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The step you highlighted is correct. If you write them as limits of partial sums, there are two independent limits creating a double sum - if this double sum converges, then adding restrictions to the parameters won't change the convergence. More directly, the function e^x is analytic in x, so we know that it has a convergent taylor series (same for e^y in terms of y), then writing them as a double sum, the double sum is independent of path (the only thing that matters is that the upper index approaches (inf,inf), regardless of path), which is a more general continuity than to pick a specific path as you did by declaring the upper index to go along the diagonal. In fact, by path independence, you could take either part of the index to be larger, and the remaining terms that don't have a partner to factor with in your factoring scheme will be negligible in the limit, same as they are in the taylor series of the original e^x (or e^y). The step that bothers me is the last one. Plugging in (x,y)=(0,0) is a statement of continuity. We can normally do this to taylor series, because they are already analytic and thus continuous. But the final series is, in a way, simply a renaming of the original function - does that make it a valid 2-variable taylor series? I don't know, but maybe it is known. I'm pretty sure it is valid here, but I don't know if it is always a given or if it requires an exchange of limits I'm too lazy to verify right now. Can anyone give me an overview on multivariable taylor series?

As others have pointed out, I think this method works fine, as long as the series you are using are absolutely convergent, which they are here.

The expression is not defined for x=y so the limit doesn't exist

Quizás se pueda demostrar con la definición epsilon - delta para límites de varías variables.

x^n - y^n can always be factored as (x-y)*(x^(n-1)+x^(n-2)*y+...+y^(n-1)). So you do this thing here and get 1 as the result

Is there an integral you could make out of this instead? I mean I think of Riemann sums when I see 0+0+0+0+0+0+yada yada

The Blackpenredpen Conjecture

I would argue as it’s the same expansion for the same base function, it would be the same number of terms, even if it is infinitely large. I guess it’s one of those ones where it depends on how it’s been defined and the convention

Lemme try. Let n(r) be the rth term of the infinite sum. From my observation, n2=(x+y) (2 terms), n3=x^2+xy+y^2 (3 terms)… lim n →0(n^p) (p=a positive integer) will get closer to true 0 as p gets bigger If x=y, n3=lim x→0 (x^2+x*x+x^2)=3x^2… Basically, n(r)=lim x→0 r*x^(r-1) This will also approach true 0 as r gets bigger, due to the decreasing rate of exponentiation when x<0 greater than the increasing rate of multiplication It is also noted that each term n(r) is divided by a constant r! Hence, the nth term will get closer to true 0 than the (n-1)th, (n-2)th, and basically every term before it So, n(∞)=lim (n,x)→(∞,0) (n*x^(n-1)/n!)=(x^(n-1)/(n-1)!), which is a term incredibly close to absolute 0 Hence, we can say with confidence that all the terms after n1 are 0. Thus, the sum=1+0+0+0+…… *=1*

I challenge you to complete the multivariable limit from hell: lim x-> 0, y-> 0 (sin(xy)/xy)^[(1)/(1-cos(xy))] LaTeX Code: \left(\frac{\sin(xy)}{xy} ight)^{\frac{1}{1-\cos(xy)}}

Both sets are obviously countable, so I guess for any number you can combine and factor out 2 members of sets. Don’t see an issue.

This step has to be legitimate. I mean, yes, those are two different sums, however, every addendum is numerated, so for every nth addendum from the first sum we can always find respective nth addendum from the second sum. In other words, for every 1/(n-1)! × x^n we can find 1/(n-1)! × y^n no matter how far n goes into infinity.

As an engineer, I'd "linearize" the problem by considering lim/x-->0 e^x = 1 + x, right up front. In other words, you consider the linear approximation of e^x in the region where x = 0. Then you have (1 + x - 1 - y)/ (x-y) = 1. I'll leave it to the mathematicians to show this is mathematically correct ;-)

That's an interesting way to solve this problem, but I'm not entirely sure what the rules for infinite series are. Related to multivariable limits, is it normal to use L'Hopital? To explain my reasoning, any f(x,y) can be written as f(x(t),y(t)), right? That's a single variable function now, so if you have f(x(t),y(t))/g(x(t),y(t))=0/0 then you should be able to use L'Hopital and the chain rule. I will write x(t) as x and x'(t) as x' for simplicity. This is how I solve this problem: (e^x-e^y)/(x-y)=(x'e^x-y'e^y)/(x'-y')=(x'-y')/(x'-y') for this, we have 2 cases if x'-y'=0, which means x=y+c, but since x and y are 0, c must be 0, so you really just have x=y, which we can see will give the limit 1 else fraction simplifies to 1 I haven't defined the functions x(t) nor y(t) or their derivatives, so I haven't added any restrictions to the problem by doing this. Do you have anything similar on the channel?

i think you ran into the 0 times infinity because i think the zero repeating infinatly many times and you just considered it as 0, love your videos sir hope to get your T-shirt <3

Also we also gotta show that the terms in the bracket have a common factor of x-y ie x⁴-y⁴, x⁵-y⁵, upto infinitely many terms. This can be shown by mathematical induction

On the path of x=y, the value is indeterminate unless you impose a value of 1 based on limits.

Stop doing easy videos, even tho I’m a kid

To be continued?

