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Sum of fourth powers of N

mathworker

Active member
May 31, 2013
118
how to find the sum,
\(\displaystyle 1^4+2^4.....n^4\)
i tried myself but couldn't find an approach , when i googled it ifound the formula but not proof
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: sum of fourth powers of N

how to find the sum,
\(\displaystyle 1^4+2^4.....n^4\)
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

$\chi$ $\sigma$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
One method to derive the formula would be to use the linear inhomogeneous recursion:

\(\displaystyle S_{n}=S_{n-1}+n^4\)

\(\displaystyle S_{n+1}=S_{n}+(n+1)^4\)

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

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

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

Subtracting again, we get:

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

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

Subtracting again, we get:

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

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

Subtracting again, we get:

\(\displaystyle S_{n+4}=5S_{n+3}-10S_{n+2}+10S_{n+1}-5S_{n}+S_{n-1}+24\)

\(\displaystyle S_{n+5}=5S_{n+4}-10S_{n+3}+10S_{n+2}-5S_{n+1}+S_{n}+24\)

Subtracting again, we get:

\(\displaystyle S_{n+5}=6S_{n+4}-15S_{n+3}+20S_{n+2}-15S_{n+1}+6S_{n}-S_{n-1}\)

We now have a homogeneous recurrence whose characteristic equation is:

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

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

\(\displaystyle S_n=k_1+k_2n+k_3n^2+k_4n^3+k_5n^4+k_6n^5\)

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

\(\displaystyle S_0=k_1=0\)

\(\displaystyle S_1=k_1+k_2+k_3+k_4+k_5+k_6=1\)

\(\displaystyle S_2=k_1+2k_2+4k_3+8k_4+16k_5+32k_6=17\)

\(\displaystyle S_3=k_1+3k_2+9k_3+27k_4+81k_5+243k_6=98\)

\(\displaystyle S_4=k_1+4k_2+16k_3+64k_4+256k_5+1024k_6=354\)

\(\displaystyle S_5=k_1+5k_2+25k_3+125k_4+625k_5+3125k_6=979\)

Solving this system, we find:

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

And so we may state:

\(\displaystyle S_n=-\frac{1}{30}n+\frac{1}{3}n^3+\frac{1}{2}n^4+\frac{1}{5}n^5=\frac{6n^5+15n^4+10n^3-n}{30}=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\)
 

mathworker

Active member
May 31, 2013
118
Re: sum of fourth powers of N

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
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: sum of fourth powers of N

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

$\chi$ $\sigma$
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Anoher way:

[tex](a)\;T:\mathbb{R}_5[x]\rightarrow{\mathbb{R}_5[x]}[/tex] defined by [tex]T(p(x))=p(x+1)-p(x)[/tex] 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 [tex]x^4\equiv(0,0,0,0,1,0)^t[/tex] we get [tex]T^{-1}(x^4)\equiv (\alpha,-1/30,0,1/3,-1/2,1/5)^t[/tex] with [tex]\alpha \in \mathbb{R}[/tex] 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 [tex]h(x)\in{T^{-1}(x^4)}[/tex] (for example, the corresponding to [tex]\alpha=0[/tex]) . Such polynomial satisfies [tex]T(h(x))=x^4[/tex] i.e. [tex]h(x+1)-h(x)=x^4\;(*)[/tex]. For [tex]x=1,2,\ldots,n[/tex] in [tex](*)[/tex] we get:

[tex]
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[/tex]

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

[tex]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}[/tex]

Simplifying:

[tex]S_4=1^4+2^4+3^4+\ldots+n^4=\dfrac{n(2n+1)(n+1)(3n^2+3n-1)}{30}[/tex]

P.S. More details here: 2. Endomorfismo y suma S_4=1^4+…+n^4 | Fernando Revilla
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: sum of fourth powers of N

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

$\chi$ $\sigma$
 

soroban

Well-known member
Feb 2, 2012
409
Hello, mathworker!

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


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

. . [tex]\begin{array}{cc}n & S_n \\ \hline 1 & 1 \\ 2 & 17 \\ 3 & 98 \\ 4 & 354 \\ 5 & 979 \\ 6 & 2275 \\ 7 & 4676 \\ 8 & 8772 \end{array}[/tex]


Take the difference of consecutive terms,
then the differences of the differences, and so on.

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

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

The general 5th-degree function is: .[tex]f(n) \:=\:An^5 + Bn^4 + Cn^3 + Dn^2 + En + F[/tex]


Substitute the first six values of the sequence:

[tex]\begin{array}{cccccc}f(1) = 1: & A+B+C+D+E+F &=& 1 \\ f(2) = 17: & 32A + 16B + 8C + 4D + 2E + F &=& 17 \\ f(3) = 98: & 243A + 81B + 27C + 9D + 3E + F &=& 98 \\ f(4) = 354: & 1024A + 256B + 64C + 16D + 4E + F &=& 354 \\ f(5) = 979: & 3125A + 625B + 125C + 25D + 5E + F &=& 979 \\ f(6) = 2275: & 7776A + 1296B + 216C + 36D + 6E + F &=& 2275 \end{array}[/tex]


Solve the system of equations: .[tex]\begin{Bmatrix}A &=& \tfrac{1}{5} && D &=& 0 \\ B &=& \tfrac{1}{2} && E &=& \text{-}\tfrac{1}{30} \\ C &=& \tfrac{1}{3} && F &=& 0 \end{Bmatrix}[/tex]

Hence: .[tex]S_n \;=\;\tfrac{1}{5}n^5 + \tfrac{1}{2}n^4 + \tfrac{1}{3}n^3 - \tfrac{1}{30}n [/tex]


Therefore: .[tex]\sum^n_{k=1} k^4 \;=\;\frac{n(6n^4 + 15n^3 + 10n^2 - 1)}{30}[/tex]
 

mathworker

Active member
May 31, 2013
118
Hello, mathworker!

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


Crank out the first few sums:

. . [tex]\begin{array}{cc}n & S_n \\ \hline 1 & 1 \\ 2 & 17 \\ 3 & 98 \\ 4 & 354 \\ 5 & 979 \\ 6 & 2275 \\ 7 & 4676 \\ 8 & 8772 \end{array}[/tex]


Take the difference of consecutive terms,
then the differences of the differences, and so on.

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

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
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What is the $n$th derivative of an $n$th degree polynomial?
 

mathworker

Active member
May 31, 2013
118
n factorial
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, if the leading coefficient is 1, but what I was getting at is that it is a constant...
 
Last edited:

mathworker

Active member
May 31, 2013
118
how do we relate differences triangle with nth derivative
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You should observe that a common theme in the various methods given above is differencing...how are differencing and differentiation related?
 

mathworker

Active member
May 31, 2013
118
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:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What is the definition of the derivative?
 

mathworker

Active member
May 31, 2013
118
What is the definition of the derivative?
change in the value of a function w.r.t an infinitesimally small change in the variable,
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What if the change in the independent variable is 1?
 

mathworker

Active member
May 31, 2013
118
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