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Why Does the Taylor Series Expansion of sin(x) Around x0 Have the Form It Does?
Why Does the Taylor Series Expansion of sin(x) Around x0 Have the Form It Does?
The Taylor series expansion of the sine function around x0 is a fundamental concept in calculus and mathematical analysis. This expansion can be represented as a power series, which allows for precise approximations of the sine function. This article explains the derivation and properties of this series, highlighting the role of derivatives and the pattern of the coefficients.
The Taylor Series Expansion of sin(x)
The Taylor series of a function f(x) around xa is defined as:
f(x) f(a) f'(a)(x-a) frac{f''(a)}{2!}(x-a)^2 frac{f'''(a)}{3!}(x-a)^3 cdots
For the sine function, we will expand it around a0, which is referred to as the Maclaurin series:
sin(x) sum_{n0}^{infty} frac{f^{(n)}(0)}{n!}x^n
Calculating the Derivatives of sin(x) at x0
To find the Taylor series expansion of sin(x), we need to calculate the derivatives of sin(x) at x0:
First derivative: sin(x), f'(0) sin(0) 0 Second derivative: cos(x), f''(0) cos(0) 1 Third derivative: -sin(x), f'''(0) -sin(0) 0 Fourth derivative: -cos(x), f''''(0) -cos(0) -1 Fifth derivative: sin(x), f'''''(0) sin(0) 0 Sixth derivative: cos(x), f'''''(0) cos(0) 1 Seventh derivative: -sin(x), f'''''''(0) -sin(0) 0From these calculations, we can observe a pattern in the derivatives. Only the odd derivatives of sin(x) contribute to the series, and they alternate in sign. The even derivatives all equal 0.
Constructing the Taylor Series for sin(x)
Using the pattern of derivatives, the Taylor series for sin(x) can be constructed as:
sin(x) 0 cdot x 1 cdot frac{x^3}{3!} - 0 cdot frac{x^4}{4!} 1 cdot frac{x^5}{5!} - 1 cdot frac{x^7}{7!} cdots
This series can be written in summation notation as:
sin(x) sum_{n0}^{infty} frac{(-1)^n x^{2n 1}}{(2n 1)!}
The series converges for all x and provides a way to compute sin(x) using its power series representation.
Reproduction of the Series Through Differentiation
Another approach to deriving the series for sin(x) involves differentiation. Starting with the general form of the series:
sin(x) a_0 a_1x a_2x^2 a_3x^3 cdots
1. At x0, we have sin(0) 0 a_0, so a_0 0.
2. Differentiating both sides with respect to x gives:
cos(x) a_1 2a_2x 3a_3x^2 4a_4x^3 cdots
3. At x0, cos(0) 1 a_1, so a_1 1.
4. Differentiating again, we have:
-sin(x) 2a_2 6a_3x 12a_4x^2 cdots
5. At x0, -sin(0) 0 2a_2, so a_2 0.
6. Differentiating a third time, we get:
-cos(x) 6a_3 24a_4x cdots
7. At x0, -cos(0) -1 6a_3, so a_3 -frac{1}{6}, which is frac{-1}{1 cdot 2 cdot 3}.
Continuing this process, we find the coefficients for higher odd powers and can substitute them back into the series:
sin(x) x - frac{x^3}{3!} frac{x^5}{5!} - frac{x^7}{7!} cdots sum_{n0}^{infty} frac{(-1)^n x^{2n 1}}{(2n 1)!}
Conclusion
This series is valid and converges for all x. It provides a powerful method for computing the sine of any angle and is an essential tool in calculus and mathematical analysis. By understanding the pattern of derivatives and using the power series expansion, we can derive the Taylor series for the sine function, which is a cornerstone of mathematical theory and practical applications in various fields.