Show that f is periodic

Added: 2026-05-11

[00:00:02]Welcome to another video. We have another functional equation that says f of x + 1 plus f of x -1 equals the square root of f ofx. We need to show that the function f is periodic.

[00:00:21]That is there is a constant we can add to x. That would not make a difference.

[00:00:26]It would just be f ofx.

[00:00:30]Let's get into the video.

[00:00:44]So the first thought that came to my mind is maybe I should write f ofx in terms of the other two and start manipulating everything. But then I realized that there is a sequence.

[00:01:00]The sequence is that you have x + one here. You have you go down one actually you go down two. From x + one you have to go down two to get to x - one and then you come back up one. Why don't we rearrange it so that we go one down and down? And that arrangement would work if we moved this to this side. So let's see what we can do. Solution you have f of x + 1 = the<unk> of 2 * fx - f of x - 1. So notice what we have right now. If I am dealing with x + one then it will be equal to the<unk> 2 * f ofx which is one down from this and then minus f of x - 1 which is one down from this. So if I change this to x + 2 this is going to become x + 1. This is going to become x. If I change this to x + 3 this is going to become x + 2. This is going to become x + um 1. So that will always be what I'm going to get. So what I'm going to say is f of x plus some k will be equal to the square<unk> of 2 * f of x + k -1 - f of x + k - 2.

[00:02:44]Actually didn't want to write that. That was not what I was thinking I was going to write. But you know what? I really don't care about this line because it doesn't help me.

[00:03:00]But that's that's the sequence. So what it means is that if I put a subscript one here to represent f of x + one. Okay, look here. Let's assume we start from because we already see this. Okay. Um if we start from one when k is one then we're going to have the original equation. But when k is we yes when k is one we're going to have the original equation but when k is two we don't have two yet. So let's try f of two. Notice that f of two I mean f sub 2 rather not of f sub 2 will be equal to square<unk> of 2 f of 2 - 1 is 1 minus f of 2 - 2 is zero. And this makes our life easier because then I don't have to write all this x plus x+ anymore. I just have to be writing this.

[00:04:02]Okay, let me clarify the notation.

[00:04:06]f of f_sub_2 rather is basically f of x + 2.

[00:04:15]Okay.

[00:04:17]And this guy f_sub_1 it's just I just don't want to write this much x + 1 and f sub0 is just f ofx. Okay, it is f of x plus 0 which is f ofx. So what we really want is we want a case where when we put any k here a certain k here it would be equal to f of0 f ofx.

[00:04:46]So we want f subk to be equal to f sub0 and whatever that k is would be the period would be the number that we added to make it equal to f ofx.

[00:05:09]Remember that's the meaning of periodic.

[00:05:12]When a function is periodic, it means you can add something to x in the argument and your answer will still be f ofx. So this is what we want. But this simplified version of this function makes our algebra easier and more compact. Okay. So already we have f of one. f of one was the problem they gave us. Not f of one. f_sub_1 was the problem they gave us. F_sub_2 is what we just created to explain this new expression from just moving things around. Let's go see what F3 is.

[00:05:48]So now we have F3 equals based on this definition square<unk> of 2* F_sub_2 minus F_sub_1.

[00:06:04]That is what F_sub_3 is.

[00:06:08]But we already know what f_sub_2 is, right? Remember, all we're doing now is we're trying to connect this to f0.

[00:06:18]So whatever we see, we start keep eliminating. How do we know there's going to be elimination? Because of the minus sign. Notice that when you're here, something is going to happen because this number is going to keep growing.

[00:06:34]it's somehow going to cancel out f1 in the future. Something in the future. So, it's more like a telescoping series. So, you're going to start canceling something out as you keep building these numbers. You see, we now have f1. This guy is going to go watch this.

[00:06:51]And this is equal to the square root of 2 * what is f_sub_2?

[00:06:58]We said f_sub_2 is rad 2 f_sub_1 minus f rad 2 f_sub_1 minus f0.

[00:07:07]Nice.

[00:07:09]That is we just replaced f_sub_2 with that. But this still has minus f_sub_1.

[00:07:18]Okay. Notice that if you clean this up, rat 2 * rat 2 is going to be two. So you have 2 f_sub_1 minus f_sub_1. That's just f_sub_1. And then this going to have rat 2 f0. So we have f_sub_3 equals 2 f_sub_1 - f_sub_1 is just f_sub_1.

[00:07:45]Let's write it. 2 f_sub_1 minus rad 2 f_sub_0 minus f_sub_1 If you clean this up, you're going to end up with f_sub_1 minus rad 2 f_sub_0 f_sub_1 minus rad 2 f0.

[00:08:09]So notice how f_sub_3 is still written in terms of 1 and zero just as f_sub_2 is written in terms of one and zero. If we keep going that way, we're going to stay within the original definition and we'll be able to connect it to f ofx. Okay, so let's put this in a box.

[00:08:31]This is our f3.

[00:08:36]Let's do f4.

[00:08:38]Again, f_sub_4 is based on the same definition. f_sub_4 will be rad 2 f_sub_3 minus f_sub_2. It's going to be rad three rad 2 f_sub_3 rad 2 f_sub_3 minus f_sub_2. Well, I'm going to try and squeeze it in here. Rad three times what is f_sub_3? We just said f_sub_3 is this. So, I'm going to plug this in.

