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3. The coefficient of friction is a constant that relates the normal force between two objects (blocks and table) and the force of friction. Based on your graph (Run 1) from Part I, would you expect the coefficient of static friction to be greater than, less than, or the same as the coefficient of kinetic friction? 4.
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Hint #2: The force of friction F F is equivalent to the normal force F N times the coefficient of friction µ. Equipment: Wooden Flat Plane Large Steel Block Hanging Mass Set Pulley Alternative surfaces: Plastic/ Wood Normal Force = F N Tension = F T Friction = F F Gravitational Force = F g To pulley → 2-Coulomb coefficient of friction. A relative motion between two bodies in contact provides a resistance to this motion which is called friction. Table 1. The measured and calculated values of RM, RT, b and µc from the simulations performed under different constant friction factors with...
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• a range of values is given for coefficients of friction. Table 1 on the next page gives several coefficients of kinetic and static friction for some common materials. coefficient of friction (m) the ratio of the force of friction to the normal force coefficient of static friction (m S) the ratio of the maximum force of static friction to the normal force coefficient of kinetic friction (m K ...
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The range of coefficient of friction in different metal forming applications is not well known and the factors affecting variation are ambiguous. stress values above 4, coefficient of friction values are approximately same for both friction models. Table of contents.
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• variation in friction is due to the variation of friction with surface roughness and BARTON (1976) has proposed that friction of rocks at low stresses can be approximated by the equation: where JRC is the joint roughness coefficient which varies between 20 for the roughest surfaces to zero for smooth surfaces. | 2,808 | 13,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-05 | latest | en | 0.816623 |
http://www.docstoc.com/docs/53530077/Convolution-Codes | 1,435,864,506,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375095668.34/warc/CC-MAIN-20150627031815-00255-ip-10-179-60-89.ec2.internal.warc.gz | 432,460,322 | 63,437 | # Convolution Codes by kjy11098
VIEWS: 288 PAGES: 26
• pg 1
``` Convolution Codes
97.478
1.0 Prologue:
Convolutional codes, why should complicate our lives with them
People use to send voice waveforms in electrical form over a twisted pair of wires. These
telephone voice signals had a bandwidth of 4KHz. If the channel polluted the signal with a bit of
noise, the only thing that happened was that the conversation got a bit noisier. As technology de-
veloped, we digitized the voice signals at 8000 samples per second (twice the highest frequency to
prevent aliasing) and transmitted the bits. If noise corrupted a few bits, the corresponding sample
value(s) would be slightly wrong or very wrong depending on whether the bad bits were near the
most-significant-bit or least-significant-bit. The conversation sounded noisier, but were still dis-
cernible. Someone saying “cat” will not be thought to have said “dog,” and probably would not
even be thought to have said “caught.”
When people started to send data files rather than voice, corrupted bits became more impor-
tant. Even one wrong bit could prevent a program from running properly. Say the noise in a chan-
nel was low enough for the probability of a bad bit to be 1x10-9 i.e. the chances of a bit being
correct is 0.999999999 (nine 9’s). The chances of 1000 bits being all correct is 0.999999 (six 9’s)
and the chances of 106 bits being all correct is 0.999 (three 9’s). A 1 megabyte file (8x106 bits) has
almost a 1% chance of being corrupted. The reliability of the channel had to be improved.
The probability of error can be reduced by transmitting more bits than needed to represent
the information being sent, and convolving each bit with neighbouring bits so that if one transmit-
ted bit got corrupted, enough information is carried by the neighbouring bits to estimate what the
corrupted bit was. This approach of transforming a number of information bits into a larger
number of transmitted bits is called channel coding, and the particular approach of convolving the
bits to distribute the information is referred to as convolution coding. The ratio of information bits
to transmitted bits is the code rate (less than 1) and the number of information bits over which the
convolution takes place is the constraint length.
For example, suppose you channel encod-
ed a message using a convolution code. Suppose Convolution
00101101 channel
you transmitted 2 bits for every information bit encoder
a b
(code rate=0.5) and used a constraint length of 3.
Then the coder would send out 16 bits for every
11 10 00 01 01 00
8 bits of input, and each output pair would de-
pend on the present and the past 2 input bits (con-
straint length =3). The output would come out at a 3 bits in the input stream generate 2 bits
twice the input speed. in the output stream.
b Take the most recent of these input bits
Since information about each input bit is plus one new input bit and generate the
spread out over 6 transmitted bits, one can usual- next 2 output bits.
ly reconstruct the correct input even with several Thus each input bit effects 6 output bits.
transmission errors. FIGURE 1
Authors, Fred Ma, John Knight January 26, 2001 1
Convolution Codes
The need for coding is very important in the use of cellular phones. In this case, the “chan-
nel” is the propagation of radio waves between your cell phone and the base station. Just by turn-
ing your head while talking on the phone, you could suddenly block out a large portion of the
transmitted signal. If you tried to keep your head still, a passing bus could change the pattern of
signal. In both cases, the SNR suddenly drops deeply and the bit error rate goes up dramatically.
So the cellular environment is extremely unreliable. If you didn’t have lots of redundancy in the
transmitted bits to boost reliability, chances are that digital cell phones would not be the success
they are today. As an example, the first digital cell system, Digital Advance Mobile Phone Service
(D-AMPS) used convolution coding of rate 1/2 (i.e. double the information bit rate) and constraint
length of 6. Current CDMA-based cell phones use spread-spectrum to combat the unreliably of the
air interface, but still use convolution coding of rate 1/2 in the downlink and 1/3 in the uplink (con-
straint length 9). What CDMA is, is not part of this lab. You can ask the TA if you are curious.
2.0 Example of Convolution Encoding
XOR Output u1
MUX Double
Shift register speed
0 output
Input information Yester- Day
Today before 1 z
bit stream x days yester-
bit
bit days
The constraint length is 3. bit
The output is effected clock
(constrained) by 3 bits.
The present input bit
and the two previous bits
Output u2
stored in the shift register. XOR
FIGURE 2
This is a convolution encoder of code rate 1/2 This means there are two output bits for each
input bit. Here the output bits are transmitted one after another, two per clock cycle.
The output u1 = x(n)⊕ x(n-1)⊕x(n-2).
Here x(n) is the present input bit, x(n-1) was the previous (yesterdays) bit, etc.
The output u2= x(n)⊕ x(n-2).
The input connections to the XORs can be written as binary vectors [1 1 1] and [1 0 1] are known
as the generating vectors for the code.
Authors Fred Ma, John Knight January 26, 2001 2
Convolution Codes
2.1 The Encoder as a Finite-State Machine
The correlation encoder can be described as a Mealy
x=0/ z=00
machine. The state is the two bits in the shift register.
Assume shift register was reset to zero and the first
data bit x(n) = 1. Then:- x=1 A=00
State= 00 = A = [x(n-1),x(n-2)] z=11
Output z=[u1,u2] x=0
u1 = x(n)⊕ x(n-1)⊕x(n-2) x=1/ z=
00
z=11
= 1⊕ 0⊕ 0 =1 B=10
u2 = x(n)⊕ x(n-2). x=0/ z=
10
= 1⊕ 0 =1 C=01
z=[u1,u2]= 11 x=1
z=01
After the clock, state bit x(n-1)=0 will shift right
into x(n-2), the input x(n)=1 will shift right into x(n-1), x=0
z=01
and the next state will be 10 = B. x = x(n) D=11
state =
[x(n-1),x(n-2)]
2.2 The Trellis Encoding Diagram
FIGURE 3 x=1/ z=10
To get the trellis diagram, squash the state diagram
NOTATION TRELLIS
so A, B, C and D are in a vertical line. This line represents
input/output
the possible states at time t=n (now). Make time the hori- x/z 0/00
A A
zontal axis. Put another line of states to show the possible 1/1
1
states at t=n+1. 1 input
11
Then add the transitions to the diagram. Make the B 0 B
0/
0 input
them all go from states @ t=n to states @ t=n+1. Thus the 1/0
self loop at state A in the state graph becomes the horizon- 0/1
0
1/
tal line from A@t=n to A @t=n+1 in the trellis diagram. C C
01
The complete trellis diagram extends past t=n+1 to as FIGURE 4 1
0/0
many time steps as are needed. 1/10
D D
Suppose the encoder is reset to state A=00, and the t=n t=n+1
input is 1,0,1,1,0,0. By following the trellis one sees that
the output is 11 10 00 01 01 11. Also it passes through
states A, B, C, B, D, C ending in A @ t=n+6.
FIGURE 5
More Complete Trellis Diagram
0/00 0/00 0/00 0/00 0/00 0/00
A A A A A A A
1/1 1/1 1/1 1/1 1/1 1/1
1 1 1 1 1 1
11
1
1
11
11
11
1
1
B 0 B 0 B 0 B 0 B
0/
0/
0/
0 B
0/
0 B
0/
0/
1/0 1/0 1/0 1/0 1/0 1/0
0/1 0/1 0/1 0/1 0/1 0/1
0 0 0 0 0 0
1/
1/
1/
1/
1/
1/
C C C C C C
01
C
01
01
0
01
01
1
1 1 1 1 1
0/0 0/0 0/0 0/0 0/0 0/0
1
D 1/10 D 1/10 D 1/10 D 1/10 D 1/10 1/10
D D
t=n t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6
Authors Fred Ma, John Knight January 26, 2001 3
Convolution Codes
2.3 First Exercise: A Convolution Encoder.
1. Problem: Encoding a number
• Take the last 4 digits in your student (the least significant digits).
• Convert them to a hexadecimal number (Matlab has a function dec2hex).
• Convert the hexadecimal number to binary (12 to 16 bits).
• Use this as data for the encoder below. Feed in the least significant bit first. Also reset the
shift register to 00 before you start.
• Calculate the output bits and states when one encodes these bits using a code rate
1/2, constraint length 3 encoder with generating vectors [111] and [101].
Tabulate how the state and output values change with each clock cycle.
Clock cycle 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
input bit
shift reg (state) 00
Output [u1,u2]
2. Problem: Draw the circuit for an encoder which has:
a code rate =1/2,
constraint length of 4,
generating vectors [1111] and [1011], where the “0” means no connection to x(n-1).
3. Problem: Draw the state graph for the above constraint length 4 encoder, and then the first
three time steps of the trellis diagram for the above constraint length 4 encoder.
2.4 First Lab: Design a Convolution Encoder in Verilog
The constraint length 4 encoder circuit is a finite-state machine. It is also a shift register.
One can code it following the standard finite-state machine model with three-bit states, or as a shift
register.
reg [2:0] State;
always @(posedge clk)
begin
State <=State >> 1; // Right shift 1 position
end
The test bench is not part of the circuit. It supplies the input sig-
nals and may compare output signals with those that are expected. Test Bench module
Test benches are easier to write and use if they are synchronous. This
means they always send out signals slightly after the clock (or at least clk reset x u1 u2
never at the same time as the clock). It also means a writing style with
few #n delay.
ConvEncode module
initial begin
... #11 x=1; Poor
#10 x=0; FIGURE 6
#10 x=1;
#20 x=0;
#10 x=1;
...
Authors Fred Ma, John Knight January 26, 2001 4
Convolution Codes
2.4.1 A Synchronous Test Bench
In a synchronous test bench, the test signals are timed by @(posedge clk) statements rather
than each having its own timing. There are only two delays here, one to set the clock period, and
the other to delay the input signal x so it’s changes are obviously past the clock edge.
Note there are things not included here, like reset.
module SyncTestBench;
reg [1:8] data; //Fill this with the data stream to be encoded.
// Note the first bit is on the left
reg x, clk;
integer I;
initial
begin
I=1;
data=8'b1010_1101; // Underscore has no meaning except
// to visually space the bits.
clk=0;
forever #5 clk=~clk;
end
// send in a new value of x every clock cycle
always @(posedge clk)
begin
if (I==9) \$finish;
// The #1 makes x change 1ns after the clock and never on the clock edge.
// The nonblocking symbol “<=” on I ensures that any other clocked module using
// I will grab the same I as this procedure, that is before I is updated to I+1.
x<= #1 data[I];
I<=I+1;
end
endmodule
For the constraint length 3 system, you must have the test bench automatically compare your
1. Write a finite-state machine encoder for the constraint length 3 system.
2. Write a shift-register based encoder for the constraint length 4 system (Sect 2.3 prob 2).
3. Write a synchronous test bench so the other two modules can be simulated.
3a) Compare the result for the data 1 1 0 1 0 0 1 0 1 1 0 0 0 0
Ans: Constraint length 3 encoded data-
11,01,01,00,10,11,11,10,00,01,01,11,00,00,00,00.
Ans: Constraint length 4 encoded data-
11,01,01,11,01,11,00,10,00,10,01,00,11,00,00,00,00.
2.5 Lab and Problem Rules
The Viterbi decoder design problem will be done by a groups of three persons.
The number of exercises is (usually) divisible by three so one person in each group does
one-third of the exercises. The three are to be submitted together with the name of the person do-
ing each part attached to the part.
All members of the group are responsible for knowing how to do each exercise. Related
problems will appear on the examination. Also the TAs will ask other members of the group to ex-
plain the submissions they did not submit. This will effect their mark as much as their submission.
Authors Fred Ma, John Knight January 26, 2001 5
Convolution Codes
3.0 Convolution Decoder
The next part of the project will be to design a convolution decoder to retrieve the informa-
tion bits from the transmitted bits. It should succeed even in the presence of some errors in the
transmitted bits. The implementation we will use is called a Viterbi decoder.
3.1 Decoding Using the Trellis Diagram
Consider a decoder that receives the transmitted signal 11 01 01 00 10 11 going from t=n to
t=n+6. Assume the trellis was reset to state A (00) at the start. One goes through the trellis as be-
fore, only for decoding the numbers are output\input. So the first input, 11, gives a decoded output
of 1 and takes the machine to state B.
At t=n+1 in state B, the next input 01 causes a 1 output and a change to state D.
Trellis Diagram for Decoding (Receiving) FIGURE 7
Received 11 01 01 00 10 11
input
Decoded 1 1 0 1 0 0
output
00/0 00/0 00/0 00/0 00/0 00/0
A A A A A A A
11/ 11/ 11/ 1/1 11/ 11/
1 1 1 1 1 1
/0
/0
/0
/0
/0
/0
11
1 B
11
B 1 B
11
1 B 11 1 B
11
1 B
11
00/ 00/ 00/ 00/ 1 B
00/ 00/
10/ 10/ 10/ 10/ 10/ 10/
0 0 0 0 0 0
01
01
01
01
01
01
C C C C C C C
/1
/1
/1
/1
/1
/1
0 0 0 0
01/ 01/ 01/ 01/ 01/
0
01/
0
D 10/1 D 10/1 D 10/1 D 10/1 D 10/1 10/1
D D
t=n t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6
3.1.1 The Hamming Distance FIGURE 8
This distance is used to show how far apart two input
binary numbers are. Compare the bits in the same posi-
00 2 A
tions in the two numbers. The number of positions that A A
11 0
are different is the Hamming distance (h). Hamming distance
Thus 11and 01 are distance 1 apart (h=1),
B B B
1001001 and 1001010 are distance 2 apart (h=2).
10
1
Applying the Hamming Distance to Decoding
01
C C C
1
Suppose the first four received bits have an error
so instead of 11 01, one receives 11 11. On the trellis
in Fig. 8, there are two choices leaving state A, one for D D D
input 11 0 and the other for input 00 2 . The number in t=n t=n+1 t=n+2
the box is the Hamming distance between the received Instead of input/output (i.e. 11/1)
input and the bits for the transition. It is clear one we now show input Hamming distance
should make the transition from A⇒B.
The next input has an error. Note there are no 11 or 00 paths leaving state B. Both possible
paths, 10 or 01, are at Hamming distance 1. At this time either transition looks equally likely, but
wait!
Authors Fred Ma, John Knight January 26, 2001 6
Convolution Codes
Trellis Diagram FIGURE 9
error
Received 11 11 01 00 10 11
input
00 2 A 00 0 00 1
A A A A A A
11 0 11 1 11
0
1
1
0
B B B
11
1 B 0 B
11
1 B
11
B
00 00 00
10 10 10 10
1 1 0 1
01
01
01
C C C C C C C
1
1
1
010 1
Outputs 01
is 0 D D D D 10 0 D 10 1 D
is 1 t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6
At t=n+2, if one starts from D, then h=0 for the path to state C. However if one starts from C
one has h=1 for either the path to A or to B.
Thus at t=n+1 the proper path was not obvious, at t=n+2, the choice is clearer. We will chose a
path through the trellis based on the cumulative Hamming distance, which is the sum of the Ham-
ming distances as one steps along a path through the trellis.
Figure 10 shows how the Hamming distances sum as one goes down various paths through
the trellis diagram. At t=n+5, one path has a total distance of 1 from the input data. The others
FIGURE 10
00 2 A 1+ 1+ 0 3 1
A A A A A
0 3 1
+1
1
Lowest cummultative
0 1
Hamming distance
B B B 1 B 0+ B 1 B
4
1+ 1+ 4
1 1+ 1
1+ 0
C C C 0 C 1 C C
1
1+ 1+
1
+1
D D D D D 3 0 D
t=n+1 t=n+2 t=n+3 t=n+4 t=n+5
The sum of the Hamming distances is shown like 1 + 1 + 0 until this gets too messy.
Then the sum, and the distance for the current step, are shown in a hexagon and a box like 2 1
have a distance of 3 or 4. Thus the most likely path is A⇒B⇒D⇒C⇒B⇒C with a cumulative
distance of 1, and the corresponding output data is 11010 (Recall trellis edges represent a
receiver output of 0, and edges represent an output of 1).
References:
Bernard Sklar, Digital Communications Fundamentals and Applications.
B.P.Lathe, Modern Digital and Analog Communication Systems, Holt, Rinehart & Winston, 1989
Prof. Ian Marsland, http://www.sce.carleton.ca/courses/94554/
Click on "Convolutional Codes." You will need Ghostview (gv) to read the Postscript file.
Authors Fred Ma, John Knight January 26, 2001 7
Convolution Codes
4.0 The Viterbi Decoder
The decoding example shown above has to follow multiple paths through the trellis, and re-
member them for future decisions. For larger decoders, such the cell phone ones with constraint
lengths (shift-register lengths + 1) of 6 and 9, the number of paths can get quite large. This means
a decoding ASIC may take considerable storage, and one may have to use slower, more compact
RAM type memory rather than the faster flip-flops.
Viterbi developed an algorithm 1n 1967, which allows the many paths to be discarded with-
out tracking them to completion. He noticed that if two paths merge to one state, only the one with
the smaller cumulative Hamming distance, “H,” need be remembered. The other one can never be
part of the most likely path.
This means that with the constraint length 3 (shift-register length 2) system in the previous
examples only have to remember 4 paths. In general a constraint length K system will have to re-
member 2K-1 paths. In theory, the path length should go for the length of the message in order to
get the true maximum likelihood path. However it turns out that path lengths of 4 to 5 times the
constraint length can almost always give the best path.
The next few figures show how the decoder picks the best path, even when there are errors.
FIGURE 11
error At t=n
input 01→11 Starting at state A, there are two possible paths.
The boxes 2 show one step Hamming distance
00 2 00 2 4 A The hexagons 4 show “H”, the cumulative
A A Hamming distance
11 0 11
0 At t=n+1
2
There are 4 paths out to t=n+2
B B B Since the input has an error,
10
1 no path has a zero Hamming distance “H”.
1
C State
01
C C C
1
4 Cumulative Hamming distance
1
2 Hamming distance for this step
D D D 00 Rec’d input to make this transition
t=n+1 Output a 0 if this transition is used
t=n t=n+2 Output a 1 if this transition is used
error
input 01→11 01
Going out to t=n+3
The paths temporarily double to 8
00 2 002 4 00 1 5
A A A 11 A There are two paths to each node.
11 0 11 1
0 5 2 One has a larger cumulative distance.
2
1 The larger “H” path can never be
B B
10 B 11 1 2 B the most likely path, hence we will
1 00
10
1 erase it.
01 2 6
01
C C C C
0
1
1
0 1
2
01
D D D 10 2 3 D
t=n t=n+1 t=n+2 t=n+3 FIGURE 12
Authors Fred Ma, John Knight January 26, 2001 8
Convolution Codes
FIGURE 13
Here just the cumulative Hamming distances Here the unneeded paths are eliminated.
are shown so it is easy to see which paths This is all important information up to
should be erased. to t=n+3.
Rec’d 11 error error
01→11 01 11 01→11 01
2 4 5 2
A A A A A A A A
0 5 2 0 2
B B B B B B B B
2 2
1 1
6
C C C 1 C C C C 1 C
1
1
2
3 2
D D D D D D D D
t=n t=n+1 t=n+2 t=n+3 t=n t=n+1 t=n+2 t=n+3
FIGURE 14
error 2nd error
Rec’d 11 01→11 01 00→01 Entering t=n+4,
input
00 Eight paths are created temporarily
A A A A 11 1 3 A
0 2 1 Again at each node, only the
2 3 2
lowest cumulative Hamming-
1 distance path can be part of the
B B B B 10 2 11 B
2 1 2 most likely path so four of the eight
1 00 paths will again be eliminated.
01 4
C C C C The second error has made
0
1 C 2
four paths all with equal chances
1
0
01 of being the most-likely paths
2
2
D D D D 10 2 4 D (four paths with 2 )
But see what happens next.
t=n t=n+1 t=n+2 t=n+3 t=n+4
FIGURE 15
error
Rec’d 11 01→11 01 2nd error
input 00→01 10
Entering t=n+5
A A A A A 00 1 3 Eight new paths are
111 A
0 2 created, keep the four
2 2 3 3
lowest H ones enter
B B B B B 1 ing each state
2 11 1 3 B
2 00 Since two paths
1 10
01 0 2 have the same H
C C C 1 C C C as they enter B,
2
we have no way
1
2 2 4 of choosing a
4
2 01 better path.
2
D D D D D 10 0 2 D Pick one randomly;
t=n t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 Here we choose
There are two most likely paths the one A⇒B.
Note path dies out.
(with H=2). Down from four at t=n+4
Authors Fred Ma, John Knight January 26, 2001 9
Convolution Codes
error 2nd error Entering t=n+6
Rec’d 11
input 01→11 01 00→01 10 11
Note how the
3 correlation has
A A A A A A 100 2 5 A continued to act.
0 10
2 2 2 3 2 We are now down
3
to one most likely
B B B B 0
2 B B
11 2 4
B path.
2
1 00
10
2
01 1 4
C C C C C C C
1
1 2
1 3
1
4
2
01
2
D D D D D 2 D 10 1 3 D
t=n t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 FIGURE 16
error FIGURE 17
Rec’d 11 2nd error
input 01→11 01 00→01 10 11 11
A A A A A 3 A A 100 2 4 A
0 10
2 2 2 3 3 2 2 3
B B B B B B B 0
2 11 2 5 B
2
1 00 0
1
2 01 1 4
C C C C C C C C
1
1 2 3
4
1
1
4
2 01
2
D D D D D D 3 D 10 1 4
2 D
t=n t=n+1 t=n+2 t=n+3 t=n+4
t=n+5 t=n+6 t=n+7
Entering t=n+7: The one best path is getting fairly clear.
