- #1

- 28

- 0

[tex]\{0\leq\tau\leq t,0\leq t<\infty\}[/tex] Is under the transformation:

[tex]t=u+v,\tau=v[/tex]

I know its just

[tex]\{0\leq u<\infty,0\leq v<\infty\}[/tex]

But i am having difficulty showing this.

Thanks.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter the1ceman
- Start date

- #1

- 28

- 0

[tex]\{0\leq\tau\leq t,0\leq t<\infty\}[/tex] Is under the transformation:

[tex]t=u+v,\tau=v[/tex]

I know its just

[tex]\{0\leq u<\infty,0\leq v<\infty\}[/tex]

But i am having difficulty showing this.

Thanks.

- #2

- 28

- 0

Anyone?

- #3

Defennder

Homework Helper

- 2,591

- 5

[tex]0 \leq v \leq u+v < \infty[/tex] after substituting u, v for tau and t and combining the inequalities.

Subtracting v from from the above inequalities give: [tex]0 \leq u < \infty[/tex]

From the first inequality you should be able to see that [tex]0\leq v<\infty[/tex] holds.

Subtracting v from from the above inequalities give: [tex]0 \leq u < \infty[/tex]

From the first inequality you should be able to see that [tex]0\leq v<\infty[/tex] holds.

Last edited:

Share: