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Taylors Series/Remainder for ln(x) about x = 1

  • Thread starter jegues
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Homework Statement


Determine whether the Taylors Series for ln(x) converges to ln(x) around x=1, for x [1/2, 2].


Homework Equations





The Attempt at a Solution



[tex]f(x) = ln(x)[/tex]

[tex]f^{1}(x) = \frac{1}{x}[/tex]

[tex]f^{2}(x) = \frac{-1}{x^{2}}[/tex]

[tex]f^{3}(x) = \frac{2}{x^{3}}[/tex]

[tex]f^{4}(x) = \frac{-6}{x^{4}}[/tex]

[tex]\vdots[/tex]

[tex]f^{n}(x) = \frac{(-1)^{n-1}(n-1)!}{x^{n}}[/tex]

So,

[tex]ln(x) = (x-1) - \frac{1}{2}(x-1)^{2} + \frac{1}{3}(x-1)^{3} - \cdots + \frac{(-1)^{n-1}}{n}(x-1)^{n} + R_{n}(1,x)[/tex]

Where,

[tex]R_{n}(1,x) = \frac{(-1)^{n}}{(n+1)} \cdot \left( \frac{(x-1)}{z_{n}} \right)^{n+1}[/tex]

We can see that the second term (The one with [tex]z_{n}[/tex] in it) is the one that causes problems. It is trivial to show that the limit as n goes to infinity of the first term is 0. (We proved it in class)

So there are 3 cases that we should consider:

Case 1: x=1, everything will converge to 0.

Case 2: [tex]\frac{1}{2} \leq x < 1[/tex]

It follows then that,

[tex] \frac{1}{2} \leq x < z_{n} <1[/tex]

We can use, [tex]\frac{1}{2} \leq x < 1[/tex] to try to gain for information about the term that is causing problems with our remainder.

[tex]\frac{1}{2} \leq x < 1 \Rightarrow \frac{1}{2} - 1 \leq x-1 < 1-1 \Rightarrow \frac{-1}{2} \leq x-1 < 0 \Rightarrow \frac{-1}{2z_{n}} \leq \frac{x-1}{z_{n}} < 0 [/tex]

Looking back at,

[tex] \frac{1}{2} \leq x < z_{n}[/tex]

we can see that,

[tex]z_{n} \geq \frac{1}{2}[/tex] or [tex]2z_{n} \geq 1[/tex]

Working with this some more we can see that,

[tex]\frac{1}{2z_{n}} \leq 1 \Rightarrow \frac{-1}{2z_{n}} \geq -1 [/tex]

Therefore we can say that,

[tex]-1 \leq \frac{-1}{2z_{n}} < \frac{x-1}{z_{n}} < 0[/tex]

i.e. [tex]-1 < \frac{x-1}{z_{n}} < 0 \Rightarrow |\frac{x-1}{z_{n}}| < 1[/tex]

This is what our professor wrote down in order to prove the following,

[tex]|R_{n}(1,x)| = \left| \frac{(-1)^{n}}{n+1} \cdot \frac{(x-1)^{n+1}}{z_{n}} \right| < \frac{1}{n+1} \cdot (1)[/tex]

Therefore,

[tex]lim_{n \rightarrow \infty} |R_{n}(1,x)| = 0 \Rightarrow lim_{n \rightarrow \infty} R_{n}(1,x) = 0 [/tex].


I'm confused about how he got rid of that unwanted term,

[tex]\left(\frac{(x-1)}{z_{n}}\right)^{n+1}[/tex]

by using the inequalities stated above. He seems to jump around alot and it's hard to follow.

EDIT: Looking at it again, he first shows that,

[tex]\frac{x-1}{z_{n}} < 0[/tex]

Now if he can prove that this is also larger than -1 it will be a small negative number inbetween (0, -1) and no matter what power we raise it to, lets say n+1, it will still be really close to 0.

Is this his attack?


Can someone try to explain his reasoning another way or offer me some insight? If I tried to do this problem without looking at my notes I don't think I'd be able to reasonably conclude that,

[tex]lim_{n \rightarrow \infty} R_{n}(1,x) = 0 [/tex]

For Case 2: [tex]\frac{1}{2} \leq x < 1[/tex]

Any advice/tips/ideas/insight would be greatly appreciated! (After I will attempt case 3)

Thanks again!
 
Last edited:

Answers and Replies

  • #2
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Case 3: [tex]1 < x \leq 2[/tex] seems to make some more sense so I'll attempt it in this post.

It follows that,

[tex]1 < z_{n} < x \leq 2[/tex]

Again, we can use, [tex]1 < x \leq 2[/tex] to gain more information about our problematic term in [tex]R_{n}(1,x)[/tex].

One can see that,

[tex]0 < x-1 \leq 1[/tex]

[tex]0 < \frac{x-1}{z_{n}} \leq \frac{1}{z_{n}}[/tex]

From here we can see that,

[tex]\frac{1}{z_{n}} < 1[/tex]

So here's where I'm at,

[tex]0 \leq |R_{n}(1,x)| < ?[/tex]

We need to figure out an upper bound so we can apply squeeze theorem.

We know that,

[tex]\frac{x-1}{z_{n}}[/tex] is less than 1 but larger than 0.

Hmm...

I'm not sure how I can find that upper bound I've been looking for. If I think of anything I'll simply edit this post.

Feel free to make any suggestions if you have any!
 

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