How Do You Calculate the Sum of Fourth Powers Up to N?

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Discussion Overview

The discussion centers on calculating the sum of the fourth powers of integers up to \( n \), expressed as \( 1^4 + 2^4 + \ldots + n^4 \). Participants explore various methods, including recursive relations, generating functions, and polynomial approaches, without reaching a consensus on a single method or formula.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose using recursive relations to derive the sum, referencing the binomial coefficients and previous sums \( s_k \).
  • Others suggest a linear inhomogeneous recursion approach, leading to a characteristic equation that indicates the closed-form solution involves polynomial terms.
  • A different method involves evaluating differences of consecutive sums to identify the degree of the polynomial generating function.
  • Some participants express confusion about the recursive relations and request further clarification on their derivation and application.
  • One participant mentions historical anecdotes related to the discovery of the formula for the sum of integers, connecting it to notable mathematicians.
  • Another approach involves defining a linear map and using it to express the fourth power sum in terms of polynomial functions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single method for calculating the sum of fourth powers. Multiple competing views and approaches are presented, with some participants seeking clarification on the methods discussed.

Contextual Notes

Some methods rely on specific assumptions about the properties of binomial coefficients and polynomial functions, which may not be universally applicable without further context. The discussion includes various mathematical steps that remain unresolved or require additional elaboration.

mathworker
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how to find the sum,
$$1^4+2^4...n^4$$
i tried myself but couldn't find an approach , when i googled it ifound the formula but not proof
 
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Re: sum of fourth powers of N

mathworker said:
how to find the sum,
$$1^4+2^4...n^4$$
i tried myself but couldn't find an approach , when i googled it ifound the formula but not proof

A slow but elementary approach consists in the use of the recursive relation... $$ \binom{k+1}{1}\ s_{1} + \binom{k+1}{2}\ s_{2} + ... + \binom{k+1}{k} s_{k} = (n+1)^{k+1} - (n+1)\ (1)$$

... where...

$$s_{k}= 1^{k} + 2^{k} + ... + n^{k}\ (2)$$

Starting with $s_{1}= \frac{n\ (n+1)}{2}$ we obtain...

$$\binom{3}{1}\ \frac{n\ (n+1)}{2} + \binom{3}{2}\ s_{2} = (n+1)^{3} - (n+1) \implies s_{2}= n\ (n+1)\ \frac{2 n + 1}{6}\ (3)$$

$$\binom{4}{1}\ \frac{n\ (n+1)}{2} + \binom{4}{2}\ n\ (n+1)\ \frac{2 n + 1}{6} + \binom {4}{3}\ s_{3} = (n+1)^{4} - (n+1) \implies s_{3}= n^{2}\ \frac{(n + 1)^{2}}{4}\ (4)$$

$$\binom{5}{1}\ \frac{n\ (n+1)}{2} + \binom{5}{2}\ n\ (n+1)\ \frac{2 n + 1}{6} + \binom {5}{3}\ n^{2}\ \frac{(n + 1)^{2}}{4}+ \binom{5}{4}\ s_{4} = (n+1)^{5} - (n+1) \implies $$

$$ \implies s_{4}= n\ (n+1)\ (2 n + 1)\ \frac{3 n^{2}+ 3 n -1}{30}\ (5)$$

Of course it is possible to proceed with $s_{5}$, $s_{6}$, etc...

Kind regards

$\chi$ $\sigma$
 
One method to derive the formula would be to use the linear inhomogeneous recursion:

$$S_{n}=S_{n-1}+n^4$$

$$S_{n+1}=S_{n}+(n+1)^4$$

Subtracting the former from the latter (called symbolic differencing), we obtain:

$$S_{n+1}=2S_{n}-S_{n-1}+4n^3+6n^2+4n+1$$

$$S_{n+2}=2S_{n+1}-S_{n}+4(n+1)^3+6(n+1)^2+4(n+1)+1$$

Subtracting again, we get:

$$S_{n+2}=3S_{n+1}-3S_{n}+S_{n-1}+12n^2+24n+14$$

$$S_{n+3}=3S_{n+2}-3S_{n+1}+S_{n}+12(n+1)^2+24(n+1)+14$$

Subtracting again, we get:

