I have a question : let I be a n-rectangle in R^n and f : I -> R be Darboux integrable. Suppose P_m be a uniform partition of I, which means P devises each edges of I to m intervals of the same lenghth. How to show that the Riemannian sum with respect to P_m is convergent to integral of f on I ? Please help, thanks.
@@franzlyonheart4362 It's much, much more complicated. There's an article by Beukers titled "On the generalized Ramanujan-Nagell equation", where he proves that there is at most one solution except when A=7, 23, or 2^k-1 for k at least 4. From skimming it, it appears to use a lot of hypergeometric series, p-adic numbers, and factorization in imaginary quadratic fields.
Ramanujam seems to be the absolute GOAT of making insane conjectures without proof. Most of them end up being proven correct after a significant amount of effort. Really shows how much intuition he seems to have about mathematics.
well in ancient india, ppl used a small blackboard thing called slate to write with chalk which was later rubbed off ..i guess ramanujan used the same for writing proofs and so he erased them...
35:40, the k was actually correct, the 2 is incorrect. In the equation right above that, it also says 2, which also is false. That should also be an exponent k, rather than the exponent 2 as written. There are quite a few calculation errors in the whole video today! And they're being glossed over, because many steps have been skipped over. Well, it already is a very long video today, so going cleanly through all steps probably isn't TH-cam-friendly.
13:24 Well, at that stage, we might still have y + omega = omega^a * omegabar^b and (y + omega)bar = omega^b * omegabar^a for othrecombinations of a and b with a+b = n-2. But a short check shows that we must have a=0 or b = 0: If both a and b are positive, then y+omega and y + omegabar are bothis divisible by omega omegabar, i.e, by 2. Then their sum y + omega + y + omegabar = 2y+1 = x is also even. But we already know that x is odd. Alternatively, the difference of y+omega and y+omegabar is sqrt(-7), which has norm 7, hence the factors cannot have a common factor of even norm (such as omega or omegabar). EDIT: Aha, I startedt commenting just a few seconds too early
This is so similar to an equation I have been looking at: Integer solutions to 2^n - n = x^2. The only solutions appear to be n = 0, 1 and 7 but I cannot prove it.
I could check that all the solutions other than (n, x) = (0, 1), (1, 1), (7, 11), if exist, would require n ≡ 7 (mod 8). Checking n ≥ 0 is trivial. Even integers (≠ 0) are easily eliminable since we need (2^m - 1)^2 < x^2 < 2^(2m) for all integers m ≥ 1. Odd integers are difficult to approach, although I suspect Pell's equation might help. We can at least reduce n ≡ 1 (mod 2) to n ≡ 7 (mod 8) by gradually checking parity.
@@pnintetr I reached the same results, but I also noticed that analysing n = 7 mod8 for sufficiently big n will force x = floor(sqrt(2^n)). Do you know of any interesting ways to look at the growth of 2^n - floor(sqrt(2^n))²? If we show it grows quicker than linear (although I'm not even sure that is true) then the problem would be over
ramanujan never needed to prove anything , he thought was waste of paper for and paper was very expensive for him , he just give answers and theorums and leave proving to future mathmathecians as their life goals
Hi Michael! I noticed quite a few calculation mistakes this time, and while you pointed some of them out in your post-edit, some of them weren't caught. This plus the skipped calculation steps for "trivial" calculations made the video slightly harder to follow, and I ended up having to look up the paper you referenced to get the full picture. If you were rushing at the end to try and meet a "TH-cam-friendly time limit" as someone else in the comments said, then I would much prefer seeing a proof like this in all of its glory with all of the calculation steps even if it's 1 hour+ long. Sometimes the best math just can't fit into a 37-minute video! Thanks for all of the great work; I'll be looking forward to the next video!
We are all brokenhearted that you had to waste your precious time to read the resource provided in the video. Use all your well intentioned advice to produce a video on this topic that meets your fastidious demands. I look forward to viewing said video on YOUR channel.
