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PS: I'm aware that I've interspersed a number of different questions and points here, I hope that doesn't cause umbrage with anyone!

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PS: I'm aware that I've interspersed a number of different questions and points here, I hope that doesn't cause umbrage with anyone!

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fresh_42

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Ok, so we might admit (or prove?) that exponential has an inverse and choose to call that the logarithm. What would we do next to complete the proof?

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Infrared

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##\lim_{h\to 0}\lim_{t\to\infty}\frac{\left(1+\frac{1}{t}\right)^{th}-1}{h}.## You can expand with the binomial theorem and take limits to get ##1.##

And once you have that ##\frac{d}{dx}e^x=e^x##, then in general ##\frac{d}{dx}a^x=\frac{d}{dx}e^{x\ln(a)}=\ln(a)a^x.##

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fresh_42

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$$

\dfrac{d}{dx} \log y = \dfrac{d}{dx}(x\log a)=\log a= \dfrac{\dfrac{d}{dx} y}{y} \Longrightarrow y'= y\log a = a^x\log a

$$

This uses the arithmetic rules of the logarithm and that the exponential function solves ##y'=y\, , \,y(0)=1.##

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And where does that definition of e come from? If I wanted to do the proof from first principles, I would need to first prove that ##\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^t## exists. I would then I would be happy to call it "e". Following that, I'm not sure about what you mean by expanding with a binomial theorem. Do you expand for integer or non-integer exponents ##th##?The usual definition of ##e## is ##e=\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^t.## So the limit you want to compute with ##a=e## is

##\lim_{h\to 0}\lim_{t\to\infty}\frac{\left(1+\frac{1}{t}\right)^{th}-1}{h}.## You can expand with the binomial theorem and take limits to get ##1.##

And then once you have that ##\frac{d}{dx}e^x=e^x##, then in general ##\frac{d}{dx}a^x=\frac{d}{dx}e^{x\ln(a)}=\ln(a)a^x.##

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I don't know how to prove that without invoking theorems I don't know the proof of. See my above reply to Infrared.I would first prove ##\dfrac{d}{dx}e^x = e^x##

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fresh_42

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Infrared

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It takes a little work to prove it, but it certainly doesn't rely on properties of the exponential function. Alternatively, you could pick integers ##n\leq th\leq n+1## and apply the standard Binomial theorem here and then bound. I'm sure this would work, but I guess it would be a little tedious.

To show that the limit exists, you want to show that ##(1+1/t)^t## is increasing in ##t## and bounded; this is a standard exercise.

Also, if you would be happy using a different definition of ##e##, choosing ##e:=\sum_{n=0}^\infty 1/n!## would probably make your life a bit easier (although it's not too hard to show these definitions are equivalent).

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fresh_42

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Out of curiousity, how would you show that it is bounded? Would you use the generalised binomial theorem again?To show that the limit exists, you want to show that ##(1+1/t)^t## is increasing in ##t## and bounded; this is a standard exercise.

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Just to be difficult, I'd define them as the ratio of the relevant two sides of a triangle. That's certainly how they get taught before we learn calculus.

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fresh_42

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The shortest way in this case is probably to use the unit circle in the complex plane and write them by Euler's formula in terms of the exponential function.Just to be difficult, I'd define them as the ratio of the relevant two sides of a triangle. That's certainly how they get taught before we learn calculus.

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Sure, although most proofs of Euler's formula use either the power series defintion of the exponential function or something similar https://en.wikipedia.org/wiki/Euler's_formula#Proofs. The proof using polar coordinates https://en.wikipedia.org/wiki/Euler's_formula#Using_polar_coordinates could be used if we allow for knowledge of how to differentiate the exponential function in the complex plane, which according to the above posts would require that we extend the binomial theorem to the complex plane. Otherwise, we could perhaps derive the power series definition of the exponential function as a corollary of the above proof sketches in this thread.The shortest way in this case is probably to use the unit circle in the complex plane and write them by Euler's formula in terms of the exponential function.

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vela

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This isn't too bad, actually. You can prove using the squeeze theorem thatI imagine it would get even more difficult if I asked how to prove the derivative of sine and cosyne.

$$\lim_{h \to 0}\frac{\sin h}{h} = 1.$$ Then it's straightforward to show that

$$\lim_{h \to 0}\frac{\cos h-1}{h} = 0.$$ Then it's just a matter of using the angle-addition formulas for sine and cosine and the definition of the derivative.

- #17

vela

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My old calculus book definedLet's take the exponential function. It's easy to show from the definition of a derivative that $$\frac{d a^x}{dx} = a^x \lim_{h\rightarrow 0}\frac{a^h -1}{h}$$ (at least for) $$a\ne0$$. However I don't know how to take that limit to complete the proof.

$$\log x = \int_1^x \frac{du}{u}.$$ Starting with this definition, you can prove ##\log ab = \log a + \log b## and ##\log a^b = b \log a## where ##b## is rational. From the fundamental theorem of calculus, it follows that ##(\log x)'= \frac 1x##.

Because the log function is one-to-one, there exists an inverse function ##\exp##, and it satisfies ##\exp \log x = x## and ##\log \exp x = x##. By differentiating the latter, it follows that ##(\exp x)' = \exp x##.

Then for ##a>0## and rational ##b##, we have ##a^b = \exp(\log a^b) = \exp(b \log a)##. Up to this point, the book had defined ##a^b## only for rational values of ##b##. Since the right-hand side is defined for all values of ##b##, this relationship gives us an obvious way to define ##a^b## for all values of ##b##. The derivative of ##a^x## then follows from the established properties of ##\exp## and the chain rule.

Finally, defining ##e## to be the value such that ##\log e = 1##, it follows that ##\exp x = e^x##.

- #18

Stephen Tashi

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That's quite possible, even if you were a diligent student. When I was reading introductory calculus texts (years ago), many texts did not prove the existence of ##\lim_{h \rightarrow 0} \frac{a^h-1}{h}##. As I recall, Johnson and Kiokemeister's, Calculus with Analytic Geometry, 3rd and 4th editions did.As far as I can tell from these responses, I was probably never actually taught via mathematical proof where these things come from.

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