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# How do we find the values of k and m that makes function continue anywhere piecewise function of (x^2) + 5 when x > 2, m(x+3) + k when -1 < x <=2 and 2(x^3) + x + 7 when x <=-1?

Nov 7, 2016

#### Answer:

$m = \frac{5}{3}$
$k = \frac{2}{3}$

#### Explanation:

Here is an example of a continuous function.

Here is an example of a discontinuous function.

So, continuous functions are where everything is connected; there are no gaps or holes.

The y-value of ${x}^{2} + 5$ when $x = 2$, (even though this would be a hole) is $9$.

We need the graph of $y = m \left(x + 3\right) + k$ to end at the point $\left(2 , 9\right)$.

The $y -$ value of $y = 2 {x}^{3} + x + 7$ at $x = - 1$ is $y = 2 {\left(- 1\right)}^{3} - 1 + 7 = - 2 - 1 + 7 = 4$.

We need the graph of $y = m \left(x + 3\right) + k$ to start at the point $\left(- 1 , 4\right)$.

Hence, we can write a systems of equations with respect to $m$ and $k$, since we know two inputs/outputs.

$4 = m \left(- 1 + 3\right) + k$
$9 = m \left(2 + 3\right) + k$

$k = 9 - 5 m$

$4 = 2 m + 9 - 5 m$

$- 5 = - 3 m$

$m = \frac{5}{3}$

$9 = 5 m + k$

$9 = 5 \left(\frac{5}{3}\right) + k$

$9 = \frac{25}{3} + k$

$k = \frac{2}{3}$

So, the function in the interval -1 < x ≤ 2 is $y = \frac{5}{3} \left(x + 3\right) + \frac{2}{3}$.

Nov 7, 2016

#### Answer:

$m = \frac{5}{3}$ and $k = \frac{2}{3}$,

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} + 5 & x > 2 \\ \frac{5}{3} \left(x + 3\right) + \frac{2}{3} & - 1 < x \le 2 \\ 2 {x}^{3} + x + 7 & x \le - 1\end{matrix}\right.$

#### Explanation:

We want to find $m$ and $k$ such that $f \left(x\right)$ is continuous:

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} + 5 & x > 2 \\ m \left(x + 3\right) + k & - 1 < x \le 2 \\ 2 {x}^{3} + x + 7 & x \le - 1\end{matrix}\right.$

Just think about what we know so far, and how it would look:

When $x = - 1 \implies f \left(- 1\right) = 2 {\left(- 1\right)}^{3} - 1 + 7 = - 2 - 1 + 7 = 4$
When $x = 2 \implies f \left(2\right) = {2}^{2} + 5 = 4 + 5 = 9$

So for the mid interval we need a straight line whose equation is $y = m \left(x + 3\right) + k$ which passes through $\left(- 1 , 4\right)$ and $\left(2 , 9\right)$

This line would have the following gradient:
$m = \frac{\Delta y}{\Delta x} = \frac{9 - 4}{2 - \left(- 1\right)} = \frac{5}{3}$

So our required line passes through $\left(- 1 , 4\right)$ (equally we could you the other coordinate and get the same answer) and has gradient $m = \frac{5}{3}$, so using $y = m \left(x + 3\right) + k$ we have:

$\frac{5}{3} \left(- 1 + 3\right) + k = 4$
$\frac{5}{3} \left(2\right) + k = 4$
$\frac{10}{3} + k = 4$
$k = 4 = \frac{10}{2} = \frac{2}{3}$

So the mid-section has equation $m \left(x + 3\right) = k$ with $m = \frac{5}{3}$ and $k = \frac{2}{3}$,

ie $y = \frac{5}{3} \left(x + 3\right) + \frac{2}{3}$,Which we can graph to confirm


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# Solving classification task with deep network

I am currently working on a classification problem:

The dataset (2000 features, 25 labels) I am using is seperable by using a 2-layered Multi-Layer-Perceptron (1 Input + 1 OutputLayer = 1 Weight Matrix) with achieving great accuracy.

Note: The Data I am using are spectrograms.

Now what I am curious about is the following: Can I use a deep network (more than 2 layers) to even increase the accuracy?

What I think about: Increasing the network structure would make the network interpret the data in a more complex way and therefore should be able to even correctly classify little "outbrakers" and "noisy elements" of the dataset.

• Whenever you add more layers, there will be vanishing and exploding gradients which may cause your network not to learn, or learning may happen so slowly. To avoid, you should use ReLU activation in order to avoid saturation of gradients. Moreover you have to use He or Xavier initialization techniques to avoid having bad random weights. There are other techniques for solving this problem which are called skip connections but at least I've never seen the use of them in MLPs Although they are really helpful for solving the mentioned problem.
• Overfitting is a problem that happens for large architectures that are not fed with enough training data. You should use regularization techniques. There are so many papers about this but in your case which is using MLP I highly recommend you using Dropout technique which is invented by the so called God Father of deep learning.


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## Dodo1 2 years ago limit comupting limit using the limit law

1. Dodo1

$\lim_{x \rightarrow 1} \frac{ x^4+x^3-2x }{ }$

2. Dodo1

not working so Lim x^4+x^3-2x/x^5+2x-3 x->1

3. ZeHanz

Do you mean l'Hôpital's Rule by limit law? In that case take the derivative of both the numerator and the denominator and then take the limit as x goes to 1.

4. Dodo1

yes!

5. Dodo1

I thought I need to factorize first

6. ZeHanz

That can help sometimes, but factorising the denominator is difficult and l'H will crack it anyway...

7. Dodo1

Whats I`H?

8. Dodo1

I have to go meeting but I will back later. thanks,

9. ZeHanz

l'H = l'Hôpital's Rule

10. Dodo1

Do i plug numbers into the equation>?

11. LABurn

No, using l'Hôpital's Rule allows you to take the limit as x goes to 1 of the top and bottom separately.

12. ZeHanz

So, after differentiating separately yo now have this:$\lim_{x \rightarrow 1}\frac{ 4x^3+3x^2-2 }{ 5x^4+2 }$No 0/0 problem anymore, just set x=1 to see the result.

