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Like in single variable, we use $f(-x)=-f(x)$ to show that a function is odd. Similarly, for two variables, we can use $f(-x,-y)=-f(x,y)$.

If we have a two variable function like this $f(x,y)=x\cos({\sqrt{x^2+(y+a)^2}})$.

So, $f(-x,y)=(-x)\cos({\sqrt{(-x)^2+(y+a)^2}})=-f(x,y)$.

Can we say this is an odd function only in $x$?

zhk
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1 Answers1

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Yes you can say such a thing: as a matter of fact, if $f:I\times J \rightarrow \mathbb{R}$ with $I, J \subseteq \mathbb{R}$ then $f(x,\bar{y})$, with $\bar{y}\in J$, is just a function in $x$, i.e. $g:I\rightarrow \mathbb{R}$ defined as $g(x):=f(x,\bar{y})$, hence it is licit to say that it can be odd or even. Now, if this property holds independently of the specific value of $y$, thus the function $f$, with $y$ fixed, is odd in $x$.

Vexx23
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