Let the function $f(x,y,t)=\frac{x^2-y^2}{2}-\frac{(x-yt)^2}{1-t^2}$ for all real values $x,y$ and $t\not=\pm1$ a) Evaluate $f(2,0,3)$ and $f(0,2,3)$. b) Show that $f(x,y,0)=f(y,x,0)$ for any values of $(x,y)$. c) Show that $f(x,y,t)=f(y,x,t)$ for any values of $(x,y)$ and $t\not=\pm1$. d) Given $$g(x,y,s)=\frac{(x^2-y^2)(1+\sin(s))}{2} -\frac{(x-y\sin(s))^2}{1-\sin(s)}$$for all real values $x,y$ and $s\not=\frac{\pi}{2}+2\pi k$, where $k$ is an integer number, show that $g(x,y,s)=g(y,x,s)$ for any values of $(x,y)$ and $s$ in the domain of $g(x,y,s)$.
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Tags: Comc
CatsMeow12
08.11.2024 06:38
a) $f(2,0,3)=\frac{4}{2}-\frac{(2)^2}{1-9}=2+\frac{4}{8}=\frac{5}{2}$
$f(0,2,3)=-\frac{4}{2}-\frac{(-6)^2}{1-9}=-2+\frac{36}{8}=\frac{5}{2}$
b) \[f(x,y,0)=\frac{x^2-y^2}{2}-(x)^2\]\[f(x,y,0)=-\frac{x^2+y^2}{2}\]This expression is symmetric, so $f(x,y,0)=f(y,x,0).$
c) \[f(x,y,0)=\frac{x^2-y^2}{2}-\frac{(x-yt)^2}{1-t^2}\]\[f(x,y,0)=\frac{(x^2-y^2)(1-t^2)-2(x^2-2xyt+y^2t^2)}{2(1-t^2)}\]\[f(x,y,0)=\frac{x^2-y^2-x^2t^2+y^2t^2-2x^2+4xyt-2y^2t^2}{2(1-t^2)}\]\[f(x,y,0)=\frac{4xyt-(x^2+y^2)(1+t^2)}{2(1-t^2)}\]Again, this expression is symmetric (in $x$ and $y$), so $f(x,y,t)=f(y,x,t)$ for any $x,y$ and all $t\neq\pm1.$
d) This is just part (c) with a different denominator (checking for a shortcut saved me some work). The numerator of the second fraction is not affected by the different denominator, while the numerator of the first fraction becomes $(x^2-y^2)(1-\sin^2(s)),$ as before, but replacing $t$ with $\sin(s).$ So the manipulated fraction is \[f(x,y,0)=\frac{4xy\sin(s)-(x^2+y^2)(1+\sin^2(s))}{2(1-\sin(s))}.\]The restriction on $s$ is obvious, otherwise there'd be division by zero. This final expression is symmetric (in $x$ and $y$), so $g(x,y,s)=g(y,x,s)$ for any $x,y$ and all $s\neq\frac{\pi}{2}+2\pi k,\:k\in\mathbb{Z}.$