13. Dodo1

Ok thank you!

14. ZeHanz

YW!


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# Applied Scientific Instrumentation

### Products

 Custom Optics diSPIM FTP FW1000: Filter Wheel MPPI MS2000, MFC2000 and RM2000 Controller Piezo Systems RAMM, MIM and VTS Portal Serial Command Software TG-1000/Tiger Controller Tracking & Stabilization XY Stage Z-Drive and Focus Stage

### Tags

You are not allowed to perform this action
tiger_ms2k_difference

# TG-1000 and MS-2000’s instruction set differences

This section is intended to quickly bring the users up to speed on the major differences in the serial commands between TG-1000 and MS-2000 controller.

Where possible the TG-1000 commands were kept close to the MS-2000 controller commands.

The Tiger reply syntax, described above, was added to Tiger starting in COMM card firmware v1.92.

Commands where the axis letter is specified work just like MS-2000, e.g. MOVE, MOVREL, and WHERE. For example, X and Y axis reside on card with address 1, and Z resides on card with address 2. When M X=## Y=### Z=### is issued the COMM card in TG-1000 controller parses the command, and redirects the command to the appropriate card automatically. These commands are called Axis-Specific Commands. A card address should not be specified with an Axis-Specific command, otherwise undefined behavior may result from any mismatch between the address specified and the card with the axis as determined by the COMM card.

The TG-1000 is able to support multiple add-ons like CRISP and Piezo stages whereas MS-2000 was only able to support one, so we were forced to alter the command set to handle commands which are intended for a certain card but do not contain the axis name. In those cases, the user must specify which card the command is intended for, and the COMM card will relay the command appropriately. For example, the SS Z command saves the settings into non-volatile memory, but the user may not want all the settings on all the cards to be saved. For the TG-1000 the user has to add the card address in front of the command to make it work. For example, if the card that drives XY axis has the address 1, to only save setting on this card user should issue 1SS Z. These commands are called Card-Addressed Commands.

Importantly, for Card-Addressed Commands the setting applies to all axes on the card. For example, the ring buffer delay can be set independently for the XY stage on one card and a piezo stage on a different card. However, if a card contains more than one independent axis (e.g. a ZF card for two linear stages, or a Micro mirror card for 2 scanners) then the same setting applies to all axes. Continuing our example, the ring buffer on the ZF card would control both Z and F axes (though one of these axes could be disconnected from the ring buffer) so Z and F could not have separate ring buffer delay times.

Various methods of determining the address of a specific card are given below.

A third category is Broadcast commands, like STATUS, HALT, RESET, and ZERO. They are routed to all cards in a TG-1000 system by default, just like the MS-2000. However, in most cases they can also be directed at a single card by simply adding the card address, so they could be considered as special type of Card-Addressed Commands sent to all cards if an address is not specified.

A final category includes commands that default to a particular card, though they can be addressed to specific cards as well. There are sub-categories of commands sent to the COMM card unless otherwise specified (e.g. BUILD and WHO), commands sent to whatever card has the X axis or else all the stage cards if the X axis doesn’t exist (e.g. JSSPD and LED), and commands sent to whatever card has the Z axis or all the stage cards if the Z axis doesn’t exist (e.g. most CRISP and autofocus commands). These sub-categories may be referred to as Comm-Default, Stage-Default, and Focus-Default commands respectively.

Beginning with COMM card firmware version 3.10 a special axis letter * was added. When this character is specified in an Axis-specific command then it applies to all the axes in the controller. For instance, all axes can be moved to zero position using M *=0. The * axis letter can also be combined with card addressing to apply the command to all the axes on the same card, e.g. 3M *=0 will move all axes on card address 3 to their zero position, and 3! * will home all axes on card 3.

## How does the user find out a card address?

When the TG-1000 is turned on or when WHO serial command is issued, the controller prints out card address, axis names, firmware version and firmware build date of all the cards installed in the system (see section Tiger Banner below for more details). Also, more complete information about the cards and corresponding axes can be accessed using the BUILD X command which is described in the section Build Command below.

As an illustrative example, the controller may have the following configuration as reported on startup:

................................................................
At 30: Comm v1.5 TIGER_COMM May 07 2013:15:42:05
At 31: X:XYMotor,Y:XYMotor v2.4 STD_XY Jun 11 2013:17:00:12
At 32: Z:Piezo v2.4 ADEPT_PIEZO Jun 11 2013:17:05:00

The card addresses are shown in ASCII code. At 31 indicates a card with X and Y axes has the address 1 (ASCII code of 1 is 31).

 To save settings in non volatile memory: To print its error dump buffer: 1SS Z Card-Addressed 1DU Y Card-Addressed M X=### Y=### Axis-Specific

Similarly the card with Z axis has address 2 (ASCII code of 2 is 32).

To move Z axis: To issue pzinfo command on this card: M Z=### Axis-Specific 2PZINFO Card-Addressed 2PZ Z=1 Card-Addressed 2SS Z Card-Addressed 2PZC Card-Addressed

TG-1000 Comm Hard coded, re-assignable 0x30 (‘0’)
Stage/ FW/Shutter Unique Address 0x31 to 0x39, then 0x81 to 0xF5
Stage Broadcast Recognized by all stage controllers 0xF6
Filterwheel Broadcast Recognized by all FW controllers 0xF7
Shutter Broadcast Recognized by all shutter controllers 0xF8
LCD Broadcast Recognized by all LCD controllers 0xF9
Broadcast Recognized by all cards 0xFD
Broadcast except Comm Recognized by all cards except TG-1000 Comm 0xFE

## Identifying Controller Configuration

### Build Command

In Controller with Tiger Comm version v1.8 and above, the BUILD X command directed at the COMM card (or without any address) can be used to query axis names, axis types and axis address. (When the build command is addressed to a specific card the build information for that card is returned.)

##### Example:
build x
TIGER_COMM
Motor Axes: X Y P Q R S 0 1
Axis Types: x x u u u u w w
Axis Addr: 1 1 2 2 2 2 3 3
Hex Addr: 31 31 32 32 32 32 33 33
Axis Props: 0 0 0 0 0 0 0 0

The above system has card address 1 with an XY stage with axes named X and Y. Card number 2 has micro mirror with axes P,Q,R,S. Then card address 3 with filter wheel IDs 0 and 1.

