Problem 24 - A Course in Calculus and Real Analysis

Question

Let \(f(x):=\cos x\) for \(x \in \mathbb{R}\). Show that the Taylor series of \(f\) is convergent for \(x \in \mathbb{R}\). Deduce that $$ \cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\cdots=\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{(2 k) !} \quad \text { for } x \in \mathbb{R} $$

Step-by-Step Solution

Answer

In conclusion, the Taylor series for the function \(f(x) = \cos x\) converges for all \(x \in \mathbb{R}\) due to the function being infinitely differentiable and its error term going to zero as the degree of the Taylor polynomial goes to infinity. Therefore, we deduce that the Maclaurin series representation of \(\cos x\) is given by: \[ \cos x = \sum_{k=0}^{\infty}(-1)^k \frac{x^{2k}}{(2k)!} \]

Step 1: Recall the Taylor Series formula

Recall that the Taylor series of a function \(f(x)\) centered at \(a\) is given by: \[ f(x) = \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x - a)^k \] In our case, we want the Taylor series centered at \(a = 0\), which is the Maclaurin series: \[ f(x) = \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}(x)^k \]

Step 2: Calculate the derivatives of \(f(x) = \cos x\)

To find the Maclaurin series, we need to find the derivatives of \(f(x) = \cos x\) and evaluate them at \(x = 0\). - \(f^{(0)}(x) = \cos x\) and \(f^{(0)}(0) = 1\) - \(f^{(1)}(x) = - \sin x\) and \(f^{(1)}(0) = 0\) - \(f^{(2)}(x) = - \cos x\) and \(f^{(2)}(0) = -1\) - \(f^{(3)}(x) = \sin x\) and \(f^{(3)}(0) = 0\) As we can observe, the derivatives follow a pattern: they alternate between \(\cos x\) and \(- \sin x\). Moreover, the function values at \(x = 0\) are either \(1\), \(-1\), or \(0\).

Step 3: Write down the Taylor series

From the derivatives and their values at \(x = 0\), we can now write the Maclaurin series for \(f(x) = \cos x\): \[ \cos x = \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \] Since the derivatives repeat after every \(4\) steps, we can write a general formula for the Taylor series as: \[ \cos x = \sum_{k=0}^{\infty}(-1)^k \frac{x^{2k}}{(2k)!} \]

Step 4: Confirm convergence

The Taylor series converges for all \(x \in \mathbb{R}\) if the function is infinitely differentiable and the error function goes to zero as the degree of the Taylor polynomial goes to infinity. Since all the derivatives exist for \(f(x) = \cos x\), it's infinitely differentiable. Recalling the remainder term in Taylor's Theorem: \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n + 1)!} (x - a)^{n+1} \] Where \(c\) is between \(x\) and \(a\). In our case, \(a = 0\), so we have: \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n + 1)!} x^{n+1} \] Considering the cyclical derivatives and the maximum value of the error at \(|x| = 1\): \[ |R_n(x)| \le \frac{1}{(n + 1)!} \] Since the expression \(\frac{1}{(n + 1)!}\) converges to \(0\) as \(n\) goes to infinity, the Taylor series converges for all \(x \in \mathbb{R}\).

Step 5: Conclusion

: We have shown that the Taylor series of the function \(f(x) = \cos x\) converges for all \(x \in \mathbb{R}\). Therefore, the Maclaurin series representation of \(\cos x\) is as follows: \[ \cos x = \sum_{k=0}^{\infty}(-1)^k \frac{x^{2k}}{(2k)!} \]

Key Concepts

Maclaurin Series

The Maclaurin series is a specific case of the Taylor series where the function is expanded around 0, simplifying the expression of a function. The general Taylor series for a smooth function at a point \(a\) is given by:

• \[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k \]

For the Maclaurin series, we set \(a = 0\), which simplifies the expression to:

• \[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k \]

This representation is particularly useful for functions like trigonometric functions, exponentials, and log functions. By using Maclaurin series, we can approximate functions just by knowing their derivatives at a single point, greatly aiding in understanding their behavior across values.

Convergence of Series

Convergence of a series refers to whether the series approaches a finite value as more terms are added. For Taylor and Maclaurin series, convergence depends on both the function and the value of \(x\).
To verify convergence, it's crucial to examine the remainder of the series, given by Taylor's theorem:

• \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n + 1)!} x^{n+1} \]

Here, \(c\) lies between 0 and \(x\). Convergence can often be confirmed if \(R_n(x)\) approaches zero as \(n\) increases.
For the Maclaurin series of \(\cos x\), given its cyclic and bounded derivatives, the series converges for all real numbers \(x\), as the remainder \(R_n(x)\) becomes negligible for large \(n\). This means the series effectively represents \(\cos x\) across its entire domain.

Infinitely Differentiable Functions

Functions that are infinitely differentiable have derivatives of all orders. This is a vital property for determining the Taylor or Maclaurin series of a function.
For \(f(x) = \cos x\), all derivatives exist and are smooth, following the pattern:

• \(f^{(n)}(x) = \cos x, -\sin x, -\cos x, \sin x \), repeating every four derivatives.

Because \(\cos x\) is infinitely differentiable, the Taylor and Maclaurin series can be continuously extended, allowing for a complete representation of the function.
In essence, any function that can be differentiated endlessly without discrepancy ensures that its behavior can be virtually captured through its Taylor series for any value of \(x\), simplifying complex calculations and analyses.

Derivatives of Trigonometric Functions

Understanding the derivatives of trigonometric functions is crucial to constructing their Taylor or Maclaurin series. For the function \(f(x) = \cos x\), the derivatives follow a periodic cycle:

• \(f^{(0)}(x) = \cos x\)
• \(f^{(1)}(x) = -\sin x\)
• \(f^{(2)}(x) = -\cos x\)
• \(f^{(3)}(x) = \sin x\)
• The cycle then repeats every four derivatives.

This cycle is due to the nature of the sine and cosine functions, which are inherently linked through their derivatives.
Notably, these derivatives evaluated at 0 determine the terms in the Maclaurin series for \(\cos x\). For instance, \(f^{(2k+1)}(0) = 0\), simplifying the series to include only even powers. Periodicity of derivatives not only helps construct power series but also highlights the harmonic nature of trigonometric functions, essential in fields like physics and engineering.