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Math Help - How to find M in tangent line error bound (taylor polynomials)

  1. #1
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    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??

    Please help someone -thanks alot!

    PS:
    / \ (absolute value)
    <= (greater than or equal to)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by adam21 View Post
    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??

    Please help someone -thanks alot!

    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
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