ganeshie8
  • ganeshie8
Find a formula for the partial sum \[\large 1+4r+9r^2+\cdots+n^2r^{n-1}\]
Mathematics
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SOLVED
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katieb
  • katieb
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UnkleRhaukus
  • UnkleRhaukus
\[\sum_{i=1}^{n}i^2r^{i-1}\]
anonymous
  • anonymous
What have you tried so far?
ganeshie8
  • ganeshie8
I actually found two different solutions, just trying to understand this problem better by looking at how others approach this

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anonymous
  • anonymous
\[ f(r) = \sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r}\\ f'(r) = \sum_{k=1}^{n}kr^{k-1}\\ rf'(r) = \sum_{k=1}^{n}kr^{k}\\ (rf'(r))' = \sum_{k=1}^{n}k^2r^{k-1} = rf''(r) + f'(r) \]
ganeshie8
  • ganeshie8
Wow! that's almost same as the first solution I have : \[\sum_{i=1}^{n}i^2r^{i-1} =\dfrac{d}{dr} r\sum_{i=1}^{n} \left(r^i\right)' =\dfrac{d}{dr} r\left(\sum_{i=1}^{n} r^i\right)' = \dfrac{d}{dr} r \left(r\dfrac{r^n-1}{r-1}\right)' \]
ganeshie8
  • ganeshie8
actually it is identical xD
ganeshie8
  • ganeshie8
The second method uses some kind of convolution of two interleaving sequences
anonymous
  • anonymous
\[ f(r) = \sum_{k=0}^{n}r^k \]And \[ S_n = \sum_{k=1}^nkr^{k-1} \]Then\[ rS_n = \sum_{k=1}^nkr^{k} \]Let \(k=m-1\):\[ rS_n = \sum_{m=2}^{n+1}(m-1)r^{m-1} = \sum_{m=2}^{n+1}mr^{m-1} -\sum_{m=2}^{n+1}r^{m-1} \]Undo the sub for the second term:\[ rS_n= \sum_{m=2}^{n+1}mr^{m-1} -\sum_{k=1}^{n}r^{k} \]Add missing terms and remove extraneous terms:\[ rS_n+1 - (n+1)r^{n+1} = S_n - f(r) \]So we get \[ S_n = \frac{1-(n+1)r^{n+1}+f(r)}{1-r} \]When we do:\[ \Psi _n = \sum_{k=1}^nk^2r^{k-1} \]We get \[ r\Psi_n = \sum_{m=2}^n(m-1)^2r^{m-1} \]And we can use previous methods to solve for it.
ganeshie8
  • ganeshie8
Looks pretty neat, so if i understand correctly we need to find \(\sum kr^{k-1}\) before finding \(\sum k^2r^{k-1}\)
ganeshie8
  • ganeshie8
\[r\Psi_n = \sum_{m=\color{red}{1}}^n(m-1)^2r^{m-1} = \Psi_n -2\sum_{m=\color{red}{1}}^nmr^{m-1} + \sum_{m=\color{red}{1}}^n r^{m-1}\]
ganeshie8
  • ganeshie8
we can simply plugin the previous result for the sum in middle, awesome!
anonymous
  • anonymous
\[ f(m,r) = \sum_{k=0}^nk^mr^{k} \]And \[ rf(m,r) = \sum_{k=0}^nk^mr^{k+1} =\sum_{k'=1}^{n+1}(k'-1)^mr^{k'} \]\[ rf(m,r)+(-1)^m-n^mr^{n+1}= \sum_{p=0}^{q}{q\choose p}(-1)^pf(q-p,r) \]This is sort of our recursive solution
anonymous
  • anonymous
Hmmm, I guess to get it explicit:\[ f(m,r) = \frac{(-1)^m-n^mr^{n+1}+\sum_{k=0}^{n-1}{n\choose k}(-1)^pf(q-p,r)}{1-r} \]
anonymous
  • anonymous
I know I made a small error in it
ganeshie8
  • ganeshie8
that was a typo in the exponent, its okay it is a beautiful solution !
anonymous
  • anonymous
\[ f(m,r) = \frac{(-1)^m-n^mr^{n+1}-\sum_{k=0}^{m-1}{m\choose k}(-1)^pf(m-k,r)}{1-r} \]This one seems to work for \(m=0\).
anonymous
  • anonymous
Hmm, I still shouldn't have that n term, they should be m most likely
sparrow2
  • sparrow2
what is r here?
ganeshie8
  • ganeshie8
I think r could be any real number
ganeshie8
  • ganeshie8
\[\large f(m,r) = \frac{1}{1-r}\left[n^m(r^n-1)+\sum\limits_{k=2}^n \sum\limits_{i=0}^{m-1} \binom{m}{i}k^i(r^k-1)\right]\]
anonymous
  • anonymous
The whole point though is to get rid of that outer sum. And you shouldn't have any \(n\) terms.
ganeshie8
  • ganeshie8
"n" is the index, nth partial sum right
anonymous
  • anonymous
oh, right, I forgot
ganeshie8
  • ganeshie8
I was just messing with your work and ended up wid above formula it is kind of recursive too as the right side part has one exponent less..
