Real Analysis 33 | Some Continuous Functions

[@angelmendez-rivera351]

The video talks about the equality exp(x) = e^x, where e = exp(1), but in this video series, it has never been explained what b^x should be taken to be for an arbitrary real x, given some base b, nor has it been explained which bases b result in b^x being well-defined for every x, or how one reconciles with the more basic definition of exponentiation. In a real analysis course, this is important to talk about, because people tend to take the expression a^x for granted, but I highly doubt that most people actually know how a^x is defined in various contexts. We know that for natural numbers n, a^n denotes the product of n copies of a, and when extending the definition to a^m for integers m, we simply say a^0 = 1, a^(m + 1) = a·a^m, or a^(m + n) = a^m·a^n in general, for integers m, n. These definitions for work for every real number a, except maybe 0 if the exponent is negative. But when q is a rational number, the definition for a^q becomes much less clear, and in many circumstances, it becomes better to leave a^q only defined for real a > 0, or not work with fractional exponents altogether. And even more difficulty emerges when q is replaced by a real number x. In light of this, I think a video on this subject may be very fruitful for this series. In many common constructions, a^x is defined as exp[ln(a)·x], with a > 0, while other constructions use a piggy-back approach, defining natural exponents, then integer exponents in terms of natural exponents, then rational exponents in terms of integer exponents, then real exponents in terms of rational exponents. These constructions come with problems, though, as they are unable to then properly account for 0^x, and also for things like (-1)^3, since for example, (-1)^3 is not equal exp[3·ln(-1)], since ln(-1) is undefined, and there is also no way of accounting for by using suprema of sets. So the approach is this very-inelegant piecewise approach that depends on whether a number is rational or not or negative or not, which makes it rather inconvenient. I find that in general, if you work with real or complex analysis, you want to just avoid using expressions such as a^x, as they are not necessarily well-defined, and just stick to expressing things, whenever necessary and possible, in terms of exp. In this series, limits with x -> ♾ for functions R -> R have not been properly defined, although this not much of an issue, since the definition can be made a straightforward generalization of the limit of a sequence. But for the purposes of this series in particular, I propose the definitions lim f(x) (x -> +♾) := lim f(1/x) (x > 0, x -> 0) and lim f(x) (x -> -♾) := lim f(1/x) (x < 0, x -> 0). In fact, for the purposes of notational elegance, I would also note that lim f(x) (x -> x0) = +♾ iff lim 1/f(x) (x -> x0) = 0 and f > 0, while lim f(x) (x -> x0) = -♾ iff lim f(x) (x -> x0) = 0 and f < 0. This just makes limits with ♾ for functions R -> R special cases of limits of functions on open intervals as previously defined in this series. I prefer this approach over the generalization, because the definition for limits of sequences do not mesh well with how limits of functions on open intervals are defined, even though one definition is related to the other, and strictly speaking, they are different analytic concepts, so I think it would be less confusing to keep them separate, rather than introduce a blurred line between the two.

[@brightsideofmaths]

Thanks for the elaborate answer. Of course, I am aware of my intentional gaps in the series. Either, I have them to induce thinking and new realisations or I have them to not lose focus. The limit to infinity is of the first kind and the exponent is of the second kind :) I have some ideas for a video about exponents but it is not quite fitting into this series yet.

[@angelmendez-rivera351]

Oops. I made a mistake. I meant write lim f(x) (x -> x0) = -♾ iff lim 1/f(x) (x -> x0) = 0 and f < 0.

[@angelmendez-rivera351]

@@brightsideofmaths I understand. I look forward to seeing the rest of your series!

[@Leslie.Green_CEng_MIEE]

2:29 *_exp()_* is strictly monotonically increasing “monotonically increasing” or “monotonically decreasing” is fair enough. “strictly monotonic” used when we either don’t know or don’t care if the function is _non-decreasing_ or _non-increasing_ . “strictly monotonically increasing” seems a bit mixed up to me. I am used to “ *_strictly increasing_* ” or “ *_strictly decreasing_* ” functions (as a naming convention). 3:23 exp: *R* → (0, ♾) is bijective Using the codomain to limit the domain seems a bit ‘ugly’ to me. I might have preferred: exp: *R⁺* → *R⁺* is bijective (where the *R* is supposed to be double struck) 😃

[@brightsideofmaths]

Different notions for increasing and non-decreasing functions are around. I use "strictly" to emphasize the strict inequality one has in the definition. On the other hand, saying that exp: R → (0, \infty) is bijective is completely standard. There is nothing ugly here. A map is given with domain and codomain and every map can be made into a surjective map by restricting the codomain.

[@Leslie.Green_CEng_MIEE]

@@brightsideofmaths Thanks 😊

[@JojiThomas7431]

Nice video

[@ahmedamr5265]

Thanks for the great video! Is the convergence of power series a uniform convergence or point-wise convergence?

[@brightsideofmaths]

Here, we just require point-wise convergence but you also have uniform convergence on the correct domains. See here: thebrightsideofmathematics.com/courses/complex_analysis/overview#part-10---uniform-convergence

[@ahmedamr5265]

@@brightsideofmaths i will check the link. Thanks so much

[@edztyMKWII]

Hallo. Is there a way to prove if for 2 functions f and g from [0,1] -> [0,infinity) with sup f(x) = sup g(x) they intersect at some point.

[@brightsideofmaths]

Are the functions bounded?

[@MrWater2]

To prove that exp: R -->(0, oo) is bijective. The injectivity is because it's monotonically increasing so if x>y then f(x) > f(y) then f(x) != f(y). But the surjectivity? Can I use the intermediate value theorem? I can take any a,b with b>a, then f:[a,b] to (0,oo) with f continous by the theorem, x in [a,b] with f(x)=y as a,b arbitrary the function is surjective?? Or I just need to use that exp is continous and monotonically increasing and try another explanation?

[@brightsideofmaths]

Nice work. IVT is a good choice!

[@DutchMathematician]

Let me scetch the proof to show that exp(x) is a bijection from R to (0, ∞). (assuming that we already know that exp is a strictly increasing function and that we know the limits of exp(x) as x → ±∞) Injectivity... Suppose therefore that exp(x) = exp(y). Since exp is a strictly increasing function on R, we can't have that x < y, since this would lead to exp(x) < exp(y). The same holds for x > y. Hence, x must equal y. (as you stated) Surjectivity... Let y be an arbitrary element of the image of exp, being (0, ∞). We must show that there exists a certain x ϵ R such that exp(x) = y. (for surjectivity, it is not needed that such a x be unique - but of course, it is: it is ln(y)!) Since lim (exp(x)) as x → -∞ equals 0, there must exist a x1 ϵ R such that exp(x1) ≤ y/2. (since, if not, all values of exp(x) would be > y/2 and thus the limit would be ≥ y/2, which is not 0) Likewise, since lim (exp(x)) as x → ∞ equals ∞, there must exist a x2 ϵ R such that exp(x1) ≥ 2*y. (since, if not, all values of exp(x) would be < 2*y and thus the limit would be ≤ 2*y, not ∞) Now apply the Intermediate Value Theorem for exp(x) on the interval [x1, x2]. This theorem ensures that there exists a x ϵ [x1, x2] such that exp(x) = w for every element w in [exp(x1), exp(x2)]. Hence, certainly for every element in [y/2, 2*y], and hence for our y. Which is exactly what we wanted to prove.