Nothing to do with programming!
Author
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.
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.
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.
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.
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.
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.
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.
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
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
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.
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.
Author
Thanks everyone, that really does clarify it for me. Thanks again for your help, now I think im going to go tell my teacher I learnt something new online. lol
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!
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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