Maclaurin series--(sinx - x)/x^3

[h2sbf7]

Hi, all. Hopefully my last question for a little while, as after this I move out of BC calculus.

The function f, defined as:

f(x) = {(sinx-x)/x^3 for x ≠ 0, 1 for x = 0

...has derivatives of all orders. Write the first three nonzero terms and the general term of the Taylor series for sinx about x = 0 [so, the Maclaurin series]. Use this series to write the first three nonzero terms and the general term of the Taylor series for f about x = 0.

The first part is easy enough. It's just one of those your teacher makes you memorize. The series for sinx is x - x^3/3! + x^5/5!... or the sum from zero to infinity of (-1)^n * [x^(2n+1)]/(2n+1)!

What I can't figure out how to do is take this and go to (sinx-x)/x^3. It's not like I can simply say, "for every x, I insert a __."

How do I make this work? Any help is appreciated. Thank you!

 

[stapel]

The function f, defined as:

f(x) = {(sinx-x)/x^3 for x ≠ 0, 1 for x = 0

...has derivatives of all orders. Write the first three nonzero terms and the general term of the Taylor series for sinx about x = 0 [so, the Maclaurin series]. Use this series to write the first three nonzero terms and the general term of the Taylor series for f about x = 0.

The first part is easy enough. It's just one of those your teacher makes you memorize. The series for sinx is x - x^3/3! + x^5/5!... or the sum from zero to infinity of (-1)^n * [x^(2n+1)]/(2n+1)!

What I can't figure out how to do is take this and go to (sinx-x)/x^3.

You have:

. . . . .$\displaystyle \sin(x)\, =\, x\, -\, \dfrac{x^3}{3!}\, +\, \dfrac{x^5}{5!}\, -\, ...$

The numerator in what you've been given is:

. . . . .$\displaystyle \sin(x)\, -\, x\, =\, \bigg(\, x\, -\, \dfrac{x^3}{3!}\, +\, \dfrac{x^5}{5!}\, -\, ...\, \bigg)\, -\, x$

How will the right-hand side simplify?

Then you're dividing by x^3. How will the summation simplify? What will the terms of the new series then be?

 

[h2sbf7]

You have:

. . . . .$\displaystyle \sin(x)\, =\, x\, -\, \dfrac{x^3}{3!}\, +\, \dfrac{x^5}{5!}\, -\, ...$

The numerator in what you've been given is:

. . . . .$\displaystyle \sin(x)\, -\, x\, =\, \bigg(\, x\, -\, \dfrac{x^3}{3!}\, +\, \dfrac{x^5}{5!}\, -\, ...\, \bigg)\, -\, x$

How will the right-hand side simplify?

Then you're dividing by x^3. How will the summation simplify? What will the terms of the new series then be?

Thanks for the quick reply. What do you mean by "right-hand side"--is that x?

 

[h2sbf7]

Oh, so the x cancels out the first term and x^3 cancels xs from the numerator! Thank you so much! That clears it up.