- Thread starter
- #1

#### mathworker

##### Well-known member

- May 31, 2013

- 119

\(\displaystyle 1^4+2^4.....n^4\)

i tried myself but couldn't find an approach , when i googled it ifound the formula but not proof

- Thread starter mathworker
- Start date

- Thread starter
- #1

- May 31, 2013

- 119

\(\displaystyle 1^4+2^4.....n^4\)

i tried myself but couldn't find an approach , when i googled it ifound the formula but not proof

- Feb 13, 2012

- 1,704

A slow but elementary approach consists in the use of the recursive relation...

\(\displaystyle 1^4+2^4.....n^4\)

i tried myself but couldn't find an approach , when i googled it ifound the formula but not proof

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

... 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

$\chi$ $\sigma$

- Admin
- #3

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

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

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:

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

- Thread starter
- #4

- May 31, 2013

- 119

thanks to both of you,A slow but elementary approach consists in the use of the recursive relation...

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

would you explain this relation.....i know nothing about it

- Feb 13, 2012

- 1,704

We start writing...thanks to both of you,

would you explain this relation.....i know nothing about it

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

... 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

$\chi$ $\sigma$

- Jan 29, 2012

- 661

[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

- Feb 13, 2012

- 1,704

A more effective and elegant formula is...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...

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

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

... 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? ...

Kind regards

$\chi$ $\sigma$

Here is a primitive procedure that should work

/ / for any polynomial function.

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

. . [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]

- Thread starter
- #9

- May 31, 2013

- 119

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

- Admin
- #10

- Thread starter
- #11

- May 31, 2013

- 119

n factorial

- Admin
- #12

- Thread starter
- #13

- May 31, 2013

- 119

how do we relate differences triangle with nth derivative

- Admin
- #14

- Thread starter
- #15

- May 31, 2013

- 119

can't figure it outYou should observe that a common theme in the various methods given above is differencing...how are differencing and differentiation related?

- Admin
- #16

- Thread starter
- #17

- May 31, 2013

- 119

change in the value of a function w.r.t an infinitesimally small change in the variable,What is the definition of the derivative?

- Admin
- #18

- Thread starter
- #19

- May 31, 2013

- 119

change in the value of function ...but change in variable should be very small as per defWhat if the change in the independent variable is 1?