Integral Formula for Natural Log (without knowing the derivative)
ฝัง
- เผยแพร่เมื่อ 15 ก.ย. 2024
- This video proves that the natural log equals the integral from 1 to x of 1/t dt under the assumption that ln(x) is the inverse function to the exponential e^x. We can do this without already knowing the derivative of the natural log!
More details on why the integral is the inverse of e^x:
We proved in the video that any right inverse to e^x must equal that integral. However, we didn't prove that e^x has a right inverse in the first place.
We know that e^x : R → R+ is a strictly increasing function whose output can be made arbitrarily large or arbitrarily small. Therefore e^x is a bijection R → R+. Every bijective function has a two-sided inverse (see [1] below). Therefore e^x has a two-sided inverse, which in particular is a right inverse. I proved in video [2] that injective functions have at most one right inverse. Therefore the right inverse to e^x is unique if it exists. But we already know that there exists one right inverse that is also a two-sided inverse. We conclude that there exists exactly one right inverse to e^x and that this right inverse is also a two-sided inverse. Hence the integral in the video is a two-sided inverse to e^x.
[1] • Bijective Functions Ha...
[2]: • Proof: Two-Sided Inver...
Calculus Problems playlist: • Calculus Problems
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Music: OcularNebula - The Lopez
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If you’re interested in complex analysis:
In this video I assumed that x is a real number. But we can also use this argument to define log functions
log z = ∫ᵧ 1/z dz
in the complex plane because the same derivative argument applies for complex numbers. This is very important because it lets us define log functions that are holomorphic on certain subsets of the complex plane!
That rocks. Also, random request: Hopf Algebras. Save me lol.
Hey, you're back! That's good. I like your approach to math.
Welcome back!
I like these compressed integrals for some reason, it’s so simple and elegant.
been a long time, welcome again :)
Bravo, Haydn! Early birthday wishes, and I don't mean William McKinley's. I wish you'd been around a half century ago so I would've had you for my professor around 1976-7. I'm back to flood out my frustrations, for I should be good in math. I scored quite high about a half century ago in aptitude tests, and I've always wondered what had happened to my skill. Seven years ago, I wrote on my LinkedIn the major reasons we don't do well in math, and you prove at least the theory that the professors don't teach it well. By the way, the beard looks good now that the acne has healed . My acne worsened when I was your age.
Omgggg you’re back!!! 😍
Glad you're back; the beard's looking righteous!
How do we know that f, defined as the right inverse of eˣ, is differentiable?
Good question! I suppose you could use some simpler version of the inverse function theorem to prove that e^x is a diffeomorphism (simpler because e^x is strictly increasing on R), although arguably that makes the original problem trivial.
Everything is differentiable if you just try hard enough.😁
@@MuPrimeMath Agreed. The inverse function theorem not only guarantees the differentiability of the inverse function in this case (as eˣ is a continuously differentiable function with nonzero derivative at every point in R) but also includes a formula for the derivative of the inverse function.
en.m.wikipedia.org/wiki/Inverse_function_theorem
Welcome back♥️
at 3:40, why can't we just use the idea that since the two sides have the same base, the exponents [f(1) and 0] have to be equal?
That's basically what I did in the video. I just wanted to explain a bit more why we can do that. The reason we can do that is that e^x is an increasing function, which means that it's injective (i.e. one-to-one)!
Nice
-without knowing the derivative
- 1:13 , takes the derivative
lol, I meant without using the limit definition of the derivative!
@@MuPrimeMath when i saw the tittle i got exited bc i thought it would be a geometric approach
You’re back - you must be in grad skool by now, right?
What would a "left inverse" of e^x be?
It would be a function f satisfying f(e^x) = x.
@@MuPrimeMath thanks! I didn't know about left and right inverses. I'm curious, about what makes them different. In both cases f = log, right?
In the case of bijective functions (such as e^x : R --> R+), every left inverse is also a right inverse and vice versa, which is why log is both the left and right inverse.
In general, if a function is not bijective, then we can have a right inverse that is not a left inverse and vice versa. For example, the positive square root is a right inverse to x^2 because (sqrt(x))^2 = x, but it is not a left inverse.
@@MuPrimeMath thanks so much. Great content!
Sir , To integrate 1/ t wrt t , put t = e^x or dt = e^ x . dx and you get the result.
I am a simple boy
I will differentiate both sides
So LHS=RHS
Because 1/x=derivative of upper limit is 1 and 1*1/x
Putting upper limit inyo value of t
And value of lower limit derivative is 0
Therefore 1/x=1/x hence proved haha
Did you invent this yourself?
It's very loosely based on a proof in Stein & Shakarchi's Complex Analysis textbook.