An inverse tangent Fibonacci sum!

in this video we're gonna look at this sum that involves reciprocals of Fibonacci numbers and the inverse tangent function and we're gonna get a pretty surprising result so let's just recall how the Fibonacci numbers are defined so we've got the 0th of the nachi number is zero the first is 1 and then we have this two-step recursion so FN equals FN minus 1 plus FN minus 2 and that's where all n bigger than or equal to 2 and so our goal is this sum from N equals 1 to infinity of the inverse tangent of 1 over the odd Fibonacci number so F sub 2 n plus 1 so notice since we start at 1 that's going to be 1 over f3 1 over F 5 and so on and so forth so we're gonna make use of 2 lemmas to prove this one is called Cassini's identity and it says that if you square the beneath Fibonacci number and subtract the product of the two on either side of that square you get minus 1 to the N minus 1 and then the second one is going to be this fact that if you take the arctan of 1 over F sub 2 n plus 1 you can split it into the arctan of 1 over F sub 2 and plus 2 minus the arctan of 1 over F sub 2 in so we're gonna prove those two limits and then deriving this sum will be pretty simple ok so let's do the proof of lemma 1 first okay so we're gonna start the proof of this first lemma by by proving the following claim so we're gonna claim that if you take the following 2x2 matrix 1 1 1 0 red column wise and raise it to the nth power you'll get this matrix of Fibonacci numbers given as follows so FN plus 1 FN FN and FN minus 1 so we'll do the proof of this claim by induction and so let's see what our base case would be notice N equals 0 doesn't make so much sense for a base case because then we would have to take the negative first Fibonacci number and yes you can't extend the Fibonacci two numbers two negatives but that's not what we're going to do here so that means our base case let's take n equals one and now notice the right-hand side in that case will become F 2 F 1 F 1 F 0 but now notice that's exactly the matrix 1 1 1 0 to the first power so in other words this formula is true for the base case for N equals 1 now let's make our induction hypothesis so let's suppose this is true for N equals K in other words if we raise this to the K power we get F sub k plus 1 FK FK and FK minus 1 and now let's see what we need to do from there so we need to make sure it's true for the k plus first power so we'll do 1 1 1 0 to the k plus first power so that's going to be 1 1 1 0 times 1 1 1 0 to the K power but then our induction hypothesis allows us to rewrite this using Fibonacci numbers so that's going to be 1 1 1 0 times F sub k plus 1 F sub K F sub K F sub K minus 1 ok good but now let's just do a plain old matrix multiplication here and that's going to give us F k plus 1 plus F K and the upper left entry that's going to give us F k + FK minus 1 in the upper right entry that's going to give us F k plus 1 in the lower left entry and then finally that's going to give us FK in the lower right entry ok but now notice by our defining recursion on the Fibonacci numbers this guy is f sub k plus 2 this guy is f sub k plus 1 so we can rewrite this whole matrix as FK plus 2 FK plus 2 sorry FK plus one goes here fk+ one goes here and then FK goes here which means this statement is true for the k plus first term and we have proven this claim by an induction okay so now what I'll do next is clean up the board and then we'll show how this claim implies Cassini's identity okay so we just got done proving this claim now we're gonna see how that implies this formula down here which is called Cassini's identity so it's as simple as just taking the determinant of both sides of this equation so notice the determinant of the left-hand side of this equation is equal to the determinant of this matrix to the nth power so and that's just from linear algebra so in other words it's going to be minus 1 to the nth power because notice the determinant of this matrix is 1 times 0 minus 1 times 1 in other words it's minus 1 but then this matrix is raised to the nth power so we get minus 1 to the N and now let's notice that over here the determinant of this matrix is FN plus 1 times FN minus 1 minus F N squared so now that is exactly this identity when you change the sign on both sides okay good so now we've established Cassini's identity now we can get to proving the next lemma okay so I'll clean up the board and we'll get to it ok so now we're ready to do the proof of this second lemma so what I'll do is I'll start with the right hand side of this equation I'll take the tangent of it and show that that equals the inside of the inverse tangent of the left hand side of these this equation