Learn AAHL 1.15.2 Proof by induction (differentiation) in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 1.15—Proof by induction, contradiction, counterexamples.
Learn AAHL 1.15.2 Proof by induction (differentiation) in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 1.15—Proof by induction, contradiction, counterexamples.
The video discusses the process of proving the nth derivative of a function using mathematical induction. The proposition states that the nth derivative of the function can be expressed as:
The proof is structured in four steps:
The video emphasizes the importance of clearly stating each step in the proof to maintain clarity and coherence throughout the process.
00:00Hi guys, okay, so the
00:02next proof prediction I'm gonna
00:03do is Differentiation, so these
00:06are actually my favorite ones
00:07probably because I like cactus,
00:10but I Think these these
00:14this makes a lot of
00:15sense what it does to
00:16me anyway So we're given
00:17this is actually a past
00:18paper question. We're given some
00:20function f of x And
00:22it says the nth derivative
00:23is f and a vector
00:25
00:27says show that proof
00:28with mathematical induction, it has
00:32to say prove by mathematical
00:33induction, but they almost always
00:35do say, prove by mathematical
00:36induction. So that the nth
00:38derivative is given by this.
00:40So that means if I
00:40was to differentiate this thing,
00:4410 times the derivative would
00:46be 2 to the power
00:46of 10 times x plus
00:4910 times 2 to the
00:51power of 9, all times
00:54e to the 2x.
00:56which is interesting in itself
01:00but we need to kind
01:01of prove that. So let's
01:02go through our routine. So
01:05our proposition, proposition is that
01:10the ant derivative, ant derivative
01:13is equal to 2 to
01:17the power of nx plus
01:19n times 2 to the
01:21power of n minus 1,
01:23all times e to the
01:24power of n
01:24part of 2x. Okay, so
01:26we're gonna, we're gonna show
01:27this step one, step one,
01:33prove true for n equals
01:40one. Okay, now to prove
01:44true for n equals one,
01:45we need to get, so
01:47f of x is equal
01:51to
01:52f of x is equal
01:55to x e to the
02:002x. Now to get f
02:01dash of x, to get
02:03f 1 are the first
02:05derivative of x. Well, here
02:08we just need to use
02:08a product rule. So f
02:09dash of x is equal
02:11to, I'm going to do
02:13this quickly guys, but don't
02:14you do this, do it
02:15to the side, but I'm
02:16going to do it in
02:16my head. U times dv
02:19dx.
02:20So u times dv dx,
02:23the derivative of this is
02:242 e to the 2x.
02:26So that's u times dv
02:27dx plus v times dv
02:29dx, which is e to
02:31the 2x. And this is
02:35equal to, um, e to
02:41the 2x times 2x plus
02:461. So this is the
02:47first derivative
02:48this is the first derivative.
02:53And then if we get,
02:56let me write here from
03:01proposition, from proposition f one
03:08of x. This is using
03:11their thing here. f one
03:14of x is the first
03:14derivative of x is equal
03:16to
03:162 to the power of
03:181 x plus 1 times
03:222 to the power of
03:241 minus 1 times, close
03:31this bracket, and I just
03:32want to bring this over,
03:34so I have space times
03:37e to the 2 x,
03:40which is 2 x plus
03:44One times two to the
03:45power of zero is one,
03:46one times one is one.
03:48This is all in the
03:49bracket. E to the two
03:50X and look this. When
03:54I sub in one there,
03:55I do get the first
03:57derivative. So I'm gonna say
03:58therefore, therefore true for n
04:03equals one. Step two. Step
04:07two. Assume, so again, this
04:10is always the same. Assume,
04:12true for n equals k.
04:19So the kth derivative if
04:21you like, the kth derivative
04:22of k of x is
04:25equal to 2 to the
04:27power of kx plus k
04:31times 2 to the power
04:32of k minus 1 times
04:35e to the power of
04:372x. So the kth derivative
04:39is this
04:40We've assumed that. Next step.
04:45Step three. Prove. Prove true
04:52for n equals k plus
04:56one. Right. This again, I
04:59want to say what are
05:00we trying to show? So
05:01trying. I don't get right
05:04in trying. I'm trying to.
05:08prove that f of k
05:16plus 1 is the k
05:17plus 1 derivative. I can
05:19say that is equal to.
05:21So this time I'm going
05:22to put k plus 1
05:24in here instead of k.
05:25So it's 2 to the
05:26power of k plus 1,
05:282 to the power of
05:29k plus 1 x plus
05:31k plus 1 times
05:362 to the power of
05:39k plus 1 minus 1,
05:41which is 2 to the
05:41power of k. Close bracket,
05:46e to the 2x. So
05:48that's what I'm trying to
05:49show. So for this step,
05:56remember, for step three, you
05:58always use step two. You
06:00use the step that you
06:03have your assumption step.
