How Does the Infinite Sum of 1/j! Equal e?

[soandos]

the common derivation of e is pretty straightforward, Lim x->oo (1+1/x)^x, but how does one prove that the infinite sum \sum\frac{1}{j!} as j goes from 0 to oo is equal to e?

p.s. how does one use LaTex for the super and subscripts in summation?

 

[mathwonk]

use taylors theorem to show that e^x equals the limit of its taylor series, in particular for x=1.

 

[Rogerio]

And use your "e" definition to prove
IF f(x) = e**x THEN f'(x) = f(x)

(your Taylor's expansion needs this result)

 

[lurflurf]

It can be done more simply.
specialize (1+1/x)^x
to a_n=(1=1/n)^n (n a positve integer)
and expand with the binomial theorem
next show that if
b_n=the sum
lim sup b_n<=e<=lim inf b_n
thus the limit of the sum is e
this is done in most calculus books

 

[soandos]

sorry to all:
i do not really understand what you are saying. in response to

It can be done more simply.
specialize (1+1/x)^x
to a_n=(1=1/n)^n (n a positve integer)
and expand with the binomial theorem

that i understand.
however, i do not get the next part.
could someone please explain it?

 

[soandos]

does this make sense:
take the taylor series for e^x, at x=1, and you get the sum?
since i unfortunately have no experience with Taylor series, what is the proof for e^x.
sorry if i seem like i want to prove everything that anyone says, it is just that i do not have a background in the area.

 

[HallsofIvy]

If y= e^x, then e^x+h- e^x= e^x(e^h-1). The derivative of e^x is
e^x\lim_{h\rightarrow 0}\frac{e^h-1}{h}
a constant times e^x. What is that constant?

Given that \lim_{x\rightarrow \infty} (1+ 1/x)^x= e, let h= 1/x. Then h goes to 0 as x goes infinity and \lim_{h\rightarrow 0} (1+h)^{1/h}= e. That means that, for h very close to 0, (1+h)^{1/h} is very close to e and so 1+ h is very close to e^h. Therefore, e^h- 1 is close to h and (e^h-1)/h is close to 1: in the limit, \lim_{h\rightarrow 0}\frac{e^h-1}{h}= 1 and so the derivative of e^x is, again, e^x. It follows that all derivatives of e^x are e^x and, at x= 0, 1. From that it follows that the MacLaurin series for e^x is
\sum_{n=0}^\infty \frac{1}{n!} x^n
and, finally, taking x= 1 that
\sum_{n=0}^\infty \frac{1}{n!}= e^1= e

I should point out that it is perfectly valid to define exp(x) to be the function satisfying "dy/dx= y with y(0)= 1" and get the derivative immediately. We could also define "ln(x)" to be
\int_0^x \frac{1}{t}dt
and then define exp(x) as its inverse function (after proving, of course, that has an inverse function).

 

[Tac-Tics]

does this make sense:
take the taylor series for e^x, at x=1, and you get the sum?
since i unfortunately have no experience with Taylor series, what is the proof for e^x.
sorry if i seem like i want to prove everything that anyone says, it is just that i do not have a background in the area.

Taylor Series are pretty simple.

Suppose that you can write some function f(x) as a polynomial of infinite degree.

f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... + a_k x^k + ...

How do you find all the coefficients a_i? Plug in 0, and you get a_0:

f(0) = a_0 + a_1 (0) + a_2 (0)^2 + a_3 (0)^3 + ... = a_0

What about the rest? Well, take the derivative of f:

f&#039;(x) = a_1 + 2 a_2 x + 3 a_3 x^2 + ... + k a_k x^{k-1} + ...

Notice that a_0 drops out. You can plug in 0 to f' and extract a_1.

f&#039;(0) = a_1 + 2 a_2 (0) + 3 a_3 (0)^2 + ... + k a_k (0)^{k-1} + ... = a_1

To find a_2, you take the derivative again and find f''(0). But be careful this time. Taking the derivative has popped a '2' from the exponent on x and thrown it into your equation.

f&#039;&#039;(x) = 2 a_2 + 6 a_3 x + ... + k (k-1) a_k x^{k-2} + ...

f&#039;&#039;(0) = 2 a_2 + 6 a_3 (0) + ... + k (k-1) a_k (0)^{k-2} + ... = 2 a_2

So a_2 = \frac{f&#039;&#039;(0)}{2}.

Continue this process, and you find that a_k = \frac{ f^{(k)}(0) }{k!} (where f^{(k)} is the k-th derivative and k! is k factorial).

So for any "well behaved" function f, we have

f(x) = \Sigma_{k=0}^\infty \frac{f^{(k)}(0)} {k!}x^k.

