Cosine Function is Absolutely Convergent
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Theorem
Let $\cos$ be the cosine function.
Then:
- $\cos x$ is absolutely convergent for all $x \in \R$.
Complex Case
The complex cosine function $\cos: \C \to \C$ is absolutely convergent.
Proof
Recall the definition of the cosine function:
- $\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
For:
- $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$
to be absolutely convergent, we want:
- $\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} } = \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n} } {\paren {2 n}!}$
to be convergent.
But:
- $\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n} } {\paren {2 n}!}$
is just the terms of:
- $\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n} {n!}$
for even $n$.
Thus:
- $\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n} } {\paren {2 n}!} < \sum_{n \mathop = 0}^\infty \frac {\size x^n}{n!}$
But:
- $\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n}{n!} = \exp \size x$
from the Taylor Series Expansion for Exponential Function of $\size x$, which converges for all $x \in \R$.
Also, the sequence of partial sums:
- $\ds \sum_{n \mathop = 0}^k \frac {\size x^{2 n} } {\paren {2 n}!}$
is increasing.
The result follows from an application of the Monotone Convergence Theorem (Real Analysis).
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.2$