Your solution has some (significant) flaws. By which I mean they're not so easy to fix. At the end, you found out the limit was gamma + sum_{k>0} (ln(1+x/k)/x - 1/k) Your argument was: I have a sum of infinetely many terms, each converging to 0, therefore the sum converges to 0. This is not true. It happened to be true in that case, and I'll show you how you'd fix it. Limits can be interpreted as functions that take functions to real values (or oo or -oo). What we all know is that limit is a linear function, this is: if limf and limg exist, then (limf + limg) = lim(f+g). Easy induction shows that if limf1, limf2, ..., limfk exist, therefore limf1 + limf2 + ... + limfk = lim(f1+f2+...+fk) But this does not apply to infinetely many terms. you cant induct your way to infinity. So what im saying is: if limf1, limf2, limf3, ... exist, this DOES NOT mean that lim(f1+f2+f3+...) exists, and even if does, it doesnt mean that lim(f1+f2+...) = limf1 + limf2 + ... Thats what you used in the solution. That "the infinite sum of the limits is the limit of the infinite sum". Here's a counter example: sum_{k>0} x/k = x+x/2+x/3+x/4+... See: if x goes to 0, then x/k goes to 0, so the sum of the limits is 0+0+... = 0. But 1+1/2+1/3+1/4 diverges. so for all x>0, x+x/2+x/3+... if infinity. so the limit is actually infinity, not 0. Another counter example: Take the sequence a_{m,n} to be 1 if m=n, and 0 otherwise. the limit as n->oo of a_{m,n} is 0 for all m. But for all n, a_{1,n} + a_{2,n} + a_{3,n} + ... = 1. So if you take n->oo, the limit is 1, because the sum is always 1. but if you take the limit on each term, the sum of the limits is 0. So these were 2 very different examples of why "sum of limits eq limit of the sum" Now here's how to fix it: If you use the taylor expansion of ln(1+x), which converges for to the actual value of ln(1+x) if |x|1, event considering the infinite sum. We'll prove that the absolute value of the sum goes to 0. |sum_{k>0} x²/3k³ - x³/4k⁴ + x⁴/5k⁵ - ...| 0} x²/3k² + x³/4k³ + x⁴/5k⁴ + ...| 0} x²/k² + x³/k³ + ...| = sum_{k>0} x²/k² . 1/(1-x/k) = sum_{k>0} x/k² . 1/(1/x - 1/k) Because x->0, we can take x such that |x|= |2 - 1| = 1, so sum_{k>0} x/k² . 1/(1/x - 1/k) 0} x/k² = x.pi²/6. And that, as x->0, goes in fact to 0. Now we deal with the terms with exponent of x equal to 1. |sum_{k>0} x/2k²| = x.sum_{k>0} 1/2k²| = x.pi²/12, which also goes to 0 as x->0. So we have that sum_{k>0} -1/k + ln(1+x/k)/x = (thing with absolute value going to 0) + (thing with absolute value going to 0), and now, using the linearity of the limit (for 2 functions, not infinetely many), the whole limit is also 0, finishing the proof. So we're left with the gamma at the start. See, it wasnt so trivial, it requires some ability to work with some algebra, but it only worked because the error of ln(1+x/k)/x - 1/k not only goes to 0, but the sum of all the errors also goes to 0. As I said, you can't say that just because one term converges to 0, the sum of all of them should. That's it. I hope I could help
This is looking good. You are right, in general the x-limit of the infinite k-sum of f_k(x) is not the same as the infinite k-sum of the x-limits of f_k(x). Thank you for your contribution. It's a great addition to the video and saves the result.
I have developed a conjecture (and written a draft paper) which is related to the limit in this video. The conjecture touches on a number of other areas as well. If any grad students/academics would be interested in co-publishing, please comment. Disclaimer: I am an independent researcher.
I got the same answer using Euler’s product definition for Gamma which I find to be more satisfying since the Euler Mascheroni constant appears at the end.