@@MyOneFiftiethOfADollar dude chill, I was just commenting to let him know what would have made the video more enjoyable for me. He doesn't have to do anything I asked for, but if people don't give feedback to content creators then they don't know if they are producing videos that people will keep enjoying. So sorry not sorry 🤷
I'm a bit confused by the step at 36:10 where we've substituted b_12 and b_13. That gives us a term of (-1)^(j+k-12) that appears to simplify, but I don't understand how it does.
He's making A LOT of calculation mistakes in those steps. You can see that he's rushing it and glancing at his notes all the time. I suppose he didn't want the video to become that long, in order to remain somewhat TH-cam friendly. The only way to "get it" is to sit down and recalculate it yourself, using the video more like a guideline than fact, and not skipping all the steps he skipped, and then you'll avoid (and fix) all these calculation errors. And perhaps in addition using the original paper with the proof that he references at the beginning.
x² + 7 = 2^n. Pre-watch: ("Ramanujan" implies it must be Diophantine, right?) Just using trial-and-error, move the "7" to the right [x² = 2^n - 7], run through some powers of 2, and see what makes the RHS a square. I quickly get these: (x, n) = (1, 3), (3, 4), (5, 5), (11, 7). I can see how a general solution could have eluded him, though...or determining whether these are the only (+ve) integer solutions. Post-watch: Oh I see, it wasn't that he couldn't solve the equation, but that he couldn't prove there were only those few solutions you gave. Fred
When I saw the thumbnail, I had the intuition to adjoin the square root and use a linear recurrence or something, but then I thought "If Ramanujan couldn't solve it, what chances do I have?" Turns out I was on the right way, even though it gets a bit more involved after that start.
4:00 To me it wasn't immediately clear that for an arbitrary complex number ω the definition Z[ω] = {aω+b with a,b integer} would define a ring. E.g. for ω=e^πi/6 and x in Z[ω], Im(x) would always be an integer multiple of 1/2, whereas Im(ω^2) would be sqrt(3)/2. So for this ω, the given definition of Z[ω] would not be a ring, because it is not closed under multiplication. For ω = (1+sqrt(7)i)/2 however, we have ω^2 = (-6 + 2sqrt(7)i)/4 = ω-2, so no problem in this special case. Also, the assertion that elements have unique factorization is not clear to me. Still any composite integer would factor in more than one way, would'nt it?
These are good observations/questions, answered by a first course in ring theory. To help you on your way: Let R be a ring without zero-divisors, that is no elements a and b such that a*b = 0. 1) What we actually mean when we write R[ω] is the set of all expressions of the form a_0 + a_1 ω + a_2 ω^2 + ....., with a_k in R, and where only finitely many of the a_k's are non-zero. This always defines a ring with multiplication given by expanding brackets and collecting like powers of ω. 2) In the cases where ω satisfies a polynomial equation of degree n, that is P(ω) = 0, this collapses down to { a_0 + a_1 ω + ... + a_{n-1} ω^{n-1} | a_k in R for all 0
I must admit I did not understand the proof. But anyway it was interesting to see the limited numbers of solutions. Can it be proven for other numbers than 7 (I mean for x^2 + p = 2^n, with p€N, or p prime)?
As far as I know there is or was no Norwegian mathematician with the name Nagelle. And that name does not sound very Norwegian. There was a Norwegian mathematician with a similar name; Trygve Nagell. He worked in this field.
I really have some issues with the solution that he demonstrates. Like, should there be a 2 before the sigma in 27:48 because omega*inverse-omega=2. Also, why does the exponent change from k in 30:01 to 2 in 30:21? Tbh, the ending equation just leaves me with confusion. Can anyone explains plz? 😢
If I've understood correctly it's because the LHS is -1 and the RHS is (-1)^k +(mult of 3) (note that the exponent of 2 is a typo), and since k is odd that reduces to -1=-1+(mult of 3), which can only be true if that multiple of 3 is 0.