[00:09:04]It's going to be f_sub_1 minus rad 2 f_sub_0.

[00:09:10]Okay. Minus f_sub_2.

[00:09:12]Oh, what is f_sub_2?

[00:09:18]Rad 2 F_sub_1 minus F0 rad 2 F_sub_1 minus F0.

[00:09:26]Okay, you see the way things are canceling out. There's going to be some number in the future that's going to just be F0.

[00:09:37]That is the expectation. So here, just be careful with your algebra. So we have f4 will be if you no this is rat two. Hey there's no rat 3. Okay. So this is going to be rat 2 f_sub_1 rat 2 f_sub_1 -2 f0 -2 f0.

[00:10:02]Then this is going to be -2 f_sub_1.

[00:10:06]This is going to be plus f_sub_0. Okay.

[00:10:09]It looks like f0 is still surviving.

[00:10:12]This minus this will cancel out and this is negative. So you're going to have one extra one. So f4 huh equals f0.

[00:10:26]So if only I could show that the function is odd, I'm done because I just use the oddness of the function. But I cannot I tried I tried to actually show this function is odd but there was no way because there was nothing that was coming out of it. So I just gave up and I said I'm just going to keep going until something plain happens.

[00:10:52]Okay. So now let's keep going. We have f5 equals following the rule rad 2 f_sub_4 rad 2 f4 minus f_sub_3.

[00:11:07]Well, this is going to be rat 2 times what did we say? F4. Oh, f_sub_4 is short. Negative f0.

[00:11:16]Okay. Minus f_sub_3. What is f_sub_3? F3 is F_sub_1 minus R2 F0 F_sub_1 F_sub_1 minus rad 2 F0 like that. So if we clean this up, we're going to get negative rad 2 f0 negative rad 2 f0 minus f_sub_1 plus rad 2 f0. And guess what? These two cancel out and we now discover that F5 equals F1.

[00:12:03]So it looks like some similarities are beginning to show and we don't have to write long things like we've been writing. Let's try F6.

[00:12:13]F6 equals Oh, that's even much easier.

[00:12:18]Equals rad 2. F5 minus f4.

[00:12:24]This is going to be equal to rat 2. What is f_sub_5? f_sub_5 is negative f_sub_1.

[00:12:31]Negative f_sub_1 and minus what is f_sub_4? f_sub_4 is negative f_sub_0 f0.

[00:12:41]Nice. So we can easily say f6 equals negative rad 2 f_sub_1 rad 2 f_sub_1 plus f0. Let's box this. Let's go. So we're going to try f7. F7. It's beginning too close to look. Oh, wait. This looks like the original.

[00:13:14]Oh, f6 is the negative of f_sub_2.

[00:13:21]f_sub_5 is the negative of f_sub_1.

[00:13:25]f_sub_4 is the negative of f0.

[00:13:32]Let's see what f7 is going to be. F7 equals rad 2 F6 minus F5.

[00:13:44]Well, that's going to be rad 2 times this is going to be negative rad 2 F_sub_1 plus F0.

[00:13:57]That's the whole of rad 6 minus f5 negative f_sub_1.

[00:14:08]Okay, let's see what this gives us.

[00:14:17]If we clean this up, this is going to be -2 f1.

[00:14:24]Oh, so we have f7 = -2 f_sub_1 -2 f_sub_1 plus rad 2 ft plus rad 2 f0 plus f_sub_1 plus f_sub_1. So f7 is equal to this. We'll take one of these out with a negative. So you have negative f_sub_1 plus rad 2 f0.

[00:15:08]Huh?

[00:15:11]Wait, that looks like the negative of the original, right?

[00:15:26]No.

[00:15:28]Yeah, it looks like the original. This is F1 actually.

[00:15:40]No, it's not F1. Ah, it's F minus one. Oh, this is F minus one.

[00:15:54]The original. This guy. Remember, if you move this guy over here, this is what you're going to get. We're close to the answer because we've done it. It looks like we've covered everything now in the original. So, let's try F8. Let's box this. If this last one works, okay, so we got F of 8, I mean F8 will be equal to RAT 2 F7 minus F6.

[00:16:28]So this is going to be minus f_sub_1 plus rad 2 f_sub_0 minus f6. What is f6 minus rad 2 f_sub_1 + f0 minus rad 2 f1 + f 0. Okay, let's do some cleaning up. This is going to give us negative rad 2 f_sub_1 plus 2 f0.

[00:17:10]This is going to be plus rad 2 f_sub_1 and this is going to give us minus f0.

[00:17:19]Okay, so um looks like these two will cancel each other. Oh, these two cancel each other out. And 2 f0 minus f0 is f0. f8 equals f0.

[00:17:41]This means f of x + 8 equals f ofx.

[00:17:52]We have found a period of the function and that means this is periodic. There is an alternative way using trig but I would have lost a lot of people once I get into trig. But this one was just algebra that anybody could follow. Just make sure you distribute correctly. And we have arrived at the correct answer.

[00:18:15]Never stop learning. Those are stop learning stop living. Bye-bye.