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

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

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

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

\delta=n+1

The Attempt at a Solution

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

Then

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

P_n(1)=1+1+\displaystyle\frac{1}{2!}+...+\displaystyle\frac{1}{n!} 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.

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

Lets call \delta=n+1

R_{\delta}=\displaystyle\frac{e^z}{\delta!}\leq{0.0005} then

\displaystyle\frac{e^z}{0.0005}\leq{\delta!} 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 6000\leq{\delta!} So I must use delta=8? 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 \delta=8

 

[Mark44]

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

 

[Telemachus]

Thanks Mark ;)