Another solution is to group the terms with k=m+n constant, and note that 1/(m^2+n^2) >= 1/(2*k^2). Since there are k-1 terms with a given value k of m+n, the sum of the k-th grouping is >= (k-1)/(2*k^2), which diverges when summed over all k>=2.
I thought of this solution, Given the diagram you drew, you can notice that within each "square" i.e for any pair of n and m that is bound by S >= n, m the smallest fraction we can get is n = m = S. so we know our nested sums are larger than the sum (from n = 1 to infinity) (2n +1) / (n+1)^2 which is larger than 1/2n for all n and thus diverges.
You can bound the inner sum with a single integral, so that, holding m constant, Σ_n 1/(m^2 + n^2) > Int 1 to inf dx/(m^2 + x^2) = 1/m * (pi/2 - atan(1/m)) Putting this back in the double sum: Σm Σn 1/(m^2 + n^2) > Σ_m 1/m * (pi/2 - atan(1/m)) > pi/4 Σ_m 1/m which is the harmonic series Note: Because atan(1/m) is always less than pi/4, (pi/2 - atan(1/m)) is always greater than pi/4
Instead of the strange summation, you can use "normal" polar coordinate but centered at (1,1). Maybe the integral would be a little more difficult (not really a problem since we are not evaluating it exactly) but the logic behind is simpler
You dont need calculus for this: 1/(m^2+n^2) is lower bounded by 1/(m+n)^2. Let k=m+n and then sum the lower bound over the positive integer grid: for each k= 2 to inf, there are exactly k-1 points where m+n =k , thus the lower bound is sum (k-1)/k^2 , for k=2 to inf, which clealy diverges because the harmonic series diverges. Since the lower bound diverges, the original double sum diverges as well.
I'd like to see something on the Madelung constant, and maybe higher dimensional. Also the integral around 6:54 can reduce down to the integral of arctan(y)/y from one to infinity, which clearly diverges. Easier that way, I think.
I think the lower bound of the last integral (with respect to r) should be 2√2. It can be 2 but we have to take the red circle with radius 2 instead of 2√2. Anyway, very nice solution!
7:44 if you are switching to polar, wouldn't it be much easier to transform the summation from 0 and then remove the term which are both 0 and then subtract the terms which one of them is 0?
You could also re-order the sum to sum_{m=1}^inf sum_{n=1}^m 1/(m²+n²) In the inner sum, n is less than or equal to m, therefore the denominator is less than or equal to 2*m², so the total sum is larger or equal to: sum_{m=1}^inf sum_{n=1}^m 1/(2m²) = sum_{m=1}^inf m/(2m²) = inf
Your proof is correct, but you should underline that the initial double sum is greater than the sum_{m=1}^inf sum_{n=1}^m 1/(m²+n²). In fact, in the second double sum your are not adding all the terms in which n > m.
@@MarcoMate87 You are absolutely right (I did not check this carefully before posting). The term in the inner sum should be 1/((m-n+1)² + n²) to be identical (essentially going through all terms in diagonals). But it is indeed even simpler just using the smaller inner sum.
My thought was that `1/(n²+m²)≥1/(n²+2nm+m²)=1/(n+m)²`. re-index `i=n+m, j=n-m`, j vanishes immediately and we're left with `(i-1)/i²` which goes to infinity.
this seems to be potentially related to the weierstrass p-function, which is a similar-looking double sum in the integers of a quantity that drops off as the inverse square of the summands, and has to be massaged a bit to get a convergent series
Isn't our double sum an iterated limit? Should it be lim m->inf lim n->inf sum 1 to m sum 1 to n 1/(m^2+n^2)? Which is different from lim m,n->inf sum 1/(m^2+n^2)? The original case is we are adding column-wise/ row wise if we rearrange them in a grid while the latter (which is what we did) is adding "radially" ie we have ordered pairs (m,n) and we add as we go outwards? From what I know, there are criterias for commutativity in infinite sums.
We can see this in definition. We say lim m,n exist if for all epsilon>0 there exist N such that if m,n>=N, |a_m,n - A|< epsilon. From what I know, existence of double limit is strong as it guarantees the iterated limits existence. But proving the double limit as divergent does not necessary mean we have divergent iterated limit.
