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ganeshie8

  • one year ago

Find a formula for the partial sum \[\large 1+4r+9r^2+\cdots+n^2r^{n-1}\]

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  1. UnkleRhaukus
    • one year ago
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    \[\sum_{i=1}^{n}i^2r^{i-1}\]

  2. anonymous
    • one year ago
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    What have you tried so far?

  3. ganeshie8
    • one year ago
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    I actually found two different solutions, just trying to understand this problem better by looking at how others approach this

  4. anonymous
    • one year ago
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    \[ f(r) = \sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r}\\ f'(r) = \sum_{k=1}^{n}kr^{k-1}\\ rf'(r) = \sum_{k=1}^{n}kr^{k}\\ (rf'(r))' = \sum_{k=1}^{n}k^2r^{k-1} = rf''(r) + f'(r) \]

  5. ganeshie8
    • one year ago
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    Wow! that's almost same as the first solution I have : \[\sum_{i=1}^{n}i^2r^{i-1} =\dfrac{d}{dr} r\sum_{i=1}^{n} \left(r^i\right)' =\dfrac{d}{dr} r\left(\sum_{i=1}^{n} r^i\right)' = \dfrac{d}{dr} r \left(r\dfrac{r^n-1}{r-1}\right)' \]

  6. ganeshie8
    • one year ago
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    actually it is identical xD

  7. ganeshie8
    • one year ago
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    The second method uses some kind of convolution of two interleaving sequences

  8. anonymous
    • one year ago
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    \[ f(r) = \sum_{k=0}^{n}r^k \]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.

  9. ganeshie8
    • one year ago
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    Looks pretty neat, so if i understand correctly we need to find \(\sum kr^{k-1}\) before finding \(\sum k^2r^{k-1}\)

  10. ganeshie8
    • one year ago
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    \[r\Psi_n = \sum_{m=\color{red}{1}}^n(m-1)^2r^{m-1} = \Psi_n -2\sum_{m=\color{red}{1}}^nmr^{m-1} + \sum_{m=\color{red}{1}}^n r^{m-1}\]

  11. ganeshie8
    • one year ago
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    we can simply plugin the previous result for the sum in middle, awesome!

  12. anonymous
    • one year ago
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    \[ f(m,r) = \sum_{k=0}^nk^mr^{k} \]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

  13. anonymous
    • one year ago
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    Hmmm, I guess to get it explicit:\[ f(m,r) = \frac{(-1)^m-n^mr^{n+1}+\sum_{k=0}^{n-1}{n\choose k}(-1)^pf(q-p,r)}{1-r} \]

  14. anonymous
    • one year ago
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    I know I made a small error in it

  15. ganeshie8
    • one year ago
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    that was a typo in the exponent, its okay it is a beautiful solution !

  16. anonymous
    • one year ago
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    \[ f(m,r) = \frac{(-1)^m-n^mr^{n+1}-\sum_{k=0}^{m-1}{m\choose k}(-1)^pf(m-k,r)}{1-r} \]This one seems to work for \(m=0\).

  17. anonymous
    • one year ago
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    Hmm, I still shouldn't have that n term, they should be m most likely

  18. sparrow2
    • one year ago
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    what is r here?

  19. ganeshie8
    • one year ago
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    I think r could be any real number

  20. ganeshie8
    • one year ago
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    \[\large f(m,r) = \frac{1}{1-r}\left[n^m(r^n-1)+\sum\limits_{k=2}^n \sum\limits_{i=0}^{m-1} \binom{m}{i}k^i(r^k-1)\right]\]

  21. anonymous
    • one year ago
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    The whole point though is to get rid of that outer sum. And you shouldn't have any \(n\) terms.

  22. ganeshie8
    • one year ago
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    "n" is the index, nth partial sum right

  23. anonymous
    • one year ago
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    oh, right, I forgot

  24. ganeshie8
    • one year ago
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    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..

  25. anonymous
    • one year ago
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    Everything goes to hell when you let \(r=1\) though.

  26. anonymous
    • one year ago
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    \[ f(m, 1) = \sum_{k=0}^nk^m \]

  27. ganeshie8
    • one year ago
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    Haha lets just say \(r\ne 1\) then

  28. anonymous
    • one year ago
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    But \(r=1\) is an interesting series.

  29. ganeshie8
    • one year ago
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    Indeed, it can be shown that that series asymptotically approaches \(\dfrac{n^{m+1}}{m+1}\) \[f(m,1) \sim \dfrac{n^{m+1}}{m+1}\]

  30. anonymous
    • one year ago
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    Is there a formula for that series?

  31. ganeshie8
    • one year ago
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    Yes, was working on some definite integral last night and ended up wid above result... let me pull up..

  32. anonymous
    • one year ago
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    I think \(f(-1,1)\) is known not to have an elementary function.

  33. ganeshie8
    • one year ago
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    Sorry it is not a formula, just a limit which gives the asymptotic sum

  34. ganeshie8
    • one year ago
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    \[\lim\limits_{n\to\infty}\dfrac{f(m,1)}{n^{m+1}} = \frac{1}{m+1}\]

  35. sparrow2
    • one year ago
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    for r=1 sum if n(n+1)(2n+1)/6 you can prove using induction

  36. ganeshie8
    • one year ago
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    http://math.stackexchange.com/questions/936924/limit-of-the-sequence-frac1k2k-nknk1

  37. ganeshie8
    • one year ago
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    right @sparrow2 thats sum of squares of first n natural numbers, fun one!

  38. sparrow2
    • one year ago
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    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

  39. sparrow2
    • one year ago
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    like in sum of squares of first n

  40. ganeshie8
    • one year ago
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    you mean explicit formula

  41. sparrow2
    • one year ago
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    i don't know what it is called

  42. sparrow2
    • one year ago
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    i don't know math's english terminology very well :)

  43. ganeshie8
    • one year ago
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    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 \[1^m+2^m+3^m+\cdots+n^m\]

  44. ganeshie8
    • one year ago
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    yeah the original question is a special case, \(m=2\)

  45. Empty
    • one year ago
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    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.

  46. ganeshie8
    • one year ago
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    Thats similar to what wio did in his recurrence relation, that works pretty nicely as it reduces the exponent.. .

  47. Empty
    • one year ago
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    Oh I thought he just did the derivatives of the geometric series approach.

  48. ganeshie8
    • one year ago
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    He did it in two ways..

  49. sparrow2
    • one year ago
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    and do we know that there is a formula for this?

  50. ganeshie8
    • one year ago
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    check this @sparrow2 https://en.wikipedia.org/wiki/Faulhaber%27s_formula

  51. sparrow2
    • one year ago
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    ok thanks

  52. ganeshie8
    • one year ago
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    generating functions might also work i think post the work @ikram002p

  53. ganeshie8
    • one year ago
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    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 \(b_n = B_{n+1}-B_n\) \[\begin{align}\color{blue}{f(n,m,r)}&=\sum\limits_{k=1}^n a_k b_k \\~\\ &=\sum\limits_{k=1}^n a_k (B_{n+1}-B_n) \\~\\ &= a_nB_n+\sum\limits_{k=2}^n (a_{k-1}-a_{k}) B_k \\~\\ &= n^m\dfrac{r^n-1}{r-1}+\frac{1}{1-r}\sum\limits_{k=1}^n [(k-1)^m-k^m] (r^k-1) \\~\\ \end{align}\]

  54. anonymous
    • one year ago
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    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

  55. anonymous
    • one year ago
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    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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