Can we use power series to evaluate the limit of a multi-variable function?
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- Опубликовано: 1 июл 2024
- I came up with the limit of (e^x-e^y)/(x-y) as (x,y) goes to (0,0) and thought about using power series expansion for e^x and e^y to help us evaluate the limit. We know both series have infinitely many terms but do they have the same amount of terms that allow us to do the factoring in the video? If you know the answer, please let me know. Thank you!
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I admire your honesty and modesty,"i am not sure if this is correct"💯👏👏
BPRP is absolutely good, but here is nothing related to honesty at all. Math is a thing that needs rigorous proof -- mathematician can't tell anything that without a solid ground. If they do, time will tell. For this video, a double variable limit will be the catch here since two variable are not necessary the same. That means they can "float" in 2 completely different dimension. We can't jump to a conclusion that the series is converge as x is nothing related to y with both of them tend to zero, though. If the question changed to "x->y and y->0", the answer should be correct! But this is not the question asked! So, I doubt! 🤔
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.
Is breaking limits in the factors also valid in limit of multiple variables?
Agreed
@@anjugour9295 it is, as long if the limits do exist, and the proof is very similar
Yeah the Absolute convergence seems enough to me
However, just to be safer in the second argument you are adopting I would fix r=x-y and the other variable can be in general R=ax+(1-a)y (so that the change of coordinates has jacobian 1)
So that the limit becomes
Lim_{(r,R)->(0.0)} [Exp(R+(1-a)r)-Exp(R-ar)]/r=Exp(R)*[Exp((1-a)r)-Exp(-ar)]/r
=Lim_{r->0} [Exp((1-a)r)-Exp(-ar)]/r=1 for any value of a
Clearly if the limit depended on a It would have meant that such a limit does not exists
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*
how do you enter this into that tool? It looks correct but it's saying "left of -> needs to be a variable" when I have (x,y)->(0,0)
edit: oh, you don't do the limit you just graph the function
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.
Can you explain what that means please? What other conditions?
@@danielarnold9042what they're getting at is not use the whole Taylor series for e^x (the polynomial with infinitely many terms that equals the function e^x) but to use a finite taylor polynomial, setting the condition that, using the same amount of finite terms, you can prove that the limit equals 1
@@charliewatch1805 Wouldn't you still have to show the limit of the remaining terms is 0?
@@vinuthomas7193 most certainly yes, but the intention here is by creating a situation where we can manipulate legitimately the Taylor expansions.
The way I understand it is to pair so many terms of e^x with so many terms of e^y to manage the factorization of (x-y) in every similar terms of 1/(n!), thus showing that the infinite terms follow the same pattern and, at the limit, the value is one.
@@vinuthomas7193 DAMN DAMN DAMN, istg, one year ago I was crying at night feeling impotent because I couldn't understand calculus 3 in university and now I'm at night doing my best to force a motherfracking limit to take a value through a stupid amount of conditions with fellow nerds of the internet, I love you @blackpenredpen for creating this space, thank you and you guys that asked and questioned me, love you all
(x^n-y^n)=(x-y)(x^n-1 +yx^n-2+.......y^n-1)for all n(use finite geometric series to prove) which makes your argument valid😊
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 ;-)
LOL, I've never seen an engineer do math before.
we can put y=kx in the equation,
lim(x,y)->(0,0) (e^x-e^kx)/(x-kx)
=> Now by L.H. Rule
lim(x)->(0) (e^x-ke^x)/(1-k)
=> 1-k/1-k = 1
You're absolutely true about uncertainty as to whether (x-y) exists in each term. It's a pretty legitimate conjecture, I'd say, but pending proof.
Nice video! Now I can understand why (e^x - 1)/x when x approaches to 0 equals 1 without L'hopital rule
That's just the limit definition of the derivative of e^x evaluated at 0
The taylor series for e^x converges absolutely, so your rearrangement of terms is valid. The rule is that sum{a_n} + sum{b_n}= sum{a_n + b_n} when the individual sums converges absolutely.
On french wikipedia, it’s written that if a function f:I→ℝ is continuously differentiable at a then f is “strongly” or “strictly” differentiable at a i.e. the limit
lim (f(x)-f(y))/(x-y)
when (x,y) goes to (a,a) exists (and equals f’(a)). Here exp function is smooth on ℝ hence f is strongly differentiable at 0 and the limit equals 1
This is super cool whether it’s right or wrong, I love your videos dude
Thanks!
