∑f(n)=1 (n=1 to n), find f(n).

  • Thread starter Crack Dragon
  • Start date
In summary, the conversation discusses the concept of anti-sums and the difference of a series. The participants also discuss the correct use of the summation symbol and the concept of discrete calculus, including the FTC for the discrete case and inverse relationships. They also touch upon the Z transform as a discrete analogue of the Laplace transform.
  • #1
Crack Dragon
3
0
Not sure if this is impossible, simply wondering if anti-sums are a thing, ie. (Anti)∑0.5n(n+1)=n.
 
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  • #2
Hi Crack Dragon! Welcome to PF! :smile:

Yes, it's a difference!

The difference of a series {an} is the series {an - an-1}

eg 0.5n(n+1) - 0.5(n-1)n = 0.5n((n+1) - (n-1)) = n :wink:
 
  • #3
Hi, thanks for that, sorry if this was in the wrong area; wasn't too sure. That's pretty handy actually, by your post count you seem to be a busy person, but do you know where I could do any reading up on summations and the like?
 
  • #4
Crack Dragon said:
… do you know where I could do any reading up on summations and the like?

sorry, no idea …

but I'm sure you'll find plenty if you google it :smile:
 
  • #5
Can you clarify, is the formula meant to hold for all n or for a fixed value of n? If it is for all n the f(n)=0 is the only solution in this case. If it is for a fixed n then there are infinite f(n) that will work.
 
  • #7
Crack Dragon said:
Fixed n? I really am new to this, just in this format though if that's what you were wondering: http://www.wolframalpha.com/input/?i=Sum+f(n),+n=1+to+n, are you saying ∑0=1?

My mistake, I missread 1 as 0. But it doesn't change much. If the sum of f(k) from k=1 to n (btw, don't use n as both the thing you are summing over as well as the number of terms you are summing) for all n, then you must have f(1)=1 and all others are 0. To see this just set n=1 so f(1)=1 then set n=2 to get f(1)+f(2)=1 so 1+f(2)=1 which gives f(2)=0. Continue inductively to get the rest being 0.
 
  • #8
[tex]\\ \sum f(n)\Delta n=1 \\ \\ \frac{\Delta\;\;}{\Delta n} \sum f(n)\Delta n = \frac{\Delta1}{\Delta n} \\ \\ f(n) = \left\{\begin{matrix} 1 & n=0\\ 0 & n\neq 0\\ \end{matrix}\right.[/tex]
 
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  • #9
Crack Dragon said:
Fixed n? I really am new to this, just in this format though if that's what you were wondering: http://www.wolframalpha.com/input/?i=Sum+f(n),+n=1+to+n, are you saying ∑0=1?
The way you're using the summation symbol is wrong crack dragon. You need to sum over a "dummy" variable such as "k" in the example below.

[tex] \sum_{k=1}^{n} f(k) = 1[/tex]

Now we can again ask the question: Do you require this to be valid just for some specific positive "n" or is it to be true for all positive "n"?

If it's for some specific "n" then there is a unique solution only for n=1, and an infinite number of solutions otherwise.

If it's for all positive "n" then there is just one solution, that given above by jsmith.
 
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  • #10
Jhenrique said:
[tex]\\ \sum f(n)\Delta n=1 \\ \\ \frac{\Delta\;\;}{\Delta n} \sum f(n)\Delta n = \frac{\Delta1}{\Delta n} \\ \\ f(n) = \left\{\begin{matrix} 1 & n=0\\ 0 & n\neq 0\\ \end{matrix}\right.[/tex]
What does Δ/Δn mean? I understand the operator d/dn, but have never seen similar notation using Δ.
 
  • #11
Mark44 said:
What does Δ/Δn mean? I understand the operator d/dn, but have never seen similar notation using Δ.

##\Delta f=f_1-f_0##

##\frac{\Delta f}{\Delta x}=\frac{f_1-f_0}{x_1-x_0}##

How in discrete calculus we not consider an infinitesimal interval but yes a discrete interval (unitary), so Δx=1, ie:

##\frac{\Delta f}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{f(x+1)-f(x)}{1}=f(x+1)-f(x)=\Delta f##

The same logic for summation:

##\sum f(x)\Delta x=\sum f(x)\cdot 1=\sum _{x}f(x)##

FTC for the discrete case:

##\sum_{x_0}^{x_1} f(x)\Delta x=\sum_{x=x_0}^{x_1}f(x)=F(x_1+1)-F(x_0)##

inverse relationships:

##\sum \frac{\Delta f}{\Delta x}(x)\Delta x=f(x)##

##\frac{\Delta}{\Delta x} \sum f(x)\Delta x = f(x)##

with infinitesimal limits...

