# Taylors Theorem

1. Apr 27, 2008

### danago

Show that the taylor series generated by $$f(x)=e^{x}$$ about x=0 converges to f(x) for ever real value of x.

Taylors theorem states that:

$$f(b) = P_{n} + \frac{e^c}{(n+1)!} b^{n+1}$$

where P_n is the taylor polynomial of order 'n' and the following term is the error term and c is some value between 0 and b. If the error term approaches 0 as n approaches infinity, then the series converges to f(x).

Does this mean that to answer the question all i need to do is show that the error term approaches zero as n gets large? If so, how would i do so?

Thanks in advance for the help,
Dan.

2. Apr 27, 2008

### exk

Well a non-rigorous approach would be to state that a factorial will grow faster than an exponential, especially since b is constant. The denominator of the error term will therefore outgrow the numerator and thereby kill of the whole term.

3. Apr 27, 2008

### HallsofIvy

Staff Emeritus
Yes, all you need to do is show that the error term approaches 0 as n gets large.

Since ex is an increasing function and c< b, ec< eb so
$$\frac{e^nb^{n+1}}{(n+1)!}< e^b\frac{b^{n+1}}{(n+1)!}$$.

Now, to expand on exk's comment, How many factors are there in bn+1? How many factors are there in (n+1)!? What happens as soon as n> b?

4. Apr 27, 2008

### danago

That inequality you posted...is that assuming that b is positive? And is the e^n in the left hand side supposed to be an e^c?

As for your second question, the bn+1 has n+2 factors? And the (n+1)! has 2n factors?

Im still a little stuck for as how id use this information to answer the question.

Thanks for the replies