# Second Order and Beyond for Multivariable Taylor Series

I am looking for reasons why an engineer might want to learn about multivariable Taylor Series beyond order one.

I have no problem seeing the value of single variable Taylor Series.

I have quite a few uses for the first order multivariable Taylor Series:

$$f(x_1,\dots,x_n)\underset{x_i\approx a_i}{\approx} f(a_1,\dots,a_n)+\sum_{i=1}^nf_{x_i}(a_1,\dots,a_n)(x-a_i).$$

I have one application of two-variable order-$n$ Taylor series: I think the derivations of higher-order Runge-Kutta formulae require it.

One of the nice applications of single variable Taylor Series to engineering is the simplification of formulae. For example, in the context of beam deflection, the exact but rather cumbersome

$$\delta=e\cdot \left(\sec\left(\sqrt{\frac{P}{EI}}\cdot \frac{L}{2}-1 \right)\right),$$ is well-approximated (using the Maclaurin series of $\sec x$), under reasonable assumptions, by the 'neat'

$$\delta\approx \frac{ePL^2}{8EI}.$$

However, I can't envision similarly nice examples using two-variable Taylor Series.

So my question is, why do engineers learn about order $n$ ($n\geq 2$) multivariable Taylor series?

I am hoping to end up with a nice exam question for my class that isn't as boring as "find the second order Taylor Series expansion of (some generic) $z=f(x,y)$ about $(a,b)$".

• To investigate local maxima and minima, for example... – Hans Lundmark Sep 15 '17 at 15:52
• @Hans do you mean once you find a critical point, go off and calculate the second order expansion to see what is going on with the second derivatives? Or something else. – JP McCarthy Sep 15 '17 at 18:37
• Yes, that's what I meant. – Hans Lundmark Sep 15 '17 at 20:34
• They can also use it to build almost every nonlinear optimization numerical method. – Koto Sep 22 '17 at 15:14
• Sure, for example, if you want to minimize $f(x)$, then using the Taylor expansion until second order, you can prove that the Newton direction $d$ such that $H_f (x) d=-\nabla f(x)$ is a descent direction from $x$ when the hessian $H_f(x)$ is positive definite at $x$, that is, $f(x+\alpha d)<f(x)$ for some $\alpha>0$. There are also tons of variations of this method, some of them approximate the inverse of the hessian for computational performance, like the BFGS. Ops, I haven't read before that this is exactly what Han de Brujin wrote, but using a different notation. – Koto Sep 22 '17 at 19:38

Apart from the obvious reason, that knowledge is a value in itself, there are indeed practical engineering application of multivariate calculus about order $n \geq 2$.

One example is the study of stability, which clearly benefits from Taylor expansions up to second order.

There are examples from computational methods too: if the engineer is using a Finite Element code and at some point along an analysis gets a "non positive definite Hessian", he would certainly benefit from knowing about second-order expansions to understand what is going on.

Another fitting example in my view is given by Laplace's method (alternatively, saddle-point) to approximate the solution of certain integrals, very common in Statistical Mechanics and not rare at all in Engineering. The method mandatorily requires a second order expansion.

Talking about thermodynamics, how would one prove the principle of minimum energy to an engineer not accustomed to second order expansions? With multivariate Taylor expansions, it takes just a moment, by expanding the entropy $S$ as function of an internal parameter $X$ and energy $U$ to second order, as $\frac{\partial S}{\partial X} = 0$ $$\mathrm{d}S \approx \frac{\partial^2 S}{\partial X^2}\mathrm{d}X^2 +\frac{\partial S}{\partial U} \mathrm{d} U$$ and the minimum nature of $U$ is on display.

Moreover, referring to elasticity as in your post, the elastic moduli are related to the strain energy function $W$ via $$\mathcal{C} = \frac{\partial^2 W }{\partial E^2}$$ where $E$ stands for a deformation tensor, so a second-order expansion of the strain energy is needed.

In $1$ D the "first-order only" engineer could avoid the problem by looking at stress-strain curves only. Indeed, in $1D$ $$\sigma = \frac{\mathrm{d} W}{\mathrm{d} \epsilon}$$ where $\epsilon$ stands for the strain. As for small enough strains $\sigma = E \epsilon$, in this case only first-order condsiderations are needed to estimate the elastic moduli (in fact, by letting the experiment perform a differentiation).

