Question

How is the series expansion of some elements like sinx, cowx, etc is proved?

Answer 1

The series expansion of elements like sinx, cosx, etc., is proved using Taylor's Theorem with a remainder term. The proof involves showing that the remainder term approaches zero as the number of terms in the series approaches infinity, thus demonstrating that the function is equal to its infinite series expansion within its radius of convergence.

Answer 2

The series expansion of functions like $\sin x$, $\cos x$, $e^x$, etc., is rigorously proved using Taylor's Theorem with a Remainder Term. This theorem states that if a function $f(x)$ has derivatives of all orders up to $(n+1)$ in an interval containing $a$, then for any $x$ in that interval, $f(x)$ can be expressed as:

Taylor's Theorem with Remainder

$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x)$$

Here, $R_n(x)$ is the remainder term, which accounts for the difference between the function and its $n$-th degree Taylor polynomial. The most common form of the remainder term is the Lagrange form:

$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$

where $c$ is some value between $a$ and $x$.

Proof of Series Expansion

To prove that a function $f(x)$ is exactly equal to its infinite Taylor series (or Maclaurin series if $a=0$) within a certain interval, we must show that the remainder term $R_n(x)$ approaches zero as $n$ approaches infinity:

$$\lim_{n \to \infty} R_n(x) = 0$$

If this condition is met for a given $x$, then the infinite Taylor series converges to $f(x)$:

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

Example: Proof for $\sin x$ (Maclaurin Series, i.e., $a=0$)

1. Find Derivatives:

   Let $f(x) = \sin x$. The derivatives are:

   • $f(x) = \sin x$

   • $f'(x) = \cos x$

   • $f''(x) = -\sin x$

   • $f'''(x) = -\cos x$

   • $f^{(4)}(x) = \sin x$

   The derivatives cycle with a period of 4.

2. Bound the Derivatives:

   For any $k$, the $k$-th derivative $f^{(k)}(x)$ will be either $\pm \sin x$ or $\pm \cos x$. Therefore, for any real $x$ and any $k$:

   $$|f^{(k)}(x)| \le 1$$

3. Apply Remainder Term:

   Using the Lagrange form of the remainder term with $a=0$:

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

   Taking the absolute value:

   $$|R_n(x)| = \left| \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \right| = \frac{|f^{(n+1)}(c)|}{(n+1)!}|x|^{n+1}$$

   Since $|f^{(n+1)}(c)| \le 1$:

   $$|R_n(x)| \le \frac{1}{(n+1)!}|x|^{n+1}$$

4. Evaluate the Limit:

   Now, we need to evaluate the limit of this upper bound as $n \to \infty$:

   $$\lim_{n \to \infty} \frac{|x|^{n+1}}{(n+1)!}$$

   For any fixed real number $x$, as $n$ becomes very large, the factorial term $(n+1)!$ grows much faster than the exponential term $|x|^{n+1}$. Therefore, this limit is 0 for all real $x$.

   For example, consider the ratio of consecutive terms: $\frac{|x|^{n+2}/(n+2)!}{|x|^{n+1}/(n+1)!} = \frac{|x|}{n+2}$. As $n \to \infty$, this ratio approaches 0, indicating that the terms rapidly decrease to zero.

5. Conclusion:

   Since $\lim_{n \to \infty} R_n(x) = 0$ for all real $x$, it proves that $\sin x$ is equal to its infinite Maclaurin series for all real numbers:

   $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$

   A similar proof applies to $\cos x$ and $e^x$, as their derivatives are also bounded (or their growth is dominated by the factorial in the denominator).

The core idea is to show that the error in approximating the function by its finite Taylor polynomial vanishes as more terms are included.