# The Jacobian of the inverse of a transformation

The Jacobian of the inverse of a transformation is the inverse of the Jacobian of that transformation

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In this post, we would like to illustrate the meaning of

the Jacobian of the inverse of a transformation = the inverse of the Jacobian of that transformation

by proving a special case.

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Consider a transformation $\mathscr{T}: \bar{x}^i=\bar{x}^i (x^1,x^2)$, which is an one-to-one mapping from unbarred $x^i$‘s to barred $\bar{x}^i$ coordinates, where $i=1, 2$.

By definition, the Jacobian matrix J of $\mathscr{T}$ is

$J= \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{pmatrix}$

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Now we consider the the inverse of the transformation $\mathscr{T}$:

$\mathscr{T}^{-1}: x^i=x^i(\bar{x}^1,\bar{x}^2)$

By definition, the Jacobian matrix $\bar{J}$ of this inverse transformation, $\mathscr{T}^{-1}$, is

$\bar{J}= \begin{pmatrix} \displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^1}{\partial \bar{x}^2}} \\ \displaystyle{\frac{\partial x^2}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^2}{\partial \bar{x}^2}} \end{pmatrix}$

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On the other hand, the inverse of Jacobian $J$ of the original transformation $\mathscr{T}$ is

$J^{-1}=\displaystyle{\frac{1}{ \begin{vmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{vmatrix} }} \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} & \displaystyle{-\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{-\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} \end{pmatrix}$

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If $\bar{J} = J^{-1}$, their (1, 1)-elementd should be equation:

$\displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}}\stackrel{?}{=}\displaystyle{\frac{1}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}-\displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} }} \bigg( \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \bigg)$

Let’s try to prove that.

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Consider equations

$\bar{x}^1 = \bar{x}^1(x^1,x^2)$

$\bar{x}^2 = \bar{x}^2(x^1,x^2)$

Differentiate both sides of each equation with respect to $\bar{x}^1$, we have:

$A := 1=\displaystyle{\frac{\partial \bar{x}^1}{\partial \bar{x}^1}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

$B := 0 = \displaystyle{\frac{\partial \bar{x}^2}{\partial \bar{x}^1}=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

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$A \times \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}:~~~~~C := \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}}$

$B \times \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}:~~~~~D := \displaystyle{0=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}}$

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$D-C:$

$\displaystyle{ \frac{\partial \bar{x}^2}{\partial x^2}= \bigg( \frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial \bar{x}^1}{\partial x^2}\bigg) \frac{\partial x^1}{\partial \bar{x}^1}}$,

results

$\displaystyle{ \frac{\partial x^1}{\partial \bar{x}^1}}=\frac{\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial \bar{x}^2}{\partial x^1}}}$

— Me@2018-08-09 09:49:51 PM

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# Chain Rule of Differentiation

Consider the curve $y = f(x)$.

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$\displaystyle{\frac{d}{dx}}$ is an operator, meaning “the slope of the tangent of”. So the expression $\displaystyle{\frac{dy}{dx}}$, meaning $\displaystyle{\frac{d}{dx} (y)}$, is not a fraction.

In order words, it means the slope of the tangent of the curve $y = f(x)$ at a point, such as point $A$ in the graph.

The symbol $dx$ has no relation with the symbol $\displaystyle{\frac{dy}{dx}}$. It means $\Delta x$ as shown in the graph. In other words,

$dx = \Delta x$

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The symbol $dy$ also has no relation with the symbol $\displaystyle{\frac{dy}{dx}}$. It means the vertical distance between the current point $A(x_0, y_0)$, where $y_0 = f(x_0)$, and the point $C$ on the tangent line $y = mx + c$, where $m$ is the slope of the tangent line. In other words,

$dy = m~dx$

or

$\displaystyle{dy = \left[ \left( \frac{d}{dx} \right) y \right] dx}$

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The relationship of $\Delta y$ and $dy$ is that

$\displaystyle{\Delta y = \frac{dy}{dx} \Delta x + \text{higher order terms}}$

$\displaystyle{\Delta y = \frac{dy}{dx} dx + \text{higher order terms}}$

$\Delta y = dy + \text{higher order terms}$

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Similarly, for functions of 2 variables:

$\displaystyle{\Delta f(x,y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \text{higher order terms}}$

$\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy}$

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For functions of 3 variables:

$\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz}$

$\displaystyle{\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt}}$

— Me@2018-07-15 09:30:29 PM

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# 多項選擇題 6

Multiple Choices 6

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（問：哪有那麼多的時間？）

— Me@2018-05-22 06:02:40 PM

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# Van der Waals equation 1.2

Whether $X_{\text{measured}}$ is bigger or smaller than $X_{\text{ideal}}$ ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

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In the ideal gas equation derivation, the volume used in the equation refers to the volume that the gas molecules can move within. So

$V_{\text{ideal}} = V_{\text{available for a real gas' molecules to move within}}$

Then, when deriving the pressure, it is assumed that there are no intermolecular forces among gas molecules. So

$P_{\text{ideal}} = P_{\text{assuming no intermolecular forces}}$

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These are the reasons that

$V_{\text{ideal}} < V_{\text{measured}}$

$P_{\text{ideal}} > P_{\text{measured}}$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

$\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) \left(V_\text{measured}-nb\right) = nRT$

— Me@2018-05-16 07:12:51 PM

~~~

… the thing to keep in mind is that the “pressure we use in the ideal gas law” is not the pressure of the gas itself. The pressure of the gas itself is too low: to relate that pressure to “pressure for the ideal gas law” we have to add a number. While the volume occupied by the real gas is too large – the “ideal volume” is less than that. – Floris Sep 30 ’16 at 17:34

— Physics Stackexchange

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# Van der Waals equation 1.1

Why do we add, and not subtract, the correction term for pressure in [Van der Waals] equation?

Since the pressure of real gases is lesser than the pressure exerted by (imaginary) ideal gases, shouldn’t we subtract some correction term to account for the decrease in pressure?

I mean, that’s what we have done for the volume correction: Subtracted a correction term from the volume of the container V since the total volume available for movement is reduced.

asked Sep 30 ’16 at 15:20

— Physics Stackexchange

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Ideal gas law:

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

However, since in a real gas, there are attractions between molecules, so the measured value of pressure P is smaller than that in an ideal gas:

$P_{\text{measured}} = P_{\text{real}}$

$P_{\text{measured}} < P_{\text{ideal gas}}$

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has:

$V_{\text{measured}} > V_{\text{real}}$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $P_{\text{measured}}$ onto the LHS, since $P_{\text{measured}} < P_{\text{ideal}}$, the LHS will be smaller than the RHS:

$P_{\text{measured}} V_{\text{ideal}} < nRT$

So in order to maintain the equality, a correction term to the pressure must be added:

$\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) V_{\text{ideal}} = nRT$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $V_{\text{measured}}$ onto the LHS, since that volume is bigger that actual volume available for the gas molecules to move, the LHS will be bigger than the RHS:

$P_{\text{ideal}} V_{\text{measured}} > nRT$

So in order to maintain the equality, a correction term to the pressure must be subtracted:

$P_{\text{ideal}} \left(V_\text{measured}-nb\right) = nRT$

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In other words,

$V_{\text{measured}} > V_{\text{real}}$

$V_{\text{ideal}} = V_{\text{real}}$

$V_{\text{measured}} > V_{\text{ideal}}$

— Me@2018-05-13 03:37:18 PM

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Why? I still do not understand.

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How come

$P_{\text{measured}} = P_{\text{real}}$

but

$V_{\text{measured}} \ne V_{\text{real}}$?

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How come

$V_{\text{real}} = V_{\text{ideal}}$

but

$P_{\text{real}} \ne P_{\text{ideal}}$?

— Me@2018-05-13 03:22:54 PM

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The above is wrong.

The “real volume” $V_{\text{real}}$ has 2 possible different meanings.

One is “the volume occupied by a real gas”. In other words, it is the volume of the gas container.

Another is “the volume available for a real gas’ molecules to move”.

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To avoid confusion, we should define

$V_{\text{real}} \equiv V_{\text{measured}}$

$P_{\text{real}} \equiv P_{\text{measured}}$

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Or even better, avoid the terms $P_{\text{real}}$ and $V_{\text{real}}$ altogether. Instead, just consider the relationship between $(P_{\text{ideal}}, P_{\text{measured}})$ and that between $(V_{\text{ideal}}, V_{\text{measured}})$.

Whether $X_{\text{measured}}$ is bigger or smaller than $X_{\text{ideal}}$ ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

— Me@2018-05-13 04:15:34 PM

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# 時空兌換率

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$E = m c^2$

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$E = c^2 m$

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$1 \text{USD} \approx 8 \times 1 \text{HKD}$

1 美元 $\approx 8 \times$ 1 港元

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（而光速 c，則是時間和空間的兌換率。）

— Me@2018-05-11 09:10:00 PM

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# Physics Question (2013 DSE MCQ29)

For convenience, I assume that the rod OP is horizontal.

1. Remember, current (positive charges) moves from high potential to low potential outside a battery, but it moves from low potential to high potential inside the battery.

2. An induced emf is a kind of battery. That is why it is called “induced emf”, not “induced potential difference”.

3. Right Hand Rule:

Motion –> downwards

B-field –> into the paper

So

current –> O to P

Then, P is at a higher potential.

4. Left Hand Rule:

B-field –> into the paper

Therefore, you, as the positive charge, experience a force, pointing to the right. So the induced emf is pointing to the right.

Inside a battery, the direction of the emf is pointing from low potential to high potential. Thus, P is at a higher potential.

— Me@2014-03-19 01:54:04 PM

# nCr, 4

（A: 那為什麼把「7P3」拆成「7P2 x 5P1」就可以？那不會「暗地裡加了次序」嗎？）

7 個蘋果中選 3 出來，即是相當於有 3 個格子要填滿：

（＿）（＿）（＿）

（7）（＿）（＿）

（7）（6）（＿）

（7）（6）（5）

3! = 6

（7）（6）（5）
—————-
（3!）

= 35

（＿）（＿）（＿）

（3）（＿）（＿）

（3）（2）（＿）

（3）（2）（1）

（7）（6）（5）
—————-
（3）（2）（1）

= 35

（7）（6）|（5）
——— ——-
（2）（1）|（1）

（7）（6）|（5）
= ——— ——
（2!） |（1!）

= 105

— Me@2014.04.21

# nCr, 3

（A: 為何把「7P3」拆成「7P2 x 5P1」就可以，而把「7C3」拆成「7C2 x 5C1」就錯誤？）

ABE

AEB

BAE

BEA

EAB

EBA

ABE

AEB

BAE

BEA

EAB

EBA

AB　E

BA　E

AE　B

EA　B

BE　A

EB　A

AB　E

BA　E

AE　B

EA　B

BE　A

EB　A

（A: 那為什麼把「7P3」拆成「7P2 x 5P1」就可以？那不會「暗地裡加了次序」嗎？）

— Me@2014.04.14

# nCr, 2

（A：我大概明白你的解釋。但是，情感上，我仍然接受不到，「7 選 3」和「7 選完 2 後再選 1」，的確有所不同。）

「nPr」即是「n 排 r」—— 如果有 n 個物件，選 r 出來排隊，總共有多少個排列方法？

7P3/(3!)

= 210/6

= 35

7P2

(7P2)(5P1)

(7P2)(5P1)/(3!)

= 35

（A: 為何把「7P3」拆成「7P2 x 5P1」就可以，而把「7C3」拆成「7C2 x 5C1」就錯誤？）

— Me@2014.04.05

# nCr

（A：那我可不可以把題目看成，分兩次抽 3 個蘋果出來？

「7C3」和「7C2 x 5C1」，所表達的情況不同。

「7C3」是指由 7 個蘋果之中，任意選 3 個出來，總共有多少個可能。

（A：但是我仍然不太明白，「7 選 3」和「7 選完 2 後再選 1」，為何有所不同。）

— Me@2014.04.01

# Indefinite and Definite Integration

Q: When should I use indefinite integration and definite integration? What is their major difference?

Indefinite integration ~ Anti-differentiation

It will give you a function.

Definite integration ~ finding the area under a curve

It will give you a number only, NOT a function.

 This is a file from the Wikimedia Commons.

— Me@2014.01.24

# Independent vs Mutually Exclusive, 2

Tree diagram 4

Mutual exclusive events are corresponding to two separate branches of the same tree diagram.

Independent events are corresponding to two independent tree diagrams.

— Me@2014-01-01 6:27 PM

# V 和 U 的分別

Electric Potential and Electric Potential Energy