I have been watching the abstract algebra series in the Math Major channel for the past 2-3 months. I didn’t completely understand the entire proof of this video, especially when it’s claimed that the solutions for b[n]=-1 has to come from n=4k+1. Anyway, it’s nice to see how ring theory can be applied to solve number theory problems. This is a really good example to show how the two areas are tied to each other.
Behind the scene there was proven a fact about b_n (mod 16) (you can find timing with b_n value table near the end, there was a circular pattern with length 4), and since 1 and -1 should stay as is after every mod we can say that b_n = -1 iff n=4k+1
Two reasons for sqrt(-7): first, it allows to nicely factor 2 (as the roots of x^2-x+2), second, unique factorization fails in the ring of integers of Q[sqrt(-5)].
I love your content! Where in norway was the confernece you attended? And what was it all about? Im going to study math in Norway soon and would love to have you hold a seminar for me sometime :)
@@HagenvonEitzen No, many times it is the process that is more important than the solution. This is a competition math channel, so computers would not be used.
That’s funny that equations like this often have some solutions in small numbers and then no solutions whatsoever. I wonder what is the simplest Diophantus equation with a couple solutions that are all over googol
n = 7 x =11 answer x^2 + 7 = 2^ n x^2 = 2^ n - 7 x^2 is odd (since 2^ n is EVEN and 7 is 0DD and EVEN - 0DD = EVEN OR vice versa) Hence x is odd Possible ending 1 * 1 = 1 3 *3 = 9 5* 5 = 25 7 * 7 - 49 9 * 9 = 81 So x^2 either ends ( or the last digit) with 1 or 9 or 5 Hence, 2^ n ends ( or last digit) is either 8 (7+ 1), 6 ( 7+ 9 =16) or 2 ( 7 + 5 =12) So we are looking for a 2^ n, which ends with last digit is either 8, 6 or 2 e.g. of 2^n = 2, 4 8, 16, 32, 64,128, 256, etc.. And when subtracting 7 from any of these numbers, it must be a square Let's try 128 (which the last digit is 8 . Recall it has to be either 8, 6, or 2) 128 -7 = 121 which is 11^2 11 is odd , so we are on the right track Answer 2^n 128 and 128 =2^7 hence n=7 and x =11 ansewr
An unfortunate result secondary education I believe. I was taught this too. It's mostly true for linear algebra but this equation has other constraints which limit the solutions.
I have a question : let I be a n-rectangle in R^n and f : I -> R be Darboux integrable. Suppose P_m be a uniform partition of I, which means P devises each edges of I to m intervals of the same lenghth. How to show that the Riemannian sum with respect to P_m is convergent to ∫_I f? Please help, thanks.
@@NH-zh8mp Can your tutors not help you? (I suppose it's a homework question from uni.) It appears quite random, because it is outside of number theory, the topic of this video here. But Penn has been known to pick up questions from many mathematical areas, maybe you'll be lucky! Thinking back to my own undergrad days, we studied Lebesgue integrals, I cannot remember Darboux, but it has been a LONG time.
My quick solution is x=the square root of {2^n-7}. going for the quickest if n=2 then the square root of -3 or x=1.73205081 i. Is this wrong and if so how?
@@조민구-h9r Ah. I didn't know they weren't allowed since complex numbers, are used in real-life applications, such as electricity, as well as quadratic equations.
Head to squarespace.com/michaelpenn to save 10% off your first purchase of a website or domain using code michaelpenn
I have a question : let I be a n-rectangle in R^n and f : I -> R be Darboux integrable. Suppose P_m be a uniform partition of I, which means P devises each edges of I to m intervals of the same lenghth. How to show that the Riemannian sum with respect to P_m is convergent to integral of f on I ?
Please help, thanks.
In fact, it turns out that the Diophantine equation x^2+A=2^n for fixed A has at most 2 solutions, except when A=7. Pretty weird...
Do you have a proof of this fact? I imagine it would use some algebraic number theory for the UFD part of the proof.
Thanks for that interesting addition. Is the proof as easy as the one shown here, or is it more complicated?