But in a more general case, dealing for example with a biaxial test (and this is not an academic example for modern engineers busy with polymers), it can often be easier to start from the energy $W$, perform a multivariate expansion up to second order, and get the elastic moduli matrix out of it. The difference stems from practical reasons: in a $1D$ test one stretches in one direction, and what happens in the other directions is easy to characterise. In a biaxial test this is far more complex: it is difficult to impose one strain value along a given direction, while the strain along the perpendicular direction is varied. In other words, it is difficult to "perform partial derivatives" experimentally.

All this could be especially useful, and releavant, when a component is loaded by a constant stress state, on top of which small oscillations are superimposed. It is then useful to re-define the constant stress-state as the "unloaded" state, and calculate "fictitious" elastic moduli.

• Could you expand a little further on the elasticity point? Could it actually involve calculating a second order expansion with two variables? – JP McCarthy Sep 19 '17 at 7:57
• You have put so much effort into this answer, I do appreciate it. – JP McCarthy Sep 21 '17 at 10:11
• Vibrations in plates could be an example. – G Cab Sep 21 '17 at 18:44
• @GCab care to elaborate? – JP McCarthy Sep 25 '17 at 10:53

Least Squares Best Fit Methods are well known for a point, a straight line, a circle, a conic section, in general: for a simple figure. But that's not really cool. I'd rather have some least squares best fit method with complicated figures, for example: two points, crossing or parallel lines, a bunch of circles, the image of a character 'A' - my ultimate dream being the least squares best fit with music score elements, a treble clef for example :-)
So far so good about dreams. But one has to start somewhere ..
Let's consider the most simple case (I think) in two dimensions: a least squares best fit with two points instead of one. Writing the minimum principle for this case is not too difficult.
Let the function M to be minimized be defined by: $$M(a,b,p,q) = \sum_k \left[ (x_k - a)^2 + (y_k - b)^2 \right].\left[ (x_k - p)^2 + (y_k - q)^2 \right]$$ Here the $(x_k,y_k)$ are the fixed points of a cloud in the plane and we are satisfied if we can find some fairly unique points $(a,b)$ and $(p,q)$ .
A procedure to accomplish this is rather straigtforward. Differentiate the above sum to $\,(a,b,p,q)\,$ and put the four outcomes of these partial differentiations equal to zero: $$F_a = \frac{\partial M(a,b,p,q)}{\partial a} = - 2 \sum_k (x_k - a)\left[ (x_k - p)^2 + (y_k - q)^2 \right] = 0\\ F_b = \frac{\partial M(a,b,p,q)}{\partial b} = - 2 \sum_k (x_k - b)\left[ (x_k - p)^2 + (y_k - q)^2 \right] = 0\\ F_p = \frac{\partial M(a,b,p,q)}{\partial p} = - 2 \sum_k (x_k - p)\left[ (x_k - a)^2 + (y_k - b)^2 \right] = 0\\ F_q = \frac{\partial M(a,b,p,q)}{\partial q} = - 2 \sum_k (x_k - q)\left[ (x_k - a)^2 + (y_k - b)^2 \right] = 0$$ The next step is to employ a four dimensional Newton-Raphson method to find the zeroes.
The one-dimensional equivalent is well known: $$x_{n+1} = x_n - \frac{F(x_n)}{F'(x_n)} \quad \Longleftrightarrow \quad F'(x_n)(x_{n+1}-x_n) = - F(x_n)$$ It generalizes as follows ( with $\Delta x = x_{n+1}-x_n$ ) and mind the symmetry: $$\large \begin{bmatrix} \frac{\partial^2 M}{\partial a^2} & \frac{\partial^2 M}{\partial a \partial b} & \frac{\partial^2 M}{\partial a \partial p} & \frac{\partial^2 M}{\partial a \partial q} \\ \frac{\partial^2 M}{\partial a \partial b} & \frac{\partial^2 M}{\partial b^2} & \frac{\partial^2 M}{\partial b \partial p} & \frac{\partial^2 M}{\partial b \partial q} \\ \frac{\partial^2 M}{\partial a \partial p} & \frac{\partial^2 M}{\partial b \partial p} & \frac{\partial^2 M}{\partial p^2} & \frac{\partial^2 M}{\partial p \partial q} \\ \frac{\partial^2 M}{\partial a \partial q} & \frac{\partial^2 M}{\partial b \partial q} & \frac{\partial^2 M}{\partial p \partial q} & \frac{\partial^2 M}{\partial q^2} \end{bmatrix}_n \begin{bmatrix} \Delta a \\ \Delta b \\ \Delta p \\ \Delta q \end{bmatrix} = - \begin{bmatrix} F_a \\ F_b \\ F_p \\ F_q \end{bmatrix}_n$$ The matrix on the left is known as the Hessian matrix, commonly denoted as $\,H$ . We need the inverse of it: $$H_n\,\Delta x = - F_n \quad \Longrightarrow \quad x_{n+1} = x_n - H_n^{-1}F_n$$ So here are the OP's second order multivariable derivatives: $$\frac{\partial^2 M}{\partial a^2} = \frac{\partial^2 M}{\partial b^2} = 2 \sum_k \left[ (x_k - p)^2 + (y_k - q)^2 \right] \\ \frac{\partial^2 M}{\partial p^2} = \frac{\partial^2 M}{\partial q^2} = 2 \sum_k \left[ (x_k - a)^2 + (y_k - b)^2 \right] \\ \frac{\partial^2 M}{\partial a \partial p} = 4 \sum_k (x_k - a)(x_k - p) \qquad \frac{\partial^2 M}{\partial a \partial q} = 4 \sum_k (x_k - a)(x_k - q) \\ \frac{\partial^2 M}{\partial b \partial p} = 4 \sum_k (x_k - b)(x_k - p) \qquad \frac{\partial^2 M}{\partial b \partial q} = 4 \sum_k (x_k - b)(x_k - q) \\ \frac{\partial^2 M}{\partial a \partial b} = \frac{\partial^2 M}{\partial p \partial q} = 0$$ Free software source implementing all this numerically and graphically is found at my website .
It turns out that, given some neatly chosen conditions, iterations converge rather quickly to a result.
Here is an animation of $10$ iterations as generated by our software. The $\color{red}{\mbox{red dots}}$ correspond with the points $(a,b)$ and $(p,q)$ that are approaching to a best fit with the two squares of black dots.