The most likely path ends at B and a close second at A. The paths to C and D have H= 4
Retrieving the Original Message FIGURE 18
error 2nd error
Rec’d 11 01→11 01 00→01 10 11 11 10
input
A A A A A A A A 4 A
0
4
B B B B B B B B B
2
C C C C C C C C C
4
Common path
D D D D D D D D D
t=n t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
Entering t=n+8: One can be fairly sure the best path at t=n+8 ends at state C.
Follow the path back from C -> B-> A-> C-> B-> C-> D-> B-> A. Also follow the paths
back from A, B and D. No matter what state you start in at t=n+8, all paths come together
when you get back to t=n+1. From the solid (for “0”)and dashed (for 1) lines along
the path one can decode the probable original message as 11010010 (travelling t=n to t=n+8).
Authors Fred Ma, John Knight January 26, 2001 10
Convolution Codes
As illustrated above, the Viterbi decoder can decode a convolution code for a message of N
bits in the presence of some errors.
If two branches entering a state have equal “H,” then the code is unable to tell if one path is
more likely than another. Pick one path at random.
FIGURE 19
4.1 Exercise 2: A Viterbi Decoder Design 0/00 h00
HA A A HA
1. Problem: Using the algorithm 1/1
11
1
h
h1
A typical step in the trellis decoder is shown. 1
11
HB B B HB
0/
The cumulative Hamming distances H at each trellis step 0
0h 0
are HA, HB, HC, and HD. 1/0 /1
0
0h
The Hamming distance for each edge are given sub- 10
1/
HC C C HC
01
scripts matching the input which makes them 0. Thus the
1
edge from A to A , and C to B both use the symbol h00. 1h 0
h 01
0/0
If the input is 00, h00=0, If the input is 10 or 01, h00=1, if HD D 1/10 h10 D HD
the input is 11, h00=2.
t=n t=n+1
a) For the first lab partner: Starting at t=n+8 with
Input = 00 FIGURE 20
an input of 00, as in Figure 20, calculate and fill
h00= in the values of hij and hence the Hk for t=n+9.
HA=4 A A HA=
h1 b) For the 2nd lab partner: Starting at t=n+9, us-
1= ing the Hk from step a) and input 10, calculate
HB=4 B B HB=
=
and fill in Figure 21.
11
=
h
h 00
h FIGURE 21
10 = Input =10
h 01
HC=2 C C HC= h00=
=
HA= A A HA=5
1
=
h0
HD=4 D h10= D HD= HB= B B HB=
t=n+8 t=n+9
FIGURE 22 HC= C C HC=
Input =11
HD= h10=
HA= A A HA'= HC + h11 = 2 D D HD=
t=n+9 t=n+10
HB= B '
B HB=
h1 c)For the 3rd partner: Start at t=n+10, with
0=
input 11 and use the Hk from step b). Let the
HC= C '
C HC= new Hk at t=n+11 be written with a prime i.e.
= HA HB HC and HD Fill in Figure 22 but put
' ' ' '.
h 01
HD= D '
D HD= in an expression, as well as a number, for
each Hk. This has already been done for HA '.
t=n+10 t=n+11 Only the better path is written here.
Pseudocode
This is Verilog in which the syntax is not critical. For example begin, end and semicolons
may be omitted if the meaning is clear to the reader. In pseudocode the comments are often
more important than the code.
Authors Fred Ma, John Knight January 26, 2001 11
Convolution Codes
2. Problem: In parts a) and b) you will write pseudocode to calculate the Hamming distances
for each step.
Let the two input bits In[1:0] be y, x.
Let the Hamming distances associated with the eight trellis edges for this step be h00, h01,
etc.
For part b) use reg /*wire*/ [1:0] h00, h01, h10, h11;
a) Calculate these distances using a case statement:
case ({y, x})
2'b00 : begin h00=0; ... h11=2; end
2'b01 : ...
b) Another partner should use Boolean algebra to calculate them as 2-bit binary numbers.
Example:
h00[1] = y&x ; h00[0] = y^x;
3. Problem: Pseudocode to update the Hk in going from step t=n to step t=n+1.
a) Use if statements to calculate the Hk' to be associated with the four states at t=n+1. Use
HAnext instead of HA' since Verilog cannot handle quotes.
if (HA+h00 < HC+h11) then HAnext= HA+h00; else ...
4. Problem: The flip-flop procedure.
Write a procedure to clock the flip-flops and store the necessary data.
Combinational logic in parts 2 and 3 calculated the D inputs for the flip-flops. For example:-
HAnext = ... HA + h00 ....
This procedure stores the data i.e it clocks HA <= HAnext.
Don’t put combinational logic in a flip-flop procedure, and don’t forget a reset.
5. Problem: When is the output correct?
Experience has shown that all backward paths converge into one if one traces them back 4 or
5 times the constraint length. Using the paths in Figure 18, you will show for this example
that if one traces back far enough it does not matter which path one follows.
a) Take a copy of Figure 18. Start at t=n+8; start at each state in turn and colour backwards un-
til you reach t=n or until you hit previous colouring. At what time (t=n+?) do the paths all
converge?
b) Look at Figure 14 in which the ending time is t=n+4.
• Using data available at t=n+4 could you say, with Test encoder decoder
Bench
confidence, what the original data bit was between
t=n and t=n+1? Why not? original data bit
encoded, 2 bits for 1
c) Take a copy of Figure 17. Start at t=n+7; start at each decoded output, back to 1 bit.
state in turn and colour backwards until you reach t=n
or until you hit previous colouring. At what time (t=n+?) do the paths all converge?
In communications latency is the term for the time difference between the time the input
signal was received and the output signal is sent out.
d) If a decoder had to wait until all paths converged before it had confidence it could send out a
correct output, what would the latency be in clock cycles?
There are two answers for c):
(i) The latency as observed for typical data stream as shown in Figure 18 and 17.
(ii) The latency, mentioned earlier, that experience has shown gives the most likely bit for
almost all cases. State the latencies for both (i) and (ii).
Authors Fred Ma, John Knight January 26, 2001 12
Convolution Codes
e) If the decoder delays the signal by 12 to 15 clock cycles, would anyone care assuming:
• The signal was a digitized phone conversation?
• The signal was a www page?
• The signal was a digital TV signal?
6. Problem: How to backtrack.
Figure 23 is the same as Figure 18 except the numbers are all removed. It still contains
enough information to trace back from t=n+8.
Retrieving the Original Message FIGURE 23
A A A A A A A A 4 A
4
B B B B B B B B B
2
C C C C C C C C C
4
D D D D D D D D D
t=n t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
Figure 24 shows a trellis decoder only. It gives no information about the data. Figure 23
shows paths, but when you trace back to the area between t=n+2 and t=n+3 you cannot tell
from the figure what the data was.
a) Use Figure 24 only. If the decoder was in state C at t=n+3, what was the original data (be-
fore encoding) between t=n+2 and t=n+3? (The obvious answer is right.)
If it was in state A at t=n+3, what was the original data between t=n+2 and t=n+3?
If it was in state B at t=n+3, what was the original data between t=n+2 and t=n+3?
If it was in state D at t=n+3, what was the original data between t=n+2 and t=n+3?
orig. data FIGURE 24
Trellis Diagram with no signal knowledge superimposed
is 0
is 1 A A A A A A A
B B B B B B B
C ? C C C C C C
D D D D D D D
t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
b) Figure 25 is the same as Figure 24 except some little squares have been drawn associated
with each state in each time step.
Authors Fred Ma, John Knight January 26, 2001 13
Convolution Codes
• Lab partner one should fill a minimum of information in each square. This information
would allow lab partner two to back trace knowing that Hc had the minimum Hamming dis-
tance at t=n+8. Thus by looking only at Figure 25 partner two, starting at state C at time
t=n+8, should be able to tell what the original bit was between t=n+2 and t=n+3. You may
establish some conventions like a 1 in the state C box means .... However they must be in-
dependent of the data.
• Using the same figure (Fill in the boxes at t=n+3 if you have not done so already) partner
three should be able to determine the original data bit between t=n+1 and t=n+2.
You should hand in the filled in Figure 25 and your list of conventions.
FIGURE 25
Trellis Diagram on which you will superimpose signal knowledge
A A A A A A A
B B B B B B B
C ? C C C C C C
D D D D D D D
t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
c) How many bits per step must be stored to allow for backtracking?
d) Describe the Verilog storage declaration for flip flops to hold the backtracking information.
One dimension of the array will be the latency.
4.2 Extracting The Original Data.
Consider Figure 26. The path with the lowest cumulative Hamming distance H , starts at
state C at t=n+8 with H=2. Backing up would take the path to B at t=n+7. The edge is a solid line
which seems to say the original data was 0. Unfortunately we can’t be sure of this. Because of the
convolution code, this path’s H of 2 could increase in the next few cycles and another path might
get the lowest H.
However if one goes back to t=n+2 and travels ahead in time, only paths that start at D or C
make it all the way to t=n+8. The others die out. Only the path from D has H=2 at t=n+8, thus we
are fairly sure the edge from D at t=n+2 to C at t=n+3 is on the most likely path and the original
data between these two clock edges was 0 (a solid line is 0, a broken line is 1).
This illustrates why we want to wait six (or usually more) cycles before sending out the out-
put. If we are at time t=n+8, we can be confident that the “0” data at t=n+2 is the most likely.
Carrying the Data Forward Instead of Tracing the Path Back
Tracing back is a long process and the full trace must be done every data cycle. The back
trace could be done as a FSM but this is convenient only if the clock runs several times faster than
the data rate. However it turns out one can avoid the back trace, and allow the data rate to equal the
clock rate, by sending the data forward through a shift register.
Authors Fred Ma, John Knight January 26, 2001 14
Convolution Codes
FIGURE 26
input error 2nd error
rec’d 11 01→11 01 00→01 10 11 11 10
A A A A A 3 A A A 100 1 4 A
0 3 2 2 11
2 2 4
3
3
B B B B B B B B B
2 2
1 1
4
2 01 00 2
C C C C C C C C C
1
2
1 2
On best path 3
4
2
D D D D D 2 D 3 D 4 D D
t=n t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
One may not be sure which data bit is best at t=n+2. However if one traces the paths forward
from t=n+2, only two paths survives to t=n+8, and one has a much smaller H.
Figure 27 shows the essence of this circuit drawn to show feeding forward the ones and ze-
ros as perceived from the data at t=n+2. This data is shifted forward through the flip flops and is
not changed by later inputs. However the path through the shift register is controlled by these later
inputs. At t=n+8, the decoder can supply the data bit originally received at t=n+2. The desired bit
is at the end of the shift register, in the flip flop corresponding to the state with the lowest H .
Note this is a schematic showing connections. However, for the data in the flip-flops and the
appropriate mux connections, each column represents a different time.
FIGURE 27
Rec’d 01 01 10 11 11 10
input
A A A A A A A
4
0 0 0 0 1 1
B B B B B
4
B B
0 1 0 0 0 0 1
C
1 C C C C C
2 C
1 0 1 0 1 0 0
4
D
0 D D D D D D
1 1 1 1 1 0
t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
The data bits at t=n+2 are shifted right each clock cycle. Thus the 0 bit here at the lowest H
at t=n+8 is the best choice for the data bit at t=n+2. The advantage of the shift register is
that one does not have to backtrack to know that the data on this path at t=n+2 was also 0.
Figure 28 shows the same thing as Figure 27 except the muxes are not specifically shown. It
shows how the 4 data bits, inserted in the four best paths at t=n+2, can be shifted from flip-flop to
flip-flop down the paths one clock cycle at a time. They arrive at the output at t=n+8.
At each cycle the data bit is shifted right into a state controlled by later input data. These
forward paths may disappear, or fork. However the minimum H path never stops, so the “0” inject-
ed into it at t=n+2 appears at t=n+8 in state C, which is the minimum H= 2 path.
Authors Fred Ma, John Knight January 26, 2001 15
Convolution Codes
Summary:
Compare Figure 28 with Figure 14. Note the path from t=n+2 to t=n+8 is the same. The dif-
ference is the data bit ‘0’was shifted through and comes out at t=n+8 without backtracking.
2nd error FIGURE 28
Rec’d 01
input 00→01 10 11 11 10
A A A A A A 00 1 4 A output
2 0 0 0 0 1 111 1
B 4 B
0 2B
1
B
0
B
0
B
0
B
0 1
output
C
1 C C C C C
10
01
0
2 C best
1 10 1 0 1 0 0
output
2
0 2D
4
D D D D D D output
1 1 1 1 1 0
t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
At t=n+3, one is probably in state C, but because of errors one might be A, B, or D.
If one is in state C, the decoded data between t=n+2 and t=n+3, would be 0. In state A
it would be 0, and in B or D it would be 1. We will shift these data bits forward through the
paths of the trellis diagram, storing the data in flip-flops during each clock cycle.
Most of the paths die out.
At t=n+8, the flip-flop beside state C indicates there was a 0 in this path between
t=n+2 and n+3. This is the same result one got in Figure 18 by tracing the path backwards.
FIGURE 29: The 0,1,0,1 between t=n+3 and n+4 is carried forward to be used at t=n+9
Rec’d 2nd error
input 00→01 10 11 11 10 00
A A A A A A A
4 00 0 4
0 0 1 1 1 1
B 0B B B B 4 B B
1 0 0 1 1 10 1
0 21
C
1C C C C
2
C
00
0
5 C
00 1 1 0 1 1
01
4
1
D
1D D D D D D
5
1 1 1 1 1 1
t=n+3 t=n+4 t=n+5 t=n+6 t=n+7
t=n+8 t=n+9
The data capture at t=n+3 is carried forward along the paths, if they don’t break. Note the
selected trellis paths are the same as in the previous figure, but the data stored in the flip-flops is
different.
Figure 29 shows the next cycle where the 0,1,0,1 data bits between t=n+4 and n+5 are inject-
ed into four paths at t=n+5. Although the paths look the same as in Figure 28, the bits travelling
down them are different unless they coincide by accident. Later we will see that this data can time-
share flip-flops with the data in the previous figures.
Figure 30 shows the bits that would be injected into flip-flops at t=n+5. This bits would trav-
el down the paths and be available for output at t=n+10.
Authors Fred Ma, John Knight January 26, 2001 16
Convolution Codes
Figure 30 shows another point. Both possible paths coming into each state at t=n+5 are
shown even though the Viterbi algorithm will discard one. This shows that no matter which path is
discarded states A and C always receive a “0” and states B and D always receive a “1.” Thus there
is no need to actually use flip-flops to store the data. It can be hard wired.
Solid lines are 0 Rec’d FIGURE 30
input 10 11 11 10
dashed lies are 1
Output will not come until t=n+10
A A A A 4 A
Note that both paths going
into state A correspond
00 0 1
4
1
B B B B B
to a data bit of “0,” thus
one can permanently
associate state A with “0.”
11 0 0 1
Also state C can be C C C C 2 C
associated with “0”
and, B and D with “1.”
00 1 0 0
1D
4
In other words the first D D D D
column of flip-flops is 1 1 1 0
redundant.
t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
FIGURE 31
Rec’d 2nd error
input 01 00→01 10 11 11 10
section 3
A A A A A A A
4
0 0 0 0
00 0
4
B B B B B B B
1 0 0 0 0 0
1 2
C C C C C C C
00 1 0 0 0 0
4
D D D D D D D
Useless ff 11 1 1 1 1 0
t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
Figure 31 is like Figure 28 except it shows how the forward path is controlled by multiplex-
ers. The state and inputs at t=n+k control the mux which selects what will be clocked into the flip-
flop at t=n+k+1. This will turn out to be the same logic used to select the path with the lowest H .
Figure 31 looks like it has 20 MUXs and 24 flip-flops just to forward 4 bits. However:
a) There is no need to use flip-flops to store the initial constants which are always 0,1,0,1.
b) The four bits injected at t=n+3 need only four flip-flops at any time. If those 4 bits are shift-
ed right the first flip-flops become available to hold data injected later.
The complete circuit can be arranged as an unconventional shift register. See Figure 33.
Figure 32 shows data injected at three different time steps travelling down the paths. At any
given time only 4 flip-flops are in use for today’s input set. However 4 more are used for the previ-
ous set, and 4 more for the set before that.
To get Figure 33 from Figure 32, note that at say t=n+6 only one set of flip-flops and MUXs
is active in each section. Thus at t=n+6 one can overlay the three sections of the figure and ignore
Authors Fred Ma, John Knight January 26, 2001 17
Convolution Codes
components operating at any time except t=n+6. The top of Figure 32 has two more sections which
were left out because the figure was too big already. When the five sections are overlaid, and one
extracts all the registers active at t=n+6, one obtains Figure 33
FIGURE 32: To show three shift registers can be compacted into one
Only 4 flip-flops in each section are active at say t=n+6, thus only the shaded flip-flops
are in use. The others only show what is going to be stored later or what was stored
earlier. This means that at t=n+6 one can erase all but the shaded flip-flops.
The data is collected between t=n+2 and n+3
Note it is always the same.
#A #B#C#D MUX control wires Sections 4 and 5 would be
4 4 4 up here but are not shown.
4
This shows how the data
0 0 0 0 1 1
collected between t=n+2
and t=n+3 goes through
1 0 0 0 0 1
the shift register.
1 0 1 0 0 To see what is in the shift
0 register at any given time
1 1 1 1 0 follow a common time
1 down as was done for the
t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8 shaded MUXs at t=n+6.
Data from t=n+3 to n+4
0 0 1 1 1 1
How the data collected
during t=n+3 to n+4 goes
section 2
1 0 0 1 1 1
through the shift register.
1 1 0 1 1 The control for the MUXs is
0 represented by the line
running through them. This
1 1 1 1 1
1 line is a cable and contains
4 wires.
t=n+4 t=n+5 t=n+6 t=n+7 t=n+8 t=n+9
Data from t=n+4 to n+5
0 0 1 1 1 1 How the data collected
during t=n+5 to n+6 goes
through the shift register.
section 1
1 0 0 1 0 1
Note all MUXs operating at a
0 1 0 0 1 0 given time, have the same
control lines. The top (state A)
1 1 0 1 0 MUX in each section all use
1 control wire #A. The bottom
t=n+5 t=n+6 t=n+7 t=n+8 t=n+9 t=n+10 MUXs all use wire #D, etc.
In Figure 33, four data bits start through the trellis paths in each clock cycle. The MUXs do
not change the bits but only place them in the right path. Since all the columns represent the same
time step all the MUXes are set the same way in each column. Here they are set for t=n+6.
Authors Fred Ma, John Knight January 26, 2001 18
Convolution Codes
FIGURE 33 WARNING: This shows hardware and data at a single time.
#A #B#C#D Control wires
4
A 4 A 4 A 4 A 4 A A
0 0 1 0 1
B B B B B B
1 0 0 0 1
C C C C C C
0 1 0 0 1
D D D D D D
1 1 1 1 0
t=n+6 t=n+6 t=n+6 t=n+6 t=n+6 t=n+6
Column (i) (ii) (iii) (iv) (v)
Shift register structure. The flip-flops contain the data for t=n+6. The 4 path edges selected
The first three columns contain the data that can be taken from Figure 26. Column (I) contains
data injected at t=n+2 and one has to go back to Figure 17 to obtain it. Can you calculate the
data in column (v)? Hypertext link to solution (Figure 45).
All 5 MUXs in a row are controlled by the same input line, i.e wire #A for the top row, #B for
the next etc.
4.3 Summary of Circuit Up Through Section 4.4
dataIn Top module
Test Bench Encoder
reset Encoder
Signal Source clk
Comparison serial_in
with original Error generator
signal
(loopback) serial_in_err
Top module serial_in_err
Decoder
Serial-to
dataOut Parallel
convSig[2:1]
Calc_Came_From_and_Ham_Dist Data_Shift_Forward
input clk, reset, convSig[2:1]; input clk, reset, came_from[4:1];
output dataOut; //maybe input HA,HB,HC,HD;
output dataOut;
Authors Fred Ma, John Knight January 26, 2001 19
Convolution Codes
4.4 Second Lab: Start of Convolution Decoder in Verilog
At this point the lab is starting to switch from a lab to a project. It is now you need to know
where you are going. You should be able to recognize each block in the block diagram. The
Data_Shift_Forward block has been described in the theory. However it will not be done until Ex-
ercise/Lab 3.
You should consider these concepts:
a) Normally the encoder and decoder are widely separated so they cannot run from a common
clock. The decoder will have a clock recovery module. This is beyond 97.478 and we will
use a common clock for both.
b) Your design will be a rate=1/2, constraint length=3; [111],[101] decoder.
This is to make it easy to check things by hand or from books.
c) To save work you will want to parameterize your design. One can do this automatically in
Verilog for some parameters. For others it too much trouble.
• First, design a distinctive comment style like
/*|Para|* comment on parameters *|Para|*/
which indicate code where there are parameters nearby which will need attention. In your
comments cross-reference all the modules that are effected by the parameters.
• Second, there are two ways to pass parameters in Verilog. Which one works for synthesis?
d) Considerable emphasis will be given to testing. One simple test is a loopback test where the
output is sent back to the input and the two compared. Other thing you can do to your circuit
to aid testing will be considered later.
e) The Error Generator block is necessary if you want to simulate to the error correction prop-
erties. First it will be done as a test bench so it is only useful during simulation. Later you
might consider making it part of the loopback test so it can be used for testing in the field.
What to do for the lab
1. Draw a block diagram somewhat like that of Section 4.3 on page 19. 1However make it big-
ger and show the arguments passed to all modules. If a module will be over a page of code,
try to divide it.
2. Write Verilog for the Decoder Top Module and Encoder Top Module (already done?)
3. Design a serial to parallel module. The serial_in signal changes at twice the clock rate. Let
the serial_in bits be labelled s, e, r, i, a, l ...
Decide how these bits will come out of a transparent-high latch.See the latch_sig below.
You will probably use a transparent latch and two flip-flops in the module.
clk
serial_in serial_in s e r i a l
or serial_in_err
1D
C1 latch_sig s
1D Transparent- ConvSig[1] s r
C1 high latch ConvSig[2] e i
1D
C1 FIGURE 34
Waveforms for serial to parallel
1. We like originality in block diagrams as long as you can give a reason for changes.
Authors Fred Ma, John Knight January 26, 2001 20
Convolution Codes
4. Write the Verilog code for the serial to parallel module.
How do you write a latch in Verilog? Hints, see Figure 46.
5. Write the Calculate Next State and Hamming Distance module. (See Exercise 2.)
6. Modify the Test Bench to handle the decoder // send in a new x every clock cycle
and the encoder. This should include a loop- always @(posedge clk)
back test which compares the dataIn (x on the begin
right) with dataOut. if (I==9) \$finish;
You will have a delay (latency) between dataIn x<= #1 data[I];
and dataOut. At the start dataOut will be the bit if(I>latncy) y<=#1 data[I-latncy];
corresponding to the present lowest H (cumula- I<=I+1;
tive Hamming distance) so the latency will be end
only that of the serial-to-parallel converter. Lat- assign err=(y!=dataOut) FIGURE 35
er you will want to add a latency of 4 to 5 times
the constraint length.
