Question

Taylor's series expansion of \(\sin z\) abeut \(z=\pi / 4\) is.

Short answer

Taylor's series expansion of \(\sin z\) about \( z=\pi/4 \) is \( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} (z-\pi/4) - \frac{1}{2\sqrt{2}}(z-\pi/4)^2 - \ldots \).

Step-by-step solution

Understand Taylor's Series Expansion Formula

Taylor's series for a function \( f(z) \) about a point \( a \) is given by:\[\sum_{n=0}^{fty} \frac{f^{(n)}(a)}{n!} (z-a)^n\]where \( f^{(n)}(a) \) is the \( n \)-th derivative of the function at \( z=a \). Our task is to find this expansion for the sine function \( \sin z \) about \( z = \pi/4 \).

Calculate Derivatives of \(\sin z\)

The derivatives of \( \sin z \) are cyclical: - First derivative: \( \cos z \)- Second derivative: \( -\sin z \)- Third derivative: \( -\cos z \)- Fourth derivative: \( \sin z \)These derivatives will repeat their cycle every four terms.

Evaluate Derivatives at \( z = \pi / 4 \)

Calculate the derivatives at \( z = \pi / 4 \):- \( \sin(\pi/4) = \cos(\pi/4) = 1/\sqrt{2} \)- Second and third derivatives: \( -1/\sqrt{2} \)- Fourth derivative: \( 1/\sqrt{2} \) again. This pattern will repeat every four derivatives.

Substitute into Taylor's Series Formula

Using the Taylor series formula, substitute the evaluated derivatives:\[\sin z \approx \sin(\pi/4) + \frac{\cos(\pi/4)}{1!} (z-\pi/4) - \frac{\sin(\pi/4)}{2!}(z-\pi/4)^2 - \frac{\cos(\pi/4)}{3!}(z-\pi/4)^3 + \frac{\sin(\pi/4)}{4!}(z-\pi/4)^4 + \ldots\]Substitute the values of the derivatives:\[\sin z \approx \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} (z-\pi/4) - \frac{1}{2\sqrt{2}}(z-\pi/4)^2 - \frac{1}{6\sqrt{2}}(z-\pi/4)^3 + \frac{1}{24\sqrt{2}}(z-\pi/4)^4 + \ldots\]

Conclusion

The Taylor series expansion of \( \sin z \) about \( z = \pi/4 \) is written as:\[\sin z \approx \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} (z-\pi/4) - \frac{1}{2\sqrt{2}}(z-\pi/4)^2 - \frac{1}{6\sqrt{2}}(z-\pi/4)^3 + \frac{1}{24\sqrt{2}}(z-\pi/4)^4 + \ldots\]This series represents \( \sin(z) \) as a polynomial approximation around the point \( \pi/4 \).

Key concepts

Sine Function

The sine function, denoted as \( \sin(z) \), is a fundamental part of trigonometry. It describes the ratio of the length of the opposite side to the hypotenuse in a right triangle. However, the sine function has interesting and complex properties beyond triangles. It oscillates between -1 and 1, repeating every \( 2\pi \).

This periodic nature makes the sine function crucial in modeling cycles and waves, such as sound waves and alternating current in electricity. Because of these properties, the sine function is widely used in various fields such as physics, engineering, and even economics. When we express \( \sin(z) \) in terms of a Taylor series, we can approximate it with polynomials, making complex calculations simpler and more manageable.

Calculus

Calculus is a branch of mathematics focused on continuous change. It is divided into differential calculus, which concerns rates of change, and integral calculus, which deals with the accumulation of quantities. The Taylor series, used to approximate functions like \( \sin(z) \), is a powerful concept from calculus. It allows us to express functions as infinite polynomials.

Differentiation is a key process in calculus, enabling us to find the derivatives of functions. For \( \sin(z) \), we cycle through its derivatives: \( \cos(z) \), \(-\sin(z)\), \(-\cos(z)\), and back to \( \sin(z) \). Calculus helps us analyze these derivatives and use them to build the Taylor series, which is essential for understanding the behavior of functions near specific points.

Polynomial Approximation

Polynomial approximation is the process of approximating complex functions like \( \sin(z) \) using polynomials. Polynomials are algebraic expressions with terms consisting of variables raised to whole number powers, which makes them simpler compared to other mathematical functions.

The idea behind using polynomial approximation is to take complex, difficult-to-compute functions and replace them with their simpler polynomial counterparts. This is where Taylor series comes into play. When you expand \( \sin(z) \) into a Taylor series around a point like \( \pi/4 \), you are creating a polynomial approximation. This makes it much easier to perform calculations and understand the function’s local behavior without the need for complex trigonometric transformations.