# Evaluating the remainder of a Taylor Series Polynomial

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## Homework Statement

The goal of this problem is to approximate the value of ln 2. We will use two different approaches: (a) First, we use the Taylor polynomial pn(x) of the function f(x) = lnx centered at a = 1.

1. Write the general expression for the nth Taylor polynomial pn(x) for f(x) = lnx centered at a = 1.
DONE

2. At x = 2, evaluate the size of the remainder Rn(2) = ln 2 − pn(2).

3. What should n be so that you are sure that pn(2) approximates ln2 to two decimal

points? What is then the approximate value of ln 2 (up to two decimal points)?

[/B]

## The Attempt at a Solution

For part 1, I netted the (should be correct) answer of
ln(x) = Σ from n=1 to infinity of (-1)^(n+1)/n (x-1)^n

Now, I am completely stuck on part 2. At x=2, evaluate the size of the remainder Rn(2) = ln(2)-Pn(2).

Are there any examples out there? I am searching the internet for examples, but not much luck.

Thank you.

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BvU
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2019 Award
If you want to evaluate Rn(2), you calculate ln(2) - Pn(2). It's straightforward !

 Correction. I looked at ln(2) - P2(2) which is too easy. Sorry.

[edit2] Check this link for an error bound expression.

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• RJLiberator
Gold Member
Thank you for the guidance.

I was starting to think it had something to do with the alternating series test.
When evaluating at x = 2 the (x-1)^n part of the summation does not matter.

I see from this link, ln(2) does not converge absolutely, but they have rearranged it in such a form with 1/2.
I guess, if I can figure out the answer to this part, then I have the value for ln(2). I will just need the value of Pn(2). Hm.

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Maybe a better question now for me to ask is, what is the meaning/definition of Pn(2)? I know that P_2(x) is taking the sum to the second degree, but I don't understand Pn(2).

BvU
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I would expect $$P_N(x) =\sum_{n=1}^N\ {(-1)^{(n+1)}\over n}\; (x-1)^n$$

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• RJLiberator
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What does Pn(x) represent?

The nth degree polynomial representation at a value "x" ?

Wouldn't that be equivalent to evaluating ln(2) in this situation?

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Okay, maybe I am looking at this wrong (well, clearly I am).

Let's say, I evaluate Pn(2) as you have prescribed in earlier post. (which can be done according to the Alternating series)
After this, I will need to find a way to evaluate Rn(2).
Once this is done, the combined values will equal = ln(2).

Is this the correct strategy to approaching this problem?
If so, how would I calculate Rn(2)

BvU
Homework Helper
2019 Award
My impression is that Pn(x) is the Taylor series up to degree n. So P1(x) = x-1, P2(x) = (1-x) - (x-2)2/2, etc.

That way R1(2) = -0.31, R2(2) = 0.19 etc. (convergence is excruciatingly slow).

There is no calculating Rn(2) unless you 1) use the actual value of ln(2) , and 2) add up all the terms up to order n with unlimited precision. All there is to evaluate is to give an upper bound for |Rn|. The link gives an expression for the error bound

However, part 3) of the exercise then poses a problem: do they want us to actually calculate all these terms and check if it rounds off to 0.70 or 0.69 ? I really can't understand what's intended there.

• RJLiberator
Gold Member
BvU, this homework was due last night. We came to the exact same conclusion. The vagueness of the question causes such confusion.

For ii) apparently they just wanted a general formula in the form of Rn(2) = ln(2) - sum (-1)^((n+1))n or something like this.
For iii) it was exactly as you stated.

I am just going to put this assignment behind me, at least I know that the vagueness of this assignment does not equate to my understanding on this topic. :) Thank you for helping.