# How to find M in tangent line error bound (taylor polynomials)

• Mar 31st 2007, 08:58 PM
How to find M in tangent line error bound (taylor polynomials)
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, right??

PS:
/ \ (absolute value)
<= (greater than or equal to)
• Mar 31st 2007, 09:58 PM
Jhevon
Quote:

Originally Posted by adam21
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, right??

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

i have no knowledge of the topic you're talking about, so i can't tell you if you're right or wrong to go in the direction you went. however, if you are right:

(M/2)|x - 1| <= 0.1
=> M/2 <= 0.1/|x - 1| ..........provided x - 1 not= 0
=> M <= 2(0.1)/|x - 1|
=> M <= 0.2/|x - 1|

so M is bounded by 0.2/|x - 1|

hope that helps