a very Fibonacci product!

here we're gonna derive a nice infinite product identity involving Fibonacci numbers so we're gonna look at the product as n goes from 1 to infinity of 1 minus negative 1 to the N over the n plus first Fibonacci number squared and we're gonna use these two tools which we will prove to keep this video self-contained the first one is so called Cassini's identity and that says F sub M minus 2 times F sub M plus 1 minus F sub M squared is equal to minus 1 to the M and the second one is the limit of the ratio of consecutive Fibonacci numbers is equal to the golden ratio in other words the limit as M goes to infinity of F sub M plus 1 divided by F sub M is fee but that's equal to 1 plus the square root of 5 over 2 okay so let's get to proving this first identity so we're gonna use the following fact which will also prove and that is if we take the matrix made up of Fibonacci numbers of this form so on the diagonal we have the impulse first and M minus first and on the off-diagonal we have the nth Fibonacci numbers so that's the same thing as the matrix 1 1 1 0 to the M power so let's go ahead and look at the base case for our induction and for our base case we'll take the M equals 1 case and notice that the left hand side will be F sub 2 and then F sub 1 F sub 1 F sub 0 where we take the 0th i Bonacci number to be 0 so notice that is exactly equal to 1 1 1 0 to the first power which is exactly what we want it to be ok so now let's go ahead and make an induction hypothesis so let's say suppose this is true for M equals K so in other words we know that this statement is true for N equals K so let's go ahead and write that down so we have F sub K plus 1 F sub K F sub K F sub K minus 1 equals 1 1 1 0 to the K and now what I'll do is I'll take this equation and I'll write multiply both sides by the same matrix so I'm hi by the matrix 1 1 1 0 and then over here 1 1 1 0 so let's see what we get so for this upper entry of the left-hand matrix we have F sub K plus 1 plus F sub K but by the recursion that's F sub K plus 2 and then this entry right here will give us F sub k plus 1 again just by matrix multiplication this entry right here gives us F sub K plus F sub K minus 1 but that's also F sub k plus 1 again by the recursion and then finally this entry right here is f sub K again by matrix multiplication but now on the right-hand side of the equation we have k plus 1 of those matrices in other words we have 1 1 1 0 to the k plus 1 but that's exactly what we needed to show in order to finish this proof by induction so in other words this equation up here is true ok so now I'll go ahead and clean up the board from here down and we will use this representation in order to prove Cassini's identity ok we just got done proving by induction that this matrix of Fibonacci numbers is the m power of our matrix 1 1 1 0 so now what we'll do is take the determinant of both sides of this equation so we know that determinant is this side of the equation will be F sub n plus 2 times F sub M minus 1 minus F sub M squared so like I said that's the determinate of this matrix made up of Fibonacci numbers but that matrix of the Bonacci numbers is the same thing as the M power of this right-hand side matrix so 1 1 1 0 to the M but now the determinant is a multiplicative function so that's the same thing as the determinant of the matrix 1 1 1 0 all raised to the M power but now we see that the determinant of this matrix 1 1 1 0 is equal to negative 1 so we end up with negative 1 to the M and now if we look at the extreme left and right hand side of the equation we have derived this formula now we're going to prove that the limit of the ratio of consecutive Fibonacci numbers is fee and we first need to prove that this limit exists we're gonna use the Leibniz test to prove this limit exists and that says the limit of a sequence a n exists if and only if the limit as n goes to infinity of the nth term minus the n minus first term is equal to zero okay so now let's go ahead and apply that to this sequence so let's look at the limit as n goes to infinity of F sub M plus 1 over F sub M minus F sub M over F sub M minus 1 okay great so now what I want to do is mash those together and I'll do that by kind of finding a common denominator and that common denominator will be the product of F M and F M minus 1 so that's going to give me the limit as M goes to infinity of F M minus 1 times F M plus 1 minus F M Squared so that's what I give get by building the denominators in each of these terms and then I have F M times F M minus 1 but now let's look at that numerator and notice that numerator exactly the left-hand side of this Cassini's identity so we can replace this numerator with minus 1 to the M and that leaves us with the limit as M goes to infinity of minus 1 to the M over F M times F M minus 1 now the next thing to notice which isn't too hard to see is that the Fibonacci numbers grow without bound in other words the denominator is approaching infinity but the numerator is just alternating between 1 and negative 1 so that means this limit is 0 so in other words our limit does exist now we just have to find the value of that limit so let's go ahead and set L equal to the limit as M goes to infinity of F M plus 1 over F M again we had to know that the limit exists before we started setting it equal to something and now the next thing that we want to do is use our recursion so notice that this is the limit as M goes to infinity of FM plus FM minus 1 over FM now we can go ahead and split this into pieces so that's equal to the limit as M goes to infinity of 1 plus FM minus 1 over FM great but you can check that this limit will not be 0 and you can do that pretty easily by noticing that this number is always bigger than 1 in other words this ratio is always bigger than 1 which means the limit cannot be 0 which means we can reasonably talk about the limit of the reciprocal which