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seen Apr 20 at 17:19

Jul
2
awarded  Curious
Apr
11
asked Finding midpoint of a cluster of points (2D)
Feb
16
awarded  Yearling
Dec
2
awarded  Popular Question
Nov
17
awarded  Popular Question
Nov
15
awarded  Popular Question
Nov
8
awarded  Popular Question
Apr
13
comment How can I solve this system of non-linear equations?
Ah, it makes perfect sense now that you point that out. Big thanks!
Apr
13
accepted How can I solve this system of non-linear equations?
Apr
13
comment How can I solve this system of non-linear equations?
Cool, I didn't know about that one.
Apr
13
asked How can I solve this system of non-linear equations?
Apr
3
comment Help with the limiting behavior of a stochastic process
Good to know. I'm slogging through this algebra text for review, and looking forward to getting to the more advanced stuff :)
Apr
2
accepted Help with the limiting behavior of a stochastic process
Apr
2
comment Help with the limiting behavior of a stochastic process
This looks like it's beyond the scope of the book, but I appreciate the help! I guess we're just supposed to observe the behavior so we'll be familiar with it when we have the necessary tools (like those you've presented) for more rigorous analysis.
Apr
2
revised Help with the limiting behavior of a stochastic process
edited title
Apr
2
asked Help with the limiting behavior of a stochastic process
Mar
30
comment Help evaluating $\lim_{\left|x\right| \to \infty}y$, given $\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$
Nice! Thanks again!
Mar
30
comment Help evaluating $\lim_{\left|x\right| \to \infty}y$, given $\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$
I think I got it now: ($\heartsuit$) says that the left side of the equation equals a constant value, even as $\left|x\right|\to\infty$, and since one factor of that left side will $\to \pm \infty$ as $\left|x\right|\to \infty$, the other must $\to 0$ to compensate and keep the product equal to that constant value. Is that about the gist of it?
Mar
30
comment Help evaluating $\lim_{\left|x\right| \to \infty}y$, given $\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$
Very cool! I wasn't able to follow you on the second approach, but the first is crystal clear. Thank you!
Mar
30
accepted Help evaluating $\lim_{\left|x\right| \to \infty}y$, given $\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$