How to Prove the Nth Derivative of f(x)=1/(sqrt(1-4x)) by Induction?

[Sam223344]

Hey
I have this problem on proof by induction that I'm struggling to do.

The problem is to prove the nth derivative of f(x)=1/(sqrt(1-4x))

I have worked out that the nth derivative is f^(n)(x)=(2n)!/n! * (1-4*x)^(-(n+1/2))

But I'm not sure how to go about completing this. My lecturer only covered induction briefly and my textbook doesn't cover it either... Also I have only done basic sequence proofs so any help would be much appreciated.

 

[CompuChip]

Generally, when you have some statement like "for all n, X is true", a proof by induction consists of two steps. First, you have to show that for some simple case (usually n = 0 or 1, depending on the question), X is true. Then you assume that X is true for all integers n up to some given value n₀, and you prove that under that assumption, X is also true for n₀ + 1.

The reasoning is then as follows: you have checked by hand that it is true for n = 1. You have proven that if it is true for n₀ = 1 (which it is), then it is true for n = n₀ + 1 = 2. So it is true for n = 2. Also, you have shown that if it is true for n = 2, it is true for n = 3. Since it is true for n = 2, it holds for n = 3. Similarly, it is true for n = 4, and for n = 5, and so on.

Now let's apply this to the proof you have. So your claim is that for f(x) = 1/\sqrt{1 - 4x}, the nth derivative is given by
f^{(n)}(x) = \frac{(2n}!}{n!} (1 - 4x)^{-(n + \tfrac12)} \qquad\qquad (*)

So first, you check that (*) holds when n = 1 (and you might want to check n = 0 as well, just to be complete) and find that it works.
Now assume that (*) is true, and show that it holds for n + 1. That is: derive (*) with respect to x, and show that you get the same expression as you'd get when plugging in n + 1 instead of n in (*).

PS Note that proof by induction is a convenient way (once you are used to it) to prove such statements "for all n", but it doesn't help you to find the statement. So if instead of "prove that the nth derivative is ...formula..." you get "derive and prove a formula for the nth derivative", you will first need to come up with a hypothesis by some other way. Once you have the hypothesis, you can make it into a theorem and try to prove it by induction.

 

[Sam223344]

I'm still a bit fuzzy on how to go about this induction. Could someone do an example of proving an nth derivative(something of similar difficulty) by induction so I can get a better understanding of my problem.

Or even post a link to where i could find some examples.

Thanks

 

[CompuChip]

I don't know if it is of similar difficulty, but here is one example:

The 2nth derivative of f(x) = cos(x) sin(x) is given by (-4)ⁿ cos(x) sin(x).

Actually this is easy to show once you see that f(x) = 1/2 sin(2x), but for instructive purposes let's do it by induction.

Proof.
First we check the base case. For n = 0 it is true, but then again, this is not really a derivative. So let's take n = 1. We need to show that the second derivative of f(x) is -4 cos(x) sin(x). We apply the product rule:
f⁽¹⁾(x) = cos(x)² - sin(x)².
Then by the chain rule (or product rule, whichever you prefer):
f⁽²⁾(x) = - 2 cos(x) sin(x) - 2 sin(x) cos(x) = - 4 cos(x) sin(x).

So the base case is checked. Now we pose the induction hypothesis, namely:
(IH:) Suppose that for all n checked so far, f⁽²ⁿ⁾(x) = (-4)ⁿ cos(x) sin(x).

We must prove that the identity is true for n + 1. So let's differentiate f(x) (2n + 1) times. That is, let's differentiate f⁽²ⁿ⁾(x) twice. By the induction hypothesis,
f⁽²ⁿ⁾(x) = (-4)ⁿ cos(x) sin(x).
Using the product rule,
f^{(2n + 1)}(x) = (-4)ⁿ cos(x)² + (-4)ⁿ sin(x)².
And differentiating again,
f^{(2n + 2)}(x) = f^{(2(n + 1))}(x) = (-4)ⁿ * (-2) cos(x) sin(x) + (-4)ⁿ * (-2) sin(x) cos(x) = (-4)ⁿ * (-4) cos(x) sin(x) = (-4)^{n + 1} cos(x) sin(x).

This concludes the proof (now explain to yourself / us, why).

 

[Sam223344]

Thanks mate this should help me out a lot.