is exactly what we have right here so notice that this is essentially the reciprocal of this starting term which means the limit is also going to be the reciprocal so we have this as 1 plus 1 over L so let's look at the extreme left and right hand side of the equation and notice we've got something that we can solve for L so I'll multiply both sides of that equation by L and that gives us L squared equals L plus 1 in other words it gives us L squared minus L minus 1 equals 0 so you can use the quadratic formula to solve this equation you'll get one positive answer and one negative answer you know that every term from the sequence is bigger than 1 so you only keep the positive solution but what you'll see is that the positive solution is equal to 1 plus the square root of 5 over 2 in other words the golden ratio and that finishes the proof of this and now we're ready for our main goal which is to evaluate this infinite product so we've got the product as n goes from 1 to infinity of the quantity 1 minus negative 1 to the N over FN plus 1 squared and so an infinite product can be seen as the limit of the partial products in other words this is the limit as capital n goes to infinity of this partial product in equals one to capital n of one minus negative one to the N over F sub n plus one squared great now what we want to do is mash these two things together again by finding a common denominator we will rewrite one as f n plus 1 squared over FN plus one squared so that's going to give us this limit as capital n goes to infinity now we have the product as little n goes from 1 to capital N and now we can rewrite the inside of this as FN plus 1 squared minus negative 1 to the N and all of that is over FN plus 1 squared great but notice this looks a lot like Cassini's identity over here where m is replaced with in plus 1 so just to be really careful let's go ahead and take this identity so if we set m equal to n plus 1 over here we'll get F sub n times F sub n plus 2 minus F sub n plus 1 squared equals minus 1 to the n plus 1 now let's go ahead and solve for a term that looks like this numerator and that will give us F sub n times F sub n plus 2 equals F sub n plus 1 squared plus minus 1 to the n plus 1 now that's not exactly the same but we can make it exactly this term if we factor a minus 1 out of that minus 1 to the n plus 1 and that's going to leave us with F sub n plus 1 squared minus negative 1 to the N great and now notice that this thing that I'm underlining in blue is exactly the same as this thing that I'm overlining in blue inside of our goal so that means I can replace the numerator of our goal with this term right here which is f sub n times F sub n plus 2 so let's go ahead and do that we can write this as the limit as capital n goes to infinity we have the product as little n goes from 1 to K Capital n and then we have F sub n times F sub n plus 2 in the numerator and then F sub n plus 1 squared in the denominator now let's go ahead and write out a bunch of terms from this partial product and see what we get so this is going to be the limit as n goes to infinity now that little N equals 1 term will give us F sub 1 F sub 3 in the numerator and then we have F sub 2 times F sub 2 in other words F sub 2 in the denominator so like I said that's the little N equals 1 term so let's see what we get for the little N equals 2 term so that's going to be F sub 2 times F sub 4 over F sub 3 times F sub 3 good and now let's look at the little N equals 3 term we have F sub 3 times F sub 5 over f sub 4 times F sub 4 and now I think we can see what's going on so let's maybe do one more the N equals 4 term so we have F sub 4 times F sub 6 over F sub 5 times F sub 5 and now let's go all the way up to let's do the little N equals capital n minus 1 term and then we'll do the little N equals capital n term for the last one so notice that's going to be F sub n minus 1 F sub n plus 1 all over F sub n times F sub n so that's what we get for the little N equals capital n minus 1 and then finally for the little N equals capital n term we get F sub n F sub n plus 2 over F sub n plus 1 times F sub n plus 1 but now notice we've got a telescoping action within this product so let's notice everything that cancels so we've got this F sub 2 cancels this F sub 2 then this F sub 3 cancels this EPS sub 3 this F sub 4 cancels this F sub 4 this F sub 3 and this F sub 3 and so on and so forth so notice everything is canceled up until we get here to the very end so this F sub n will be cancelled by something before this F sub M minus one will be canceled by something before this F sub n plus one is cancelled with this one this F sub n is cancelled with this one and so all we're left with is f sub 1 over F sub 2 both of those are 1 so that's just equal to 1 and then we have F sub n plus 2 over F sub n plus 1 in other words this boils down to the limit as n goes to infinity everything cancels except for F sub capital n plus 2 over F sub capital n plus 1 so now let's go ahead and bring that to the top and we'll summarize and finish it off now we're ready to finish it off in summarize so we had our goal infinite product which we wrote as the limit of a partial product we used Cassini's identity to rewrite that as this product of F sub n times F sub n plus 2 over F sub n plus 1 squared then we notice that that thing telescoped down to just the limit as capital n goes to infinity of F sub n plus 2 over F sub n plus 1 and now finally we can apply this limit of the ratio of consecutive Fibonacci numbers to see that this is equal to fie in other words the golden ratio 1 plus the square root of 5 over 2 so that means our goal infinite product is equal to the golden ratio and that finishes this problem and that's a good place to stop

Channel: Michael Penn

Views: 15,266

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: wyMFmx9mgcs

Length: 15min 2sec (902 seconds)

Published: Thu Jun 18 2020