7. Add to the FIGURE 36
input 11
Calc_Nxt_State_and_Ham_Dist
module to generate a four-bit signal 3 A 00 2 5 A came_from[0]=down
called came_from. These signals in- 110 2
dicate whether the trellis lines lead- 3
3 B 0 B came_from[1]=up
ing back from the next states to the
11 0 2 5
present-time-step states, came from a 0 10 0
higher state or a lower state. 01 3
2 C 2 C came_from[2]=up
5
Thus it would show whether the 3
next state A came from the present A 1
01
(up) or C (down). Figure 36 shows 2 D 10 1 3 D came_from[3]=down
these bits and their meaning. These t=now t=next step (now + 1)
will be the MUX control signals to
shift the data forward.
8. Then write a preliminary Data_shift_Forward module “stub” which has input declarations
for these four signals but no code to do anything with them, at least not yet.
9. Consider the Error Generator. For the moment treat it FIGURE 37
like a test bench so you can use nonsynthesizable reg [4:0] randy;
constructs like \$random. This gives a new random always @(clk) begin
integer every time it is called. Try to make it so there //Run at twice clock rate
is an error every 16 time steps (every 32 serial_in bits) //randy gets the 5 lsb of \$random
on average. A random 5-bit number will be 01110 (or randy <= \$random;
any single value) one time out of 32 on average. if (randy = = 4’b01110)
Alternately you might try a pseudorandom generator serial_in_err<= ~serial_in;
as used to be done in 97.350. There is a lot about
pseudorandom generators in the notes.
Authors Fred Ma, John Knight January 26, 2001 21
Convolution Codes
4.4.1 Feed Forward, an Alternate Viewpoint
D C B A Input 11 11 01 01 10 11 11 10 00
n+0 n+1 n+2 n+3 n+4 n+5 n+6 n+7 n+8
time
“State” Inputs
1 0 1 0 1010 1010 1010 1010 1010 1010 1010 1010 1010
1100 1100 1100 1100 1100 1100 1100 1100
0001 0111 0111 1000 1101 1001 0000
0000 1011 0111 1100 1011 0011
space
0000 1111 0011 1001 1111
0000 1111 0110 0011
0000 1111 1100
Output 0 1 1
Direction of \\/\ ///\ /\\\ /\\\ \//\ \\/\ \//\ //// //\/
“came_from”
FIGURE 38 The horizontal axis is time. the vertical axis shows the circuit (space). The table
shows how the “state” input bits shift through the flip-flops with time.
Boxes in the same row show the contents of the same four flip-flops, each box shows a
different time.
Boxes in the same column show the flip-flop contents at a given time.
The “\” and “/” symbols at the bottom show the direction the mux inputs came from. In each
time step the six muxs in a column have the same control signal.
4.5 Third Exercise: Forward Tracing
1. Problem: Walk-through of the data-shift-forward algorithm
Figure 39 through Figure 44 show one step of the trellis on the left, and the circuit to imple-
ment the shift forward on the right. You will fill in the data as it exists in the circuit.
It is important to note the circuit only shows data at one time, t=now. The trellis diagram
shows more than one time. In the circuit you will show the data at the flip-flop’s D inputs ready to
be stored and the likely different data stored inside the flip-flop.
2. Problem: Assign a group member to clean up unfinished parts from Lab 2.
3. Problem: Write pseudocode for the Data_Shift_Forward module
a) The latency will be a parameter which will effect this module and the loopback test.
You will want to define this parameter from the top module (or maybe the testbench) and
pass it down. Write down how to do this.
Authors Fred Ma, John Knight January 26, 2001 22
Convolution Codes
decoded bit between t=now and t=now+1 for state A FIGURE 39
input
rec’d 01 01 If you know the present and next trellis
A A o A A states, then you know what the proper
2 2 decoded output bit is.
You can guess the states by choosing the
B 02 B B 0 B lowest H but often you must wait several
2
steps to be sure.
1 1
C
1
C C 13C
0 0
2
D 13 D D
t=now t=now+1
D
t=now t=now+1
a) Fill in the blanks showing the candidate decoded bits between t=now and now+1
for each state in these two typical trellis steps.
#1 #2 #3 #4 FIGURE 40
Here we inject the data associated input
with each path into the initial flip rec’d 01
A 4 A
flops of the feed-forward shift
A A X
register. 2
0 Unknown
0 We haven’t
b) At the D-input to each B B put anything
flip-flop, show the bit that B 02 B
1 X is in these
will be stored on the next flip-flops
clock after t=now. 1 1 C 1 C
at the start
C C 0 X
State C has it written in 1 0
0 0
D 13 D D
1 D D input
t=now t=now+1 X to this
(t=n+2) (t=n+3) 1
t=now
flip-flop
TRELLIS CIRCUIT
The clock has advanced one FIGURE 41
#1 #2 #3 #4 Control wires
cycle from Figure 38. input 2nd error
rec’d 00→01
A 4 A 4 A
c) Fill in the data on the D inputs 2 A A
in column (i), and also what 2 0
was clocked into the column B B B
(i) flip-flops (Q). 2 B 0 B
Fill the column (ii) D in- 1
2 1
puts.These will be shifted from 1 C C C C
3 C 1
X
column (i) along the best four 0 0
paths.
2
3 D 1 D D D D
1
The first row of muxes is shown t=now t=now+1 1 1
with dashes because they are (t=n+3) (t=n+4) t=now
redundant. D (i) (ii)
Q
Authors Fred Ma, John Knight January 26, 2001 23
Convolution Codes
the up or down position UP #1 Control wires FIGURE 42
input #2
rec’d 10 #3 A A 4 A 4 A
3 A #4
The first column 2 A 0 0
#1
#1
of muxes is 3 1
slowly being B B B B
removed from the 2 B B
1
#2
0
#2
picture. 2
3 C C C C C C
1
0
#3
#3
2 D 2 D
1t=now+1 D
1
D D D
t=now 1 1
#4
#4
(t=n+4) (t=n+5)
t=now (i) (ii) (iii)
d) Continue writing in D and Q for the flip-flops.
Also write in whether the control wires will switch the muxes to up or down.
#1 #2 #3 #4 Control wires FIGURE 43
input
rec’d 11
4 4 4 A
The 1st column A A A A
of MUXes has 3 A 2
A
0
been removed. 1 3 B B B B
We don’t need B
3 B 0 B
ed flip-flops will 03 C C C C
be removed in 2 C C C
the next figure. 0
2 D 13 D D D D D
D
t=now t=now+1 1
(t=n+5) (t=n+6)
(i) (ii) (iii) (iv)
e) Continue writing in D and Q for the flip-flops.
t=now
#1 #2 #3 #4 Control wires FIGURE 44
input
rec’d 11
4 4 4 A
A A A A
2 A 3
A
0
1 2 B B B B B
3 B 0 B 1
04 C
2 C C C C C C
0
2 D 14 D D D D D D
t=now t=now+1 1
(t=n+6) (t=n+7) t=now
f) Complete filling in the Q outputs.
At now which is also t+6, we know the best path ends on B.
g) How do you find what the data value on that path was between t=n+2 and n+3?
Authors Fred Ma, John Knight January 26, 2001 24
Convolution Codes
b) Develop pseudocode to implement the forward shift registers. A hint as to a possible meth-
od follows:
Define four registers in the form of:
reg [1:latency-1] Areg, Breg, Creg, Dreg;
Then consider-
if (came_from[A]) AregNxt ={0,Areg}>>1; else AregNxt={0,Creg}>>1;
Watch out whether you shift in a 1 or a 0. on the left.
The following code may give you some ideas:
parameter lat=12; // latency
parameter A=0, B=1, C=2, D=3;
reg [1:lat] Areg, Breg, Creg, Dreg; //Storage
reg [1:lat] Anxt, Bnxt, Cnxt, Dnxt; //Not storage
always @(came_from or A reg or Breg or Creg or Dreg)
begin
if (came_from[A])Anxt={1'b0,Areg}>>1;
else Anxt={1'b0,Creg}>>1;
if (came_from[B])Bnxt= whatever;
if (came_from[C]) Cnxt= ...;
if (came_from[D]) Dnxt= ...;
end
always @(posedge clk or posedge reset)
begin
Areg<=Anxt;
Breg<=Bnxt;
Creg ...
Dreg ...
if (reset) begin
Areg<=0; Breg<=0; Creg<=0; Dreg<=0;
end
end
c) Decide a strategy to get the dataOut bit. Do you take the one on the best path? Do you take
the majority of the four? Do you take the first one and assume the others will be the same?
Write some pseudocode to implement your strategy to get the dataOut bit.
Authors Fred Ma, John Knight January 26, 2001 25
Convolution Codes
Solution: to Figure 33. (back)
FIGURE 45
PARTIAL COPY OF FIGURE 17
A A A A A 3 A A 1 A
0 10
2
2 2 3 2 3
3
B B B B B B B 0 B
2 11
2
1 10 1
2
4
C C C C C C C C
1 3
3
1
2
2
D D D D D 2 D 3 D 10 1 4 D
t=n t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7
0 0 1 1 1 1 1 1
1 0 1 1 1 1 1 1
1 1
0 0 1 1 1 1
1 0
1 1 1 1 1 1
t=n+1 t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
0 0 1 1 1 0 0
1 0 1 1 1 1 0
1
0 1 1 0 1 1
1
1 0 0 0 0 1
t=n+2 t=n+3 t=n+4 t=n+5 t=n+6 t=n+7 t=n+8
FIGURE 46. How to code a latch (Back)
Latches have something happen on both edges. Do not use posedge clock.
When transparent, latches must follow the data. You need more than @(clock).
Put the reset for the latch in the procedure for the latch, not with some stray flip-flop.
Latches cannot be put in the same procedure as flip-flops.
Do not put logic in the same procedure as the latch.
Authors Fred Ma, John Knight January 26, 2001 26
```
To top | 19,366 | 85,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2015-27 | longest | en | 0.956801 |
https://www.experts-exchange.com/questions/28301517/Big-IF.html | 1,481,126,557,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00093-ip-10-31-129-80.ec2.internal.warc.gz | 936,039,101 | 28,382 | Solved
# Big IF
Posted on 2013-11-22
114 Views
=IF(ED2="SCC","SCC",IF(LARGE(V2:\$AG2,3)>Analysis!\$D\$32,"High",IF(LARGE(V2:AG2,3)<Analysis!\$E\$32,"Low","Medium")))
Hi,
this IF works nicely for me but i have just realised a flaw, i can only check against the large value of D32 + E32 if column D2 = 2
And to add, if column D2 = 1,3,4,5 or 6, i must check against D31 + E31
if column D2 <> 2,1,3,4,5 OR 6 then i must check against D33 + E33
---------------------
So, my formula must first check to see the client type in column D, then get the large number from the array, then based on client type, check the large number against
D31/E31
D32/E32
D33/E33
Thanks
Seamus
0
Question by:Seamus2626
• 3
• 2
Author Comment
ID: 39669440
Attached sample file
Analysis-Template-EE.xlsm
0
LVL 50
Expert Comment
ID: 39669518
so why not create an extra column which produces the correct sum ...
e.g. az2 =iif(d2 = 2 , d32+e32,iif(d2=1 or d2=3 or d2=4 or d2=5 or d2=6 , d31+e31,d33+e33))
and then just use the az2 value in your existing formula...
0
LVL 23
Expert Comment
ID: 39669520
Do you mean?
=IF(ED2="SCC","SCC",IF(LARGE(V2:\$AG2,3)>IF(ISNUMBER(MATCH(D2,{1,3,4,5,6},0)),Analysis!\$D\$31,IF(D2=2,Analysis!\$D\$32,Analysis!\$D\$33)),"High",IF(LARGE(V2:AG2,3)<>IF(ISNUMBER(MATCH(D2,{1,3,4,5,6},0)),Analysis!\$E\$31,IF(D2=2,Analysis!\$E\$32,Analysis!\$E\$33)),"Low","Medium")))
0
Author Comment
ID: 39669555
Hi NB_VC, the first column produces a score of Low, which is incorrect, it should be medium, i must go now but will look through monday morning, thanks
0
LVL 23
Accepted Solution
NBVC earned 500 total points
ID: 39669714
Sorry Seamus, not sure how I inserted a <>, should be <
here is the updated formula:
=IF(ED2="SCC","SCC",IF(LARGE(V2:\$AG2,3)>IF(ISNUMBER(MATCH(D2,{1,3,4,5,6},0)),Analysis!\$D\$31,IF(D2=2,Analysis!\$D\$32,Analysis!\$D\$33)),"High",IF(LARGE(V2:AG2,3)<IF(ISNUMBER(MATCH(D2,{1,3,4,5,6},0)),Analysis!\$E\$31,IF(D2=2,Analysis!\$E\$32,Analysis!\$E\$33)),"Low","Medium")))
0
Author Closing Comment
ID: 39674025
poiffect, many thanks :-)
0
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https://fr.mathworks.com/help/dsp/ref/idct.html | 1,716,504,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00785.warc.gz | 216,774,824 | 23,055 | # IDCT
Inverse discrete cosine transform (IDCT) of input
Libraries:
DSP System Toolbox / Transforms
## Description
The IDCT block computes the inverse discrete cosine transform (IDCT) of the input signal u. When the input is an M-by-N matrix, the block computes the IDCT of each channel in the matrix.
Here is the equivalent MATLAB® code.
`y = idct(u) `
## Ports
### Input
expand all
Specify the input signal as a vector, matrix, or an N-D array.
For all N-D input arrays, the block computes the IDCT across the first dimension. The size of the first dimension (frame size) must be a power of two. To work with other frame sizes, use the Pad block to pad or truncate the frame size to a power-of-two length.
When the input is an M-by-N matrix, the block treats each input column as an independent channel containing M consecutive samples.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `uint8` | `uint16` | `uint32` | `fixed point` | `bus`
Complex Number Support: Yes
### Output
expand all
The block outputs the IDCT of the input signal as a vector, matrix, or an N-D array.
When the block outputs an M-by-N matrix, the lth column contains the length-M IDCT of the corresponding input column.
`$y\left(m,l\right)=\sum _{k=1}^{M}w\left(k\right)u\left(k,l\right)\mathrm{cos}\frac{\pi \left(2m-1\right)\left(k-1\right)}{2M},\begin{array}{ccc}& & m=1,...,M\end{array}$`
where
`$w\left(k\right)=\left\{\begin{array}{l}\frac{1}{\sqrt{M}}\\ \sqrt{\frac{2}{M}}\end{array}\begin{array}{c}\begin{array}{l}\\ ,\\ \\ ,\end{array}\\ \end{array}\begin{array}{ll}\hfill & \hfill \\ \hfill & k=1\hfill \\ \hfill & \hfill \\ \hfill & 2\le k\le M\hfill \\ \hfill & \hfill \end{array}$`
When the input is fixed-point (signed and unsigned), the output is signed fixed-point.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `fixed point` | `bus`
Complex Number Support: Yes
## Parameters
expand all
### Main Tab
Set the block to compute sines and cosines by either looking up sine and cosine values in a speed-optimized table (`Table lookup`), or by making sine and cosine function calls (```Trigonometric fcn```).
To compute sines and cosines, set the parameter to one of these values:
• `Table lookup` –– The block computes and stores the trigonometric values before the simulation starts, and retrieves them during the simulation. When you generate code from the block, the processor running the generated code stores the trigonometric values computed by the block in a speed-optimized table, and retrieves the values during code execution.
The block runs much more quickly, but requires extra memory for storing the precomputed trigonometric values.
• `Trigonometric fcn` –– The block computes sine and cosine values during the simulation. When you generate code from the block, the processor running the generated code computes the sine and cosine values while the code runs.
The block runs more slowly, but does not need extra data memory. For code generation, the block requires a support library to emulate the trigonometric functions, increasing the size of the generated code.
### Data Types Tab
Select the rounding mode for fixed-point operations. The sine table values do not obey this parameter; they always round to `Nearest`.
When you select this parameter, the block saturates the result of its fixed-point operation. When you clear this parameter, the block wraps the result of its fixed-point operation. For details on `saturate` and `wrap`, see overflow mode for fixed-point operations.
Note
The Rounding mode and Saturate on integer overflow parameters have no effect on numeric results when these conditions are met:
• Product output data type is ```Inherit: Inherit via internal rule```.
• Accumulator data type is ```Inherit: Inherit via internal rule```.
With these data type settings, the block operates in full-precision mode.
Choose how you specify the word length of the values of the sine table. The fraction length of the sine table values always equals the word length minus one. You can set this parameter to:
• A rule that inherits a data type, for example, ```Inherit: Same word length as input```
• An expression that evaluates to a valid data type, for example, `fixdt(1,16)`
The sine table values do not obey the Rounding mode and Saturate on integer overflow parameters. The block always saturates and rounds off the sine table values to `Nearest`.
Specify the product output data type. See Fixed-Point Data Types and Multiplication Data Types for illustrations depicting the use of the product output data type in this block. You can set this parameter to:
• A rule that inherits a data type, for example, ```Inherit: Inherit via internal rule```. For more information on this rule, see Inherit via Internal Rule.
• An expression that evaluates to a valid data type, for example, `fixdt(1,16,0)`
Click the button to display the Data Type Assistant, which helps you set the Product output parameter.
See Specify Data Types Using Data Type Assistant (Simulink) for more information.
Specify the accumulator data type. See Fixed-Point Data Types for illustrations depicting the use of the accumulator data type in this block. You can set this parameter to:
• A rule that inherits a data type, for example, ```Inherit: Inherit via internal rule```. For more information on this rule, see Inherit via Internal Rule.
• An expression that evaluates to a valid data type, for example, `fixdt(1,16,0)`
Click the button to display the Data Type Assistant, which helps you set the Accumulator parameter.
See Specify Data Types Using Data Type Assistant (Simulink) for more information.
Specify the output data type. See Fixed-Point Data Types for illustrations depicting the use of the output data type in this block. You can set this parameter to:
• A rule that inherits a data type, for example, ```Inherit: Inherit via internal rule```.
When you select ```Inherit: Inherit via internal rule```, the block calculates the output word length and fraction length automatically. The internal rule first calculates an ideal output word length and fraction length using the following equations:
`$W{L}_{idealoutput}=W{L}_{input}+floor\left({\mathrm{log}}_{2}\left(DCTlength-1\right)\right)+1$`
`$F{L}_{idealoutput}=F{L}_{input}$`
Using these ideal results, the internal rule then selects word lengths and fraction lengths that are appropriate for your hardware. For more information, see Inherit via Internal Rule.
• An expression that evaluates to a valid data type, for example, `fixdt(1,16,0)`
Click the button to display the Data Type Assistant, which helps you set the Output parameter.
See Control Data Types of Signals (Simulink) for more information.
Specify the minimum value that the block should output. The default value is `[]` (unspecified). Simulink® software uses this value to perform:
• Simulation range checking (see Specify Signal Ranges (Simulink))
• Automatic scaling of fixed-point data types
Specify the maximum value that the block should output. The default value is `[]` (unspecified). Simulink software uses this value to perform:
• Simulation range checking (see Specify Signal Ranges (Simulink))
• Automatic scaling of fixed-point data types
Select this parameter to prevent the fixed-point tools from overriding the data types you specify in the block dialog box.
## Block Characteristics
Data Types `bus` | `double` | `fixed point` | `integer` | `single` Direct Feedthrough `no` Multidimensional Signals `yes` Variable-Size Signals `no` Zero-Crossing Detection `no`
expand all
## Version History
Introduced before R2006a | 1,830 | 7,660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.67528 |
https://www.trueachievements.com/a41124/i-like-360-achievement | 1,544,773,236,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825495.60/warc/CC-MAIN-20181214070839-20181214092339-00116.warc.gz | 1,071,741,364 | 12,934 | 1,432
(1,000)
Superstars V8: Next Challenge
There are 50 Superstars V8: Next Challenge achievements worth 1,432 (1,000)
1,562 tracked gamers have this game, 400 have completed it (25.61%)
I Like 36054 (36)
Score exactly 360 points in a Championship.
• Unlocked by 704 tracked gamers (45% - TA Ratio = 1.48) 1,562
Achievement Guide for I Like 360
Solution
For this achievement you will need to finish a Championship with exactly 360 points.
Here is a breakdown of how many points you get for each finishing position:
Qualifying 1st - 1point
1st - 21points
2nd - 15points
3rd - 12points
4th - 10points
5th - 8points
6th - 6points
7th - 4points
8th - 3points
9th - 2points
10th - 1point
11-19 - 0points
This will be easier to achieve by setting up a Championship with Easy AI, Penalties and Car Damage OFF
For this you will need to race a typical Championship, which consists of 9 tracks (Qualifying, Race 1 and Race 2)
You can do this however you wish but this is the way i did it.
Firstly Qualify 1st on every track then finish 1st on 8 tracks (both Race 1 and Race 2) then on the 9th track you place 2nd on Race 1 and 11-19 in Race 2.
So this works out at 8 tracks (race 1 and 2) placed 1st, 9 qualifications at 1st and a second place race.
(21 x 2) x 8 = 336 + (1 x 9) = 345 + 15 = 360.
This way you will also earn
The Always in front achievement in Superstars V8: Next Challenge worth 21 pointsAchieve a Pole Position in every race of a championship. | 452 | 1,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-51 | latest | en | 0.95662 |
https://se.mathworks.com/matlabcentral/answers/1755615-alternative-to-using-multi-level-field-struct?s_tid=prof_contriblnk | 1,713,907,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00405.warc.gz | 470,901,460 | 31,550 | # Alternative to using Multi-level field struct
10 views (last 30 days)
Science Machine on 7 Jul 2022
Edited: Bruno Luong on 8 Jul 2022
I am taking two populated structs and re-ordering them, so that I can do an operation(such as fft), on the end-most field's elements, eg
bigNum=1e4;
for b=1:bigNum
for a=1:bigNum
val=Struct_A(a).dat(b);
Struct_B(b).dat(a)=val;
val2 = fft(Struct_B(b).dat(:);
end
end
I do this also with 3-layered field structures such as Struct_C(f).fielde(e).fieldd(d).dat.
• Is there a more efficient way to handle this reordering?
• Can I vectorize this or otherwise speed this up? I am using parfor loops, but, it seems just the idea of re-ordering structures is a bad paradigm when considering large data
Michael Van de Graaff on 7 Jul 2022
I gave it shot.
n = 1000;
s(n) = struct();
for ii = 1:n
for jj = 1:n
s(ii).a(jj) = ii;%this is just building the original struct
end
end
tic
for ii = 1:n
for jj = 1:n
val = s(ii).a(jj);
s2(jj).a(ii) = val;% do the loop like you have
end
end
toc
Elapsed time is 0.660588 seconds.
