## Taylor polynomial approximation- Help!

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.
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 Recognitions: Homework Help 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
 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 ???

Recognitions:
Homework Help

## Taylor polynomial approximation- Help!

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
 Recognitions: Homework Help 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)!

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