Taylor expanding $f(y+\epsilon U1 + {\epsilon}^{2} U2,t,\epsilon)$ in $\epsilon$

Question

How would one Taylor expand $\epsilon f(y+\epsilon U1 + {\epsilon}^{2} U2,t,\epsilon)$ in $\epsilon$? Somehow the professor obtained the first few terms to be: $\epsilon f(y+\epsilon U1 + {\epsilon}^{2} U2,t,\epsilon)$ $\approx$ $\epsilon$ $f(y,t,0)$ + ${\epsilon}^{2} f_{y}(y)U1 + {\epsilon}^{2} f_{\epsilon}(y,t,0)$?

Answer 1

The Taylor expansion of a function $g(\epsilon)$ around $\epsilon = 0$ to second order is given by \begin{equation} g(\epsilon) \approx g(0) + \epsilon \, \frac{\text{d} g}{\text{d} \epsilon}(0) + \frac{\epsilon^2}{2}\,\frac{\text{d}^2 g}{\text{d} \epsilon^2}(0). \tag{1} \end{equation} The problem here is that the function $g$ depends in a rather complicated way on $\epsilon$, namely \begin{equation} g(\epsilon) = \epsilon\,f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon). \tag{2} \end{equation} The first term in $(1)$ is rather easy to calculate: we see that $g(0) = 0 \cdot f(y,t,0) = 0$. Now, for the second term in $(1)$, we need to calculate the derivative of $g$ to $\epsilon$. Using $(2)$, we see that \begin{equation} \frac{\text{d} g}{\text{d} \epsilon} = f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon) + \epsilon \frac{\text{d}}{\text{d}\epsilon} \Big[f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon)\Big] \tag{3} \end{equation} by the product rule. The derivative of $f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon)$ to $\epsilon$ is still quite complicated. For the moment, let's denote it as \begin{equation} \frac{\text{d}}{\text{d}\epsilon} \Big[f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon)\Big] =: h(\epsilon).\tag{4} \end{equation} Luckily, we don't need to calculate it at this stage: whatever it is, we see that \begin{equation} \frac{\text{d} g}{\text{d} \epsilon}(0) = f(y,t,0) + 0\cdot h(0) = f(y,t,0). \end{equation}

For the third term in $(1)$ though we do have to get our hands a bit dirty. We need to calculate the second derivative of $g$ to $\epsilon$. This can be done by taking the $\epsilon$-derivative of $(3)$: \begin{align} \frac{\text{d}^2 g}{\text{d} \epsilon^2} = \frac{\text{d}}{\text{d} \epsilon} \Big[\frac{\text{d} g}{\text{d} \epsilon}\Big] &= \frac{\text{d}}{\text{d} \epsilon} \Big[f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon) + \epsilon\, h(\epsilon)\Big] \\ &= \frac{\text{d}}{\text{d} \epsilon} \Big[f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon)\Big] + h(\epsilon) + \epsilon \frac{\text{d} h}{\text{d}\epsilon}\\ &= 2\, h(\epsilon) + \epsilon \frac{\text{d} h}{\text{d}\epsilon}. \tag{5} \end{align} When evaluated at $\epsilon = 0$, we obtain \begin{equation} \frac{\text{d}^2 g}{\text{d} \epsilon^2}(0) = 2\,h(0) + 0\cdot \frac{\text{d} h}{\text{d} \epsilon}(0) = 2\,h(0). \end{equation}

Now, what is $h(\epsilon)$? For this, we need the multidimensional chain rule. Namely, we have function of three variables: $f(x_1,x_2,x_3)$. In addition, in the place of these variables, we substitute functions of $\epsilon$. Denoting \begin{align} z_1(\epsilon) &= y + \epsilon U_1 + \epsilon^2 U_2,\\ z_2(\epsilon) &= t,\\ z_3(\epsilon) &= \epsilon, \end{align} we see that \begin{equation} f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon) = f\big(z_1(\epsilon),z_2(\epsilon),z_3(\epsilon)\big). \end{equation} Now we can use the multidimensional chain rule to calculate $h$: we see that \begin{equation} h(\epsilon) = \frac{\text{d}}{\text{d}\epsilon} \Big[ f\big(z_1(\epsilon),z_2(\epsilon),z_3(\epsilon)\big)\Big] = \frac{\text{d} z_1}{\text{d}\epsilon}\cdot f_{x_1}(z_1(\epsilon),z_2(\epsilon),z_3(\epsilon)) + \frac{\text{d} z_2}{\text{d}\epsilon}\cdot f_{x_2}(z_1(\epsilon),z_2(\epsilon),z_3(\epsilon)) + \frac{\text{d} z_3}{\text{d}\epsilon}\cdot f_{x_3}(z_1(\epsilon),z_2(\epsilon),z_3(\epsilon)), \end{equation} where $f_{x_1} = \frac{\partial f}{\partial x_1}$ is the partial derivative of $f$ to its first variable, $x_1$. In our case, we have \begin{align} \frac{\text{d}z_1}{\text{d}\epsilon} &= U_1 + 2\epsilon U_2,\\ \frac{\text{d}z_2}{\text{d}\epsilon} &= 0,\\ \frac{\text{d}z_3}{\text{d}\epsilon} &= 1, \end{align} so \begin{equation} h(\epsilon) = \left(U_1 + 2 \epsilon U_2\right) f_{x_1}(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon) + f_{x_3}(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon), \end{equation} hence \begin{equation} h(0) = U_1 f_{x_1}(y,t,0) + f_{x_3}(y,t,0). \end{equation} Combining the above expressions, we see that \begin{equation} \epsilon\,f(y+\epsilon U_1 + \epsilon^2 U_2,t,\epsilon) \approx 0 + \epsilon\,f(y,t,0) + \epsilon^2 \Big(U_1 f_{x_1}(y,t,0) + f_{x_3}(y,t,0)\Big). \end{equation}