How could I find the $y=f(x)$ for each $x$ in an implicit function base on Newton-Raphson ?
For example -
Given the implicit function - $$3x^7+2y^5-x^3+y^3-3=0 $$ how could I find the $f(0)$ base on Newton-Raphson ?
Any idea or hint would be welcome .
First of all, not necessarily $y$ in implicit expression is actually a function of $x$. For instance, implicit equation $$ x^2+y^2 = 1 $$ doesn't define a function, but two functions (top and bottom semicircles). But, you can say, what points $(x,y)$ satisfy your implicit equation, if $x = 0$.
SO, basically, you have to solve $$ 2y^5+y^3-3=0 $$ From that point, you can apply Newton-Raphson, or any other known solvers, to find those $y$'s.
To find $f(0)$ you are trying to solve $g(y)=2y^5+y^3-3=0$. Now you can take a derivative to get $g'(y)=10y^4+3y^2$. You have a function value and its derivative, so you can use Newton-Raphson just fine. If I start at $y=0.9$ it converges quickly:$$\begin {array}{r r r} y&g(y)&g'(y)\\ 0.9&-1.09002&8.991\\ 1.021234568&0.286623354&14.00558228\\ 1.000769631&0.01001884&13.03544037\\ 1.000001047&1.36057E-05&13.00004814\\ 1&2.51941E-11&13 \end {array}$$
I only knew to start there because I could see that $y=1$ was a root. It converges from $0.5$ as well, but takes a bit longer. In this case this is the only real root in $y$.
