Understanding Euler's identity

[Qacer]

Hey all,

Referring to this link: link

I can't seem to understand For a=e, we thus have (a^x)' = (e^x)'.... and it goes on to say that "we have proved that the derivative of e^x is e^x."

Can someone please explain this to me in a way that I can understand easily? I'm not seeing the proof that the derivative of e^x is e^x.

Thanks!

 

[uec0]

For this derivation, they define "e" such that "lim delta->0 of (e^delta - 1) / delta == 1". They use some facts about "(a^delta - 1) / delta" as a function of "a" to show such an "e" exists.

Therefore, if you plug in "a = e" into the "expanded" derivative of a^x, then you get
"derivative of e^x = e^x * lim delta->0 of (e^x - 1) / delta = e^x * 1" by simple subsitution.

Obviously, "deriviative of e^x = e^x" for this "e".

 

[itachi]

sounds like hw..
i'm assuming you meant napier's constant? cos euler's identity is for solving equations with imaginary components..

anyways.. proof that the derivative of e^x is itself.
f'(x) = lim{h->0} [(a^x)(a^h - 1) / h] = (a^x) lim{h->0}[(a^h - 1) / h]

a = e => f'(x) = (e^x) lim{h->0}[(e^h - 1) / h]
..by definition, e = lim{k->inf} (1 + 1/k)^k
e = a = lim{k->inf} a = lim{k->inf} (1 + 1/k)^k
lim{k->inf}a^(1/k) = lim{k->inf}(1 + 1/k) => lim{k->inf}(a^(1/k) - 1) = lim{k->inf}(1 / k)

divide the left with the right..
lim{k->inf} (a^(1/k) - 1) / lim{k->inf}(1 / k) = 1

h = 1 / k => lim{k->inf}(1 / k) = 0 => lim{k->inf} = lim{h->0}
dx = lim{k->inf}((a^(1/k) - 1) / (1 / k)) = lim{h->0}((a^h - 1) / h) = 1

f'(x) = f(x) dx = f(x) * 1 = f(x) = e^x


p.s. you may want to look elsewhere for information on this kinda stuff.. i'm not sure what you're trying to learn, but that site that you gave a link to is a bit advanced. the topics covered will be extremely difficult to understand without prior knowledge on solving partial differential equation models.

 

[TuxDave]

I dunno... I don't really like this proof unless the 'e' that we know it was defined in this manner. I was always shown that 'e' was defined as the limit as x approaches infinity of (1+1/x)^x

 

[eLiu]

This definition of e is no good imo. I've seen a couple of rigorous proofs on the nature of e and the exponential function...and none of them define e in this way. iirc one of the favored ways is sum of x/k! TuxDave's post is a classic example, but its convergence is horridly slow & it's quite awkward to work with in my experiences.

 

[Gustavus]

Euler's identity is e^i*pi + 1 = 0, that is e raised to the power pi times the imaginary unit i is equal to -1. This is the special case (for x = pi) of his more general (and more easily seen) identity e^i*x = cos(x) + i*sin(x). Euler's identity is often said to be one of the most beautiful relations in all of mathematics since it combines in one terse statement the imaginary unit, i, the real unit, 1, the base of the natural logarithms, e, and one of the fundamental constants in mathematics, pi.

 

[itachi]

Originally posted by: TuxDave
I dunno... I don't really like this proof unless the 'e' that we know it was defined in this manner. I was always shown that 'e' was defined as the limit as x approaches infinity of (1+1/x)^x

that's what i based it off of.. i didn't try to redefine e or anything, i just used properties that are known. like the one you stated.

mm.. maybe it was a bit too messy, cos yall are pointin out the use of e.. when i used e in the manner that it's defined, as tuxdave (and me.. 3rd paragraph, 2nd line) described. so.. i'll clean it up a bit.

1. a = e => e = a -- if e is 'a' in the function, a can be interchanged with e (but only if a = e).
2a. a = lim{x->inf}(1 + 1/x)^x -- if 2e is valid, then so is 2a.. take the x root of both sides
2e. e = lim{x->inf}(1 + 1/x)^x
3a. lim{x->inf} a^(1/x) = lim {x->inf}(1 + 1/x) -- subtract 1 from both sides
3e. lim{x->inf} e^(1/x) = lim {x->inf}(1 + 1/x)
4a. lim{x->inf} a^(1/x) - 1 = lim{x->inf}(1 / x) -- multiply both sides by x
4e. lim{x->inf} e^(1/x) - 1 = lim{x->inf}(1 / x)
5a lim{x->inf} (a^(1/x) - 1) * x = 1 -- find substitution for x.. x = 1 / h
5e lim{x->inf} (e^(1/x) - 1) * x = 1 -- find value of h as x approaches infinity.. h = 0.
6a. lim{x->inf}(a^(1/x) - 1) * x = lim{h->0}(a^h - 1) / h) = 1
6e. lim{x->inf}(e^(1/x) - 1) * x = lim{h->0}(a^h - 1) / h) = 1

which is the simplest way to prove it that i can think of without going into geometric ratios..

oh yea and btw.. when i write lim{x->z}.. i mean limit as x approaches z.

 

[TuxDave]

Originally posted by: itachi

Originally posted by: TuxDave
I dunno... I don't really like this proof unless the 'e' that we know it was defined in this manner. I was always shown that 'e' was defined as the limit as x approaches infinity of (1+1/x)^x

that's what i based it off of.. i didn't try to redefine e or anything, i just used properties that are known. like the one you stated.

mm.. maybe it was a bit too messy, cos yall are pointin out the use of e.. when i used e in the manner that it's defined, as tuxdave (and me.. 3rd paragraph, 2nd line) described. so.. i'll clean it up a bit.

1. a = e => e = a -- if e is 'a' in the function, a can be interchanged with e (but only if a = e).
2a. a = lim{x->inf}(1 + 1/x)^x -- if 2e is valid, then so is 2a.. take the x root of both sides
2e. e = lim{x->inf}(1 + 1/x)^x
3a. lim{x->inf} a^(1/x) = lim {x->inf}(1 + 1/x) -- subtract 1 from both sides
3e. lim{x->inf} e^(1/x) = lim {x->inf}(1 + 1/x)
4a. lim{x->inf} a^(1/x) - 1 = lim{x->inf}(1 / x) -- multiply both sides by x
4e. lim{x->inf} e^(1/x) - 1 = lim{x->inf}(1 / x)
5a lim{x->inf} (a^(1/x) - 1) * x = 1 -- find substitution for x.. x = 1 / h
5e lim{x->inf} (e^(1/x) - 1) * x = 1 -- find value of h as x approaches infinity.. h = 0.
6a. lim{x->inf}(a^(1/x) - 1) * x = lim{h->0}(a^h - 1) / h) = 1
6e. lim{x->inf}(e^(1/x) - 1) * x = lim{h->0}(a^h - 1) / h) = 1

which is the simplest way to prove it that i can think of without going into geometric ratios..

oh yea and btw.. when i write lim{x->z}.. i mean limit as x approaches z.

Oh.. I didn't see your proof. The Stanford proof didn't include the transformation which shows that the 'e' that they define is the same 'e' that we all know it. That's why I didn't think it was a complete proof. Your addition helps it (unless that transformation is commonly known)

 

[eLiu]

Say E(z) = sum z^n/n! from n=0 to infinity.

Easy to show with series product that E(z+w)=E(z)*E(w)

Then use limit definition of derivative: lim x->h of [E(x+h)-E(x)]/h = lim E(x)*(E(h)-1)/h = E(x)

So d/dx exp(x) = exp(x).

edit: this is my favorite...quick and simple