2

Let $$F(r,\mu,t)=r- \frac{2\mu\left ( \dfrac{e^t(t-1)-1}{t^2} \right )}{(1-2\mu)+2\mu \left ( \dfrac{e^t-1}{t} \right )}=0,$$ I would like to find a linear approximation$$ t(r,\mu) \approx t(0,0) + r\frac{\partial t(r,\mu)}{\partial r}+\mu \frac{\partial t(r,\mu)}{\partial \mu},$$ where$$ \frac{\partial t(r,\mu)}{\partial r}=-\left.\frac{\partial F(r,\mu,t)}{\partial r}\middle/\frac{\partial F(r,\mu,t)}{\partial t}\right., \quad \frac{\partial t(r,\mu)}{\partial \mu}=-\left.\frac{\partial F(r,\mu,t)}{\partial \mu}\middle/\frac{\partial F(r,\mu,t)}{\partial t}\right..$$

I want to find $t(0,0)$, which I do not know how to find it because if I have $t(r,\mu)$ I do not need any more linear approximation.

Any idea or hint for a better approximation of $t(r,\mu)$ would help me a lot.

Ѕᴀᴀᴅ
  • 34,263
Melina
  • 937
  • Are you saying that you care about nothing else if you know the point value $t(0,0)$? – Lee David Chung Lin Apr 10 '18 at 09:18
  • @LeeDavidChungLin what I wanted to say is that I am not sure if I can do the above approximation because I need $t(0,0)$ (but if I knew what it is $t(r,\mu)$ at $(0,0)$ - then I don't need anymore lin. approximation). – Melina Apr 10 '18 at 09:29
  • What region of $t$ are you interested in? $|t|<0.2?$ What region of $\mu$ is relevant to you? $| \mu | < 1/\pi$? What region of $r$ do you care? $r < 0.4$? These numbers I just listed are just arbitrary examples. I'd just like to point out that this surface is rather simple within a small region, but it quickly gets "interesting" as the range expands. Do you have Mathematica, Maple, Matlab, or anything that allows you to plot it? – Lee David Chung Lin Apr 11 '18 at 08:12
  • @LeeDavidChungLin $\mu \in [0,1/2]$ and I think $r$ in the neighborhood of $1/2$ and $t$ whatever it would be. – Melina Apr 11 '18 at 10:03
  • Shouldn't it be $t(r,\mu) \approx t(0,0) + r\dfrac{\partial t(0,0)}{\partial r}+\mu \dfrac{\partial t(0.0)}{\partial \mu}$? – zhw. Apr 12 '18 at 14:48
  • 2
    Since the set ${(0,0,t): t\ne 0}$ is contained in the level surface $F(r,\mu,t)=0,$ there is no unique value of $t(0,0)$ possible. – zhw. Apr 12 '18 at 15:08
  • @zhw. what about $r=1/2$ and $\mu=1/2$? – Melina Apr 12 '18 at 15:18
  • 1
    I thought you wanted $t(0,0)?$ Also, can you answer my first comment? – zhw. Apr 12 '18 at 15:39
  • I am afraid you will not get useful information about $t(0,0)$; it might well does not exist at all. As per the definition, $t(r,\mu)$ would depend on the factor $r/\mu$ as a whole, just like unbounded example $f(x,y)=x/y$ and bounded example $f(x,y)=\arctan\left(x/y\right)$. You may see from your equation that $\lim_{r\to 0}t(r,\mu)$, if it exists, is independent from $\mu$, the value of which is always the very $t$ that satisfies $e^t\left(t-1\right)=1$, numerically around $1.2785$. Unfortunately, as $r/\mu$ appears as a whole, this value might not be what you expect. – hypernova Apr 12 '18 at 18:02
  • @zhw. All I want is to find a good approximation for $t(r,\mu)$. For your first question I think that's true what you asked. – Melina Apr 12 '18 at 20:31