~~~~~~~~~~

electric potential = electric potential energy per unit charge

 V=\frac{U}{Q}

~~~~~~~~~~

 \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r}

 \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}

— Me@2014.01.04

# 機會率驗算 1.2

（問：在運算機會率題目時，怎樣可以知道，自己的思路有沒有錯呢？）

— Me@2013.12.24

# 機會率驗算 1.1

「P 方法」的意思是 Probability（機會率）方法，即是將幾個 probability 分數乘在一起，從而得到最終的機會率分數。

「S 方法」的意思是 Statistics（統計學）方法，即是透過 counting（點數）去運算；由此至終，只寫一個分數 —— 將所有可能性放在分數，然後再將你想要的可能性，放在分子。

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P 方法：

P（三顆骰抽兩顆，然後兩顆都擲到 2）

= (1/3)(1/6)(1)(1/3) + (2/3)(1/3)[(1/2)(1/6) + (1/2)(1/3)]

= …

= 2/27

~~~

S 方法：

= (分子)/(分母)

= 想要的可能性/所有的可能性

（「3C2」即是「3 選 2」；「3 選 2」有 3 個可能性。

= 1×2 （抽到一顆骰子正常，一顆不正常）+ 1×2（抽到一顆正常，和抽到另一顆的不正常骰子）+ 2×2（兩顆骰子也不正常）

= 8

P（三顆骰抽兩顆，然後兩顆都擲到 2）

= 8/108

= 2/27

「S 方法」所得出的答案

= 2/27

= 「P 方法」所得出的答案

— Me@2013.12.20

# 逃避問題 1.2

 \frac{d}{dx} \left( \frac{\sin x}{x} \right)

 \frac{d}{dx} \left[ (\sin x) \left( \frac{1}{x} \right) \right]

— Me@2013.12.04

# Factorial

n! = n(n-1)(n-2)…(3)(2)(1)

n(n-1)(n-2)…(3)(2)(1)

— Me@2013.10.12

# Distance vs Displacement

distance = length of the path travelled

displacement = change of position

— Me@2013-10-03 11:03 AM

Price is what you pay; value is what you get.

— Ben Graham

HH
HT
TH
TT

HH
HT <
TH <
TT

2
_

4

2
_

4

— Me@2013.07.27