%now lets try something without loops, should be faster for large n
tic
tmp = cell2mat(squeeze(struct2cell(s))).';
tmp2=mat2cell(tmp,ones(1,n),[n]);
s3 = cell2struct(tmp2,fieldnames(s),4);
toc
Elapsed time is 0.033685 seconds.
% and let's do a loop to ensure s2 and s3 are the same
s_diff = zeros(n);
for ii = 1:n
for jj = 1:n
s_diff(ii,jj) = s2(ii).a(jj)-s3(ii).a(jj);
end
end
if max(abs(s_diff(:)))==0
disp('success') %I got success
else
disp('failure')
end
success
##### 3 CommentsShow 1 older commentHide 1 older comment
Science Machine on 8 Jul 2022
Edited: Science Machine on 8 Jul 2022
'While' keeping the data in its natual organiztion as eg
struct(index 1).field2(index2).field3(index3).dat(array or matrix)
each field is generally a different # of entries.
• I was reading maybe multi-dimensional array?
Michael Van de Graaff on 8 Jul 2022
"would be possible to eschew structures altogether?"
Are you familiar with cell arrays? That seems like precisely what you are looking for, as they can be n-dimsionsal and each cell can contain whatever data type, including arrays of differeing sizes. You don't get the built in clarity of the fieldnames, but a GPU don't need no fieldnames :D (I have literally no knowledge of how GPU implementations work fyi)
Bruno Luong on 8 Jul 2022
Edited: Bruno Luong on 8 Jul 2022
It seems you just burry an array into nested structures for no apparent reason other than using struct with a fieldname for a sake of it speaks to you.
Undo that and think like a computer:
% Generate dummy data
a = 10; % your big number
for i=1:a
for j=1:b
STRUCTA.A(i).B(j) = rand();
end
end
% Undo the thing to getback numerical array
a = length(STRUCTA.A);
b = length(STRUCTA.A(1).B);
AB=reshape([STRUCTA.A.B],[b,a]); % deepest nested length first, ...
% Simply work on the array, that how MATLAB should be used
% If you want to swap a/b dimension, transpose AB, no need for
% the ugly double for-loop as you do.
val2 = fft(AB,[],2)
val2 =
4.0661 + 0.0000i 0.0323 - 0.7934i -0.2520 - 0.6357i 0.4981 + 0.0308i -0.4295 - 0.9440i -1.0016 + 0.0000i -0.4295 + 0.9440i 0.4981 - 0.0308i -0.2520 + 0.6357i 0.0323 + 0.7934i 5.5655 + 0.0000i 0.4685 + 0.8790i 0.3279 + 0.0606i 0.1125 + 0.1375i 0.4377 - 0.4593i -1.2876 + 0.0000i 0.4377 + 0.4593i 0.1125 - 0.1375i 0.3279 - 0.0606i 0.4685 - 0.8790i 4.8333 + 0.0000i -0.5814 - 0.5989i 0.4646 - 0.5380i -0.1544 - 0.4535i -0.5136 - 0.3355i 1.9391 + 0.0000i -0.5136 + 0.3355i -0.1544 + 0.4535i 0.4646 + 0.5380i -0.5814 + 0.5989i 5.2862 + 0.0000i 0.7492 - 1.0385i 1.3908 + 0.8551i 0.2308 - 0.0047i -0.6972 + 0.4188i -0.5792 + 0.0000i -0.6972 - 0.4188i 0.2308 + 0.0047i 1.3908 - 0.8551i 0.7492 + 1.0385i 4.8983 + 0.0000i -0.4043 + 0.3373i 0.2988 - 0.9046i -0.4060 + 0.4049i 0.7049 - 0.6535i 0.2248 + 0.0000i 0.7049 + 0.6535i -0.4060 - 0.4049i 0.2988 + 0.9046i -0.4043 - 0.3373i 4.8601 + 0.0000i -0.2928 - 0.1440i -0.3565 - 0.5095i 0.4126 - 0.2177i -0.3226 + 0.0414i 1.4404 + 0.0000i -0.3226 - 0.0414i 0.4126 + 0.2177i -0.3565 + 0.5095i -0.2928 + 0.1440i 4.4551 + 0.0000i -1.6476 - 0.0270i 0.7631 - 0.6674i -0.2366 - 0.8680i -0.1395 - 0.1165i -0.4810 + 0.0000i -0.1395 + 0.1165i -0.2366 + 0.8680i 0.7631 + 0.6674i -1.6476 + 0.0270i 4.1921 + 0.0000i -0.6224 - 0.1522i 1.5197 - 0.6028i 0.5938 + 0.2190i 1.0945 - 0.7803i -0.1243 + 0.0000i 1.0945 + 0.7803i 0.5938 - 0.2190i 1.5197 + 0.6028i -0.6224 + 0.1522i 5.2778 + 0.0000i 0.2595 + 0.7150i -0.2160 - 0.3907i 1.0669 + 0.0340i 0.3649 - 0.8534i -0.3927 + 0.0000i 0.3649 + 0.8534i 1.0669 - 0.0340i -0.2160 + 0.3907i 0.2595 - 0.7150i 5.5209 + 0.0000i 1.3374 + 0.5652i -0.5742 + 0.1283i -0.1109 + 0.2116i 0.7891 + 0.7053i -0.8280 + 0.0000i 0.7891 - 0.7053i -0.1109 - 0.2116i -0.5742 - 0.1283i 1.3374 - 0.5652i 4.2459 + 0.0000i 0.6400 - 0.5771i 0.4029 + 0.8782i -0.1877 + 0.6045i -0.5498 - 0.5591i -1.6546 + 0.0000i -0.5498 + 0.5591i -0.1877 - 0.6045i 0.4029 - 0.8782i 0.6400 + 0.5771i
Science Machine on 8 Jul 2022
Can this be done with 3 or more levels? i have a more complicated structure like myStr(i).angle(theta).radius(r).dat(1:xLocation) and since there are so many parameters, I needed to organize that somehow. Several times, depending on the operation, I have to juggle the entries.
function strTest3()
% Gen data
a = 3; %
b = 5; %
c = 7; %
for i=1:a
for j=1:b
for k=1:c
%STRUCTA.A(i).B(j) = i*j;
STRUCTA.A(i).B(j).C(k) = i*j*k;
end
end
end
% Undo
a = length(STRUCTA.A);
b = length(STRUCTA.A(1).B);
c = length(STRUCTA.A(1).B(1).C);
AB=reshape([STRUCTA.A.B],[b,a]); % this works
ABC=reshape([STRUCTA.A.B.C],[c,b,a]); % this produces error
val2 = fft(AB,[],2)
end
Bruno Luong on 8 Jul 2022
Edited: Bruno Luong on 8 Jul 2022
With three (four) levels you have to do 2-step
...
AB = [STRUCTA.A.B];
ABC = reshape([AB.C],[c b a]);
...
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Start Hunting! | 2,569 | 5,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-18 | latest | en | 0.831983 |
https://wiraelectrical.com/digital-multimeter-diagram/ | 1,719,167,130,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00273.warc.gz | 540,449,517 | 17,309 | # Digital Multimeter Diagram – How it Works
Contents
Looking at a digital multimeter diagram will greatly help you to understand why it is called a multimeter. Just as its name states, multimeter means a meter with a multi function, multi measurement, and multi purpose. Of course all of them need to be in the scope of electrical and electronics.
Below are the uses of digital multimeter, measure various electrical variables such as:
• Voltage (AC or DC),
• Current (AC or DC),
• Resistance,
• Transistor gain,
• Continuity,
• Diode,
• Frequency,
• Capacitance,
• Inductance, and
• Temperature.
Some expensive and advanced multimeters are able to measure power as well. Keep in mind not every multimeter has the same measurement choices. The most common measurements to find in every multimeter are voltage, current, continuity, diode, and resistance.
There are two types of multimeter:
• Analog multimeter
• Digital multimeter
Digital multimeters are widely used nowadays so we will focus on digital multimeters instead. But it is not a bad thing to at least learn what an analog multimeter is and what makes it different and lose popularity.
## What is an Analog Multimeter
Just as an analog voltmeter and analog ammeter, this multimeter has a printed background with several numbers and symbols to indicate the measurements and their values.
The values will be displayed by an arrow, moved by magnetic force generated by a coil. The arrow will move when there is:
• Current flows through the coil, or
• Internal electrical power to measure resistance, or
• Electrical pressure.
Even though its popularity is less than the digital one, it still has some advantages. Its size is relatively small and you can observe the changes in current and voltages real-time even if it is only by small movement.
What makes it lose popularity is you need a quick and precise mathematical calculation in your head before getting the measurement values. Analog multimeter uses scales in its probe terminal and thus you need to do a quick calculation. And we already know that time is short in the practical field and of course we don’t like math very much. As long as we get the value, we are good. This is where a digital multimeter kicks in.
## What is a Digital Multimeter
A digital multimeter uses an LCD for displaying the measurement value. This display helps us a lot because it allows us to read the value immediately without doing any calculation.
For its measurement capability, a digital multimeter doesn’t differ much from an analog multimeter. The basic measurement such as voltage, current, and resistance will be the same and they have their own scale we can choose with a rotary switch.
Nowadays, a common digital multimeter also has additional measurements such as diode, continuity, capacitance, inductance, and temperature along with their scales.
A more advanced digital multimeter is capable of measuring power and even has auto-ranging measurement. But in our opinion, having an auto-ranging measurement won’t help us very far, an adjustable scale still works very fine for most people.
One thing to remember, manual scale needs us to take extra caution to the maximum value we will get. If you think it will be over 500V then it is wise to use 1000V scale first. It will be disastrous if we use 100V to measure 500V.
Well, in conclusion, a digital multimeter is the absolute winner because of the convenience, speed, and practicality.
## Digital Multimeter Diagram
Observe the digital multimeter block diagram below.
The block diagram above is for the common digital multimeter to measure basic quantities: resistance, current, and voltage.
The line connected to the input probe + is also connected to the rotary switch to select the electrical variables:
1. Resistance,
2. AC voltage (ACV),
3. AC current (ACI),
4. DC current (DCI),
5. DC voltage (DCV),
The resistance line is connected to the constant current source to produce voltage that will be used as resistance measurement. After data passes the buffer amplifier, it will be converted by ADC (Analog to Digital Converter) before it is displayed by the digital display.
ACV and DCV are connected to the “calibrated attenuator”. This attenuator is used to reduce the signal’s power without disturbing its waveform. This step is taken to prevent power surge by the voltage measurement.
ACI and DCI are simply connected to the “current to voltage converter” or “I-V Converter”. This converter produces voltages proportional to the input current. The circuit is pretty simple, only an op-amp and a feedback resistor as shown below.
Observe further to the right, we find a rectifier circuit connected to the ACV and ACI. Why do we need a rectifier? The rectifier will convert the AC signals into DC signals for easier conversion.
All our measurement data from resistance, ACV, ACI, DCV, and DCI are converted by the “Analog to Digital Converter” (ADC) then displayed by the digital display consisting of their number, symbol, and unit.
Auto-ranging or auto-scaling a digital multimeter only needs us to use proper probe connection and let the multimeter handle the scaling, measuring, and displaying the result.
## How Does a Multimeter Work
A digital multimeter can measure different electrical quantities such as:
• Voltage and current (DC and AC),
• Resistance, inductance, capacitance,
• Diode and continuity,
• Additional measurement (transistor gain, temperature, etc).
This multimeter comes with a pair of probes (red and black) for the hot line (or active line) and ground line (or negative line).
Digital multimeter is constructed with:
• Sensing element connected to probes,
• Analog to digital converter,
• Rectifier circuit,
• Range selector and converter,
• Digital display, and
• Amplifier.
Most digital multimeters are supplied by a pair of AA batteries.
### How a Multimeter Measures Resistance
When the multimeter is used to measure resistance, it uses its reference voltage source inside it. The voltage is applied to the measured point and the voltage drop is generated. This voltage drop is calculated using a calibrated value to produce resistance on the measured points.
Sometimes a current source is used instead of voltage source. The result will be the same.
This current source with a fixed value will be used to measure the desired resistance. The current will flow through the resistor and produce a voltage. Using the basic Ohm’s Law (R=V/I), it will result in the resistance value and be displayed on the digital display.
### How a Multimeter Measures Current
While doing current measurement in series, the probes will sense the current. If the measured current is DC then we will not have any problem, but if it is AC then it is converted to DC first by the AC to DC converter (rectifier circuit).
This measurement will convert the current into equivalent voltage from its internal resistance in the multimeter. Keep in mind that every measurement given by the multimeter is processed in the form of voltage.
Digital multimeter has a low resistance resistor with the set value. This low resistor acts as close as a conductor wire to minimize resistance that can disturb the measurement.
Later on, the current flows through this resistor and the multimeter measures the voltage across this resistor. This value is then calculated and calibrated by the multimeter using Ohm’s Law (I = V/R) and displayed digitally.
### How a Multimeter Measures Voltage
Just as stated before, calibration beside the voltage will be calculated in the form of equivalent voltage, converted into a desired value equivalent to the voltage.
Now we are measuring voltage, then the process will be much simpler. The voltage measured by the multimeter will be calibrated and converted by the ADC (analog to digital converter) to show in the digital display.
## How to Use Digital Multimeter
A digital multimeter or DMM is a device to measure various electrical quantities in a single handheld-size device. DMM is light, relatively small, operated by batteries, and multipurpose.
Even a DMM is a single device and some sensing probes, we still need to learn how to operate it based on what we want to measure.
Below is the complete step on how to use a digital multimeter to measure voltage, current, resistance, and diode.
The measurement for capacitance and inductance will be similar to the resistance measurement.
### How to Measure Resistance
1. Prepare the resistor or a circuit. If you need to measure a single resistor connected in a circuit then it is impossible to do because resistance in a circuit is already combined with other components.
2. Plug the red probe’s terminal to the multimeter’s socket indicated by the symbol “Ω”. Plug the black probe’s terminal to the multimeter’s socket indicated by the ground symbol “⏚” or COM.
3. Rotate the rotary switch to the symbol “Ω”. This symbol is Omega and used to represent Ohm in electricity.
4. Use the bigger scale first (MΩ or kΩ) if you have no idea how big the resistance will be. You may lower it if the result shows zero in the first digit. That indicates that the scale is too big. Don’t worry, using a small scale won’t harm your multimeter.
5. Turn off or cut off the power supply from the circuit.
6. Remove the resistor you want to measure if it is already in the circuit to get an accurate result. Or if it has not been installed in the circuit, then we can measure it immediately.
7. Touch one end of the resistor with the red probe and another end with the black probe. Since resistance doesn’t have polarity then we don’t need to think about which one should be connected to the red probe.
8. Read the number in the digital display and take note of the scale you use. A measurement of 30 may be 30Ω, 300Ω, 3kΩ, or 3MΩ.
### How to Measure Voltage
1. Plug the red probe’s terminal to the multimeter’s socket indicated by the symbol “V”. Plug the black probe’s terminal to the multimeter’s socket indicated by the ground symbol “⏚” or COM.
2. Determine whether the voltage is AC or DC. Some multimeters have auto-ranging indicated by DC and AC symbols in a single mode but some of them only measure mV.
3. Assume that we don’t have auto-ranging then we need to manually choose DC or AC.
4. Rotate the rotary switch to the voltage symbol.
5. We start with the highest scale for safety if we don’t have an idea how high the voltage is.
6. Connect the red probe to the hot line or positive line and black probe to the ground line or negative line.
7. If the measured voltage is AC then we don’t need to specifically determine the polarity. If we measure DC voltage then connecting the wrong probe to the wrong polarity will show a negative value.
8. Read the displayed value. Reduce the scale if the result is too small, indicating the scale is too big.
### How to Measure Current
1. Plug the red probe’s terminal to the multimeter’s socket indicated by the symbol “mA” for milliamps or “A” for Amps. Plug the black probe’s terminal to the multimeter’s socket indicated by the ground symbol “⏚” or COM.
2. If you are not sure how high the measured current is, just use the “Amps” first. If the result is too low, maybe it is within milliamps and you need to use the mA.
3. Determine whether the voltage is AC or DC. Some multimeters have auto-ranging indicated by DC and AC symbols in a single mode but some of them only measure mA.
4. Assume that we don’t have auto-ranging then we need to manually choose DC or AC.
5. Rotate the rotary switch to the “A” symbol.
6. We start with the highest scale for safety if we don’t have an idea how high the voltage is.
7. Turn off the circuit or cut off its power supply.
8. Break the circuit at the point you want to measure the current.
9. Connect the red probe to the positive side and black probe to the negative side. These two sides we connect with the probes are the result when we broke the circuit. Remember that an ammeter should be connected in series right?
10. If the measured voltage is AC then we don’t need to specifically determine the polarity. If we measure DC voltage then connecting the wrong probe to the wrong polarity will show a negative value.
11. Turn on the circuit or connect the power supply.
12. Read the displayed value. Reduce the scale if the result is too small, indicating the scale is too big.
### How to Test Diode
1. Plug the red probe’s terminal to the multimeter’s socket indicated by the symbol diodes (an arrow). Plug the black probe’s terminal to the multimeter’s socket indicated by the ground symbol “⏚” or COM.
2. Rotate the rotary switch to the diode symbol.
3. Turn off the circuit or cut off its power supply.
4. To test the forward bias, connect the red probe to the positive terminal of the diode and black probe to the negative terminal.
5. If the displayed number is more than 0 then the forward bias is good (more than 0 less than 1). If the displayed number is OL (OverLoad) or 0 then the forward bias is bad.
6. To test the reverse bias, connect the red probe to the negative terminal of the diode and black probe to the positive terminal.
7. If the displayed number is 0 or OL then the reverse bias is good. If the displayed number is more than 0 and less than 1 then the reverse bias is bad.
8. To conclude whether the diode is good or bad, the results for both forward bias and reverse bias should be good.
9. Turn on the circuit or connect the power supply.
10. Read the displayed value. Reduce the scale if the result is too small, indicating the scale is too big.
### How to Test Continuity
We can use the same probe configuration as the resistance measurement. We simply connect the probes to the desired connection and if it shows any value in Ohm then it is good. If it shows OL then the continuity is bad.
Most of the multimeter will beep if the continuity is good. But if it is not beep that doesn’t mean the connection is bad, maybe the impedance is too high.
This test is useful for checking the fuse, switch, conductors, and many more. | 3,035 | 14,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.926829 |
uyxw.civitavecchia-cruise-transfer.it | 1,638,354,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00627.warc.gz | 657,785,777 | 9,465 | Harman absolute pellet stove
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• The simultanous equation calculator helps you find the value of unknown varriables of a system of linear, quadratic, or non-linear equations for 2, 3,4 or 5 unknowns. A system of 3 linear equations with 3 unknowns x,y,z is a classic example. This solve linear equation solver 3 unknowns helps you solve such systems systematically.
Part 1 of the instructional video series on how to use the calculators on calculator academy.
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Algebra-cheat.com delivers great material on two step equations calculator, algebra i and matrix algebra and other algebra subject areas. In cases where you will need assistance on graphing linear equations or perhaps dividing polynomials, Algebra-cheat.com is always the excellent place to check-out!
Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to solve system of linear equations using Entering data into the Cramer's rule calculator. You can input only integer numbers or fractions in this online calculator.
give the equation instead of having to find it yourself. [7] 2021/04/26 20:07 Under 20 years old / Elementary school/ Junior high-school student / Useful / ... I need to find a linear regression calculator where I can see the exact values of the points on the line.Pocketmath.net includes helpful resources on standard form of a linear equation calculator, graphing and greatest common factor and other math subject areas. In the event that you need help on equivalent fractions or even trigonometry, Pocketmath.net is going to be the right site to head to! Solving Linear Equations by Graphing Solving Linear Equations by Graphing ... Enter the left side of the equation as a linear equation of the form y = ax + b in Step 4.
The graph above is an example. When writing the equation, cos(x²)=sin(y²) the following graph is plotted. I find it mesmerizing that an equation can give amazing results. Thank you for making Transum free and available on the internet. I am very grateful for this site. Thank you, Soumi. " Akshat, India. Saturday, March 6, 2021
The calculator calculates linear equation. Write it according to the mentioned general form in the form. If there is a minus sign in the equation, write the corresponding variable as a negative number. Graph. Calculator.
How to Use a Graphing Calculator to Solve an Advanced System of Linear Equations. Step 1: Make sure the linear equations are in the form of y = mx + b. If the x term or the constant term is on the ...
Linear Equation System Solver. This app is a free math calculator which is able to solve systems of linear equations. Best mathematical tool for school and college! If you are a student, it will help you to learn linear algebra! Note: Linear equation systems play an important role in engineering, physics, chemistry, computer science, and ...
It is a method of solving linear system of equations. In this method, we graph the given equations on the coordinate plane and look for the points of intersection. The point of intersection determines whether the system of linear equations has no solution, one solution, or infinitely many solutions.
Graph linear functions by plotting points, using the slope and y-intercept, and using transformations. Write the equation of a linear function given its graph. Match linear functions with their graphs. Find the x-intercept of a function given its equation. Find the equations of vertical and horizontal lines. Linear diophantine equation in two variables is equation of the form: This calculator is based on the extended Euclidean algorithm written as a To solve a linear diophantine equation in two variables, one first need to find a particular solution and , and then find the general solution using the formulasCheck to make sure that it is the correct equation you typed. (7.) Review the answers. Using the Solve Equations Calculator; All outputs/answers are written as both integers and/or decimals; and integers and/or fractions. Linear Equations: Type: \$5p - 4 = 21\$ as 5 * p - 4 = 21; Linear Equations: Type: \$-1.2p + 1.25 = 0.77\$ as-1.2 * p + 1.25 = 0.77
Graphing calculator allows to shift, zoom and center the graph using the control buttons below the graph pane. Alternatively mouse drag can be used to shift the graph and mouse wheel to resize it. Graphing calculator could be used to visualize the results of other computations (e.g. to draw graphs of a function and its derivative).
How to Use a Graphing Calculator to Solve an Advanced System of Linear Equations. Step 1: Make sure the linear equations are in the form of y = mx + b. If the x term or the constant term is on the ...Best Free Online Graphing Calculator. Math Calculators. GraphCalc is the best free online graphing calculator that almost completely replaces the TI 83 and TI 84 plus calculators. GraphCalc allows you to graph 2D and 3D functions and equations as well as find intersects and create table values. y = x2.
The above application is a simplified version of our graphing method calculator available to students who have a membership with us; however, it has all the basic functionality required to graph most linear programming exercises in your school.Linear Regression (Line of Best Fit) Calculator. Linear regression is a simple statistics model describes the relationship between a scalar dependent variable and other explanatory variables. If there is only one explanatory variable, it is called simple linear regression, the formula of a simple regression is y = ax + b, also called the line ... Graphing and Systems of Equations Packet 1 Intro. To Graphing Linear Equations The Coordinate Plane A. The coordinate plane has 4 quadrants. B. Each point in the coordinate plain has an x-coordinate (the abscissa) and a y-coordinate (the ordinate). The point is stated as an ordered pair (x,y). C. Horizontal Axis is the X - Axis. (y = 0)
Differential Equation Calculator. The calculator will try to find the solution of the given ODE: first-order, second-order, nth-order, separable, linear, exact, Bernoulli, homogeneous, or inhomogeneous. Initial conditions are also supported.
Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more.
Algebra-cheat.com delivers great material on two step equations calculator, algebra i and matrix algebra and other algebra subject areas. In cases where you will need assistance on graphing linear equations or perhaps dividing polynomials, Algebra-cheat.com is always the excellent place to check-out! Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to solve system of linear equations using Entering data into the Cramer's rule calculator. You can input only integer numbers or fractions in this online calculator.
Solutions. Graphing. Practice. Geometry. Calculators. Middle School Math Solutions - Equation Calculator. Welcome to our new "Getting Started" math solutions series.Dec 29, 2014 · To find a linear model for a scatterplot (which is what I assume you want), you just need to do a couple of things. Firstly, you need to enter your data into the calculator. To do this, hit your "STAT" key, and select "EDIT". You should see a table with lists. Enter all your x values into one list, and all your y values into the other. Once you have done this, hit 2ND and QUIT (normally mode ... TI-84 Plus and TI-83 Plus graphing calculator program shows step by step solutions to linear equations, logarithms, exponents, functions, trigonometry, domain and range problems. Requires the ti-83 plus or a ti-84 model.(
Free linear first order differential equations calculator - solve ordinary linear first order differential equations step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy. out of 100. IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more. Graphing calculator allows to shift, zoom and center the graph using the control buttons below the graph pane. Alternatively mouse drag can be used to shift the graph and mouse wheel to resize it. Graphing calculator could be used to visualize the results of other computations (e.g. to draw graphs of a function and its derivative).
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Using the linear equation (labeled A in Figure 5), a spreadsheet cell can have an equation associated with it to do the calculation for us. We have a value for y (Absorbance) and need to solve for x (Concentration). Below are the algebraic equations working out this calculation: y = 2071.9x + 0.111. y - 0.0111 = 2071.9x (y - 0.0111) / 2071.9 = x | 2,002 | 9,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-49 | latest | en | 0.894296 |
https://www.airmilescalculator.com/distance/bau-to-cnq/ | 1,606,737,565,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141213431.41/warc/CC-MAIN-20201130100208-20201130130208-00617.warc.gz | 570,324,523 | 33,305 | # Distance between Bauru (BAU) and Corrientes (CNQ)
Flight distance from Bauru to Corrientes (Bauru Airport – Doctor Fernando Piragine Niveyro International Airport) is 703 miles / 1131 kilometers / 611 nautical miles. Estimated flight time is 1 hour 49 minutes.
Driving distance from Bauru (BAU) to Corrientes (CNQ) is 865 miles / 1392 kilometers and travel time by car is about 17 hours 38 minutes.
## Map of flight path and driving directions from Bauru to Corrientes.
Shortest flight path between Bauru Airport (BAU) and Doctor Fernando Piragine Niveyro International Airport (CNQ).
## How far is Corrientes from Bauru?
There are several ways to calculate distances between Bauru and Corrientes. Here are two common methods:
Vincenty's formula (applied above)
• 702.961 miles
• 1131.306 kilometers
• 610.856 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 702.730 miles
• 1130.935 kilometers
• 610.656 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Bauru Airport
City: Bauru
Country: Brazil
IATA Code: BAU
ICAO Code: SBBU
Coordinates: 22°20′41″S, 49°3′13″W
B Doctor Fernando Piragine Niveyro International Airport
City: Corrientes
Country: Argentina
IATA Code: CNQ
ICAO Code: SARC
Coordinates: 27°26′43″S, 58°45′42″W
## Time difference and current local times
There is no time difference between Bauru and Corrientes.
-03
-03
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 125 kg (275 pounds).
## Frequent Flyer Miles Calculator
Bauru (BAU) → Corrientes (CNQ).
Distance:
703
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
703
Round trip? | 522 | 1,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-50 | latest | en | 0.767991 |
https://hitchesguide.com/how-many-miles-is-14-tank-of-gas/ | 1,669,962,752,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710898.93/warc/CC-MAIN-20221202050510-20221202080510-00074.warc.gz | 325,851,021 | 22,895 | # How Many Miles Is 1/4 Tank Of Gas
How Many Miles Is 1/4 Tank Of Gas. This calculator allows you to compute the miles per gallon of your car via the miles travelled and the number of gallons used. Also, if you enter the cost per gallon and how many miles you. How many miles can you drive on a full tank of gas calculator?
This comes out to 23 mpg times 20 gallons which equals 460 miles. Just looking for a rough estimate of my 90 fj62 gas gauge. It's actually brand new cause my original one froze up with rust.
## The Gas Tank Is 1/4 Full, Or 3/4 Empty – semi-rad.com
According to Google, the tank will hold 17 gallons of gas, and should get between 25-31 mpg. If you get 25-31 MPG that's 106.25-131.75 mile range off 1/4 tank of gas. Presuming you mean 40 miles each way. Total 80 miles, if consumption rate be at 20 mpg, then 4 gallons of fuel would be required.According to the chart, you can expect to get anywhere from 30 miles to more than 100 miles on a nearly empty tank, depending on the car. Of course, the real-life numbers will vary based on how you…. How Many Miles Is 1/4 Tank Of Gas
Calculating your car’s range to calculate your car’s total range, multiply its average highway miles per gallon by its fuel. We need to know fuel capacity and gas mileage. If you have 1/4 tank of gas, thats 4. 25 gallons. To do this, we take our average mpg and multiply it by the number of gallons our car can hold in its tank.
I noticed that there is no gas light that comes on. How far will a 1/4 tank of gas? You should be able to cover about 120 150 miles on that. how far can i drive on 1/4 tank of gas? Gallon ↔ miles conversion table.
## I rented a car, and the gas gauge tells you if the tank is 1/4 full, 1/
Here’s Why You Should Never Let Your Car Get Below a 1/4 Tank of Gas Subscribe for more daily vids ► youtube.com/channel/UCuxpxCCevIlF-k-K5YU8XPA?sub_confirmation=1 ⬇️Scotty’s Top DIY Tools: 1. Bluetooth Scan Tool: amzn.to/2nfvmaD 2. Mid-Grade Scan Tool: amzn.to/33dKI0k 3. My Fancy (Originally \$5,000) Professional Scan Tool: amzn.to/31khBXC 4. Cheap Scan Tool: amzn.to/2D8Tvae 5. Dash Cam (Every Car Should Have One): amzn.to/2YQW36t 6. Basic Mechanic Tool Set: amzn.to/2tEr6Ce 7….
Replacing your fuel pump is expensive. Avoid costly repairs; watch the video. SUBSCRIBE TODAY! ► youtube.com/c/smartdrivetest INTRODUCTION Hi there smart drivers. Rick with Smart Drive Test – quick tip for driving smart: Keep your fuel tank above a quarter tank! If you constantly run it below a quarter tank–down to the empty mark–eventually you're going to burn the fuel pump out. And to replace that fuel pump is about \$650 to \$1,000. I had a friend the other day, her son was driving…
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Miles returns this week as Alex and Ethan see how far they can get with only a litre (0.26 gallons) of fuel… SUBSCRIBE: bit.ly/CTSubscribe —– Follow Car Throttle —– Subscribe to Car Throttle: bit.ly/CTSubscribe On our website: carthrottle.com On Facebook: facebook.com/carthrottle On Twitter: twitter.com/carthrottle —– Music by —– Epidemic Sound epidemicsound.com
In this video, I tell you how much gas is left after the gas light comes on. This seems to be something that many drivers wonder, and I wanted to answer it in this video. I provide you with a general answer that will help you understand the amount of gas that is left in the gas tank, but it can vary from one car to the other. This means that your specific car might have a different amount or let you drive further once the gas light comes on, or the complete opposite. But it is good to know…
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BUY A GAS CAN: amzn.to/2mItSCj Today, I talk about why I NEVER let my tuck go below a 1/4 tank. Unfortunately, I had the experience of running out of diesel in one of my past diesel trucks. As you can probably imagine, it is a horrible experience. Wasted time, energy, and of course embarrassment, as you sit on the side of the road waiting for someone to bring you fuel. An easy tip to avoid this, is to always fill up when your vehicle hits a 1/4 tank. This tip works for gas, diesel, and…
## Final Words
Size of gas tank. The gas tank size can also affect how far a car can go on a full tank. If the gas tank is larger, the gas tank capacity will be higher.
How Many Miles Is 1/4 Tank Of Gas. For example, a car with a 45. How many miles are you guys getting from 1/4 a tank of gas? All ford mustangs | model.
I have a passion for anything automotive. I love that there is always an answer to a problem with vehicles, and I enjoy the challenge of finding it. John enjoys recreational motorsports and spending time with his family outside of work. | 1,237 | 4,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-49 | latest | en | 0.87212 |
https://shirleenjeweliciouss.com/goldsmith/best-answer-how-many-diamonds-are-in-64-blocks.html | 1,653,362,661,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00330.warc.gz | 597,962,638 | 18,646 | # Best answer: How many diamonds are in 64 blocks?
Contents
## How many diamonds are in 64 diamond blocks Minecraft?
This is 27,648 Diamonds in total.
## How many diamonds is 64 blocks?
A stack of diamond blocks contains 64 diamond blocks, which are equivalent to 9 diamonds each.
## How many diamond are in a block?
A block of diamond is a precious mineral block equivalent to nine diamonds.
## How many diamond blocks is 10000 diamonds?
This is 10,000 diamonds visualised (equal to 17 stacks + 23 blocks and 1 diamond) good luck Iskall!! : r/HermitCraft.
## Can diamond blocks be turned into diamonds?
Diamond Blocks can only be mined successfully with an Iron, Diamond, or Netherite Pickaxe. To get the Diamonds back, simply put the Diamond Block in the Crafting Grid to split it back into nine Diamonds. Diamond Blocks can be used to power a Beacon.
## Can diamond break iron?
A block of diamond would scratch a block of iron, and diamond vise would slowly flatten an iron block, while a fast impact would rather shatter diamond. Also, you can easily flatten or break a small piece of steel with a steel hammer.
## What is M in Minecraft?
Metric units
Officially [1], Minecraft uses the metric system, and each block is considered to be 1 cubic meter. … Every second marker (ignoring the zero-mark) is then a 10-meter mark.
## How many blocks are in Minecraft?
There are over 150 different types of blocks in Minecraft, including environmental features like air and water, which cannot be obtained, various types of wood, various colors of wool, various, slabs, stairs, utility blocks, and the list goes on.
## How many diamonds are in a world of Minecraft?
There is on average 3.7 diamonds in a chunk, with a chunk being a 16×16 area. The total amounts of chunks is 14,062,500,000,000. This means there’s 14,062,500,000,000 * 3.7 = 52,031,250,000,000 diamonds in a world!
## How much is a diamond block worth in real life?
The current assessed value of a one carat diamond (US) is currently listed as any where in between, \$1,910-\$15,650, with a recommended average value of \$4,280. With these figures in mind, one can quickly see that owning a real life version of this tool would be an incredible investment and endeavor.
## Can you find a 10 vein of diamonds in Minecraft?
Due to how Minecraft vein generation works, it is possible to find veins with up to 10 diamonds even though the max vein size is 8. As the Minecraft Wiki stated in Meantub’s answer: Diamond ore attempts to generate 1 time per chunk in veins of 0-10 ore, in layers 1 to 16 in all biomes.
## What is tuff Minecraft?
Tuff is mostly used for decorative purposes in Minecraft. So you can add it to a wall and use it for any particular decoration project. It is an ornamental rock that can replace Stone, Diorite, Andesite, Granite, and Deepslate. Also, any ore that generates in Tuff becomes its Deepslate variety.
IT IS SURPRISING: Why was narrator not keen on buying diamonds from him the diamond maker? | 720 | 3,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | latest | en | 0.934735 |
https://nrich.maths.org/public/leg.php?code=-395&cl=4&cldcmpid=320 | 1,508,797,099,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826840.85/warc/CC-MAIN-20171023221059-20171024001059-00677.warc.gz | 797,589,212 | 9,151 | # Search by Topic
#### Resources tagged with physics similar to Growing:
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### There are 80 results
Broad Topics > Applications > physics
### Approximately Certain
##### Stage: 4 and 5 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Cobalt Decay
##### Stage: 5 Challenge Level:
Investigate the effects of the half-lifes of the isotopes of cobalt on the mass of a mystery lump of the element.
### Big and Small Numbers in Chemistry
##### Stage: 4 Challenge Level:
Get some practice using big and small numbers in chemistry.
### Sweeping Satellite
##### Stage: 5 Challenge Level:
Derive an equation which describes satellite dynamics.
### The Not-so-simple Pendulum 1
##### Stage: 5 Challenge Level:
See how the motion of the simple pendulum is not-so-simple after all.
### Big and Small Numbers in Physics - Group Task
##### Stage: 5 Challenge Level:
Work in groups to try to create the best approximations to these physical quantities.
### Earth Orbit
##### Stage: 5 Challenge Level:
Follow in the steps of Newton and find the path that the earth follows around the sun.
### Universal Time, Mass, Length
##### Stage: 5 Short Challenge Level:
Can you work out the natural time scale for the universe?
### Resistance
##### Stage: 5 Challenge Level:
Find the equation from which to calculate the resistance of an infinite network of resistances.
### Escape from Planet Earth
##### Stage: 5 Challenge Level:
How fast would you have to throw a ball upwards so that it would never land?
### Whoosh
##### Stage: 5 Challenge Level:
A ball whooshes down a slide and hits another ball which flies off the slide horizontally as a projectile. How far does it go?
### Alternative Record Book
##### Stage: 4 and 5 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### The Amazing Properties of Water
##### Stage: 4 and 5 Challenge Level:
Find out why water is one of the most amazing compounds in the universe and why it is essential for life. - UNDER DEVELOPMENT
### Stemnrich - Technology
##### Stage: 3 and 4 Challenge Level:
This is the technology section of stemNRICH - Core.
### Lunar Leaper
##### Stage: 5 Challenge Level:
Gravity on the Moon is about 1/6th that on the Earth. A pole-vaulter 2 metres tall can clear a 5 metres pole on the Earth. How high a pole could he clear on the Moon?
### Cannon Balls
##### Stage: 5 Short Challenge Level:
How high will a ball taking a million seconds to fall travel?
### Which Twin Is Older?
##### Stage: 5
A simplified account of special relativity and the twins paradox.
### Motion Capture
##### Stage: 3 and 4 Challenge Level:
Explore displacement/time and velocity/time graphs with this mouse motion sensor.
### Gravity Paths
##### Stage: 5 Challenge Level:
Where will the spaceman go when he falls through these strange planetary systems?
### Moving Stonehenge
##### Stage: 5 Challenge Level:
A look at the fluid mechanics questions that are raised by the Stonehenge 'bluestones'.
### Mach Attack
##### Stage: 5 Challenge Level:
Have you got the Mach knack? Discover the mathematics behind exceeding the sound barrier.
### Constantly Changing
##### Stage: 4 Challenge Level:
Many physical constants are only known to a certain accuracy. Explore the numerical error bounds in the mass of water and its constituents.
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A look at a fluid mechanics technique called the Steady Flow Momentum Equation.
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When a mixture of gases burn, will the volume change? | 1,581 | 7,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-43 | longest | en | 0.835272 |
https://cpromptgames.com/.tmb/forum/sfx152t.php?page=5-cm-to-inches-4270aa | 1,618,433,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00049.warc.gz | 299,795,787 | 11,352 | among others. Centimeter [cm] Inch [in] 0.01 cm: 0.0039370079 in: 0.1 cm: 0.0393700787 in: 1 cm: 0.3937007874 in: 2 cm: 0.7874015748 in: 3 cm: 1.1811023622 in: 5 cm: 1.968503937 in: 10 cm: 3.937007874 in: 20 cm: 7.874015748 in: 50 cm: 19.6850393701 in: 100 cm: 39.3700787402 in: 1000 cm: 393.7007874016 in The inch is a popularly used customary unit of length in the United States, Canada, and the United Kingdom. How far is 5 centimeters in inches? 5.5 centimeters equal 2.1653543307 inches (5.5cm = 2.1653543307in). An inch (symbol: in) is a unit of length. To convert 5.5 cm to in multiply the length in centimeters by 0.3937007874. Thus, for 5.5 centimeters in inch we get 2.1653543307 in. Thus, for 5.5 centimeters in inch we get 2.1653543307 in. A centimeter is equal to 0.01(or 1E-2) meter. There are 2.54 centimeters in an inch. It is defined as 1⁄12 of a foot, also is 1⁄36 of a yard. Please, choose a physical quantity, two units, then type a value in any of the boxes above. centimeters to The centimeter practical unit of length for many everyday measurements. What is the formula to convert from The Cm to Inches Conversion Calculator is used to convert centimeters to inches. In this case you will have: Value in inches? To convert 5.5 cm to in multiply the length in centimeters by 0.3937007874. 5 cm to in conversion. Supose you want to convert 5 cm into In this case we should multiply 5 Centimeters by 0.39370078740157 to get the equivalent result in Inches: 5 Centimeters x 0.39370078740157 = 1.9685039370079 Inches inches. Fractions are rounded to the nearest 8th fraction. 5 Centimeters is equal to how many Inches. A centimeter, or centimetre, is a unit of length equal to one hundredth of a meter. Here is the formula: Value in inches = value in centimeters × 0.39370078740157 Five Centimeters is equivalent to one point nine six nine Inches. 5.5 Centimeter to Inches, 5.5 Centimeter in Inches, 5.5 cm to in, 5.5 cm in in, 5.5 Centimeters to in, 5.5 Centimeters in in, 5.5 Centimeter to Inch, 5.5 Centimeter in Inch, 5.5 cm to Inches, 5.5 cm in Inches, 5.5 Centimeter to in, 5.5 Centimeter in in, 5.5 cm to Inch, 5.5 cm in Inch, 5.5 εκαÏοÏÏÏμεÏÏο Ïε ίνÏÏα, 5.5 সà§à¦¨à¦à¦¿à¦®à¦¿à¦à¦¾à¦° মধà§à¦¯à§ à¦à¦à§à¦à¦¿, 5.5 सà¥à¤à¤à¥à¤®à¥à¤à¤° सॠà¤à¤à¤, 5.5 à¹à¸à¸à¸à¸´à¹à¸¡à¸à¸£à¸à¸´à¹à¸§, 5.5 સà«àª¨à«àªà«àª®à«àªàª° àªàªàª. How long is 5 centimeters? The conversion factor from Centimeters to Inches is 0.39370078740157. It is also the base unit in the centimeter-gram-second system of units. What is the There are 12 inches in a foot, and 36 inches in a yard. centimeters to the corresponding value in How to transform inches? An inch is a unit of length equal to exactly 2.54 centimeters. inches, just multiply the quantity in inches? To calculate 5 Centimeters to the corresponding value in Inches, multiply the quantity in Centimeters by 0.39370078740157 (conversion factor). The 5.5 cm in in formula is [in] = 5.5 * 0.3937007874. inches = 5 × 0.39370078740157 = 1.9685039370079. centimeters to The centimeter (symbol: cm) is a unit of length in the metric system. Though traditional standards for the exact length of an inch have varied, it is equal to exactly 25.4 mm. To calculate a value in 5 Centimeters is equivalent to 1.9685039370079 Inches. To calculate 5 Centimeters to the corresponding value in Inches, multiply the quantity in Centimeters by 0.39370078740157 (conversion factor). Simply use our calculator above, or apply the formula to change the length 5.5 cm to in. The 5.5 cm in in formula is [in] = 5.5 * 0.3937007874. centimeters by 0.39370078740157 (the conversion factor). To find out how many Centimeters in Inches, multiply by the conversion factor or use the Length converter above. In this case we should multiply 5 Centimeters by 0.39370078740157 to get the equivalent result in Inches: 5 Centimeters x 0.39370078740157 = 1.9685039370079 Inches. How to convert 5 cm: 1.97 in: 6 cm: 2.36 in: 7 cm: 2.76 in: 8 cm: 3.15 in: 9 cm: 3.54 in: 10 cm: 3.94 in: 11 cm: 4.33 in: 12 cm: 4.72 in: 13 cm: 5.12 in: 14 cm: 5.51 in: 15 cm: 5.91 … inches conversion factor? centimeters to Centimeters to Inches Conversion Formula To convert from cm to inches, use the following conversion equation: inches = cm / 2.54 Note: Values are rounded to 4 significant figures. centimeters in How to convert 5 centimeters to inches To calculate a value in centimeters to the corresponding value in inches, just multiply the quantity in centimeters by 0.39370078740157 (the conversion factor). 5 cm: 1.9685 in: 1 31/32 in: 6 cm: 2.3622 in: 2 23/64 in: 7 cm: 2.7559 in: 2 3/4 in: 8 cm: 3.1496 in: 3 5/32 in: 9 cm: 3.5433 in: 3 35/64 in: 10 cm: 3.9370 in: 3 15/16 in: 20 cm: 7.8740 in: 7 7/8 in: 30 cm: 11.8110 in: 11 13/16 in: 40 cm: 15.7840 in: 15 3/4 in: 50 cm: 19.6850 in: 19 11/16 in: 60 cm: 23.6220 in: 23 5/8 in: 70 cm: 27.5591 in: 27 9/16 in: 80 cm: 31.4961 in: 31 1/2 in: 90 cm: 35.4331 in: 35 7/16 in: 100 cm: … Using this converter you can get answers to questions like: 5 centimeters equals 1.969 inches because 5 times 0.3937 (the conversion factor) = 1.969. Converting 5.5 cm to in is easy. | 1,914 | 5,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-17 | latest | en | 0.81901 |
https://waverz.info/viewtopic.php?t=3242 | 1,550,536,828,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489282.7/warc/CC-MAIN-20190219000551-20190219022551-00277.warc.gz | 736,396,457 | 9,571 | # Long division homework help - Marist parking assignments
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Long and synthetic division. Math Learning Video. It is the preferred method when dividing by a number with two more digits particularly if the division is not exact.
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## Division Custom
Homework Help Division Remainders - backstreet- bar. com Long Division Calculator- Shows all work and help with write a dissertation 2 weeks steps for any numbers. 1 and 2- digit by homework help division remainders 2- digit multiplication.
Thursday, December 18,. Division Worksheets | Long Division Worksheets - Math Aids This Long Division Worksheet the number of digits for the divisors and quotients may be varied between 1 & 3.
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Soal bahasa inggris report text essay | 3,874 | 19,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-09 | latest | en | 0.868804 |
http://mathhelpforum.com/calculus/193439-acumulated-present-value.html | 1,480,870,278,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00118-ip-10-31-129-80.ec2.internal.warc.gz | 177,783,660 | 9,346 | ## Acumulated Present Value
Find Acumulated Present Value of an investment over 10 years of a continous money flow of 2700 per year and interest rate is .1 compounded continously.
