This is not homework, I am not enrolled at any school. I am trying to brush up on my Math skills and I have come accross a problem that I should know how to work out.
The equation is e^x - (10/7)x - 1 = 0
Using Newton’s method I have detemrined that the two solutions are 0 and 0.675471592932108.
How can I get that by manipulating the equation?
There’s no closed-form solution in terms of elementary functions. You can solve it in terms of the Lambert W-function, but you’re probably better off doing it numerically. Or just ask Wolfram Alpha…
Thanks, I have a long way to go before I will be casually using the “Lambert W-function”!
You can re-arrange the equation and take the natural logarithm of each side, giving
x = ln(ax + 1)
where, in your case, a = 10/7.
If you expand the right hand side, you get ( see Natural logarithm - Wikipedia , or a book of tables or infinite series)
ln(ax + 1 ) = ax - ((ax)[sup]2[/sup])/2 + ((ax)[sup]3[/sup])/3 - …
Equating this to x, you immediately see that ** x = 0** is one solution. You can get your other solution by taking successively more terms of the expansion. Note that you must have ** |ax| < 1 **, which you do in both cases. That’s not closed-form for the second solution, but I don’t think you’re going to find one.
Writing in terms of parameter “a”: e^x - a*x - 1 = 0, a good approximation can be obtained by expressing in terms of the “a”, although I only did it terms of b = 1-a.
x_0=b*(12b^3 - 100b^2 +180b -90) / (2b^4 - 32b^3 + 105b^2 - 120*b +45)
Inserting b=-3/7 (a=10/7) gives a result within 4e-6 of the exact value.
Using a higher order version of Newton’s method allows that result to be improved:
Let g=1-(1+ax_0)/exp(x_0) and t=1-a/exp(x_0), then a better approximation is:
x=x_0 + 3g*(g-2t^2)/(6t^3-6tg+g^2)
This gives a results within 8e-22 of the exact solution.
Not exact, but pretty close.