a bigger product rule!

[Music] here we're going to look at one of my favorite induction problems to give to students just after they've learned the principle of mathematical induction and what i like about this problem other than the fact that it's a great exercise for learning induction is that it ties back into calculus and generally when you're learning about induction you've just taken some calculus classes and so this has to do with something that i'll just call a bigger product rule so let's recall the standard product rule and that is the derivative of f of x times g of x so i'll denote that by f of x g of x prime so that equals f prime g plus f g prime so in other words we sum the product of f and g taking the derivative of one at a time you might say well is there a product rule for a second derivative and yes there is and you can in fact calculate it pretty easily just by taking the derivative of both sides of this first line i'll let you guys do all the details if you need to but what you end up with is f of x times g of x double prime is equal to f double prime g plus two f prime g prime plus g double prime so this looks mysteriously familiar let's compare that to maybe a well-known formula that you learn like maybe before calculus even and that is a plus b quantity squared equals a squared plus 2 a b plus b squared so what it looks like is happening here is that this expansion of the second derivative in terms of derivatives of the component functions is occurring just like squaring out a binomial and in fact that's what happens in general and that's sometimes called the leibniz rule and so how it goes is this if we take the nth derivative of the product f times g obviously the these need to be differentiable or something what we get is the sum as k goes from 0 to n of n choose k the kth derivative of f times the n minus kth derivative of g so that looks exactly like binomial expansion if we were to write out a plus b to the nth power it's just our exponents are being replaced by derivatives okay so let's prove this and like i alluded to earlier this is a good exercise to prove with induction although i bet there's some trick to prove this without induction if you maybe know a trick to use to prove this without induction maybe post it in the comments so that means we need to start with a base case now the base case should probably be n equals 1 because that's the most illustrative i think you can probably use n equals 0 at the base case but that doesn't really give you much motivation for what's going on but we don't even need to do that n equals 1 clearly is just the normal product rule which we are assuming to be true already okay so now let's make an induction hypothesis so let's suppose for some m bigger than or equal to one we have this statement is true for the nth derivative we'll just write that down just so that we have it so we have f of x times g of x the nth derivative which i'll write like that is equal to the sum as k goes from 0 up to m of m choose k the k derivative of x and then the m minus kth derivative of g okay so we're supposing that that is true and then we want to consider the m plus first derivative so let's write that down so consider like i said the m plus first derivative so i'll write that as f of x times g of x to the m plus one parentheses meaning that we're taking the m plus first derivative but notice the m plus first derivative is just the derivative of the nth derivative so i can write that as the derivative of the nth derivative like i just said but now we can apply our induction hypothesis to this thing which maybe i will square in red we can replace that with this thing here which i've also squared in red so let's see that's going to give us the derivative with respect to x of this sum as k goes from 0 up to m of m choose k and then we've got the kth derivative of x and then the m minus kth derivative of g okay then for our next step we'll use the fact that the derivative is a linear transformation on the space of differentiable functions so that means we can bring this derivative inside of the sum and in fact we can also bring it inside of this multiplication by the binomial coefficient but then we can apply this derivative to our product of the kth derivative of f and the m minus kth derivative of g and in fact that's really just doing the base case on that product instead of having f times g we've got this kth derivative of f and the m minus kth derivative of g so let's just like point out that we're doing that because it's a linear transformation bringing it inside the sum and then bringing it and applying it to those things we will use the base case so let's see that's going to give us this sum as k goes from 0 up to m of m choose k and now maybe in blue parentheses i'll write what happens to this guy right here when we take the derivative so we'll have the k plus first derivative of x and then the m minus k derivative of g plus the kth derivative of f and then the m plus first minus k derivative of g so that's just the standard product rule we take the derivative of that which brings this k to k plus one and then we take the derivative of that which brings that m minus k to m minus k plus one which i've just written in this other way okay so let's maybe bring a summary of the situation to the top and then we can finish off this inductive proof and then look at a little bit of a cheeky application okay on the last board we got to this point where we had the m plus first derivative of our product f times g as a sum of two terms but we can take that sum of two terms and break it into two sums and i've done that off the board that's really obviously true because we've got a finite sum here so we don't have to worry about convergence or anything like that okay so now keeping in mind where we're trying to go that should motivate how we should re-index this so notice this starts at the first derivative of f and it ends at the zeroth derivative of g so it's missing the zeroth derivative of f and then likewise it'll be missing the m plus first derivative of g just keeping in mind the formula that we're