I didn't quite understand this passage after 5:30. Why is it that by including what's between the lattice points (going from a sum to an integral) you're creating something smaller? I'm guessing it's because of some shenanigans with infinity, but I'm not sure.
It's because f is decreasing along the x and y axes, so the area under the function between say (x,y) and (x+1,y) is smaller than the area of a rectangle with height f(x,y), and width 1 (this is true in all directions)
@@andrycraft69 He is comparing the value of the double sum to the value of an integral, which is why he introduces a delta-x and delta-y in the double sum, which have a value of 1 (which represents the width of the rectangle, or if you will the size of the base square of a box since it is 2D sum/integrals). The double integral of f is the volume under the curve of f within a specified region; the double sum here is the sum of the volume of 1 by 1 boxes with height equal to the value of f at each integer pair (the dots in the lattice he drew). Since f is decreasing along x and y, the curve of f is always contained within these boxes, so the double sum is greater than the double integral.
i remember seeing this in the romanian mathematical magazine and i also thought this was kinda interesting tho i got solution in a more straight forward manner
I did almost the same, using f(x,y) =1/(x^2+y^2) over R×R, but I did not do the ray thing. Instead, I divided up the first quadrant into five regions: Region 0, (unit disk) 0 < x, 0 < y, x^2+y^2
@@popodori That is also not correct. Under "Related Facts" you see two infinite sums of numbers each bigger than 0 which do converge to finite values. Another example is 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2ᵏ + ... = 1.
I don't get how the green part between α and β is bigger than the whole [1,∞)^2 purple part. It seems like it's a subset except this small quadrilateral near the origin so it should be smaller, no?
![แฉกลโกงงานวัด 2024 [ โกงมั้ยครับ ep.106 ] | DOM](http://i.ytimg.com/vi/taC5dpP3HXA/mqdefault.jpg)
Another solution is to group the terms with k=m+n constant, and note that 1/(m^2+n^2) >= 1/(2*k^2). Since there are k-1 terms with a given value k of m+n, the sum of the k-th grouping is >= (k-1)/(2*k^2), which diverges when summed over all k>=2.
Crack !!
Best
A basil problem is what have when your pesto separates. You mean the Basel problem.
I thought of this solution,
Given the diagram you drew, you can notice that within each "square" i.e for any pair of n and m that is bound by S >= n, m the smallest fraction we can get is n = m = S. so we know our nested sums are larger than the sum (from n = 1 to infinity) (2n +1) / (n+1)^2 which is larger than 1/2n for all n and thus diverges.
You can bound the inner sum with a single integral, so that, holding m constant, Σ_n 1/(m^2 + n^2) > Int 1 to inf dx/(m^2 + x^2) = 1/m * (pi/2 - atan(1/m))
Putting this back in the double sum:
Σm Σn 1/(m^2 + n^2) > Σ_m 1/m * (pi/2 - atan(1/m)) > pi/4 Σ_m 1/m which is the harmonic series
Note: Because atan(1/m) is always less than pi/4, (pi/2 - atan(1/m)) is always greater than pi/4
Instead of the strange summation, you can use "normal" polar coordinate but centered at (1,1).
Maybe the integral would be a little more difficult (not really a problem since we are not evaluating it exactly) but the logic behind is simpler
Ho pensato esattamente a questo
@@aronbucca6777 è decisamente la cosa più logica da fare
I was literally thinking about the same.
You dont need calculus for this: 1/(m^2+n^2) is lower bounded by 1/(m+n)^2. Let k=m+n and then sum the lower bound over the positive integer grid: for each k= 2 to inf, there are exactly k-1 points where m+n =k , thus the lower bound is sum (k-1)/k^2 , for k=2 to inf, which clealy diverges because the harmonic series diverges. Since the lower bound diverges, the original double sum diverges as well.
I'd like to see something on the Madelung constant, and maybe higher dimensional.
Also the integral around 6:54 can reduce down to the integral of arctan(y)/y from one to infinity, which clearly diverges. Easier that way, I think.
