tag:blogger.com,1999:blog-6837159629100463303.post7046147659574304104..comments2023-06-18T01:25:08.748-07:00Comments on Information Transfer Economics: Effective information equilibrium theoryJason Smithhttp://www.blogger.com/profile/12680061127040420047noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6837159629100463303.post-86790196033555668182016-03-26T22:03:43.296-07:002016-03-26T22:03:43.296-07:00Typo. Thanks Tom.
Can see from units of [c] = [A/...Typo. Thanks Tom.<br /><br />Can see from units of [c] = [A/B] so need [1/A] constant to give overall dimensionless formula.<br /><br />I'll probably re-write it in a different way, though.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-87227784294426557352016-03-26T21:47:23.316-07:002016-03-26T21:47:23.316-07:00Multiply both sides of (2) by $A_0$ to get an expr...Multiply both sides of (2) by $A_0$ to get an expression for $A$, and call that (3). Then differentiate (3) wrt B, and call that (4). Then compare that with dividing (3) by B, multiplying by k and then adding c. The result should be (4).Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-17824193003091588332016-03-26T21:42:04.961-07:002016-03-26T21:42:04.961-07:00Jason, instead of your 3rd expression:
$$\frac{A}{...Jason, instead of your 3rd expression:<br />$$\frac{A}{A_{0}} = \frac{c}{1-k} \frac{B}{B_{0}} + \left( \frac{B}{B_{0}} \right)^k \tag 1$$<br />I get this:<br />$$\frac{A}{A_{0}} = \frac{c}{1-k} \frac{B}{A_{0}} + \left( \frac{B}{B_{0}} \right)^k \tag 2$$<br />Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.com