solve x*tan(x)-0.5=0

I am again stuck with similar question which I asked earlier. I am trying to solve x*tan(x)-0.5=0,but when i cross checked the result I am getting very different results.

syms x
n=10;
for n = 1:2:n

x1 = (n-1/2)*pi;
x2 = (n)*pi;
w(n) = double(vpasolve(x*tan(x)-0.5 == 0,x,[x1 x2]))
check(n)=w(n)*tan(w(n))-0.5
end

I am getting

w =

1.5708 0 7.8540 0 14.1372 0 20.4204 0 26.7035

check =

1.0e+16 *

2.5653 0 2.5653 0 2.5653 0 -2.0830 0 -3.6311. I am unable to find my mistake.

Answer 1

[Torsten]

x1=0:0.1:1.3;
y1=x1.*tan(x1)-0.5;
x2=2:0.1:4.1;
y2=x2.*tan(x2)-0.5;
x3=5.5:0.1:6.9;
y3=x3.*tan(x3)-0.5;
plot(x1,y1,x2,y2,x3,y3)

So was your choice of x1 and x2 resonable ?

Here you get the asymptotic distribution of the zeros of your function in order to make a proper choice of x1 and x2:

https://math.stackexchange.com/questions/1238520/how-to-find-asymptotic-expansions-of-all-real-roots-of-x-tanx-epsilon-1

[Sam Chak]

Hi @Torsten

I'm unfamiliar with this problem. Was the issue caused by numerical sensitivity? I compared plot with fplot. fplot shows the true graph because the asymptotes are not part of .

x = -3*pi:0.001:3*pi;
y = x.*tan(x) - 0.5;
plot(x, y)
axis([min(x) max(x) -40 40])

fplot(@(x) x*tan(x) - 0.5, [-3*pi 3*pi])

[Torsten]

I just meant to say that there should be no singularity in between the x1 and x2 values given to "vpasolve".

[rakesh kumar]

thanks for your replies and correcting my mistakes. the valus of x1 is (n-1)*pi and x2 is (n-1/2)*pi. However, I am gettting similar kind of answer. could you suggest an interval for odd n to solve xtanx-0.5

thanks to all of you

[Torsten]

But I gave you the link which suggests choosing something like

x1 = sqrt(0.5) - 1/6*0.5^1.5 - 0.1  x2 = sqrt(0.5) - 1/6*0.5^1.5 + 0.1

for the first root and

x1 = n*pi + 0.5/(n*pi) - 0.1 and x2 = n*pi  + 0.5/(n*pi) + 0.1

for the other roots.

Didn't you look into it ?

syms x
n = 10;
x1 = sqrt(0.5) - 1/6*0.5^1.5 - 0.1;
x2 = sqrt(0.5) - 1/6*0.5^1.5 + 0.1;
w(1) = double(vpasolve(x*tan(x)==0.5,x,[x1,x2]))
for i=1:n
    x1 = i*pi + 0.5/(i*pi) - 0.1;
    x2 = i*pi + 0.5/(i*pi) + 0.1;
    w(i+1) = double(vpasolve(x*tan(x)==0.5,x,[x1,x2]))
end
% Check
w.*tan(w)-0.5

[rakesh kumar]

Thanks once again. I ran your code and its giving the perfect answer. You people are really great!

[Sam Chak]

@Torsten

Learned something today. Thank you.

@rakesh kumar

Also thanks for showing an interesting question.