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DreamWeaver
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The aims of this tutorial are threefold:
(1) By assuming two values of the Cosine function - which will be proven later on - we develop a large number of multiply-nested radicals to express \(\displaystyle \cos(\pi/2^kn),\, \sin(\pi/2^kn), \,\) and \(\displaystyle \tan(\pi/2^kn) \) in closed form, for ever smaller arguments \(\displaystyle (k, n \in \mathbb{Z}^{+})\).
(2) By considering the limiting values of \(\displaystyle \cos(\pi/2^kn),\, \sin(\pi/2^kn), \,\) and \(\displaystyle \tan(\pi/2^kn) \) as \(\displaystyle k\to \infty\), we evaluate a number of infinitely-nested radicals, such as:\(\displaystyle \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } } = 2\)(3) Finally, we consider three new (?) trigonometric functions, which I will henceforth refer to as the Fractional Sine, Cosine, and Tangent functions respectively:\(\displaystyle \mathscr{Fs}(\theta) = \sum_{k=0}^{\infty} \sin\left( \frac{\theta}{2^k} \right)\)\(\displaystyle \mathscr{Fc}(\theta) = \sum_{k=0}^{\infty} (-1)^k\cos\left( \frac{\theta}{2^k} \right)\)\(\displaystyle \mathscr{Ft}(\theta) = \sum_{k=0}^{\infty} \tan\left( \frac{\theta}{2^k} \right)\)Naturally, some restrictions will need to be applied to \(\displaystyle \theta\) in each case, to ensure convergence. Also, note that the Fractional Cosine function is an alternating series; this is due to the fact that \(\displaystyle \lim_{\theta \to 0}\, \cos (\theta) = 1\).
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For now, we will assume the following two values of the Cosine:\(\displaystyle (01) \quad \cos\left(\frac{\pi}{2}\right) = 0\)\(\displaystyle (02) \quad \cos\left(\frac{\pi}{5}\right) = \frac{1}{2} \sqrt{ \frac{3+\sqrt{5} }{2} }\)In addition, the trigonometric identities used herein are elementary, so proofs will be omitted. The main identities used, aside from \(\displaystyle \cos^2 x+ \sin^2 x = 1\), are essentially three different forms of the Cosine Double Angle formula:\(\displaystyle (03)\quad \sin\frac{x}{2} = \pm \sqrt{\frac{1-\cos x}{2}}\)\(\displaystyle (04)\quad \cos\frac{x}{2} = \pm \sqrt{\frac{1+\cos x}{2}}\)\(\displaystyle (05)\quad \tan^2x= \frac{1-\cos 2x}{1+\cos 2x}\)
In identity (03), we have \(\displaystyle \text{sgn}= + \Leftrightarrow 0 < x < 2\pi\), whereas in identity (04) we have \(\displaystyle \text{sgn}= + \Leftrightarrow -\pi < x < \pi\).
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Entering \(\displaystyle \cos(\pi/2)= 0\) in the RHS of identity (04) gives:\(\displaystyle \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}\)If we repeat this process, each time entering our new value in the RHS of identity (04), we obtain the series of values:\(\displaystyle \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)\(\displaystyle \cos\left(\frac{\pi}{8}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{2} }\)\(\displaystyle \cos\left(\frac{\pi}{16}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }\)\(\displaystyle \cos\left(\frac{\pi}{32}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }\)\(\displaystyle \cos\left(\frac{\pi}{64}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 } } } } }\)\(\displaystyle \cos\left(\frac{\pi}{128}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 } } } } } }\)
We introduce the operator \(\displaystyle \Delta_n(x)\), where the operation is "add 2 and then square-root the sum" \(\displaystyle n\)-times. Here, \(\displaystyle x\) is the operand, that is the initial value that is to be operated on. Then we man write:\(\displaystyle \cos\left(\frac{\pi}{2^{n+1}}\right) = \frac{1}{2}\, \Delta_n(0)\quad \quad \quad n=0, 1, 2, \cdots \)Now, on the one hand, we have\(\displaystyle \text{limit}_{n\to \infty} \, \Delta_n(0) = \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } }\)whereas on the other, \(\displaystyle \text{limit}_{n\to \infty} \cos\left(\frac{\pi}{2^{n+1}}\right) = \text{limit}_{\theta \to 0}\, \cos \theta = \cos 0 = 1\)Equating the two, and multiplying both sides by a factor of \(\displaystyle 2\) then gives the following infinitely nested radical:
\(\displaystyle \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } } = 2\)
Comments and/or questions should be posted here:
http://mathhelpboards.com/commentary-threads-53/commentary-sines-cosines-infinitely-nested-radicals-8283.html
(1) By assuming two values of the Cosine function - which will be proven later on - we develop a large number of multiply-nested radicals to express \(\displaystyle \cos(\pi/2^kn),\, \sin(\pi/2^kn), \,\) and \(\displaystyle \tan(\pi/2^kn) \) in closed form, for ever smaller arguments \(\displaystyle (k, n \in \mathbb{Z}^{+})\).