I think the lower bound of the last integral (with respect to r) should be 2√2. It can be 2 but we have to take the red circle with radius 2 instead of 2√2. Anyway, very nice solution!
7:44 if you are switching to polar, wouldn't it be much easier to transform the summation from 0 and then remove the term which are both 0 and then subtract the terms which one of them is 0?
4:30 If you're still wondering about the "decreasing property" of *f(x, y) := 1/(x^2 + y^2)* on shifted squares *Q_{m, n}:*
*(x; y)^T ∈ Q_{m, n} := [m; m+1) ⨯ [n; n+1) => 0 < f(x, y) ≤ f(m, n)*
6:10 Instead of polar coordinates, we may rotate the integration domain *D := [1; N]^2* by *-𝜋/4* around the *z* -axis:
*I := ∫_D f(x, y) d(x, y) = ∫_R(D) f(u, v) d(u, v) // (u, v)^T := Rot_z(-𝜋 /4) * (x, y)^T*
*// d(u, v) = | 1 | * d(x, y)*
*// f(x, y) = f(u, v)*
After transformation, we can use mirror-symmetry *f(u, -v) = f(u, v)* to simplify the rotated integration domain *R(D).* We ignore everything before *u = 2* and after *u = (N+1)/√2* to get an under-estimate (sketch the rotated domain to find the borders):
*I > 2 ∫_{u = 2}^{ (N+1)/√2 } ∫_{v = 0}^{u-1} f(u, v) dv du*
*= 2 ∫_{u = 2}^{ (N+1)/√2 } 1/u * atan(1 - 1/u) du*
*> 2 ∫_{u = 2}^{ (N+1)/√2 } 1/u * atan(1 - 1/2) du*
*= 2 * atan(1/2) * [ ln( (N+1)/√2 ) - ln(2) ] -> ∞ as N -> ∞*
You could also re-order the sum to
sum_{m=1}^inf sum_{n=1}^m 1/(m²+n²)
In the inner sum, n is less than or equal to m, therefore the denominator is less than or equal to 2*m², so the total sum is larger or equal to:
sum_{m=1}^inf sum_{n=1}^m 1/(2m²) = sum_{m=1}^inf m/(2m²) = inf
Your proof is correct, but you should underline that the initial double sum is greater than the sum_{m=1}^inf sum_{n=1}^m 1/(m²+n²). In fact, in the second double sum your are not adding all the terms in which n > m.
@@MarcoMate87 You are absolutely right (I did not check this carefully before posting).
The term in the inner sum should be 1/((m-n+1)² + n²) to be identical (essentially going through all terms in diagonals). But it is indeed even simpler just using the smaller inner sum.
Can we compute at what exponent this goes from divergent to convergent? And what if it were a triple sum? At what exponent would that converge?
N>2, in general is N>(#dimensions)
My thought was that `1/(n²+m²)≥1/(n²+2nm+m²)=1/(n+m)²`. re-index `i=n+m, j=n-m`, j vanishes immediately and we're left with `(i-1)/i²` which goes to infinity.
8:41 Wouldn't that have to be 2sqrt(2) instead of just 2?
this seems to be potentially related to the weierstrass p-function, which is a similar-looking double sum in the integers of a quantity that drops off as the inverse square of the summands, and has to be massaged a bit to get a convergent series
Isn't our double sum an iterated limit? Should it be lim m->inf lim n->inf sum 1 to m sum 1 to n 1/(m^2+n^2)? Which is different from lim m,n->inf sum 1/(m^2+n^2)? The original case is we are adding column-wise/ row wise if we rearrange them in a grid while the latter (which is what we did) is adding "radially" ie we have ordered pairs (m,n) and we add as we go outwards? From what I know, there are criterias for commutativity in infinite sums.
We can see this in definition.
We say lim m,n exist if for all epsilon>0 there exist N such that if m,n>=N, |a_m,n - A|< epsilon.
From what I know, existence of double limit is strong as it guarantees the iterated limits existence. But proving the double limit as divergent does not necessary mean we have divergent iterated limit.
I was wondering if there is the possibility of using m^2+n^2 = (m + i n) (m - i n) and therefore
= 1 / ( 2m (m+i n)) + 1 / ( 2m (m-i n)) ...
