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Tan x = 1/x

  • Thread starter Dustinsfl
  • Start date
  • #1
699
5
How do I find the first 10 positive solutions numerically?
$$
\tan x = \frac{1}{x}
$$
 

Answers and Replies

  • #3
699
5
I tried using NSolve in Mathematica but that didn't work.

I also tried
syms x
solve(tan(x)==1/x)
in Matlab and that just gave -263.

How can I do this in one of these programs?
 
Last edited:
  • #4
Ray Vickson
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How do I find the first 10 positive solutions numerically?
$$
\tan x = \frac{1}{x}
$$
How do you find the first positive solution? You tell US; do not ask us to tell you: read the forum rules.

RGV
 
Last edited:
  • #5
SammyS
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How do I find the first 10 positive solutions numerically?
$$
\tan x = \frac{1}{x}
$$
Where are you stuck?

What have you tried?

After 490 posts, you should know the drill here at PF .

But here's one hint: One thing I would try is to take the reciprocal of both sides of that equation, giving you:
[itex]\displaystyle \cot(x)=x[/itex]​

Graph each side.
 
  • #6
699
5
How do you find the first positive solution? You tell US; do not ask us to tell you: read the forum rules.

RGV
I am not good at numerical analysis. I haven't taken a course in it yet as well as in Matlab. I will be taking Matlab this spring and Numerical Analysis next fall. So when I know how to do it and use Matlab efficiently, I won't even ask you.
 
  • #7
699
5
Where are you stuck?

What have you tried?

After 490 posts, you should know the drill here at PF .

But here's one hint: One thing I would try is to take the reciprocal of both sides of that equation, giving you:
[itex]\displaystyle \cot(x)=x[/itex]​

Graph each side.
Read post 3.
 
  • #8
699
5
Numerically means approximately since this is a transcendental equation. Only some computer software would give you the results, I think.

http://www.wolframalpha.com/input/?i=Plot+y(x)+=+tan+x+-+1/x
I have a graph of it. I don't know if the forum can handle tikz but here it is
$$
\begin{tikzpicture}[>=stealth',x = .5cm,y = .5cm,scale = .60]
\def\npi{3.1416}
\def\periods{4}
\draw[->] (-\npi/2,0) -- ({(\periods + .5)*\npi},0) node[above] {$\lambda_n$};
\draw[->] (0,-10) -- (0,10) node
{$f(\lambda_n)$};
\clip (-\npi/2,-9.8) rectangle ({(\periods + .5)*\npi},9.8);
\draw[thick, domain = 0.05:{(\periods + .4)*\npi},samples = 300,smooth,color = red] plot (\x,1/\x);
\foreach \n in {0,...,4}
\draw[thick,shift = {(\npi*\n,0)},domain = -\npi/2+.1:\npi/2-.1,samples = 100,smooth] plot (\x,{tan(\x r)});
%draw the ticks
\foreach \x in {1,...,10} \draw (\x*\npi/2,2pt) -- (\x*\npi/2,-2pt);
%draw labels n\pi/2 for odd n >= 3
\foreach \x in {3,5,...,7} \node[below] at (\x*\npi/2,0) {$\frac{\x\pi}{2}$};
%draw labels n\pi for n >= 2
\foreach \x in {2,...,4} \node[below] at (\x*\npi,0) {$\x\pi$};
\node[below] at (\npi/2,0) {$\frac{\pi}{2}$};
\node[below] at (\npi,0) {$\pi$};
\end{tikzpicture}
$$​
 
  • #9
Ray Vickson
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I am not good at numerical analysis. I haven't taken a course in it yet as well as in Matlab. I will be taking Matlab this spring and Numerical Analysis next fall. So when I know how to do it and use Matlab efficiently, I won't even ask you.
You don't need to know how to use Matlab, etc. Just apply Newton's method, which you can do using a hand-held scientific calculator. If you have not ever seen Newton's method before, I would be very surprised.

RGV
 
  • #10
SammyS
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Here are the graphs of y=x and y=cot(x), superimposed. (from WolframAlpha)

attachment.php?attachmentid=51374&stc=1&d=1349045971.gif
 

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  • #11
699
5
Here are the graphs of y=x and y=cot(x), superimposed. (from WolframAlpha)

attachment.php?attachmentid=51374&stc=1&d=1349045971.gif
I tried using NSolve in Mathematica but that didn't work.

I also tried
syms x
solve(tan(x)==1/x)
in Matlab and that just gave -263.

How can I do this in one of these programs?
This is what I wrote in post 3.
I can make graphs of this with the Tikz package and in Mathematica (that isn't the challenge here). How can I use Mathematica or Matlab to generate the first 10 positive solution?
 
  • #12
Ray Vickson
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I tried using NSolve in Mathematica but that didn't work.

I also tried
syms x
solve(tan(x)==1/x)
in Matlab and that just gave -263.

How can I do this in one of these programs?
This is what I wrote in post 3.
I can make graphs of this with the Tikz package and in Mathematica (that isn't the challenge here). How can I use Mathematica or Matlab to generate the first 10 positive solution?
I don't have access to Mathematica (except through Wolfram Alpha) or to Matlab, but in Maple it is easy:
S:=fsolve(tan(x)=1/x,x=0..Pi/2),seq(fsolve(tan(x)=1/x,x=Pi/2+(i-1)*Pi..Pi/2+i*Pi),i=1..9);
S := 0.8603335890, 3.425618459, 6.437298179, 9.529334405,

12.64528722, 15.77128487, 18.90240996, 22.03649673,

25.17244633, 28.30964285
I'm sure it must be possible, even easy, to do the same thing in Mathematica.

RGV
 
Last edited:
  • #13
699
5
I don't have access to Mathematica (except through Wolfram Alpha) or to Matlab, but in Maple it is easy:
S:=fsolve(tan(x)=1/x,x=0..Pi/2),seq(fsolve(tan(x)=1/x,x=Pi/2+(i-1)*Pi..Pi/2+i*Pi),i=1..9);
S := 0.8603335890, 3.425618459, 6.437298179, 9.529334405,

12.64528722, 15.77128487, 18.90240996, 22.03649673,

25.17244633, 28.30964285
I'm sure it must be possible, even easy, to do the same thing in Mathematica.

RGV
Thanks. Hopefully in the spring when I take programming in Matlab I will be better at this.
 
  • #14
vela
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I tried using NSolve in Mathematica but that didn't work.
Try using FindRoot instead.
 
  • #15
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I tried using NSolve in Mathematica but that didn't work.
...
How can I do this in one of these programs?
In[7]:= Plot[Tan[x]-1/x,{x,.01,30}]

Out[7]= <snip graphic showing approximate positions of solutions>

In[8]:= FindRoot[Tan[x]-1/x,{x,.9}]

Out[8]= {x->0.860334}

In[9]:= FindRoot[Tan[x]-1/x,{x,3.5}]

Out[9]= {x->3.42562}

etc,etc,etc
 

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