
On September 14 2017 00:34 AbouSV wrote:Show nested quote +On September 13 2017 14:49 FiWiFaKi wrote:On September 13 2017 14:48 Acrofales wrote: @fiwi: a mathematically elegant way of doing it is in the answer above yours. Oh, I'm just too dumb to understand it, I see. edit: Oh nevermind, I get it. I read the problem and I got too excited about trying to solve it rather than properly reading the comments that followed. A right answer is a right answer Shows the difference between how an engineer vs how a mathematician approaches a problem. i.e. Python vs Excel :p
Matlab if I have an idea about the method of the solution, Excel when I have no idea how to start, Mathematica if I want to be a smart ass instead of being practical ^^.
But that was all abou being clever and finding how to start it, not about the software.


Hey y'all.
I've been messing around with differentiation of exponential functions last few days and there's something that I cant seem to figure out so I thought I'd ask TeamLiquid for advise/help/inspiration.
By the definition of differentiation, the derivative of a^x is a^x times the limit of ((a^h))1/h) as h goes to 0. Evaluating that limit is trivial by applying L'H BUT that rests upon already knowing the derivative of a^x so assuming that isnt known how does one figure out that limit?
the brute approach obviously is just convincing oneself that the derivate of 2^x is 2^x itself times a constant which appears to be approximately 0.7
then one could do the same for 8^x and come to the realization that the derivative of 8^x is 8^x times a constant which is about 3 times as great as 0.7  which suggests we're looking at something logarithmic  the base of which can be approximated by a Taylor Series etc etc
this approach seems so bruteish and nonelegant so my question is: is there a cleverer and more elegant way to evaluate the limit which holds the key to differentiation of exponential functions?
(help me know if my wording was too poor/confusing  English is not my first language)

a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x

On September 29 2017 02:37 Liebig wrote: a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x you're missing the point my friend (maybe I explained poorly)
Im asking how to approach that limit question if one does NOT know beforehand what the derivative of a^x (including e^x) is.

meeh :/ i tried using latex to bbcode converter but it doesnt work the way i want it to So in short and without formality:
You want the proof that the limit f(x), x > 0 with f(x) = (b^x  1)/x exists. Did i get that right? You proof this by showing that f is strictly monotonic increasing in R+, which you can do by using the mean value theorem. Hence, the limit f(x), x > 0 equals inf{ f(x)  x in R+} =: s. (1) Now you take a convergent seqeuence in R\{0}, h_n, with limit h_n > 0 and show limit h_n = s (via monotonic increase and infimum property). You can easily show that f(h) = b^(h) f(h) with h in R+. Also x > b^x is continuos in 0, so limit b^(h_n) = b^0 = 1. Hence, limit b^(h_n) f(h_n) = 1 s = s. (2)
d in R+. From (1) follows theres N_1 in Naturals with f( h_n )  s < d for all n > N_1. From (2) follows theres N_2 in Naturals with  b^(h_n) f( h_n )  s  < d for all n > N_2. For all n in Naturals is f(h_n) either f( h_n ) or b^(h_n) f( h_n ), so  f(h_n)  s  < d for all n in Naturals with n > max{N_1 , N_2}.
Hope this is helpful. Never did do any math in english so im lacking all the vocabulary. however, i hope you can follow my "proof".
Edit: There is an elegant way to determine the derivate of a^x (and hence the limit you are looking for). Which is excactly what the dude wrote before me: a^x = exp( x ln(a) ) and it tells you the limit is ln(a). I dont think you need to look any further for a clever and elegant way. Because you already use it all the time The common derivation rules and the fact that exp is derivable on the entire definition range. (Btw. you dont need L'H to proof any of this.)

Oh sorry, that was my bad.
Can't really answer on how to approach the limit question without knowing exponential before, but my 2 cents:
So if you assume you don't know exponential, the only way (i know of) that you can define a^x is as the unique solution to the functional equation f(x+y)=f(x)*f(y) and f(1)=a for a > 0.
So f'(x) must be such that f'(x+y)=f'(x)*f(y) So, now you have f'(x)=f'(0)*f(x)
And now I'm stuck cause one of the definition of e^x is as the solution of differential equation f'(x)=f(x)
But that gives an answer to why there is some constant times the function itself.