(2) By considering the limiting values of \(\displaystyle \cos(\pi/2^kn),\, \sin(\pi/2^kn), \,\) and \(\displaystyle \tan(\pi/2^kn) \) as \(\displaystyle k\to \infty\), we evaluate a number of infinitely-nested radicals, such as:\(\displaystyle \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } } = 2\)(3) Finally, we consider three new (?) trigonometric functions, which I will henceforth refer to as the Fractional Sine, Cosine, and Tangent functions respectively:\(\displaystyle \mathscr{Fs}(\theta) = \sum_{k=0}^{\infty} \sin\left( \frac{\theta}{2^k} \right)\)\(\displaystyle \mathscr{Fc}(\theta) = \sum_{k=0}^{\infty} (-1)^k\cos\left( \frac{\theta}{2^k} \right)\)\(\displaystyle \mathscr{Ft}(\theta) = \sum_{k=0}^{\infty} \tan\left( \frac{\theta}{2^k} \right)\)Naturally, some restrictions will need to be applied to \(\displaystyle \theta\) in each case, to ensure convergence. Also, note that the Fractional Cosine function is an alternating series; this is due to the fact that \(\displaystyle \lim_{\theta \to 0}\, \cos (\theta) = 1\).
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For now, we will assume the following two values of the Cosine:\(\displaystyle (01) \quad \cos\left(\frac{\pi}{2}\right) = 0\)\(\displaystyle (02) \quad \cos\left(\frac{\pi}{5}\right) = \frac{1}{2} \sqrt{ \frac{3+\sqrt{5} }{2} }\)In addition, the trigonometric identities used herein are elementary, so proofs will be omitted. The main identities used, aside from \(\displaystyle \cos^2 x+ \sin^2 x = 1\), are essentially three different forms of the Cosine Double Angle formula:\(\displaystyle (03)\quad \sin\frac{x}{2} = \pm \sqrt{\frac{1-\cos x}{2}}\)\(\displaystyle (04)\quad \cos\frac{x}{2} = \pm \sqrt{\frac{1+\cos x}{2}}\)\(\displaystyle (05)\quad \tan^2x= \frac{1-\cos 2x}{1+\cos 2x}\)
In identity (03), we have \(\displaystyle \text{sgn}= + \Leftrightarrow 0 < x < 2\pi\), whereas in identity (04) we have \(\displaystyle \text{sgn}= + \Leftrightarrow -\pi < x < \pi\).
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Entering \(\displaystyle \cos(\pi/2)= 0\) in the RHS of identity (04) gives:\(\displaystyle \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}\)If we repeat this process, each time entering our new value in the RHS of identity (04), we obtain the series of values:\(\displaystyle \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)\(\displaystyle \cos\left(\frac{\pi}{8}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{2} }\)\(\displaystyle \cos\left(\frac{\pi}{16}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }\)\(\displaystyle \cos\left(\frac{\pi}{32}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }\)\(\displaystyle \cos\left(\frac{\pi}{64}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 } } } } }\)\(\displaystyle \cos\left(\frac{\pi}{128}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 } } } } } }\)
We introduce the operator \(\displaystyle \Delta_n(x)\), where the operation is "add 2 and then square-root the sum" \(\displaystyle n\)-times. Here, \(\displaystyle x\) is the operand, that is the initial value that is to be operated on. Then we man write:\(\displaystyle \cos\left(\frac{\pi}{2^{n+1}}\right) = \frac{1}{2}\, \Delta_n(0)\quad \quad \quad n=0, 1, 2, \cdots \)Now, on the one hand, we have\(\displaystyle \text{limit}_{n\to \infty} \, \Delta_n(0) = \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } }\)whereas on the other, \(\displaystyle \text{limit}_{n\to \infty} \cos\left(\frac{\pi}{2^{n+1}}\right) = \text{limit}_{\theta \to 0}\, \cos \theta = \cos 0 = 1\)Equating the two, and multiplying both sides by a factor of \(\displaystyle 2\) then gives the following infinitely nested radical:
\(\displaystyle \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } } = 2\)
Comments and/or questions should be posted here:
http://mathhelpboards.com/commentary-threads-53/commentary-sines-cosines-infinitely-nested-radicals-8283.html
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