The Taylor series is one of the most powerful tools in calculus, allowing us to represent complex functions as infinite sums of polynomial terms. When we encounter a function like cos x 2, the notation introduces a critical ambiguity: does it mean $\cos(x^2)$ (the cosine of $x$ squared) or $\cos^2(x)$ (the square of the cosine of $x$)?
Both interpretations appear frequently in physics, engineering, and advanced mathematics, yet they require distinctly different approaches to expand. This article provides a complete, step-by-step derivation for both cases, starting from the fundamental Maclaurin series of $\cos(x)$. We will explore the substitution method for composite functions and the trigonometric identity method for powers, ensuring you can tackle either problem with confidence.
The Foundation: Maclaurin Series of $\cos(x)$
Before diving into the variations, we must establish the baseline. The Maclaurin series (a Taylor series centered at $a=0$) for $\cos(x)$ is derived from the derivatives of cosine evaluated at zero.
The derivatives cycle as follows:
- $f(x) = \cos(x) \implies f(0) = 1$
- $f'(x) = -\sin(x) \implies f'(0) = 0$
- $f''(x) = -\cos(x) \implies f''(0) = -1$
- $f'''(x) = \sin(x) \implies f'''(0) = 0$
- $f^{(4)}(x) = \cos(x) \implies f^{(4)}(0) = 1$
Because the odd derivatives are zero at $x=0$, only even powers remain. The general formula is:
$ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} $
Writing out the first few terms explicitly:
$ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!
Radius of Convergence: This series converges for all real numbers ($-\infty < x < \infty$). This infinite radius is crucial because it guarantees that any algebraic manipulation we perform (substitution or squaring) will also result in a series valid for all real $x$.
Case 1: Taylor Series of $\cos(x^2)$ (Composition)
If "cos x 2" implies $\cos(x^2)$, we are dealing with a composite function. The standard and most efficient method to find this series is direct substitution.
The Substitution Method
Since the series for $\cos(u)$ converges for all real $u$, we can simply replace the variable $u$ with $x^2$ in the standard formula.
Let $u = x^2$. Then: $ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} (x^2)^{2n} $
Applying the power rule $(x^a)^b = x^{ab}$:
$ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{4n} $
Expanded Form
Writing out the terms reveals a distinct pattern: only powers that are multiples of 4 appear.
- $n=0$: $\frac{(-1)^0}{0!} x^0 = 1$
- $n=1$: $\frac{(-1)^1}{2!} x^4 = -\frac{x^4}{2}$
- $n=2$: $\frac{(-1)^2}{4!} x^8 = \frac{x^8}{24}$
- $n=3$: $\frac{(-1)^3}{6!} x^{12} = -\frac{x^{12}}{720}$
$ \cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \frac{x^{16}}{40320} - \dots $
Why This Works (Theoretical Justification)
You might ask: *Is it always valid to substitute a polynomial into a power series?And ** If a power series $\sum a_n u^n$ has a radius of convergence $R = \infty$ (entire function), the composition $\sum a_n (g(x))^n$ converges for all $x$ where $g(x)$ is defined. Practically speaking, * **Yes. Since $g(x) = x^2$ is a polynomial defined everywhere, the resulting series for $\cos(x^2)$ converges for all $x \in \mathbb{R}$ Still holds up..
Application: The Fresnel Integral
A primary reason $\cos(x^2)$ appears in advanced calculus is the Fresnel C Integral: $ C(x) = \int_0^x \cos\left(\frac{\pi t^2}{2}\right) dt $ Because $\cos(x^2)$ has no elementary antiderivative, we integrate the Taylor series term-by-term: $ \int \cos(x^2) dx = \int \left( 1 - \frac{x^4}{2} + \frac{x^8}{24} - \dots \right) dx = x - \frac{x^5}{10} + \frac{x^9}{216} - \dots + C $ This allows for precise numerical approximation of diffraction patterns in optics.
Case 2: Taylor Series of $\cos^2(x)$ (Power Reduction)
If "cos x 2" implies $\cos^2(x)$ (often written as $\cos^2 x$), we are dealing with the square of the function. Substitution does not work here directly. Instead, we use a trigonometric identity to linearize the power before expanding.
The Power-Reduction Identity
The double-angle formula for cosine gives us: $ \cos(2x) = 2\cos^2(x) - 1 $ Rearranging for $\cos^2(x)$: $ \cos^2(x) = \frac{1 + \cos(2x)}{2} $
This transforms a nonlinear problem (squaring a series) into a linear one (scaling and shifting a series).
Substitution into the Identity
We already know the series for $\cos(u)$. Substitute $u = 2x$: $ \cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} (2x)^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n}}{(2n)!} x^{2n} $ $ \cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{(2n)!
Now plug this into the identity: $ \cos^2(x) = \frac{1}{2} + \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{(2n)!} x^{2n} $
Handling the Constant Term
Notice that when $n