Because the terms of the sum are [ln(1+x/k)/x -1/k] so it dissolves. But yeah, I would like more in TH-cam videos a reason why he can interchanges limit and series
In order to put a limit inside a infinite sum you have to show that the sum uniformly converges. So at 6:30 you cannot just take the limit of the inside of the sum. I am not saying that your sum is not uniformly convergent but you didnt show that work so the solution is lacking that proof and maybe it is wrong.
@@maths_505 yea 😂 i wanted to calculate a limit of a sum too but when i searched the internet, stack exchange had other ideas. Luckily it pushed me to find another way to solve my problem so i was happy in the end
I saw the form 1 power inf So went with e^ (gamma (x +1 ) -1)/ x I saw inf by inf form Used lhopital Found the answer I thought I must be wrong but didn't had any reason for that But I got it right
Yup. As soon as we are considering x → 0+ it should be obvious we have to replace x! with the gamma function. Then, of course, we have a continuous function and can take the limit inside. There's no way we can make progress without that substitution.
Huh? As we are going to take the limit as x approaches 0, it is obvious that we are talking here about a *continuous* version of the factorial function, the natural choice being the well-known Pi function investigated by Euler and others, defined as Pi(x) = Gamma(x + 1) for all real numbers except for negative integers, in modern terms.
@@Grecks75 yes but x!=x(x-1)…2.1 is only true for integers, that was my point. Also the purpose of the pi function is to make the factorial function continuous
Do you accept integral suggestions? Because if so I have an interesting one one I came up with that you might like to make a video on. Also great video
@@maths_505 Don’t have insta but I can show you here: Int (0, inf) sum (n=0, inf) sum (k=0, inf) x^(2+2n) ln(x) ln^k(lnx) (-1)^n /(n!k!) A little hint is that part of the solution allows you to plug your merch as well
@@maths_505 Don’t have insta but I can show you here: Int (0, inf) sum (n=0, inf) sum (k=0, inf) x^(2+2n) ln(x) ln^k(lnx) (-1)^n /(n!k!) A little hint is that part of the solution allows you to plug your merch as well
Don’t have insta but I can show you here: Int (0, inf) sum (n=0, inf) sum (k=0, inf) x^(2+2n) ln(x) ln^k(lnx) (-1)^n /(n!k!) A little hint is that part of the solution allows you to plug your merch as well
What do you get when you substitute x = 0? You get 1^inf, right. Now, what do you know about the form 1^inf? It's an indefinite form, its limit can be *anything*!
Nice! I'm not too surprised the Oily-Macaroni constant came out
The most delicious constant in math
Oily-Macaroni, lol 🤣🤣🤣 Never heard that before!
Hi,
"Ok, cool" : 0:13 , 2:20 , 5:38 ,
"terribly sorry about that" : 1:56 .
Thank you for your service 🫡
Not the hero we deserve, but the one we need
Deep thoughts... from a deep guy
lim (ln x! - ln 0!) / (x-0) is literally the definition of ψ'(1)
Ψ(1)*
@@Mathguy1729 oh right, thanks
Good observation. 👍
Your solution has some (significant) flaws. By which I mean they're not so easy to fix.
At the end, you found out the limit was
gamma + sum_{k>0} (ln(1+x/k)/x - 1/k)
Your argument was:
I have a sum of infinetely many terms, each converging to 0, therefore the sum converges to 0.
This is not true. It happened to be true in that case, and I'll show you how you'd fix it.
Limits can be interpreted as functions that take functions to real values (or oo or -oo). What we all know is that limit is a linear function, this is:
if limf and limg exist, then (limf + limg) = lim(f+g).
Easy induction shows that
if limf1, limf2, ..., limfk exist, therefore
limf1 + limf2 + ... + limfk = lim(f1+f2+...+fk)
But this does not apply to infinetely many terms. you cant induct your way to infinity. So what im saying is:
if limf1, limf2, limf3, ... exist, this DOES NOT mean that lim(f1+f2+f3+...) exists, and even if does, it doesnt mean that lim(f1+f2+...) = limf1 + limf2 + ...
Thats what you used in the solution. That "the infinite sum of the limits is the limit of the infinite sum". Here's a counter example:
sum_{k>0} x/k = x+x/2+x/3+x/4+...
