MHB Solve Homogeneous Function f(x,y) w/ Euler's Rule

[Yankel]

Hello,

I need some help with this question here, I'll explain why in a second. The question is:

f(x,y) is a homogeneous function of order 3. It is known that:

\[f_{x}(2,1)=7\]

and

\[f_{y}(8,4)=5\]

find f(12,6).

Now, I know of Euler's rule, which includes the partial derivatives, but in the question the (x,y) point if different between f, fx, and fy, I am confused. Can you assist please ?

Thanks !

(Final answer should be 1030.5)

 

[chisigma]

Hello,

I need some help with this question here, I'll explain why in a second. The question is:

f(x,y) is a homogeneous function of order 3. It is known that:

\[f_{x}(2,1)=7\]

and

\[f_{y}(8,4)=5\]

find f(12,6).

Now, I know of Euler's rule, which includes the partial derivatives, but in the question the (x,y) point if different between f, fx, and fy, I am confused. Can you assist please ?

Thanks !

(Final answer should be 1030.5)

You can reasonably suppose that the unknown function has the form...

$\displaystyle f(x,y)= a\ x^{2}\ y + b\ x\ y^{2}\ (1)$

... so that is...

$\displaystyle f_{x} = 2\ a\ x\ y + b\ y^{2}$

$\displaystyle f_{y} = a\ x^{2} + 2\ b\ x\ y\ (2)$

Now a and b can be found inserting in (2) the conditions $f_{x} (2,1)=7$ and $f_{y}(8,4)=5$ and solving the 2 x 2 linear system...

Kind regards

$\chi$ $\sigma$

 

[Yankel]

Thanks.

Can I take you back to the first stage ? How did you know / decide that this is the form of the function ? Based on what ?

 

[chisigma]

Thanks.

Can I take you back to the first stage ? How did you know / decide that this is the form of the function ? Based on what ?

Particular homogeneous functions of degree 3 are $f_{1}=x^{2}\ y$ and $f_{2}=x\ y^{2}$, so that any linear combination of them like $f(x,y) + c_{1}\ f_{1} + c_{2}\ f_{2}$ is also a homogeneous function of degree 3. But why don't include also $f_{3}= x^{3}$ and $f_{4} = y^{3}$?... that's right, and that means that the general case is $f(x, y) = c_{1}\ f_{1} + c_{2}\ f_{2} + c_{3}\ f_{3} + c_{4}\ f_{4}$ and Your problem has infinity solutions...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

Hello,

I need some help with this question here, I'll explain why in a second. The question is:

f(x,y) is a homogeneous function of order 3. It is known that:

\[f_{x}(2,1)=7\]

and

\[f_{y}(8,4)=5\]

find f(12,6).

Now, I know of Euler's rule, which includes the partial derivatives, but in the question the (x,y) point if different between f, fx, and fy, I am confused. Can you assist please ?

Thanks !

(Final answer should be 1030.5)

A homogeneous function $f$ of order $k$ has the properties (i) its partial derivatives are homogeneous of order $k-1$; (ii) $f(\alpha \vec{v}) = \alpha^kf(\vec{v})$ (for any scalar $\alpha$ and vector $\vec{v}$); (iii) [Euler's theorem] $\vec{v}\cdot \nabla f(\vec{v}) = k f(\vec{v})$.

The useful feature of this question is that all the vectors are multiples of each other. In fact, $(12,6) = 6(2,1) = \frac32(8,4)$. By (i), the partial derivatives of $f$ are homogeneous of order $2$. It follows from (ii) that $f_x(12,6) = 6^2f_x(2,1)$ and $f_y(12,6) = \bigl(\frac32\bigr)^2f_y(8,4).$ You can then use (iii) to find $f(12,6).$