Solving Equation with Summation Properties

[Jhonny]

Hello, I can not find the way to solve the following equation:
sum of k^2 f(k) from k=1 to n.

In Particular, k^2 * (1/k)

 

[mfb]

There is no general formula for arbitrary functions f.
In your specific example, there is a very easy one that you'll find in every collection of sum formulas.

 

[Jhonny]

I have found how to solve ∑{k=1,n} [k·f(k)], for example:

∑{k=1,n} [k·f(k)] = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ]
Dem:
∑{k=1,n} [k·f(k)] = 1·f(1) + 2·f(2) + 3·f(3) + ... + n·f(n) = f(1) + f(2) + f(2) +
+ f(3) + f(3) + f(3) + ... + f(n) + ... + f(n) = [f(1) + f(2) + f(3) + ... + f(n)] +
+ [f(2) + f(3) + ... + f(n)] + [f(3) + ... + f(n)] + ... + f(n) =
= ∑{j=1,n} [f(j)] + ∑{j=2,n} [f(j)] + ∑{j=3,n} [f(j)] + ... + ∑{j=n,n} [f(j)] =
= ∑{k=1,n} [ ∑{j=k,n} [f(j)] ]

However, I need to solve ∑{k=1,n} [k^2·f(k)]

 

[mfb]

That is not "solving", it is transforming one expression into another, here a sum into a double sum.

You can do the same scheme here. Let g(k)=k*f(k) and apply the formula you found for k*g(k).
Do the same thing again and you get a triple sum over f(m)

 

[Jhonny]

I already tried that, but it didn't work.
For example:

∑{k=1,n} [k·f(k)] = 1·f(1) + 2·f(2) + 3·f(3) + ... + n·f(n) = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ]

But,

∑{k=1,n} [k^2·f(k)] = 1·f(1) + 4·f(2) + 9·f(3) + ... + n^2·f(n) ≠ ∑{k=1,n} [ ∑{j=k,n} ∑{i=j,n}[[f(i)]] ]

 

[Jhonny]

It seems to be that:
Although: ∑{k=1,n} [k·f(k)] = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ] holds,
∑{k=1,n} [ ∑{j=k,n} [j*f(j)] ] = ∑{k=1,n} [ ∑{j=k,n} ∑{i=j,n}[[f(i)]] ] does not hold.

 

[mfb]

I already tried that, but it didn't work.
For example:

∑{k=1,n} [k·f(k)] = 1·f(1) + 2·f(2) + 3·f(3) + ... + n·f(n) = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ]

But,

∑{k=1,n} [k^2·f(k)] = 1·f(1) + 4·f(2) + 9·f(3) + ... + n^2·f(n) ≠ ∑{k=1,n} [ ∑{j=k,n} ∑{i=j,n}[[f(i)]] ]

That's not what I suggested.

It seems to be that:
Although: ∑{k=1,n} [k·f(k)] = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ] holds,
∑{k=1,n} [ ∑{j=k,n} [j*f(j)] ] = ∑{k=1,n} [ ∑{j=k,n} ∑{i=j,n}[[f(i)]] ] does not hold.

Right, as you start the multiplications with j=k, not with 1. There is a way to avoid that.

 

[Jhonny]

That's not what I suggested.

Right, as you start the multiplications with j=k, not with 1. There is a way to avoid that.

How can I solve: ∑{k=1,n} [k^2· 2^(-k)]

 

[Mark44]

How can I solve: ∑{k=1,n} [k^2· 2^(-k)]

As you have already been told, you are not "solving" these summations. Instead, you are writing them in a different form.

Please ask your question in such a way that makes it clear what you want to happen.

 

[Jhonny]

I need to find an equivalent expression without the sigma notation of the expression: ∑{k=1,n} [k^2· 2^(-k)]

 

[mfb]

It is possible with the ideas given in the thread.

In case you are just interested in the result, WolframAlpha can calculate it - I copied your equation 1:1, didn't even need reformatting. I just changed the brackets so the forum displays the link correctly.

 

[mfb]

Sigma ($\Sigma$) is the greek symbol used for sums. The question is about properties of sums.

 

[Mark44]

hello,
what is sigma properties??

Sigma ($\Sigma$) is the greek symbol used for sums. The question is about properties of sums.

I have changed the title of this thread to "Summation properties".

 

[fresh_42]

I have changed the title of this thread to "Summation properties".

What a pitty. ∑ properties would have been far more interesting.

 

[Mark44]

What a pitty. ∑ properties would have been far more interesting.

Are you referring to sigma-algebras ($\sigma-\text{algebras}$) and the like? If so, lowercase $\sigma$ (sigma) is used.
In any case, and interesting or not, the title now reflects what the OP asked.

 

[fresh_42]

Are you referring to sigma-algebras ($\sigma-\text{algebras}$) and the like? If so, lowercase $\sigma$ (sigma) is used.
In any case, and interesting or not, the title now reflects what the OP asked.

I meant hyperons and yes, it was meant to be both funny and a food for thought. But hey, I'm new and I'am trying to get used to the language here. Jokes don't seem to rank very high. Ok, lesson learned.