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Taylor polynomial approximation- Help!

  1. Apr 13, 2008 #1
    Use Taylor's theorem to determine the degree of the Maclaurin polynomial required for the error in the approximation of the function to be less than .001.

    e^.3

    I really, really don't know what to do for this one, and I have a quiz tomorrow. I have read through the section in the book, but I am frustrated and can't figure out what kind of method I should use to solve these types of problems.
     
  2. jcsd
  3. Apr 13, 2008 #2

    lurflurf

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    Homework Helper

    Do you know the remainder?
    exp(x)=1+x+x^2/2+x^3/6+x^4/24+...+x^n/n!+R
    R is the error and can be written different ways for different situations
    ie
    R=f^(n+1)(u)/(n+1)!
    ie the formula can be made exact by evaluating the final term at x* with 0<=u<=x for some x* in the case of exp we have terms like x^k/k! so the remainder looks like
    u^(n+1)/(n+1)!
    it is usually hard to find u so we instead chose n so that u des not mater ie
    we want n so that
    u^(n+1)/(n+1)!<.001
    we do not know u but we know (since exp is increasing on [0,.3])
    u^(n+1)/(n+1)!<.3^(n+1)/(n+1)!
    so that if
    .3^(n+1)/(n+1)!<.001
    then
    u^(n+1)/(n+1)!<.001
    so we do not need u, but we may end up with n larger than needed
    solve
    .3^(n+1)/(n+1)!<.001
    for n
    then the maclaurin polynomial of degree n has error<.001
     
  4. Apr 13, 2008 #3
    So is the procedure to take the derivatives and plug in 0 (since c=0) and find an expression for the n+1 derivative?

    f'(c) = 1 f''(c)=1 f'''(c) =1 ......

    so the n+1 derivative is 1

    So Rn= 1/(n+1)! * (.3) ^(n+1)

    Then I set up an equality to find n so that Rn < .001

    and n = 3 ???
     
  5. Apr 13, 2008 #4

    lurflurf

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    Homework Helper

    yes
    we also want n an integer, so just round up
    that is take the first n so that
    .3^(n+1)/(n+1)!<.001

    do you see how you might do
    say
    approximate
    1/(1-x)
    at x=1.1 to an error less than .005
    or aproximate
    sin(x) at x=.001 to an error less than .00001
     
  6. Apr 15, 2008 #5

    lurflurf

    User Avatar
    Homework Helper

    sorry we should have

    Do you know the remainder?
    exp(x)=1+x+x^2/2+x^3/6+x^4/24+...+x^n/n!+R
    R is the error and can be written different ways for different situations
    ie
    R=f^(n+1)(u)/(n+1)!
    ie the formula can be made exact by evaluating the final term at x* with 0<=u<=x for some x* in the case of exp we have terms like x^k/k! so the remainder looks like
    e^u/(n+1)!
    it is usually hard to find u so we instead chose n so that u des not mater ie
    we want n so that
    e^u/(n+1)!<.001
    we do not know u but we know (since exp is increasing on [0,.3])
    u^(n+1)/(n+1)!<e^.3/(n+1)!
    so that if
    e^.3/(n+1)!<.001
    then
    e^u/(n+1)!<.001
    so we do not need u, but we may end up with n larger than needed
    solve
    e^u/(n+1)!<.001
    for n
    then the maclaurin polynomial of degree n has error<.001
     
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