# Calculate y^(p+q) using the Leibniz formula

## Homework Statement

Statement:[/B] Let y = u(x)v(x).
a) Find y' , y'', and y'''
b) The general formula for yn, the n-th derivative, is called Leibniz’ formula: it uses the same coefficients as the binomial theorem , and looks like https://i.gyazo.com/53728964c6b3ef142fd70f600c29e037.png
Use this to check your answers in part (a), and use it to calculate y(p+q) , if y = xp(1 + x)q

## Homework Equations

Supposedly, only the Leibniz formula https://i.gyazo.com/53728964c6b3ef142fd70f600c29e037.png

## The Attempt at a Solution

I have done a) correctly and I believe the fist part of the b) statement aswell, but I can't figure out how to calculate the new y.
I'm not even in college yet, this is the first time I ever see this 'Leibniz formula', so, the only thing I have done, is naming
u=xp and v=(1+x)q, because I feel that I can do something with y=uv. But so far I can't come up with any ideas.

## Answers and Replies

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andrewkirk
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It'll be easier if we write the Leibniz formula in summation form as
$$(uv)^{(n)}=\sum_{k=0}^n\left(\substack{n\\k}\right)u^{(n-k)}v^{(k)}$$
To use this we need to get expressions for the last two factors inside the sum.

Do you think you can get an expression for $u^{(n-k)}$, given that $u=x^p$? If you do the first few differentiations of $u$ you'll see a pattern that you can then write as a general formula.

Do you mean I need to differentiate u (n-k) times? If so, I can make an expression like:
D(n-k)un=(n(n-1)(n-2)...k)⋅uk.
Which I don't think it's what you meant.
If I differentiate u alone, It would go like: D(xp)=px(p-1)
So Dp(u(p))=(p(p-1)(p-2)...1)⋅x(p-p)=p!
I think the problem relies in the fact that I'm not really sure what I'm looking for here...

andrewkirk
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Do you mean I need to differentiate u (n-k) times? If so, I can make an expression like:
D(n-k)un=(n(n-1)(n-2)...k)⋅uk.
Which I don't think it's what you meant.
Actually, that's exactly what I meant.
You're almost there. You just need to substitute that right-hand side into the formula, then do the same sort of thing for $v^{(k)}$.

What we are seeking to obtain is a formula for $y^{(n)}$ that uses only variable names $x,p,q$.
By the way, in the OP, where it says "The general formula for $y^n$ ", I think it should have written $y^{(n)}$.

Trying to get an expression for v(k), and being v=(1+x)q ,I get: v(k)=(q(q-1)(q-2)...(q-k+1))(1+x)(q-k).
Which I believe is wrong, mostly because you said we are seeking for a formula with only x,p and q as variables.
If I plug the expressions I'm getting so far into the formula, it looks messy and with the variable k everywhere.
Maybe I'm just trying to do something way too advanced to me..

andrewkirk
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Trying to get an expression for v(k), and being v=(1+x)q ,I get: v(k)=(q(q-1)(q-2)...(q-k+1))(1+x)(q-k).
Which I believe is wrong, mostly because you said we are seeking for a formula with only x,p and q as variables.
If I plug the expressions I'm getting so far into the formula, it looks messy and with the variable k everywhere.
Sorry, I should have written 'a formula with only x,p and q as unbound variables'. A bound variable is a variable that is used as index for something like a sum, as in $\sum_{k=0}^n$. So the k is a bound variable in your formula, and that's fine.
Maybe I'm just trying to do something way too advanced to me..
Not at all. You're doing just fine. Write out the expression you get when making those two substitutions. There may be some cancelling and general simplification possible. Then see what's left.

But before you do that, there's a subtlety that needs to be dealt with. That is that $D^{(k)}(1+x)^q$ is equal to zero if $k>q$. So we can only write
$$v^{(k)}=q(q-1)(q-2)...(q-k+1)(1+x)^{q-k}\equiv \frac{q!}{(q-k)!}(1+x)^{q-k}$$
if $k\leq q$, and otherwise it is zero.
So instead define a function $f$ of two integer variables such that $f(a,b)=\frac{a!}{(a-b)!}$ if $a\geq b$ and is otherwise zero. You can then use that function to write general formulas for $v^{(k)}$ and $u^{(p+q-k)}$ and to then write the overall formula.

