E

[Aki]

Why is e=2.71828183 ?

I know it comes from lim x-> infinity (1+1/x)^x , but the function (1+1/x)^x keeps increasing to infinity, so how do we determine the limit?

 

*** correction, the limit of (1+1/x)^x is not infinity

[Bo]

this is a similar question as to 'why is pi=3.14....' in other words: there is no answer :friday:

 

i can only quote Euler:

 

e^(pi*i)=-1, therefore god exists.

 

Bo

[Qfwfq]

but the function (1+1/x)^x keeps increasing to infinity

No, it keeps increasing to e. If it kept increasing to infinity, the limit would be infinity. It is always increasing because it approaches the limit from below.

 

so how do we determine the limit?

Mathematicians have worked it out, I haven't. It is one case of "1 raised to infinity" which, by having a bound between the base and the exponent, has a limit (the limit of the base is 1). No doubt it can be shown as equivalent to some convergent series but I've never looked it up exactly, there's probably some outline on the Mathworld site.

[C1ay]

Why is e=2.71828183 ?

 

See e...

[zadojla]

Why is e=2.71828183 ?

If I remember my math history correctly, the number we call "e" was the one with the property:

e^e = 10

And then it was discovered to have all sorts of other cool and useful properties. It is a transcendental number like pi, not just irrational like sqrt 2.

[Bo]

e^e=10 is not true,

 

one definition of e is that

d/dx e^x=e^x

 

Bo

[Kent]

Why is e=2.71828183 ?

I know it comes from lim x-> infinity (1+1/x)^x , but the function (1+1/x)^x keeps increasing to infinity, so how do we determine the limit?

 

 

You must expend it using the binomial therom, and set x->inf.

 

Expend ( a+:friday:^ n .....

 

substitude a=1 and b=(1/n)

[Kent]

My formula beat you all

 

e^( 2*pi*i) =1

[Qfwfq]

e^(2*pi*i)=1 is quite equivalent to e^(pi*i)=-1 already posted by Bo. If you want to really win the prize you ought to show us how the actual value 2.71828... ensues.

 

What you say about the binomial expansion seems to be on the right track and I used it to confirm that the limit is convergent. Of course it means considering a sequence for natural values of x, sure enough I had usually seen it written with an n. I have so far only had a quick go at it but I found it can be majored by the summation of:

 

A_n = (n*(2^n))/((n!)^2) < 1/(n - 1)!

 

and I scarcely doubt it is convergent. But how about the actual limit?

 

It looks like you haven't quite beat us all, not yet anyway. :)

[tom]

e = lim ( 1 + 1/1! + 1/2! + ... + 1/n!)

e = lim ( 1 + 1/x)^(x+1)

[Qfwfq]

The first of the two limits you give Tom is a majoring series, its limit won't be the actual value of e.

 

At home I worked out the equivalence between the criteria, including the derivative of a logarithm which must be 1/x and not a/x. The one about the limit of (1 + 1/x)^x is somewhat arbitrary, it is obvious that many an expression could have the same limit, but I worked out that if we want one of the form A(x) raised to the x then A(x) must have limit 1 and derivative -1/(x^2). This almost determines A(x) as being (1 + 1/x).

 

Note Bo, I managed to do this without God's intervention. :)

[Bo]

Tom wrote:

e = lim ( 1 + 1/1! + 1/2! + ... + 1/n!)

Qfwfq wrote:

The first of the two limits you give Tom is a majoring series, its limit won't be the actual value of e.

are you objecting the notation here? (somewhere should be stated that n goes to infinitity...), because this sum most definitely goes to e. (proof by Taylor's series of e^x and put x=1)

 

then A(x) must have limit 1 and derivative -1/(x^2). This almost determines A(x) as being (1 + 1/x).

 

Note Bo, I managed to do this without God's intervention.

 

The criteria you give determine A(x) exactly, not almost :o

 

Euler also prooved his formula without the help of god, only the fact that all these strange numbers (e, pi, i, -1) appear in such a neat way in 1 formula was 'proof' to him that god existed (in the same way creationists give arguments on the beauty of life...). (the story does't tell if Euler was serious, or poetically...) (in my personal opinion Euler's argument makes more sense, because it doesn't need statistics; every argument that uses statistics is very arguable at least.)

 

Bo

[Qfwfq]

(proof by Taylor's series of e^x and put x=1)

Goodness, I hadn't thought of that! And yet I knew it, but it didn't come to mind! :o I was reasoning only on the limit of (1 + 1/x)^x. I'll do some more equivalence reckoning this evening...

 

The criteria you give determine A(x) exactly, not almost

I'd like to see proof of that! According to my reckoning the asymptotic behaviour (implicitly for x-->inf) is sufficient to determine equivalence. So let's see proof of the following theorem:

 

Hyp: A(x) has limit 1 and derivative -1/(x^2).

 

Th: A(x) = (1 + 1/x).

 

Note: If I remember my argument well enough, I meant the limit of the derivative too, it's sufficient, it needn't be -1/(x^2) for finite x. The first addend must have limit 1, the second must be asymptotically -1/(x^2).

 

As for Euler's ontological argument, I quite disagree. Not as a serious one. If there were no way to show that the numbers satisfying the different criteria for e are the same numbers, but they yet turn out to be the exact same number, one and the same, I'd say someone must be behind the scheme. :o However beautiful (and I certainly agree with that), "the fact that all these strange numbers (e, pi, i, -1) appear in such a neat way in 1 formula" is quite consequential to the definition of each of these "strange numbers", no more, no less. The formula e^(i*pi) = -1 is beautiful in it's simplicity.

[Qfwfq]

Thinking again of the Talor series of e^x I realise that's what a friend meant when he gave me the quick advice: calculate the reciprocal of e. Indeed this gives a series with alternating sign and decreasing modulus, the handiest way of knowing how near a partial result is to the actual limit. This gives a way to know how many decimals of the partial result are right. I should have thought of the Taylor series!

 

I'll still try to work out if there is some other way to show that the two limits are equal, I mean without seeing the series as being the Taylor expansion of e^x in 1.

 

Apart from this, the whole picture is clear, we can answer Aki's question:

 

Consider the Taylor expansion with x = -1, which is:

 

1/e = 1/0! - 1/1! + 1/2! - 1/3! ... + ((-1)^n)/n!

 

The first two terms are just 1, apart from the sign, the third is just 1/2, the next is -1/6... For any two successive partial sums, e is somewhere between their reciprocals; so in particular if these have the same first n decimals then those are also the first n decimals of e.

[Qfwfq]

lim (1 + 1/n)^n

lim ( 1 + 1/1! + 1/2! + ... + 1/n!)

 

Yes, now on lunch hour I worked out that, although the finite sum of the binomial expansion is less than the finite sum of 1/n!, they can be shown convergent to the same limit by posing it in terms of fixed a and increasing n and manipulating it into:

 

lim ((n - a)!n^a)/n! = 1

 

which is obvious enough. Now everything matches up. Again, without Divine Intervention! :o

[Monoxdifly]

this is a similar question as to 'why is pi=3.14....' in other words: there is no answer

 

Because everyone accepted that pi is used as the ratio of the perimeter of a circle to its diameter?