MHB Can $S_5$ be written as a multiple of $S_3$ and $S_2$?

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion centers on proving the equation $\frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}$ given that $S_n=x^n+y^n+z^n$ and $S_1=0$, which implies $x+y+z=0$. The proof involves manipulating the expressions for $S_5$, $S_3$, and $S_2$ using algebraic identities and relationships derived from the condition $S_1=0$. The calculations show that $S_5$ can be expressed in terms of $S_3$ and $S_2$, ultimately confirming the desired equation. The proof concludes successfully with the established relationship.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $$S_n=x^n+y^n+z^n$$. If $$S_1=0$$, prove that $$\frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}$$.
 
Mathematics news on Phys.org
Re: Prove S_5/5= S_3/3. S_2/2

My solution:

If we view $S_n$ as a recursive algorithm, we see that it must come from the characteristic equation:

$$(r-x)(r-y)(r-z)=0$$

$$r^3-(x+y+z)r^2+(xy+xz+yz)r-xyz=0$$

Since $x+y+z=S_1=0$, we obtain the following recursion:

$$S_{n+3}=-(xy+xz+yz)S_{n+1}+xyzS_{n}$$

Now, observing we may write:

$$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$$

$$0=S_2+2(xy+xz+yz)$$

$$-(xy+xz+yz)=\frac{S_2}{2}$$

Also, we find:

$$(x+y+z)^3=-2\left(x^3+y^3+z^3 \right)+3\left(x^2+y^2+z^2 \right)(x+y+z)+6xyz$$

$$0=-2S_3+6xyz$$

$$xyz=\frac{S^3}{3}$$

And so our recursion may be written:

$$S_{n+3}=\frac{S_2}{2}S_{n+1}+\frac{S_3}{3}S_{n}$$

Letting $n=2$, we then find:

$$S_{5}=\frac{S_2}{2}S_{3}+\frac{S_3}{3}S_{2}$$

$$S_{5}=\frac{5}{6}S_2S_{3}$$

$$\frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}$$

Shown as desired.
 
Re: Prove S_5/5= S_3/3. S_2/2

MarkFL said:
My solution:

If we view $S_n$ as a recursive algorithm, we see that it must come from the characteristic equation:

$$(r-x)(r-y)(r-z)=0$$

$$r^3-(x+y+z)r^2+(xy+xz+yz)r-xyz=0$$

Since $x+y+z=S_1=0$, we obtain the following recursion:

$$S_{n+3}=-(xy+xz+yz)S_{n+1}+xyzS_{n}$$

Now, observing we may write:

$$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$$

$$0=S_2+2(xy+xz+yz)$$

$$-(xy+xz+yz)=\frac{S_2}{2}$$

Also, we find:

$$(x+y+z)^3=-2\left(x^3+y^3+z^3 \right)+3\left(x^2+y^2+z^2 \right)(x+y+z)+6xyz$$

$$0=-2S_3+6xyz$$

$$xyz=\frac{S^3}{3}$$

And so our recursion may be written:

$$S_{n+3}=\frac{S_2}{2}S_{n+1}+\frac{S_3}{3}S_{n}$$

Letting $n=2$, we then find:

$$S_{5}=\frac{S_2}{2}S_{3}+\frac{S_3}{3}S_{2}$$

$$S_{5}=\frac{5}{6}S_2S_{3}$$

$$\frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}$$

Shown as desired.

Thanks for participating, MarkFL! And your method is neat and elegant!

We are given $S_n=x^n+y^n+z^n$ and $S_1=0$ which implies $x+y+z=0$.

From this given information we then have

[TABLE="class: grid, width: 500"]
[TR]
[TD]1.[/TD]
[TD]2.[/TD]
[/TR]
[TR]
[TD]$\small(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$

$0=x^2+y^2+z^2+2(xy+yz+xz)$

$x^2+y^2+z^2=-2(xy+yz+xz)$

$x^2+y^2+z^2=-2(xy+z(x+y))$

$x^2+y^2+z^2=-2xy-2z(-z)$

$x^2+y^2-z^2=-2xy$

$xy=-\left(\dfrac{x^2+y^2-z^2}{2} \right)$[/TD]
[TD]$\small(x+y+z)^3=x^3+y^3+z^3+3(xy(x+y)+yz(y+z)+xz(x+z))+6xyz$

$0=x^3+y^3+z^3+3(xy(-z)+yz(-x)+xz(-y))+6xyz$

$0=x^3+y^3+z^3+3(xy(-z)+yz(-x)+xz(-y))+6xyz$

$x^3+y^3+z^3=3xyz$[/TD]
[/TR]
[/TABLE]

We're asked to prove $\dfrac{S_5}{5}=\dfrac{S_3}{3}\cdot\dfrac{S_2}{2}$.

We see that

$S_5=x^5+y^5+z^5$

$\;\;\;\;\;=x^5+y^5+(-x-y)^5$

$\;\;\;\;\;=x^5+y^5-(x+y)^5$

$\;\;\;\;\;=x^5+y^5-(x^5+5x^4y+10x^3y^3+10x^2y^3+5xy^4+y^5)$

$\;\;\;\;\;=-(5x^4y+10x^3y^3+10x^2y^3+5xy^4)$

$\;\;\;\;\;=-(5xy(x^3+y^3)+10x^2y^2(x+y))$

$\;\;\;\;\;=-5xy((x^3+y^3)+2xy(x+y))$

$\;\;\;\;\;=-5xy((x+y)(x^2-xy+y^2)+2xy(x+y))$

$\;\;\;\;\;=-5xy((x+y)(x^2-xy+y^2+2xy))$

$\;\;\;\;\;=-5xy((x+y)(x^2+xy+y^2))$

$\;\;\;\;\;=-5\left(\dfrac{(x^3+y^3+z^3}{3z} \right)(-z)(x^2-\left(\dfrac{x^2+y^2-z^2}{2} \right)+y^2))$

$\;\;\;\;\;=5\left(\dfrac{x^3+y^3+z^3}{3} \right)\left(\dfrac{x^2+y^2+z^2}{2} \right)$

and therefore we obtain

$\dfrac{S_5}{5}=\dfrac{S_3}{3}\cdot\dfrac{S_2}{2}$ and we're done.(Emo)
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top