A=0 has an inf # of solutions (n even, x = 2^(n/2))
@@bamdadtorabi2924 It's mentioned on the wikipedia page for this function. en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation
@@franzlyonheart4362 It's much, much more complicated. There's an article by Beukers titled "On the generalized Ramanujan-Nagell equation", where he proves that there is at most one solution except when A=7, 23, or 2^k-1 for k at least 4. From skimming it, it appears to use a lot of hypergeometric series, p-adic numbers, and factorization in imaginary quadratic fields.
Ramanujam seems to be the absolute GOAT of making insane conjectures without proof. Most of them end up being proven correct after a significant amount of effort. Really shows how much intuition he seems to have about mathematics.
He probably saw the answers before he finished reading the question, wrote down the answers and went on to something more interesting.
And most tragic of all, he was killed by the low quality of early 20th century British food.
well in ancient india, ppl used a small blackboard thing called slate to write with chalk which was later rubbed off ..i guess ramanujan used the same for writing proofs and so he erased them...
maybe so, but he still couldn't prove them - now could he ...
@@extreme4180 guess they couldn't get a piece of paper and pencil, Uh ?
35:40, the k was actually correct, the 2 is incorrect. In the equation right above that, it also says 2, which also is false. That should also be an exponent k, rather than the exponent 2 as written.
There are quite a few calculation errors in the whole video today! And they're being glossed over, because many steps have been skipped over. Well, it already is a very long video today, so going cleanly through all steps probably isn't TH-cam-friendly.
13:24 Well, at that stage, we might still have y + omega = omega^a * omegabar^b and (y + omega)bar = omega^b * omegabar^a for othrecombinations of a and b with a+b = n-2.
But a short check shows that we must have a=0 or b = 0: If both a and b are positive, then y+omega and y + omegabar are bothis divisible by omega omegabar, i.e, by 2. Then their sum y + omega + y + omegabar = 2y+1 = x is also even. But we already know that x is odd.
Alternatively, the difference of y+omega and y+omegabar is sqrt(-7), which has norm 7, hence the factors cannot have a common factor of even norm (such as omega or omegabar).
EDIT: Aha, I startedt commenting just a few seconds too early
16 omega's
@ 27:48 The sum should have a two in the front, and the b_n^k term should be b_n^j.
The name of this norwegian guy was actually "Nagell", no trailing "e" (and therefor also not pronounced french/italian).
This is so similar to an equation I have been looking at: Integer solutions to 2^n - n = x^2. The only solutions appear to be n = 0, 1 and 7 but I cannot prove it.
I could check that all the solutions other than (n, x) = (0, 1), (1, 1), (7, 11), if exist, would require n ≡ 7 (mod 8).
Checking n ≥ 0 is trivial.
Even integers (≠ 0) are easily eliminable since we need (2^m - 1)^2 < x^2 < 2^(2m) for all integers m ≥ 1.
Odd integers are difficult to approach, although I suspect Pell's equation might help.
We can at least reduce n ≡ 1 (mod 2) to n ≡ 7 (mod 8) by gradually checking parity.
@@pnintetr I reached the same results, but I also noticed that analysing n = 7 mod8 for sufficiently big n will force x = floor(sqrt(2^n)). Do you know of any interesting ways to look at the growth of 2^n - floor(sqrt(2^n))²? If we show it grows quicker than linear (although I'm not even sure that is true) then the problem would be over
I checked all the values of n from 0 to 400 and only 0, 1, 7 work in that range, as well. Really makes it seem like they're the only answers
ramanujan never needed to prove anything , he thought was waste of paper for and paper was very expensive for him , he just give answers and theorums and leave proving to future mathmathecians as their life goals
Hi Michael! I noticed quite a few calculation mistakes this time, and while you pointed some of them out in your post-edit, some of them weren't caught. This plus the skipped calculation steps for "trivial" calculations made the video slightly harder to follow, and I ended up having to look up the paper you referenced to get the full picture.