The next line contains the axis type short code. For example, x means xy stage, u means micro mirror and w means filter wheel. A complete listing of axis types with designations is as follows:

The next two lines contain the addresses in two forms, first the form that is used to prefix Card-Addressed commands and second in hex format.

Finally, any special axis properties are printed (e.g. CRISP or SCAN capabilities) starting with firmware version 2.8. The decimal equivalent of a byte (0-255) is printed with the bits of the byte representing the following

##### Axis Type List
Axis Type Short Axis Type Long Description
x XYMotor XY stage
z ZMotor Z focus motor drive. LS50s, Z scopes etc
p Piezo Piezo Focus. ASIs ADEPT, Piezo DAC etc
o Tur Objective Turret
f Slider Filter Changer
t Theta Theta Stage
l Motor Generic linear motorized stage, TIRF, SISKIYOU etc
a PiezoL Generic linear piezo stage
m Zoom Zoom magnification motor axis
u MMirror Micro Mirror, Scanner 75 etc
w FW Filter Wheel
s Shutter Shutter
g Logic Programmable logic card
i LED card Multi LED Driver card
b Lens Tunable Lens
u Unknown Unknown axis type
##### Axis Properties
 Bit 0: CRISP auto-focus firmware Bit 1: RING BUFFER firmware Bit 2: SCAN firmware Bit 3: ARRAY firmware or MM_TARGET firmware Bit 4: SPIM firmware (v2.81+) Bit 5: SINGLEAXIS and/or MULTIAXIS firmware (v2.81+) Bit 6: LED illumination (v2.87+) Bit 7: Reserved

### Tiger Banner

A banner is printed by the controller on Startup and by the WHO command. It tells the user (and also to any scripts and programs) all the available cards in the system, with their axis characters, axis types etc.

At 30: Comm v1.5 TIGER_COMM May 07 2013:15:42:05
At 31: X:XYMotor,Y:XYMotor v2.4 STD_XY Jun 11 2013:17:00:12
At 32: Z:Piezo v2.4 ADEPT_PIEZO Jun 11 2013:17:05:00

It is a multiline reply, each line is terminated by a <Carriage Return> and final line by a <Carriage Return> +<Line Feed> to designate end of Transmission.

Each line can be sub divided into strings using a white space as delimiter. Then the 2nd string is the card address in Hex. 0x30 is 0 in ascii, 0x31 is 1 and 0x32 is 2.All possible address are 0x30 to 0x39 and then 0x81 to 0xF5. 0x81 is ü, 0xF5 is ⌡.

The second string can be further subdivided with comma as delimiter. Then the first character of the string is the axis character, colon and what kind of an axis it is. A to Z are all possible axis names, system is case insensitive. Except in case of filter wheels, IDs are integers 0 to 9

##### Example:

X:XYMotor,Y:XYMotor . X and Y are axis names of a XY stage. Z:Piezo . Z is the axis names of a Piezo focus.

Then the 4th string is the firmware version number. 5th String is our build name, which is an internal designation. The last set of strings are the firmware compile date and time.

### A Note Regarding Units

The most common commands including MOVE, MOVEREL, HERE, and WHERE use axis units, which can be changed using the UM command. However, some commands such as SETUP, SETLOW, and PCROS always use units of mm. By default, axis units where position is given in distance (e.g. for motorized stages and piezo stages) are represented in 0.1um increments, or 10,000 units per millimeter of travel. For mirror scanner axes, default axis units are 0.001 degrees (uncalibrated), or 1000 units per degree of travel. When a time is required, the unit is generally milliseconds. Some commands use integer codes or other input units as described below.


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# Solve the following initial value problem (DiracDelta function)

1. Apr 27, 2012

### IHave

1. The problem statement, all variables and given/known data
y'' +4y = 2 delta(t - pi/4)

where y(0)=0 and y'(0)=0

2. Relevant equations
Laplace transform
Inverse Laplace transform

3. The attempt at a solution
after applying laplace tranform
Y(s)=2e^((-pi/4)*s) / s^(2)+4

as the final answer i have
y(t) = U(t-pi/4)*2sin(t-pi/4)

I'm not very comfortable with these types of problems yet, is my answer correct?

2. Apr 27, 2012

### dikmikkel

I get the forward transform as:
$s^2\tilde{y}(s) +4\tilde{y}(s) =2e^{-s\pi /4}$
Then isolating i get the same as you. But you got something wrong, the product of two functions cant be inverse transformed generally. The result should be:
$-\text{Heaviside}(t-\pi/4)\cos(2t)$


text:
# GATE2015-55

A heterozygous tall plant $(Tt)$ was crossed with a homozygous dwarf plant $(tt)$. The resultant seeds were collected. If five seeds are chosen at random, then the probability (in $\%$) that exactly two of these seeds will yiIeld dwarf plants is __________


text:
Two events are independent if the outcome of one doesn't affect the outcome of the other. Otherwise they are dependent.

Examples:

• When tossing a fair coin twice, the result of the first toss doesn't affect the probability of the outcome of the second toss, and vice versa. The event 'getting a head' in the second toss is independent of the event 'getting a head' in the first toss.
• When drawing two cards from a deck of 52 playing card, the event 'getting a King' on the first card and the event 'getting a black card' are not independent. The probability of the second card change after the first card is drawn. The two events would be independent if after drawing the first card, the card is returned to the deck (thus the deck is complete 52 again).

For two independent events, $A$ and $B$, the probability of both occuring, $\mathbb{P}\left(A\cap B\right)$, is the product of the probability of each event. $\cap$ is the mathematical symbol for and or intersection.

$\mathbb{P}\left(A\cap B\right)=\mathbb{P}\left(A\right)×\mathbb{P}\left(B\right)$

For example, when tossing a fair coin twice, the probability of getting a 'Head' (let's call it event $H$) on the first and then getting a 'Tail' (let's call it event $T$) on the second is

$\begin{array}{rl}\mathbb{P}\left(H\cap T\right)& =\mathbb{P}\left(H\right)×\mathbb{P}\left(T\right)\\ & =0.5×0.5\\ & =0.25\end{array}$

Two events are mutually exclusive if they cannot occur at the same time.

