How to merge the sum and ##x^n##?

[MevsEinstein]

TL;DR

What the title says

How do I merge $x^n + \displaystyle\sum^x_{k=1} \frac{d^k}{d^kx} \frac{x^ny^k}{k}$? I tried changing the denominator of the summand to $k+1$ and make $k$ go from zero, but I had to divide by zero when k equaled one.

 

[LCSphysicist]

Summary: What the title says

How do I merge $x^n + \displaystyle\sum^x_{k=1} \frac{d^k}{d^kx} \frac{x^ny^k}{k}$? I tried changing the denominator of the summand to $k+1$ and make $k$ go from zero, but I had to divide by zero when k equaled one.

Is the upper bound index of your sum right? With to merge both terms, you mean to write everything as a unique sum?

 

[Mark44]

I'm not able to guess what you mean by this expression -- $x^n + \displaystyle\sum^x_{k=1} \frac{d^k}{d^kx} \frac{x^ny^k}{k}$. The upper limit of the summation should not be x, and it should be something other than n.

Also, and this is minor, your derivative is not formed correctly. The usual notation for the k-th derivative operator with respect to x is $\frac {d^k}{dx^k}$.

Once you get the summation written correctly, the $x^n$ term outside the summation would get added to the corresponding term(s) of the summation. You'll probably need to expand the summation to get the addition right. Sums like this show up in power series solutions of differential equations.

 

[mfb]

Assuming your sum goes from 1 to n: While it's possible to write this as $\displaystyle \sum_{k=0}^n ...$ it will look ugly. What's the point of this transformation?

 

[MevsEinstein]

Hold up the upper bound is $n$. I couldn't edit though.

 

[MevsEinstein]

you mean to write everything as a unique sum?

I wanted to get rid of $x^n$ and put something in replacement inside the summand of the sum so that the expression is the same. And my upper bound was supposed to be $n$, sorry for the misconvenience. I couldn't edit for some reason.

 

[mfb]

Ah, I missed that the y^k term disappears on its own for k=0 in my previous post. Then it's only mildly ugly.
$$x^n + \sum^n_{k=1} \frac{d^k}{dx^k} \frac{x^ny^k}{k} = \sum^n_{k=0} \frac{d^k}{dx^k} \frac{x^ny^k}{|k-1/2|+1/2}$$

The denominator can be replaced by max(1,k).

 

[Mark44]

$$x^n + \sum^n_{k=1} \frac{d^k}{d^kx} \frac{x^ny^k}{k} = \sum^n_{k=0} \frac{d^k}{d^kx} \frac{x^ny^k}{|k-1/2|+1/2}$$

Minor nit: the derivatives in the above should be

$$x^n + \sum^n_{k=1} \frac{d^k}{dx^k} \frac{x^ny^k}{k} = \sum^n_{k=0} \frac{d^k}{dx^k} \frac{x^ny^k}{|k-1/2|+1/2}$$

 

[pbuk]

But
$$ x^n + \sum^n_{k=1} \frac{d^k}{dx^k} \frac{x^ny^k}{k} $$
is simpler than
$$ \sum^n_{k=0} \frac{d^k}{dx^k} \frac{x^ny^k}{|k-1/2|+1/2} $$
so why not stick with that?

Anyway, what is $y$ ?

 

[MevsEinstein]

But
$$ x^n + \sum^n_{k=1} \frac{d^k}{dx^k} \frac{x^ny^k}{k} $$
is simpler than
$$ \sum^n_{k=0} \frac{d^k}{dx^k} \frac{x^ny^k}{|k-1/2|+1/2} $$
so why not stick with that?

Anyway, what is $y$ ?

Actually let's just stick with that. The expression is equal to [math](x+y)^n[/math]. Turns out that this formula is harder to use than the one we have now.

 

[mfb]

Minor nit: the derivatives in the above should be

Copy&paste error, fixed.

@pbuk: That's my point, the original expression is easier.

The expression is equal to \(\displaystyle (x+y)^n\).

Only if you make it a k! in the denominator.