$$S_{n+3}=4S_{n+2}-6S_{n+1}+4S_{n}-S_{n-1}+24n+36$$

$$S_{n+4}=4S_{n+3}-6S_{n+2}+4S_{n+1}-S_{n}+24(n+1)+36$$

Subtracting again, we get:

$$S_{n+4}=5S_{n+3}-10S_{n+2}+10S_{n+1}-5S_{n}+S_{n-1}+24$$

$$S_{n+5}=5S_{n+4}-10S_{n+3}+10S_{n+2}-5S_{n+1}+S_{n}+24$$

Subtracting again, we get:

$$S_{n+5}=6S_{n+4}-15S_{n+3}+20S_{n+2}-15S_{n+1}+6S_{n}-S_{n-1}$$

We now have a homogeneous recurrence whose characteristic equation is:

$$r^6-6r^5+15r^4-20r^3+15r^2-6r+1=(r-1)^6=0$$

Since we have the root $r=1$ of multiplicity 6, we know the closed-form is:

$$S_n=k_1+k_2n+k_3n^2+k_4n^3+k_5n^4+k_6n^5$$

Now we may use initial values to determine the parameters $k_i$:

$$S_0=k_1=0$$

$$S_1=k_1+k_2+k_3+k_4+k_5+k_6=1$$

$$S_2=k_1+2k_2+4k_3+8k_4+16k_5+32k_6=17$$

$$S_3=k_1+3k_2+9k_3+27k_4+81k_5+243k_6=98$$

$$S_4=k_1+4k_2+16k_3+64k_4+256k_5+1024k_6=354$$

$$S_5=k_1+5k_2+25k_3+125k_4+625k_5+3125k_6=979$$

Solving this system, we find:

$$k_1=0,\,k_2=-\frac{1}{30},k_3=0,\,k_4=\frac{1}{3},\,k_5=\frac{1}{2},\,k_6=\frac{1}{5}$$

And so we may state:

$$S_n=-\frac{1}{30}n+\frac{1}{3}n^3+\frac{1}{2}n^4+\frac{1}{5}n^5=\frac{6n^5+15n^4+10n^3-n}{30}=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
 
Re: sum of fourth powers of N

chisigma said:
A slow but elementary approach consists in the use of the recursive relation... $$ \binom{k+1}{1}\ s_{1} + \binom{k+1}{2}\ s_{2} + ... + \binom{k+1}{k} s_{k} = (n+1)^{k+1} - (n+1)\ (1)$$
thanks to both of you,
would you explain this relation...i know nothing about it
 
Re: sum of fourth powers of N

mathworker said:
thanks to both of you,
would you explain this relation...i know nothing about it

We start writing...

$$\displaystyle a_{j}= (j+1)^{k+1} - j^{k+1} = \sum_{i=0}^{k} \binom{k+1}{i}\ j^{i}\ (1)$$

... and then we evaluate...

$$\displaystyle \sum_{j=0}^{n} a_{j} = (n+1)^{k+1} = \sum_{i=0}^{k} \binom{k+1}{i} s_{i} \implies \sum_{i=1}^{k} \binom{k+1}{i}\ s_{i} = (n+1)^{k+1} - (n+1)\ (2)$$

Kind regards

$\chi$ $\sigma$
 
Anoher way:

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

$(b)\;$ If $Y=AX$ is the equation of $T$ with respect to the canonical basis of $\mathbb{R}_5[x]$, using x^4\equiv(0,0,0,0,1,0)^t we get T^{-1}(x^4)\equiv (\alpha,-1/30,0,1/3,-1/2,1/5)^t with \alpha \in \mathbb{R} that is, $$T^{-1}(x^4)=\left\{{\alpha -x/30+x^3/3-x^4/2+x^5/5:\alpha \in{\mathbb{R}}}\right\}$$ $(c)$ Choose any polynomial h(x)\in{T^{-1}(x^4)} (for example, the corresponding to \alpha=0) . Such polynomial satisfies T(h(x))=x^4 i.e. h(x+1)-h(x)=x^4\;(*). For x=1,2,\ldots,n in (*) we get:

<br /> h(2)-h(1)=1^4\\<br /> h(3)-h(2)=2^4\\<br /> h(4)-h(3)=3^4\\<br /> \ldots\\<br /> h(n+1)-h(n)=n^4
​

That is, h(n+1)-h(n)=1^4+2^4+\ldots+n^4=S_4, hence

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}
​

Simplifying:

