Symmetrical Summation with Central Point | Solving for a(0) to a(N-1/2)

[JeeebeZ]

Homework Statement

I need a summation where the answer is 1 2 2 2 2 2 2 2

Homework Equations

a(0) + sum(2*a(1) + 2*a(2) +2*a(3))

The Attempt at a Solution

I unfortunately have no idea where to start, basically it is taking a symmetrical function from 0 to N-1. where the function is symmetric from the center point.

I have an answer for it which is

a(N-1/2) + sum_n=1^(N-1)/2 (2*a[((N-1)/2) - n])

However I wanted to put it all inside the summation

 

[haruspex]

I've no idea what you mean.
In what sense does a summation have an answer like "1 2 2 2 2 2 2 2" rather than a single number?
What does a(0) + sum(2*a(1) + 2*a(2) +2*a(3)) mean? Do you mean a(0) + sum(2*a(1) + 2*a(2) +2*a(3) + ...)? If so, how is that different from 2(∑a_n) - a₀?
From the mention of symmetry, I'm guessing this comes from $\Sigma_{n=-N}^Na_n$, where a_-n = a_n, but I still don't understand what you wish to achieve.
What quantity does your final equation represent?

 

[JeeebeZ]

Sorry, the "1 2 2 2 2 ..." bit was basically was how I wanted each part of the summation to equate to, so essentially it would be "1 + 2 + 2 + 2 ..."

"2(∑an) - a0" is the exact same as "a0 + 2(∑an)" with the exception that n goes from 0->N for the first one and 1->N for the second one.

Basically I want to achieve this, "2(∑an) - a0", without having to subtract the a0 at the end.

 

[haruspex]

Still unclear.. are you looking for an algebraic function f(n) that takes the values f(0) = 1, f(n) = 2 for n > 1? So that you can write ∑_n>=0a_nf(n)?

 

[JeeebeZ]

yes that's exactly what I'm looking to do

 

[LCKurtz]

You want $f(n) = 2-\textrm{sinc}(n)$, where sinc is the normalized sinc function:$$
\textrm{sinc} (x) = \begin{cases}
\frac {\sin(\pi x)}{\pi x}&x \ne 0\\
1 & x = 0
\end{cases}$$

 

[haruspex]

You want $f(n) = 2-\textrm{sinc}(n)$, where sinc is the normalized sinc function:$$
\textrm{sinc} (x) = \begin{cases}
\frac {\sin(\pi x)}{\pi x}&x \ne 0\\
1 & x = 0
\end{cases}$$

Cute, but I don't see the value in this. As you note, that function has to have a separate definition at x=0. So why not use Kronecker delta? I see nothing in the thread to suggest a continuous function is required.
JeeebeZ, why do you want such a function? What ultimately are you trying to achieve?

 

[LCKurtz]

Cute, but I don't see the value in this.

Agreed. But then, I don't see any point to the original question either.