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The original problem was showing that this infinite sum converges to $\tan\theta$: $$\sum_{n=1}^\infty \frac{\tan\dfrac{\theta}{2^n}}{\cos\dfrac{\theta}{2^{n-1}}}$$ One hint was given: the series telescopes. I figured this meant I could factorize the denominator if I just apply some identities, and indeed it does: $$\frac{\tan\dfrac{\theta}{2^n}}{\cos\dfrac{\theta}{2^{n-1}}}=\frac{\sec\dfrac{\theta}{2^n}}{\cos\dfrac{\theta}{2^n}-\sin\dfrac{\theta}{2^n}}-\frac{\sec\dfrac{\theta}{2^n}}{\cos\dfrac{\theta}{2^n}+\sin\dfrac{\theta}{2^n}}$$ I plugged in the first few $n$ to see if a pattern emerges, and indeed it does. Mathematica gives me this output:

enter image description here

This gives me the identity, $$\dfrac{1}{\cos\dfrac{\theta}{2^n}-\sin\dfrac{\theta}{2^n}}+\dfrac{1}{\cos\dfrac{\theta}{2^n}+\sin\dfrac{\theta}{2^n}}=\dfrac{\sin\left(1-\dfrac{1}{2^n}\right)\theta}{\cos \dfrac{\theta}{2^n}}$$ where $x=\theta$, and this converges to $\tan \theta$ as $k\to\infty$. So I'm curious, how would I come to derive something like this?

A simpler approach was expected, I'll have to check to see if I still remember it. Something to do with a tangent identity, if I recall correctly. If I find it, I'll include it in an edit.

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hint

Call $\tan{\frac{\theta}{2^n}}=t_n$ and use the identity $\cos{\frac{\theta}{2^{n-1}}}=\frac{1-t_n^2}{1+t_n^2}$ the term of the series is

$$\frac{t_n(1+t_n^2)}{1-t_n^2}$$

And you decompose that into a sum of rational fractions

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This is not an answer (neither a proof of anything) but it is too long for a comment.

Consider $$A_n=\tan \left( 2^{-n}\,\theta\right) \sec \left( 2^{1-n}\,\theta\right)$$ and expand it as a Taylor series built at $\theta=0$ up to order $p$ ; this let you with infinite sums of terms in geometric progression. After summations over $n$, you end with a sum which is strictly identical to the Taylor expansion of $\tan(\theta)$ to the same order $p$.

Since you use Mathematica, I suppose that the following should work.

s[n_, p_] := Normal[Series[Tan[x/2^n]/Cos[x/2^(n - 1)], {x, 0, p}]]
sum[p_]   := Expand[Sum[s[n, p], {n, 1, Infinity}]]
diff[p_]  := sum[p] - Normal[Series[Tan[x], {x, 0, p}]]
Table[{p, diff[p]}, {p, 1, 100}]
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