If you were rushing at the end to try and meet a "TH-cam-friendly time limit" as someone else in the comments said, then I would much prefer seeing a proof like this in all of its glory with all of the calculation steps even if it's 1 hour+ long. Sometimes the best math just can't fit into a 37-minute video!
Thanks for all of the great work; I'll be looking forward to the next video!
We are all brokenhearted that you had to waste your precious time to read the resource provided in the video.
Use all your well intentioned advice to produce a video on this topic that meets your fastidious demands.
I look forward to viewing said video on YOUR channel.
@@MyOneFiftiethOfADollar dude chill, I was just commenting to let him know what would have made the video more enjoyable for me. He doesn't have to do anything I asked for, but if people don't give feedback to content creators then they don't know if they are producing videos that people will keep enjoying. So sorry not sorry 🤷
I'm a bit confused by the step at 36:10 where we've substituted b_12 and b_13. That gives us a term of (-1)^(j+k-12) that appears to simplify, but I don't understand how it does.
He's making A LOT of calculation mistakes in those steps. You can see that he's rushing it and glancing at his notes all the time. I suppose he didn't want the video to become that long, in order to remain somewhat TH-cam friendly.
The only way to "get it" is to sit down and recalculate it yourself, using the video more like a guideline than fact, and not skipping all the steps he skipped, and then you'll avoid (and fix) all these calculation errors. And perhaps in addition using the original paper with the proof that he references at the beginning.
I think that k-12 should be k-j, which means that term becomes (-1)^k, but since k is odd this reduces to -1
x² + 7 = 2^n. Pre-watch: ("Ramanujan" implies it must be Diophantine, right?)
Just using trial-and-error, move the "7" to the right [x² = 2^n - 7], run through some powers of 2, and see what makes the RHS a square. I quickly get these:
(x, n) = (1, 3), (3, 4), (5, 5), (11, 7).
I can see how a general solution could have eluded him, though...or determining whether these are the only (+ve) integer solutions.
Post-watch: Oh I see, it wasn't that he couldn't solve the equation, but that he couldn't prove there were only those few solutions you gave.
Fred
i think it's worth pointing out that omega_bar is also an element of your ring since omega_bar=1-omega which is clearly an element of Z[omega]
When I saw the thumbnail, I had the intuition to adjoin the square root and use a linear recurrence or something, but then I thought "If Ramanujan couldn't solve it, what chances do I have?" Turns out I was on the right way, even though it gets a bit more involved after that start.
4:00 To me it wasn't immediately clear that for an arbitrary complex number ω the definition Z[ω] = {aω+b with a,b integer} would define a ring.
E.g. for ω=e^πi/6 and x in Z[ω], Im(x) would always be an integer multiple of 1/2, whereas Im(ω^2) would be sqrt(3)/2. So for this ω, the given definition of Z[ω] would not be a ring, because it is not closed under multiplication.
For ω = (1+sqrt(7)i)/2 however, we have ω^2 = (-6 + 2sqrt(7)i)/4 = ω-2, so no problem in this special case.
Also, the assertion that elements have unique factorization is not clear to me. Still any composite integer would factor in more than one way, would'nt it?
These are good observations/questions, answered by a first course in ring theory. To help you on your way:
Let R be a ring without zero-divisors, that is no elements a and b such that a*b = 0.
1) What we actually mean when we write R[ω] is the set of all expressions of the form a_0 + a_1 ω + a_2 ω^2 + ....., with a_k in R, and where only finitely many of the a_k's are non-zero. This always defines a ring with multiplication given by expanding brackets and collecting like powers of ω.
2) In the cases where ω satisfies a polynomial equation of degree n, that is P(ω) = 0, this collapses down to { a_0 + a_1 ω + ... + a_{n-1} ω^{n-1} | a_k in R for all 0
@@reijerboodt8715 sorry but i did not mean to ask any questions, it was more a comment on the way it was presented.
