Evaluating the remainder of a Taylor Series Polynomial

[RJLiberator]

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)?
   

Homework Equations

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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.

 

[BvU]

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

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

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

 

[RJLiberator]

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.

 

[RJLiberator]

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]

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

 

[RJLiberator]

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?

 

[RJLiberator]

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]

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]

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.