1 Answers1

3

$\def\e{\mathrm{e}}$Define$$ G(r, μ, t) = rt ((1 - 2μ)t + 2μ (\e^t - 1)) - 2μ ((t - 1)\e^t - 1), %\tag{1} $$ then $F(r, μ, t) = 0 \Leftrightarrow G(r, μ, t) = 0$. Note that\begin{align*} \frac{\partial G}{\partial r}(r, μ, t) &= t ((1 - 2μ)t + 2μ (\e^t - 1)),\\ \frac{\partial G}{\partial μ}(r, μ, t) &= -2rt + 2rt (\e^t - 1) - 2((t - 1) \e^t - 1),\\ \frac{\partial G}{\partial t}(r, μ, t) &= 2r (1 - 2μ)t + 2rμ ((t + 1)\e^t - 1) - 2μ t\e^t. \end{align*} Suppose $t = f(r, μ)$ satisfies $G(r, μ, t) \equiv 0$, then from (1) there is\begin{align*} 0 &= \frac{\partial}{\partial r}(G(r, μ, f(r, μ))) = \frac{\partial G}{\partial r}(r, μ, f(r, μ)) + \frac{\partial G}{\partial t}(r, μ, f(r, μ)) · \frac{\partial f}{\partial r}(r, μ),\\ 0 &= \frac{\partial}{\partial μ}(G(r, μ, f(r, μ))) = \frac{\partial G}{\partial μ}(r, μ, f(r, μ)) + \frac{\partial G}{\partial t}(r, μ, f(r, μ)) · \frac{\partial f}{\partial μ}(r, μ). \end{align*} For specific $t_0 = f(r_0, μ_0)$, if $\dfrac{\partial G}{\partial t}(r_0, μ_0, t_0) \neq 0$, then by the implicit function theorem, there is a uniquely-defined $f$ in a neighborhood of $(r_0, μ_0)$ such that $f(r_0, μ_0) = t_0$ and$$ \frac{\partial f}{\partial r}(r_0, μ_0) = -\frac{\dfrac{\partial G}{\partial r}(r_0, μ_0, t_0)}{\dfrac{\partial G}{\partial t}(r_0, μ_0, t_0)}, \quad \frac{\partial f}{\partial μ}(r_0, μ_0) = -\frac{\dfrac{\partial G}{\partial μ}(r_0, μ_0, t_0)}{\dfrac{\partial G}{\partial t}(r_0, μ_0, t_0)}. $$

Now, note that when $(r_0, μ_0) = \left( \dfrac{1}{2}, \dfrac{1}{2} \right)$,$$ G\left(\frac{1}{2}, \frac{1}{2}, t\right) = 0 \Longleftrightarrow \frac{1}{2}t (\e^t - 1) - ((t - 1) \e^t - 1) = 0 \Longleftrightarrow (t - 2)(\e^t + 1) = 0, $$ thus $t_0 = 2$ is the unique solution. Because$$ \frac{\partial G}{\partial t}\left( \frac{1}{2}, \frac{1}{2}, 2 \right) = -\frac{1}{2}(\e^2 + 1) \neq 0, $$ then\begin{align*} \frac{\partial f}{\partial r}\left( \frac{1}{2}, \frac{1}{2} \right) &= -\frac{\dfrac{\partial G}{\partial r}\left( \dfrac{1}{2}, \dfrac{1}{2}, 2 \right)}{\dfrac{\partial G}{\partial t}\left( \dfrac{1}{2}, \dfrac{1}{2}, 2 \right)} = 4 · \frac{\e^2 - 1}{\e^2 + 1},\\ \frac{\partial f}{\partial μ}\left( \frac{1}{2}, \frac{1}{2} \right) &= -\frac{\dfrac{\partial G}{\partial μ}\left( \dfrac{1}{2}, \dfrac{1}{2}, 2 \right)}{\dfrac{\partial G}{\partial t}\left( \dfrac{1}{2}, \dfrac{1}{2}, 2 \right)} = -\frac{4}{\e^2 + 1}, \end{align*} and in a neighborhood of $(r_0, μ_0) = \left( \dfrac{1}{2}, \dfrac{1}{2} \right)$, there is\begin{align*} f(r, μ) &\approx 2 + \frac{\partial f}{\partial r}\left( \frac{1}{2}, \frac{1}{2} \right) \left( r - \frac{1}{2} \right) + \frac{\partial f}{\partial μ}\left( \frac{1}{2}, \frac{1}{2} \right) \left( μ - \frac{1}{2} \right)\\ &= 2 + 4 \left( \frac{\e^2 - 1}{\e^2 + 1} \left( r - \frac{1}{2} \right) - \frac{1}{\e^2 + 1} \left( μ - \frac{1}{2} \right) \right)\\ &= \frac{2}{\e^2 + 1} (2(\e^2 - 1) r - 2μ + 3). \end{align*}

Ѕᴀᴀᴅ
  • 34,263
  • Now, I see that you have modified the original function instead of $(e^t(t-1)+1)$ you changed to $(e^t(t-1)-1)$ which is not what I was looking for. – Melina May 20 '18 at 11:32
  • @Melina I just checked the revision history and what you wrote was exactly $\mathrm e^t(t-1)-1$. – Ѕᴀᴀᴅ May 20 '18 at 11:39
  • it might be but I see that $(e^t(t−1)+1)$ is what I really need and that's doesn't have unique solution. – Melina May 20 '18 at 15:52
  • if you could edit or make another post with $e^t(t-1)+1$ because it is giving me so much stress and I have to finish my project by Monday evening. I would be extremely thankful. – Melina May 20 '18 at 16:04
  • @Melina I'll update the answer soon. – Ѕᴀᴀᴅ May 20 '18 at 16:08
  • THANK YOU SO MUCHH... – Melina May 20 '18 at 16:11
  • @Melina I just found that for $μ=\dfrac12$, the equation always has a solution $t=0$ no matter what $r$ is, which implies the implicit function theorem fails for the root $t=0$. However, the other root $t=2$ can still be linearly approximated. – Ѕᴀᴀᴅ May 20 '18 at 16:23
  • that's why I saw it very difficult. Do you have any other idea how can I do this for example by expanding $e^t$. – Melina May 20 '18 at 16:50