The comment i searched for
I must admit I did not understand the proof. But anyway it was interesting to see the limited numbers of solutions. Can it be proven for other numbers than 7 (I mean for x^2 + p = 2^n, with p€N, or p prime)?
Anything can be proven in maths. Otherwise it would not exist.
It actually sounds like a classic Ramanujan. He just made a conjecture he didn't care to prove which turned out to be true.
As far as I know there is or was no Norwegian mathematician with the name Nagelle. And that name does not sound very Norwegian. There was a Norwegian mathematician with a similar name; Trygve Nagell. He worked in this field.
good motivator for abstract algebra, number theory and recurrence relations
I really have some issues with the solution that he demonstrates. Like, should there be a 2 before the sigma in 27:48 because omega*inverse-omega=2. Also, why does the exponent change from k in 30:01 to 2 in 30:21?
Tbh, the ending equation just leaves me with confusion. Can anyone explains plz? 😢
37:01
13:08 There's not that much to check: The norm omega * omegabar is a rational prime.
Wow..sounds like a poem. Absolute pleasure!!!
Michael Pennis my favourite matematician!
33:10 how does the second equation imply the third equation? b4=-3 and b5=-1, so how do we get cancelation down to 0?
If I've understood correctly it's because the LHS is -1 and the RHS is (-1)^k +(mult of 3) (note that the exponent of 2 is a typo), and since k is odd that reduces to -1=-1+(mult of 3), which can only be true if that multiple of 3 is 0.
You're a super mathematician
Please post video of you and spouse debating household budgeting process
Lol !😄
I have been watching the abstract algebra series in the Math Major channel for the past 2-3 months. I didn’t completely understand the entire proof of this video, especially when it’s claimed that the solutions for b[n]=-1 has to come from n=4k+1. Anyway, it’s nice to see how ring theory can be applied to solve number theory problems. This is a really good example to show how the two areas are tied to each other.
Behind the scene there was proven a fact about b_n (mod 16) (you can find timing with b_n value table near the end, there was a circular pattern with length 4), and since 1 and -1 should stay as is after every mod we can say that b_n = -1 iff n=4k+1
At 27:53, I think your sum is missing a factor of 2 due to omega omega bar... ditto for 28:39.
However, I still can't figure out where the 2 pre-factor on the sum arose at 30:17 (not to mention the exponents of 2 rather than k)
Not that the pre-factor matters when that whole sum is 0 !
I came here after two years and hardly got to follow the explanation till the end. Im so proud of my mediocre maths ❤❤
Seeing Michael smile at the beginning is the highlight of the video.
He solved it, but on a slate with chalk, as he didn't have much paper and wrote the result only in his notes
Why mod 16? Also, why root -7 (though this I assume because +-5 didn't work)?
Two reasons for sqrt(-7): first, it allows to nicely factor 2 (as the roots of x^2-x+2), second, unique factorization fails in the ring of integers of Q[sqrt(-5)].
Tha's because the equation is x^2+7.
They never do the UFD part..
I love your content! Where in norway was the confernece you attended? And what was it all about? Im going to study math in Norway soon and would love to have you hold a seminar for me sometime :)
It is interesting that this formula breaks numerical investigation. You cannot just plug it in and test all integers.
Isn't that always the case with general statements?
@@HagenvonEitzen No, many times it is the process that is more important than the solution. This is a competition math channel, so computers would not be used.
@@adamwho9801I think his point was that you can never "test all integers"...
@@gerryiles3925 In this case it is true, but it isn't always the case. Sometimes numerical solutions can serve as a proof.
can you solve a problem like this using binary nums?
