# Summation+Differentiation=Disaster. I with a summation problem I'm having.

PEZenfuego
My logic is flawed somewhere, but I can't figure out where or why.

So I've been playing with summation a bit and figured out a way to make equations for Ʃ$^{n}_{k=1}$K and Ʃ$^{n}_{k=1}$K$^{2}$ That looks odd, so I'll just use Ʃ from now on, but realize that it is always from k=1 to n.

ƩK is a series like 1+2+3+4...+(n-2)+(n-1)+n

and Ʃk$^{2}$ is a series like1$^{2}$+2$^{2}$+3$^{2}$+4$^{2}$...+(n-2)$^{2}$+(n-1)$^{2}$+n$^{2}$

Anyway, the equation for Ʃk$^{2}$ is (2n$^{3}$+3n$^{2}$+n)/6

and the one for ƩK is (n$^{2}$+n)/2

Now here is what I'm stepping in:

Ʃk$^{2}$=1$^{2}$+2$^{2}$+3$^{2}$+4$^{2}$...

So the derivative of this with respect to n would be

dƩk$^{2}$/dn=2(1)+2(2)+2(3)+2(4)...

another way to write this would be

dƩk$^{2}$/dn=2(1+2+3+4...)

Or...

dƩk$^{2}$/dn=2(ƩK)

So since Ʃk$^{2}$ is (2n$^{3}$+3n$^{2}$+n)/6 it stands to reason that the derivative of this equation would be equal to 2(ƩK), but it isn't.

Ʃk$^{2}$=(2n$^{3}$+3n$^{2}$+n)/6=

Ʃk$^{2}$=(1/3)n$^{3}$+(1/2)n$^{2}$+(1/6)n=

dƩk$^{2}$/dn=(n$^{2}$+n+1/6) Which we said equals 2Ʃk

dƩk$^{2}$/dn=(n$^{2}$+n+1/6)=2Ʃk

dƩk$^{2}$/dn=(n$^{2}$+n+1/6)/2=Ʃk

We earlier said that Ʃk=(n$^{2}$+n)/2 Which renders the above equation untrue.

The funny part of this is that this holds true for Ʃk$^{3}$ and a few others I have tried. Can anyone explain why this doesn't work?

## Answers and Replies

Staff Emeritus
Gold Member

So the derivative of this with respect to n would be

dƩk2/dn=2(1)+2(2)+2(3)+2(4)...

No, this is wrong. You seem to have forgotten that you're differentiating with respect to n and not with respect to k. Therefore the expression is $$\frac{d}{dn}\sum_{k=1}^n k^2$$I couldn't think of a a way to compute this "term-wise" (i.e. without first just evaluating the sum and then differentiating) until I realized that I could re-write the summation as$$\sum_{i=0}^{n-1} (n-i)^2$$So now we have$$\frac{d}{dn}\sum_{i=0}^{n-1} (n-i)^2 = \sum_{i=0}^{n-1} \frac{d}{dn}\left[(n-i)^2\right] = \sum_{i=0}^{n-1} 2(n-i)$$$$= 2\sum_{i=0}^{n-1} n - 2\sum_{i=0}^{n-1}i = 2n^2 - 2\frac{n(n-1)}{2}$$$$=2n^2 - (n^2 - n)$$$$= n^2 + n$$

Hmm...looks like I'm missing a 1/6 term somewhere. But you get the basic idea.

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Homework Helper

You can't actually define the derivative of a discrete series with respect to the upper limit of the sum. The definition of the derivative,
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\mathrm{d}x\to 0}\frac{y(x + \mathrm{d} x) - y(x)}{\mathrm{d}x}$$
involves an interval $\mathrm{d}x$ which has to become arbitrarily close to zero, while still allowing you to evaluate the function $y(x)$ at two points separated by that interval. When you only have discretely spaced "points," you can't do that. There is a lower limit on how small $\mathrm{d}x$ can become, namely 1, which means the normal definition of a derivative doesn't work.

What you have to do is analytically continue the series into a function - in other words, you have to find a differentiable function which reproduces the terms of the series when evaluated at integer values, but which is also defined at non-integer values. The easiest way to do this is just to sum the series analytically. In this case, when you do that you get
$$\sum_{k=1}^{n}k^2 = \frac{2n^3+3n^2+n}{6}$$
which is a perfectly fine function to differentiate.

cepheid, I think the reason you're missing your 1/6 term is that you aren't accounting for the increase in the number of terms in the series as $n$ increases. I actually can't think of a way to do that offhand (which is why people tend to prefer just summing the series before differentiation).

Homework Helper
Hi PEZenfuego! (try using the X2 icon just above the Reply box )
… Anyway, the equation for Ʃk$^{2}$ is (2n$^{3}$+3n$^{2}$+n)/6

and the one for ƩK is (n$^{2}$+n)/2

Do it the easy way don't go for ∑n2, go for ∑n(n-1) [or ∑n(n-1)…(n-r)] …

what does that look like the derivative of? :tongue2:

PEZenfuego

Oh wow...how did I not see that?

The derivative of Ʃk2 is 2Ʃk with respect to Ʃk...obviously and not n. Thanks guys.

Mentor

Oh wow...how did I not see that?

The derivative of Ʃk2 is 2Ʃk with respect to Ʃk
...with respect to k, not Ʃk
...obviously and not n. Thanks guys.

PEZenfuego

...with respect to k, not Ʃk

Salt in my wound...lol