Find the MacLaurin polynomial of degree 5 for F(x).

[McAfee]

Homework Statement

Homework Equations

The Attempt at a Solution

I keep getting the answer 1- 96/4!*x^4. Can't find where I'm going wrong.

 

[Dick]

It's really hard to tell what you are doing from the blurry photograph. Can you explain? What are F(0), F'(0), F''(0) etc?

 

[McAfee]

It's really hard to tell what you are doing from the blurry photograph. Can you explain? What are F(0), F'(0), F''(0) etc?

Sorry for the blurriness. This should be clearer.

As you can see I get the answer in the box. I can't figure out what I'm doing wrong here.

 

[Dick]

Much clearer, thanks! Now I see where you are going wrong. f(0)=0, isn't it? It's f'(0) that is 1. And if you think about it, you didn't have to compute a lot of those terms involving high powers of x. But that's in retrospect.

 

[McAfee]

Much clearer, thanks! Now I see where you are going wrong. f(0)=0, isn't it? It's f'(0) that is 1. And if you think about it, you didn't have to compute a lot of those terms involving high powers of x. But that's in retrospect.

I'm still not getting the correct answer.
I think f(0)=1. 0^4=0 0 times -4 equals 0 and then e to the power of 0 equals 1.

 

[Dick]

I'm still not getting the correct answer.
I think f(0)=1. 0^4=0 0 times -4 equals 0 and then e to the power of 0 equals 1.

Nooo. f(0) is the integral from 0 to 0 of e^(-4t^4). That's zero! f'(0) is exp(-4t^2) evaluated at t=0. Your derivative count is off by one!

 

[McAfee]

Nooo. f(0) is the integral from 0 to 0 of e^(-4t^4). That's zero! f'(0) is exp(-4t^2) evaluated at t=0. Your derivative count is off by one!

So it should be x-(96/5!)*x^5 ?

 

[Dick]

So it should be x-(96/5!)*x^5 ?

Yes, it should. But I don't like that '?' in the answer. You do see why, don't you?

 

[McAfee]

Yes, it should. But I don't like that '?' in the answer. You do see why, don't you?

Honestly, it is the right answer but I don't see why. Is there something special with finding a maclaurin polynomial of a function that includes e?

 

[Dick]

Honestly, it is the right answer but I don't see why. Is there something special with finding a maclaurin polynomial of a function that includes e?

No, nothing special about e. You don't agree f(0)=0? Suppose the function in your integral was just 1. What are f(0) and f'(0)?

 

[McAfee]

No, nothing special about e. You don't agree f(0)=0? Suppose the function in your integral was just 1. What are f(0) and f'(0)?

f(0) would be 1 and f'(0) would be 0

Looking at it again is it because of the integral would be from 0 to 0

 

[Dick]

f(0) would be 1 and f'(0) would be 0

Looking at it again is it because of the integral would be from 0 to 0

If you are integrating 1 from 0 to x, then f(x)=x. I can say pretty definitely that f(0)=0. f'(0)=1.

 

[McAfee]

If you are integrating 1 from 0 to x, then f(x)=x. I can say pretty definitely that f(0)=0. f'(0)=1.

Can it basically be a rule if you integrating from 0 to x, that the first term is 0 followed by the second term being x?

 

[Dick]

Can it basically be a rule if you integrating from 0 to x, that the first term is 0 followed by the second term being x?

You can't simplify it that much. Integrate 2 from 0 to x. The point of the example is just to show what you were doing wrong. f(0) is always zero, sure. f'(0) depends on the value of the integrand at 0. You just had your derivatives shifted by one. That's all.

 

[McAfee]

You can't simplify it that much. Integrate 2 from 0 to x. The point of the example is just to show what you were doing wrong. f(0) is always zero, sure. f'(0) depends on the value of the integrand at 0. You just had your derivatives shifted by one. That's all.

OK. I understand it now. Thanks for sticking with me until I understood. I have a final for college coming up so understanding this was very pivotal. Thanks, once again.