06:04So you assume this. So
06:07f, let's write this out
06:09again. The kth derivative of
06:12x is equal to 2kx
06:16plus k2k minus 1 e
06:20to the 2x. So this
06:24is the kth derivative. So
06:26how do I get, if
06:27this is the kth derivative,
06:28how do I get the
06:29k plus 1th derivative? So
06:31we're assuming this is true,
06:32this is the
06:33So to get the cables
06:35one derivative, the next derivative,
06:37I just need to differentiate
06:39this. So if I differentiate
06:42this, I should get that.
06:45And then I've, then I
06:46have proven it. So this
06:49I got from sub -in
06:50-k plus one into my
06:52proposition. I sub -in -in
06:54here instead of n. And
06:57this one, I'm going to
06:58get what I have got
06:59from, what this I've
07:01got from the assumption, but
07:03when I differentiate this, I
07:04should get this. And if
07:06I get it, I've proven
07:07it. Okay, so let's do
07:10it. I'm going to say
07:12the K plus one derivative
07:19is equal to U times
07:24dv, dx, this one I
07:25actually am going to write
07:26it out. I'm going to
07:27write it out properly over
07:29using
07:29the product row. So u
07:31is equal to 2 to
07:34the k x plus k
07:37times 2 to the k
07:39minus 1. And v is
07:46equal to e to the
07:472x. d u dx is
07:50equal to 2 to the
07:52k because this is just
07:53a constant, which is k.
07:55And d v dx
07:57is equal to 2e to
07:59the 2x chain rule. So
08:02the k plus 1 derivative
08:03is u times u times
08:07u times u times dv
08:14dx, I think you've actually
08:15let's bring this up, let's
08:20bring this up here and
08:23this down.
08:25Here. So it's u times
08:29dv dx to eat the
08:322x plus v times v
08:36times d u dx to
08:40the k. Now, at this
08:42point I look at this,
08:44right? One more, actually, what
08:45am I trying to show?
08:47So there's an e to
08:48the 2x outside. So yeah,
08:50I'm going to have that.
08:51Let's bring that outside. And
08:53then
08:53Right, look, I think maybe
08:55we can see something, right?
08:58So I'll take out my
08:59e to the 2x, I'll
09:00eat the 2x and inside
09:04this square bracket I will
09:07put, there's a 2 here,
09:10so it's 2 times 2
09:14times 2 to the kx
09:18plus k times 2 to
09:21the k
09:21plus 2 to the k.
09:28So it's not clear I've
09:28taken out this two, I've
09:30taken it out here. This
09:31is just this and this
09:33two k is here and
09:34the e to the 2x
09:34is out there, right? So
09:35maybe, just make sure you
09:39know how I've got from
09:39there to there. Okay here
09:41now I have my e
09:41to the 2x outside. Let's
09:44multiply this out. So I've
09:45two times 2 times 2
09:49to the k
09:49x, right? Two times two
09:53to the k, this is
09:54two to the one if
09:55you like. Two to the
09:57one times two to the
09:58k is actually two to
10:00the k plus one. x.
10:04And even better, even better.
10:11Two times two to the
10:13k minus one is just
10:14going to be two to
10:15the k because one
10:17one plus k minus one
10:19is just k. So this
10:21becomes k times 2k times
10:262 to the k. And
10:30then I have this guy
10:32here, which is plus 2
10:35to the k. Okay, that's
10:39not of k's. Nearly there,
10:43I see this is working
10:44out now. Look, I have
10:45my e to the 2x,
10:48I have my 2 to
10:51the k plus 1x. So
10:54here's my e to the
10:552x. Here's my 2k plus
10:581x. And then here, what
11:01factors is, I can take
11:03out the 2 to the
11:04k, and I'm left with
11:06k plus 1 times 2
11:11to the k, close bracket.
11:13This is what I was
11:15trying to prove. And this
11:17is why, again, why I
11:20like to write this, I
11:21like to write this step
11:23here, this step here, because
11:25when you have that, you
11:27know what you're working towards,
11:28you're working towards that, you're
11:29trying to turn this into
11:30this. If you don't write
11:32that down, I find it
11:33far more difficult. Okay, so
11:36done. That's it. But am
11:39I done? No, I have
11:41to write my step.
11:41step four, you have to
11:44write step four. Now, guess
11:45what I'm gonna do, guys?
11:47Obviously, you don't have the
11:47luxury of doing this, but
11:49I'm gonna go to my
11:51previous example. I'm going to
11:55copy it. I'm gonna paste
11:58it. So, I'm doing this
12:00for two reasons. One, I
12:03don't have to write it
12:04out again. That's quite quick.
12:05But two, to show you
12:07that it's literally the exact
12:09same thing.
12:09I don't need to change
12:14it for the different types
12:21of proof -brand -duction. It's
12:24literally the exact same thing.
12:25The exact same statement. True
12:26-friend equals 1. If true
12:28-friend equals k, then true
12:29-friend equals k equals 1.
12:31Therefore, by the principle, my
12:32data -modern -conduction, true for
12:33all n is an element
12:34of the positive integers.
12:37That's differentiation. I need to
12:40do divisibility. And actually there
12:42was one other one I
12:43think that I didn't mention
12:44that I'm going to do,
12:45which is inequality. So see
12:47you in those next two
12:48lessons.