Numerically:

( 0.030000000, 0.030000000)  ;  ( 3.970000000, 3.970000000)
( 0.660367766, 0.655184806)  ;  ( 3.346246442, 3.338009121)
( 1.067931194, 1.059662172)  ;  ( 2.947311493, 2.933165294)
( 1.224322703, 1.215456066)  ;  ( 2.610372656, 2.591778103)
( 1.483524248, 1.473141085)  ;  ( 2.587165222, 2.567701858)
( 1.492857336, 1.482684318)  ;  ( 2.507701174, 2.488004719)
( 1.500161598, 1.489958755)  ;  ( 2.505384218, 2.485704103)
( 1.500206864, 1.490003939)  ;  ( 2.505303084, 2.485623596)
( 1.500206877, 1.490003953)  ;  ( 2.505303074, 2.485623586)
( 1.500206877, 1.490003953)  ;  ( 2.505303074, 2.485623586)


• A great help so far considering I teach this class least squares also this semester! Thank you. – JP McCarthy Sep 21 '17 at 13:59

The result of an observation of a function $\,f(x,y,z)\,$ with Gaussian blur $\,G(x,y,z)\,$ for example is in general a convolution integral: $$\overline{f} = \iiint G(\xi,\eta,\zeta)\,f(x-\xi,y-\eta,z-\zeta)\,d\xi\,d\eta\,d\zeta$$ Here $\,G\,$ is the kernel function of the blur and $\,f\,$ is the function to be observed. The function $\,f\,$ is developed into a Taylor series around $\,(\xi,\eta,\zeta)=(0,0,0)$ : $$f(x-\xi,y-\eta,z-\zeta) \; \approx \; f(x,y,z)$$ $$- \xi \frac{\partial f}{\partial x} - \eta \frac{\partial f}{\partial y} - \zeta \frac{\partial f}{\partial z}$$ $$+ \frac{1}{2} \xi^2 \frac{\partial^2 f}{\partial x^2} + \frac{1}{2} \eta^2 \frac{\partial^2 f}{\partial y^2} + \frac{1}{2} \zeta^2 \frac{\partial^2 f}{\partial z^2}$$ $$+ \xi \eta \frac{\partial f}{\partial x \partial y} + \eta \zeta \frac{\partial f}{\partial y \partial z} + \zeta \xi \frac{\partial f}{\partial z \partial x}$$ In this expression the partial differential quotients are no longer dependent on $\,(\xi,\eta,\zeta)$ , because they are calculated for $\,(\xi,\eta,\zeta)=(0,0,0)$ . Therefore the convolution integral is approximately equal to: $$f(x,y,z) \iiint \, G \, d\xi\,d\eta\,d\zeta \\ \, - \, \frac{\partial f}{\partial x} \iiint \xi \, G \, d\xi\,d\eta\,d\zeta \, - \, \frac{\partial f}{\partial y} \iiint \eta \, G \, d\xi\,d\eta\,d\zeta \, - \, \frac{\partial f}{\partial z} \iiint \zeta \, G \, d\xi\,d\eta\,d\zeta \\ + \frac{1}{2} \frac{\partial^2 f}{\partial x^2}\iiint \xi^2 \, G \, d\xi\,d\eta\,\zeta + \frac{1}{2} \frac{\partial^2 f}{\partial y^2}\iiint \eta^2 \, G \, d\xi\,d\eta\,d\zeta + \frac{1}{2} \frac{\partial^2 f}{\partial z^2}\iiint \zeta^2\, G \, d\xi\,d\eta\,d\zeta \\ + \frac{\partial f}{\partial