Infinite Sum Extra Stuff - Numberphile

0:03 So this gets a little bit hairy but but hang in there because we don't

0:08 have to use I have to write it in a way that something really nice happens.

0:16 Okay.

0:15 And and [snorts] reaching that stage can be can look worse than it is.

0:21 Okay.

0:22 All right.

0:23 So let let's try and prove it.

0:25 So let's just remind ourselves.

0:27 So we've got this sum to do, right?

0:30 I'm I'm wanting to take the limit of this f going to zero of this sum.

0:35 So the sum is n is 1 to infinity of n sin

0:40 n f over alpha 2 sin^ 2<unk>i n/ alpha +<unk> 2 n^ 2.

0:50 It's worth pointing out at this really we're thinking

0:53 of alpha as being a small number in in this calculation.

0:58 So the first thing I'm going to do is

0:59 I'm going to get this denominator onto the numerator.

1:03 And so let me just first of all simplify

1:06 a little bit by I'll I'll drop this limit.

1:10 We we know we're going to be taking um the limit at the end.

1:14 So what I'm going to do is I'm going to pull the pi squ and n squ out.

1:18 You you'll see what I mean.

1:20 So I pull the pi squ out and then what I have is the sum from n is

1:27 1 to infinity and I've got then n sin nf over n 2 into and it becomes 1.

1:36 So I've pulled that pi^ 2 n^ 2 out that's what I'm

1:39 doing 1+ and then I've got to account for it again here.

1:43 So it becomes alpha^ 2 over<unk> 2 n^ 2 times by sin^ 2<unk>i n/ alpha.

1:52 It's exactly the same expression but by pulling the pi^ 2 n n

1:56 n n n n n n n n n n n n n n

1:56 n n n n n n n n n n n n n n n n n n n n n n n n squ 2 out,

1:57 I have to divide by it in in this term here.

2:01 Now I'm going to use the binomial expansions, right?

2:04 So if you if you recall or a tailor expansion you might for when x is small.

2:10 So this is 1+ x^2 to the -1.

2:15 So 1 this is what I'm thinking of as being x^2.

2:18 This thing here is actually equal to you can you know

2:23 it's 1- x^2+ x 4- x 6 it just keeps alternating.

2:31 So the way you write this as the sum of m

2:34 is 0 to infinity of -1 to the m x^ 2 m.

2:41 So when m is zero anything to the raised to the power 0 is 1.

2:46 So the first term becomes 1.

2:48 When x when m is 1 that's the minus sign* x^ 2 when x is 2- 1 2 is+ 1.

2:56 So you get that.

2:58 So now I can write what i is.

3:00 So I now becomes 1 /<unk>^2 times the sum from N is 1

3:07 to infinity of sin NF/ and I can cancel one of the ns.

3:13 There's an n in the numerator and there's an n squ on the denominator.

3:17 So I can cancel one.

3:18 So I get that and then all the m terms I'm just going to put together.

3:23 So this is m= n to infinity of -1 to the m sin 2 m of 2<unk> n/ alpha times

3:33 by and then I've got this thing also there

3:35 to that power alpha over pi n to the power of 2n.

3:41 So so we're here now but this is not a well one thing to notice this sum

3:46 is from n is 1 to infinity and this is from m equals 0 to infinity.

3:50 So I'm just going to take out the zero bit and remember

3:54 what we said anything raised to the power 0 is one.

3:58 So -1 to the^ 0 is one sine to the^ 0 is 1.

4:03 Alpha to the power 0 is 1.

4:05 So the zeroth term here is 1.

4:07 So what I can do I can write this now as 1 /<unk>

4:12 2 times the sum from n is 1 to infinity sin nf/ n.

4:20 and then plus and then I'm going to do this sum from m= 1 to zero.

4:24 And I'm I'm going to switch these around,

4:26 which I can do because they're both going to infinity.

4:29 So it becomes 1 /<unk>^ 2 times the sum from m is 1 to infinity

4:36 of -1^ m* and I'll pull out this alpha over pi to the power 2 m.

4:45 And then I'm left with this.

4:46 I'm putting together all the n terms.

4:49 That's my what I'm trying to do here.

4:51 So I've got this sine nf here.

4:54 I've got a sine 2 m* 2<unk>i n/ alpha.

5:00 There's the n in there.

5:02 And I've got a 1/ n here.

5:05 And I've got a 1/ n to the 2 m.

5:07 So this becomes n to the 1+ 2 n.

5:11 I mean, it's looking a nightmare.

5:13 I I admit it.

5:14 But this this we've already seen this is this and we

5:23 we saw in the limit of f goes to 0.

5:27 We saw that this was 1 over 2 pi.

5:29 So this term gives me 1/ 2 pi.

5:31 So in the limit of f going to zero that gives me 1/ 2 pi.

5:36 And then I'm plus this thing.

5:38 And let me just call this thing i1 plus i1.

5:41 So I don't have to keep writing it.

5:43 This is what I'm trying to prove, right?

5:44 I'm trying to prove that I equals this.

5:47 So, can I show that this is zero?

5:51 If you didn't already have a hunch what the solution was,

5:55 this is something you probably wouldn't have thought to do, but

5:57 no, no, I wouldn't have had I wouldn't have been brave

6:01 enough because I'm going to end up with yet another sum.

6:04 I would have ended up with a three infinite sums.

6:06 And in what planet replacing one infinite sums does

6:10 by three infinite sums does it make it easier?

6:12 But it turns out this one does.

6:15 And in fact, I don't have to do any of the sums.

6:17 Now, I don't have to do any of them.

6:19 So, we're kind of reaching the final set of sums

6:22 because I don't want this s time sign to some power.