Examples:

• When tossing a fair coin, the event 'getting a head' and the event 'getting a tail' are mutually exclusive because they can't occur at the same time.
• When throwing a fair die, the event 'getting a 1' and the event 'getting a 4' are mutually exclusive because they can't occur at the same time. But the event 'getting a 3' and the event 'getting an odd number' are not mutually exclusive since it can happen at the same time (i.e. if you get 3)

For two mutually exclusive events, $A$ and $B$, the probability of either one occuring, $\mathbb{P}\left(A\cup B\right)$, is the sum of the probability of both events. $\cup$ is the mathematical symbol for union. $\mathbb{P}\left(A\cup B\right)=\mathbb{P}\left(A\right)+\mathbb{P}\left(B\right)$

For example, when choosing a ball at random from a bag containing 3 blue balls, 2 green balls, and 5 red balls, the probability of getting a blue (let's call it event $B$) or red ball (let's call it event $R$) is

$\begin{array}{rl}\mathbb{P}\left(B\cup R\right)& =\mathbb{P}\left(B\right)+\mathbb{P}\left(R\right)\\ & =\frac{3}{10}+\frac{5}{10}\\ & =\frac{8}{10}\\ & =0.8\end{array}$

For non mutually exclusive events the probability of either one or both occuring is

$\mathbb{P}\left(A\cup B\right)=\mathbb{P}\left(A\right)+\mathbb{P}\left(B\right)-\mathbb{P}\left(A\cap B\right)$

where $\mathbb{P}\left(A\cap B\right)$ is the probability of event $A$ and event $B$ happening at the same time.

For example, when drawing a card from a deck of 52 playing cards, the probability of getting a red card (let's call it event $R$) or a King (let's call it event $K$) is

$\begin{array}{rl}\mathbb{P}\left(R\cup K\right)& =\mathbb{P}\left(R\right)+\mathbb{P}\left(K\right)-\mathbb{P}\left(R\cap K\right)\\ & =\frac{26}{52}+\frac{4}{52}-\frac{2}{52}\\ & =\frac{28}{52}\\ & =\frac{7}{13}\end{array}$

This is so because a card can either be red, king, or both (i.e. red king). So that's why we need to subtract the probability of a card being both red and king because it has already been accounted for in the probability of the card being red and the probability of the card being king.


text:
Question

# Silicon doped with electron-rich impurity forms ________.

A
ptype semiconductor
B
ntype semiconductor
C
intrinsic semiconductor
D
insulator

Solution

## The correct option is B $$n-type$$ semiconductorAn N-Type semiconductor is created by adding pentavalent impurities like phosphorus ($$P$$), arsenic ($$As$$), antimony ($$Sb$$), or bismuth ($$Bi$$). A pentavalent impurity is called a donor because it is ready to give a free electron to a semiconductor. A pentavalent impurity is electron rich impurity.So, the correct answer is option $$B$$.Chemistry

Suggest Corrections

0

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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 corrected English technical term

A really elementary one :

A linear application map between two vector spaces of the same finite dimension is an isomorphism if and only if its kernel is zero.

As an application, I like the proof of the existence of Lagrange interpolation polynomials.


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# Equivalence of tail sigma-fields between two triangular arrays of Brownian motions

Let $$\{B^{(i)}_t : 0 \leq t \leq T, i \geq 1\}$$ be a sequence of i.i.d. standard Brownian motions. For each $n,$ let $V^{(n)}_t, \,0 \leq t \leq T$ be a continuous bounded process adapted to the filtration $\mathcal{F}_t^n$ generated by the first $n$ Brownian motions.

Let $P_n$ denote the law of $(B^{(1)}, \dots, B^{(n)})$ on $C := C([0, T], \mathbb{R}^n).$

Since $V^{(n)}$ satisfies conditions for a change of measure, we can apply a Girsanov transform to yield a measure $Q_n$ with $$\frac{dQ_n}{dP_n} = \exp\Big( \sum_{i = 1}^n \int_0^\cdot V^{(n)}_sdB^{(i)}_s - \frac{1}{2}\int_0^\cdot [V_s^{(n)}]^2 ds\Big).$$

So $Q_n$ has the same law on $C$ as $(B^{(1)} + V^{(n)}, \dots, B^{(n)} + V^{(n)}),$ an $n$ dimensional Brownian motion with a drift of $V^{(n)}$ in each coordinate.

If we define two triangular arrays whose row for each $n$ are $$R^P_n : =(B^{(1)}, \dots, B^{(n)}),$$ and $$R^Q_n := (B^{(1)} + \int_0^\cdot V^{(n)}, \dots, B^{(n)} + \int_0^\cdot V^{(n)}),$$

we immediately see the laws of the rows $R^P_n$ and $R^Q_n$ are equivalent for each finite $n,$ and in fact under the finite triangular array $R_1^P \times \cdots \times R_n^P \equiv R_1^Q \times \cdots \times R_n^Q$ via the canonical product measure.

We may embed each $R^P_n$ to induce a measure on $C^\infty = C([0, T], \mathbb{R}^{\mathbb{N}})$ and can place a standard metric on this to make it Polish. Similarly with $R^Q_n.$ Now that each $R^P_n$ has a measure on the same space, consider its filtration $\mathcal{F}_n^P$ and tail $\sigma$-field $\mathcal{F}_\infty^P.$ (Similar for $R^Q_n$). Notice that because $V^{(n)}$ is adapted, we know $\mathcal{F}_n^Q \subset \mathcal{F}^P_n$ so for our purposes here we restrict to $\mathcal{F}^P_n.$

By the Girsanov transform, we see that null sets in $\mathcal{F}_n^P$ under $R^P$ are also null sets under $R^Q.$

Let us further assume that $V^{(n)}$ converges a.s. and in $L_p$ to some bounded process $V.$

Can we assert the null sets in $\mathcal{F}_\infty$ under $R^P$ are also null under $R^Q?$

In essence, we wish to extend the $0-1$ law of a triangular array of i.i.d. Brownian motions to that of a triangular array of BMs with an adapted row-wise convergent drift.

