Ytukyg

I'm asked to find the second order taylor series expansion of $cos(x+2y)e^{2x}$ around $(2, -1)$. It's from an old exam of a professor who doesn't give solutions to them, so I don't know if my results are correct. So I woud like that you tell me if my results are correct, and if not, where I made a mistake. Thank you.

As first partial derivatives, I get $$f_x=-sin(x+2y)e^{2x}+2cos(x+2y)e^{2x}$$ $$f_y=-2sin(x+2y)e^{2x}$$ As second partial derivatives, I get
$$begin{align}
f_{xx}&=-cos(x+2y)e^{2x}-2sin(x+2y)e^{2x}-2sin(x+2y)e^{2x}+4cos(x+2y)e^{2x}\
&=3cos(x+2y)e^{2x}-4sin(x+2y)e^{2x}
end{align}$$
$$f_{yy}=-4cos(x+2y)e^{2x}$$ $$f_{xy}=f_{yx}=-2cos(x+2y)e^{2x}-4sin(x+2y)e^{2x}$$

So now, if I plug everything into the formula, I get
$$e^{4}+2e^{4}(x-2)+frac{1}{2!}(3e^{4}(x-2)^2-4e^{4}(x-2)(y+1)-4e^{4}(y+1)^2)=e^{4}(1+2(x-2)+frac{1}{2!}(3(x-2)^2-4(x-2)(y+1)-4(y+1)^2))$$

Are my results correct ? I just proceeded as usual, using the common two-variables taylor series expansion. Thanks for your help !

Yes, it is correct. An easy way to make sure is just to plot things: blue is the original function, red is your expansion

And this is the code to draw it

from mpl_toolkits.mplot3d import Axes3D

import matplotlib.pyplot as plt

from matplotlib import cm

from matplotlib.ticker import LinearLocator, FormatStrFormatter

import numpy as np



fig = plt.figure()

ax = fig.gca(projection = '3d')

ax.set_xlabel('x')

ax.set_ylabel('y')

ax.set_zlabel('z')



def f(X, Y):

    return np.cos(X + 2 * Y) * np.exp(2 * X)



def taylor(x, y):

    return np.exp(4) * (1 + 2 * (x - 2) + 0.5 * (3 * (x - 2)**2 - 4 * (x - 2) * (y + 1) - 4 * (y + 1)**2))



# Make data.

X = np.arange(1, 3, 0.25)

Y = np.arange(-2, 0, 0.25)

X, Y = np.meshgrid(X, Y)

Z = f(X, Y)

W = taylor(X, Y)



# Plot the surface.

surf = ax.plot_surface(X, Y, Z, cmap = cm.autumn, linewidth = 0, antialiased = False, alpha = 0.5)

sur1 = ax.plot_surface(X, Y, W, cmap = cm.winter, linewidth = 0, antialiased = False, alpha = 0.7)



# Add a color bar which maps values to colors.

ax.xaxis.set_major_locator(LinearLocator(4))

ax.yaxis.set_major_locator(LinearLocator(4))

ax.zaxis.set_major_locator(LinearLocator(4))



plt.show()

It is correct. You can also obtain the expansion by using the Taylor series of $sin, cos,exp$:

begin{align}
cos(x+2y)e^{2x} &= cos((x-2) + 2(y+1))e^{4+2(x-2)}\
&= e^{4}big[cos(x-2)cos(2(y+1)) - sin(x-2)sin(2(y+1))big]e^{2(x+2)}\
&= e^{4}left[left(sum_{n=0}^infty frac{(-1)^n}{(2n)!}(x-2)^{2n}right)left(sum_{n=0}^infty frac{(-1)^n}{(2n)!}2^{2n}(y+1)^{2n}right)right.\
&- left.left(sum_{n=0}^infty frac{(-1)^n}{(2n+1)!}(x-2)^{2n+1}right)left(sum_{n=0}^infty frac{(-1)^n}{(2n+1)!}2^{2n+1}(y+1)^{2n+1}right)right]left(sum_{n=0}^infty frac{2^n}{n!}(x-2)^{n}right)\
&= e^{4}left[1+2(x-2) + left(frac{(-1)^1}{2!} - frac{(-1)^1}{1!}frac{2^1}{1!}right)(x-2)^2 - frac{(-1)^0}{0!}2^1cdot frac{2}{2!}(x-2)(y+1) + frac{(-1)^1}{2!}2^2(y+2)^2 + cdotsright]\
&= e^4left[1+2(x-2)+frac32 e^{4}(x-2)^2-2e^{4}(x-2)(y+1)-2e^{4}(y+1)^2 + cdotsright]
end{align}