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Surg AKA Kunark

Nothing to do with programming!

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I know that this has nothing to do with programming, but I was just talking to a friend online about Calculus. And we were talking about eulas number. Can anyone show me the proof that the derivative of the function e^x is also e^x. Where e^x is the symbol for eulas number. If so I would greatly appreciate it, and no this is not school homework. Just something I''m wondeirng about lol.

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I think that''s the definition of e; ie that d/dx of e^x is e^x for e != 0 is how e is defined.

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One possible definition of e^x is
e^x := 1 + x + x^2/2! + x^3/3! + x^4/4! + ...

The derivative of x^i/i! is x^(i-1)/(i-1)!, so taking the derivative of the expression that defines e^x converts each term in the previous term, with the term 1 disappearing because it''s derivative is 0.

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We''ve had a very similar discussion before, but what alvaro just described is exp(x), not e^x. e is a number, which can be defined as exp(1). Now e^x is just that number to the power of x. This could be multi-valued, it could have infinitely many values(if x is irrational). However in many situations e^x is written instead of exp(x).

I think what you''re wanting to look at is exp(x), for which you want alvaro''s definition. You can probably just ignore my point, it will never be very important for simple calculus.

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e is actually dedefined as:
e=(1+1/N)^N lim N->Infinity

Thus e^x=(1+1/N)^(Nx) lim N->Infinity

This expands to 1/0!+x/1!+x^2/2!... in the limit.
proving that exp(x) is e^x. Then you can run the proof above that exp''(x)=exp(x).

There is also some niggles in showing it coverges for all x, which I will not go into.

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e^x is defined as the inverse logarithm, which extends the concepts of powers to all real numbers, rather than rational numbers.

The logarithm was defined as ∫1/t dt from 1 to x (IIRC), which means that its derivative is 1/x

Now, let y = e^x. Then, x = ln y
Differentiating this equation implicitly with respect to x,
1 = 1/y dy/dx

Multiplying both sides by y, we have
y = dy/dx

Substituting y = e^x back in,
d(e^x)/dx = e^x

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Actually you''re wrong, e is a number, e^x is not a function, exp(x) is. However in calculus it really doesn''t matter, you just use exp(x), using e^x notation.

Also sadwanage: that ''proof'' that exp(x) = e^x is flawed, you''ve only shown that exp(x) belongs to the ''set of solutions'' of e^x.

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First, we''ll show that ∫1x 1/t dt is loge x for some unknown e, and then we''ll solve for e to define it. To abbreviate, we''ll write ln x instead of loge x So, using the definition of a derivative,

d/dx (ln x) = limit((ln(x + Δx) - ln x) / Δx, Δx -> 0)
= limit((ln(1 + Δx/x) / Δx, Δx -> 0)
= limit(1/x * x/Δx * ln(1 + Δx/x), Δx -> 0)
= limit(1/x * ln((1 + Δx/x)x/Δx), Δx -> 0)

Write w = Δx / x, so the above equals

limit(1/x * ln((1 + w)1/w), w -> 0)
= 1/x * ln( limit((1 + w)1/w, w -> 0))

Now if we define e as limit((1 + w)1/w, w -> 0), then ln e = 1, so

d/dx (ln x) = 1/x, we expected, and it also has the initial condition that ln 1 = 0, so it is the solution to the definite integral ∫1x 1/t dt.

Now that we have a definition for e, we find the derivative of ex using implicit differentiation:

y = ex
ln y = x
d/dx (ln y) = d/dx (x)
1/y * dy/dx = 1
dy/dx = y
dy/dx = ex

And that''s it!

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Guest Anonymous Poster
quote:
Original post by higherspeed
Actually you''re wrong, e is a number, e^x is not a function, exp(x) is.

You can define a^(p/q) where a is a positive real number and p and q are integers, q!=0. The only continuous function F(x) from R to R such that F(p/q)=a^(p/q) can be called a^x.

Where do you see a problem with using e^x instead of exp(x)? They are the exact same thing, as far as I know.

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You''re viewing e^x as a function, in fact as exp(x). That''s fine for calculus, however it isn''t e to the power of x, although it is connected and shares many properties.

eg

e^1/2 could be equal to exp(1/2) or -exp(1/2).

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Not at all, e^x is fine to use as a notation for exp(x), but when talking about definitions you have to be exact or it leads to confusion when the differences do matter. When doing calculus I certainly use e^x to mean exp(x). But I understand the situation and when it comes to proofs use the exact notation.

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I''m closing the thread. Although the original poster claimed this is not homework, they did not follow the rules in the Forum FAQ for questions that sound like homework and/or are not obviously about game development.

Review the Forum FAQ for the full requirements on posting this kind of question.

Graham Rhodes
Principal Scientist
Applied Research Associates, Inc.

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