# Need help checking my answers on derivatives and proof by induction

1. Jan 3, 2014

### sooyong94

1. The problem statement, all variables and given/known data
Hi, I'm new here, so pardon me for my mistakes. ;)
I need someone good at maths to check my answer:

Let y=x sinh x

(a) Show that d^2 y/dx^2 = x sinh x+2 cosh x, and find d^4 y/dx^4
(b) Write down a conjecture for d^2n y/dx^2n.
(c) Use induction to establish a formula for d^2n y/dx^2n

2. Relevant equations
Product rule and proof by induction

3. The attempt at a solution

(a) I used product rule and I have managed to prove the first part, and
d^4 y/dx^4 =x sinh x+4 cosh x

(b) The conjecture is d^2n y/dx^2n =x sinh x+ 2n cosh x

(c) *Sigh* This section is too long for me to type on... :(
https://www.dropbox.com/s/vq16c32kdra8xnt/Let y.pdf

2. Jan 3, 2014

### Curious3141

Hi, welcome to PF.

In induction, you start by *proving* the base case (n=1 in this example), and then *assuming* the case for $n=k$ then *proving* that the proposition holds for $n=k+1$, then finally *concluding* it holds for all positive integral n.

You shouldn't *assume* the base case, as you wrote ("It is assumed that this is true for n=1"). In this case, the base case has already been established based on the previous result in part a). Just quote that result.

Subsequently, you wrote "assume this is true for n=k+1". This is wrong. You should prove that the assumption that it holds for n=k implies that it holds for n=k+1. You actually did this, but writing the assumption before the proof is wrong. It is acceptable to start this off by stating: "We then have to show that $\displaystyle \frac{d^{2(k+1)}y}{dx^{2(k+1)}} = x\sinh x + 2(k+1)\cosh x$", so it is clear what you're hoping to end up with. But you shouldn't state it as an assumption before proving it. Semantics, but these are important in a proof.

A minor point: you did a bit of unnecessary work in your derivation by going through the odd derivatives of the function. E.g. to get the fourth derivative, you calculated the third derivative. This is not necessary. You can just apply the second derivative directly based on your previous result, getting $\displaystyle \frac{d^4y}{dx^4} = \frac{d^2}{dx^2}(\frac{d^2y}{dx^2}) = \frac{d^2}{dx^2}(x\sinh x + 2\cosh x) = x\sinh x + 2\cosh x + 2\cosh x = x\sinh x + 4\cosh x$. The same applies in your inductive step. Simplifies your working quite a bit.

3. Jan 3, 2014

### sooyong94

So how's the induction step look like?

4. Jan 3, 2014

### Curious3141

Outline of proof:

Show the base case holds true.

Inductive step:

Assume proposition is true for some n=k. That is, $\displaystyle \frac{d^{2k}y}{dx^{2k}}= x\sinh x + 2k\cosh x$. ---equation (1)

***We now need to prove that $\displaystyle \frac{d^{2(k+1)}y}{dx^{2(k+1)}}= x\sinh x + 2(k+1)\cosh x$.

Differentiate equation (1) twice wrt x:

$\displaystyle \frac{d^{2(k+1)}y}{dx^{2(k+1)}}= \frac{d^2}{dx^2}(x\sinh x + 2k\cosh x) = ...$.

Now your job is to fill in the "..." to make sure you get a result identical to what you wanted to show (stated in the "***" statement).

Note that the "***" statement is optional, but good for clarity. Otherwise you might lose track of where you're going, especially in more complicated induction proofs (this is a really simple one, though).

Finally, conclude by stating what you've proven. That is, "Hence $\displaystyle \frac{d^{2n}y}{dx^{2n}} = x\sinh x + 2n\cosh x$ for all positive integral n."

5. Jan 4, 2014

### sooyong94

So, in order to prove the base case, all I have to do is to plug n=1? :P

6. Jan 4, 2014

### Curious3141

Yes, but you already did it in part a). Just quote that result as having already been shown.

7. Jan 4, 2014

### sooyong94

So, algebraically, I didn't do any errors right? :P

8. Jan 4, 2014

### Curious3141

No. But it wasn't the quickest way, as I said. You could just have applied the double differential operator directly to the expression on the strength of your previously established result.

Let's represent the double derivative $\frac{d^2}{dx^2}$ as $D^2$. You've already established that $D^2(x\sinh x) = x\sinh x + 2\cosh x$.

Now, when working out $D^2(x\sinh x + f(x))$, where $f(x)$ is any function of $x$, it is easiest to split it up to $D^2(x\sinh x) + D^2f(x)$. You can immediately get the result for the first term from what you've established. The second term for the inductive step is trivial because it's just a constant ($2k$) multiplied by $\cosh x$. The hyperbolic trig functions are cool because they just "flip" from $\cosh$ to $\sinh$ and vice-versa with each derivative. So a double derivative would just get it back to where it started, i.e. $D^2(\cosh x) = \cosh x$. This way, you can more elegantly do your inductive step in c). As I mentioned before, this helps in b) as well, when you need to work out $D^4y$. You simply don't have to work out $D^3y$ as an intermediate step. Not wrong, but not necessary either.

9. Jan 5, 2014

### sooyong94

10. Jan 5, 2014

### Curious3141

I'm sorry, but it's not at all clear what you're *assuming* and what you're *showing*. That has to be very clear in a proof (as I said before).

First of all, the base case. The "therefore" symbol (∴) is misplaced because it is not a conclusion of anything that has gone on just before it. This trivial example is a legitimate use of the "therefore" notation:

$\sqrt{x} = 2$

$\therefore x = 2^2 = 4$

One mathematical statement leads to another. In intermediate steps, some people would use an "implies" notation ($\implies$), followed by the "therefore" symbol for a final conclusion.

The base case would make more sense if you flipped it around. "It's already been established in part a) that $D^2y = x\sinh x + 2\cosh x$. Therefore, the base case for n = 1 is proven". May seem like semantics, but as I said, semantics are critical in a valid proof.

For the inductive step, you need to be very clear that what you've stated initially is an *assumption*. Use the words "Assume that... for some n = k".

The next step is again unclear. You need to double-differentiate the equation in the assumption to get to $D^{2(k+1)}y$. It's not clear that you did this. No point just stating that it's true for n = k+1, that's what you have to *show*!

As I said, if you want to write out what the result you want to arrive at, it should be made clear that it's not already a proven result, just something you have to show. I've already given you all the words you can use. I've also said that part (stating "Now we need to show this..." etc.) is strictly optional. There should be no issue as long as you don't lose track of where you're going in the inductive step. The critical part is that you show (prove) by valid mathematical steps that you're going from your *assumption* for n = k to the *end result* for n = k+1.

Finally, I know you're new, but could you please learn LaTex and write your derivation in this format. Very tedious for me otherwise - can't quote you, and I end up having to write out all the stuff myself. Thanks.

Last edited: Jan 5, 2014
11. Jan 5, 2014

### sooyong94

Then does it look something like this?

12. Jan 5, 2014

### Curious3141

Suggestions/comments in red. The working you showed in the first post is acceptable, even though (as I said) it's not the fastest/most elegant way. But it's not acceptable to just jump to the statement for n=k+1 as you did in the latest post.

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13. Jan 5, 2014

### sooyong94

Sorry about that one... Forgot to edit that part... :P

14. Jan 5, 2014

### Curious3141

Yeah, I think that's good enough. Not the quickest way (since you're differentiating sequentially), but sufficient.

15. Jan 6, 2014

### sooyong94

Huge thanks then. :D

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