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I've been trying to find a way to get an idea about the root for the implicit function:

$$ \frac{1}{\sqrt{x}}=4\log_{10}(\sqrt{x}\,Re) - 0.4 $$ where $Re>0$ is some constant (Reynolds number).

I've tried to express this function to a function of $x$ and that makes the $Re$ value the function value. However, I'm interested to know what value $x$ is around when $Re = 10^4$.

Any ideas?

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What's reason for the down vote? – Curtain Dec 19 '12 at 9:06
First, I shall upvote your question to erase that downvote. Some people sometimes just down vote for the sport or the hell knows why. About your question: what does $\,Re*\sqrt x\,$ mean?? – DonAntonio Dec 19 '12 at 9:17
@DonAntonio: Multiplication, but I edit that I think. – Curtain Dec 19 '12 at 9:19
$Re$ probably means Reynold Number? What context is this? – k1next Dec 19 '12 at 9:22
@macydanim: It's a Matlab exercise, and yep, it's a Reynold number. – Curtain Dec 19 '12 at 9:30
up vote 1 down vote accepted

For this example simple fixed-point iteration works. Reformulate the problem, such that $x=F(x)$. One(!) example of doing so would be \begin{align} x = \frac{1}{(4\log_{10}(Re\cdot \sqrt{x})-0.4)^2}=F(x) \end{align} Fixed-point iteration works like this: $x^{k+1}=F(x^k)$. That means you simply plug in the result of $F(x^k)$ once more into $F$. Now you have to prove that this does converge, see i.e. Banach fixed-pint theorem.

Another method to use would be Newton's Method, but it turns out that the fixed-point iteration just works fine. With the Matlab code below, you reach a fixed-point in 5 iterations.

function fixedPointIteration

x = 1;
while abs(x-f(x))>tol
sprintf('Result of fixed-point iteration is x=%s',x)

function y=f(x)
Re = 1e4;
y =1/(4*log10(Re*sqrt(x)-0.4))^2;

Which gives $x=0.0072742$. It should be easy to adapt the code into other languages. Hope this helps.

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Thanks for your post. Although I'm currently solving it in Matlab, I only need a good start guess for fzero. Do you think I should du Newton's method manually to achieve this? – Curtain Dec 19 '12 at 9:31
I don't think you need Newton's method. Bisection method works also and you need no derivatives. My method converges for each $x_0$ in linspace(1e-7,1e-3,100); and linspace(100000,1e-3,10001); is also no problem. – k1next Dec 19 '12 at 9:48
Yes, of course! Thank you! – Curtain Dec 19 '12 at 9:49

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