# Taylors Theorem

Gold Member
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.

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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.

HallsofIvy
Homework Helper
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?

Gold Member
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?
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