going for and then likewise this one right here is missing the m plus first derivative of f so let's see how that can motivate a change of index for each of these so for this term what i want to do is maybe take out the nth term so taking out the nth term we'll have the m plus first derivative of f and then the zeroth derivative of g which is just g of x so just let's notate that by taking out the nth term to go right here and then over on this branch we'll make a change of index so let's maybe change our index where we replace k with k minus one so that's going to change our starting point and our ending point so it'll change our starting point so that we start at k equals one and it'll change our ending point so that we end at m plus one but we won't worry about that because we've already taken that top point out so here that means we're going to stop at m keeping in mind that the m plus first term is over here now and now we're going to have m choose k minus 1 and then the k derivative of f and then the m plus first minus k derivative of g keeping in mind that we really have the m minus the quantity k minus 1 but that minus turns that into a plus so we've got exactly that right there okay and now let's play the same gain here but instead we'll bring out the zeroth term so let's bring out the k equals zero term and then for the rest of the terms here we do not need to re-index so let's notice that bringing out the zeroth term will leave us with m choose zero so that's just one i should have pointed out that this term gives us m choose m which is one and then we'll have just f of x with the zeroth derivative so that's just f of x and then we'll have the m plus first derivative of g like that okay and then let's see what we get for the rest of these terms nothing really changes except the starting point of our sum so this is going to go from k equals 1 up to m of m choose k the kth derivative of f and then the m plus first minus k derivative of g okay nice but now let's look at what we have what we have we have these two terms which i've squared in purple that are outside of the sums and then these which i'll square in green that are inside of the sun and they're inside of the sum with the same indexing and the derivative terms are the same so that means we can start putting this together so let's start with this guy right here so we've got the m plus first derivative of g of f and the zeroth derivative of g and then we'll have plus the sum as k goes from one up to m of m choose k minus 1 plus m choose k the kth derivative of f and then the m plus first minus kth derivative of g so again the indices on these sums are the same and then these derivatives of f and g match up so that's how we can combine all of those terms okay so now we'll be left with this last term f of x and then the m plus first derivative of g okay so now let's bring that up here and then we can finish this proof off so so far we have the following equality and now we just want to put this into the correct form so i'm going to take advantage of the fact that there exists some nice binomial coefficient identities but i won't prove them i'll let you guys look those up if you need to see proofs um or perhaps i'll make a video later where we prove these kind of things although i think i have some on the channel already so first of all there's a nice binomial coefficient identity involving those two so let's see that allows us to take this stuff in yellow and replace it with m plus one choose k okay then we can sneak something in here we can sneak in m plus one choose zero in here because anything choose zero is one and then likewise we can sneak in m plus one choose m plus one here because anything choose itself is one but now that allows us to slam all of these things together into a single sum this is the zeroth term of the sum this is the k term for the sum for k equals one to m and then that's the m plus first term of the sum so we're left with this sum as k goes from zero up to m plus one of m plus one choose k and then the kth derivative of f and then the m plus first minus k derivative of g and if we put this m plus 1 here in parentheses we can see that assuming our induction hypothesis we have proven the other case in other words we've proven that this equation holds okay so now to finish this video off i want to do like a sneaky little application of this proposition it's most definitely not the easiest way to prove the identity that we're going to get at but i think it's kind of cute okay so like i said we're going to finish this off with a neat little application of this so i'm going to start with taking the number 2 to the n and notice that that is exactly the same thing as 2 to the n times e to the 2x where we've evaluated that at x equals but then 2 to the n times e to the 2x is exactly the nth derivative of e to the 2x and then again we've evaluated this thing at x equals 0. okay but notice that e to the 2x is e to the x times e to the x so we can write this as e to the x times e to the x the nth derivative of that evaluated at x equals 0. then where can we go from here well we can apply this leibniz rule to our setup here where this is the function f and this is the function g so that's going to give us this sum as k goes from 0 up to n of n choose k and then we have the kth derivative of e to the x well that's just going to be e to the x and then the n minus kth derivative of e to the x that's just going to be another e to the x and then finally we need to evaluate this at x equals 0 because that was part of our setup but now evaluating that at x equals zero we'll just get one and that leaves us with this sum as k goes from zero up to n of n choose k and now looking at the extreme left and right hand side of this equation we see this nice binomial coefficient identity which can most definitely be proven a lot of other ways which are easier and that's a good place to stop

Channel: Michael Penn

Views: 6,848

Rating: 4.9380531 out of 5

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

Id: x4gLwWPhko4

Length: 16min 19sec (979 seconds)

Published: Tue Sep 14 2021