With your videos, students all over the world don't need uni professors anymore!!!!
You can use Lagrange theorem (F(x)-F(y))/(x-y)=F'(c) with c in [x y] interval and when x and y go to 0, it doesn't matter how, alco c go to 0, and so with F(x) = e^x, the limit is always e^0 = 1.
In your expansion, every term will be either A x^m,B y^n or C(x^m)(y^n). Thus when you take the limit, each term becomes zero.
It helps to change coordinates in the (x,y) plane to u=(y+x)/2 and v=(y-x)/2. Then the function (e^x-e^y)/(x-y) = e^u sinh(v)/v, which is separable, and hence there is no quandary when taking the limit to (0,0). If you wish, you can consider this a proof that everything will be alright with the higher powers of x-y, which correspond to powers of v.
I think the best way to attack something like this is to convert it to a single variable limit; let x(t) and y(t) be such that x(p) = y(p) = 0 for some real value p, both continuous with continuous derivatives of all orders on an open interval containing p, and not identically equal to each other within the interval. Then the limit becomes LIM(t->p; (e^x(t) - e^y(t))/(x(t) - y(t))), and you can use L'hopital's rule to compute it. In this specific case, you can calculate the limit without knowing anything about x(t) or y(t) except for the continuity and differentiability constraints. Since x(p) = y(p), you have the same 0/0 indeterminate form, so you can take derivatives to get LIM(t->p; (e^x(t) - e^y(t))/(x(t) - y(t))) = LIM(t->p; (x'(t)*e^x(t) - y'(t)*e^y(t))/(x'(t) - y'(t)). If x'(p) is not equal to y'(p), this simplifies to (x'(p)-y'(p))/(x'(p)-y'(p)) = 1. For the case where x'(p) = y'(p) and similar, just keep taking derivatives. Yes, this will be messy, but it's pretty straightforward to show that, denoting the n'th derivative of x(t) or y(t) as x[n](t) or y[n](t), if A(t)*e^x(t) - B(t)*e^y(t) is the n'th derivative of e^x(t) - e^y(t), and for all j < n, x[j](p) is equal to y[j](p), then A(p)*e^x(p) - B(p)*e^y(p) = x[n](p) - y[n](p). And, since x(t) is not identically equal to y(t), there must be an n such that x[n](p) is not equal to y[n](p). So, this also leads to a limit of one.
sir you really make sense with the method that you used
As an engineer I say this is totally legit lol
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?
It's not exactly a renaming of the original function since the denominator got cancelled out, so the domain grew. Look up "removable singularity" for more detail.
@@decare696 How do you know it was removed? To say that the discontinuity was removed is to say that the original limit exists.
We can show that the numerator is continuous and 0 along x=y. I don't believe this is the case, but suppose for example that the degree of that 0 is less than 1, then when we factored out that (x-y), the taylor series at the end would be of a function with a pole at x=y.
@@SlipperyTeeth I think I get what you mean now. It certainly feels like it should work, but I haven't put in the effort to verify it either.
a cooler way (and a way that convinces your way is good) is taking out the term e^y in the numerator to get e^y[e^(x-y)-1]. then expanding the same way but for the term e^(x-y).
then the taylor expansion becomes 1+(x-y)+1/2!*(x-y)^2... etc. the 1's cancel out and u can take out the term x-y and cancel it out for an easier solution of 1.
Since the expansion of e^x is convergent, I think your method is valid
More specifically, it is absolutely convergent. Because if the two series are only conditionally convergent you cannot reorder and regroup the terms.
It is true, just make a finite sum to N, collect and cancel terms and take limit as N goes to infinity.
Below is my instinct, I haven't done the rigor to show it.
If you put the sums as N->Infinity and M->infinity I believe you can change the order of the limits so that the limits to M,N are on the outside. then you have 3 scenarios, M>N, M
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).
Also anyone who sees this, let me know if there's anything wrong and sorry for the notation here!
@@carlosp.2898imo this is good. However, we also need to prove that we can rearrange the terms in the Taylor serie (because bprp factors the (x-y)) but this can be proven easily because the sum converges absolutely
@@matonphare you mean in the Riemann product right? Yes, i kinda skipped that and took it for granted because of what you said!
