Taylor Expansion: E^x Sin(y) At (ln(3), -6π)

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When we delve into the fascinating world of calculus, especially multivariable calculus, we often encounter functions that, while beautiful in their exact form, can become quite complex to work with. This is where the power of Taylor's formula comes into play. It's like having a secret decoder ring for complicated functions, allowing us to approximate them with simpler polynomials in a specific region. Today, we're going to tackle the function f(x,y)=ex⋅sin⁡(y)f(x, y) = e^x \cdot \sin(y) and expand it using Taylor's formula up to the second-order terms around the point (ln⁡(3),−6π)(\ln(3), -6\pi). This process is crucial for understanding the local behavior of functions, which has wide-ranging applications in physics, engineering, economics, and many other fields. By approximating a complex function with a polynomial, we can often simplify calculations, analyze stability, and make predictions about how the function will behave in the vicinity of a particular point. The beauty of Taylor's formula lies in its ability to capture the essence of a function's behavior – its value, its slope, its curvature – using a series of increasingly refined polynomial terms. We'll be breaking down this expansion step-by-step, making sure to illuminate each part of the process so you can follow along with confidence. So, buckle up, grab your favorite thinking beverage, and let's dive into the intricate yet rewarding world of Taylor expansions!

Understanding Taylor's Formula for Multivariable Functions

Before we begin the actual expansion, let's refresh our understanding of Taylor's formula for a function of two variables, f(x,y)f(x, y), around a point (a,b)(a, b). The general form of the Taylor expansion up to the second-order terms is given by:

f(x,y)≈f(a,b)+∂f∂x(a,b)(x−a)+∂f∂y(a,b)(y−b)+12!(∂2f∂x2(a,b)(x−a)2+2∂2f∂x∂y(a,b)(x−a)(y−b)+∂2f∂y2(a,b)(y−b)2)f(x, y) \approx f(a, b) + \frac{\partial f}{\partial x}(a, b)(x-a) + \frac{\partial f}{\partial y}(a, b)(y-b) + \frac{1}{2!} \left( \frac{\partial^2 f}{\partial x^2}(a, b)(x-a)^2 + 2 \frac{\partial^2 f}{\partial x \partial y}(a, b)(x-a)(y-b) + \frac{\partial^2 f}{\partial y^2}(a, b)(y-b)^2 \right)

This formula is essentially an extension of the single-variable Taylor series. The first term, f(a,b)f(a, b), gives us the value of the function at the point of expansion. The next two terms, involving the first partial derivatives, represent the linear approximation of the function – essentially, the tangent plane at (a,b)(a, b). The terms within the parentheses, involving the second partial derivatives, capture the curvature of the function, providing a more accurate quadratic approximation. Each partial derivative is evaluated at the point (a,b)(a, b), and the terms are weighted by powers of (x−a)(x-a) and (y−b)(y-b), representing the distance from the expansion point. This structure allows the polynomial to closely mimic the function's behavior as we move away from (a,b)(a, b), with higher-order terms providing progressively finer adjustments. The factorial denominators ensure that the coefficients are consistent with the general Taylor series coefficients for functions of a single variable when one of the variables is held constant. It's a beautiful piece of mathematical machinery that provides a powerful tool for approximation and analysis.

Step 1: Evaluating the Function and Its Derivatives at the Given Point

Our function is f(x,y)=ex⋅sin⁡(y)f(x, y) = e^x \cdot \sin(y), and the point of expansion is (a,b)=(ln⁡(3),−6π)(a, b) = (\ln(3), -6\pi). The first step is to evaluate the function itself at this point:

f(ln⁡(3),−6π)=eln⁡(3)⋅sin⁡(−6π)=3⋅0=0f(\ln(3), -6\pi) = e^{\ln(3)} \cdot \sin(-6\pi) = 3 \cdot 0 = 0

Now, let's find the first partial derivatives. First, with respect to xx:

∂f∂x=∂∂x(exsin⁡(y))=exsin⁡(y)\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(e^x \sin(y)) = e^x \sin(y)

Evaluate this at our point:

∂f∂x(ln⁡(3),−6π)=eln⁡(3)sin⁡(−6π)=3⋅0=0\frac{\partial f}{\partial x}(\ln(3), -6\pi) = e^{\ln(3)} \sin(-6\pi) = 3 \cdot 0 = 0

Next, the partial derivative with respect to yy:

∂f∂y=∂∂y(exsin⁡(y))=excos⁡(y)\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(e^x \sin(y)) = e^x \cos(y)

Evaluate this at our point:

∂f∂y(ln⁡(3),−6π)=eln⁡(3)cos⁡(−6π)=3⋅1=3\frac{\partial f}{\partial y}(\ln(3), -6\pi) = e^{\ln(3)} \cos(-6\pi) = 3 \cdot 1 = 3

So far, we have f(a,b)=0f(a, b) = 0, ∂f∂x(a,b)=0\frac{\partial f}{\partial x}(a, b) = 0, and ∂f∂y(a,b)=3\frac{\partial f}{\partial y}(a, b) = 3. These will form the linear part of our Taylor approximation.

Step 2: Calculating the Second Partial Derivatives

Now we need to compute the second partial derivatives. Let's start with the second partial derivative with respect to xx:

∂2f∂x2=∂∂x(∂f∂x)=∂∂x(exsin⁡(y))=exsin⁡(y)\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial x}(e^x \sin(y)) = e^x \sin(y)

Evaluating at (ln⁡(3),−6π)(\ln(3), -6\pi):

∂2f∂x2(ln⁡(3),−6π)=eln⁡(3)sin⁡(−6π)=3⋅0=0\frac{\partial^2 f}{\partial x^2}(\ln(3), -6\pi) = e^{\ln(3)} \sin(-6\pi) = 3 \cdot 0 = 0

Next, the mixed second partial derivative:

∂2f∂x∂y=∂∂y(∂f∂x)=∂∂y(exsin⁡(y))=excos⁡(y)\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial y}(e^x \sin(y)) = e^x \cos(y)

Evaluating at (ln⁡(3),−6π)(\ln(3), -6\pi):

∂2f∂x∂y(ln⁡(3),−6π)=eln⁡(3)cos⁡(−6π)=3⋅1=3\frac{\partial^2 f}{\partial x \partial y}(\ln(3), -6\pi) = e^{\ln(3)} \cos(-6\pi) = 3 \cdot 1 = 3

And finally, the second partial derivative with respect to yy:

∂2f∂y2=∂∂y(∂f∂y)=∂∂y(excos⁡(y))=−exsin⁡(y)\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial}{\partial y}(e^x \cos(y)) = -e^x \sin(y)

Evaluating at (ln⁡(3),−6π)(\ln(3), -6\pi):

∂2f∂y2(ln⁡(3),−6π)=−eln⁡(3)sin⁡(−6π)=−(3⋅0)=0\frac{\partial^2 f}{\partial y^2}(\ln(3), -6\pi) = -e^{\ln(3)} \sin(-6\pi) = -(3 \cdot 0) = 0

We now have all the components needed for the second-order terms: ∂2f∂x2(a,b)=0\frac{\partial^2 f}{\partial x^2}(a, b) = 0, ∂2f∂x∂y(a,b)=3\frac{\partial^2 f}{\partial x \partial y}(a, b) = 3, and ∂2f∂y2(a,b)=0\frac{\partial^2 f}{\partial y^2}(a, b) = 0. These values will help us understand how the function bends and curves around our point of interest.