I think one of the working definitions of e is that the derivative of e^x is e^x. If you choose any of the different ways to define e, you can prove from there that that sentence is true. I don't remember the specifics, but it works.
If you have that as set, and also have the chain rule of calculus proven, you can take the way that Liebig explained above.
And that is an absolutely fine way to prove the thing you wanted to prove, or to approach the series. You know other things, and derive the result from them. So you don't know what the derivative of a^x is, but you know what the derivative of e^x is, either by definition or by something that follows from that definition.
You are trying to take the way in the opposite direction. You don't start with proving what the derivative of a^x is, and go to e^x from there, you go the other way and start with e^x, and go to a^x.

exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n > infty.
But besides that i agree with you, simberto. It feels like HKTPZ goes about it the wrong way. Thats also what i was trying to say in some way: You already know d(a^x)/dx = ln(a) a^x because of the chain rule and exp properties you use daily, d(a^x)/dx = d(exp(x ln(a)))/dx = exp(x ln(a)) ln(a) = a^x ln(a). And d(a^x)/dx = lim ( a^(x+h)  a^x )/h = a^x lim (a^h  1)/h with h > 0. In short a^x ln(a) = d(a^x)/dx = a^x lim (a^h  1)/h.
Commonly you calculate d(a^x)/dx = a^x ln(a) via derivation rules. These are no rules of thumb but proven generalization that work always for the right parameters. So the clever and elegant way you are looking for to determine the limit lim (a^h  1)/h, is the set of derivation rules and exp properties.

On September 29 2017 06:16 jodljodl wrote: exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n > infty.
Exponential can also be defined as the only solution of y' = y, with y(0) = 1, implying exp'(x) = exp(x).

On September 29 2017 06:46 AbouSV wrote:Show nested quote +On September 29 2017 06:16 jodljodl wrote: exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n > infty. Exponential can also be defined as the only solution of y' = y, with y(0) = 1, implying exp'(x) = exp(x). You need a lot more mathematical background to prove that this differential equation has a unique solution than you need to define the exponential function by it's series representation. Also the series representation can be generalized more easily (say to the case of bounded operators acting on possibly infinitedimensional banach spaces) and it allows you to verify properties of the (real) exponential function like its limit behaviour when concatenated with polynomial functions in a very simple way.. I dont really see the advantage of using your proposed definition, as long as it's not part of a class for say engineers that are primarily interested in solving differential equations.
Mind you I dont want to say that either definition is really more "correct" (at least in the case of real exponential functions), I just think that using the series representation fits more naturally into the way an analysis class is structured. At least where I'm from, maybe it's different if you do several classes of "calculus" before developing an axiomatic theory..
Edit: The more I think about it the less I like your way of defining the exponential function. It really only works properly in the case of functions defined on at least some open neighbourhood of 0 on the real axis. And at the very least as soon as you try to generalize to the complex plane you will inevitably need the series representation to prove uniqueness of the solution anyway, bc you will need some form of an identity theorem for holomorphic functions... Just feels like it really only is useful in a setting where the people being taught are interested only in the theory of ordinary differential equations on the set of real functions and their strong solutions....

On September 29 2017 07:23 Joni_ wrote:Show nested quote +On September 29 2017 06:46 AbouSV wrote:On September 29 2017 06:16 jodljodl wrote: exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n > infty. Exponential can also be defined as the only solution of y' = y, with y(0) = 1, implying exp'(x) = exp(x). You need a lot more mathematical background to prove that this differential equation has a unique solution than you need to define the exponential function by it's series representation. Also the series representation can be generalized more easily (say to the case of bounded operators acting on possibly infinitedimensional banach spaces) and it allows you to verify properties of the (real) exponential function like its limit behaviour when concatenated with polynomial functions in a very simple way.. I dont really see the advantage of using your proposed definition, as long as it's not part of a class for say engineers that are primarily interested in solving differential equations. Mind you I dont want to say that either definition is really more "correct" (at least in the case of real exponential functions), I just think that using the series representation fits more naturally into the way an analysis class is structured. At least where I'm from, maybe it's different if you do several classes of "calculus" before developing an axiomatic theory.. Edit: The more I think about it the less I like your way of defining the exponential function. It really only works properly in the case of functions defined on at least some open neighbourhood of 0 on the real axis. And at the very least as soon as you try to generalize to the complex plane you will inevitably need the series representation to prove uniqueness of the solution anyway, bc you will need some form of an identity theorem for holomorphic functions... Just feels like it really only is useful in a setting where the people being taught are interested only in the theory of ordinary differential equations on the set of real functions and their strong solutions....
I remember at least three different definitions of e or exp. One via the infinite series, one via the differential equation, and one as the limit of (1+1/n)^n. Though i admit that i had to look that last one up. It has been a bit since my introductory maths classes, but i am rather certain that you can prove that they are all equivalent anyways. So basically, just use whichever is most useful in the situation you are in. The differential equation is probably not the most useful for introductory maths classes for maths students due to the problems that you mention. And except for maths students, noone cares about the exact way something can be derived from other stuff and whether that is actually correct.
Edit: Also, from the infinity series of the exp function to d/dx exp(x) = exp(x) you just need to derive polynoms. And derivatives of polynoms are easily proven by the definition of the derivative. So with regards to the original question, it doesn't really matter with which one we start. The way to d/dx a^x (and the limit in question) from not a lot of other stuff is: Prove d/dx x^n = n x^n1 prove chain rule define exp = infinite series x^n/n! do some work to get a reasonable definition of ln(x) derive d/dx exp(x) = exp(x) define a^x as exp(x lna) derive via ruleset that has been proven.

