Lecture 34 Two variable Taylor Series

Text References: Course notes pp. 151-157 & Rogawski N/A

34.1 Recap

Last time, we learned how to use Big-\(\mathcal{O}\) notation to simplify our calculations of Taylor polynomials and limits.

Exercise 34.1 Given \(\sin(x)=x+\mathcal{O}(x^3)\), find \(\displaystyle \lim_{x\to 0}\frac{\sin(x)}{x}\)

Solution. \[\begin{align*} \lim_{x\to 0}\frac{\sin(x)}{x} &= \lim_{x\to 0}\frac{x+\mathcal{O}(x^3)}{x}\\ &= \lim_{x\to 0} 1+\mathcal{O}(x^2)\\ &= 1+0 =1 \end{align*}\]

34.2 Learning Objectives

  • Given a function of two variables, find its \(n\)th order Taylor polynomial.

34.3 Two-Variable Taylor Series

Now that we’ve spent a lot of time with Taylor series, it’s time to move back into multivariable calculus and consider the \(n\)th order Taylor series of a function of two variables.

Consider a function of two variables \(f(x,y)\). We want to approximate the value of \(f\) which will be valid for values of \(x\) and \(y\) near the point \((x_0,y_0)\).

Let’s reduce this problem to a single-variable Taylor polynomial, which we’re already familiar with by fixing a value for \(y\). Now our function becomes a function of \(x\) only, and we can do our usual Taylor series expansion centred at \(x_0\):

\[f(x,y)=f(x_0,y)+f_x(x_0,y)\cdot(x-x_0)+\dfrac{1}{2!}f_{xx}(x_0,y)(x-x_0)^2+\cdots\]

Note that we’re using partial derivatives with respect to \(x\) here since we’re dealing with a multivariate function.

We still have to take the variable \(y\) into account. We’ll do so by expanding each of the terms \(f(x_0,y)\), \(f_x(x_0,y)\), \(f_{xx}(x_0,y)\) etc into their own Taylor series centred at \(y_0\):

  • \(f(x_0,y)=f(x_0,y_0)+f_y(x_0,y_0)\cdot(y-y_0)+\dfrac{1}{2!}f_{yy}(x_0,y_0)\cdot(y-y_0)^2+\cdots\)
  • \(f_x(x_0,y)=f_x(x_0,y_0)+f_{xy}(x_0,y_0)\cdot(y-y_0)+\dfrac{1}{2!}f_{xyy}(x_0,y_0)\cdot(y-y_0)^2+\cdots\)
  • \(f_{xx}(x_0,y)=f_{xx}(x_0,y_0)+f_{xxy}(x_0,y_0)\cdot(y-y_0)+\dfrac{1}{2!}f_{xxyy}(x_0,y_0)\cdot(y-y_0)^2+\cdots\)

As you might imagine, substituting each of these expressions back into our original expansion would be rather daunting. The good news is that we know what we’re after: an \(n\)-th degree approximation.

  • For \(n=1\), we want a linear approximation, so we should collect the constant term and the linear terms.
    • The constant term is \(f(x_0,y_0)\)
    • There are two linear terms: \(f_x(x_0,y_0)(x-x_0)\) and \(f_y(x_0,y_0)(y-y_0)\)
  • For \(n=2\), we want a quadratic approximation, so we should collect the constant term, the linear terms, and the quadratic terms.
    • The constant and linear terms are the same as for \(n=1\)
    • There are three quadratic terms: \(f_{xx}(x_0,y_0)(x-x_0)^2\), \(f_{xy}(x_0,y_0)(x-x_0)(y-y_0)\) (twice), and \(f_{yy}(y-y_0)^2\)

Let’s use these observations to write our Taylor series out: \[\begin{align*} f(x,y)&= f(x_0,y_0) \\ &+ f_x(x_0,y_0)\cdot (x-x_0) + f_y(x_0,y_0)\cdot (y-y_0)\\ &+ \dfrac{1}{2!}\left [f_{xx}(x_0,y_0)\cdot(x-x_0)^2 + 2f_{xy}\cdot(x-x_0)\cdot(y-y_0)+f_{yy}(x_0,y_0)\cdot(y-y_0)^2\right ]\\ &+ \dfrac{1}{3!} [f_{xxx}(x_0,y_0)(x-x_0)^3+3f_{xxy}(x_0,y_0)(x-x_0)^2\cdot(y-y_0)\\ & \quad +3f_{xyy}(x_0,y_0)(x-x_0)\cdot (y-y_0)^2+f_{yyy}(x_0,y_0)(y-y_0)^3 ] & + \cdots \end{align*}\]

Exercise 34.2 Find the values of second order Taylor approximation to the function \(f(x,y)=\sin(x^2y)\) centred at \(\left(1,\frac{\pi}{2}\right )\).

Solution. We have \[\begin{align*} f(x,y) = \sin(x^2y) \quad & \Rightarrow \quad f\left(1,\frac{\pi}{2}\right )=1\\ f_x(x,y) = 2xy\cos(x^2y) \quad & \Rightarrow \quad f_x\left(1,\frac{\pi}{2}\right )=0 \\ f_y(x,y) = x^2\cos(x^2y) \quad & \Rightarrow \quad f_y\left(1,\frac{\pi}{2}\right )=0 \\ f_{xx}(x,y) = 2y\cos(x^2y)-4x^2y^2\sin(x^2y) \quad & \Rightarrow \quad f_{xx}\left(1,\frac{\pi}{2}\right )= -\pi^2\\ f_{xy}(x,y) = 2x\cos(x^2y)-2x^3y\sin(x^2y) \quad & \Rightarrow \quad f_{xy}\left(1,\frac{\pi}{2}\right )= -\frac{\pi^2}{2}\\ f_{yy}(x,y) = -x^4\sin(x^2y) \quad & \Rightarrow \quad f_{yy}\left(1,\frac{\pi}{2}\right )= -1 \end{align*}\]

Substituting the values into our formula, we get \[\begin{align*} \sin(x^2y) & \approx 1+ 0\cdot(x-1)+0\cdot \left(y-\frac{\pi}{2}\right)\\ & \quad +\frac{1}{2!}\left[-\pi^2(x-1)^2+2\left(-\frac{\pi^2}{2}\right)(x-1)\left(y-\frac{\pi}{2}\right) - \left(y-\frac{\pi}{2}\right)^2\right]\\ &= 1 -\frac{\pi^2(x-1)^2}{2}-\frac{\pi^2}{2}(x-1)\left(y-\frac{\pi}{2}\right) -\frac{1}{2}\left(y-\frac{\pi}{2}\right)^2 \end{align*}\]