- #1
PEZenfuego
- 48
- 0
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?
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?