That’s funny that equations like this often have some solutions in small numbers and then no solutions whatsoever. I wonder what is the simplest Diophantus equation with a couple solutions that are all over googol
this is crazy
n = 7 x =11 answer
x^2 + 7 = 2^ n
x^2 = 2^ n - 7
x^2 is odd (since 2^ n is EVEN and 7 is 0DD and EVEN - 0DD = EVEN OR vice versa)
Hence x is odd
Possible ending
1 * 1 = 1
3 *3 = 9
5* 5 = 25
7 * 7 - 49
9 * 9 = 81
So x^2 either ends ( or the last digit) with 1 or 9 or 5
Hence, 2^ n ends ( or last digit) is either 8 (7+ 1), 6 ( 7+ 9 =16) or 2 ( 7 + 5 =12)
So we are looking for a 2^ n, which ends with last digit is either 8, 6 or 2
e.g. of 2^n = 2, 4 8, 16, 32, 64,128, 256, etc..
And when subtracting 7 from any of these numbers, it must be a square
Let's try 128 (which the last digit is 8 . Recall it has to be either 8, 6, or 2)
128 -7 = 121 which is 11^2
11 is odd , so we are on the right track
Answer 2^n 128 and 128 =2^7 hence n=7
and x =11 ansewr
Very cool video
And here I thought the proof to Bertrand's conjecture was horrible.
this video got me messed up
Can anyone use this and prove that for n>3, there is no solution for 5^n-121 = x^2?
Think x² + 11² and Pythagorean triples. Is it possible that the hypotenuse in this case is divisible by 5?
i could figure this out in seconds.
Were there any equations he could solve?
For the first time I'm commenting that the place stopped wasn't a good one; the contradiction was not very clear.
I'm no mathematician, but looking at that equation I see 2 variables and 1 equation - so that must be insolvable?
An unfortunate result secondary education I believe. I was taught this too. It's mostly true for linear algebra but this equation has other constraints which limit the solutions.
Video title: "The equation Ramanujan couldn't solve",
Me noticing that video is 37 minutes long. Me: "Makes sense".
if x = 1, 3, and 5, then n = 3, 4, and 5
I have a question :
let I be a n-rectangle in R^n and f : I -> R be Darboux integrable. Suppose P_m be a uniform partition of I, which means P devises each edges of I to m intervals of the same lenghth. How to show that the Riemannian sum with respect to P_m is convergent to ∫_I f?
Please help, thanks.
You want us to do your homework for you ? 😂
@@franzlyonheart4362 nah, I just tried every way I know and find on sites, but I’m still struggle, so I really need help
@@NH-zh8mp Can your tutors not help you? (I suppose it's a homework question from uni.)
It appears quite random, because it is outside of number theory, the topic of this video here. But Penn has been known to pick up questions from many mathematical areas, maybe you'll be lucky! Thinking back to my own undergrad days, we studied Lebesgue integrals, I cannot remember Darboux, but it has been a LONG time.
Well, this is out of My reach :(
Out of mine too
You are in very good company - with Ramanujan at least
Which part is confusing
Out of my reach too. It is pretty simple to at least prove that n=4 is the only even solution.
not looking at comments, x is 3 n is 4
x=3, n=4 is a solution
My quick solution is x=the square root of {2^n-7}. going for the quickest if n=2 then the square root of -3 or x=1.73205081 i. Is this wrong and if so how?
x is natural number, which means x cannot be an imaginary number.
@@조민구-h9r Ah. I didn't know they weren't allowed since complex numbers, are used in real-life applications, such as electricity, as well as quadratic equations.
1^2+7=2^3 x=1; n=3
In B4 8 year old kids in the comment section claim they can solve it
But Michael Penn could finally do it.
X=5i n=1
It's fashinating! But.. I guess... Could this be a question of a calculus ex? I do not think so...
There's ton of solution bro. First x = 1 and n = 3. And so on
He is not "bro", have some respect
Yeah that was pretty
Kindly change that thumbnail
No one can judge legend Ramanujan
yes we can, and yes we will do
Michael Penn ,you are better than Ramanujan -you know it,😉
😂😂😂😂
x=1 and n=3. I must br smarter than ramunijan!
Hummm. The thumbnail and the title made me lose a lot of the respect I had for Mr. Penn. They are so clickbaity. What a bad decision.
shut up kiddo where is the clickbait?
I hate this... Combinatorics please