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Is it possible to re-express the function $$ f(t+x_1,t+x_2,x_1,x_2)=x_1+x_2+t $$ as $f(y_1,y_2,y_3,y_4)=???$

user103828
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We could have $f(y_1, y_2, y_3, y_4) = y_1 + y_4$ or $f(y_1, y_2, y_3, y_4) = y_2 + y_3$ so there is no unique solution. Infinitely many solutions can be found of the form $ay_1 + (1-a)y_2 + (1-a)y_3 + ay_4$ for constant $a$. However, there exist solutions that are not of this form.

Karolis Juodelė
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  • Thanks. Can this be related to something from linear algebra? For example, does it have to do with the relationship between the dimension of the kernel or range. (I'm trying to relate this to something that I know). – user103828 Jul 30 '14 at 07:30
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    @user103828, certainly, you can consider the function $f(y_1, y_2, y_3, y_4) = ay_1 + by_2 + cy_3 + dy_4$, derive from the given equation, a system of constraints $a, b, c, d$ must satisfy, and observe that the corresponding matrix is not invertible. – Karolis Juodelė Jul 30 '14 at 08:08