Question

Calculus help - Taylor polynomials

Answer 1

Interesting question! I don't have my favorite Calculus book with me at present (for reference), but can nevertheless help you get started.

Where would you like to start, Jared?

Answer 2

Haha, yeah it is pretty interesting. I guess we can start with part a. I really just have no idea where/how to even start...I couldn't find anything in my notes that looked like this problem for reference either

Answer 3

Any Taylor polynomial is in infinite series. Prior to starting Taylor series, you most likely got some experience determining whether a given series converges or diverges. (Is that correct?)

One of the most commonly used tests for convergence of a power series is the Ratio Test. If you recall this Test, or could look it up, apply it to that general term of the Taylor series. Just let me know if you need any guidance with that.

Answer 4

Yes, that is correct. And I know of the ratio test, I will try it out on this. What about the (-1)^k? What will happen to that in the test?

Answer 5

Since the absolute value of (-1)^k is 1, ignore it completely.

Answer 6

Do I even have to write it as part of the ratio test and show that it becomes 1? Or not even bother with that?

Answer 7

Perhaps a better rationale would be the recognition that the given T(x) is an alternating series, so you could focus on the part of the nth term that does not include that (-1)^k. Suggest that you review "alternating series" and especially the criteria for the convergence of an alternating series. Still, the Ratio Test ought to work fine also.

Answer 8

Ask yourself what happens as k approaches infinity in

k+1
--------
k!

You might want to write out the first four or five terms, beginning with that for k=1.

Answer 9

Let me show you the first few steps I wrote out and see if they're correct

Answer 10

Throughout all this work, our end goal is to show that the given Taylor series is convergent for all x.

So, if (k+1)x^(3k) / (k!) is always decreasing, and if the limit as k goes to infinity is 0, what conclusion would be appropriate?

Do share those first steps.

Answer 11

Jared: Have you come to any conclusion yet?
I promise to help you through this entire problem, so long as I don't need my Calculus book as a reference, but I have unexpected visitors and need to get off the Internet. I will resume our conversation as soon as possible.
I'm sorry for the interruption.

Answer 12

No, I was just hoping for a check on those first few steps.
It's alright! I will be here when you get back :)

Answer 13

your last line should show (k+1)^2

Answer 14

\[ \frac{k!}{(k+1)!} = \frac{1}{(k+1)} \]
but there is a second 1/(k+1) which gives 1/(k+1)^2

Answer 15

Ah, thank you!
From here, is it correct that it equals abs(x^3) ?

Answer 16

Jared: please take the limit, as k approaches infinity, of and see if you can then come to some conclusion regarding the convergence or divergence of the given series.

Answer 17

of...\[\frac{ (k+2)x^3 }{ k+1)^2 }\]

Answer 18

I believe that the limit equals abs(x^3), and since f(x) is about x=0, abs(x^3) = 0.
So, T(x) converges because the result of the Ratio Test = 0 <1.

Correct?

Answer 19

But, Jared, we'd be focusing on that k+1 in the denominator and letting k approach infinity, wouldn't we?

The limit of 1 over (k+1) as k approaches infinity is clearly zero, leaving us with zero times x^3.

Answer 20

Where do you get a 1 in the numerator from?

Answer 21

In your shoes I'd also test this given series for convergence using the Alternating Series Test for verification.

Are you OK with the rest of this question, or would you care to discuss what's happening in Part 2?

Answer 22

To answer your question:

We want to find the limit as k approaches infinity of \[\frac{ k+2}{(k+1)^2 }x^3\]
Please look at the k terms. As k approaches infinity, k will be so much larger than 2 / so much larger than 1, that in effect we have \[\frac{ k }{ k^2 } = 1/k\]

Answer 23

Oh I see now. Okay, thank you.

Yes, if you could help with the other parts, too, it would be much appreciated :)

Answer 24

Not sure I'll have time to do all four parts right now, but let's move on to Part 2. As you read part 2, what particular action or actions come to mind through which you'd derive a series for x^2 * f '(x)?

Answer 25

For example, supposing I were to give you a Taylor series about x=0 that looked like this: (Infinite sum of) 2x^k over k^2. What would the series become if you were to multiply that by x^2?

Answer 26

(Just looking ahead for a second, I should be fine on part c. All you need to do is plug 0.5 and 1 into the four nonzero terms provided, right?)
I believe it would be (infinite sum of) 2x^(2k) over k^2

Answer 27

Actually scratch that it would be 2x^(k+2) ?

Answer 28

Recall the rule for multiplication of exponential expressions involving the same base:

a^b * a^c = a^(b+c) Yes. You caught this before I could finish typing my suggestion. Nice going.

Answer 29

So, in the problem you've posted, in Part 2, the effect of multiplying the function f(x) by x^2 is simply to add 2 to the exponent of x in the expression for the general term of the power series. This is precisely what you did here: you added 2 to k (the exponent of x). Nice work.

Answer 30

Now let me pose another question for you:

Suppose that a power series expansion for f(x) is the infinite sum of terms 2x^k over k^2. Now suppose you want the power series for the derivative of f(x).

What would the terms of your new power series look like? Which rule of differentiation would you be applying here?

Answer 31

Thank you :) I believe you would use the product rule on the numerator, right?

Answer 32

Specifically, you'd use the product rule on x^k alone (but keep the remaining k terms).

OK: We've done 2 things: (1) we multiplied the function and its corresponding Taylor series by x^2. (2) we differentiate the original f(x) and its corresponding Taylor series with respect to x (not with respect to k).

As for your actual, posted problem:

First, Take the given Taylor series term.
Differentiate it with respect to x.
Multiply the result by x^2.
Simplify.
You'll then have the desired T. series for (x^2) * f '(x).

Answer 33

Okay, I will do that! Just give me a minute or two and I'll see if I get an answer

Answer 34

Jared, I may have to excuse myself for a while again. But so long as you respond to this, I will receive notification that you typed something in and will then respond myself. Advise you to move on to another problem and to return to this one later.