You are enjoying the maths in better way, You are very energetic dude🤟🫶

I am afraid there is some problem for this approach. The reason behind is that there is NO correlation between x and y. The limit will totally different if you select different path for x and y to move toward (0,0). A path x=y will come up with the answer 1, while x=10y path will come up to a limit e^10/9. 🤔

Thanks PROF UR VIDEOS are the BEST

i havent done multivariable calc but cant you solve it using a substitution? lim x,y --> 0,0 of e^y(e^(x-y) - 1)/(x-y) and if the normal rules are still valid, limit of a product = product of a limit so we can isolate the e^y we factored and for the remaining limit make a substitution of x-y = t, and as x and y approach the origin, t will approach 0 [lim (x,y)--> (0,0) of e^y ] * [lim t-->0 of (e^t - 1)/t] which is just e^0 * 1 since the 2nd limit is just a standard limit so the final answer still ends up to be 1 but since i dont know multivariable calc i dont know if the substitution i made i valid or not

Does cancelling our the x-y not just mean you're dividing by zero? Similar to the 1=2 'proof'?

Why not make the substitution y=x-z, Then you can turn it into a single-variable limit.

The value of the limit is indeed 1 (proof by 3D desmos lol). However, by this reasoning, we use Taylor series for two analytic functions, so we do: lim (x,y) -> (0,0) then (lim n -> ∞ of "first sum" + lim m -> ∞ of "second sum"). (I will put A="first sum" and B = "second sum" for the rest) So the question is whether we can do: lim n -> ∞ of A + lim m -> ∞ of B = lim n-> ∞ of A+B. And seen like that the answer is obvious: lim n -> ∞ of (A+B) = lim n -> ∞ of A + lim n -> ∞ of B. (Indeed, we can use the linearity of the limit because the two limits exist because exp is an analytic function). So the reasoning is good and the limit is equal to 1 □. Edit: I just proved that lim of A + lim of B = equal lim of (A+B) in this case. Moreover we can rearrange the terms (thus factor like you’ve done) because *exp converges absolutely*

Isn’t the possibility of constructing bijection between terms of the infinite series a proof of them having the same „number” of terms? I would this bijection as „swap x with y” and the reverse one as „swap y with x”.

maybe, the life is better in the U world....

y=x+h lim (h -> 0) (e^x - e^(x+h))/(x-(x+h)) = lim (h -> 0) (e^(x+h) - e^x)/h = e^x lim (x -> 0) e^x = 1

as long as x and y are never equal this holds true, even if x and y have a big difference, its still around 1

What about highlighting the infinite nature of the series explicitly: Lim(x,y)->(0,0) (e^x- e^y )/ (x-y) Lim(x,y)->(0,0) limN->infty Sum(n=0)N (x^n - y^n)/ n!(x-y) Now I believe your question is : what allows us to set a single N to both expansions instead of doing two limits with M and N for x and y respectively? I believe this becomes clear for a more general case if we use expansion of multivariete functions! Set f(x,y) = e^x -e^y. We now do (taking the limits implicitly) f(x,y) = sum(nx)N sum(ny)M 1/(nx!ny!)f^(nx,ny) x^nx y^ny. We now do l=nx+ny and do L=M+N. We get sum(l=0)L sum(n=0)l f^(l-n,n) x^l-n y^n Where now we only take the limit of M+N=L . We basically use a Riemann product of series. Now this function in particular is such that mixed derivatives yield 0. Therefore we have the series: Sum(n=0)infty 1/n! (x^n-y^n).

By MVT there exist c between x and y such that (e^x-e^y)/(x-y)=e^c. Hence if (x,y) approach (0,0) then c will approach 0 (by squeeze). Thus the limit is simply e^c as c approaches 0, that is 1

Since the expansion of e^x is convergent, I think your method is valid

I feel it's wrong. If we approach along a path where x goes to 0 much faster than y, then we only need to expand till like 3 terms of e^x to get a good approximation and many terms of y, but if they have unequal number of terms it doesn't work But I am not sure

The point here is not to use Taylor expansion, but truncated expansion until order 3. Then, due the guaranteed conditions of the remaining functions you'll be able to continue.

I think this is fine as the series both converge absolutely, so we are free to rearrange them without changing their value. Another way to look at it is to change variables to u=x-y, v=y. Then the expression becomes lim e^v (e^u-1)/u, decoupling the variables into two separate factors. The limit of the e^v term is 1, and limit of the (e^u-1)/u term is also 1 by definition of the derivative of e^u. Using algebra of limits gives the result.

I would suggest transferring this problem into complex analysis to make things simpler where we have analytic functions of multiple complex variables meaning that we can use the techniques of complex analysis to simplify things. I think we would end up with a function with coefficients being other complex analytic functions, which should probably be well defined.

Seems legit to be, would look nicer if you would have used sigma notation

With your videos, students all over the world don't need uni professors anymore!!!!

I admire your honesty and modesty,"i am not sure if this is correct"💯👏👏

Nice video! Now I can understand why (e^x - 1)/x when x approaches to 0 equals 1 without L'hopital rule

(sorry for my bad English and sorry for my bad math, im still at school) i think this step is correct because we have “equals” things: in left of side ”x”, in right side “y” and if we know how look expression in left side we know how it look right side and they close to be equals, because of that we have 1/n!* (x-y)^n, i think its very easy to imagine please tell me if i have error, i really want understand why we have problem in this step

Idk if im correct, but i think we can. Since, we do something like this only while proving the euler's formula of exp(iθ) = cos(θ) + isin(θ)

I do think you are correct here, because the power series (by definition) would have infinite terms, no? And every term is going to have a factor of (x - y). Whether you do this over 3 terms, 30 terms, or 300 terms, there will always exist a factor of (x - y), which means that it can be factored out. I can't say with absolute certainty that it is correct, but I also cannot think of any reason why it would not be correct. Interesting video!

I think is for sure right

This is super cool whether it’s right or wrong, I love your videos dude

Nice way.

can you make a video on very fun derivatives? Really love your way of teaching !😁

Sleep

Can you make video about Hartley transform or some unusual integral transforms