I am stuck on this one... i got a negative answer of -427.134, and im pretty sure i am wrong. Any help?
i had 2700 = Rt((e^-1/.1)- (1/.1)) | 91 | 320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-50 | longest | en | 0.924286 |
https://www.physicsforums.com/threads/dynamics-newtons-second-law.791168/ | 1,709,489,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00804.warc.gz | 929,772,992 | 17,479 | Dynamics: Newton's Second Law
• yandereni
In summary, the conversation discusses a problem involving a racecar covering a quarter mile track in 6.40 seconds and the driver's acceleration during this time. The first equation is used to solve for the "g's" and the second equation is used to solve for the horizontal force, with the correct answer being 2g or approximately 19.62m/s^2. The direction of the car's movement is important when setting up the equations.
Homework Statement
Here's the problem:
A particular racecarcan cover a quartermile track (402m) in 6.40s starting from standstill. Assuming that the acceleration is constant, how many "g's"(all i know is that it means accleration due to gravity) does the driver experience? If the combined mass of the driver and the racecar is 485kg, what horizontal force must the road exert on the tires?
Homework Equations
1. Δy(displacement) = [(Velocityinitial)(Δt)] + ((1/2)acceleration)(Δt)2
2. Forcegrav + (acceleration)(mass)= Forcetotal
I'm not even sure if the second equation even exists but that's what i made out from the problem.
The Attempt at a Solution
I tried solving for the "g's" using the first equation and i substituted the possible things:
402m = [(0m/s)(6.4s)] + [((1/2)a)(6.4s)]
i don't know if this is the right equation but its the closest that i can get
then i tried solving for the horizontal force with the second equation and substituted the given:
[(-9.8m/s2)(485kg)] + [(19.6m/s2)(485kg)] = Forcetotal
and in case oyu are wondering where i got the acceleration(19.6m/s2) its from the first equation i solved (which i think is wrong)
-yandereni
Hi, yandereni.
You did well in the first part. It looks correct.
yandereni said:
then i tried solving for the horizontal force with the second equation and substituted the given:
[(-9.8m/s2)(485kg)] + [(19.6m/s2)(485kg)] = Forcetotal
What are the directions of the two forces you've included here? What is the direction the question is concerned with?
I
Bandersnatch said:
Hi, yandereni.
You did well in the first part. It looks correct.What are the directions of the two forces you've included here? What is the direction the question is concerned with?
Bandersnatch said:
Hi, yandereni.
You did well in the first part. It looks correct.What are the directions of the two forces you've included here? What is the direction the question is concerned with?
I looked at the problem again and there were no directions stated. And when i told my other classmates about the first part if i did it right, they said no and the answer should be 2.00m/s2.
yandereni said:
I looked at the problem again and there were no directions stated.
what's this then, eh?
yandereni said:
what horizontal force must the road exert on the tires?
yandereni said:
when i told my other classmates about the first part if i did it right, they said no and the answer should be 2.00m/s2.
That's wrong. The car would cover some 41m at that acceleration.
Maybe they meant 2g, not 2m/s^2?
oh, now i understand. but can i solve the horizontal force without any directions given?
yandereni said:
without any directions given?
But you do know which direction the car is moving, don't you? All you need to make sure when setting up your equation is not to include forces that are not acting (i.e., have no component) in the direction you're interested in.
Thank you very much! I got the right answer and i get the concept now. Thanks a lot! :)
1. What is Newton's Second Law of Motion?
Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. In simpler terms, this means that the more force applied to an object, the greater its acceleration will be, and the more mass an object has, the slower its acceleration will be.
2. How is Newton's Second Law mathematically represented?
The mathematical representation of Newton's Second Law is F = ma, where F represents the net force acting on an object, m represents the mass of the object, and a represents the acceleration of the object.
3. What is the relationship between force, mass, and acceleration according to Newton's Second Law?
According to Newton's Second Law, force, mass, and acceleration are directly related. This means that as force increases, acceleration also increases, and as mass increases, acceleration decreases.
4. How does Newton's Second Law apply to real-life situations?
Newton's Second Law can be seen in action in various real-life situations. For example, when a car accelerates, the engine applies a force to the car, causing it to accelerate. Similarly, when a person pushes a shopping cart, the force applied causes it to accelerate. This law also explains why it is harder to push a heavier object than a lighter one, as the heavier object has more mass and therefore requires more force to accelerate.
5. What are some common misconceptions about Newton's Second Law?
One common misconception about Newton's Second Law is that objects always move in the direction of the net force applied. However, this is not always the case, as other factors such as friction can also affect an object's motion. Another misconception is that the acceleration of an object is only dependent on the net force applied, when in reality, it also depends on the mass of the object. | 1,255 | 5,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-10 | latest | en | 0.943764 |
https://softwareengineering.stackexchange.com/questions/337993/find-transformation-function-for-point-on-an-image | 1,726,784,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00841.warc.gz | 484,865,380 | 41,975 | # Find transformation function for point on an image
I am trying to develop an algorithm, but I barely can describe it so I'm not sure what to look for. I have two images img_A and img_B. img_A is the original, while img_B is an image of the same item but slightly different. Let's say one is a photo of a person, while the other is a drawing of that same person. This means that img_B is very similar, to img_A, but it could be slightly rotated and maybe a bit bigger/smaller.
What I am trying to do is : from a point from img_A a_a get the corresponding a_b on the img_B. I think I need to define two points on img_A and their correspoinding points on img_B to define the axis correspondance. Once I have those 4 points, I calculate a_b. The 4 "axis points" will be defined by hand once, then the User will select a point "a_a" by hand and the correspoinding "a_b" will be calculated.
On a visual example: I select the center of the left eye and the lower point of the chin from both images. I then select the right eye from img_A and as a result I get the coordinates of a_b. (if the drawing was perfectly done, the coordinate of a_b will correspond to the right eye on img_b).
How I develop this? I need a matrix to multiply a_a to obtain a_b? how?
• Are you looking for an algorithm to find corresponding points in those images automatically? Or do you intend to let a human pick some corresponding points, and you are just looking for the related coordinate transformation? Please clarify! Commented Dec 12, 2016 at 16:28
• You would map that with a vector. Though constructing that vector is extremely complex and definitely too broad as a question.
– user188153
Commented Dec 12, 2016 at 16:52
• @DocBrown I edited the question. I will do all by hand. Pick the first 2+2 reference points and then select one on the original image and see where it should be on the second one. Commented Dec 12, 2016 at 16:56
• @ThomasKilian Why too broad? I think it's not more complex that what a Photoshop/Gimp does when you resize an image, only done in a different portion of the screen. Commented Dec 12, 2016 at 17:01
• @mobinoob: it would have probably been too complex if it you were looking for an algorithm to find the points automatically. Commented Dec 12, 2016 at 17:07
You are looking for an affine transformation of the form
``````f(v) = s * R * v + t
``````
where v is a 2D vector in the coordinate system of image A, `s` is a scaling factor (positive real value), `R` is a 2x2 rotation matrix of the form
`````` [cos(alpha) sin(alpha)]
[-sin(alpha) cos(alpha)]
``````
and `t` is a 2D vector, the translation. `f(v)` then gives you the corresponding point in image B. So what you need to calculate is s, alpha and t.
Lets assume you have two points `p1` and `p2` in the coordinate system of image A, and two corresponding points `q1` and `q2` in the coordinate system of image B. Then you can easily calculate `s = |q2-q1|/|p2-p1|`. The angle `alpha` is the angle difference between the direction angles of the vector from p1 to p2 and the vector from q1 to q2. The direction angle for p1 to p2 is calculated by `atan(dy/dx)` where dx and dy are the x and y coordinates of `p2-p1` (for q1 to q2 it works similar). To avoid problems of a zero denominator and the sign, use the `atan2` function, which is available in many programming languages.
So now you have `s` and `R`, and from `f(p1)=q1` you can now deduce
``````t= q1- s * R * p1
``````
which gives you finally the translation `t`.
• That works for a simple transformation, but not for a photo to a drawing as the OP requires.
– user188153
Commented Dec 12, 2016 at 17:37
• @ThomasKilian: honestly, we don't know what the OP really requires. He said he wants to use two control points, and used the photo/drawing example as - just an example. Lets see if my answer helps him. Commented Dec 12, 2016 at 17:41
• I believe this is exactly what I need. I am going to test it soon. see other comment below Commented Dec 13, 2016 at 18:28
• @ThomasKilian This is what I want to do: imgur.com/a/4XxHg note that f(v) does not correspond to the chin just to explain that the right version of the monalisa does not have the same proportion as the left one. This is the easiest way to see if a reproduction is similar to the original, even if a bit rotated. Commented Dec 13, 2016 at 18:30
• @mobinoob Indeed that's what Doc Brown suggested. From your description it sounded like photo vs. drawing which is not really possible to achieve with an affine transformation.
– user188153
Commented Dec 13, 2016 at 22:33 | 1,220 | 4,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-38 | latest | en | 0.942493 |
https://www.mrexcel.com/board/threads/help-with-a-formula.802780/ | 1,696,354,857,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511170.92/warc/CC-MAIN-20231003160453-20231003190453-00455.warc.gz | 955,408,435 | 16,898 | # Help With a Formula
#### OdysseusVII
##### New Member
I'm not sure if this is even possible but I'm trying to populate a series of different answers depending on the value of previous cells. The different formulas are below, but I would need any given cell in the 5th column to be able to return any of these four answers based on the previous 4 cells. Right now we are populating the 5th column manually and it is hugely time consuming.
1 #N/A 1 1 Complete 1 #N/A #N/A #N/A No Cert #N/A #N/A #N/A #N/A Not Started 1 #N/A 1 #N/A Not Compliant
<tbody>
</tbody>
### Excel Facts
Which lookup functions find a value equal or greater than the lookup value?
MATCH uses -1 to find larger value (lookup table must be sorted ZA). XLOOKUP uses 1 to find values greater and does not need to be sorted.
Hi,
You don't actually give the criteria, though I presume it's based on the number of 1s in a given row:
=INDEX({"Not Started";"No Cert";"Not Compliant";"Complete"},1+COUNTIF(A1:D1,1))
You also don't mention what should be the result if all four cells contain 1s, so I presume this is not a possibility.
Regards
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Go back | 594 | 2,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-40 | latest | en | 0.857618 |
deemples.com | 1,606,926,059,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00365.warc.gz | 259,433,178 | 20,735 | # 15 May How to calculate a USGA Golf Handicap Index
See the “HCP Index” next to your name on the Deemples app? Wonder how this number if calculated from the scores you have submitted on the app. Here we’ll show you how a USGA Golf Handicap Index (HCP Index) is calculated.
## Formula to Calculate Handicap Index
Handicap Index = Average of your qualified Handicap Differentials * 0.96
2 questions here:
1. How do you calculate Handicap Differentials?
2. Which Handicap Differentials qualify for Handicap Index calculations?
### 1. Formula to calculate Handicap Differential per game
Handicap Differential = (Adjusted Gross Score – Course Rating) * 113 / Slope Rating
More questions here:
1. What is Adjusted Gross Score (AGS)?
2. What is Course Rating?
3. What is Slope Rating?
#### What is Adjusted Gross Score
Gross Score = Sum of your scores for each hole.
Adjusted Gross Score however, takes into account Equitable Stroke Control (ESC) where the maximum allowable score per hole for handicapping purposes is dependent on your course handicap. See the following table:
Course Handicap = Slope rating / 113 * Handicap Index
#### Course Rating and Slope Rating
Course Rating and Slope Rating are numbers that are available on the score card that show a difficulty of a golf course from a specific tee box in relation to another tee box of the same golf course or any other golf courses using the same USGA rating system.
They usually look like this:
• Course Rating – the higher the number, the more difficult the golf course
• Slope Rating – the higher the number, the more difficult the golf course
If a woman plays off the White Tee box, they wouldn’t be using the Men’s course and slope rating numbers for the white tee, but the women’s rating instead. If these ratings are not available on the score card, check with the golf course.
Going back to how a Course Handicap is calculated, and let’s say for this example, we’re playing off the men’s blue tee box with a 13.9 HCP Index:
Course Handicap = 132 / 113 * 13.9
= 16.237 or 16 (rounding to the nearest whole number)
So since my Course Handicap is 16 from the blue tees at Kinrara Golf Club, for my Adjusted Gross Score, based on the table above, the maximum allowable score per hole for handicapping purposes, would be 7.
My Gross Score for the game is the sum of all my scores for each hole. Let’s say for this game my score for each hole was:
As you can see on hole 2 and hole 14, I had a 9, which both are 2 over the maximum allowable score for handicapping purposes. Therefore, my Adjusted Gross Score for this game is 96 for handicapping purposes, whilst my Gross Score is still 100 for competition purposes.
Net Score = Gross Score – Course Handicap
In a competition, my net score would be 100 – 16 = 84.
#### Back to Handicap Differentials
Handicap Differential = (Adjusted Gross Score – Course Rating) * 113 / Slope Rating
Now that we know what Adjusted Gross Score, Course Rating, and Slope Ratings are, for this example we use the numbers above.
The Handicap Differential for this game
= (96 – 71.7) * 113 / 132
= 20.8 (round to 1 decimal point from 20.802)
### 2. Which Handicap Differentials qualify for Handicap Index Calculations
Now that we know how to calculate the handicap differential for each game, it’s important to know which of the handicap differentials qualify for Handicap Index calculations.
Depending on the number of handicap differentials from games you have posted results for, the number of handicap differentials you use are different:
Let’s say for this example, we have played 7 games, and have 7 differentials to use. Including the 20.8 we had above, we also have:
Based on 7 available differentials, and based on the table above that, the number of differentials used for handicap index calculation would be the lowest 2 = 14.8 and 15.9.
So based on:
Handicap Index = Average of your qualified Handicap Differentials * 0.96
My new Handicap Index
= (14.8 + 15.9)/2 * 0.96
= 14.736
Do not round this number, but remove the digits up to the tenth decimal.
So 14.736 becomes a handicap index of 14.7.
If the number ended up being 14.787, DO NOT round handicap index to 14.8, but again, remove digits to the tenth decimal, which will still give you a handicap index of 14.7.
DO NOT ROUND, but DROP digits to the nearest tenth decimal.
## Migrating to World Handicap System (WHS)
Starting in 2020, some countries in the world are on a 2 year plan to migrate from their existing handicap systems to the World Handicap System (WHS). This effort is to standardise golf handicaps around the world, so everyone can compete on a level playing field.
For this to happen, there’s a few variables that have to be standardised, which mostly includes the difficulty levels of the golf courses. 2 measurements used a) Course Rating and b) Slope Rating, would have to be implemented for golf courses all around the world. With the USGA rated golf courses already have course and slope rating implemented, which is more than 50% of the world, the rest will follow suit.
## Deemples Handicap System
This transition to WHS will take 2 years, until end 2021. For the time being Deemples will continue using the USGA method since it requires way less variables. Only 1) Adjusted Gross Score 2) Course Rating, and 3) Slope Rating for it to be accurately calculated.
We’ll do another update again when we switch over to WHS, which might be later down the road. Meanwhile, we kinda agree with: https://www.golfdigest.com/story/voices-the-flaw-in-the-new-world-handicap-system-dean-knuth | 1,347 | 5,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-50 | longest | en | 0.851482 |
http://www.enotes.com/homework-help/what-answer-question-2-http-postimg-org-image-446086 | 1,461,987,749,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111592.84/warc/CC-MAIN-20160428161511-00003-ip-10-239-7-51.ec2.internal.warc.gz | 497,881,101 | 11,968 | # What is the answer for question 2) ? http://postimg.org/image/zft18m0fb/ (Reminder) : This is 1 question.
Posted on
The function `g(x) = (-3x^3)/4 + 3x` . A secant line to any curve is one that intersects two points on it.
To determine the slope of the secant line between points representing x = a and x = b, evaluate g(a) and g(b). The slope is given by `(g(b) - g(a))/(b - a)` .
For the given function, the slope of the secant line in the interval x = 1 and x = 2 is: `S = (g(2) - g(1))/(2-1)`
= `((-3*2^3)/4 + 3*2 + (3*1^3)/4 - 3*1)/(2 - 1)`
= `((-3*8)/4 + 6 + 3/4 - 3)/1`
= `(-6 + 6 + 3/4 - 3)/1`
= `(3/4 - 3)/1`
= -2.25
Similarly the slope of the secant line in the other intervals can also be determined. | 274 | 723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-18 | longest | en | 0.843819 |
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oltre un anno ago
Solved
The Goldbach Conjecture
The <http://en.wikipedia.org/wiki/Goldbach's_conjecture Goldbach conjecture> asserts that every even integer greater than 2 can ...
oltre un anno ago
Solved
Find the alphabetic word product
If the input string s is a word like 'hello', then the output word product p is a number based on the correspondence a=1, b=2, ....
oltre un anno ago
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Binary numbers
Given a positive, scalar integer n, create a (2^n)-by-n double-precision matrix containing the binary numbers from 0 through 2^n...
oltre un anno ago
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Counting Money
Add the numbers given in the cell array of strings. The strings represent amounts of money using this notation: \$99,999.99. E...
oltre un anno ago
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Cell joiner
You are given a cell array of strings and a string delimiter. You need to produce one string which is composed of each string fr...
oltre un anno ago
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Remove all the words that end with "ain"
Given the string s1, return the string s2 with the target characters removed. For example, given s1 = 'the main event' your ...
oltre un anno ago
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Pascal's Triangle
Given an integer n >= 0, generate the length n+1 row vector representing the n-th row of <http://en.wikipedia.org/wiki/Pascals_t...
oltre un anno ago
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Which values occur exactly three times?
Return a list of all values (sorted smallest to largest) that appear exactly three times in the input vector x. So if x = [1 2...
oltre un anno ago
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Add damping to a mass spring system
Model an ideal mass-spring-damper system shown below where the spring is initially stretched. <<http://blogs.mathworks.com/im...
oltre un anno ago
Solved
Model a mass spring system
Model an ideal mass-spring system shown below where the spring is initially stretched. <<http://blogs.mathworks.com/images/se...
oltre un anno ago | 1,401 | 5,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-26 | longest | en | 0.849587 |
https://www.jiskha.com/computers/operating_systems/?page=3 | 1,503,388,352,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110485.9/warc/CC-MAIN-20170822065702-20170822085702-00079.warc.gz | 882,359,109 | 13,312 | Homework Help: Computers: Operating Systems
Recent Homework Questions About Operating Systems
SS. help
What impact did this have on the presidents who were trying to get rid of the spoils system and higher people based on their qualifications (aka the Pendelton Act) i have looked up different sites but it still don't help me.
Algebra
Check my answer- Solve the system of equations algebraically. Show all of your steps. Y=x^2+2x Y=3x+20 My answer: X^2+2x+3x+20 X^2-x-20 (x-5),(x+4)=0 x=5, x=-4 y=35, y=8 Thank you!
Ethics
What are the fundamental values of each of these health systems: The system In Mexico, the system in Canada, the system in the United Kingdom, and the system in the United States? I think it has to do with comparison of the healthcare systems in Canada and the United States is...
probability and statistics
Let Θ1, Θ2, W1, and W2 be independent standard normal random variables. We obtain two observations, X1=Θ1+W1,X2=Θ1+Θ2+W2. Find the MAP estimate θ^=(θ^1,θ^2) of (Θ1,Θ2) if we observe that X1=1, X2=3. (You will have to solve a system of two linear equations.)
Math
particle P of mass m kg is attached to two fixed points A and B by two identical model springs, each of stiffness k and natural length l . The point A is at a height / l above the point B. The particle is free to oscillate vertically under gravity. The stiffness of each spring...
Physics
A Micro –Hydro turbine generator rotor is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity of 837 rpm. The radius of the rotor is 0.62 m and its rotational acceleration is 5.9 rad/s2. What is the rotor’s angular velocity (in rad/...
PE
your digestive system has three main functions:digestion, absorption and elimination. in which organ dose absorption begin? a. mouth b.stomach c.(small intestine) d.gallbladder
physics
a man weighs 162 pounds, and is 69 inches tall. his forearm is parallel to the ground. he holds an 8 pound weight in his hand. the biceps has an angle of insertion of 85 degrees with a distance of 1.4 inches at insertion. the system is equilibrium. what force is the biceps ...
Chemistry
A 152 g sample of ice at –37 degree C is heated until it turns into liquid water at 0 degree C. How do you find the change in heat content in the system? I've found H(I) [Change in heat of the ice] to be 11416.72 but im still stuck on the change of heat for the water.
Biology
The p53 enzyme system prevents progression to which stage of the cell cycle? choices are: 1. G1 2.Cytokineses 3.S phase 4. Mitosis 5. G2 * I think it's S phase, It's not G1 right?
A 2.0kg block of wood starts at rest and slides down a ramp. Its initial height is 12.0m. If the final velocity of the block is 13m/s, determine the energy of this system that has been turned into heat.
Physics
3 particles of masses m1 = 1.5 kg , m2 = 2.5 kg and m3 = 3.5 kg form an equilateral triangle of edge length l= 135 cm. Determine the center of mass of the system.
Accounting help
Cost flow assumptions-FIFO and LIFO using a periodic system. Beginning inventory was 600 units at a cost of \$20 per unit. Goods available for sale during the year 2600 units at a total cost of \$57,600. In May 1200 units were purchased at a total cost \$26,400. The only other ...
Accounting
Cost flow assumptions for FIFO and LIFO using a periodic system. Question 1. Sales during the year were 700 units. Beginning inventory was 400 units at a cost of \$10.00 per unit. Purchase 1 was 500 units at \$12.00 per unit. Purchase 2 was 300 units at \$14.00 per unit. Required...
Physics
In the figure, the meterstick's mass is 0.130 kg and the string tension is 2.30 N . The system is in equilibrium. A) Find the unknown mass m. B) Find the upward force the fulcrum exerts on the stick. I found part A and the answer is .634kg but I can not figure out part B.
Math
Solve the system equations y=4x+8,x^2+7x-20 I'm truly lost
What did the Mayan caldenders track? 1. Religous celebrations and battles 2. Religous celebrations and the seasons ** 3. Games of pok-ta-tok and the seasons 4. Games of pok-ta-tok and battles The Aztech capital of Tenochtitlan stood on the site of present-day_______. 1. Los ...
Algebra 1
A high school club has a performance, with 800 tickets to sell. Tickets are \$6 before the day of the show, and \$9 on the day of the show. They must sell \$5000 in tickets. How do I write a system of inequalities to represent the equation?
Chemistry
Consider 1.00 L of the buffer system that contains 0.200 M hydrocyanic acid(HCN) and 0.150 M sodium cyanide (NaCN). The pKa of hydrocyanic acid is 9.31. What is the [HCN] after 0.020 mol of HCl is added?
Chemistry
A fixed amount of oxygen gas is held in a .500 L tank at a pressure of 4.09 atm. The tank is connected to an empty 1.50 L tank by a tube with a valve. After this valve has been opened and the oxygen is allowed to flow freely between the two tanks at a constant temperature, ...
physics
A typical magnitude of the external magnetic field in a cardiac catheter ablation procedure using remote magnetic navigation is B = 0.080 T. Suppose that the permanent magnet in the catheter used in the procedure is inside the left atrium of the heart and subject to this ...
chemistry
A fixed amount of oxygen gas is held in a .500 L tank at a pressure of 4.68 atm. The tank is connected to an empty 1.50 L tank by a tube with a valve. After this valve has been opened and the oxygen is allowed to flow freely between the two tanks at a constant temperature, ...
