# What is a Jacobian? [duplicate]

I understand how to calculate the Jacobian for any function. What I don't get is, what does it actually achieve? what does it mean? Are there any real life examples where we use a Jacobian?

## marked as duplicate by Hans Lundmark, Sahiba Arora, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 2 '17 at 10:52

• The Jacobian can also be thought of as describing the amount of "stretching", "rotating" or "transforming" that a transformation imposes locally – AspiringMat Aug 2 '17 at 0:26
• ok. but can anyone cite an actual example of what this means in practice? – Nirvana Guha Aug 2 '17 at 0:30
• Jacobians are used in any double or tripple integration where a change of corrodinate system might produce an easier form to evaluate. Cartesian to Polar coordinate transformation for examples, or vice versa. Evaluating surface integrals. And so forth. – Graham Kemp Aug 2 '17 at 1:04
• The Jacobian matrix should really just be called the "derivative" of a function $f:\mathbb R^n \to R^m$ at a point $x$. math.stackexchange.com/a/1127350/40119 – littleO Aug 2 '17 at 1:09
• The question here asked about "real life" applications, which IMO made it significantly different from the previous question. – Philip Roe Aug 3 '17 at 15:12

The very important topic of Conservation Laws makes great use of the Jacobian matrix. A set of conservation laws in one dimension takes the form of partial differential equations $$\partial_t\mathbf{u}+\partial_x\mathbf{f}(\mathbf{u})=0$$ where $\mathbf{u}$ is a vector containing the conserved variables. In the canonical example of compressible fluid flow these are mass density, momentum density and energy density. $\mathbf{f}(\mathbf{u})$ is another vector that contains, loosely, the rates at which these quantities flow. This is often written as $$\mathbf{u}_t+\mathbf{A}\mathbf{u}_x=0$$ where $\mathbf{A}$ is the Jacobian matric relating changes in $\mathbf{f}$ to changes in $\mathbf{u}$.

The eigenvalues of $\mathbf{A}$ are the speeds with which various waves propagate with or through the flow. The right eigenvectors are the patterns of disturbance carried by a wave, and the left eigenvectors yield quantities that are constant along a wave path.

EDIT

Another very important application is to computational geometry. Consider a mapping in $R^3$ where an initial point with position vector ${\mathbf x}$ is taken to a point with position vector $\mathbf{X}$. The Jacobian $\mathbf{J}=\frac{\partial {\mathbf X}}{\partial{\mathbf x}}$ is called the distortion gradient tensor. Assume the mapping is smooth enough that small parallelopipeds are taken to small parallelopipeds. Then , starting with a control volume oriented with the $\mathbf{x}$ cordinates

1. The rows of $\mathbf{J}$ are the vectors that represent the new edges.

2. The rows of the cofactor matrix of $\mathbf{J}$ are the normals to the new faces.

3. The determinant of $\bf J$ is the new volume, assuming the (now nonplanar) faces are closed by bilinear surfaces.

This used in computing the behavior of materials undergoing large distortion.

• whoa! best answer ever! – Nirvana Guha Aug 4 '17 at 19:16

Given a function $f : \mathbb{R}^n \to \mathbb{R}^m$ and $\mathbf{x} \in \mathrm{R}^n$, the definition of differentiability of $f$ at $\mathbf{x}$ is that there exists an $m \times n$ matrix $J$ and function $\eta$ such that $$f(\mathbf{y}) = f(\mathbf{x}) + J(\mathbf{y} - \mathbf{x}) + \eta(\mathbf{y})$$ and $\frac{\lVert \eta(\mathbf{y}) \rVert}{\lVert \mathbf{y} - \mathbf{x} \rVert} \to 0$ as $\mathbf{y} \to \mathbf{x}$. Intuitively, that means that $f$ has a good linear approximation $f(\mathbf{y}) \approx f(\mathbf{x}) + J(\mathbf{y} - \mathbf{x})$ for $\mathbf{y}$ near $\mathbf{x}$. This matrix $J$ is exactly the Jacobian.

One possible application of the Jacobian gives a generalization of the Newton-Raphson method to functions of multiple variables. Say $f : \mathbb{R}^n \to \mathbb{R}^n$, and you want to find a vector $\mathbf{x}$ such that $f(\mathbf{x}) = \mathbf{0}$. Then, by applying the approximation above, if the Jacobian matrix $J$ at $\mathbf{x}_0$ is invertible, you can find the zero of the approximation function as $\mathbf{x}_1 = \mathbf{x}_0 - J^{-1} f(\mathbf{x}_0)$. By iterating this, you will often get convergence to a zero of the original function. (In fact, a refinement of this procedure can be used to prove the Inverse Function Theorem, which states that if the Jacobian at $\mathbf{x}_0$ is a nonsingular matrix, then $f$ has an inverse function defined on some neighborhood of $f(\mathbf{x}_0)$.)

I am not sure what you mean by "real life application".

The Jacobian describes the amount of "stretching", "rotating" or "transforming" that a transformation imposes locally.

For example, if $\boldsymbol{f}(x,y)=(g(x,y), h(x,y))$ then the jacobian $\boldsymbol{J_{\boldsymbol{f}}(x,y)}$ describes how the neighborhood of that point is transformed.

• i mean to say, does this have any application in physics, or chemistry, or maybe some other field? Because as far as I understand it so far, calculus is turning out to be a tool to solve other problems. So, what problems can a Jacobian help me simplify? – Nirvana Guha Aug 2 '17 at 0:35
• Well, the only use outside of helping you to evaluate integrals (which is useful in physics and chemistry) that I know of is to solve a system of differential equations at an equilibrium point or approximate solutions near an equilibrium point. Would this count as an application for you? A lot of differential equations come up in physics so does that count? – AspiringMat Aug 2 '17 at 0:39
• i see. thank you. yes, this clears up a lot for me. – Nirvana Guha Aug 2 '17 at 0:41