A Hyperbolic Take on the Fibonacci Sequence

Ajouté : 2026-05-31

[00:00:00]Okay, so we're going to explore how to express the Fibonacci numbers in terms of hyperbolic trig functions. And we're going to use our starting point Benet's formula, which is a really useful formula that tells us that the nth Fibonacci number. So we can work with this as a formula rather than having to keep adding the previous terms. We can write 1 over<unk> 5 multiplied by then we've got in brackets 1 +<unk> 5 / 2 to the n and 1us<unk> 5 / 2 to the n being subtracted. So this gives us the formula for the nth Fibonacci number. So you can verify this algebraically using proof by induction for example if you're interested. But we'll use this as our starting point for this video. So you see how we've got the 1 +<unk> 5 / 2 and 1 minus<unk> 5 /2. So we're going to use these to try and extract some relationship between the Fibonacci numbers and the hyperbolic trig functions. So if we just start by setting e to the x equal to 1 +<unk> 5 / 2 then you'll see that x is going to be we can just take natural logs here. So x is the natural logarithm of this number but 1 plus roo<unk> 5 /2 is actually the golden ratio. So it's ln fi where phi is the golden ratio. So this is quite interesting already. And then if we consider e to thex let's see what happens here. We've got e to the x and e to thex in the definition of our hyperbolic trig functions. So this just becomes the reciprocal. So we've got 2 over 1 +<unk> 5. And then if we want to rationalize the denominator, so multiply by 1 minus<unk> 5 on the top and bottom of the fraction, we're going to get on the numerator 2 - 2<unk> 5. The denominator becomes 1^ 2 -<unk> 5^ 2 or 1 - 5 gives us -4. and then dividing through by two on the top and bottom we get 1 -<unk> 5 over -2. So then we could just say that e to thex is equal to this with the negative sign or even better we could say negative ex = 1 -<unk> 5 / 2.

[00:02:02]So these two bits of information now alongside the fact that x is ln 5 are going to be really useful for connecting this Fibonacci sequence to the hyperbolic trig functions. So now we can see in Bernay's formula we've got this 1 +<unk> 5 / 2 and 1 minus<unk> 5 / 2. So we can express these in terms of our e to the x and e to the x terms now. So we've still got 1 over<unk> 5 and then we've got first of all this is just e to the x being raised to the power of n. So we can write this as e to the nx. Then this next term requires a little bit more care. We've got the negative sign.

[00:02:36]So it's minus e to thex being raised to the power of n. Then we can deal with this negative sign just by taking this out as a factor of -1 to the power of n. So this is going to give us still e to the nx take away -1 to the n.

[00:02:53]And then we've got e to thex to the^ of n gives us e to the n x. So now from here we're going to get slightly different behavior depending on if n is even or odd. So if n is even negative -1 to an even power is just going to give us positive one. So we're going to get a hyperbolic sign kind of identity. And if n is odd this will turn into a minus and a minus. So we get a plus there. So that we have hyperbolic cause. So if we split into these two cases we can say that the nth Fibonacci number is first of all if we deal with the even case. So when it's even, we've got 1 /<unk> 5 * e nx - e to the - nx. Then we can divide through by two here. And we just multiply. Let's turn this one into a two. So then you can see we've got 2 over<unk> 5 times.

[00:03:44]This is just hyperbolic s of e to the n x. So this equals 2 over<unk> 5 hyperbolic s of nx. And this is exactly the case where n is even. And then similarly when n is odd, we've got 1 /<unk> 5. And then in our brackets here, we're going to have because n is odd, we're going to get a plus sign there. So e to the nx plus e to the nx. And again, we can just divide through by two and multiply by two here to cancel that out.

[00:04:16]So then you can see we've got 2 over<unk> 5 * we've now got the definition of hyperbolic cos of nx. And this is where n is odd, we get hyperbolic because cos, we get the plus sign. When it's even, we end up with the minus sign. There's also another nice way of presenting this is for our even numbers, we could actually go from n to 2 n. So we could write f of 2n is going to be 2 over<unk> 5 hyperbolic sign of instead of nx, it's 2 n * x. And similarly when we've got f of for an odd number we could do 2 n + 1. So we get 2 over<unk> 5 * hyperbolic cos of 2n + 1 * x just to give an alternative way that we don't necessarily need to write it like this. We could also write it a little bit more compactly but we're still having to split into the two different cases depending on n being odd or even.

[00:05:14]So what can we do with this? Well, there's a lot of identities related to the hyperbolic trig functions. There's also a lot of identities related to the Fibonacci numbers. And we can see how the two of them are connected now. So, let's say if we were interested in this identity, the 2n Fibonacci number is the square of the n +1 Fibonacci number minus the square of the n minus one Fibonacci number. We could actually prove this identity and get quite an elegant proof using some of our hyperbolic trig function identities and see the connection between the two. Just as quite a fun little exercise really.

[00:05:48]So we could just before we get started on this, we could rewrite this difference of two squares expression as fn + 1 + fn minus1 multiplied by fn + 1 minus fn minus one. Then you see here we've got fn + 1 minus fn minus one.