On September 29 2017 08:02 Simberto wrote:Show nested quote +On September 29 2017 07:23 Joni_ wrote:On September 29 2017 06:46 AbouSV wrote:On September 29 2017 06:16 jodljodl wrote: exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n > infty. Exponential can also be defined as the only solution of y' = y, with y(0) = 1, implying exp'(x) = exp(x). You need a lot more mathematical background to prove that this differential equation has a unique solution than you need to define the exponential function by it's series representation. Also the series representation can be generalized more easily (say to the case of bounded operators acting on possibly infinitedimensional banach spaces) and it allows you to verify properties of the (real) exponential function like its limit behaviour when concatenated with polynomial functions in a very simple way.. I dont really see the advantage of using your proposed definition, as long as it's not part of a class for say engineers that are primarily interested in solving differential equations. Mind you I dont want to say that either definition is really more "correct" (at least in the case of real exponential functions), I just think that using the series representation fits more naturally into the way an analysis class is structured. At least where I'm from, maybe it's different if you do several classes of "calculus" before developing an axiomatic theory.. Edit: The more I think about it the less I like your way of defining the exponential function. It really only works properly in the case of functions defined on at least some open neighbourhood of 0 on the real axis. And at the very least as soon as you try to generalize to the complex plane you will inevitably need the series representation to prove uniqueness of the solution anyway, bc you will need some form of an identity theorem for holomorphic functions... Just feels like it really only is useful in a setting where the people being taught are interested only in the theory of ordinary differential equations on the set of real functions and their strong solutions.... I remember at least three different definitions of e or exp. One via the infinite series, one via the differential equation, and one as the limit of (1+1/n)^n. Though i admit that i had to look that last one up. It has been a bit since my introductory maths classes, but i am rather certain that you can prove that they are all equivalent anyways. So basically, just use whichever is most useful in the situation you are in. The differential equation is probably not the most useful for introductory maths classes for maths students due to the problems that you mention. And except for maths students, noone cares about the exact way something can be derived from other stuff and whether that is actually correct. Your last one should read (1+x/n)^n if you want to define the actual exponential map, I think.
Of course they are "equivalent" as long as you stay in the realm of real functions. But that was precisely the point I was trying to make. Understanding why the differential equation a) has a solution and b) the solution is unique requires way more machinery than covering convergent series. The simplestway of proving that the differential exquation in question has a solution is to give one. But if you are going to define the exponential function that way, then you can hardly use the exponential function to show that a solution exists. Using the series expression would also defeat the purpose of using the differential equation to actually define the exponential function. So all that is left is relying on more sophisticated tools like some kind of existence/uniqueness result for differential equations, I guess. That does seem rather 'high level' for such simple endavours.
I mean assuming we want to define the expfunction as the unique solution to the differential equation... How would I go about proving that the differential equation has a solution that is unique? Even attempting to go for some fix point iteration seems a bit ridiculous because that relies on an understanding of convergent sequences (which correspond 1:1 to convergent series  the exact thing we were trying to avoid).
Concerning your edit: You cant just assume that the derivative of a series (i.e. the limit of finite sums) is the same as the limit of the derivative of the finite sums. There are certain setting where limits and derivatives can be exchanged but this is NOT true in General and you should always make very sure why exactly you are allowed to interchange limits (after all taking a derivative is just another limit). In this case the uniform convergence of the series on compact sets and the pointwise convergence of the (continuous) derivatives saves the day, but we should try to always make sure we know the reasons why certain operations yield correct results.
Of course after making sure that all this is allowed, you're perfectly right. Then all that remains to do is exchange the series' limit and the d/dx and differentiate dem juicy polynomials. :>

On September 29 2017 08:25 Joni_ wrote:Show nested quote +On September 29 2017 08:02 Simberto wrote:On September 29 2017 07:23 Joni_ wrote:On September 29 2017 06:46 AbouSV wrote:On September 29 2017 06:16 jodljodl wrote: exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n > infty. Exponential can also be defined as the only solution of y' = y, with y(0) = 1, implying exp'(x) = exp(x). You need a lot more mathematical background to prove that this differential equation has a unique solution than you need to define the exponential function by it's series representation. Also the series representation can be generalized more easily (say to the case of bounded operators acting on possibly infinitedimensional banach spaces) and it allows you to verify properties of the (real) exponential function like its limit behaviour when concatenated with polynomial functions in a very simple way.. I dont really see the advantage of using your proposed definition, as long as it's not part of a class for say engineers that are primarily interested in solving differential equations. Mind you I dont want to say that either definition is really more "correct" (at least in the case of real exponential functions), I just think that using the series representation fits more naturally into the way an analysis class is structured. At least where I'm from, maybe it's different if you do several classes of "calculus" before developing an axiomatic theory.. Edit: The more I think about it the less I like your way of defining the exponential function. It really only works properly in the case of functions defined on at least some open neighbourhood of 0 on the real axis. And at the very least as soon as you try to generalize to the complex plane you will inevitably need the series representation to prove uniqueness of the solution anyway, bc you will need some form of an identity theorem for holomorphic functions... Just feels like it really only is useful in a setting where the people being taught are interested only in the theory of ordinary differential equations on the set of real functions and their strong solutions.... I remember at least three different definitions of e or exp. One via the infinite series, one via the differential equation, and one as the limit of (1+1/n)^n. Though i admit that i had to look that last one up. It has been a bit since my introductory maths classes, but i am rather certain that you can prove that they are all equivalent anyways. So basically, just use whichever is most useful in the situation you are in. The differential equation is probably not the most useful for introductory maths classes for maths students due to the problems that you mention. And except for maths students, noone cares about the exact way something can be derived from other stuff and whether that is actually correct. Your last one should read (1+x/n)^n if you want to define the actual exponential map, I think. Of course they are "equivalent" as long as you stay in the realm of real functions. But that was precisely the point I was trying to make. Understanding why the differential equation a) has a solution and b) the solution is unique requires way more machinery than covering convergent series. The simplestway of proving that the differential exquation in question has a solution is to give one. But if you are going to define the exponential function that way, then you can hardly use the exponential function to show that a solution exists. Using the series expression would also defeat the purpose of using the differential equation to actually define the exponential function. So all that is left is relying on more sophisticated tools like some kind of existence/uniqueness result for differential equations, I guess. That does seem rather 'high level' for such simple endavours. I mean assuming we want to define the expfunction as the unique solution to the differential equation... How would I go about proving that the differential equation has a solution that is unique? Even attempting to go for some fix point iteration seems a bit ridiculous because that relies on an understanding of convergent sequences (which correspond 1:1 to convergent series  the exact thing we were trying to avoid). Concerning your edit: You cant just assume that the derivative of a series (i.e. the limit of finite sums) is the same as the limit of the derivative of the finite sums. There are certain setting where limits and derivatives can be exchanged but this is NOT true in General and you should always make very sure why exactly you are allowed to interchange limits (after all taking a derivative is just another limit). In this case the uniform convergence of the series on compact sets and the pointwise convergence of the (continuous) derivatives saves the day, but we should try to always make sure we know the reasons why certain operations yield correct results. Of course after making sure that all this is allowed, you're perfectly right. Then all that remains to do is exchange the series' limit and the d/dx and differentiate dem juicy polynomials. :> If you define e^x as an infinite series, then the derivative is obviously the derivative of that infinite series. You don't need the general case, you need this very specific case. You should of course prove f(x) = f'(x) for f(x) = e^x, using whatever definition of the derivative has your preference.

On September 29 2017 15:19 Acrofales wrote:Show nested quote +On September 29 2017 08:25 Joni_ wrote:On September 29 2017 08:02 Simberto wrote:On September 29 2017 07:23 Joni_ wrote:On September 29 2017 06:46 AbouSV wrote:On September 29 2017 06:16 jodljodl wrote: exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n > infty. Exponential can also be defined as the only solution of y' = y, with y(0) = 1, implying exp'(x) = exp(x). You need a lot more mathematical background to prove that this differential equation has a unique solution than you need to define the exponential function by it's series representation. Also the series representation can be generalized more easily (say to the case of bounded operators acting on possibly infinitedimensional banach spaces) and it allows you to verify properties of the (real) exponential function like its limit behaviour when concatenated with polynomial functions in a very simple way.. I dont really see the advantage of using your proposed definition, as long as it's not part of a class for say engineers that are primarily interested in solving differential equations. Mind you I dont want to say that either definition is really more "correct" (at least in the case of real exponential functions), I just think that using the series representation fits more naturally into the way an analysis class is structured. At least where I'm from, maybe it's different if you do several classes of "calculus" before developing an axiomatic theory.. Edit: The more I think about it the less I like your way of defining the exponential function. It really only works properly in the case of functions defined on at least some open neighbourhood of 0 on the real axis. And at the very least as soon as you try to generalize to the complex plane you will inevitably need the series representation to prove uniqueness of the solution anyway, bc you will need some form of an identity theorem for holomorphic functions... Just feels like it really only is useful in a setting where the people being taught are interested only in the theory of ordinary differential equations on the set of real functions and their strong solutions.... I remember at least three different definitions of e or exp. One via the infinite series, one via the differential equation, and one as the limit of (1+1/n)^n. Though i admit that i had to look that last one up. It has been a bit since my introductory maths classes, but i am rather certain that you can prove that they are all equivalent anyways. So basically, just use whichever is most useful in the situation you are in. The differential equation is probably not the most useful for introductory maths classes for maths students due to the problems that you mention. And except for maths students, noone cares about the exact way something can be derived from other stuff and whether that is actually correct. Your last one should read (1+x/n)^n if you want to define the actual exponential map, I think. Of course they are "equivalent" as long as you stay in the realm of real functions. But that was precisely the point I was trying to make. Understanding why the differential equation a) has a solution and b) the solution is unique requires way more machinery than covering convergent series. The simplestway of proving that the differential exquation in question has a solution is to give one. But if you are going to define the exponential function that way, then you can hardly use the exponential function to show that a solution exists. Using the series expression would also defeat the purpose of using the differential equation to actually define the exponential function. So all that is left is relying on more sophisticated tools like some kind of existence/uniqueness result for differential equations, I guess. That does seem rather 'high level' for such simple endavours. I mean assuming we want to define the expfunction as the unique solution to the differential equation... How would I go about proving that the differential equation has a solution that is unique? Even attempting to go for some fix point iteration seems a bit ridiculous because that relies on an understanding of convergent sequences (which correspond 1:1 to convergent series  the exact thing we were trying to avoid). Concerning your edit: You cant just assume that the derivative of a series (i.e. the limit of finite sums) is the same as the limit of the derivative of the finite sums. There are certain setting where limits and derivatives can be exchanged but this is NOT true in General and you should always make very sure why exactly you are allowed to interchange limits (after all taking a derivative is just another limit). In this case the uniform convergence of the series on compact sets and the pointwise convergence of the (continuous) derivatives saves the day, but we should try to always make sure we know the reasons why certain operations yield correct results. Of course after making sure that all this is allowed, you're perfectly right. Then all that remains to do is exchange the series' limit and the d/dx and differentiate dem juicy polynomials. :> If you define e^x as an infinite series, then the derivative is obviously the derivative of that infinite series. You don't need the general case, you need this very specific case. You should of course prove f(x) = f'(x) for f(x) = e^x, using whatever definition of the derivative has your preference. Not sure what youre trying to say? Or even what part of my post you are referring to exactly? If it's about the last part..: "The derivative of the series" would be d/dx \lim_n\sum_{k=0}^n x^k/k! and all that I said is that you need a reason for why this would equal \lim_n d/dx \sum_{k=0}^n x^k/k!, i.e., a reason for why you are allowed to exchange the order of derivative and limit. This is NOT trivial or true for arbitrary series. It is however exactly what you need to do to reduce the case of differentiating the exponential function to the case of differentiating polynomials, i.e., applying the differentiation to the finite sums.
One possible way to go about this is using this: https://math.stackexchange.com/questions/409178/caniexchangelimitanddifferentiationforasequenceofsmoothfunctions
I know that in some cases in courses for engineers (or even physicists) people pretend like limits can be arbitrarily exchanged. Time to accept that this is just wrong and only ever done because it would be too timeconsuming to give the full picture.

