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I would like to ask if anyone can explain to me how can I find the closed form of the sum $$\sum_{k=0}^n k^4$$ with the expansion contraction method of the sum

Like this example : $$Y_n=\sum_{k=0}^n k^2$$ Yet another way to discover a closed form for Cl, is to replace the original sum by a seemingly more complicated double sum that can actually be simplified if we massage it properly $$\sum_{k=0}^n k^2 =\sum_{0\le j \le k \le n } k= \sum_{0\le j \le n} \sum k=\sum_{0\le j \le n}\frac{(j+n)(n-j+1)}{2} =\frac{1}{2}\sum_{0\le j \le n}(n(n+l)+j-j2)= \frac{1}{2}n^2(n+1)+\frac{1}{4}n(n+1)-\frac{1}{2}Yn$$

Nosrati
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  • @juniven Yes .Thank you for the edit.It was easy to deal with it with other methods like the repertoire method of by replacing the sum with an intergral.But the expansio contraction method it is too difficult to understand –  Jan 12 '17 at 09:18
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    What is the expansion contraction method? Google shows no results. –  Jan 12 '17 at 09:23
  • @OpenBall with this "name" is presented at the book "concrete mathematics second american publication" page 61 –  Jan 12 '17 at 09:26
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    Well, for those of us who don't have access to that book, please define this method or give some example of how it works. –  Jan 12 '17 at 09:33
  • @OpenBall it is free to download it .I will try to bring the example –  Jan 12 '17 at 09:36
  • this means also that we prove the answer by induction – Dr. Sonnhard Graubner Jan 12 '17 at 09:44
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    Downvoting and voting to close because you have received a couple of good answers but you still insist on a vaguely named method from a specific reference, which you have so far been unable to produce in detail. You cannot expect others to be familiar with that exact same book, and no one claims to be a mind reader here. – Deepak Jan 12 '17 at 09:49
  • right now i try to right the example with the mathjax –  Jan 12 '17 at 09:51
  • Is it the Graham–Knuth–Patashnik book? As far as I can tell, the word “contraction” is never used in this book, and certainly not on page 61. – Hans Lundmark Jan 12 '17 at 10:04
  • Might it be this method: http://math.stackexchange.com/questions/980843/proof-of-a-summation-of-k4? – Hans Lundmark Jan 12 '17 at 10:04
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    @HansLundmark page 46 in the original version .I read a translated version –  Jan 12 '17 at 10:11
  • @HansLundmark this mehtod is the disturbance of the sum –  Jan 12 '17 at 10:14
  • $\sum_{k=0}^n k^2 =\sum_{0\le j \le k \le n } k^2$ already looks wrong to me. – Martin R Jan 12 '17 at 10:26
  • @MartinR Sorry I update it –  Jan 12 '17 at 10:55
  • OK, now we're talking! They call it “expand and contract”, to be precise. – Hans Lundmark Jan 12 '17 at 14:42
  • Can't you just take the corresponding double sum of $k^3$, and then use the formula for $\sum k^3$ (which must be known in order for this method to work, it seems)? – Hans Lundmark Jan 12 '17 at 14:50
  • @HansLundmark Yes i have tried it but a i had problem handling the js and ns –  Jan 12 '17 at 17:59

4 Answers4

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An approach via linear maps:

1. $T:\mathbb{R}_5[x]\rightarrow{\mathbb{R}_5[x]},\quad T(p(x))=p(x+1)-p(x)$ is a linear map.

2. The matricial equation of $T$ with respect to the canonical basis of $\mathbb{R}_5[x]$ is$$\begin{bmatrix}{y_1}\\{y_2}\\{y_3}\\{y_4}\\{y_5}\\{y_6}\end{bmatrix}=\begin{bmatrix}{0}&{1}&{1}&{1}&{1}&{1}\\{0}&{0}&{2}&{3}&{4}&{5}\\{0}&{0}&{0}&{3}&{6}&{10}\\ {0}&{0}&{0}&{0}&{4}&{10}\\{0}&{0}&{0}&{0}&{0}&{5}\\{0}&{0}&{0}&{0}&{0}&{0}\end{bmatrix}\begin{bmatrix}{x_1}\\{x_2}\\{x_3}\\{x_4}\\{x_5}\\{x_6}\end{bmatrix}\quad (*)$$