There are some interesting ideas in this video, but it was very hard for me to understand. I'll need to review it if I want to get them.
Thank you.
Could we use 1/((x+1)^2 + (y+1)^2) and calc integral from 0 to inf? No tricks needed for polar coordinates.
Great video as always!
I notice it’s been a while since we had a backflip transition. Perhaps they could make a comeback?
Back Comeback? 😊
I think he said once that he no longer does that because of the mic placement
@@farfa2937 Oh, that makes sense but is also kinda sad
@@farfa2937 I guess once in a while he could use a different mic placement.
I didn't quite understand this passage after 5:30. Why is it that by including what's between the lattice points (going from a sum to an integral) you're creating something smaller? I'm guessing it's because of some shenanigans with infinity, but I'm not sure.
It's because f is decreasing along the x and y axes, so the area under the function between say (x,y) and (x+1,y) is smaller than the area of a rectangle with height f(x,y), and width 1 (this is true in all directions)
@@nolanrata7537 Why width 1? Isn't it supposed to be "width 0", since you're using single values of the function?
@@andrycraft69 He is comparing the value of the double sum to the value of an integral, which is why he introduces a delta-x and delta-y in the double sum, which have a value of 1 (which represents the width of the rectangle, or if you will the size of the base square of a box since it is 2D sum/integrals).
The double integral of f is the volume under the curve of f within a specified region; the double sum here is the sum of the volume of 1 by 1 boxes with height equal to the value of f at each integer pair (the dots in the lattice he drew). Since f is decreasing along x and y, the curve of f is always contained within these boxes, so the double sum is greater than the double integral.
@@nolanrata7537 Oh ok. That makes sense. I think I was visualizing it in the wrong way. Thanks!
i remember seeing this in the romanian mathematical magazine and i also thought this was kinda interesting tho i got solution in a more straight forward manner
I did almost the same, using f(x,y) =1/(x^2+y^2) over R×R, but I did not do the ray thing. Instead, I divided up the first quadrant into five regions:
Region 0, (unit disk) 0 < x, 0 < y, x^2+y^2
I think sum(1/(m^3+n^3)) should converge. Does sum(1/(m^3+n^2)) converge?
Cool video! I wonder what exponent we'd need instead of 2 to have convergence.
3 should work
@@JosuaKrause I'll raise you two plus epsilon...
wow, i'm surprised that it would diverge.
Unexpected result. Interesting.
I was wondering whether a counting argument on primes p = 4K+1 would work
I dont understand the sign > between thé double sum and the integral you add positive things
10:15
Nice!
my 2cent. if Sum 1/n2 converges, then Sum 1/(c+n2) also converges, thus Sum Sum 1/(m2+n2) is the sum of converging nrs, thus also converges
That's valid if you only sum in one of the variables.
@@md2perpe i think you are right. the infinite sum of numbers each bigger than 0 can only be infinite, ie Sum Sum 1/(m2+n2) has to be infinite
@@popodori Not necessarily. Sum(n=1..inf)[1/2^n] = 1, even though each term is bigger than 0.
@@popodori That is also not correct. Under "Related Facts" you see two infinite sums of numbers each bigger than 0 which do converge to finite values. Another example is 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2ᵏ + ... = 1.
I don't get how the green part between α and β is bigger than the whole [1,∞)^2 purple part. It seems like it's a subset except this small quadrilateral near the origin so it should be smaller, no?
It is smaller, and that's the point. It's what makes the integral over the displaced quadrant >= the integral in polar coordinates.
@@TJStellmach Ohh yeah. I'm just stupid then. I got my inequality directions mixed up as usual.
Hi,
ok, great! I really thought this would converge !
Fun, thanks.🙂
thé somme goes to infinité....thème is no raison ....thanks for the video
slick
Butthead:
"Uh-h'huh-h'huh-h'huh, he said 'pee test'!!!"
;^}
Sorry, I couldn't help being sophomoric
COOL !
Bruh what is that I'm scared
Great video as always! But was anyone else weirdly disappointed that it didn't converge? I was looking forward to a cool sum value :[
I think, you could be more didactic