Social Studies Help!!
1, what was the main purpose Roosevelt made in the 1932 election that helped him win? He promised to reorganize the economy He promised to help the jobless, poor farmers,and elderly ******* He promised to regulate big banks more heavily to prevent another depression He ...
Chemistry
In which of the following reactions is work done by the system to the surroundings? A) 2CH3OH(l) + 3O2(g) --> 4H2O(l) + 2CO2(g) B) S(s, rhombic) + O2(g) --> SO2(g) C) 2AgNO3(s) --> 2AgNO2(s) + O2(g) D)4Fe(s) + 3O2(g) --> 2Fe2O3(s) My guess is B because according to...
Social Studies
1. Which option describes a pull factor that influenced European immigration to the United States? 1.Land Scarcity 2.Industrial jobs*** 3.Political Unrest 4.Religious Persecution 2. What role did nativism play in federal policy? Concern for immigrants' children led to ...
Social Studies
Which of these are part of the estate's water system? aviary wrestling grounds baths colonnade
ICT
consider an organization as a computer system and identify its components
maths , science
Two point charges q and -2q are kept 'd' distance apart . Find the location of point relative to charge 'q' at which potential due to this system of charges is zero.
Physics/Mechanics
A particle A of mass 150g lies at rest on a smooth horizontal surface. A second particle B of mass 100g is projected along the surface with a speed of u m/s and collides directly with A. On collision the masses coalesce and moves on with a speed of 4 m.s . find the value of u ...
Physics
At what acceleration would you expect the blood pressure in the brain to drop to zero for an erect person? Why? (Assume there are no body mechanisms operating to compensate for such conditions)
biophysics
At what acceleration would you expect the blood pressure in the brain to drop to zero for an erect person ?why? (Assume there are no body mechanisms operating to compensate for such conditions)
micro economics theory 2
a watch making firm operating in a competitive market.given by c= 100+ q2., where q is the level of output & c is total cost. a.the price of watches is birr is 60, how many watches produce to maximize profit? b.what will your profit level be? c.at what minimum price will you ...
ELECTRICITY AND MAGNETISM
2. Find the charge in potential energy of the system of two charges plus a third charge q3= 3.00µC as the latter charge moves from the infinity to point P.
Human resource
Suppose you are a staff nurse in a hospital that uses an incentive compensation system. Do you have an obligation to disclose the nature of the compensation arrangement to patients? if so, how should this information be communicating, and by whom?
math
Mrs. Morton has a special reward system for her class. When all her students behave well, she rewards them by putting 333 marbles into a marble jar. When the jar has 100100100 marbles or more, the class has a party. Right now, the jar contains 242424 marbles. Let rrr represent...
physical science
a cars radio draws 0.25 A of current in the autos 12 volt electrical system. How much power in watts does the radio use?
math
Tickets to a museum cost \$3 and \$8 for adults. a group of four visitors to the museum spent a total of \$22 on tickets. Write and solve a system of equations to represent this situation. Interpret the solution
U.S History
Hi I took a test that I am not too confident in so if someone could help me out I would love it. Here are the questions and my answers to them. ______________________________________ In the text, you read about the healthcare issue during Bill Clinton's presidency. In 1994, ...
physics
A pulley system of 4 pulleys is used to raise a loan of mass 50kg vertically,if its effeciency is 80%,determine the minimum effort required to raise the load(g=10ms-2)?
infomation system
is a videon great to infom peple about traveling?
math
You have a total of 52 dimes and quarters. You have 10 more quarters than dimes. Which system of equations can you use to find the number x of quarters and number y of dimes you have? Use the system to determine how much money you have in quarters and dimes.
Health and Physical Education
The Nose, The Diagram and lungs are apart of which system ?
World History
Which of the following best explains how the alliance system contributed to the outbreak of World War I? A. Alliances had limited trade, causing economic problems and spurring unrest. B. Rival alliances had promoted tension rather than security. C. Alliances had failed to ...
Software Engineering
Create an automatic question paper generator system in JAVA which involves Artificial Intelligence with the following requirements: The system should generate a test paper automatically based on the difficulty level. For this create a fake data for 25 questions and collect...
Astronomy
Please help! I'm really behind and need to catch up. 1. The rings orbiting the Jovian planets may have been small ____ that got too close to the planets long ago. A. Comets B. Asteroids C. Moons D. Meteors 2. What is the composition of the rings around the Jovian planets? A. ...
Science
Which of the following boxes below represent a stable system? Describe how a disturbance to this box would affect it. Use the terms center of mass and equilibrium in your response. Box A: Shaped like a diamond on a flat surface with an arrow pointing downwards. Box B: A square...
math
Given: ∆PQR, m∠R = 90° m∠PQR = 75° M ∈ PR , MP = 18 m∠MQR = 60° Find: RQ M is on segment PR 9sqrt3 is wrong(the system did not take it)
Java
Which of the following classes best represents rare and serious internal system errors? Error Throwable RuntimeException Exception I think it may be Error or Throwable. Been having alot of trouble with this class lately, help would be appreciated. ~Thank you.
Math
Find the exact values for the lengths of the labeled segments a, b and p drawn in green, red, and blue, respectively. Note that r=3 is the radius of the circle, and s=2 is the arc length from the point (3,0) around the circle to the indicated point. (So imagine a triangle ...
math
Conversions in the Metric System
finance
The Switch division of Tornax Inc. produces a small switch that is used by various companies as a component part in their products. Tornax operates its divisions as autonomous units, giving its divisional managers great discretion in pricing and other decisions. Each division ...
physics
a point mass of 12.5 kg is hung from an 18.3 m long rope and swings back and forth with an amplitude of 2.34m. what is the oscillation period for the pendulum system?
Really quick question
Science
The following sphere is responsible for forming the solid shell of Earth that holds up life? -Hydrosphere -Lithosphere -Atmosphere -Biosphere** Explain the interaction between the four main components of Earth's physical systems. Be sure to include the name of each system and ...
physics
Explain in a few words why, when operating a garden hose fitted with a trigger-operated nozzle, the water jet is momentarily more powerful at the instant the trigger is squeezed than later, when there is a continuous stream of water. Show that the amount of power in the water ...
Accounting
Jim Hurley is an accountant for a local manufacturing company. Jim’s good friend, Mike Kotowski, has been operating a retail sporting goods store for about a year. Mike has proposed that the fee he will pay for Jim’s accounting work should be contingent upon his receiving ...
Social Studies
which of the following was an effect of nativism in the United States in the 1920s? A. the Great Migration** B. the Scopes Trial C. the Jones Act of 1917 D. the quota system
Information system
Is video conferencing really a viable green alternative to travel for most companies
Gen Chem 125
A process takes place at constant pressure. The volume changes and the temperature increases by 91.0oC. The heat capacity of the system at constant volume, CV, is 695.6 kJ/oC. What is ΔE in kJ? (Careful--pay attention to the units of the heat capacity.)
math
Consider these two tables for Line 1 and Line 2. Line 1: x | y 0 1 3 4 Line 2: x | y 0 0 2 4 (a) Graph Line 1 (already did this) (b) Graph Line 2 on the same graph (already did this) (c) What is the solution to the system? Use graphing to solve. Explain your answer.
Accounting Help (with answers for checking)
Ben Guslists, vice-president of sales, has recommended adding a new product line. A market study and cost analysis show that the new line should yield the following annual results: New Sales \$2,800,000.00 Cost of Sales \$1,600,000.00 Operating Expenses \$200,000.00 Total ...
natural science/bio
flow diagram of circulatory and respiratory system combined
physics
A golf ball of mass 125g hits a billards ball of mass 172.5g that is initially at rest. after the colision, the golf ball moves to the right at a speed of 10cm/s. the billards ball moves to the left with a speed of 58 cm/s. (a) What was the velocity of the golf ball before the...
Social studies
Which of the following was the main contributor to finally bringing the United States out of the Great Depression? A the National Recovery Administration B the Good Neighbor Policy C wartime spending D the election of Roosevelt D or B ----------------------------------- Which ...
PE
Extra Credit Opportunity Health and PE Project: Body Systems 10 Extra Credit Points! (5 points for the project, 5 points for presenting during LL) What: Choose a Body System from Unit 6 and Create a PowerPoint Slide or a Poster! Why: Teach us about the body system you ...
maths
Mere, Sam and Nathan go to a bookshop to buy supplies for their school. Mere bought two erasers and four pencils and paid \$2.40 for it. Sam bought six erasers, one pencil and two pens and paid \$3.95 while Nathan paid \$4.15 for an eraser, a pencil and four pens. Write a system ...
Science
In a heat engine, if 700 J of heat enters the system, and the piston does 400 J of work, what is the final internal (thermal) energy of the system if the initial energy is 1,200?
physics
In using a pulley system, an effort of 100N is needed to raise steadily a load of 800N (a) what is the mechanical advantage?. Data effort=100N load=800N M.A=load/Effort M.A=800N/100N M.A=8
ap physics
Determine the change in rotational kinetic energy when the rotational velocity of the turntable of a stereo system increases from 0 to 33 rpm. Its rotational inertia is 6.8×10−3kg⋅m2 .
Science
Can Someone explain this? Should other states wait to see if the high-speed rail system works well before starting their own plans for similar systems?
social studies thinking works
When several stores in the same neighborhood sell the same video game system, what will likely happen? A. The price in the video game will increase B.the cost of production for video games will increase C.the price of video games will decrease** D.the supply of video games ...
History
1. Which of the following happened in south carolina during reconstruction? A. The state developed an oil industry. B. The state stopped cotton production. C. The state built an iron industry. D. The state rebuilt its railroad system.
Algebra
Slove each system graphically. X+y=8 and x-y=4
Physics
In a rectangular coordinate system, a positive point charge 5.0 nC is placed at the point x=0, y=2.0 cm, and a negative point charge -5.0 nC is placed at x=0, y=-2.0 cm. Both charges have masses 4.0x10^-3 g. Point P is at x=3.0 cm, y=2.0 cm. The electric potential at infinity ...
english
Which label best fits the following text? Sir Isaac Newton, the first person who studied gravity seriously, discovered what is known as the law of universal gravitation. It defines the amount of attraction of one particle of matter on another particle of matter. The formula ...
MATH
se a graphing calculator or Excel to find the solution of the system of equations. (If the system is dependent, enter DEPENDENT. If there is no solution, enter NO SOLUTION.) 5x + 3y = 2 3x + 7y = −4
Biology
I am doing a project on the lymphatic system. I need ideas as a activity the class can do. I don't want there to be a lot of materials needed. Help?
physics
1.00L piston with 1.00 mole of an ideal gas at 298.0K and 1.00bar is isothermally and reversibly compressed to a final volume of 0.200L, the irreversibly expanded in first step to its original volume with an applied pressure of 1.00bar. Calculate the change in heat for the ...
Physics
An ideal gas in a moveable piston is allowed to reversibly expand by slowly heating at constant pressure. a. Derive the equation for the work done reversibly as a function of the initial and final temperatures.( i have used w=-nrtln(v final/v intial) i dont know how can I take...
Chemistry
Thanks to Dr. Bob for answering all these questions! I always do poorly on conceptual problems. Consider the following equilibrium system at 25°C and select the true statement below: H2PO4−(aq) + SO42−(aq) ↔ HSO4−(aq) + HPO42−(aq) ΔG°rxn > 0 at 25°C (1) this ...
social studies
1. which of these was a result of the invention of the cotton gin. A. It greatly reduced the need to plant cotton seeds by hand. B.It provided a faster method for separating seeds from cotton fibers C.It enabled slaves to work in factories operating textile machines D.It ...
MINNESOTA STUDIES
Multiple Choice 1. Which of the following most contributed to the formation of labor unions? (1 point) cuts to wages rising railroad rates grain elevator measures women working in factories 2. How did the Grange and the Farmers’ Alliance both work to help farmers? (1 point) ...
chemisrty
I took 6.22 g of potassium hypochlorite and dissolved it up in a 100.0 mL flask. I then added 8.00 mL of 2.00 M nitric acid. The solution should be able to buffer that addition. How much more (mLs) of the nitric acid can the system take before it is no longer in the buffer zone?
Algebra
A rectangle's length is 6 less than twice its width. The perimeter of the rectangle is 36 units. Set up and solve a system of equations to determine the length and width?
Algebra
You walk at a pace of 3 miles per hour, and jog at a pace of 6 miles per hour. You want to cover a distance of more than 18 miles in less than 5 hours. Write a system of inequalities to represent the situation. What is one possible combination of the number of hours you can ...
social studies
what was major cause of problems with the sharecropping system
Algebra
You walk at a pace of 3 miles per hour, and jog at a pace of 6 miles per hour. You want to cover a distance of more than 18 miles in less than hours. Write a system of inequalities to represent the situation. What is one possible combination of the number of hours you walk/ ...
Social Studies
1. Which of the following apply to fascism? Select all that apply. A rooted in militarism B extreme nationalism C respect for minorities D blind loyalty to the state I think its A,B, and c ----------------------------------- 2. Which of the following best describes attitudes ...
Algebra
You want to purchase at least 10 tickets to a baseball game, but can spend at most \$200. Lower level tickets &8 each, while upper level tickets cost \$5. Write and graph a system of inequalities to represent the situation. Give two possible combinations of the number of lower ...
Algebra
A rectangle's length is 6 less than twice its width. The perimeter of the rectangle is 36 units. Set up and solve a system of equations to determine the length and width.
Algebra
A boat traveled 180 miles downstream and back. The trip downstream took 10 hours. The trip back took 30 hours. Find the speed of the boat in still water and the speed of the current. Solve each system using any method
Science
Which of these is found inside the solar system? A) Moon B) Nebula C) Galaxy D) Black hole
History
Please check my answers. Question 1(Multiple Choice Worth 5 points) [06.06 MC] Which answer best describes how the Supreme Court viewed Maryland's taxing of the national bank created after the War of 1812? The Supreme Court decided Maryland was within its rights as a state. ...
Social studies
What was one of the major failures of the New Deal program? A It did not help the poor. B It never reformed the banking system. C It did not end the Great Depression. D The jobs it created did not last. A?
physics
A 10kg box is attached to a 7kg box which rests on a 30 degrees incline the coeficient of kinetic friction is mk=0.1 between the box and surface find the rate of acceleration of the system
Physics
When a mass is hung from a spring stretches it by 15 cm. What is the period of the oscillations of this system?
Math 115
While solar energy powered home systems are quite expensive to install, adding some passive energy features to conventional construction can make a substantial reduction in heating and cooling costs. Passive solar requires no machinery and uses the building itself to collect, ...
Math
While solar energy powered home systems are quite expensive to install, adding some passive energy features to conventional construction can make a substantial reduction in heating and cooling costs. Passive solar requires no machinery and uses the building itself to collect, ...
math 1
system of linear equations of substitution. a=2/b-3 a=2b-8
algebra
system of equations c-3d=27 4d+10c=120 | 5,603 | 22,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-34 | latest | en | 0.927655 |
http://www.jiskha.com/members/profile/posts.cgi?name=lomas | 1,495,649,904,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607849.21/warc/CC-MAIN-20170524173007-20170524193007-00300.warc.gz | 540,444,321 | 3,509 | # Posts by lomas
Total # Posts: 15
math
THANKS!
math
Help with factoring this one: x^9+1 (x^3+1) (x^3-x^3+1) Is this correct?
math
Great...i was adding the exponents that were on the outside instead of multiplying it. Took me a while to remember the rule. Thanks Ron.
math
Did some rethinking..is it 128x^16y^8 ?
math
multiply: (2x^4y^2)^3 (4x^2y)^2 Is this right? 128x^11y^8 Not sure.
math
aaarrggh....ok thanks!
math
last one on this topic i promise ;-) 4X-14+2X^3-7x^2 rearrange by degree 2X^3-7X^2+4X-14 and them im stuck..
Earth/Environmental Science
surface currents
History
Google "battle of bunker hill" and you will get a ton of information. It's now up to you to compact/summarize what you have learned into two hundred words.
math
Gotta remember those damn steps. I immediately assumed it was four terms so I factor by grouping. I forgot that you first look for common factor, then squares etc. Thanks.
math
having trouble factoring this particular problem x^2-12x+36-49y2 factored by grouping x(x-12) (7y-6) (7y+6) the answer in the book has (x-6+7y)(x-6-7y) confused.
math
Need help with factoring this problem. Just want to make sure I followed the right steps. X^3-2X^2-X+2 there are four terms so i used grouping method X^2(X-2)(-1)(X-2) (X-2) (X^2-1) (X-2) (X-1) (X+1) Correct?
math
thank you.
math
sorry. it's 4 more girls
math
there are 4 girls in Mrs. Changs class than Mr. Blackwell's. 5 girls moved from Mrs. Changs to Mr. Blackwells. Now there are twice as many girls in Mr. Blackwells class as there are in Mrs. Changs, How many girls were in Mr. Blackwells class to begin with? Help, stuck with...
1. Pages:
2. 1
Post a New Question | 544 | 1,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-22 | latest | en | 0.924145 |
https://metanumbers.com/30155 | 1,686,079,851,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00621.warc.gz | 431,006,638 | 7,706 | # 30155 (number)
30,155 (thirty thousand one hundred fifty-five) is an odd five-digits composite number following 30154 and preceding 30156. In scientific notation, it is written as 3.0155 × 104. The sum of its digits is 14. It has a total of 3 prime factors and 8 positive divisors. There are 23,328 positive integers (up to 30155) that are relatively prime to 30155.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 14
• Digital Root 5
## Name
Short name 30 thousand 155 thirty thousand one hundred fifty-five
## Notation
Scientific notation 3.0155 × 104 30.155 × 103
## Prime Factorization of 30155
Prime Factorization 5 × 37 × 163
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 30155 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 30,155 is 5 × 37 × 163. Since it has a total of 3 prime factors, 30,155 is a composite number.
## Divisors of 30155
1, 5, 37, 163, 185, 815, 6031, 30155
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 37392 Sum of all the positive divisors of n s(n) 7237 Sum of the proper positive divisors of n A(n) 4674 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 173.652 Returns the nth root of the product of n divisors H(n) 6.45165 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 30,155 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 30,155) is 37,392, the average is 4,674.
## Other Arithmetic Functions (n = 30155)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 23328 Total number of positive integers not greater than n that are coprime to n λ(n) 324 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3264 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 23,328 positive integers (less than 30,155) that are coprime with 30,155. And there are approximately 3,264 prime numbers less than or equal to 30,155.
## Divisibility of 30155
m n mod m 2 3 4 5 6 7 8 9 1 2 3 0 5 6 3 5
The number 30,155 is divisible by 5.
## Classification of 30155
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (30155)
Base System Value
2 Binary 111010111001011
3 Ternary 1112100212
4 Quaternary 13113023
5 Quinary 1431110
6 Senary 351335
8 Octal 72713
10 Decimal 30155
12 Duodecimal 1554b
20 Vigesimal 3f7f
36 Base36 n9n
## Basic calculations (n = 30155)
### Multiplication
n×y
n×2 60310 90465 120620 150775
### Division
n÷y
n÷2 15077.5 10051.7 7538.75 6031
### Exponentiation
ny
n2 909324025 27420665973875 826870182442200625 24934270351544559846875
### Nth Root
y√n
2√n 173.652 31.1257 13.1777 7.86814
## 30155 as geometric shapes
### Circle
Diameter 60310 189469 2.85673e+09
### Sphere
Volume 1.14859e+14 1.14269e+10 189469
### Square
Length = n
Perimeter 120620 9.09324e+08 42645.6
### Cube
Length = n
Surface area 5.45594e+09 2.74207e+13 52230
### Equilateral Triangle
Length = n
Perimeter 90465 3.93749e+08 26115
### Triangular Pyramid
Length = n
Surface area 1.575e+09 3.23156e+12 24621.5
## Cryptographic Hash Functions
md5 ae8970f7d83581427155bc725c003594 f4ba6a647926b04170594c1de71539c029a25587 4ed8df78a1a326951e4fbfb32ee2c036ab1cb88fcccd7aca85f8112ebe3f5cc9 fd87971ffb0c3a4a0e2b6d9768dd791bdfc77a955cd0bb0f3bd4de8106ed0947ed1d3eb0a0f4b2e2931b81656be8cdff735a50154ec56e6bee0e53ac0cde7eff 82a64809f035a366944403b4363cce3b490f795e | 1,458 | 4,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-23 | latest | en | 0.808302 |
https://www.meritnation.com/ask-answer/question/1-two-poles-of-height-6m-and-11m-stand-vertically-on-a-plane/triangles/15356091 | 1,656,199,636,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036176.7/warc/CC-MAIN-20220625220543-20220626010543-00480.warc.gz | 932,372,135 | 8,196 | 1. Two poles of height 6m and 11m stand vertically on a plane ground . If the distance between their feets is 12m; find the distance between their tips.
Dear Student,
Regards
• 0
Dear Student,
The figure below shows two poles AB and CD such that AB = 6 m and CD = 11 m;
And the distance between the poles = AC = 12 m
So; BE = AC = 12 m and CE = AB = 6 m
And DE = AC - CE = 11 m - 6 m = 5 m
Now considering right triangle BED we have;
(BD)2 = (BE)2+(DE)2 {using Pythagoras theorem}⇒(BD)2 = (12)2+52⇒(BD)2 = 169⇒BD = 169−−−√ = 13BD2 = BE2+DE2 {using Pythagoras theorem}⇒BD2 = 122+52⇒BD2 = 169⇒BD = 169 = 13
Therefore the distance between their tops is 13 m.
Regards
• 1
What are you looking for? | 273 | 705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-27 | latest | en | 0.778067 |
https://www.jiskha.com/display.cgi?id=1159674763 | 1,503,149,521,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105451.99/warc/CC-MAIN-20170819124333-20170819144333-00012.warc.gz | 952,868,632 | 6,110 | # math problem
posted by .
My ID number is quite remarkable.Its a 9 digit number with the digits 1-9 appearing only once. The entire 9 digit nuymber is divisible by 9. If you remove the last digit, the remaining 8 digit number is divisible by 8. If you remove the last digit again, you are left with a 7 digit number dicisible by 7. Again remove the last digit to have a 6 digit number divisible by 6. This process continues all the way down to one digit. What is my ID number?
987654321
I'm sorry, but that's not the correct answer, 7 does not divide 9876543, check for yourself. All the other numbers did work however.
Ok, we have the number 1-9 appearing once in some order. Let abcdefghi denote the number, then we're told
9|abcdefghi
8|abcdefgh
7|abcdefg
6|abcdef
5|abcde
4|abcd
3|abc
2|ab
1|a
Here is what you should be able to tell by inspection alone:
(1) b,d,f, and h are all even since evens do not divide odds.
(2) since 0 in not included, a,c,e,g and i are the odds.