@@carlosp.2898 yeah I meant that my bad
Thanks PROF UR VIDEOS are the BEST
If your concern is with the infinite amount of terms then I don't think there is a problem because it really looks like taking the limit of the truncated sums and using measure theory theorems allows you to move everything around to get to the expression you found. Besides, I personally wouldn't even have considered this an issue in the first place for two reasons :
1) Both have a countably many terms and each has a formal definition, so we can associate each term to its counterpart without issues
2) We could decide to use truncated Taylor series expansionss and remove the problem of the number of terms
(Also, after a litttle research, I think we could just claim "I used the mutlivariate version of the Taylor theorem" and call it a day if we really are feeling a bit lazy)
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.
can you make a video on very fun derivatives? Really love your way of teaching !😁
I think it is correct using series, it is converg;
btw: we can also use 2-nd special limit for this problem
lim (x,y ->0, 0) e^y*(e^(x-y)-1)/(x-y) = lim (x,y ->0, 0) (e^(x-y)-1)/(x-y) , x-y = t
lim (t->0) (e^t-1)/t; e^t - 1 = v, t = ln(1+v);
lim (v->0) v/ln(1+v) = lim (v->0) 1/ln((1+v)^(1/v)) = lim (v->0) 1/ln(e) = 1
السؤال الاقوى هنا
هل
Lim(x,y)-->(a,a) [f(x)-f(y)/x-y]=f'(a)
?????
أو يمكن إعتبار xعلى يمين a و y على يسار a و ان f قابلة للإشتقاق في a، فإن المعادلة أعلاه صحيحة؟
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
Can you make video about Hartley transform or some unusual integral transforms
It would be awesome if youd upload a video solving the equation e^(a*x^2) + e^(b*x) + e^c = e^0
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
The essential question is this. This is a question of a limit of two variable function. Thus the limit is dependent upon the pass of (x, y) to (0, 0). Besides the function is undefined at (0, 0).
They both absolutely converge for any x (and y) so you may swap the around however you like, so long as every term appear at some point. Where does the nth term of e^x lie? At the (2n-1)th place. and 2n for the nth term of e^y. So every term is accounted for and the step is legitimate
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
There are actually two important questions to address:
1. Can we combine the two infinite sums into one sum? (This is what you asked in the video.)
2. Are we allowed to substitute x=0, y=0 into infinitely many terms of the final sum?
The answer to both questions is yes. Let's address question 1 first.
THEOREM 1: Given any sequences a_n, b_n for which \sum_{n=0}^{\infty} a_n and \sum_{n=0}^{\infty} b_n converge, we have that \sum_{n=0}^{\infty} a_n - \sum_{n=0}^{\infty} b_n = \sum_{n=0}^{\infty} (a_n - b_n).
PROOF: Recalling that an infinite sum is a limit of finite sums, we have that
\sum_{n=0}^{\infty} (a_n - b_n)
= \lim_{N\to\infty} \sum_{n=0}^N (a_n - b_n)
= \lim_{N\to\infty} (\sum_{n=0}^N a_n - \sum_{n=0}^N b_n)
= \lim_{N\to\infty} \sum_{n=0}^N a_n - \lim_{N\to\infty} \sum_{n=0}^N b_n
= \sum_{n=0}^{\infty} a_n - \sum_{n=0}^{\infty} b_n.
QED.
(Remark: absolute convergence is not required! As you can see, all we are using is the difference law for limits.)
This basically answers question 1 -- take a_n = x^n/n! and b_n = y^n/n!. Then e^x = \sum_{n=0}^{\infty} a_n and e^y = \sum_{n=0}^{\infty} b_n, so by Theorem 1 we can say
e^x - e^y = \sum_{n=0}^{\infty} (a_n - b_n)
= \sum_{n=0}^{\infty} [1/n! * (x^n-y^n)]
= \sum_{n=1}^{\infty} [1/n! * (x^n-y^n)]
= \sum_{n=1}^{\infty} [1/n! * (x-y) * \sum_{k=0}^{n-1} x^ky^{n-k}]
= (x-y) * \sum_{n=1}^{\infty} [1/n! * \sum_{k=0}^{n-1} x^ky^{n-k}].
