Implicit function theorem problem maybe? Want to solve $f(x)$ in $f(x) + 1 = \sin(xf(x))$.

Question

Consider $f(x) + 1 = \sin(xf(x))$. I want to solve for $f(x)$ or at least approximate it. This question I have revolves around the Implicit Function Theorem because we can rewrite it as $y = -1 + \sin(xy)$. Here $\phi(x,y) = \sin(xy)$.

Answer 1

Let's use the local linearization $L$ of $f(x)$ at $x=0$ to approximate $f(x)$ near $x=0$. To do so, recall that $$ L(x)=f^\prime(0)\cdot x+f(0) $$ Of course, taking $x=0$ in the formula $$ f(x)+1=\sin(x\cdot f(x)) $$ gives $f(0)=-1$ so we need only compute $f^\prime(0)$ to find $L(x)$. To do so, differentiate the formula to obtain $$ f^\prime(x)=\cos(x\cdot f(x))\cdot(f(x)+x\cdot f^\prime(x)) $$ Now, taking $x=0$ gives $$ f^\prime(0)=\cos(0\cdot(-1))\cdot(-1+0\cdot f^\prime(0))=-1. $$ Hence our approximation of $f(x)$ near $x=0$ is $$ f(x)\approx L(x)=-x-1. $$

Answer 2

Using the same philosophy as proposed by Brian Fitzpatrick, we can assume that function $f(x)$ can be approximated by a polynomial of $x$. This means basically that we are looking for the Taylor series of $f(x)$ built around $x=0$.

This being assumed, "just" expand $f(x) + 1 - \sin(x f(x))$ as a Taylor series around $x=0$. We so obtain a polynomial expression in which appear all the coefficients of the polynomial which we assumed to be $f(x)$. Now, cancel as many terms as we can since we want $$f(x) + 1 - \sin(x f(x))=0$$ Being very patient, we arrive to
$$f(x)\approx -1 -x -x^2 -\frac{5 x^3}{6}-\frac{x^4}{3}+\frac{79 x^5}{120} +\frac{11 x^6}{5} + ... $$

What is interesting is that if $f(x) + 1 - \sin(x f(x))=0$ is solved for $x$, the solution is $$x=\frac{\sin ^{-1}(f(x)+1)}{f(x)}$$ If we plug in the rhs the polynomial expansion obtained for $f(x)$ and develop again as a Taylor expansion around $x=0$, we obtain for the rhs $$x+\frac{20539 x^7}{5040}+O\left(x^8\right)$$

For sure, and this was our initial assumption, $x$ needs to be small. If this has not to be the case, I definitely prefer to use
$$x=\frac{\sin ^{-1}(y+1)+2 \pi k}{y}$$ where $y$ stands for $f(x)$.

Answer 3

You may want to use this approach, in addition to the other excellent answers.

From the original functional equation, by differentiating wrt. to $x$, it yields the non-linear 1st order differential equation with initial condition (substitute $x$ by $0$ to obtain it):

$$f' = \cos{(x f)}(xf'+f), \quad f(0) = -1.$$

This equation can be solved numerically using, for example, Matlab, with the following code:

f0 = -1;
t_span =[0 10];

[x,f] = ode23(@problem,t_span,f0);

%Plot solution:
plot(x,f);
hold on 

%Verifying if f satisfies the original equation:
plot(x,f+1-sin(x.*f),'k');

xlabel('x','FontSize',20);
legend('f(x)','f(x)+1-sin(x f(x))');
set(gca,'FontSize',20);

where problem.m is given by:

function df = problem(x,f)    
df = cos(x.*f)*f./(1-x.*cos(x.*f));    
end

This should give you the figure:

Note that $f(x) \approx -1-x$ near $x = 0$ as @Brian Fitzpatrick pointed out in his comment.

I hope this may be useful to you.

Cheers!