How Many Terms Needed in Maclaurin Polynomial for Error Below 0.001?

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The discussion focuses on determining the number of terms required in the Maclaurin polynomial for the function e^0.3 to ensure that the error in approximation is below 0.001. The approach involves using Taylor's theorem, where the (n+1)th derivative of e^x at c=0 is consistently 1. The remainder term Rn is expressed as Rn = 1/(n+1)! * (0.3)^(n+1). By solving the inequality Rn < 0.001, it is concluded that n must be at least 3 to meet the error requirement.

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Taylor Polynomial Error--Please 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



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 ?


I want to be sure I am taking the right approach on these problems, so is this the way to do it?
 
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Personally I think the best way to do these is to first find the series for the function and then to plug in a few values. The summation for the exponential function is:
[tex]e^{x} = \sum^{\infty}_{n=0}\frac{x^{n}}{n!}[/tex]

if you compute a few setting x = .3, you will see how many terms you need

[tex]e^{.3} = \sum^{\infty}_{n=0}\frac{.3^{n}}{n!} = 1+ \frac{.3}{1!}+\frac{.3^{2}}{2!}+...[/tex]
 
Last edited:
Solving the inequality Rn < .001 will give the answer. It is easy to solve this by just plugging in values of n until it is satisfied.
 

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