Approximation e number using taylors polynomial

[Telemachus]

Homework Statement

Well, this problem is quiet similar to the one I've posted before. It asks me to approximate to the e number using taylors polynomial, but in this case tells me that the error must be shorter than 0.0005

Homework Equations

[tex]R_{n+1}=\displaystyle\frac{f^{n+1}(z)(x-x_0)^{n+1}}{(n+1)!}\leq{\epsilon}[/tex], [tex]0<z<1[/tex]

[tex]x_0=0[/tex] [tex]x=1[/tex] [tex]f^{n+1}(x)=e^x[/tex] [tex]\epsilon=0.0005[/tex]

[tex]P_n(x)=1+x+\displaystyle\frac{x}{2!}+...+\displaystyle\frac{x^n}{n!}[/tex]

[tex]\delta=n+1[/tex]

The Attempt at a Solution

I've tried using the function [tex]f(x)=e^x[/tex] at [tex]x_0[/tex] [tex]\epsilon=0.0005[/tex]
[tex]f^n(x)=e^x[/tex]

Then

[tex]P_n(x)=1+x+\displaystyle\frac{x}{2!}+...+\displaystyle\frac{x^n}{n!}[/tex]

[tex]P_n(1)=1+1+\displaystyle\frac{1}{2!}+...+\displaystyle\frac{1}{n!}[/tex] which tends to e as n tends to infinity.

But now I don't know how to get the n, so I get the degree for the polynomial with the error it asks me.

[tex]R_{n+1}=\displaystyle\frac{f^{n+1}(z)(x-x_0)^{n+1}}{(n+1)!}[/tex] [tex]0<z<1[/tex]

Lets call [tex]\delta=n+1[/tex]

[tex]R_{\delta}=\displaystyle\frac{e^z}{\delta!}\leq{0.0005}[/tex] then

[tex]\displaystyle\frac{e^z}{0.0005}\leq{\delta!}[/tex] I know between which values I can find z, but I don't know how to work the factorial in the inequality.

Am I proceeding right?

Bye there.

 

[Mark44]

On the interval [0, 1], e¹ < 3.

 

[Telemachus]

Right, I must use the bigger value on the interval. Then I've got [tex]6000\leq{\delta!}[/tex] So I must use [tex]delta=8[/tex]? but if I use e on the calculator I think I get that seven is a better approximation for it, and I think it works for seven. What you say Mark?

Sorry, I was wrong. You're right, so [tex]\delta=8[/tex]

 

[Mark44]

I wouldn't introduce another variable, [itex]\delta[/itex]. Just work with n + 1.

 

[Telemachus]

Thanks Mark ;)