Proof of Taylor's Theorem for e^x Convergence

[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:

<br /> f(b) = P_{n} + \frac{e^c}{(n+1)!} b^{n+1}<br />

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.

 

[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]

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

Since e^x is an increasing function and c< b, e^c< e^b so
\frac{e^nb^{n+1}}{(n+1)!}&lt; 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?

 

[danago]

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

Since e^x is an increasing function and c< b, e^c< e^b so
\frac{e^nb^{n+1}}{(n+1)!}&lt; 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