1
$\begingroup$

I need to solve this equation (find $\lambda$) using numerical methods:

$\displaystyle N_0e^{\lambda}+v\frac{e^{\lambda}-1}{\lambda}-N_1 = 0$

All other terms are constant and known.

N0 = 1000000; v = 435000; N1 = 1564000;

I need to solve it by using bisection, false position, newton Raphson and fixed point. I have already solved it by using the three first approaches, however I am stuck in the fixed point. I cannot find a function, such that finding the roots of that function represents the equation and also $|f'(\alpha)| < 1$ where alpha is near the root (Condition that must be satisfied so that the method converges. I have tried several combinations, like dividing by N0, changing the starting approximation (I know that the root is around 0.1). Could somebody help me?

$\endgroup$
  • $\begingroup$ Treat the $e^\lambda$ as a separate variable and solve for $\lambda$. I don't have the time at the moment to check the work, but this is a method that does converge to machine precision pretty quickly. The function looks like $$\lambda_{i+1}=\frac{v(e^{\lambda_i}-1)}{n_1-n_0e^{\lambda_i}}$$ $\endgroup$ – Terra Hyde Sep 3 '15 at 5:06
  • 1
    $\begingroup$ It actually converges to zero, which is the solution to $\lambda(N_0 e^\lambda - N_1) + \nu (e^\lambda - 1) = 0$. It actually runs away from the root since $$ \left|\frac{d}{d\lambda}\left(\frac{\nu (e^\lambda - 1)}{n_1 - n_0 e^\lambda}\right)\right| \approx 1.28 > 1. $$ Sure, you can solve for $e^\lambda$ and do the inverse iterations. $\endgroup$ – uranix Sep 3 '15 at 7:35
  • $\begingroup$ And that is what I get for working while tired. Thanks. :) $\endgroup$ – Terra Hyde Sep 3 '15 at 14:34
3
$\begingroup$

If you know that the root is around $0.1$ then $$ e^\lambda - 1 \approx \lambda $$ so the solution is close to the solution of $$ N_0 e^\lambda + \nu = N_1 $$ Solving it for $\lambda$ yields $$ \lambda = \log \frac{N_1 - \nu}{N_0}. $$ Let's use that as the initial guess. $$ \lambda_0 = \log \frac{1.564 - 0.435}{1} \approx 0.121332. $$ Recall that we approximated $\frac{e^\lambda - 1}{\lambda}$ by $1$ to obtain the approximate solution. Now let's bring that term back $$ \lambda = \log \left(N_1 - \nu \frac{e^{\lambda} - 1}{\lambda}\right) - \log{N_0}. $$ Instead of approximating the term with $1$, let's use the last known approximate value for $\lambda$. That would give the following iterative method: $$ \lambda_{n+1} = \log \left(N_1 - \nu \frac{e^{\lambda_n} - 1}{\lambda_n}\right) - \log{N_0}. $$ Here are the first 10 iterations of the method: $$ \begin{array}{ccc} n & \lambda_n & \text{error}\\ 0 & 1.21332000000000\times 10^{-1} & 2.03341\times 10^{-2} \\ 1 & 9.66817916349767\times 10^{-2} & 4.31614\times 10^{-3} \\ 2 & 1.01904141090938\times 10^{-1} & 9.06211\times 10^{-4} \\ 3 & 1.00807223823552\times 10^{-1} & 1.90706\times 10^{-4} \\ 4 & 1.01038042978185\times 10^{-1} & 4.01133\times 10^{-5} \\ 5 & 1.00989491349787\times 10^{-1} & 8.43834\times 10^{-6} \\ 6 & 1.00999704757899\times 10^{-1} & 1.77507\times 10^{-6} \\ 7 & 1.00997556283298\times 10^{-1} & 3.73402\times 10^{-7} \\ 8 & 1.00998008234252\times 10^{-1} & 7.85485\times 10^{-8} \\ 9 & 1.00997913162376\times 10^{-1} & 1.65234\times 10^{-8} \\ 10 & 1.00997933161588\times 10^{-1} & 3.47584\times 10^{-9} \end{array}. $$ The method has $|f'(\alpha)| \approx 0.210$ near the root (gives $\approx 0.7$ correct digits per iteration).

NB. In fact Newton's method also is a fixed point method.

The other method, based on inverse iterations proposed by Terra Hyde may be obtained if you solve the equation for $e^\lambda$ treating $\lambda$ as independent variable $$ e^\lambda = \frac{N_1 + \nu \lambda}{N_0 + \nu \lambda}\\ \lambda_{n+1} = \log \frac{N_1 + \nu \lambda_n}{N_0 + \nu \lambda_n}. $$ The method has $|f'(\alpha)| \approx 0.771$ near the root (gives $\approx 0.11$ correct digits per iteration).

$\endgroup$
  • $\begingroup$ awesome answer! I am speechless, thank you! $\endgroup$ – dpalma Sep 3 '15 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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