How to solve for "M" in error bound calculations???

Question

No i'm not asking for sum1 2do my homework, i have about 10probs to do and i'm running up against same roadblock. 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)}.

Answer 2

I don't know what ur doin with M it looks to me like M is just some constant that gives <= in the inequality you set up. does it have to be exact or not really? I can get some M using the next term of the series

| Ln(x) - (x-1) | = | -(x-1)^2 + etc...| <= |x-1|^2 + etc... <= 2|x-1|^2

and put this in

2|x-1|^2 <= M/2|x-1|

divide by |x-1| and get 2|x-1|<=M/2

|x-1|<2 because the interval containing x has to be inside the bigger interval [0,3] whose endpoints
| log 0 + 1 |= minus infinity
| log 3 - 2 | > 1
(as you can see this is way off from <.01)

So put in 2 for |x-1| and we get 4<=M/2 or M>=8.

At least u can get a suitable interval with M=8.