Step 3: Assembling the Taylor Expansion

Now, let's plug all the calculated values into Taylor's formula up to the second-order terms:

f(x,y)≈f(a,b)+∂f∂x(a,b)(x−a)+∂f∂y(a,b)(y−b)+12(∂2f∂x2(a,b)(x−a)2+2∂2f∂x∂y(a,b)(x−a)(y−b)+∂2f∂y2(a,b)(y−b)2)f(x, y) \approx f(a, b) + \frac{\partial f}{\partial x}(a, b)(x-a) + \frac{\partial f}{\partial y}(a, b)(y-b) + \frac{1}{2} \left( \frac{\partial^2 f}{\partial x^2}(a, b)(x-a)^2 + 2 \frac{\partial^2 f}{\partial x \partial y}(a, b)(x-a)(y-b) + \frac{\partial^2 f}{\partial y^2}(a, b)(y-b)^2 \right)

Substitute the values we found:

f(x,y)≈0+(0)(x−ln⁡(3))+(3)(y−(−6π))+12((0)(x−ln⁡(3))2+2(3)(x−ln⁡(3))(y−(−6π))+(0)(y−(−6π))2)f(x, y) \approx 0 + (0)(x - \ln(3)) + (3)(y - (-6\pi)) + \frac{1}{2} \left( (0)(x - \ln(3))^2 + 2(3)(x - \ln(3))(y - (-6\pi)) + (0)(y - (-6\pi))^2 \right)

Simplifying this expression:

f(x,y)≈3(y+6π)+12(6(x−ln⁡(3))(y+6π))f(x, y) \approx 3(y + 6\pi) + \frac{1}{2} \left( 6(x - \ln(3))(y + 6\pi) \right)

f(x,y)≈3(y+6π)+3(x−ln⁡(3))(y+6π)f(x, y) \approx 3(y + 6\pi) + 3(x - \ln(3))(y + 6\pi)

This is our Taylor expansion up to the second-order terms. This polynomial approximation gives us a good sense of how the function f(x,y)=ex⋅sin⁡(y)f(x, y) = e^x \cdot \sin(y) behaves in the immediate neighborhood of the point (ln⁡(3),−6π)(\ln(3), -6\pi). The linear term 3(y+6π)3(y + 6\pi) indicates the primary direction of increase around the point, while the quadratic term 3(x−ln⁡(3))(y+6π)3(x - \ln(3))(y + 6\pi) refines this approximation by accounting for the interplay between changes in xx and yy and the function's curvature. The fact that the second derivatives with respect to x2x^2 and y2y^2 are zero at this point simplifies the quadratic part significantly, leaving only the cross-derivative term. This means that along the xx and yy axes passing through (ln⁡(3),−6π)(\ln(3), -6\pi), the function behaves linearly to the second order. The significant contribution comes from the interaction term, highlighting how the function's value changes as both xx and yy deviate from the expansion point.

Conclusion: The Power of Local Approximation

We have successfully expanded the function f(x,y)=ex⋅sin⁡(y)f(x, y) = e^x \cdot \sin(y) at the point (ln⁡(3),−6π)(\ln(3), -6\pi) up to the second-order terms using Taylor's formula. The resulting approximation is:

f(x,y)≈3(y+6π)+3(x−ln⁡(3))(y+6π)f(x, y) \approx 3(y + 6\pi) + 3(x - \ln(3))(y + 6\pi)

This approximation is incredibly useful for understanding the local behavior of the function. Near the point (ln⁡(3),−6π)(\ln(3), -6\pi), the complex exponential and sine functions can be reasonably well-represented by this simpler polynomial. This local approximation is the foundation for many numerical methods and theoretical analyses in various scientific disciplines. For instance, in physics, it could be used to linearize a system around an equilibrium point to study its stability. In engineering, it might help in designing control systems by approximating system dynamics. The process of Taylor expansion, while involving careful calculation of derivatives, ultimately provides a powerful lens through which we can view and manipulate complex mathematical objects. It’s a testament to the elegance and utility of calculus in modeling and understanding the world around us. The ability to transform complicated functions into simpler polynomials is a cornerstone of applied mathematics, enabling us to solve problems that would otherwise be intractable.

For further exploration into the concepts of Taylor series and multivariable calculus, you can refer to the following resources: