The Math Thread - Page 11

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

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.

On September 29 2017 06:46 AbouSV wrote:


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 07:23 Joni_ wrote:

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 infinite-dimensional 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 08:02 Simberto wrote:


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.

On September 29 2017 08:25 Joni_ wrote:

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 exp-function 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 15:19 Acrofales wrote:

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 16:24 Joni_ wrote:

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/can-i-exchange-limit-and-differentiation-for-a-sequence-of-smooth-functions

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 time-consuming to give the full picture.

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 non-elegant 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)

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

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)

On September 29 2017 18:34 Shalashaska_123 wrote:
+ Show Spoiler [long] +
I hope this helped you out.

Sincerely,
Shalashaska_123