6:27 I want them so that it's s time cosine or s

6:30 time s because then I can use my trigonometric identities.

6:35 So there's a useful formula which tells me that sin^ 2 m of x is

6:41 equal to -1^ m over 2^ 2 m* 2 star I'll explain what that is

6:51 in a minute times a third sum right the sum from k= to m

6:57 of 2 m this is the binomial coefficient times by cosine of 2k- 2 m* x.

7:06 Right?

7:07 So, I've got a formula which tells me what sign

7:10 to some high power is in terms of a single cosine,

7:15 but the price I've paid, I've got to have lots of them.

7:18 That's what this sum of a k is.

7:20 And this is the binomial coefficient.

7:23 So, so the bracket a simply means a factorial

7:28 over b factorial over a minus b factorial.

7:32 And this star is 2* is equal to 2 unless k=

7:38 m and then it's equal to just the binomial 2 m/ m.

7:43 So it's just a form simplifying thing but we don't need it.

7:47 We're not going to need it.

7:48 This formula was useful.

7:49 I remember you told me this and you so what

7:52 we've done is this is the sum the key sum.

7:55 There's a sign and sign to the 2m and I don't want that.

7:58 I' I'd like it to get so it looks like a sign time

8:01 a cosine or a s times a sign not raised to this big power.

8:06 So I can use the fact that this sign to this power is a sum of cosiness.

8:13 So I've got it so that it's now

8:15 a sum of individual cosiness and that really helps.

8:19 So now having having done that and so now I can sort of go a little bit

8:24 further and and what and what I've ended up with now is so this I1 which is what

8:30 I'm trying to show is zero because we've seen

8:32 that the leading term gives me what I think

8:34 is the answer that this this I1 is and now I'm not going to be using these sums.

8:41 So let me just say there's going to be a sum

8:44 from n= 1 to infinity of some function of m.

8:49 I don't we don't need to worry about it.

8:51 It's it's here.

8:52 People can and then there's a sum from [snorts] k equals 0 to m

9:00 of a of a some function um let me call it h of k and m.

9:06 But the key one that I've got is this sum over n.

9:10 And I'm not even in the end going to use the sum of n.

9:13 But this is where I've collected all the n dependent terms.

9:16 So I have sine of nf.

9:20 I have this cosine of 2 k- 2 m*

9:25 and it's actually 4 k- 4 m sorry *<unk> n/ alpha.

9:33 And then that's all divided by this n to the 1+ 2 m.

9:38 Okay.

9:38 And I'm just going to call this bit here I1A.

9:43 So this is now all I'm going to concentrate on.

9:46 In fact, all I'm going to concentrate on is this bit here,

9:50 the sign times the cosine.

9:52 I'm not even going to have to try and do the sum.

9:54 And this is where my high school comes back.

9:56 Okay?

9:57 Because one of the things we learned

9:58 at high school were various trigonometric identities.

10:04 And one of them is the following.

10:07 that if you have sin of a* cosine of b sin of a* cosine of b that's what I've

10:17 got in mind then that's equal to a half into sin of a+ b plus s of a minus b.

10:31 So we did do this at school but now look what I've done.

10:35 It means where I've got the sign and cosine,

10:37 I can replace by a sign and have two terms.

10:42 So what do I have to do?

10:43 I just have to read off what my a and b are.

10:45 So I just look across.

10:47 You see that a is equal to nf and I see

10:52 that b is equal to 4 into k minus m pi n/ alpha.

11:02 So that means if I these two together give me the a plus

11:09 and minus b cuz that's what I need for these two is just the following.

11:14 It's just f plus and minus 4 into k- m pi over alpha* n because n is common.

11:29 And I'm nearly there because now what I've got for this sum I

11:34 want a it's equal to well there's a half I pull the half out

11:38 times the sum from n is 1 to infinity of sine of and it's

11:46 the first one f+ 4 into k- m<unk>/ alpha* n/ n 1+ 2 a.

11:58 So there's that term and then I add to it sine of f minus

12:07 4 into k- m<unk>/ alpha all* n/ the same factor n^ 1+ 2 m.

12:17 Okay, so this is what I've got.

12:19 But all I need to remember now is we want the limit of f going to zero.

12:26 And because f is by itself, I can just set f equal to zero.

12:31 And look what I have.

12:32 If I just concentrate on these two signs, I get the following.

12:38 I get s of 4 k- m<unk>/ alpha* n plus sin of- 4 k- m<unk>/ alpha* n.

12:52 The sign function does this.

12:54 This is what sign does.

12:55 as a function of x sin of x it's goes

12:58 up and then comes down oscillates and then it if I

13:02 go negativex it's minus so sin of x equ= minus

13:09 sin of minus x that's the result for signs that's sin

13:13 of x that's sine of minus x so these cancel so I get s of 4 into k- n pi n/

13:28 alpha minus so I replace this sign of minus 4

13:35 by minus sign so it just gives me the same thing

13:41 but with a minus sign there's me zero there's me zero

13:45 so I've shown this is equal to zero so it means that up here

13:50 up here somewhere i1 equals zero which means that I which is

13:56 the sum of I1 and this thing is 1 over 2 pi.

14:01 We already knew that but well done.

14:03 We already knew it.

14:04 [laughter] So there you go.

14:06 And it's just fun.

14:08 Fun fun.

14:09 Look at that.

14:10 Fun for all the family.

14:14 Yeah.

14:13 Is F R 2 over big M* little M.

14:16 But there's one more equation.

14:19 What about the What about F?

14:20 Well, possibly the most famous equation alongside E= MC² is F= MA.

14:26 We learned that at school.