S_4=1^4+2^4+3^4+\ldots+n^4=\dfrac{n(2n+1)(n+1)(3n^2+3n-1)}{30}
​

P.S. More details here: http://www.fernandorevilla.es/docencia-problemas/problemas-2/2-s_414-n4-y-endomorfismo
 
Re: sum of fourth powers of N

chisigma said:
A slow but elementary approach consists in the use of the recursive relation... $$ \binom{k+1}{1}\ s_{1} + \binom{k+1}{2}\ s_{2} + ... + \binom{k+1}{k} s_{k} = (n+1)^{k+1} - (n+1)\ (1)$$

... where...

$$s_{k}= 1^{k} + 2^{k} + ... + n^{k}\ (2)$$

A more effective and elegant formula is...

$$ \binom {k+1}{0}\ s_{0} + \binom{k+1}{1}\ s_{1} + \binom{k+1}{2}\ s_{2} + ... + \binom{k+1}{k} s_{k} = (n+1)^{k+1}\ (1)$$

... where...

$$s_{k}= 0^{k} + 1^{k} + 2^{k} + ... + n^{k}\ (2)$$

Starting from $s_{0}= n+1$ [remember that $0^{0}$ is not an 'indeterminate form' but is $0^{0}=1$...] we obtain all the successive $s_{k}$ and in particular is $ n+1 + 2\ s_{1} = (n+1)^{2} \implies s_{1}= n\ \frac{n+1}{2}$. According to a very known anectode '... in 1787, when Gauss was only ten years old, his teacher had the students add up all the numbers from one to a hundred, with instructions that each should place his slate on a table as soon as he had completed the task. Almost immediately Gauss placed his slate on the table. The teacher looked at Gauss scornfully while the others worked diligently. When the instructor finally looked at the results, the slate of Gauss was the only one to have the correct answer, 5050, with no further calculation...', so that Gauss is considered as the 'discover' of the numerical value to $s_{1}$. Very well!... it could be interesting for someone of You to know that the general formula (1) was 'discovered' in the year 1654 [!] by the French matematician and philosopher Blaise Pascal... no comments? (Wasntme)...

Kind regards

$\chi$ $\sigma$
 
Hello, mathworker!

Here is a primitive procedure that should work
/ / for any polynomial function.

\text{Find the sum: }\:S_n \:=\;1^4+2^4+3^4+\cdots+n^4
Crank out the first few sums:

. . \begin{array}{cc}n &amp; S_n \\ \hline 1 &amp; 1 \\ 2 &amp; 17 \\ 3 &amp; 98 \\ 4 &amp; 354 \\ 5 &amp; 979 \\ 6 &amp; 2275 \\ 7 &amp; 4676 \\ 8 &amp; 8772 \end{array}Take the difference of consecutive terms,
then the differences of the differences, and so on.

\begin{array}{cccccccccccccccccc}\text{Sequence} &amp; 1 &amp;&amp; 17 &amp;&amp; 98 &amp;&amp; 354 &amp;&amp; 979 &amp;&amp; 2275 &amp;&amp; 4676 &amp;&amp; 8772 \\ \text{1st diff.} &amp;&amp; 16 &amp;&amp; 81 &amp;&amp;256 &amp;&amp; 625 &amp;&amp; 1296 &amp;&amp; 2401 &amp;&amp; 4096 \\ \text{2nd diff.} &amp;&amp;&amp; 65 &amp;&amp; 175 &amp;&amp; 369 &amp;&amp; 671 &amp;&amp; 1105 &amp;&amp; 1695 \\ \text{3rd diff.} &amp;&amp;&amp;&amp; 110 &amp;&amp; 194 &amp;&amp; 302 &amp;&amp; 434 &amp;&amp; 590 \\ \text{4th diff.} &amp;&amp;&amp;&amp;&amp; 84 &amp;&amp; 108 &amp;&amp; 132 &amp;&amp; 156 \\ \text{5th diff.} &amp;&amp;&amp;&amp;&amp;&amp; 24 &amp;&amp; 24 &amp;&amp; 24 \end{array}

We see that the 5th differences are constant.
Hence, the generating function is of the 5th degree.

