Issues with linear and quadratic multivariate Taylor series

**AmberLamps** · February 21st 2010, 02:18 PM

Hey guys

I am having a few issues with my linear algebra course. Me and my friends are in a study group and we are working on this problem for one of our seminars:

3. Find the linear and quadratic approximations to the funtion f(x,y)=e^(ax-by) around the point (0,0)

We think we have found the linear approximation of this as:

1 + xae^(ax-by) - ybe^(ax-by)

Is this correct? How do we get the quadratic approximation?

Hope you can help guys, it would be much appreciated.

**TheEmptySet** · February 21st 2010, 02:51 PM

Originally Posted by AmberLamps

Hey guys

I am having a few issues with my linear algebra course. Me and my friends are in a study group and we are working on this problem for one of our seminars:

3. Find the linear and quadratic approximations to the funtion f(x,y)=e^(ax-by) around the point (0,0)

We think we have found the linear approximation of this as:

1 + xae^(ax-by) - ybe^(ax-by)

Is this correct? How do we get the quadratic approximation?

Hope you can help guys, it would be much appreciated.

The taylor series for
$g(x)=e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}$

This is valid for any real number. So define

$f(x,y)=g(ax-by)=\sum_{n=0}^{\infty}\frac{(ax-by)^n}{n!}$

Now just crank out as many terms as you need.

So when $n=1$ you get the linear approx is

$L(x,y)=\frac{(ax-by)^0}{0!}+\frac{(ax-by)^1}{1!}=1+ax-by$

**HallsofIvy** · February 22nd 2010, 02:40 AM

Originally Posted by AmberLamps

Hey guys

I am having a few issues with my linear algebra course. Me and my friends are in a study group and we are working on this problem for one of our seminars:

3. Find the linear and quadratic approximations to the funtion f(x,y)=e^(ax-by) around the point (0,0)

We think we have found the linear approximation of this as:

1 + xae^(ax-by) - ybe^(ax-by)

Is this correct? How do we get the quadratic approximation?

Hope you can help guys, it would be much appreciated.

Close- You have to take the value of the derivatives at (0,0):
since e^(a0- b0)= 1, the linear approximation is 1+ ax- by.

The quadratic approximation will be have the second derivatives and second powers added:

1+ ax- by+ $\frac{\partial^2 f}{\partial x^2}(0,0)x^2+ \frac{\partial^2 f}{\partial x\partial y}(0,0) xy+ \frac{\partial^2 f}{\partial y^2}(0,0) y^2$

**AmberLamps** · February 22nd 2010, 12:24 PM

Thanks for the help guys! it is much appreciated. So following that for the quadratic you get the formula below when evaluated at (0,0), correct?

$1+ ax - by + a^2x^2 + b^2y^2 - 2baxy$

The reason I ask is because the way this material was presented in lectures was in vector/matrices form and following that method my answer is subtly different due to as it becomes

$1/2 [1+ ax - by + a^2x^2 + b^2y^2 - 2baxy]$

due to a $1/2!$ term.

**HallsofIvy** · February 23rd 2010, 05:10 AM

You may have misunderstood that "1/2". If you have the quadratic form $ax^2+ bxy+ cy^2$ , you can write that as the matrix multiplication
$\begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}a & \frac{1}{2}b \\ \frac{1}{2}b & c \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}$ .

Note that the "1/2" is only in the off diagonal terms. It is there so that when you multiply that out and add the two "xy" terms, it sums to bxy.

**AmberLamps** · February 23rd 2010, 12:59 PM

yeh apologies I meant to change my answer to:

$1+ax + by + 1/2a^2x^2+1/2b^2x^2+abxy$

So I understand how it comes to bxy in your example. But do not understand why the 1/2 is only in the off diagonal terms; because all the examples I am able to find online seem to have 1/2 for all terms? Thanks again for your help.

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