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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

a. Find the macluarin polynomial of sin (x)

b. Find the maclaurin polynomial of sin (x^2)

c. Substitue x^2 for x in the solution to problem a. How does it compare to the solution of problem b?

Ok, a is easy enough. C is the easy way, but in b, I take it they mean to find the maclaurin series the traditional way, by taking the 1st, 2nd, 3rd.... derivatives, evaluating it at zero, and inserting it in the maclaurin polynomial.

Do you have any idea what a nightmare it is to find the maclaurin series of sin(x^2)? f(0) is 0, f'(0 is 0, f''(0) is 2, everything else is zero all the way up to the 6th derivative. Try finding the 6th derivative of sin(x^2) it takes FOREVER! I did it twice and I've now got a screaming headache! Both times the sixth derivative evaluated at zero came to -24.

so the second derivative evaluated at zero is 2, and the 6th derivative evaluated at zero is -24. All others below 6 are zero.

So these would be inserted in the second and 6th term in the maclaurin polynomial respectively.

2(x^2)/2! (-24X^6)/6!

So the first term in the macluarin series would be x^2, the next term would be -X^6/30.

The first term x^2 is correct, but the last term is supposed to be x^6/3! and obviously its not.

TWICE I did this nightmarish problem and twice I got the same result. Any idea's?

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I'll right, I just had some crackers and milk so I'm ready to relive the nightmare again. I'll Find the derivatives one at a time and post them in separate replies, so you guys can check each one individually, (if you care to).

Ok. Here we go:

f(x) = sin(x^2)

f'(x) = 2xcos(x^2)

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

f'(x) = 2xcos(x^2)

f''(x) = -4x^2 sin (x^2) + 2 cos (x^2)

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

f''(x) = -4x^2 sin (x^2) + 2 cos (x^2)

f'''(x) = -8x^3 cos (x^2) - 12x sin (x^2)

*Last edited by mikau (2006-01-03 16:45:14)*

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

f'''(x) = -8x^3 cos (x^2) - 12x sin (x^2)

f''''(x) = (16x^4 - 12) sin (x^2) - 48x^2 cos (x^2)

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

f''''(x) = (16x^4 - 12) sin (x^2) - 48x^2 cos (x^2)

f'''''(x) = (160x^3) sin (x^2) + (32x^5 -120x) cos (x^2)

*Last edited by mikau (2006-01-03 17:39:26)*

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

f'''''(x) = (160x^3) sin (x^2) + (32x^5 -120x) cos (x^2)

f''''''(x) = (720x^2 - 64x^6) sin (x^2) + (480x^4 - 120) cos (x^2)

WHEW! Ok, I think I did it right this time. We have to evaluate all these functions at zero. Any term that contains sin(x) or x as a factor will have a value of zero.

f(0) = 0

f'(0) 0

f''(0) = 2

f'''(0) = 0

f''''(0) = 0

f'''''(0) = 0

f''''''(0) = -120

We insert them into the appropriare places and get:

(2x^2)/2! - (120 x^6)/6!

= x^2 - x^6/6

= x^2 - x^6/3!

HORRAY! I DID IT! (wonders if it was really worth it...)

Thanks! You guys are the best!!! ;-)

A logarithm is just a misspelled algorithm.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Yes, we are!!!

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

By this formula we get also this:

*Last edited by krassi_holmz (2006-01-04 06:06:58)*

IPBLE: Increasing Performance By Lowering Expectations.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

forgive me if I'm wrong but isn't there a x^5/5! missing there?

A logarithm is just a misspelled algorithm.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

you're wrong.

IPBLE: Increasing Performance By Lowering Expectations.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

lol. I was just really confused for a split second.

A logarithm is just a misspelled algorithm.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

See the

"last edited tag"...

IPBLE: Increasing Performance By Lowering Expectations.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

yeah thats why I was only utterly perplexed for a moment.

A logarithm is just a misspelled algorithm.

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