##\int f(x)dx=\lim_{\Delta x \rightarrow 0} \sum f(x) \Delta x##

##\frac{df}{dx}=\lim_{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}##

computing...

##\frac{d}{dx}\left (\frac{1}{2}kx^2 \right )=kx##

##\frac{\Delta}{\Delta x}\left (\frac{1}{2}kx^2 \right )=kx+\frac{1}{2}k##

##\sum f(x)\Delta x=\frac{1}{6}kx^3-\frac{1}{4}kx^2+\frac{1}{12}kx + C##

##\int f(x)dx=\frac{1}{6}kx^3 + C##

And you still have a Z transform that is the discrete analogue of Laplace transform. All this is the discrete calculus.
 
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  • #12
Jhenrique said:
##\Delta f=f_1-f_0##

##\frac{\Delta f}{\Delta x}=\frac{f_1-f_0}{x_1-x_0}##

How in discrete calculus we not consider an infinitesimal interval but yes a discrete interval (unitary), so Δx=1, ie:

##\frac{\Delta f}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{f(x+1)-f(x)}{1}=f(x+1)-f(x)=\Delta f##

The same logic for summation:

##\sum f(x)\Delta x=\sum f(x)\cdot 1=\sum _{x}f(x)##

FTC for the discrete case:

##\sum_{x_0}^{x_1} f(x)\Delta x=\sum_{x=x_0}^{x_1}f(x)=F(x_1+1)-F(x_0)##

inverse relationships:

##\sum \frac{\Delta f}{\Delta x}(x)\Delta x=f(x)##

##\frac{\Delta}{\Delta x} \sum f(x)\Delta x = f(x)##

with infinitesimal limits...

##\int f(x)dx=\lim_{\Delta x \rightarrow 0} \sum f(x) \Delta x##

##\frac{df}{dx}=\lim_{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}##

computing...

##\frac{d}{dx}\left (\frac{1}{2}kx^2 \right )=kx##

##\frac{\Delta}{\Delta x}\left (\frac{1}{2}kx^2 \right )=kx+\frac{1}{2}k##
This (above) is what I've never seen before. Where does the k/2 on the right side come from?
Jhenrique said:
##\sum f(x)\Delta x=\frac{1}{6}kx^3-\frac{1}{4}kx^2+\frac{1}{12}kx + C##

##\int f(x)dx=\frac{1}{6}kx^3 + C##

And you still have a Z transform that is the discrete analogue of Laplace transform. All this is the discrete calculus.
 
  • #13
Mark44 said:
This (above) is what I've never seen before. Where does the k/2 on the right side come from?

From the definition of discrete differentiation: ##\frac{\Delta f}{\Delta x} = f(x+1)-f(x)=f'(x)##

If f(x)=1/2kx² so f'(x)=kx+1/2k

Ironically or not, It's so that math works...
 
  • #14
Now, my 2nd answer (considering the interval of summation):

The "graphic" of question below is:

##F[n]=\sum _{1}^{n}f[m]\Delta m=1##
\begin{matrix}
n & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\
F[n] & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\
F'[n] & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{matrix}

Thus we have ##F'[n]=\frac{\Delta F[n]}{\Delta n}=F[n+1]-F[n]=f[n]## ([...] the colchets says that the domain is discrete). Remember that:
http://en.wikipedia.org/wiki/Kronecker_delta
http://en.wikipedia.org/wiki/Dirac_delta_function

##\delta [n]:=\left\{\begin{matrix}
1 & n=0\\
0 & n\neq 0
\end{matrix}\right.##

So, f[n] in terms of discrete delta is:
[tex]f[n]= \delta [n-1]=\left\{\begin{matrix} 1 & n=1\\ 0 & n\neq 1 \end{matrix}\right.[/tex]

I found f(n) to you. Capiche?
 

1. What does the notation ∑f(n) mean?

The ∑ symbol represents a summation, which means that the values of f(n) for all values of n from 1 to n are added together.

2. How do I find the value of f(n) for a specific value of n?

To find the value of f(n) for a specific value of n, you need to plug in that value for n in the given equation and then solve for f(n).

3. Can the value of n be any number?

No, the value of n must be a positive integer since the summation starts at 1 and goes up to n.

4. What if I don't know the value of n?

If the value of n is not specified, then you cannot find the value of f(n) since you need a specific value of n to plug into the equation.

5. Can the summation symbol be replaced with a different symbol?

Yes, the summation symbol can be replaced with any other letter or symbol, such as ∑f(n)=1 (k=1 to k), as long as it is consistent throughout the equation.

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