I would like to get an approx. solution to the equation: $ \cos x = 2x$, I don't need an exact solution just some approx. And I need a solution using elementary mathematics (without derivatives etc).
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2http://www.wolframalpha.com/input/?i=solve+cos%28x%29%3D2x – Hans Lundmark Jan 10 '13 at 15:32
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@HansLundmark I need to present some solution not just giving some numbers from some computer. – gen Jan 10 '13 at 15:33
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Hint: Do a Taylor series expansion of $\cos x$ (say, up to order 6) on the left and then solve for x in the remaining equation. Now, explain this approach and figure out how to select the right value and what the error is. Regards – Amzoti Jan 10 '13 at 15:49
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try using the fact that $ |\cos x| \le 1 $ and $ \cos (-x) = \cos (x)$ – S L Jan 10 '13 at 15:49
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@gen: The word "solution" is ambiguous. It can mean both the value of $x$ which satisfies the equation, and the procedure for finding that value. When you say "an approx. solution", I think most people will interpret this as just the number, so if you want the procedure you should perhaps have said so in the question. – Hans Lundmark Jan 10 '13 at 18:30
3 Answers
Take a pocket calculator, start with $0$ and repeatedly type [cos], [$\div$], [2], [=]. This will more or less quickly converge to a value $x$ such that $\frac{\cos x}2=x$, just what you want.
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At small angles $\alpha ≈ sin\alpha$, so at the equation we can write $2 sin\alpha≈ cos \alpha$. From that $tan \alpha ≈ 1/2$. The solution to the last equation is less than a degree off the exact solution to the original equation.
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if $\cos x=2x$ which means $$x=\frac {1}{2.22131587} \,\mathrm{rad},$$
$$x=28.65957881\,\mathrm{Grad},$$
$$x=25.79362093\,\mathrm{Degree}$$
point: $$\cos x=1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!}+\cdots$$
$0 < x < +\frac {1}{2.5}\,\mathrm{rad}$ because $\cos x=2x$
This is also a mathematical method, i mean Estimate the amount through math without using a computer or calculator!.
$$\cos x \simeq 1-\frac {x^2}{2!}=2x,$$
$$x^2+4x-2=0,$$
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
where the $a=1$ and $b=4$ and $c=-2$ then $x\simeq 0.45$
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4How is this supposed to help? The two upvoters might want to explain, as well. – Did Jan 10 '13 at 16:26
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@did Forget computers and calculators, I add new answer through a mathematical method – Neo Jan 11 '13 at 20:25
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For this approach to be helpful, it should provide bounds on the error between the number proposed and the true solution. – Did Jan 12 '13 at 09:17