and that will show that this equation is indeed true so in other words I'm going to set theta equal to the arc tan of 1 over F 2n minus the arc tan of 1 over F 2 n plus 2 and then our goal is to calculate the tangent of and show that that equals 1 over F 2 n plus 1 so that's our goal here and that will show that this equation is true okay good so now what I'm going to use is the angle difference formula for tangent so let's recall that real quick so let's recall that the tangent of a minus B equals this quotient tan a minus tan B over 1 plus tan a times tan B so that's what we'll use and we'll use that where a is equal to this term and then B is equal to this term okay good so now if we take the tangent of theta notice the tangent and the inverse tangent will cancel and so that's going to give us 1 over F 2 n minus 1 over F 2 n plus 2 over 1 plus this product so 1 over F 2 N and then F 2 n plus 2 ok great so the next thing that we're going to do is take this quotient multiply the numerator and the denominator by the same thing and so we'll multiply by F 2 n F 2 n plus 2 upstairs and downstairs ok good so let's see what that gives us so in the numerator that's going to give us F 2 n plus 2 minus F 2 n and then in the denominator that's going to give us F 2 n times F 2 n plus 2 and then plus 1 and now we can notice two things first of all we can notice that this numerator is exactly equal to F 2 n plus 1 because of the defining Fibonacci recursion so these two are equal and you can see that because if you add F sub 2 into both sides you get exactly this recursion so this is just a different version of that recursion and then we can also notice that the denominator of this using Cassini's identity is equal to f sub 2 n plus 1 squared which means this cancels down to exactly what we want it to 1 over F 2 n plus 1 so now if we take the inverse tangent of both sides we'll get theta equals the arc tangent of that so let's write that down so taking the inverse tangent of both sides we'll get theta which is our expression equals arc tan of 1 over F 2 n plus 1 which is exactly what we wanted okay so that's the end of the proof of this second lemma ok so I'll clean up the board and then we're ready to finish the whole thing off ok so we've proven our two limits as tools and now we're ready to evaluate this sum so we're going to do this by taking the limit of partial sums so in other words we're going to take the limit as K goes to infinity as a sum from N equals 1 to K of arctan of 1 over F sub 2 n plus 1 so the first thing that we'll do is we'll use this second limit down here to split that 1 over F 2 n plus 1 inverse tangent term into two so notice that's going to give us the limit as K approaches infinity of the sum N equals 1 to K of now we have the arc tan of 1 over F 2n minus the arc tan of 1 over F 2 n plus 2 ok great now the next thing we're going to do is split this into two sums so that's going to give us the limit as K approaches infinity of now we have the sum N equals 1 to K of arc tan of 1 over F 2n minus the sum N equals 1 to K of arc tan of 1 over F 2 n plus 2 okay good the next thing that we're going to do is take the first term out of this first sum the last term out of the last sum so let's see what that gives us so we have the limit as K goes to infinity of so we'll have arc tan of 1 over f2 so that's the very first term plus the sum N equals 2 to K of arc tan of 1 over F 2 n ok good minus the sum N equals 1 to K minus 1 of arc tan of 1 over F 2 n plus 2 and then minus the last term being taken out which is going to be the arc tan of 1 over F 2 k plus 2 okay great now from here we can reindex so we can either re-index this one or we can reindex that one it doesn't really matter one way or the other so maybe we'll reindex this one so let's send in 2n plus 1 and notice that's going to change this to start at 1 and end at K minus 1 and that's going to change this to be 2n plus 2 okay but the magic is that now this sum and this sum cancel great and that leaves us with the following expression so now we have the limit as K goes to infinity of arctan but we know F sub 2 is 1 so we have arc tan of 1 minus arc tan of 1 over F sub 2 k plus 2 ok now we're almost done now as K goes to infinity the inside of this inverse tangent goes to 0 but we know the inverse tangent of 0 is 0 so all of this cancels down to 0 and we're with just the inverse tangent of 1 which is equal to PI over 4 and that's our final answer

Channel: Michael Penn

Views: 3,094

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn, fibonacci numbers, combinatorics, fibonacci, arctan, series, infinite series

Id: pn0J0p0R_GM

Length: 13min 26sec (806 seconds)

Published: Sun Dec 08 2019