See: if x goes to 0, then x/k goes to 0, so the sum of the limits is 0+0+... = 0.
But 1+1/2+1/3+1/4 diverges. so for all x>0, x+x/2+x/3+... if infinity. so the limit is actually infinity, not 0.
Another counter example:
Take the sequence a_{m,n} to be 1 if m=n, and 0 otherwise.
the limit as n->oo of a_{m,n} is 0 for all m.
But for all n,
a_{1,n} + a_{2,n} + a_{3,n} + ... = 1.
So if you take n->oo, the limit is 1, because the sum is always 1. but if you take the limit on each term, the sum of the limits is 0.
So these were 2 very different examples of why "sum of limits
eq limit of the sum"
Now here's how to fix it:
If you use the taylor expansion of ln(1+x), which converges for to the actual value of ln(1+x) if |x|1, event considering the infinite sum. We'll prove that the absolute value of the sum goes to 0.
|sum_{k>0} x²/3k³ - x³/4k⁴ + x⁴/5k⁵ - ...| 0} x²/3k² + x³/4k³ + x⁴/5k⁴ + ...| 0} x²/k² + x³/k³ + ...| = sum_{k>0} x²/k² . 1/(1-x/k) = sum_{k>0} x/k² . 1/(1/x - 1/k)
Because x->0, we can take x such that |x|= |2 - 1| = 1, so
sum_{k>0} x/k² . 1/(1/x - 1/k) 0} x/k² = x.pi²/6. And that, as x->0, goes in fact to 0.
Now we deal with the terms with exponent of x equal to 1.
|sum_{k>0} x/2k²| = x.sum_{k>0} 1/2k²| = x.pi²/12, which also goes to 0 as x->0.
So we have that
sum_{k>0} -1/k + ln(1+x/k)/x = (thing with absolute value going to 0) + (thing with absolute value going to 0), and now, using the linearity of the limit (for 2 functions, not infinetely many), the whole limit is also 0, finishing the proof. So we're left with the gamma at the start.
See, it wasnt so trivial, it requires some ability to work with some algebra, but it only worked because the error of ln(1+x/k)/x - 1/k not only goes to 0, but the sum of all the errors also goes to 0. As I said, you can't say that just because one term converges to 0, the sum of all of them should.
That's it. I hope I could help
This is looking good. You are right, in general the x-limit of the infinite k-sum of f_k(x) is not the same as the infinite k-sum of the x-limits of f_k(x). Thank you for your contribution. It's a great addition to the video and saves the result.
Thanks for filling in this big gap in the proof!
what my doubt is even if that is correct if lim x-->0 ln(1+x/k)/x = 1/k , then ∑k>1 ( lim x-->0 ln(1+x/k)/x = 1/k) = 1 didn't it?
It can be easily solved using L'Hopital's rule and using a Digamma function
Yes,this is good for a short video,math505 you can get 2 videos for one problem🙃
You can also use definition of differentiability to express gamma function as 1-yx+o(x), and then everything cancels out nicely
And the first derivative at 0 approaches 1/12*e^(-γ)*π².
I think using the stirling approxmiatjon can also solve this quickly, though not as elegant.
i thought of using gamma(x)~1/x-Eulergamma approximation near 0
Then shift x -> x + 1, and then take the limit. Cheeky series expansion riddled throughout QFT
I have developed a conjecture (and written a draft paper) which is related to the limit in this video. The conjecture touches on a number of other areas as well. If any grad students/academics would be interested in co-publishing, please comment. Disclaimer: I am an independent researcher.
I got the same answer using Euler’s product definition for Gamma which I find to be more satisfying since the Euler Mascheroni constant appears at the end.
I'm confused as to why the summation dissolved at 7:30. Shouldn't it be sum(1/k)?
Because the terms of the sum are [ln(1+x/k)/x -1/k] so it dissolves.
But yeah, I would like more in TH-cam videos a reason why he can interchanges limit and series
Gorgeous result
Beautiful result as mentioned
In order to put a limit inside a infinite sum you have to show that the sum uniformly converges. So at 6:30 you cannot just take the limit of the inside of the sum. I am not saying that your sum is not uniformly convergent but you didnt show that work so the solution is lacking that proof and maybe it is wrong.