Clearly, I didnt manage to ask my question the right way so I will try to rephrase ^^
I know a^x can be rewritten as e^(x*lna) and I know the derivative of e^x is e^x times ln e which simplifies to e^x and I know Leibniz' chain rule.
Im asking: WITHOUT using that knowledge, is there a clever and elegant way to show that lim ((a^h)1)/h as h goes to 0 is exactly 1/n of lim (((a^n)^h)1)/h as h goes to 0 implying there is a connection to log.
I can show that lim ((a^h)1)/h as h goes to 0 exists but Im at a loss trying to show the limit is ln a (what the base is is trivial  the point is to show the connection to logarithmic behaviour)

On September 29 2017 16:24 Joni_ wrote:Show nested quote +On September 29 2017 15:19 Acrofales wrote:On September 29 2017 08:25 Joni_ wrote:On September 29 2017 08:02 Simberto wrote:On September 29 2017 07:23 Joni_ wrote:On September 29 2017 06:46 AbouSV wrote:On September 29 2017 06:16 jodljodl wrote: exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n > infty. Exponential can also be defined as the only solution of y' = y, with y(0) = 1, implying exp'(x) = exp(x). You need a lot more mathematical background to prove that this differential equation has a unique solution than you need to define the exponential function by it's series representation. Also the series representation can be generalized more easily (say to the case of bounded operators acting on possibly infinitedimensional banach spaces) and it allows you to verify properties of the (real) exponential function like its limit behaviour when concatenated with polynomial functions in a very simple way.. I dont really see the advantage of using your proposed definition, as long as it's not part of a class for say engineers that are primarily interested in solving differential equations. Mind you I dont want to say that either definition is really more "correct" (at least in the case of real exponential functions), I just think that using the series representation fits more naturally into the way an analysis class is structured. At least where I'm from, maybe it's different if you do several classes of "calculus" before developing an axiomatic theory.. Edit: The more I think about it the less I like your way of defining the exponential function. It really only works properly in the case of functions defined on at least some open neighbourhood of 0 on the real axis. And at the very least as soon as you try to generalize to the complex plane you will inevitably need the series representation to prove uniqueness of the solution anyway, bc you will need some form of an identity theorem for holomorphic functions... Just feels like it really only is useful in a setting where the people being taught are interested only in the theory of ordinary differential equations on the set of real functions and their strong solutions.... I remember at least three different definitions of e or exp. One via the infinite series, one via the differential equation, and one as the limit of (1+1/n)^n. Though i admit that i had to look that last one up. It has been a bit since my introductory maths classes, but i am rather certain that you can prove that they are all equivalent anyways. So basically, just use whichever is most useful in the situation you are in. The differential equation is probably not the most useful for introductory maths classes for maths students due to the problems that you mention. And except for maths students, noone cares about the exact way something can be derived from other stuff and whether that is actually correct. Your last one should read (1+x/n)^n if you want to define the actual exponential map, I think. Of course they are "equivalent" as long as you stay in the realm of real functions. But that was precisely the point I was trying to make. Understanding why the differential equation a) has a solution and b) the solution is unique requires way more machinery than covering convergent series. The simplestway of proving that the differential exquation in question has a solution is to give one. But if you are going to define the exponential function that way, then you can hardly use the exponential function to show that a solution exists. Using the series expression would also defeat the purpose of using the differential equation to actually define the exponential function. So all that is left is relying on more sophisticated tools like some kind of existence/uniqueness result for differential equations, I guess. That does seem rather 'high level' for such simple endavours. I mean assuming we want to define the expfunction as the unique solution to the differential equation... How would I go about proving that the differential equation has a solution that is unique? Even attempting to go for some fix point iteration seems a bit ridiculous because that relies on an understanding of convergent sequences (which correspond 1:1 to convergent series  the exact thing we were trying to avoid). Concerning your edit: You cant just assume that the derivative of a series (i.e. the limit of finite sums) is the same as the limit of the derivative of the finite sums. There are certain setting where limits and derivatives can be exchanged but this is NOT true in General and you should always make very sure why exactly you are allowed to interchange limits (after all taking a derivative is just another limit). In this case the uniform convergence of the series on compact sets and the pointwise convergence of the (continuous) derivatives saves the day, but we should try to always make sure we know the reasons why certain operations yield correct results. Of course after making sure that all this is allowed, you're perfectly right. Then all that remains to do is exchange the series' limit and the d/dx and differentiate dem juicy polynomials. :> If you define e^x as an infinite series, then the derivative is obviously the derivative of that infinite series. You don't need the general case, you need this very specific case. You should of course prove f(x) = f'(x) for f(x) = e^x, using whatever definition of the derivative has your preference. Not sure what youre trying to say? Or even what part of my post you are referring to exactly? If it's about the last part..: "The derivative of the series" would be d/dx \lim_n\sum_{k=0}^n x^k/k! and all that I said is that you need a reason for why this would equal \lim_n d/dx \sum_{k=0}^n x^k/k!, i.e., a reason for why you are allowed to exchange the order of derivative and limit. This is NOT trivial or true for arbitrary series. It is however exactly what you need to do to reduce the case of differentiating the exponential function to the case of differentiating polynomials, i.e., applying the differentiation to the finite sums. One possible way to go about this is using this: https://math.stackexchange.com/questions/409178/caniexchangelimitanddifferentiationforasequenceofsmoothfunctionsI know that in some cases in courses for engineers (or even physicists) people pretend like limits can be arbitrarily exchanged. Time to accept that this is just wrong and only ever done because it would be too timeconsuming to give the full picture.
Thank you for talking down to engineers and physicists. I am actually a mathematician (well, at least I got my propaedeuse in math, never completed my BSc as I moved to CS which got less hooked up on infinitessimal details )
Anyway, I see your problem, and I think you are simply misunderstanding what I am saying. I do see that there is a more fundamental problem with Simberto's approach, and it's all in the definition of e^x as an infinite series, but it's not in the proof of the derivative. So lets clarify that first:
We define:
f(x) = sum x^n/n! (with n 0 to infinity)
Now:
df /dx = d / dx sum x^n/n! (with n 0 to infinity) (by definition)
Now we have to prove that df / dx = f.
Lemma: d / dx sum x^n/n! (with n 0 to N) = sum x^n/n! (with n 0 to N  1).
Proof:
Base step: d / dx sum x^n/n! (with n 0 to 2) = d / dx (1 + x) = 1 = sum x^n/n! (with n 0 to 1). (little square box)
Inductive step: Assume d / dx sum x^n/n! (with n 0 to M) = sum x^n/n! (with n 0 to M  1).
d / dx sum x^n/n! (with n 0 to M + 1) = d / dx (x^(M + 1)/(M + 1)! + sum x^n/n! (with n 0 to M) ) = d / dx (x^(M + 1)/(M + 1)! + d / dx sum x^n/n! (with n 0 to M) = (M + 1)x^M / (M + 1)! + d / dx sum x^n/n! (with n 0 to M) = x^M / M! + d / dx sum x^n/n! (with n 0 to M) = x^M / M! + sum x^n/n! (with n 0 to M  1) = sum x^n/n! (with n 0 to M) (little square box)
QED.
Theorem:
d / dx sum x^n/n! (with n 0 to infinity) = sum x^n/n! (with n 0 to infinity).
Proof (using Lemma above): lim(N > inf) d / dx sum x^n/n! (with n = 0 to N) = lim(N  > inf) sum x^n/n! (with n = 0 to N  1) = lim(N  > inf) sum x^n/n! (with n = 0 to N) = lim(N > inf) sum x^n/n!
QED
So there we have it: f(x) = f'(x) for f defined as above.
Now, of course, there is a problem, and that is that we haven't defined a^x, only e^x. I believe the usual solution is to simply define a^x = e^(x ln a). And from then onwards it's easy. But I'm not sure that this satisfies what the original question was asking.

