Math Problem Statement

Expand sin(x+h)(y+k) by Taylor's theorem as far as terms of second degree.

Solution

To expand the function f(x,y)=sin((x+h)(y+k))f(x, y) = \sin((x+h)(y+k)) using Taylor's theorem up to the second degree, we proceed as follows:

Step 1: Taylor Series Expansion

The Taylor series expansion of a function f(x,y)f(x, y) around the point (x0,y0)(x_0, y_0) up to the second degree is given by:

f(x+h,y+k)f(x0,y0)+hfx(x0,y0)+kfy(x0,y0)+h22fxx(x0,y0)+k22fyy(x0,y0)+hkfxy(x0,y0)f(x+h, y+k) \approx f(x_0, y_0) + h f_x(x_0, y_0) + k f_y(x_0, y_0) + \frac{h^2}{2} f_{xx}(x_0, y_0) + \frac{k^2}{2} f_{yy}(x_0, y_0) + h k f_{xy}(x_0, y_0)

where:

  • fx(x0,y0)f_x(x_0, y_0) is the first partial derivative of ff with respect to xx at (x0,y0)(x_0, y_0),
  • fy(x0,y0)f_y(x_0, y_0) is the first partial derivative of ff with respect to yy at (x0,y0)(x_0, y_0),
  • fxx(x0,y0)f_{xx}(x_0, y_0) is the second partial derivative of ff with respect to xx,
  • fyy(x0,y0)f_{yy}(x_0, y_0) is the second partial derivative of ff with respect to yy,
  • fxy(x0,y0)f_{xy}(x_0, y_0) is the mixed partial derivative.

Step 2: Compute the Partial Derivatives

Given f(x,y)=sin(xy)f(x, y) = \sin(xy), let's calculate the partial derivatives up to the second order:

  1. First Partial Derivatives: fx(x,y)=xsin(xy)=ycos(xy)f_x(x, y) = \frac{\partial}{\partial x} \sin(xy) = y \cos(xy) fy(x,y)=ysin(xy)=xcos(xy)f_y(x, y) = \frac{\partial}{\partial y} \sin(xy) = x \cos(xy)

  2. Second Partial Derivatives: fxx(x,y)=x(ycos(xy))=y2sin(xy)f_{xx}(x, y) = \frac{\partial}{\partial x}(y \cos(xy)) = -y^2 \sin(xy) fyy(x,y)=y(xcos(xy))=x2sin(xy)f_{yy}(x, y) = \frac{\partial}{\partial y}(x \cos(xy)) = -x^2 \sin(xy) fxy(x,y)=y(ycos(xy))=cos(xy)xysin(xy)f_{xy}(x, y) = \frac{\partial}{\partial y}(y \cos(xy)) = \cos(xy) - xy \sin(xy)

Step 3: Taylor Expansion at (x0,y0)=(x,y)(x_0, y_0) = (x, y)

Substitute these into the Taylor expansion formula:

\sin((x+h)(y+k)) &\approx \sin(xy) + h y \cos(xy) + k x \cos(xy) \\ &\quad + \frac{h^2}{2} (-y^2 \sin(xy)) + \frac{k^2}{2} (-x^2 \sin(xy)) \\ &\quad + h k (\cos(xy) - xy \sin(xy)). \end{aligned}$$ Simplifying the expression: $$\begin{aligned} \sin((x+h)(y+k)) &\approx \sin(xy) + (hy + kx)\cos(xy) \\ &\quad - \frac{h^2 y^2 + k^2 x^2}{2} \sin(xy) + hk (\cos(xy) - xy \sin(xy)). \end{aligned}$$ This is the expansion of $$\sin((x+h)(y+k))$$ up to the second degree in $$h$$ and $$k$$. --- Would you like any further details or explanations on this expansion? Here are some related questions: 1. How does Taylor's theorem apply to multivariable functions? 2. What are the conditions for a Taylor series expansion to be valid? 3. How do higher-order terms in the expansion affect the accuracy of the approximation? 4. What is the geometric interpretation of the first and second derivatives in this context? 5. How would the expansion change if we considered terms up to the third degree? **Tip:** Taylor series expansions are powerful tools for approximating functions near a point, making them especially useful in numerical analysis and approximation methods.

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Math Problem Analysis

Mathematical Concepts

Taylor Series
Multivariable Calculus
Partial Derivatives

Formulas

Taylor series expansion formula

Theorems

Taylor's theorem

Suitable Grade Level

Advanced Undergraduate