Define f(x, y) = (e^x-e^y)/(x-y). Then
f(x, y) = \sum_{n=1}^{\infty} [1/n! * \sum_{k=0}^{n-1} x^ky^{n-k}]
= 1 + \sum_{n=2}^{\infty} [1/n! * \sum_{k=0}^{n-1} x^ky^{n-k}].
Now we are ready to move on to Question 2. This question is more subtle and is about *interchange of limits*.
For all integers N >= 2, define S_N(x, y) = \sum_{n=2}^N [1/n! * \sum_{k=0}^{n-1} x^ky^{n-k}], so that
f(x, y) = 1 + \lim_{N\to\infty} S_N(x, y).
When computing \lim_{(x, y)\to (0, 0)} f(x, y), we are faced with the double limit
\lim_{(x, y)\to (0, 0)}\lim_{N\to\infty} S_N(x, y).
When we substitute x=0, y=0 into each term of the infinite sum, we are claiming that we can move the \lim_{(x, y)\to (0, 0)} inside the infinite sum. This is equivalent to changing the order of the limits -- in other words, we are claiming that
\lim_{(x, y)\to (0, 0)}\lim_{N\to\infty} S_N(x, y) = \lim_{N\to\infty}\lim_{(x, y)\to (0, 0)} S_N(x, y).
Interchanging limits isn't always valid. However, it is valid when we have *uniform convergence*. We will prove that S_N(x, y) converges uniformly in the disk D = {(x, y) | x^2 + y^2 = 2, let t_n(x, y) = 1/n! * \sum_{k=0}^{n-1} x^ky^{n-k}, so that S_N(x, y) = \sum_{n=2}^N t_n(x, y). If (x, y)\in D, then |x|
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. 🤔
We can only tell the answer is correct if and only if we set a criteria x tend to y and y->0.
x and y are independent, yet this limit will always be 1, it's that simple. Yeah, in lots of cases (like x/y where (x,y)→0) it doesn't exist but here it does for any path (other than x=y)
how did you get lim(y→0) (e^10y - e^y)/y = e^10 ???
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 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!
Yes, but you can can expand both to a different number of terms, so I don't think it's correct
@@thatapollo7773 As long as the number of terms it is expanded for it clearly defined to be equal, then I think it is still valid.
@@terryhand5871 Yes, but the point here is --- the question didn't express clearly that x tend to y. This means x and y are two independent plane which different limits will be obtained through different path toward the (0,0) point. We can't assume this unless the questioned stated explicitly x tend to y. Or, in another way round, x and y approach zero on the straight line x=y.
I think this is absolutely because both the series if an infinite series convergence you can rearrange it's term however you like and the answer was still with the same under the condition that that the positive and negative parts of the series is not infinite. So by keeping that in mind (e^x - e^y) has always a factor of (x-y) and x and y both are approaching 0. So e^x and e^y both are finite.
I'm not at calc3 yet but is there a specific reason why we can't split and then throw L'H's rule at it until the problem disappears?
The Blackpenredpen Conjecture
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
That does not prove anything
I think your specific concern has no reasons to be there and in that sense your answer is right.
But i would point another concern that you basically get 1+ infinite sum of (number * 0). If we can factor 0 out of there, the infinite sum of 1/n! converges to a non infinite number, therefore 1+0 = 1?
(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^n-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
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
Induction? I mean we can factor these functions in terms of their solutions (=0), and x^n - y^n = 0 will always have solution x = y, so we can factor that out easily. Am I wrong?
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.
Quizás se pueda demostrar con la definición epsilon - delta para límites de varías variables.
On the path of x=y, the value is indeterminate unless you impose a value of 1 based on limits.
Probably wrong but what about (e^y*e^-y)((e^x-e^y)/(x-y) = e^y((e^(x-y) - 1)/(x - y)) = e^y((e^r - 1)/r) if r = x-y and r --> 0 if (x,y) --> 0.
For me the "possible" error is more between the 2 last liens: You have an infinity of terms converging to zero. as they are infinite, you can't say taht the summe is converginig to Zero.
Did you ask Dr. πm about the integral in German language?
@@cdkw2 no, not yet
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
Let x=5t and y=t, what will be limit equal? and if x=t and y=t?