I apologize in advance for any errors or typos! Any references would be much appreciated!

No, the measures are not necessarily equivalent on the tail $\sigma$-field.
For a concrete example, take $V_t^{(n)} = t$, a deterministic continuous process of bounded variation. Then trivially the $V_t^{(n)}$ converge to $V_t=t$ in whatever sense you want. Consider the tail random variable $X = \limsup_{n \to \infty} \frac{1}{n} \sum_{k=1}^n B^{(k)}_1$. Under $P$, each $B^{(k)}_1$ has an $N(0,1)$ distribution, and they are iid, so by the strong law of large numbers we have $P(X=0)=1$, where we note that $\{X=0\} \in \mathcal{F}_\infty$. But under $Q$, each $B^{(k)}_1$ has the law of a standard Brownian motion plus $V_1=1$, so it is $N(1,1)$. Thus we have $Q(X=1)=1$ and $Q(X=0)=0$. As such, $P,Q$ are mutually singular on $\mathcal{F}_\infty$.
You need some sort of summability criterion on the $V^{(n)}$; this is some corollary of the Cameron-Martin theorem for Gaussian measures. They should not only converge, but converge to 0, and reasonably fast. I would have to work it out, but I think you might want something like $\sum_n \mathbb{E}\left[\sup_{[0,T]} |V_t^{(n)}|^2\right] < \infty$. Bogachev's book Gaussian Measures is probably the ultimate resource here, though there may be lighter-weight sources also.
• Thanks, Nate. In your example, the slln still holds for the sequence with drift. Do you know if under these conditions a slln holds for $n^{-1}\sum_{i=1}^nl(B^{(i)} + V^{(n)})$ where $l$ is a functional on $C([0, T], \mathbb{R})?$ – user253775 Feb 22 '17 at 3:21
• I mean $n^{-1}\sum_{i=1}^nl(B^{(i)} + \int_0^\cdot V^{(n)}).$ – user253775 Feb 22 '17 at 4:48
• If it's a linear functional, and if SLLN holds for $V$, then it should. – Nate Eldredge Feb 22 '17 at 5:05


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## Saturday, January 16, 2010

### Chaotic Method of Manufactured Solutions: Lorenz 63

I quickly brushed over the method of manufactured solutions (MMS) aspects in the previous post on Lorenz 63, you might wonder why. Well, it turns out that applying the MMS to chaotic systems includes an additional consideration that is not normally of concern. Here is the ’standard’ sort of checklist for choosing manufactured solutions (taken from Verification of Computer Codes in Computational Science and Engineering, emphasis mine):

1. Manufactured solutions should be sufficiently smooth on the problem domain so that the theoretical order-of-accuracy can be matched by the observed order-of-accuracy obtained from the test. A manufactured solution that is not sufficiently smooth may decrease the observed order-of-accuracy (however, see # 7 in this list).
2. The solution should be general enough that it exercises every term in the governing equation. For example, one should not choose temperature T in the unsteady heat equation to be independent of time. If the governing equation contains spatial cross-derivatives, make sure the manufactured solution has a non-zero cross derivative.
3. The solution should have a sufficient number of nontrivial derivatives. For example, if the code that solves the heat conduction equation is second order in space, picking T in the heat equation to be a linear function of time and space will not provide a sufficient test because the discretization error would be zero (to within round-off) even on coarse grids.
4. Solution derivatives should be bounded by a small constant. This ensures that the solution is not a strongly varying function of space or time or both. If this guideline is not met, then one may not be able to demonstrate the required asymptotic order-of-accuracy using practical grid sizes. Usually, the free constants that are part of the manufactured solution can be selected to meet this guideline.
5. The manufactured solution should not prevent the code from running successfully to completion during testing. Robustness issues are not a part of code order verification. For example, if the code (explicitly or implicitly) assumes the solution is positive, make sure the manufactured solution is positive; or if the heat conduction code expects time units of seconds, do not give it a solution whose units are nanoseconds.
6. Manufactured solutions should be composed of simple analytical functions like polynomials, trigonometric, or exponential functions so that the exact solution can be conveniently and accurately computed. An exact solution composed of infinite series or integrals of functions with singularities is not convenient.
7. The solution should be constructed in such a manner that the differential operators in the governing equations make sense. For example, in the heat conduction equation, the flux is required to be differentiable. Therefore, if one desires to test the code for the case of discontinuous thermal conductivity, then the manufactured solution for temperature must be constructed in such a way that the flux is differentiable. The resultant manufactured solution for temperature is nondifferentiable.
8. Avoid manufactured solutions that grow exponentially in time to avoid confusion with numerical instability.

As emphasized in # 4 we often have to use the free constants in our manufactured solution to bound the derivatives in the domain. For chaotic systems we’ll need to use not only the free constants, but the initial conditions as well, to achieve another end: ensure that the chosen solution is in the convergent region at all times. This is a sufficient but not necessary condition for the source term to be in the basin of entrainment. When the source term is in the basin of entrainment then we can expect to be able to verify the ordered convergence of our implementation as we would in the non-chaotic case. Otherwises we can expect the truncation error to eventually lead to an arbitrarily large departure from the manufactured solution. This terminology comes from the nonlinear dynamics and control literature [1] [3]. A worked example for the Lorenz ’63 system showing the boundary of the convergent region is in Chapter 6 of [2].

The convergent region is defined by the region of phase space for which the eigenvalues of the system Jacobian has negative real parts. As in any other application of the MMS, we start by choosing some convenient solution consistent with the desiderata enumerated above. Trigonometric functions are convenient in this case:

 (1)

 (2)

 (3)

Now we need the Jacobian of the system (and then its eigenvalues), since our governing equations are definied in Maxima, that is easy enough:

/* Need the Jacobian of the system to find the convergent region so
we can select a useful IC/amplitude for our manufactured solutions */
sys_eqns : [eqn_1, eqn_2, eqn_3] $J_cr : zeromatrix(3,3)$
for j : 1 thru 3 do
for i : 1 thru 3 do
(
J_cr[i,j] : diff(rhs(sys_eqns[i]), unknowns[j])
)
) $[J_cr_eigvals, J_cr_eigmult] : eigenvalues(J_cr)$
J_cr_eigvals : fullratsimp(J_cr_eigvals)

As before these expressions are output in Fortran90 format for compilation. Here are the eigenvalues in the complex plane for bx = 1.0, by = -1.0, bz = 10.0.