On September 29 2017 01:39 HKTPZ wrote: Hey y'all.
I've been messing around with differentiation of exponential functions last few days and there's something that I cant seem to figure out so I thought I'd ask TeamLiquid for advise/help/inspiration.
By the definition of differentiation, the derivative of a^x is a^x times the limit of ((a^h))1/h) as h goes to 0. Evaluating that limit is trivial by applying L'H BUT that rests upon already knowing the derivative of a^x so assuming that isnt known how does one figure out that limit?
the brute approach obviously is just convincing oneself that the derivate of 2^x is 2^x itself times a constant which appears to be approximately 0.7
then one could do the same for 8^x and come to the realization that the derivative of 8^x is 8^x times a constant which is about 3 times as great as 0.7  which suggests we're looking at something logarithmic  the base of which can be approximated by a Taylor Series etc etc
this approach seems so bruteish and nonelegant so my question is: is there a cleverer and more elegant way to evaluate the limit which holds the key to differentiation of exponential functions?
(help me know if my wording was too poor/confusing  English is not my first language)
Hello, HKTPZ.
Honestly, I would do it the way Liebig suggested.
On September 29 2017 02:37 Liebig wrote: a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x
If we can't use that approach, though, then we have to go back to the definition of the derivative.
The function we have is f(x) = a^x. Plug this into the definition.
Use one of the properties of exponents.
Factor a^x and bring it in front of the limit. This can be done because it doesn't depend on h.
We have here the indeterminate form 0/0, so we can apply L'hopital's rule.
The denominator is just 1, so it disappears. In the numerator, the constant disappears. We're just left with what we set out to find from the beginning.
This is just f'(0).
We stated at the beginning that f(x) = a^x. Substitute it here.
We thus have a differential equation for f(x). Divide both sides by f(x).
The left side is just the derivative of ln f.
Now integrate both sides with respect to x.
Exponentiate both sides so that we just have f(x) on the left side.
Use one of the properties of exponents.
Use a new constant for e^C.
Now substitute a^x for f(x).
We can determine A by setting x = 0. Doing so gives us 1 = A.
All that's left to do is to solve for f'(0). Take the natural log of both sides.
Use the property of logs that allows us to bring the power of the argument to the coefficient.
ln e = 1 and x cancels from both sides. We're left with the value of f'(0).
f'(x) = a^xf'(0). Therefore,
I hope this helped you out.
Sincerely, Shalashaska_123

