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How to find the roots of the following equation:

$10.9-11.4 x^2+0.5 *\ln (17 x)=0 \ $ using computer program or else .

I need the roots only but I have no idea to find the roots. Can anyone help me with roots of the above equation .

Jean Marie
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MAS
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    according to wolframalpha the only root is about 1.04 – Vasili Sep 08 '17 at 12:35
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    Now, how can you refine (get more decimals of) this root ? Try dichotomy, or (much better) Newton's method. – Jean Marie Sep 08 '17 at 12:44
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    There is another root near $2\cdot 10^{-11}$ – Peter Sep 08 '17 at 12:52
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    @Vasya Once more, Wolfram Alpha is not trustable (in fact, you guess there can exist a second root when looking at the graphical representation) – Jean Marie Sep 08 '17 at 12:56
  • Well, using the command "roots" I found both roots, moreover, using the original equation a picture with two points appeared (but only one root was mentioned) – Peter Sep 08 '17 at 12:58
  • @JeanMarie As a programmer, I know all too well that human brain cannot be replaced by a machine. My first thought was that there should be two roots because of $x^2$ – Vasili Sep 08 '17 at 12:59
  • The second derivate is negative for all $x>0$, hence we cannot have more than $2$ real roots – Peter Sep 08 '17 at 13:08

3 Answers3

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You need numerical methods (Apparantly, mathworld does not find a closed-form-solution, so the Lambert-W-function seems to be useless here). The two roots are (calculated with PARI/GP)

? solve(x=2*10^(-11),2.01*10^(-11),10.9-11.4*x^2+0.5*log(17*x))
%4 = 2.004155307311702078727439326 E-11
? solve(x=1,1.1,10.9-11.4*x^2+0.5*log(17*x))
%5 = 1.040257251066025097928117626
?

Using the first two derivates should help to prove that there are only two roots.

Peter
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  • On the contrary, in this case Lambert-W-function gives a closed form solution
    • see the other answer.
     – g.kov Sep 08 '17 at 16:14
  • @g.kov Strange that Wolfram alpha does not find the closed form solution of the concrete function! Note that I formulated "seems" because I had some doubts that Lambert-W does not apply here. – Peter Sep 09 '17 at 12:23
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Wolfram Alpha gives an exact solution to \begin{align} a+b\,x^k+c\,\ln(d\,x)=0 \end{align}

in terms of Lambert W function:

\begin{align} x&=\left(\frac{c}{k\,b}\, \operatorname{W}\left(\frac{k\,b}{c}\,\left(\tfrac1d\,\exp(-\frac{a}{c})\right)^k\right)\right)^{1/k}, \end{align} which for $a=10.9,\ k=2,\ b=-11.4,\ c=0.5, d=17$ gives two real solutions,

$x_1\approx 2.0041553073117\cdot10^{-11}$ for the branch $\operatorname{W_0}$ and $x_2\approx 1.04025725106996$ for the branch $\operatorname{W_{-1}}$.

Edit

The solution in terms of $W$ can be derived manually as follows:

\begin{align} a+b\,x^k+c\,\ln(d\,x) &=0,\\ -k\,\tfrac{a}c-k\,\tfrac{b}c\,x^k+\ln(d^{-k}\,x^{-k}) &=0,\\ \ln\left(\exp(-k\,\tfrac{a}c)\,d^{-k}\,x^{-k}\right) &=k\,\tfrac{b}c\,x^k ,\\ \exp(-k\,\tfrac{a}c)\,d^{-k}\,x^{-k}\,\ln\left(\exp(-k\,\tfrac{a}c)\,d^{-k}\,x^{-k}\right) &=k\,\tfrac{b}c\,\exp(-k\,\tfrac{a}c)\,d^{-k} ,\\ \ln\left(\exp(-k\,\tfrac{a}c)\,d^{-k}\,x^{-k}\right) &=\operatorname{W}(\,\tfrac{b}c\,\exp(-k\,\tfrac{a}c)\,d^{-k}) ,\\ \exp(-k\,\tfrac{a}c)\,d^{-k}\,x^{-k} &=\exp\left(\operatorname{W}(\,\tfrac{b}c\,\exp(-k\,\tfrac{a}c)\,d^{-k})\right) ,\\ \exp(\tfrac{a}c)\,d\,x &=\exp\left(-\tfrac1k\,\operatorname{W}(\,\tfrac{b}c\,\exp(-k\,\tfrac{a}c)\,d^{-k})\right) ,\\ x&=\tfrac1d\, \exp\left(-\tfrac1k\,\operatorname{W}(\,\tfrac{b}c\,\exp(-k\,\tfrac{a}c)\,d^{-k}) -\tfrac{a}c \right) . \end{align}

For the given set of constants, the argument of $W$ is negative, but greater than $-\tfrac1e$, hence there are two real solutions corresponding to the branches $W_0$ and $W_{-1}$.

g.kov
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If you do not use Lambert function, consider the function $$f(x)=10.9-11.4 x^2+0.5 \,\log (17 x)$$ $$f'(x)=-22.8x+\frac{0.5} x$$ $$f''(x)=-22.8-\frac{0.5} {x^2}$$ So, the first derivative cancels at $x_*$ such that $$x_*=\sqrt{\frac 5 {228}}\approx 0.148087$$ At this point $f(x_*)\approx 11.1116$ and $f''(x_*)<0$. So there are two roots $0 < x_1 <x_*$ and $x_*< x_2$.

The largest root is easy to find by inspection (or graphing). Since it is close to $1$, we could approximate it using Taylor series $$f(x) =0.5 (\log (17)-1)-{22.3 (x-1)}+O\left((x-1)^2\right)$$ leading to $x_2\approx 1.04110$.

For $x_1$, close to $0$,we have $$f(x)=10.9+ 0.5 *\log (17 x)+O\left(x\right)$$ leading to $x_1\approx 2.00416\times 10^{-11}$.

As you can see, we can get good approximation at the price of solving one linear equation (the first one in $x$, the second one in $\log(x)$.

Claude Leibovici
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