They both have countably infinitely many terms, since there is a bijection between them
sorry to interrupt, or state something stupid, but is x-y when the limit in both cases is approaching 0, equal to 0? in short, are you dividing by 0?
Hey please explain this equation
3^(2sinx)² +3^(2cosx)² = Re((3 + 4i)^(2 + 3i))+e^(πi)
Can you really cancel (x-y)f / (x-y) to f??? Isn't that the case you have to watch out for, when x=y? Or can you just confirm the limit is the same along that path "x=y"😊
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.
(x^n - y^n) = (x-y) * sum( x^(n-i)*y^(i-1) ) for starting_index i=1 , end_index= n
YOU BETTER READ THAT BLACK PEN RED PEN!
YOU CAN ALWAYS FACTOR OUT X-Y of the difference between (x^n -y^n) for n >= 1
So you can divide this by (x-y) ... you will get a limit for your limes
Also you can use the sandwich theorem, and calculate the the limit. 1 time with no denominater which basically gives you a bigger limit, and with a bigger demonitor, giving you a smaller limit
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
Why do you always take simple integrals ?!!, tan^2x÷(1+sec^4x )dx ,, can you try this one
I came at this with a more simplistic approach.
LIM (x,y)->(0,0) [e^x-e^y]/[x-y]
Let T(u,r) the taylor series expansion for e^u, but starting off at r.
LIM (x,y)->(0,0) [(1+x+T(x,2))-(1+y+T(y,2)]/(x-y)
LIM (x,y)->(0,0) 1 + (T(x,2)-T(y,2))/(x-y)
T(x,2)-T(y,2) would vanish to zero quicker then x-y would
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”.
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
I suppose one should technically use the Lagrange Remainder to deal with this.
wouldn't the limit not exist because it's indeterminate along the path (x,x)?
the value along that entire path would be 0/0
I think is for sure right
Nice way.
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.
It's correct
The expression is not defined for x=y so the limit doesn't exist
So to make the questionable step, if somebody wants to prove this, we just need to prove the equivalance of the cardinality of each set containings the power series as elements. I'll come back if I can prove that
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)}}
To be continued?
@@scottleung9587 hopefully!
It is so logical that it has to be right.
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 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
While that’s true for an approximation, the series expansion is supposed to act more as a limit to infinity approaching the exact value, so I think that the number of terms should both always be approaching infinity since we’re talking about an exact value which is only obtained at the limit to infinity. At least that is what I think but it does always get tricky with infinite series so you might be right
Man, you hit the nail on the head! This series isn't converge as x and y are not related!
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
Very interesting. Unfotunately, I can barely add 2 and 2 together
Seems legit to be, would look nicer if you would have used sigma notation
I'm concerned of your action at 6:06. If this is legal, then you should be able to prove that lim(x,y)->(0,0) [(x-y)/(x-y)] is 1 and not undefined. Are you sure it's not equivalent to 0/0?
I think the last step is fishy as well.
Adding values that tend to 0 for infinite elements yields 0xinf format which is indeterminate and hence cannot be done.
Abuse the MVT. WLG and you prove the limit exists. We don’t have to prove for infinitely terms be the difference is a tail end of a convergent sequence. So all that’s left is prove x - y is a factor of start of sequence by induction.
It is indeterminate
x=y does not give u limit
8:27
@@blackpenredpengot it 👍
@@blackpenredpen i think their power series is the same, based on alpha wolfram, e^x-e^y is equal to summation_{k=0}^inf (x^k-y^k)/k!
Which means their infinity is the same...
After that u can factorise out (x-y) and do cancellation
Why not make the substitution y=x-z, Then you can turn it into a single-variable limit.
Does cancelling our the x-y not just mean you're dividing by zero?
Similar to the 1=2 'proof'?
they arent equal. its a limit not a computation. If you and usain bolt are both running a 100m dash, you're both approaching the finish line but you are NOT the same.
The fact that the denominator is zero along a curve (x=y in this case) suggests that the limit doesn't exist. So we try approaching (0,0) along paths like x=y + ky, x=y+ky^2, etc.
You'll find that the limit along x=y+ky^2 depends on k, hence the limit doesn't exist.
Yo the j
Sleep
Stop doing easy videos, even tho I’m a kid