How to Expand and Approximate the sinc Function Using Taylor Series

The sinc function, denoted by (text{sinc}(x)), is defined as:

[ text{sinc}(x) frac{sin(x)}{x} quad text{for} quad x eq 0, quad text{sinc}(0) 1 ]

In this article, we will explore how to expand the product (xtext{sinc}(x^2sqrt{pi}x)) using the Taylor series of trigonometric functions. We will employ a series of mathematical transformations and approximations, providing a comprehensive guide for SEO purposes.

Introduction to the Problem

The task at hand involves expanding the expression (xtext{sinc}(x^2sqrt{pi}x)). We will utilize the Taylor series of trigonometric functions around zero, making appropriate substitutions, and then simplifying the expression step by step.

Step 1: Completing the Square

First, we complete the square on the argument of the sine function within the sinc function:

[ x^2sqrt{pi}x left(frac{x}{2}right)^2 cdot sqrt{pi}x - frac{pi}{4} ]

This can be simplified to:

[ x^2sqrt{pi}x frac{x^2sqrt{pi}}{4} - frac{pi}{4} cdot x^2 ]

Step 2: Applying Goniometric Formula

Next, we use the goniometric identity for the sine of a sum of angles:

[ sin(a - frac{pi}{4}) frac{1}{sqrt{2}} (sin(a) - cos(a)) ]

Here, let (a x^2sqrt{pi}x). We then substitute (u xfrac{sqrt{pi}}{2}), which results in:

[ sinleft(x^2sqrt{pi}x - frac{pi}{4}right) frac{1}{sqrt{2}} left(sin u^2 - cos u^2right) ]

Step 3: Substitution and Simplification

Using the substitution (u xfrac{sqrt{pi}}{2}), we get:

[ xtext{sinc}(x^2sqrt{pi}x) x cdot frac{sin left(frac{x^2sqrt{pi}}{4} - frac{pi}{4} cdot x^2right)}{frac{x^2sqrt{pi}}{4} - frac{pi}{4} cdot x^2} ]

Substituting in the goniometric identity:

[ xtext{sinc}(x^2sqrt{pi}x) x cdot frac{frac{1}{sqrt{2}} left(sin u^2 - cos u^2right)}{u - frac{sqrt{pi}}{2}} ]

Since (u xfrac{sqrt{pi}}{2}), we can further simplify this expression:

[ xtext{sinc}(x^2sqrt{pi}x) x cdot frac{frac{1}{sqrt{2}} left(sin u^2 - cos u^2right)}{u - frac{sqrt{pi}}{2}} cdot frac{2}{sqrt{pi}} ]

This simplifies to:

[ xtext{sinc}(x^2sqrt{pi}x) frac{x}{sqrt{2pi}} left(sin u^2 - cos u^2right) ]

Step 4: Using Taylor Series

To further simplify, we need the Taylor series for (sin(x)) and (cos(x)) around zero:

[ sin(x) x - frac{x^3}{3!} frac{x^5}{5!} - cdots ]

[ cos(x) 1 - frac{x^2}{2!} frac{x^4}{4!} - cdots ]

Applying these to (sin(u^2)) and (cos(u^2)) where (u xfrac{sqrt{pi}}{2}), we get:

[ sin(u^2) u^2 - frac{(u^2)^3}{3!} frac{(u^2)^5}{5!} - cdots ]

[ cos(u^2) 1 - frac{(u^2)^2}{2!} frac{(u^2)^4}{4!} - cdots ]

Thus, the expression becomes:

[ xtext{sinc}(x^2sqrt{pi}x) frac{x}{sqrt{2pi}} left(u^2 - frac{(u^2)^3}{3!} frac{(u^2)^5}{5!} - cdots - 1 frac{(u^2)^2}{2!} - frac{(u^2)^4}{4!} cdots right) ]

Substituting back (u xfrac{sqrt{pi}}{2}), we get the final simplified form:

[ xtext{sinc}(x^2sqrt{pi}x) frac{x}{sqrt{2pi}} left( left(xfrac{sqrt{pi}}{2}right)^2 - frac{left(xfrac{sqrt{pi}}{2}right)^6}{3!} frac{left(xfrac{sqrt{pi}}{2}right)^{10}}{5!} - cdots - 1 frac{left(xfrac{sqrt{pi}}{2}right)^4}{2!} - frac{left(xfrac{sqrt{pi}}{2}right)^8}{4!} cdots right) ]

Conclusion

In this article, we have successfully expanded the sinc function using the Taylor series of trigonometric functions and several mathematical transformations. This approach is particularly useful in signal processing and approximation theory, making this a versatile method for various applications.

Remember, the sinc function is fundamental in signal processing and the Taylor series approximation can help in understanding and manipulating more complex expressions.