Now, applying this to f(x) = e^x, what do we get? Well, e^x is magical, because no matter how many times you take the derivative, it stays the same. That is, f^{(k)}(x) = f(x) = e^x. And knowing that e^0 = 1, we know that f^{(k)}(0) = f(0) = e^0 = 1. So finally, for our grand finale, we have:

e^x = \Sigma_{k=0}^\infty \frac{1} {k!} x^k.

So if we want to know what the number "e" itself is equal to, we just set x = 1:

e = e^1 = \Sigma_{k=0}^\infty \frac{1} {k!}.

 

[yasiru89]

As a side note, there's usually an error term for Taylor expansions, but elementary functions like the exponential, sine and cosine have an infinite radius of convergence so this is disregarded in those cases.

The LaTeX goes \sum_{n = 1}^{\infty} a_{n} to get \sum_{n = 1}^{\infty} a_{n}[/tex], you might find tutorials somewhere on here or elsewhere!

 

[Gib Z]

I should point out that it is perfectly valid to define exp(x) to be the function satisfying "dy/dx= y with y(0)= 1"

This, at least to me, seems the easiest way. Then all we must do is to consider the series

\sum_{n=0}^{\infty} \frac{x^n}{n!}.

and we see immediately that it converges for all values of x by the ratio test, and that it fulfills the definitions requirements.

 

[HallsofIvy]

As a side note, there's usually an error term for Taylor expansions, but elementary functions like the exponential, sine and cosine have an infinite radius of convergence so this is disregarded in those cases.

No. Error terms have nothing to do with "radius of convergence". Error terms apply only to Taylor "polynomials" where we cut the Taylor series off after a fixed power "n".

And it has nothing to do with "elementary" functions. The function ln(x) is as "elementary" as exponential but its Taylor series, around x=1, has radius of convergence 1.

 

[lurflurf]

sorry to all:
i do not really understand what you are saying. in response to
that i understand.
however, i do not get the next part.
could someone please explain it?

so you have (with n large)
(1+1/n)^n
by the binomial theorem we have
(1+1/n)^n=Σn!/k!/(n-k)!/n^k
1/k! is what we have in the alternate sum
so a two part proof would be
lim (1+1/n)^n=lim Σn!/k!/(n-k)!/n^k=Σlim n!/k!/(n-k)!/n^k=Σ1/k!
lim Σn!/k!/(n-k)!/n^k=Σlim n!/k!/(n-k)!/n^k
is the hard part since
lim n!/(n-k)!/n^k=1 is obvious
we should first show both limits exist
next we use a common calculus method to show two number are equal it is often easier to show that they are not unequal for example if we wish to show x=y we might show
x<=y
y<=x

now let
{s_i} be the sequence of partial sums of Σ1/k!
{t_i} be the sequence of values of (1+1/n)^n
with s and t there respective limits (both equal to e in the end)
it can be seen that
t_i<s_i<=s
because the terms in t_i when considered a sum by the binomial theorm
a {s_n} is a sequence of partial sums of positive numbers hence increasing
terms of t_i are the terms of s_i multiplied by
n!/(n-k)!/n^k (are all less than or equal 1)
that is
{t_i}={(1+1/n)^n}={Σn!/k!/(n-k)!/n^k}=={Σ[n!/(n-k)!/n^k][1/k!]}
so
t<=s
so the harder step will be
s<=t
suppose we make two approximations
one where n>N
one where n>M>N
the idea is when n is very large early terms of t_n look like terms of s_N and while later terms do not, they do not matter (they are small)
s-2eps<s_n-eps<t_n<=t whenever n>M>N
so we first estimated t_n<s_N-eps then estimated s_n<s-eps and combined them to get
s-2eps<=t
hence
s<=t
hence
s=t
qed

 

[yasiru89]

No. Error terms have nothing to do with "radius of convergence". Error terms apply only to Taylor "polynomials" where we cut the Taylor series off after a fixed power "n".

And it has nothing to do with "elementary" functions. The function ln(x) is as "elementary" as exponential but its Taylor series, around x=1, has radius of convergence 1.

Sorry, run-ins with asymptotic truncations have given me a bit of a twisted vocabulary. In my defence, the 'error term' was meant as the difference between the value of the function and the limiting sum of the series whenever (and however) they might be summed (this being zero is not always the case, but often is) and not the 'remainder' of a Taylor polynomial approximation (which is the difference between the approximation and the funtion); given convergence throughout the plane and coinciding results for these functions, this can be disconsidered in this case. Even then, it is worth keeping in mind that a Taylor series is simply the limiting case of a Taylor polynomial, and thus the behaviour of that very same remainder can't be so easily discounted.

I wouldn't consider all elementary functions of course, only those under consideration, which are entire with the Taylor series coinciding with the function at every point.