I was thinking about Maclaurin Series to compute sin, cos and tan. It was in a Python programming group and we started to think about simplification of tan, based on sin/cos formula. As both have summations, but their interval are the same ($n=1$ to $\infty$), I thought that I could write them in single summation of the quotient of both.

$$\sin x= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(2n+1)} x^{2n}x$$

$$\cos x= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$$

$$\frac{\sin x}{\cos x} = \frac{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(2n+1)} x^{2n}x}{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}x}{(2n)!(2n+1)} \times \frac{(2n)!}{(-1)^n x^{2n}} = \sum_{n=0}^{\infty} \frac{x}{2n+1}$$

But the result is far from correct. Can anyone help me understand why I cannot join these two summations in a single one (as they have the same period/interval)?

I understand the result is incorrect, but I could not figure out why.

  • 7
    $\begingroup$ $(a+b)/(c+d) \neq a/c + b/d$. You are confusing sums with products. $\endgroup$ Apr 12 '17 at 13:42
  • $\begingroup$ I once did this exact same thing. Easy mistake to make. Good for you for catching it and asking about it. $\endgroup$
    – The Count
    Apr 12 '17 at 13:47

Your error is more obvious if we write it in a different way. You make the following incorrect simplification on the third line of your answer

$$\dfrac{a+b}{c+d} = \dfrac{a}{b} + \dfrac{c}{d}$$

I understand the confusion - even though it is a sum, the notation makes it look like one term that can be inverted over a division, but that is not the case.

  • $\begingroup$ Yes, I went directly to the term and I did not saw it as the whole expression. $\endgroup$
    – nmenezes
    Apr 12 '17 at 13:53

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