The general 5th-degree function is: .f(n) \:=\:An^5 + Bn^4 + Cn^3 + Dn^2 + En + FSubstitute the first six values of the sequence:

\begin{array}{cccccc}f(1) = 1: &amp; A+B+C+D+E+F &amp;=&amp; 1 \\ f(2) = 17: &amp; 32A + 16B + 8C + 4D + 2E + F &amp;=&amp; 17 \\ f(3) = 98: &amp; 243A + 81B + 27C + 9D + 3E + F &amp;=&amp; 98 \\ f(4) = 354: &amp; 1024A + 256B + 64C + 16D + 4E + F &amp;=&amp; 354 \\ f(5) = 979: &amp; 3125A + 625B + 125C + 25D + 5E + F &amp;=&amp; 979 \\ f(6) = 2275: &amp; 7776A + 1296B + 216C + 36D + 6E + F &amp;=&amp; 2275 \end{array}Solve the system of equations: .\begin{Bmatrix}A &amp;=&amp; \tfrac{1}{5} &amp;&amp; D &amp;=&amp; 0 \\ B &amp;=&amp; \tfrac{1}{2} &amp;&amp; E &amp;=&amp; \text{-}\tfrac{1}{30} \\ C &amp;=&amp; \tfrac{1}{3} &amp;&amp; F &amp;=&amp; 0 \end{Bmatrix}

Hence: .S_n \;=\;\tfrac{1}{5}n^5 + \tfrac{1}{2}n^4 + \tfrac{1}{3}n^3 - \tfrac{1}{30}nTherefore: .\sum^n_{k=1} k^4 \;=\;\frac{n(6n^4 + 15n^3 + 10n^2 - 1)}{30}
 
soroban said:
Hello, mathworker!

Here is a primitive procedure that should work
/ / for any polynomial function.

Crank out the first few sums:

. . \begin{array}{cc}n &amp; S_n \\ \hline 1 &amp; 1 \\ 2 &amp; 17 \\ 3 &amp; 98 \\ 4 &amp; 354 \\ 5 &amp; 979 \\ 6 &amp; 2275 \\ 7 &amp; 4676 \\ 8 &amp; 8772 \end{array}Take the difference of consecutive terms,
then the differences of the differences, and so on.

\begin{array}{cccccccccccccccccc}\text{Sequence} &amp; 1 &amp;&amp; 17 &amp;&amp; 98 &amp;&amp; 354 &amp;&amp; 979 &amp;&amp; 2275 &amp;&amp; 4676 &amp;&amp; 8772 \\ \text{1st diff.} &amp;&amp; 16 &amp;&amp; 81 &amp;&amp;256 &amp;&amp; 625 &amp;&amp; 1296 &amp;&amp; 2401 &amp;&amp; 4096 \\ \text{2nd diff.} &amp;&amp;&amp; 65 &amp;&amp; 175 &amp;&amp; 369 &amp;&amp; 671 &amp;&amp; 1105 &amp;&amp; 1695 \\ \text{3rd diff.} &amp;&amp;&amp;&amp; 110 &amp;&amp; 194 &amp;&amp; 302 &amp;&amp; 434 &amp;&amp; 590 \\ \text{4th diff.} &amp;&amp;&amp;&amp;&amp; 84 &amp;&amp; 108 &amp;&amp; 132 &amp;&amp; 156 \\ \text{5th diff.} &amp;&amp;&amp;&amp;&amp;&amp; 24 &amp;&amp; 24 &amp;&amp; 24 \end{array}

We see that the 5th differences are constant.
Hence, the generating function is of the 5th degree.
yeah, i got it thank you, but why should that happen
 
  • #10
What is the $n$th derivative of an $n$th degree polynomial?
 
  • #11
n factorial
 
  • #12
Yes, if the leading coefficient is 1, but what I was getting at is that it is a constant...
 
Last edited:
  • #13
how do we relate differences triangle with nth derivative
 
  • #14
You should observe that a common theme in the various methods given above is differencing...how are differencing and differentiation related?
 
  • #15
MarkFL said:
You should observe that a common theme in the various methods given above is differencing...how are differencing and differentiation related?
can't figure it out:confused:
 
  • #16
What is the definition of the derivative?
 
  • #17
MarkFL said:
What is the definition of the derivative?

change in the value of a function w.r.t an infinitesimally small change in the variable,
 
  • #18
What if the change in the independent variable is 1?
 
  • #19
MarkFL said:
What if the change in the independent variable is 1?

change in the value of function ...but change in variable should be very small as per def
 

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