Anyone willing to help me with a taylor polynomial problem?

Question

It is just one little step in this problem which is confusing me all the time.
Please sum 1 help.
I posted the problem earlier but didnt get amny replies, --the thing is, i dont need help with whole problem just a part of it.
Anyway heres the problem (posted earlier)

How to solve for "M" in error bound calculations?
Just need help on solving for M and then i'm Ok (i think)

heres the prob:
f(x) = Ln(x), b=1, and i need to use tang line error bound to find interval so that error is at most .01 on J.

So... heres what i set up:

f(x) = / Ln(x) - [x-1] \ <= M/2 /x-1\ <= .01

So i need to find "M" right?

So.. M/2 /x-1\ <= .01

And thats where i get stuck.. can't seem to solve for M
Correct me if wrong, but if i could solve for M then i could find a suitable interval??

Please help -thanks alot!

PS:
/ \ (absolute value)
<= (greater than or equal to)

Answer 1

Ln(x) means the natural log of x here, doesn't it? I'm going to assume that's what you mean. Right, the Taylor polynomial of first degree for f at 1 (ie. its tangent line) is given by

f(1) + f '(1) (x - 1) = Ln(1) + (1/1)(x - 1) = 0 + x -1 = x.

Taylor's theorem states that the error between f and its tangent line is given by

½f''(y)(y - 1)² = ½(-1/y² )(y - 1)² = -½(1 - 1/y)²

for some y between 1 and x, and the absolute value of this is bounded by

½(1 - 1/x)².

Thus, we want to find the range of x for which

½(1 - 1/x)² ≤ 1/100.

½(1 - 1/x)² ≤ 1/100 ⇔ |1 - 1/x| ≤ √2 / 10

⇔ |x - 1| ≤ x√2 / 10

⇔ - x√2 / 10 ≤ x - 1 ≤ x√2 / 10

⇔ x ≥ 10 / (10 + √2) and x ≤ 10 / (10 - √2).

Hence, the interval we're looking for is

[10 / (10 + √2), 10 / (10 - √2]

ie. {x∈R | 10 / (10 + √2) ≤ x ≤ 10 / (10 - √2)}.