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

  • Thread starter PEZenfuego
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  • #1
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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[itex]^{n}_{k=1}[/itex]K and Ʃ[itex]^{n}_{k=1}[/itex]K[itex]^{2}[/itex] 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[itex]^{2}[/itex] is a series like1[itex]^{2}[/itex]+2[itex]^{2}[/itex]+3[itex]^{2}[/itex]+4[itex]^{2}[/itex]...+(n-2)[itex]^{2}[/itex]+(n-1)[itex]^{2}[/itex]+n[itex]^{2}[/itex]

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

and the one for ƩK is (n[itex]^{2}[/itex]+n)/2

Now here is what I'm stepping in:

Ʃk[itex]^{2}[/itex]=1[itex]^{2}[/itex]+2[itex]^{2}[/itex]+3[itex]^{2}[/itex]+4[itex]^{2}[/itex]...

So the derivative of this with respect to n would be

dƩk[itex]^{2}[/itex]/dn=2(1)+2(2)+2(3)+2(4)...

another way to write this would be

dƩk[itex]^{2}[/itex]/dn=2(1+2+3+4...)

Or...

dƩk[itex]^{2}[/itex]/dn=2(ƩK)

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

Ʃk[itex]^{2}[/itex]=(2n[itex]^{3}[/itex]+3n[itex]^{2}[/itex]+n)/6=

Ʃk[itex]^{2}[/itex]=(1/3)n[itex]^{3}[/itex]+(1/2)n[itex]^{2}[/itex]+(1/6)n=

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

dƩk[itex]^{2}[/itex]/dn=(n[itex]^{2}[/itex]+n+1/6)=2Ʃk

dƩk[itex]^{2}[/itex]/dn=(n[itex]^{2}[/itex]+n+1/6)/2=Ʃk

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

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

Answers and Replies

  • #2
cepheid
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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 [tex] \frac{d}{dn}\sum_{k=1}^n k^2[/tex]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[tex]\sum_{i=0}^{n-1} (n-i)^2 [/tex]So now we have[tex]\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)[/tex][tex] = 2\sum_{i=0}^{n-1} n - 2\sum_{i=0}^{n-1}i = 2n^2 - 2\frac{n(n-1)}{2}[/tex][tex]=2n^2 - (n^2 - n)[/tex][tex] = n^2 + n [/tex]

Hmm...looks like I'm missing a 1/6 term somewhere. But you get the basic idea.
 
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  • #3
diazona
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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,
[tex]\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\mathrm{d}x\to 0}\frac{y(x + \mathrm{d} x) - y(x)}{\mathrm{d}x}[/tex]
involves an interval [itex]\mathrm{d}x[/itex] which has to become arbitrarily close to zero, while still allowing you to evaluate the function [itex]y(x)[/itex] 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 [itex]\mathrm{d}x[/itex] 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
[tex]\sum_{k=1}^{n}k^2 = \frac{2n^3+3n^2+n}{6}[/tex]
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 [itex]n[/itex] 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).
 
  • #4
tiny-tim
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Hi PEZenfuego! :smile:

(try using the X2 icon just above the Reply box :wink:)
… Anyway, the equation for Ʃk[itex]^{2}[/itex] is (2n[itex]^{3}[/itex]+3n[itex]^{2}[/itex]+n)/6

and the one for ƩK is (n[itex]^{2}[/itex]+n)/2
Do it the easy way :wink:

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:
 
  • #5
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Oh wow...how did I not see that?

The derivative of Ʃk2 is 2Ʃk with respect to Ʃk...obviously and not n. Thanks guys.
 
  • #6
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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.
 
  • #7
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...with respect to k, not Ʃk
Salt in my wound...lol
 

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