x \partial y}\iiint \xi \eta \, G \, d\xi\,d\eta\,d\zeta + \frac{\partial f}{\partial y \partial z}\iiint \eta \zeta \, G \, d\xi\,d\eta\,d\zeta + \frac{\partial g}{\partial z \partial x}\iiint \zeta \xi \, G \, d\xi\,d\eta\,d\zeta$$ The first integral is by definition equal to $\,1$ . The next three integrals are equal to the expectation value of the Gaussian kernel function, and therefore equal to $\,0$ . The next three integrals are equal to the spreads of the Gaussian kernel function in the different coordinate directions. The last three integrals, at last, are zero. Thus: $$\overline{f} \approx f + \frac{1}{2} \sigma_x^2 \frac{\partial^2 f}{\partial x^2} + \frac{1}{2} \sigma_y^2 \frac{\partial^2 f}{\partial y^2} + \frac{1}{2} \sigma_z^2 \frac{\partial^2 f}{\partial z^2} = f + \frac{1}{2} \sigma^2 \nabla^2 f$$ An asymptotic approximation for large distances, which can also be found in any decent book about Statistics, where the last equality is for the isotropic case only.

One possible application is the renormalization of singularities, for example the Electric field of a point charge: $$E(r) = \frac{q}{4\pi\epsilon_0 r^2} \quad \mbox{with} \quad \begin{cases} q = \mbox{(point) charge} \\ \epsilon_0 = \mbox{dielectric constant} \\ r = \mbox{distance to point} \end{cases}$$ The far away field can be calculated without even knowing about the convolution integral: $$\overline{E} \approx E + \frac{1}{2} \sigma^2 \nabla^2 E \qquad \mbox{with} \qquad \nabla^2 \frac{1}{r^2} = \frac{1}{r} \frac{\partial^2}{\partial r^2} r \frac{1}{r^2} = \frac{2}{r^4}$$ Hence: $$\overline{E} \approx E \left( 1 + \frac{\sigma^2}{r^2} \right) \quad \mbox{for} \quad r \gg \sigma$$ The Electric field with Gaussian blur is: $$\overline{E}(x,y,z) = \left( \frac{1}{ \sigma \sqrt{2\pi} } \right)^3 \iiint \frac{q}{4 \pi \epsilon_0 ( \xi^2 + \eta^2 + \zeta^2 ) } \, e^{\, - [ (\xi - x)^2 + (\eta - y)^2 + (\zeta - z)^2 ] / 2\sigma^2 } \; d\xi\,d\eta\,d\zeta$$ A special case is the value of the renormalized field at the origin, which turns out to be nicely finite : $$\overline{E}(r=0) = \left( \frac{1}{ \sigma \sqrt{2\pi} } \right)^3 \frac{q}{\epsilon_0} \int_0^\infty e^{- r^2 / 2 \sigma^2 } d r = \left( \frac{1}{ \sigma \sqrt{2\pi} } \right)^3 \frac{q}{\epsilon_0} \frac{1}{2} \sigma \sqrt{2\pi} = \frac{q}{4 \pi \epsilon_0 \sigma^2 } = E(\sigma)$$ Anything in between is somewhat more difficult ..

• Thanks very much for this. – JP McCarthy Sep 25 '17 at 11:08