3. Using $(*)$ we get: $$T^{-1}(x^4)=\left\{{\alpha -x/30+x^3/3-x^4/2+x^5/5:\alpha \in{\mathbb{R}}}\right\}$$

4. Consider $h(x)\in{T^{-1}(x^4)}$ with $\alpha=0$. Then $T(h(x))=x^4$ i.e. $h(x+1)-h(x)=x^4$.

5. For $x=1,2,\ldots,n$ we get: $$\begin{aligned} &h(2)-h(1)=1^4\\ &h(3)-h(2)=2^4\\ &h(4)-h(3)=3^4\\ &\ldots\\ &h(n+1)-h(n)=n^4. \end{aligned}$$

6. Simplifying: $h(n+1)-h(n)=1^4+2^4+\ldots+n^4=S_4$. That is $$S_4=h(n+1)-h(1)=$$ $$-\displaystyle\frac{n+1}{30}+\displaystyle\frac{(n+1)^3}{3}-\displaystyle\frac{(n+1)^4}{2}+\displaystyle\frac{(n+1)^5}{5}+\displaystyle\frac{1}{30}-\displaystyle\frac{1}{3}+\displaystyle\frac{1}{2}-\displaystyle\frac{1}{5}$$ $$=S_4=1^4+2^4+3^4+\ldots+n^4=\dfrac{n(2n+1)(n+1)(3n^2+3n-1)}{30}.$$

Fernando Revilla
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We have $$ (k+1)^5-k^5 = 5 k^4 + 10 k^3 + 10 k^2 + 5 k + 1 $$ and so $$ (n+1)^5= \sum_{k=0}^n (k+1)^5-\sum_{k=0}^nk^5 =\\ = \sum_{k=0}^n 5 k^4 + 10 k^3 + 10 k^2 + 5 k + 1 = 5 S_4 + 10 S_3 + 10 S_2 + 5 S_1 + S_0 $$ If you know the formulas for $S_3, S_2, S_1, S_0$, then you can solve for $S_4$.

lhf
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As $S_4(n)-S_4(n-1)=n^4$ is a quartic polynomial in $n$, $S_4(n)$ must be quintic.

You can obtain it as the Lagrangian interpolator of the six points

$$(0,0),(1,1),(2,17),(3,98),(4,354),(5,979).$$

The computation is quite laborious, but gives the correct

$$S_4(n)=\frac{n^5}5+\frac{n^4}2+\frac{n^3}3-\frac n{30}.$$

  • $S_4(n)$ must be quintic if you know that $S_4(n)$ must be a polynomial... – lhf Jan 12 '17 at 10:49
  • @lhf: what else could it be, when the first order difference is a polynomial ? (Note that it is easy to show the linear dependence of $S_0,S_1,\cdots S_d,n^{d+1}$.) –  Jan 12 '17 at 11:12
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Let $S_d(n):=\sum_{k=1}^n k^d$.

Then as $$k^d-(k-1)^d=dk^{d-1}-\binom d2k^{d-2}+\binom d3k^{d-3}-\cdots(-1)^d$$ we have $$S_1(n)-S_1(n-1)=S_0(n)=n,$$ $$S_2(n)-S_2(n-1)=2S_1(n)-S_0=n^2,$$ $$S_3(n)-S_3(n-1)=3S_2(n)-3S_1(n)+S_0=n^3,$$ $$S_4(n)-S_4(n-1)=4S_3(n)-6S_2(n)+4S_1-S_0=n^4,$$ $$S_5(n)-S_5(n-1)=5S_4(n)-10S_3(n)+10S_2-5S_1+S_0=n^5.$$

Solving, we obtain

$$S_1(n)=\frac{n^2}2+\frac n2,$$ $$S_2(n)=\frac{n^3}3+\frac{n^2}2+\frac n6,$$ $$S_3(n)=\frac{n^4}4+\frac{n^3}2+\frac{n^2}4,$$ $$S_4(n)=\frac{n^5}5+\frac{n^4}2+\frac{n^3}3-\frac n{30}.$$