anonymous
  • anonymous
Everything goes to hell when you let \(r=1\) though.
anonymous
  • anonymous
\[ f(m, 1) = \sum_{k=0}^nk^m \]
ganeshie8
  • ganeshie8
Haha lets just say \(r\ne 1\) then
anonymous
  • anonymous
But \(r=1\) is an interesting series.
ganeshie8
  • ganeshie8
Indeed, it can be shown that that series asymptotically approaches \(\dfrac{n^{m+1}}{m+1}\) \[f(m,1) \sim \dfrac{n^{m+1}}{m+1}\]
anonymous
  • anonymous
Is there a formula for that series?
ganeshie8
  • ganeshie8
Yes, was working on some definite integral last night and ended up wid above result... let me pull up..
anonymous
  • anonymous
I think \(f(-1,1)\) is known not to have an elementary function.
ganeshie8
  • ganeshie8
Sorry it is not a formula, just a limit which gives the asymptotic sum
ganeshie8
  • ganeshie8
\[\lim\limits_{n\to\infty}\dfrac{f(m,1)}{n^{m+1}} = \frac{1}{m+1}\]
sparrow2
  • sparrow2
for r=1 sum if n(n+1)(2n+1)/6 you can prove using induction
ganeshie8
  • ganeshie8
http://math.stackexchange.com/questions/936924/limit-of-the-sequence-frac1k2k-nknk1
ganeshie8
  • ganeshie8
right @sparrow2 thats sum of squares of first n natural numbers, fun one!
sparrow2
  • sparrow2
i think the point is that we just should write the sum in the form so when you give me a natural number for ex n=39 i will give you the answer not the sum written in other form
sparrow2
  • sparrow2
like in sum of squares of first n
ganeshie8
  • ganeshie8
you mean explicit formula
sparrow2
  • sparrow2
i don't know what it is called
sparrow2
  • sparrow2
i don't know math's english terminology very well :)
ganeshie8
  • ganeshie8
got you :) basically we're trying to find a recursive relation for \[\large 1^m + 2^mr + 3^mr^2+\cdots+n^mr^{n-1}\] when \(r=1\), we get the series \[1^m+2^m+3^m+\cdots+n^m\]
ganeshie8
  • ganeshie8
yeah the original question is a special case, \(m=2\)
Empty
  • Empty
I am a little late to the party, but I thought I'd try to challenge myself to come up with a different approach. I think it doesn't help much, but is there anything to be gained by writing the squares as: \[k^2 = \sum_{n=1}^k (2n-1) \] My thought was to switch the order of summation signs, however it won't work here since the upper limit k is summed over in the other summation, oh well.
ganeshie8
  • ganeshie8
Thats similar to what wio did in his recurrence relation, that works pretty nicely as it reduces the exponent.. .
Empty
  • Empty
Oh I thought he just did the derivatives of the geometric series approach.
ganeshie8
  • ganeshie8
He did it in two ways..
sparrow2
  • sparrow2
and do we know that there is a formula for this?
ganeshie8
  • ganeshie8
check this @sparrow2 https://en.wikipedia.org/wiki/Faulhaber%27s_formula
sparrow2
  • sparrow2
ok thanks
ganeshie8
  • ganeshie8
generating functions might also work i think post the work @ikram002p
ganeshie8
  • ganeshie8
Let \[f(n,m,r)=\sum\limits_{k=1}^n k^mr^{k-1} \] Let \(a_k = k^m\) and \(b_k = r^{k-1}\), then \(B_n=\sum\limits_{k=1}^n b_k = \dfrac{r^n-1}{r-1}\) It is easy to see that \(b_n = B_{n+1}-B_n\) \[\begin{align}\color{blue}{f(n,m,r)}&=\sum\limits_{k=1}^n a_k b_k \\~\\ &=\sum\limits_{k=1}^n a_k (B_{n+1}-B_n) \\~\\ &= a_nB_n+\sum\limits_{k=2}^n (a_{k-1}-a_{k}) B_k \\~\\ &= n^m\dfrac{r^n-1}{r-1}+\frac{1}{1-r}\sum\limits_{k=1}^n [(k-1)^m-k^m] (r^k-1) \\~\\ \end{align}\]
anonymous
  • anonymous
Liam did an investigation to see how water temperature affects the amount of salt that will dissolve in the water. He filled 4 beakers with exactly 100 milliliters of water each. He then heated each beaker to a different temperature and tested salt solubility at each different temperature. What was Liam's independent variable? amount of water in each beaker the number of beakers the amount of salt that dissolved in each beaker the temperature of the water in each beaker
anonymous
  • anonymous
A student conducts an experiment to determine how the additional of salt to water affects the density of the water. The student fills 3 beakers with equal amounts of water. He then adds 1 cup of salt to the first beaker, 2 cups of salt to the second beaker and no salt to the third beaker. What is the dependent variable in the student's investigation? the amount of salt in the water the temperature of the water the density of the water the ph of the water

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