We can immediately see the problem from Figure 1: our manufactured solution is in the region of phase space where one of the Jacobian’s eigenvalues has a positive real part. We can adjust the initial conditions of our solution to get all of our eigenvalues on the left half-plane.

Being limited to only certain portions of the phase space for conducting verification is fine, because a single well-designed manufactured solution verifies the correctness of the entire implementation (as long as all of the terms are activated). This is the same reason that it is often emphasized in the V&V literature that there is no need for the manufactured solution to be physically meaningful.

### References

[1]   Jackson, E.A., “Controls of dynamic flows with attractors,” Physics Review A, Vol. 44, Is. 8, 1991.

[2]   Wu, W., “Analytical and Numerical Methods Applied to Nonlinear Vessel Dynamics and Code Verification for Chaotic Systems,” Disertation, Virginia Polytechnic Institute and State University, Dec, 2009.

[3]   W. Wu, L.S. McCue, and C.J. Roy. “The method of manufactured solutions applied to chaotic systems,” Nonlinear Dynamics, 2009, under review.

1. Hello

I am visiting here for the first time because Dan Hughes sent me a mail pointing to your blog .
First . My chaos interest comes from the mathematical physics point of view so I have difficulty to read and interpret the pages of code that you post .
If I understood well , you began to look at 2 aspects of deterministic chaos - convergence and stochasticity .
As it happens I have a rather good knowledge of both so if you continue being interested by the chaos theory , I could perhaps show you some avenues .

- The convergence question is easy .
There is none .
A fast qualitative argument bases on the Lyapounov coefficient . The size of the step used in the numerical solution can be considered as the difference between 2 initial conditions . But as any chaotic system has at least one positive Ltapounov coefficient , we know that 2 trajectories diverge exponentially .
Now the Heisenberg uncertainty relation gives us the lower bound for the accuracy with which we can distinguish 2 different initial conditions .
Follows that the greatest accuracy achievable corresponding to this lower bound leads to trajectories that still diverge (exponentially).
In other words a quantum fluctuation is enough to give an upper bound on the predictability time . Beyond that time no converging solution is computable even in principle .

2. - stochasticity . Obviously a chaotic system cannot be predicted deterministically be it numerically or otherwise (see above) . Therefore a legitimate question is whether there can be a statistical description of the trajectories . Or of the frequency of places that the trajectory will visit what is a similar question .
Well here the answer is : sometimes yes .
The key word is ergodicity and that's why it is (almost) impossible to study the chaos theory without studying the ergodic theory too .
Indeed an ergodic system fulfills the conditions of the ergodic theorem which says broadly that "the time average of a dynamical parameter taken along a trajectory converges to the phase space average weighted by a suitable probability density function when time goes against infinity" .
The ergodicity is the condition of existence because it guarantees that a pdf exists such as it permits to connect the phase space states to the temporal evolutions .
Not all chaotic systems are ergodic . F.ex colliding hard spheres are chaotic and ergodic while planetary orbits are chaotic and non ergodic . The use of ergodicity in the former leads to statistical thermodynamics which works very well . On the other hand the unpredictability of planetary orbits is both in the deterministic and the stochastical sense .

How does the climate fit in ?
Well it doesn't fit in . The chaos theory is mathematically a problem of non linear ODE while the climate (and weather) is a problem of non linear PDE . Space matters . We deal here with SPATIO-temporal chaos . The best way to intuitively grasp teh spatio temporal chaos is to look at the cigarette smoke . You clearly see that chaotic structures exist both in time and in space . The ergodic theorem applies only on dynamical systems depending on time . Finding an equivalent of ergodicity for spatio-temporal systems is cutting edge science - it is just beginning .
What is known and what the cigarette smoke also demonstrates is that while temporal chaos is often ergodic in physical systems of interest , it is not the case for spatio-temporal systems .
One consequence of that is that there is no equivalent of statistical thermodynamics for the Navier Stokes equations - you can't find "statistical" properties of solutions .
We don't even know if there is a unique smooth solution for given boundary/initial conditions . Kolmogorov turbulence theory is a kind of attempt at statistical fluid dynamics but it doesn't work well and it fails completely for lower Reynolds numbers .
So when one sees that in the so called "climatology" people get rid of the space by simply ... AVERAGING (!) and then "analyse" the time series of spatially averaged parameters (like GMST) , one wonders what kind of physics they do . Etc .
Well I think I have been too long already so I hope it shed some light .

3. Hi Tom,
Thank you for leaving such thoughtful comments. Dan emailed me too and gave me a 'nudge' about actually doing some demonstration calculations based on the ideas in this post: they are at the next post, which I think you'll find interesting.

You said: The size of the step used in the numerical solution can be considered as the difference between 2 initial conditions . But as any chaotic system has at least one positive Ltapounov coefficient , we know that 2 trajectories diverge exponentially .

I understand that, but my background is engineering, so I'm interested in the behaviour at finite time. So I'm happy as long as the time-step size or the numerical method still gives me a 'knob' I can tweak to get arbitrarily close to the solution for arbitrary (but finite) times. The convergence results in this post and this post seem to show that this is quite possible.

You said:Kolmogorov turbulence theory is a kind of attempt at statistical fluid dynamics but it doesn't work well and it fails completely for lower Reynolds numbers.

While that's true, it's also quite understandable considering the assumptions that go in to deriving the scaling relations of Kolmogorov's 'energy cascade', we shouldn't expect it to work at low Re.

You said: So when one sees that in the so called "climatology" people get rid of the space by simply ... AVERAGING (!) and then "analyse" the time series of spatially averaged parameters (like GMST) , one wonders what kind of physics they do .

