My Math Forum Direct method to solve for x in (cos x)/x=0.25

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 April 12th, 2017, 10:22 PM #1 Member   Joined: Nov 2016 From: USA Posts: 33 Thanks: 1 Direct method to solve for x in (cos x)/x=0.25 I can plug in numbers for x and then get closer & closer approximations. What is a quick, direct method to solve for x in: (cos x)/x = 0.25
 April 12th, 2017, 10:35 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,692 Thanks: 860 There's no closed form answer. You have to find a numeric solution. You can use Newton's method $f(x) = \dfrac{\cos(x)}{x}-0.25$ $f^\prime(x) = \dfrac{-x\sin(x) -\cos(x)}{x^2}$ $x_{n+1} = x_n - \dfrac{f(x_n)}{f^\prime(x_n)}$ and run this until it converges close enough for your satisfaction I get $x=1.25235$ Thanks from Seventy7 Last edited by romsek; April 12th, 2017 at 11:21 PM.
 April 12th, 2017, 11:03 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 881 Thanks: 353 Umm, for Newton's Method $f(x) = \dfrac{\cos(x)}{x} - 0.25?$ Thanks from romsek Last edited by skipjack; April 13th, 2017 at 04:00 PM.
April 13th, 2017, 11:19 AM   #4
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Quote:
 Originally Posted by JeffM1 Umm, for Newton's Method $f(x) = \dfrac{\cos(x)}{x} - 0.25?$
Yes, Newton's method applies to equations of the form f(x) = 0. Here the problem was to solve $\dfrac{\cos(x)}{x}= 0.25$ which is the same as $\dfrac{\cos(x)}{x}- 0.25= 0$.

Last edited by skipjack; April 13th, 2017 at 03:59 PM.

April 13th, 2017, 05:09 PM   #5
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Quote:
 Originally Posted by Country Boy Yes, Newton's method applies to equations of the form f(x) = 0. Here the problem was to solve $\dfrac{\cos(x)}{x}= 0.25$ which is the same as $\dfrac{\cos(x)}{x}- 0.25= 0$.
Romsek missed the 0.25 in his original post, but edited it in after JeffM1 pointed it out .

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