Series Help: Finding ∫qk(x)dx for k = 2,6,10,14 in Approximating ∫sin(x^2)dx

[Grew Gore]

Homework Statement

Find ∫qk(x) dx where the upper bound is 1 and the lower bound is 0. g is some function and we are finding for k = 2,6,10 and 14, hence the first four non-zero terms of a series that can be used to calculate approximations to I = ∫sin(x^2) dx were the upper bound is 1 and the lower bound is 0.

Homework Equations

The Attempt at a Solution

I am struggling to figure out how to incorporate k and get the terms we are after like in a Taylor Polynomial, but what I got so far is q/2 but that's assuming that q is just a constant. I'm stuck.

 

[LCKurtz]

Homework Statement

Find ∫qk(x) dx where the upper bound is 1 and the lower bound is 0. g is some function and we are finding for k = 2,6,10 and 14, hence the first four non-zero terms of a series that can be used to calculate approximations to I = ∫sin(x^2) dx were the upper bound is 1 and the lower bound is 0.

Homework Equations

The Attempt at a Solution

I am struggling to figure out how to incorporate k and get the terms we are after like in a Taylor Polynomial, but what I got so far is q/2 but that's assuming that q is just a constant. I'm stuck.

I would start by writing the first four nonzero terms of the series for $\sin x$. Then replace $x$ by $x^2$. Then you might see where the values of $k$ come from and be on your way.

 

[Grew Gore]

I would start by writing the first four nonzero terms of the series for $\sin x$. Then replace $x$ by $x^2$. Then you might see where the values of $k$ come from and be on your way.

Ok thanks. So the series for sin(x^2) is x^2 - (x^6)/3! + (x^10)/5! - (x^14)/7! + ...
And when its asks to find ∫qk(x) dx for 2, is that just the integral of x^2 which is (x^3)/3, and the integral of (x^6)/3! which is (x^7)/42

I see how the powers relate to the k values, but am unsure what the q is about. I'm sorry if I'm asking too much, just really need to get this done.

 

[LCKurtz]

Ok thanks. So the series for sin(x^2) is x^2 - (x^6)/3! + (x^10)/5! - (x^14)/7! + ...
And when its asks to find ∫qk(x) dx for 2, is that just the integral of x^2 which is (x^3)/3, and the integral of (x^6)/3! which is (x^7)/42

I see how the powers relate to the k values, but am unsure what the q is about. I'm sorry if I'm asking too much, just really need to get this done.

The $q_k$'s are just names for the four terms you are integrating. They have been named by their exponents. $q_2(x) = \frac{x^2}{1}$, etc... If you had expanded $\sin(x^2)$ by its Taylor series the long way, the $k$ values would be the usual subscripts for the nonzero terms you get.

 

[Ray Vickson]

Ok thanks. So the series for sin(x^2) is x^2 - (x^6)/3! + (x^10)/5! - (x^14)/7! + ...
And when its asks to find ∫qk(x) dx for 2, is that just the integral of x^2 which is (x^3)/3, and the integral of (x^6)/3! which is (x^7)/42

I see how the powers relate to the k values, but am unsure what the q is about. I'm sorry if I'm asking too much, just really need to get this done.

It is not clear from what you wrote whether $q_k = (-1)^k x^{k}/k!$ of if $q_k = \sum_{j=0,2, \ldots k} (-1)^j x^j/j!.$ In other words, is $q_k$ the $k$th term, or is it the $k$th partial sum?