I agree, boiling the climate down to a single globally averaged number is foolish, and most likely politically motivated. Averaging on smaller scales in space and time and introducing models for the extra terms you introduce is practically useful though (RANS / LES). I'm not really familiar with the internals of the climate models, do they do LES (wasn't Smagorinsky a climate modeler)?

4. Hello

Only fast today .
It is not for ANY finite time . There is an upper bound given by the Heisenberg relation . But even the Heisenberg accuracy/step is completely out of reach of any computer .
So what is practically feasible leads to very short times and if you wanted to go beyond , you'd need to decrease the steps exponentially .
That gives an upper time limit for numerical solutions very fast .
In short there is no converged numerical solution .
Actually you can not even be sure that a limit when the step goes to 0 is near to what you get with non 0 steps .
That's why some people call all these curves numerical artefacts because nobody knows if those curves are parts of solutions restricted to some finite [0,T] interval .
Not really surprising for a chaotic system where even the beginning may change when the values reach , pass some thresholds :)

I don't do numerical models so have little experience with climate models too . But they use LES in the dynamical core .
However one must not think that the climate models "solve" N-S in any sense of the word !
What they basically do is to compute 2 different equilibrium solutions and compare them .
It's 2 snapshots , it is not a movie .

5. I buy the upper-bound part, I kind of experienced that with this little toy problem, trying to converge the solutions further and further out required really cranking down the time-step size to the point that it became impractical (this is a blog after all, it's not worth waiting a week for solutions). And the ensemble where the ICs were perturbed on the order of the round-off error all eventually diverged (at about the same time) no matter what the time-step.

That's why some people call all these curves numerical artefacts because nobody knows if those curves are parts of solutions restricted to some finite [0,T] interval .

This is the part I have trouble with, if I can show that my implementation converges to a manufactured solution, and I can show time-step convergence for an ensemble (with IC's perturbed on the order of the machine precision), then haven't I shown that the trajectory is not a numerical artifact?

I haven't done much reading on 'chaos' per se, do you have any references you could point me at that would clear this up (no hurry, this is just my idle curiosity)?

6. Interesting discussion. My thanks to both of you.

7. Ok, I thought about it some more; I guess you could argue that even if you put your ensemble member's ICs as close as possible given the limits of your machine, there could still be ICs 'between' those that are in different basins of attraction and you'd never know it.

8. Since Tom got me thinking, I found this article on attractors which discusses the effect of uncertainty in initial conditions has on determining which basin an IC is in. Basins can have fractal boundaries, so it could have 'interesting' features even at the scale of the machine precision.

9. Well you are rediscovering the chaos theory ;)
I already mentioned it but would like to stress again the importance of ergodicity .
The Lorenz system is ergodic that's why eventual results obtained with this system can't be generalized to all chaotic systems that might be non ergodic .
Ergodicity means broadly that the trajectories go everywhere in the attractor . In other words with any initial conditions if you wait long enough the trajectory will visit all points .
Hence the notion of pdf and the possibility to treat the system stochastically (of course the individual trajectories can't still be predicted beyond some finite time) .

10. But if the system is NOT ergodic , it means that there are "holes" and "traps" . For such system you can be never sure what it will do for a given initial condition . Even 2 initial conditions separated by the smallest possible interval (the Heisenberg limit) might lead to 2 very different trajectories .
You are right that fractal sets (strange attractors) are typically an example .
I will take an example of convergence difficulty - the Newton-Rhapson iteration .
If you apply it to some complex polynomial (f.ex z^4 - 1) and vary the initial conditions you will obtain a fractal boundary between the points that lead to a convergence to one of the 4 roots .
That means that there will be an infinity of points such as that you can make an interval I as small as you want around these points and for points interior to this interval you will converge to any of the 4 roots .
So actually you are unable to say to what root will converge the Newton Raphson iteration for a given initial point regardless of the smallness of the interval between 2 initial conditions .

The Lorenz attractor is a "well behaved" chaos .
While the attractor itself is strange , its strangeness only translates the already known "banal" chaos feature - unpredictability of individual trajectories .
But its boundary is not fractal so you will always finish on the attractor if you are in the chaotic regime .
Once you are on it , the ergodicity will make sure that you will run through it by visiting every part but the trajectories will NEVER intersect (demonstrated very important result for all chaotic systems) .
This property allows me to mention an amusing paradox showing the problems with numerical integrations .
You choose the smallest possible integration step that the best computer allows .
You take some initial condition and begin to compute . Given the computer accuracy there is a FINITE number of different points it can represent .
Even if this number is very large , if you compute long enough your trajectory will eventually find a point which is EXACTLY the same that it has already visited and from that point on your computer will only periodically cycle this same trajectory .
The computer just violated in the major way the fundamental rule that trajectories never intersect !
And I have just proven that it will always happen with any computer with finite accuracy :)

As for the books , there are of course many of varying difficulty .
Personnaly I like and usually recommend the R.Hilborn's Introduction to chaos and non linear dynamics . Comprehensive , rigorous and gives all the references and ways to go deeper in the matters if one wishes to do so .

11. You gents might appreciate the discussion over here on this paper about the effects of different hardware / software on ensemble results.

From the abstract:
[...] We demonstrate that effects of parameter, hardware and software variation are detectable, complex and interacting. However we find most of the effects of parameter variation are due to a small subset of parameters. [...] We demonstrate the effect of hardware and software is small relative to the effect of parameter variation and, over the wide range of systems tested, may be treated as equivalent to that caused by changes in initial conditions.

12. Interesting comment on control of chaotic systems, which is very closely related to verification with MMS (since control terms show up as forcings in the governing equations).

13. The point Tomas raises here about averages being chaotic deserves a post in the Lorenz63 series; that's something we can show with the toy model.


text:
# Counterexample for Corneil's efficient algorithm for Graph Isomorphism

In the paper An Efficient Algorithm for Graph Isomorphism by Corneil and Gotlieb, 1970 a conjecture was stated upon which the stated algorithm relied for solving GI in polynomial time. Namely:

that the representative graphs exhibit the automorphism partitioning of the given graph

Obviously, this conjecture isn't proven until now (otherwise we would know that GI is in P). My question is whether it was already shown to be false and possibly a counter example was given?


text:
# Time-dependent angular acceleration problem

1. ### Ceenaya19

8
1. The problem statement, all variables and given/known data

As a result of friction, the angular speed of a
wheel changes with time according to
dθ/dt=ω0*e^-σt ,
where ω0 and σ are constants. The angular
speed changes from an initial angular speed
Determine the magnitude of the angular
acceleration after 2.44 s.

