# Approximation e number using taylors polynomial

1. Aug 12, 2010

### Telemachus

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

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

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

$$\delta=n+1$$

3. 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.

Last edited: Aug 12, 2010
2. Aug 12, 2010

### Staff: Mentor

On the interval [0, 1], e1 < 3.

3. Aug 12, 2010

### 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 $$\delta=8$$

4. Aug 12, 2010

### Staff: Mentor

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

5. Aug 12, 2010

### Telemachus

Thanks Mark ;)