[00:06:05]Just to make this really clear, we know that fn plus one is the sum of the previous two terms because they're Fibonacci numbers. So then if we subtract fn minus one, we get fn + one minus fn minus one is equal to fn. So you can see then that this whole term in the bracket is actually just equal to fn. So if we want to show that f_sub_2n is equal to this difference of squares identity expression, then it's equivalent to show and this is what we'll proceed with for our proof. We want to show that f2n is equivalent to fn multiplied by fn + 1 plus fn minus1.

[00:06:45]So unfortunately here we've got to do some splitting into cases depending on if n is odd or even because this n + one and n minus one term we could have hyperbolic sign or hyperbolic cosine depending on whether n is odd or even.

[00:06:59]So we'll show the proof just in the case where n is even. And you can do something very similar in the case where n is odd. So if n is even, we can say n= 2k. So then what we're really trying to show is let's start with our right hand side and we'll try and turn it into the left hand side using some hyperbolic trig identities along the way. So our right hand side is fn becomes f2k multiplied by f 2k + 1 plus f 2k minus one. So you can see here we've got an even term. This is going to be hyperbolic sign and these two terms are going to have hyperbolic co in them. So then we can rewrite all of this as 2 over<unk> 5 * hyperbolic s of 2k * x.

[00:07:44]And then in the brackets we've got 2 over<unk> 5 * hyperbolic cos of 2k + 1 * x plus 2 over<unk> 5 hyperbolic cos 2k -1 * x. So then we've got all of these expressions. You can see we can multiply by the 2 over<unk> 5. That's a common factor. Take that out. So then we get a factor of 4 over 5. We've still got this hyperbolic sin 2k x. And then the term in the bracket is hyperbolic cos 2k + 1 * x plus hyperbolic cos 2k -1 * x. Now we can actually use an identity now related to hyperbolic cause. So if we have hyperbolic co of a plus hyperbolic co of b. This is like our angle sum formula but in reverse almost. This is equivalent to having 2 * hyperbolic cos of a + b / 2. So you might recognize this like our sum to product identity which we also have for regular trig functions as well. Hyperbolic cos a minus b / 2 there. So now we can use this identity to rewrite these two terms in quite a simple form. So we've got 4 over 5. We've still got our hyperbolic sin 2 kx, but now this is all being multiplied by 2 * hyperbolic cos of our a + b / 2. So we have 2k + 1 * x plus 2k -1 * x. This gives us 4k x, but then divided by 2. So this goes back to just 2k * x. That's our first term, our a + b / 2. Then when we do a minus b / 2, we've got 2k + 1 * x - 2k -1 * x. So this gives us 2x. Then we divide by 2.

[00:09:39]This just gives us x. We've got this multiplied by it's just hyperbolic co of x. Now, and if we remember what x actually is, the natural logarithm of the golden ratio, we've got hyperbolic cos x. We can just write this as it's a half e to x plus e x. We'll write it as minus e^x. We saw earlier, didn't we, that e to x, we've chosen x specifically so that we've got 1 +<unk> 5 / 2. And similarly, we chose x so that egx is going to give us 1 -<unk> 5 / 2. So minus 1 -<unk> 5 / 2. And if we calculate this in the bracket, we ones cancel and we've got roo<unk> 5 minus<unk> 5. So all of this simplifies actually just to a half *<unk> 5.

[00:10:31]So<unk> 5 / 2. So hyperbolic cos x the upshot here is that this is just roo<unk> 5 / 2. So then we can rewrite all of the right hand side. We're going to write this as you can see we've got this factor of two here as well. So we've got 4 fths *<unk> 5 / 2. Then we've also got this factor of two. Then perhaps you can see now where we're going to go next with our hyperbolic trig identities. We've got hyperbolic sin 2 kx multiplied by hyperbolic cos of 2kx. And now we're going to use the identity that if we have hyperbolic s of let's say 2 y, this is always equivalent to 2 * hyperbolic sin y * hyperbolic cos y. So then we can replace all of this just by hyperbolic s of 2 * 2 kx. So this gives us a really nice expression.

[00:11:20]Then so the 4<unk> 5 * 5 and 2. You can see the root fives cancel and the four and the two cancel slightly as well giving us a 2 over<unk> 5. And then all of this remaining term just turns into hyperbolic sign of it's double this. So it's 4k x. And then if we remember that n is equal to 2k. So 4k is just 2 n.

[00:11:43]We've got 2 over<unk> 5 * hyperbolic s of 2 n * x. And this is exactly what f2n is as we saw earlier. So the 2n even fibonacci number. So we know that this is then equal to f 2n which is exactly what we were trying to show is we started with this right hand side. We're trying to show this is equivalent to f2n. And we've done this at least in the case where n is even. And you could do as an exercise if you're interested try and do very similar kind of argument where n is odd. And if you're really keen to see what more we can do with this as well, we could try and do a very similar looking identity that f 2n + one to where we've now got an odd term here.

[00:12:27]This is going to be equivalent to fn + 1 squar but now it's plus fn minus1^ 2. So again I would start with the right hand side here split into the two cases according to n being odd or even because this n + one and n minus one. You don't know whether this is going to be hyperbolic sign or hyperbolic cosine.

[00:12:46]And take the right hand side. Try and use some hyperbolic trig identities to get it to come out nicely to be the left hand side.