Trivial 😂
@@maths_505 yea 😂 i wanted to calculate a limit of a sum too but when i searched the internet, stack exchange had other ideas. Luckily it pushed me to find another way to solve my problem so i was happy in the end
@@maths_505Not really, I wish it was! 😭
I was expecting you to use the digamma function. Cool as always.
I saw the form 1 power inf
So went with e^ (gamma (x +1 ) -1)/ x
I saw inf by inf form
Used lhopital
Found the answer
I thought I must be wrong but didn't had any reason for that
But I got it right
Beautiful stuff
I screamed at 1:37. x! is NOT a finite product of integers.
expanding x! as x(x-1)…2.1 was also just wrong cuz u assumed that x! was a continuous function
Because x(x-1)…1 is only true for positive integers so its a step function
Yup. As soon as we are considering x → 0+ it should be obvious we have to replace x! with the gamma function. Then, of course, we have a continuous function and can take the limit inside. There's no way we can make progress without that substitution.
Yeah... Before seeing vid I was thinking: lim x! = 0! =1 so I = lim 1^(1/x) which is just... 0.
Huh? As we are going to take the limit as x approaches 0, it is obvious that we are talking here about a *continuous* version of the factorial function, the natural choice being the well-known Pi function investigated by Euler and others, defined as Pi(x) = Gamma(x + 1) for all real numbers except for negative integers, in modern terms.
@@Grecks75 yes but x!=x(x-1)…2.1 is only true for integers, that was my point. Also the purpose of the pi function is to make the factorial function continuous
Digamma function?
Thanks a lot✨
pls what is the app do you use to write this
Looks like microsoft onenote
@@nestorv7627looks like samsung notes
6:08 THE GOOFY SOUND EFFECTS ARE BACK
Hearing this sound at midnight will freak me
What is the app you are using for notes? I've been trying to find it since it looks very convenient
It's samsung notes, you can use notein if you don't have a samsung device
@senpai12349 Thanks!
Thanks.
It's about 0.561459483566885169824143214790880786765710386925 FYI
hi cro ! i fw you heavy just know that
Is this gen alpha speech cuz I've only just familiarised myself with gen z speech
Kamal are you coming to NUST?
Yup
Wouldn’t you have been left with 1+1/k and therefore gotten the answer + 1?
He remains me Mr Mackey
Do you accept integral suggestions? Because if so I have an interesting one one I came up with that you might like to make a video on.
Also great video
Sure
You can DM me on Instagram
@@maths_505
Don’t have insta but I can show you here:
Int (0, inf) sum (n=0, inf) sum (k=0, inf) x^(2+2n) ln(x) ln^k(lnx) (-1)^n /(n!k!)
A little hint is that part of the solution allows you to plug your merch as well
@@maths_505
Don’t have insta but I can show you here:
Int (0, inf) sum (n=0, inf) sum (k=0, inf) x^(2+2n) ln(x) ln^k(lnx) (-1)^n /(n!k!)
A little hint is that part of the solution allows you to plug your merch as well
Don’t have insta but I can show you here:
Int (0, inf) sum (n=0, inf) sum (k=0, inf) x^(2+2n) ln(x) ln^k(lnx) (-1)^n /(n!k!)
A little hint is that part of the solution allows you to plug your merch as well
@@Nolinlc yo that is sick 🔥🔥
❤❤❤
Why can't you just substitute x = 0 into (x!)^(1/x) to get the limit equal to 1?
What do you get when you substitute x = 0? You get 1^inf, right. Now, what do you know about the form 1^inf? It's an indefinite form, its limit can be *anything*!
1
How can this linit exist like if x! Tends to 0
can i be the gamma to your X ????
w^{In^x^e^x^x^0+x^0 ➖In^x^e^x}^w=x^1(w^{In^x^e^x➖ 1In^x^e^x+1}^w)
Hiiiii❤
Hey
no L'Hopital 😭
First comment hehe
First reply
@@maths_505First Thread now.
Please don't OKCool your videos
ok cool, I will OKCool your comment