2. Relevant equations
dω/dt = $$\alpha$$

3. The attempt at a solution
I've tried differentiating the given expression for omega in an attempt to get the angular acceleration, but that didn't work because $$\sigma$$ is undefined in the problem. I've also tried taking the ln of both sides, but that didn't work either. I tried solving for $$\sigma$$ in terms of ω and ω0, but that didn't work. Finally, I tried just assuming that the acceleration is just constant from t0 to t, but that also wasn't the right answer. So, I have no idea what to try next......

Last edited: Mar 28, 2010
2. Science news on Phys.org
3. ### Ceenaya19

8
I still can't get it O_O


text:
Today, I met the following problem. I wrote an article, and use rake generate to generate the site. Then the console reports:

/home/chico/.rvm/rubies/ruby-1.9.2-p290/lib/ruby/1.9.1/psych.rb:148:in `parse’: couldn’t parse YAML at line 3 column 17 (Psych::SyntaxError)

why?
Then I found the reason:

“\“ should not be written in title of your article. If you really want to write it, use “\\“


text:
# RC circuit frequency response

I am developing a project where I must analyze an incoming signal that was acquired from a microcontroller. The objective is to obtain the main frequency of the incoming signal.

At first, I’m analyzing an RC step response signal. So, basically, I want to obtain the frequency response, from the characteristic charging curve of an RC circuit.

My mentor told me that I could achieve this with the FFT. He told me that if I applied it twice at the incoming signal, I would obtain the frequency response curve. I think I did, but my only issue is the frequency axis. They’re completely messed up. Even when I apply the method to obtain the frequencies, it goes bad. (I know it's wrong because my cut-off frequency is around 11 Hz)

1. Why is it that applying the FFT twice to the signal give the frequency response?

2. Does anyone know how I can obtain the correct frequency axis?

Taking the FFT twice is similar to a method called "cepstrum". It finds the spacing between harmonics. You need to measure an oscillating signal to apply this. I would go back to your mentor for more clarification.

Here is a solution for accurately measuring your charging rate in your diagram: least square fitting to inverse exponential function

Ced

Followup:

I was very impressed by these articles.

From RC Charging Circuit in the section labeled: "RC Time Constant, Tau"

$$V_c = V_s ( 1 - e^{-t/(RC)} )$$

So when $t = R C$ then $\frac{V_c}{ V_s} \approx .63$

Since you already know $V_s$, the technique I referenced earlier is not necessary. Looking at your graph $R C \approx .07$.

From Passive Low Pass Filter in the section labeled: "Cut-off Frequency and Phase Shift"

$$f_c = \frac{ 1 }{ 2 * \pi * R C }$$

Your cutoff frequency can then be found: $f_c \approx 2.27$

From the section labeled: "RC Low Pass Filter Circuit"

$$X_c = \frac{1}{ 2 \pi f C }$$

$$V_{out} = V_{in} \cdot \frac{X_c}{ \sqrt{ R^2 + X_c^2 } }$$

Divide the numerator and denominator by $X_c$:

$$\frac{ V_{out} }{ V_{in} } = \frac{1}{ \sqrt{ \left( \frac{R}{X_c} \right)^2 + 1 } }$$

Substitute in $X_c$ and simplify:

$$\frac{ V_{out} }{ V_{in} } = \frac{1}{ \sqrt{ \left( 2 \pi f R C \right)^2 + 1 } }$$

Now take the log (base 10):

$$\log \left( \frac{ V_{out} }{ V_{in} } \right) = -\frac{1}{2} \log \left( \left( 2 \pi f R C \right)^2 + 1 \right)$$

From the section labeled: "Low Pass Filter Summary"

$$Gain_{db} = 20 * \log \left( \frac{ V_{out} }{ V_{in} } \right)$$

Plug in the log of your voltage ratio to get:

$$Gain_{db} = -10 \log \left( \left( 2 \pi f R C \right)^2 + 1 \right)$$

Your cutoff frequency is below the audible range of Hz, so you are going to attenuate all your frequency. The higher the frequency, the greater the attenuation. To find you fundamental frequency I recommend using a FFT and the frequency calculations I present in my blog articles. You can find the link on my profile page.

• I will. Just to clarify it for me, do you think there is a way to obtain a Bode plot of an RC step response, without having its transfer function? I think, ultimately, this is my question Mar 25 '18 at 12:44
• I'm a little outside my comfort zone answering that. Check this reference out: electronics-tutorials.ws/filter/filter_2.html Mar 25 '18 at 13:06
• Thank you anyways! I've been fighting with this for a long time, but now I have a starting point! Mar 25 '18 at 13:39
• I have worked the math. Your RC is about .07 and your cutoff frequency is about 2.27 Hz (assuming your time scale is in second on your first chat). Do you want me to post the equations or would that be a spoiler? Mar 25 '18 at 15:58
• I'd be very much grateful,if you could Mar 26 '18 at 11:13


text:
# Bloch sphere visualization using Cirq, ProjectQ, PyQuil

Is there a way to visualize the Bloch sphere using Cirq, ProjectQ, and/or PyQuil, as it is possible using plot_bloch_vector in Qiskit?

• TensorFlow Quantum, which primarily uses Cirq (since they are both owned by Google), used qutip.Bloch in this binary classification tutorial instead of any Cirq function, implying that Cirq does not have an equivalent.
• Rigetti, which owns PyQuil, used qutip.Bloch in this training material instead of any PyQuil function, and in this intro tutorial, implying that PyQuil does not have an equivalent.
• ProjectQ does not have any publicly visible Bloch sphere visualization tools in their Github, either. But their software already uses the IBM quantum experience, so I believe they would default to qiskit.plot_bloch_vector.