PhysicsCat
New member
- Joined
- May 1, 2023
- Messages
- 5
Hello!
I have the following task:
"[math]f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} &\text{if } x \in A_{\alpha} = \{(x,y) \in \mathbb{R}^2 : |y| < x^{\alpha} \} \\ 0 &\text{if } x = (0,0) \end{cases}[/math]For which [imath]\alpha > 0[/imath] is [imath]f[/imath] continuous?"
My ideas:
We have shown in the lectures, that [imath]f[/imath] is continuous for [imath]\alpha = 2[/imath]. We did this with the following inequality:
[math]\left| \frac{xy}{x^2+y^2} - 0 \right| = |x| \cdot \frac{|y|}{x^2+y^2} \leq |x| \frac{x^2}{x^2+y^2} \leq |x| \rightarrow 0 = f(0,0)[/math]If we do the same steps with [imath]\alpha > 0[/imath], we get
[math]|x| \cdot \frac{x^{\alpha}}{x^2+y^2} \leq |x|[/math]In order for this inequality to be true, we need
[math]\frac{x^{\alpha}}{x^2+y^2} \leq 1[/math]This is the point, where I get stuck. I cant figure out, how I can determine an expression for [imath]\alpha[/imath] out of this. Does someone have an idea for me?
(Also excuse my english please, its not my first language)
Thank you very much and have a nice day
I have the following task:
"[math]f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} &\text{if } x \in A_{\alpha} = \{(x,y) \in \mathbb{R}^2 : |y| < x^{\alpha} \} \\ 0 &\text{if } x = (0,0) \end{cases}[/math]For which [imath]\alpha > 0[/imath] is [imath]f[/imath] continuous?"
My ideas:
We have shown in the lectures, that [imath]f[/imath] is continuous for [imath]\alpha = 2[/imath]. We did this with the following inequality:
[math]\left| \frac{xy}{x^2+y^2} - 0 \right| = |x| \cdot \frac{|y|}{x^2+y^2} \leq |x| \frac{x^2}{x^2+y^2} \leq |x| \rightarrow 0 = f(0,0)[/math]If we do the same steps with [imath]\alpha > 0[/imath], we get
[math]|x| \cdot \frac{x^{\alpha}}{x^2+y^2} \leq |x|[/math]In order for this inequality to be true, we need
[math]\frac{x^{\alpha}}{x^2+y^2} \leq 1[/math]This is the point, where I get stuck. I cant figure out, how I can determine an expression for [imath]\alpha[/imath] out of this. Does someone have an idea for me?
(Also excuse my english please, its not my first language)
Thank you very much and have a nice day