Manipulating the Maclaurin Series for e^(-4x^4) to Solve an Integral

[HolyDesperado]

Homework Statement

Assume that e^x equals its Maclaurin series for all x.
Use the Maclaurin series for e^(-4 x^4) to evaluate the integralYour answer will be an infinite series. Use the first two terms to estimate its value.

Homework Equations

The Attempt at a Solution

I've tried using the e^x series above to solve the given series. From what I understand, I am suppose to manipulate the e^x series so that it becomes e^(-4x^4). I am new to these forums, and I don't know another way to show what I'm doing, so I'll post an image of what I have so far (which is incorrect):Can anyone point me in the right direction or tell me what my mistakes are?
Thanks a lot and sorry if I broke any forum etiquette rules here! It's my first time here. =)

 

[Dick]

No rules broken. Welcome to the forums! You want (-4*x^4)^n. Not -4*(x^4)^n. You get powers of the -4 as well.

 

[HolyDesperado]

Hey Dick sorry for the late reply, I had to run out real quick, but I tried your suggestion of (-4x^4)^n. However, I am still getting an incorrect answer. I brought out -1/n! and then integrated (4x^4)^n. Which should give me 4^nx(x^4)^n/4n+1?

 

[Dick]

Then you get powers of the -1 as well. (-1)^n. Your power on the x isn't right either in the integral. You had it right before.

 

[HolyDesperado]

Then you get powers of the -1 as well. (-1)^n. Your power on the x isn't right either in the integral. You had it right before.

I'm afraid I don't understand. Can you elaborate on this please?

 

[Dick]

(-4)^1=-4. (-4)^2=16. (-4)^3=-64. (-4)^4=256. They aren't all the same sign. You can't factor the (-1) out.

 

[HolyDesperado]

so you're saying I should get (-4x)^(4n+1)/(4n+1)n! ... ?

 

[Dick]

No. (-4x^4)^n=(-4)^n*(x^4)^n=(-4)^n*x^(4n). Now integrate the power.

 

[HolyDesperado]

No. (-4x^4)^n=(-4)^n*(x^4)^n=(-4)^n*x^(4n). Now integrate the power.

By integrate the powers you mean -4^n*x^(4n) = -4^(n+1)*x^(4n+1)/(4n+1)n! ...??

Sorry, English is not my native tongue.

 

[Dick]

By 'power' I meant the x^(4n). Sorry. (-4)^n is a constant in each term. It doesn't depend on x. You only apply the power law to x^(4n).

 

[HolyDesperado]

By 'power' I meant the x^(4n). Sorry. (-4)^n is a constant in each term. It doesn't depend on x. You only apply the power law to x^(4n).

Ah, I see it more clearly now, but once I plug in x, which is .1, I should get (-4)^n*(0.1)^(4n+1)/(4n+1)n! I then plugged in 0 and 1 for n to test the first two terms of the series, but my answer came out wrong again. =( Please help.

 

[Dick]

Can you show us the first two terms of the series you used and the numbers you got? Remember the first term is n=0. That term is pretty simple.

 

[HolyDesperado]

Sure, here is what I've done below:

http://img261.imageshack.us/img261/9651/mathhelp.jpg

 

[Dick]

I get 0.099992. I didn't round off. Could that be the problem? Otherwise, it looks fine.

 

[HolyDesperado]

I get 0.099992. I didn't round off. Could that be the problem? Otherwise, it looks fine.

I left off the two at the end when I entered my solution, and webwork counted that as incorrect. I just tried your solution with the extra 2 and now it is correct. Silly webwork. =) Thanks for you patience throughout helping me find this solution. Hopefully we can do business again in the future!

 

[Dick]

You're welcome. You know where to find me. BTW your English is excellent. You even spell words correctly. Native speakers don't do that any more.