On September 29 2017 17:23 HKTPZ wrote: Clearly, I didnt manage to ask my question the right way so I will try to rephrase ^^
I know a^x can be rewritten as e^(x*lna) and I know the derivative of e^x is e^x times ln e which simplifies to e^x and I know Leibniz' chain rule.
Im asking: WITHOUT using that knowledge, is there a clever and elegant way to show that lim ((a^h)1)/h as h goes to 0 is exactly 1/n of lim (((a^n)^h)1)/h as h goes to 0 implying there is a connection to log.
I can show that lim ((a^h)1)/h as h goes to 0 exists but Im at a loss trying to show the limit is ln a (what the base is is trivial  the point is to show the connection to logarithmic behaviour)
As in the post above. Please define a^x in a suitably formal manner, and we can take it from there

On September 29 2017 18:34 Shalashaska_123 wrote:+ Show Spoiler [long] +On September 29 2017 01:39 HKTPZ wrote: Hey y'all.
I've been messing around with differentiation of exponential functions last few days and there's something that I cant seem to figure out so I thought I'd ask TeamLiquid for advise/help/inspiration.
By the definition of differentiation, the derivative of a^x is a^x times the limit of ((a^h))1/h) as h goes to 0. Evaluating that limit is trivial by applying L'H BUT that rests upon already knowing the derivative of a^x so assuming that isnt known how does one figure out that limit?
the brute approach obviously is just convincing oneself that the derivate of 2^x is 2^x itself times a constant which appears to be approximately 0.7
then one could do the same for 8^x and come to the realization that the derivative of 8^x is 8^x times a constant which is about 3 times as great as 0.7  which suggests we're looking at something logarithmic  the base of which can be approximated by a Taylor Series etc etc
this approach seems so bruteish and nonelegant so my question is: is there a cleverer and more elegant way to evaluate the limit which holds the key to differentiation of exponential functions?
(help me know if my wording was too poor/confusing  English is not my first language) Hello, HKTPZ. Honestly, I would do it the way Liebig suggested. On September 29 2017 02:37 Liebig wrote: a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x If we can't use that approach, though, then we have to go back to the definition of the derivative. The function we have is f(x) = a^x. Plug this into the definition. Use one of the properties of exponents. Factor a^x and bring it in front of the limit. This can be done because it doesn't depend on h. We have here the indeterminate form 0/0, so we can apply L'hopital's rule. The denominator is just 1, so it disappears. In the numerator, the constant disappears. We're just left with what we set out to find from the beginning. This is just f'(0). We stated at the beginning that f(x) = a^x. Substitute it here. We thus have a differential equation for f(x). Divide both sides by f(x). The left side is just the derivative of ln f. Now integrate both sides with respect to x. Exponentiate both sides so that we just have f(x) on the left side. Use one of the properties of exponents. Use a new constant for e^C. Now substitute a^x for f(x). We can determine A by setting x = 0. Doing so gives us 1 = A. All that's left to do is to solve for f'(0). Take the natural log of both sides. Use the property of logs that allows us to bring the power of the argument to the coefficient. ln e = 1 and x cancels from both sides. We're left with the value of f'(0). f'(x) = a^xf'(0). Therefore, I hope this helped you out. Sincerely, Shalashaska_123 Holy wowzerz. You generated a million different images from latex, uploaded them to imgur and included them here. You're my hero.
Also, a super elegant solution, but there is one slight hitch:
"The left side is just the derivative of ln f."
It's easy to prove, but you'd still have to return to first principles. I think you don't escape needing to define e^x in order to be able to define ln x as its inverse. And then you'd need to prove that sentence there. Which means a lot of work.



