1
$\begingroup$

I'm trying to solve an equation of the form $$ R(\theta)f(\theta)+hg(\theta) = 0 $$ for small $h$, where $R$, $f$, and $g$ are functions. I've assumed a power series expansion for $\theta$ in terms of $h$: $$ \theta = \sum_{i=0}^\infty \theta_i h^i $$ and I'm wondering how I can solve this in Mathematica in sort of an automated way. One complication is that I don't have an explicit expression for $R(\theta)$. Instead, I have $R'(\theta) = m(\theta)R(\theta)$ for some function $m$, and I know $R(\theta_0)\equiv R_0 \neq 0$. I do, however, have explicit expressions for $f(\theta)$ and $g(\theta)$. Assuming the following expansions $$ R(\theta) = R_0 +R'(\theta_0)(\theta-\theta_0) + \frac{1}{2}R''(\theta_0)(\theta-\theta_0)^2 + ...\\ f(\theta) = f_0 + f'(\theta_0)(\theta-\theta_0) + \frac{1}{2}f''(\theta_0)(\theta-\theta_0)^2 + ...\\ g(\theta) = g_0 + g'(\theta_0)(\theta-\theta_0) + \frac{1}{2}g''(\theta_0)(\theta-\theta_0)^2 + ... $$ where $f_0 \equiv f(\theta_0)$ and $g_0 \equiv g(\theta_0)$. I can find the first few terms by hand. For instance, at $O(1)$: $$ R_0f_0 = 0 \implies f_0=0 $$ This gives me an expression for $\theta_0$. Then, at $O(h)$, I obtain $$ (R'(\theta_0)f_0 + R_0f'(\theta_0))(\theta-\theta_0) + g_0 = 0 $$ Since we already determined $f_0 = 0$, this becomes $$ R_0f'(\theta_0)(\theta-\theta_0) + g_0 = 0 $$ Then, using the expansion for $\theta$, and remembering this is the $O(h)$ term, we obtain $$R_0f'(\theta_0)\theta_1 + g_0 = 0\implies \theta_1 = -\frac{g_0}{R_0 f'(\theta_0)}$$ and I can continue this. The question is how can I do this in Mathematica. I have tried several versions of the following:

    θw[n_] := Sum[θ[i] h^(i - 1), {i, 1, n}] + O[h]^n;
    f[θ_, n_] := Series[f[θw[n]], {h, θ0, n}]
    g[θ_, n_] := Series[g[θw[n]], {h, θ0, n}]
    R[θ_, n_] := Series[R[θw[n]], {h, θ0, n}]
    j[θ_, n_] := R[θ, n] f[θ, n] + h*g[θ, n]

    Solve[j[θ, 2] == 0]
    {{g[θ[1]] -> -R[θ[1]] θ[2] 
    Derivative[1][f][θ[1]], 
    f[θ[1]] -> 0}, {g[θ[1]] -> -f[θ[1]] θ[2]     Derivative[1][
 R][θ[1]], R[θ[1]] -> 0}}

The part in the curly braces is the output of the Solve command. The $n$ in the functions is to set the order in $h$ to which the expansions are taken. This seems close to working. It does give me something like the expressions I derived. However, I would like for it to solve for the $\theta_i$s automatically, and I would like for it to ignore the $R_0=0$ solution, because I'm not interested in that one. A further complication is that for higher order corrections, I will need to implicitly differentiate my expression for $R$: $$R'(\theta) = m(\theta)R(\theta)$$ Further, is it possible to automatically use the information from lower orders to find explicit expressions for the higher order $\theta_i$s?

$\endgroup$
1
  • $\begingroup$ @Mr.Wizard Thank you for the edit. That looks much better. $\endgroup$
    – Mike Bell
    Jun 15, 2016 at 16:10

1 Answer 1

1
$\begingroup$

You made your question to complicated. It is sufficient to consider the following functional relation $A(\theta)=h$, where $A(\theta)=R(\theta)f(\theta)/g(\theta)$. Now we can solve this equation, i.e. $\theta=A^{-1}(h)$, where $A^{-1}$ denotes the inverse function, $A^{-1}(A(\theta))=\theta$. You are seeking now the series expansion of $\theta(h)$:

 Series[InverseFunction[A][h], {h, 0, 3}]

yielding $A^{(-1)}(0)+\frac{h}{A'\left(A^{(-1)}(0)\right)}-\frac{h^2 A''\left(A^{(-1)}(0)\right)}{2 A'\left(A^{(-1)}(0)\right)^3}+\frac{h^3 \left(3 A''\left(A^{(-1)}(0)\right)^2-A^{(3)}\left(A^{(-1)}(0)\right) A'\left(A^{(-1)}(0)\right)\right)}{6 A'\left(A^{(-1)}(0)\right)^5}+O\left(h^4\right)$

$\endgroup$
2
  • $\begingroup$ Thank you. This looks reasonable to me, though I'm still not sure how to deal with the $R(\theta)$, since I currently have an expression in the form of $R'(\theta) = m(\theta)R(\theta)$. Do I have to solve for $R(\theta)$, or is there a way I can use the expression $$ R(\theta) = R_0 + R'(\theta_0)(\theta - \theta_0) + \frac{1}{2}R''(\theta_0)(\theta-\theta_0)^2+... ? $$ $\endgroup$
    – Mike Bell
    Jun 15, 2016 at 18:28
  • $\begingroup$ Equation $R'(\theta)=m(\theta)R(\theta)$ has analytic solution $R(\theta)=R(\theta_0)\exp(\int_{\theta_0}^\theta m(\tau)d\tau)$. This solution can be substituted in the expression for $A(\theta)$ $\endgroup$
    – yarchik
    Jun 15, 2016 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.