is given by the integral, to get the degree 4 and lower terms in f(x), we need only the degree 3 and lower terms in the integrand. Then, since sin(z) = z + O(z^3), we have sin(3t^2) = 3t^2 + O(t^6), and so we need only keep the first term.

- I understand what you're saying about only needing to go to the degree 3, but I'm not quite sure I get the sin(z) = z +O(t^6). Can you please explain that a little more?

 

[Avodyne]

sin(z) = z - z^3/3! + ... is the standard series for the sine function. If we substitute 3t^2 for z, we get sin(3t^2) = 3t^2 - 9t^6/3! + ... The t^6 and higher terms will not be needed, as they integrate to x^7 and higher, and we only want terms up to x^4.