Maclaurin approximation to approximate e^5 to within 0.00001

[mooshupork34]

i was confused by the following problem, so any help would be appreciated!

To approximate e^5 to within 0.00001, which Maclaurin polynomial for e^x would you need? Write the terms of that polynomial, then express it using sigma notation.

 

[galactus]

Just use the expansion for e^x and use x=5.

\(\displaystyle \L\\e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+...........\)

Take it out about 20 places or so to get that accurate.

Is this what you mean?.

 

[mooshupork34]

Just use the expansion for e^x and use x=5.

\(\displaystyle \L\\e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+...........\)

Take it out about 20 places or so to get that accurate.

Is this what you mean?.

Yes, thanks! I'm just unsure as to the procedure for approximating because my book isn't very clear. Meaning, how do you know how many derivatives to take or how far to expand?

 

[galactus]

Do you have a calculator?. Expand out the series and use trial and error to get within the required accuracy.

I used \(\displaystyle \L\\e^{5}-(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+....+\frac{x^{20}}{20!})=0.000012\)


You can look up Lagranges form of the Remainder, but it's a little beyond what I am willing to get into here. Go with the trial and error thing with the calculator.

 

[mooshupork34]

Do you have a calculator?. Expand out the series and use trial and error to get within the required accuracy.

I used \(\displaystyle \L\\e^{5}-(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+....+\frac{x^{20}}{20!})=0.000012\)


You can look up Lagranges form of the Remainder, but it's a little beyond what I am willing to get into here. Go with the trial and error thing with the calculator.

AH! I see how you do it now! Thank you!