ln(40.00000000*x)*x+.4000000000-15.35417296*x = 0 need to solve for x
J jmj5148 New member Joined Sep 6, 2010 Messages 6 Sep 6, 2010 #1 ln(40.00000000*x)*x+.4000000000-15.35417296*x = 0 need to solve for x
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 6, 2010 #2 One way is Newton's method. It simplifies to: f(x)=xln(x)−11.665293x+0.4\displaystyle f(x)=xln(x)-11.665293x+0.4f(x)=xln(x)−11.665293x+0.4 f′(x)=ln(x)−10.6652935059\displaystyle f'(x)=ln(x)-10.6652935059f′(x)=ln(x)−10.6652935059 xn+1=x−xln(x)−11.665293x+0.4ln(x)−10.6652935059\displaystyle x_{n+1}=x-\frac{xln(x)-11.665293x+0.4}{ln(x)-10.6652935059}xn+1=x−ln(x)−10.6652935059xln(x)−11.665293x+0.4 Try an initial guess of, say, 0.5 Several iterations gives: .0261265....\displaystyle .0261265.....0261265.... There is another solution as well. Much larger. Try finding it by using a much larger initial guess.
One way is Newton's method. It simplifies to: f(x)=xln(x)−11.665293x+0.4\displaystyle f(x)=xln(x)-11.665293x+0.4f(x)=xln(x)−11.665293x+0.4 f′(x)=ln(x)−10.6652935059\displaystyle f'(x)=ln(x)-10.6652935059f′(x)=ln(x)−10.6652935059 xn+1=x−xln(x)−11.665293x+0.4ln(x)−10.6652935059\displaystyle x_{n+1}=x-\frac{xln(x)-11.665293x+0.4}{ln(x)-10.6652935059}xn+1=x−ln(x)−10.6652935059xln(x)−11.665293x+0.4 Try an initial guess of, say, 0.5 Several iterations gives: .0261265....\displaystyle .0261265.....0261265.... There is another solution as well. Much larger. Try finding it by using a much larger initial guess.
J jmj5148 New member Joined Sep 6, 2010 Messages 6 Sep 6, 2010 #3 i used Maple and this is the answer it gave me, i just couldn't think how to solve it.....thanks alot
i used Maple and this is the answer it gave me, i just couldn't think how to solve it.....thanks alot
