MHB Showing $XF_{X}+YF_{Y}+ZF_{Z}=nF$ with a Homogeneous Polynomial

[evinda]

Hi! (Smile)

Let $F(X,Y,Z) \in \mathbb{C}[X,Y,Z]$ a homogeneous polynomial of degree $n$. Could you give me a hint how we could show the following? (Thinking)

$$XF_{X}+YF_{Y}+ZF_{Z}=nF$$

 

[Fallen Angel]

Hi evinda,

This is a particular case of Leibniz's identity.

Call $F=\displaystyle\sum_{j=0}^{k}\lambda_{j}X^{e_{x,j}}Y^{e_{y,j}}Z^{e_{z,j}}$ where $e_{x,j}+e_{y,j}+e_{z,j}=n$ for all $j$.

Then $F_{X}=\displaystyle\sum_{j=0, \ e_{x,j}\geq 1}^{k}e_{x,j}\lambda_{j}X^{e_{x,j}-1}Y^{e_{y,j}}Z^{e_{z,j}}$

And so for $Y, Z$, now it's just a computation.

Actually, in this way you can prove that this kind of equality holds for every homogeneus polynomial (no matter how many variables) over any field $K$ while $n$ is not a divisor of $ch(K)$

 

[evinda]

Hi evinda,

This is a particular case of Leibniz's identity.

Call $F=\displaystyle\sum_{j=0}^{k}\lambda_{j}X^{e_{x,j}}Y^{e_{y,j}}Z^{e_{z,j}}$ where $e_{x,j}+e_{y,j}+e_{z,j}=n$ for all $j$.

Then $F_{X}=\displaystyle\sum_{j=0, \ e_{x,j}\geq 1}^{k}e_{x,j}\lambda_{j}X^{e_{x,j}-1}Y^{e_{y,j}}Z^{e_{z,j}}$

And so for $Y, Z$, now it's just a computation.

Actually, in this way you can prove that this kind of equality holds for every homogeneus polynomial (no matter how many variables) over any field $K$ while $n$ is not a divisor of $ch(K)$

This is the original exercise:

If $F(x,y,z) \in \mathbb{C}[x,y,z]$ is a homogeneous polynomial of degree $n$, prove the Euler's formula.

$$XF_X+YF_Y+ZF_Z=nF$$

Is the last equality Euler's formula? (Thinking)

 

[evinda]

Hi evinda,

This is a particular case of Leibniz's identity.

Call $F=\displaystyle\sum_{j=0}^{k}\lambda_{j}X^{e_{x,j}}Y^{e_{y,j}}Z^{e_{z,j}}$ where $e_{x,j}+e_{y,j}+e_{z,j}=n$ for all $j$.

Then $F_{X}=\displaystyle\sum_{j=0, \ e_{x,j}\geq 1}^{k}e_{x,j}\lambda_{j}X^{e_{x,j}-1}Y^{e_{y,j}}Z^{e_{z,j}}$

And so for $Y, Z$, now it's just a computation.

Actually, in this way you can prove that this kind of equality holds for every homogeneus polynomial (no matter how many variables) over any field $K$ while $n$ is not a divisor of $ch(K)$

Could we also take this polynomial: $F(X,Y,Z)=\sum_{i+j+k=n} a_{ijk} X^i Y^j Z^k$ ? (Thinking)

Could you explain me why we choose such a polynomial, that there is for example no constant term? (Thinking)

 

[Fallen Angel]

Hi evinda,

Both polynomials (yours and mine) are the same with different notation.

We choose such a polynomial because we want a homogeneus one of degree $n$.And this is more known as Euler's formula, but once a professors told me that it was originally proved by Leibniz, I haven't really found reliable information about this but I trust in my professor.