That's an interesting point, I didn't realize the k>q condition.
So we have: y(p+q)=(n(n-1)...k)u(k)·[q!/(q-k)!]·(1+x)(q-k)
I don't know how to put (n(n-1)...k) in a simplified way, or (q-k)!.
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I have a different approach today that might be better
If v=(1+x)q, v(q)=q!, and v(q+1)=0.
The same way: If u=xp, u(p)=p!, and u(p+1)=0.
y(p+q)=u(p+q)v + uv(p+q) + A·u(p)·v(q).
The first terms are equal to 0, so I have left y(p+q)= A·p!q!
I named that A because I don't know what it is.
Maybe A should be a function of two integer variables as you said in the last post?
Sorry about my ignorance on factorial calculations, in Spain we don't study them before college.

Which approach do you think I should follow?

andrewkirk
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Your different approach is good. The challenge is to work out what A is. We can do that through the formula we had earlier:
$$(uv)^{(p+q)}=\sum_{k=0}^{p+q}\left(\substack{p+q\\k}\right)u^{(p+q-k)}v^{(k)}$$
I have replaced the $n$ by $p+q$ here, as that is the degree of differentiation required.

Now, given that $u=x^p$ and $v=(1+x)^q$ we know that differentiaing $u$ more than $p$ times gives 0, and so does differentiating $v$ more than $q$ times. Whenever either of $u^{(p+q-k)}$ or $v^{(k)}$ is zero, that term of the sum above will be zero. That enables us to put some quite tight constraints around what values of $k$ in that sum will have nonzero terms.

What are those constraints?

For u(p+q-k), k≥q, otherwise it's zero.
For v(k), k≤p, otherwise it's zero.
Therefore, we have this: q≤k≤p.
What's the goal of setting this constraints or limitations to the values of k?
I thought that the terms between parenthesis after the Σ were a matrix, now I think I was wrong, what is that factor called? I think that that's one of the things that was giving much such trouble, being wrong about the notation.

andrewkirk
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For u(p+q-k), k≥q, otherwise it's zero.
For v(k), k≤p, otherwise it's zero.
The first is correct. The second is not. If you correct the second one, you will get a tighter and more helpful set of constraints. Once you have that, the goal should, I hope, become apparent, as it will allow you to use the idea you raise in post 7.

$\left(\substack{p+q\\k}\right)$ is not a matrix here. It means the number of Combinations of $k$ objects selected out of a set of $p+q$, without regard to the order of selection. It is also sometimes written ${}^{p+q}C_k$ or $C^{p+q}_k$. It is equal to
$$\frac{(p+q)!}{(p+q-k)!k!}$$

How is the second constraint not correct?
If I differentiate v more than q times, I get zero. Therefore, for v(k)not to be zero, k must be smaller than q, or equal, I used the same logic procedure for the first constraint.
Shouldn't it be more appropriate to have a combination like C(p+q, q)? This way I'd have an expression like
(p+q)!/(p+q-q)q!=(p+q)!/p!q!
And ultimately, y(p+q)=(p+q)!/p!q! · p!q! = (p+q)!

By the way, sorry about how I write down the formulas, I don't know how to show them like you do.

andrewkirk
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How is the second constraint not correct?
If I differentiate v more than q times, I get zero. Therefore, for v(k)not to be zero, k must be smaller than q, or equal, I used the same logic procedure for the first constraint.
That is all correct. You have concluded $k\leq q$. What was incorrect was what you wrote above in post #9, which was $k\leq p$. Perhaps that was just a typo, typing p instead of q.

Now you can put together the two inequalities you have worked out about $k$ and $q$ and replace them with an equality, which will allow you to greatly simplify things.
By the way, sorry about how I write down the formulas, I don't know how to show them like you do.
They are written using a system called latex. The post here tells how to use it. Latex is used for typesetting mathematics all around the world in universities and mathematical professions, so it is very much worth learning if you think you will be using maths a lot in the future. The formulas you have written are fine by the way. They are easy to read. But once you know latex, they can be written much more quickly using that, as well as looking neater.

I've been out for a few days, but now it's done, thanks for taking your time to help me!