(3) e=5 since 5 only divides numbers ending in 5.
(4) if 8 divides a number, then it divides the last 3 digits of the number.
(5) 4 divides a number if and only if it divides the last 2 digits.
Here is what you can conclude right now: there are 4! ways to permute 1,3,7 and 9 for a,c,g and i; there are 4! ways to permute 2,4,6, and 8 in b,d,f and h. Thus there are 576 potential numbers to check.
I did a check of all of them and there is exactly one solution. That's all I'll give you right now.
Lets take the easy road and see where it takes us.
Divisibility rules:
Odd numbers can only be divided evenly by another odd number.
Even numbers can be evenly divided by either odd or even numbers.
2---Numbers that end in 0, 2, 4, 6, or 8 are evenly divisible by 2.
3---If the sum of a number's digits is evenly divisible by 3, the number is divisible by
4---If the last two digits are both zero or they form a two digit number evenly divisible by 4, then the whole number is evenly divisible by 4.
5--Any number ending in 5 or 0 is evenly divisible by 5.
6--If the number is even and the sum of the digits is evenly divisible by 3, the whole number is divisible by 3.
7---Double the last digit and subtract from the number without the last digit. If the result is evenly divisible by 7, so is the original number.
8---If the last 3 digits are zero or if they form a number that is evenly divisible by 8, then the whole number is evenly divisible by 8.
9---If the sum of the digits is evenly divisible by 3, the number is evenly divisible by 9.
10---Any number ending in 0 is divisible by 10.
11--- Add up the odd position digits. Add up the even position digits. Calculate the difference between the two sums. If the difference is divisible by 11, the original number is divisible by 11
Lets portray our number, with the digits identified, by
_.._.._.._.._.._.._.._.._.._
1..2..3..4..5..6..7..8..9.10
Clearly, the last digit must be 0 in order for the whole number to be divisible by 10.
Similarly, the 5th digit must be 5 in order for the 5 digit number to be divisible by 5.
This makes our number so far _.._.._.._..5.._.._.._.._..0
.............................................1..2..3..4..5..6..7..8..9.10
The sum of the digits 1 - 9 is 45, which makes any nine digit number using all nine digits divisible by 9. Thus, we need not concern ourselves with the 9th digit.
By the definition of our problem, all even numbers of digits must be divisible by an even number and all odd digit numbers must be divisible by an odd number, making the even digits even, and the odd digits odd.
The 2nd, 4th, 6th, and 8th digits can then only be 2, 4, 6, or 8 and the 1st, 3rd, 5th, 7th, and 9th digits can only be 1, 3, 5, 7, or 9.
For the 8 digit number to be divisible by 8, the last 3 digits must be divisible by 8, making these three 3 digit numbers start, and end, with 2, 4, 6, or 8. These three digits therefore form numbers in the 200+, 400+, 600+ or 800+ family, each of which is, itself, divisible by 8, thus forcing the 7th and 8th digits together to be evenly divisible by 8. The only possibilities for these two digits then becomes 16, 32, 72 and 96.
This makes our possible numbers so far....._.._.._.._..5.._..1..6.._..0
.............................................................. _.._.._.._..5.._..3..2.._..0
.............................................................. _.._.._.._..5.._..7..2.._..0
.............................................................. _.._.._.._..5.._..9..6.._..0
..............................................................1..2..3..4..5..6..7..8..9.10
From our divisibility rules, a number is divisible by 4, if, and only if, the last 2 digits are divisible by 4 forcing our 3rd plus 4th digit number to be divisible by 4 with the 3rd digit being odd. The possible 2 digit numbers are therefore 12, 16, 32, 36, 52, 56, 72, 76, 92 and 96. Since all of these candidates end in 2 or 6, the 4th digit can be safely said to be 2 or 6.
This makes our possible numbers so far....._.._.._..2..5.._..1..6.._..0
.............................................................. _.._.._..6..5.._..3..2.._..0
.............................................................. _.._.._..6..5.._..7..2.._..0
.............................................................. _.._.._..2..5.._..9..6.._..0
..............................................................1..2..3..4..5..6..7..8..9.10
From our divisibility rules, a number is divisible by 3 if the sum of its digits is divisible by 3. Similarly, a number is divisible by 6 if the sum of its digits is divisible by 3. SInce the first 3 digits must be divisible by 3, as must the first 6 digits, the number formed by the 4th, 5th, and 6th digits must be divisible by 3. Having the 4th and 5th digits, and knowing that the 6th digit must be even, we know that this 6th digit can only be 4 or 8.
This makes our possible numbers so far... _..8.._..2..5..4..1..6.._..0
..............................................................._..4.._..2..5..8..1..6.._..0
.............................................................. _..8.._..6..5..4..3..2.._..0
.............................................................. _..4.._..6..5..8..3..2.._..0
.............................................................. _..8.._..6..5..4..7..2.._..0
.............................................................. _..4.._..6..5..8..7..2.._..0
.............................................................. _..8.._..2..5..4..9..6.._..0
.............................................................. _..4.._..2..5..8..9..6.._..0
..............................................................1..2..3..4..5..6..7..8..9.10
However, the digits in 254 and 658 do not add up to a number divisible by 3.
This makes our possible numbers so far.. _..4.._..2..5..8..1..6.._..0 A
.............................................................. _..8.._..6..5..4..3..2.._..0 B
.............................................................. _..8.._..6..5..4..7..2.._..0 C
.............................................................. _..4.._..2..5..8..9..6.._..0 D
..............................................................1..2..3..4..5..6..7..8..9.10
We know that any number placed in the 9th digit will make the total number divisible by 9 so lets work with the 1st and 3rd digits.
The numbers remaining in the 4 cases are 3-7-9 for case A, 1-7-9 for case B, 1-3-9 for case C, and 1-3-7 for case D.
We have already pointed out from our divisibility rules that a number is divisible by 3 if the sum of its digits is divisible by 3. Which of the three candidate groups of 3 remaining digits will make our first three digits divisible by 3?
I think this final screening will lead you to the remaining possibilities, with the divisibility by 7 forcing the final answer to fall out.
Try 381654729
What is the answer 3 2/5 - 1 3/5.
Thanks
• math problem -
Your answer works, but it does not encompass all possible answers. The solution is far from unique.
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http://files.righto.com/fractals/neg.html | 1,550,486,050,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484928.52/warc/CC-MAIN-20190218094308-20190218120308-00117.warc.gz | 108,340,595 | 2,093 | # Negative-exponent fractals
To compute a point in the Mandelbrot set, the equation z = z^2+c is repeated many times. If z exceeds 4, it will grow to infinity, and the point is outside the Mandelbrot set. Similar sets can be generated for exponents of 3, 4, etc.
What happens if you want to use a negative exponent such as -2?
You can use a similar mechanism, and declare that a point has escaped if it exceeds a bound. However, unlike the positive exponent case, large values don't really escape, since z^-2 is very small if z is very large, so the z values come back. Thus, the results depend on the bound selected and show artifacts, which appear as concentric circles.
A better solution is to see if the sequence of z values forms a cycle. For some values of c, z will converge to a fixed value. For others, z will oscillate between two or more values. In other cases, z will chaotically wander. By coloring each point according to the cyclic behavior, we can see the true structure of the fractal without artifacts that depend on the escape bound, as shown on the left. (The result is reminiscent of Mobius fractals.)
With c fixed and z varying, a Julia set is created. Negative-exponent fractals show a variety of Julia sets, depending on which c value is selected. To compute the Julia set, again the cycles must be detected. However, for the Julia set, every point will converge to the same cycle. The figure can be colored according to the "phase" of the cycle, though. This yields the following images.
For more details, see my paper, which was published as ``An Investigation of z -> 1/z^n+c,'' Computers & Graphics, 17(5), Sep. 1993, pp 603-607.
Ken Shirriff: shirriff@eng.sun.com | 407 | 1,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-09 | longest | en | 0.917296 |
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XP Math - Forums Algebra 2 work help with sequences?
08-02-2007 #1 Lonely Girl Guest Posts: n/a Algebra 2 work help with sequences? 1.35/8, 245/32, 1715/128, 12205/512 9th term geometric and equation for this?2.Geometric means of 5 and 3203.Positive geometric means of 5 and 3204.2+4+22+4+6+4+22+4+ . . . + (2N-4+2N)+(2N-2) + .... +2Find the 15th row5.3+ 3/2 +3/4 + 3/8 + 3/16 + ....Sum infinite Geometric Sequencegot 1 ****** question
08-02-2007 #2 b8k3p Guest Posts: n/a yup. . . you're screwed.
08-02-2007 #3 PaeKm Guest Posts: n/a 5.) 3+3/2+3/4+3/8+3/16r = 1/2 and a = 3S inf = a/1-rS inf = 3 /(1-1/2) = 6 ANS1.)35/8, 245/32, 1715/128, 12205/512 r = (245/32) / (35/8) = 7/49th is ar^8 = (35/8)(7/4)^8 ANS2.) 5,x, 320r^2 = 320/5 = 64 then r = +-8x = 5*(+-8) = +- 40 ANS3.) + 40 ANS4.) 2+4+ . . . + (2N-4+2N)+(2N-2) + .... +2 in mid is 6N-6if N =15 then 2+4+6+12+....+84+...+12+6+4+2
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Topic: Why can no one in sci.math understand my simple point?
Replies: 658 Last Post: Feb 17, 2013 10:15 AM
Messages: [ Previous | Next ]
Posts: 822 Registered: 9/1/10
Re: Musatov responds to WM on Cantor: intelligence,
real, judge for yourself: Cantor’s idea seems to me to ass
ume because two sets converge to infinity the nature of infi
nity and the number of elements in each set must become equa
l.
Posted: Feb 17, 2013 9:19 AM
On Sunday, February 17, 2013 6:16:24 AM UTC-8, Musatov wrote:
> On Thursday, June 17, 2010 5:50:29 PM UTC-7, Transfer Principle wrote:
>
> > On Jun 17, 2:37 am, Simplane Simple Plane Simulate Plain Simple
>
> > <marty.musa...@gmail.com> wrote:
>
> > > COMMENTARY
>
> > > To contrast my conception to my perception of Cantor?s conception of
>
> > > the infinite. Cantor?s idea seems to me to assume because two sets
>
> > > converge to infinity the nature of infinity and the number of elements
>
> > > in each set must become equal.
>
> > > Requiring one to one ?mapping? anything other than a one to one
>
> > > correspondence is excluded.
>
> > > My conception is more as infinite quantities are not mathematically
>
> > > expressible, in the sense we can only conceive of infinity as
>
> > > something consisting forever. You start with something finite and then
>
> > > multiply or divide it forever. In this way two infinite sets are not
>
> > > equal but, as shown in they can be ratio-ed by the finite expression
>
> > > defines the set, but both extend infinitely.
>
> > > Cantor?s reversal of conceptions, for some sets (non-d enumerable)
>
> > > uses conception #2. This is a contradiction to conception #1, and it
>
> > > would seem to me it can?t be both ways.
>
> >
>
> > Musatov might be interested in knowing about Tony Orlow,
>
> > who is working on a set size that is very similar to
>
> > Conception #1 above.
>
> >
>
> > In particular, Musatov describes how he uses ratios to
>
> > determine that the size of {2,4,6,8,...} is exactly half
>
> > that of {1,2,3,4,...}, since the elements of the sets
>
> > are in the ratio 2:1. TO uses a similar argument to
>
> > conclude that if {1,2,3,4,...} has the set size (or
>
> > "Bigulosity") tav, then {2,4,6,8,...} would have a
>
> > Bigulosity of tav/2.
>
> >
>
> > TO uses the name "Post-Cantorian" to describe those who
>
> > use what Musatov calls "Conception #1," as opposed to the
>
> > "Cantorian" Conception #2. Thus, Musatov and TO are
>
> > natural allies wrt set size.
>
> >
>
> > On the other hand, Herc and WM are "Anti-Cantorian" in
>
> > that they don't believe in different sizes of infinity. So
>
> > I'm glad that Herc and WM are in this thread together. So
>
> > these two reject both Conceptions #1 and #2.
>
> >
>
>
> > he can click on the current TO thread (warning -- this
>
> > thread now exceeds 500 posts).
>
> 0
>
> -1 +2 -4 +8
>
> +1 -2 +4 -8
>
> 0
>
> 1 -2 4 -8
>
> -1 2 -4 8
>
> On Thursday, June 17, 2010 5:50:29 PM UTC-7, Transfer Principle wrote:
>
> > On Jun 17, 2:37 am, Simplane Simple Plane Simulate Plain Simple
>
> > <marty.musa...@gmail.com> wrote:
>
> > > COMMENTARY
>
> > > To contrast my conception to my perception of Cantor?s conception of
>
> > > the infinite. Cantor?s idea seems to me to assume because two sets
>
> > > converge to infinity the nature of infinity and the number of elements
>
> > > in each set must become equal.
>
> > > Requiring one to one ?mapping? anything other than a one to one
>
> > > correspondence is excluded.
>
> > > My conception is more as infinite quantities are not mathematically
>
> > > expressible, in the sense we can only conceive of infinity as
>
> > > something consisting forever. You start with something finite and then
>
> > > multiply or divide it forever. In this way two infinite sets are not
>
> > > equal but, as shown in they can be ratio-ed by the finite expression
>
> > > defines the set, but both extend infinitely.
>
> > > Cantor?s reversal of conceptions, for some sets (non-d enumerable)
>
> > > uses conception #2. This is a contradiction to conception #1, and it
>
> > > would seem to me it can?t be both ways.
>
> >
>
> > Musatov might be interested in knowing about Tony Orlow,
>
> > who is working on a set size that is very similar to
>
> > Conception #1 above.
>
> >
>
> > In particular, Musatov describes how he uses ratios to
>
> > determine that the size of {2,4,6,8,...} is exactly half
>
> > that of {1,2,3,4,...}, since the elements of the sets
>
> > are in the ratio 2:1. TO uses a similar argument to
>
> > conclude that if {1,2,3,4,...} has the set size (or
>
> > "Bigulosity") tav, then {2,4,6,8,...} would have a
>
> > Bigulosity of tav/2.
>
> >
>
> > TO uses the name "Post-Cantorian" to describe those who
>
> > use what Musatov calls "Conception #1," as opposed to the
>
> > "Cantorian" Conception #2. Thus, Musatov and TO are
>
> > natural allies wrt set size.
>
> >
>
> > On the other hand, Herc and WM are "Anti-Cantorian" in
>
> > that they don't believe in different sizes of infinity. So
>
> > I'm glad that Herc and WM are in this thread together. So
>
> > these two reject both Conceptions #1 and #2.
>
> >
>
>
> > he can click on the current TO thread (warning -- this
>
> > thread now exceeds 500 posts).
>
> On Thursday, June 17, 2010 5:57:58 PM UTC-7, Transfer Principle wrote:
>
> > On Jun 17, 12:59 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > On Jun 17, 1:54 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > Does ZFC not prove that all constructible numbers are countable?
>
> > > I don't know. What is the definition IN THE LANGUAGE of ZFC of
>
> > > 'constructible number'?
>
> >
>
> > I know that the word "constructible" occurs in the context
>
> > of V=L, where L is called the "constructible universe."
>
> >
>
> > So it is possible to call the elements of L intersect R
>
> > "constructible reals"?
>
> >
>
> > (Of course, even if we can, I'm not sure whatOn Thursday, June 17, 2010 5:57:58 PM UTC-7, Transfer Principle wrote:
>
> > On Jun 17, 12:59 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > On Jun 17, 1:54 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > Does ZFC not prove that all constructible numbers are countable?
>
> > > I don't know. What is the definition IN THE LANGUAGE of ZFC of
>
> > > 'constructible number'?
>
> >
>
> > I know that the word "constructible" occurs in the context
>
> > of V=L, where L is called the "constructible universe."
>
> >
>
> > So it is possible to call the elements of L intersect R
>
> > "constructible reals"?
>
> >
>
> > (Of course, even if we can, I'm not sure what effect, if
>
> > any, this would have on the cardinality of R.)
>
>
>
> effect, if
>
> > any, this would have on the cardinality of R.)
.=
Date Subject Author
6/15/10 |-|ercules
6/15/10 jaimie
6/15/10 Peter Webb
6/15/10 |-|ercules
6/15/10 Peter Webb
6/15/10 |-|ercules
6/15/10 Daryl McCullough
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 Daryl McCullough
6/15/10 Virgil
6/15/10 |-|ercules
6/15/10 Daryl McCullough
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 |-|ercules
6/15/10 |-|ercules
6/15/10 Peter Webb
6/15/10 Dingo
6/15/10 |-|ercules
6/15/10 Derek Holt
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 Virgil
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 Virgil
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 Daryl McCullough
6/15/10 Virgil
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 Daryl McCullough
6/15/10 Virgil
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 Virgil
6/16/10 mueckenh@rz.fh-augsburg.de
6/16/10 Virgil
6/16/10 Tim Little
6/17/10 mueckenh@rz.fh-augsburg.de
6/17/10 Virgil
6/18/10 Newberry
6/18/10 mueckenh@rz.fh-augsburg.de
6/18/10 Virgil
6/18/10 Newberry
6/20/10 Newberry
6/20/10 mueckenh@rz.fh-augsburg.de
6/20/10 Virgil
6/20/10 mueckenh@rz.fh-augsburg.de
6/20/10 Virgil
6/20/10 Newberry
6/20/10 |-|ercules
6/20/10 Peter Webb
6/20/10 Virgil
6/20/10 |-|ercules
6/20/10 Newberry
6/21/10 Sylvia Else
6/22/10 Newberry
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6/22/10 Newberry
6/22/10 Virgil
6/23/10 Newberry
6/23/10 Virgil
6/23/10 Newberry
6/23/10 Virgil
6/23/10 Newberry
6/24/10 Virgil
6/24/10 Newberry
6/24/10 Virgil
6/24/10 Newberry
6/21/10 mueckenh@rz.fh-augsburg.de
6/21/10 Sylvia Else
6/21/10 mueckenh@rz.fh-augsburg.de
6/21/10 Mike Terry
6/21/10 mueckenh@rz.fh-augsburg.de
6/21/10 Mike Terry
6/22/10 mueckenh@rz.fh-augsburg.de
6/22/10 Sylvia Else
6/22/10 mueckenh@rz.fh-augsburg.de
6/22/10 Virgil
6/22/10 mueckenh@rz.fh-augsburg.de
6/22/10 Sylvia Else
6/22/10 Mike Terry
6/22/10 mueckenh@rz.fh-augsburg.de
6/22/10 Virgil
6/22/10 mueckenh@rz.fh-augsburg.de
6/22/10 Mike Terry
6/22/10 mueckenh@rz.fh-augsburg.de
6/22/10 Mike Terry
6/22/10 Virgil
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 Virgil
6/23/10 Mike Terry
6/24/10 mueckenh@rz.fh-augsburg.de
6/24/10 Virgil
6/24/10 mueckenh@rz.fh-augsburg.de
6/24/10 Virgil
6/24/10 mueckenh@rz.fh-augsburg.de
6/24/10 Virgil
6/24/10 Mike Terry
6/25/10 mueckenh@rz.fh-augsburg.de
6/25/10 Mike Terry
6/25/10 Virgil
6/25/10 mueckenh@rz.fh-augsburg.de
6/25/10 Mike Terry
6/25/10 Virgil
6/26/10 mueckenh@rz.fh-augsburg.de
6/26/10 Virgil
6/26/10 Mike Terry
6/26/10 mueckenh@rz.fh-augsburg.de
6/26/10 Virgil
6/27/10 Newberry
6/27/10 mueckenh@rz.fh-augsburg.de
6/27/10 Virgil
6/27/10 Mike Terry
6/27/10 mueckenh@rz.fh-augsburg.de
6/27/10 Virgil
6/22/10 Virgil
6/22/10 Virgil
6/22/10 Virgil
6/22/10 mueckenh@rz.fh-augsburg.de
6/22/10 Sylvia Else
6/22/10 Virgil
6/22/10 Sylvia Else
6/22/10 Sylvia Else
6/22/10 Tim Little
6/23/10 Virgil
6/23/10 Virgil
6/23/10 Sylvia Else
6/23/10 Virgil
6/23/10 Sylvia Else
6/23/10 Virgil
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 Virgil
6/23/10 Mike Terry
6/23/10 Virgil
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 Virgil
6/23/10 Sylvia Else
6/24/10 mueckenh@rz.fh-augsburg.de
6/24/10 Virgil
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 Sylvia Else
6/23/10 Virgil
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 Sylvia Else
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6/23/10 Sylvia Else
6/23/10 Virgil
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6/23/10 Sylvia Else
6/23/10 Newberry
6/23/10 Virgil
6/23/10 Virgil
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 Virgil
6/23/10 Newberry
6/23/10 Virgil
6/23/10 Virgil
6/21/10 Virgil
6/21/10 Virgil
6/20/10 lwalke3@lausd.net
6/20/10 Newberry
6/21/10 lwalke3@lausd.net
6/22/10 Newberry
6/24/10 Charlie-Boo
6/24/10 Charlie-Boo
6/25/10 Charlie-Boo
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6/25/10 Newberry
6/26/10 Charlie-Boo
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6/27/10 Virgil
6/27/10 Charlie-Boo
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6/27/10 Newberry
6/27/10 Charlie-Boo
6/27/10 Charlie-Boo
6/26/10 Charlie-Boo
6/27/10 Charlie-Boo
6/26/10 Charlie-Boo
6/26/10 Virgil
6/27/10 Charlie-Boo
6/26/10 Newberry
6/26/10 Virgil
6/26/10 Peter Webb
6/26/10 Virgil
6/27/10 Peter Webb
6/27/10 Virgil
6/27/10 Owen Jacobson
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6/28/10 Owen Jacobson
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6/28/10 Tim Little
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6/28/10 Virgil
6/28/10 Owen Jacobson
6/29/10 Peter Webb
6/30/10 Owen Jacobson
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6/28/10 Virgil
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6/27/10 Peter Webb
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6/27/10 Newberry
6/15/10 fishfry
6/15/10 |-|ercules
6/15/10 Peter Webb
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6/16/10 Tim Little
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6/16/10 mueckenh@rz.fh-augsburg.de
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6/16/10 The Raven
6/16/10 mueckenh@rz.fh-augsburg.de
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6/16/10 |-|ercules
6/16/10 Virgil
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6/15/10 Peter Webb
6/15/10 Daryl McCullough
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 Daryl McCullough
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6/15/10 Daryl McCullough
6/15/10 Daryl McCullough
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6/15/10 Virgil
6/15/10 mueckenh@rz.fh-augsburg.de
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6/15/10 Virgil
6/16/10 mueckenh@rz.fh-augsburg.de
6/16/10 Virgil
6/15/10 mueckenh@rz.fh-augsburg.de
6/15/10 Daryl McCullough
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6/16/10 Peter Webb
6/16/10 mueckenh@rz.fh-augsburg.de
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6/15/10 Aatu Koskensilta
6/15/10 mueckenh@rz.fh-augsburg.de
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6/16/10 J. Clarke
6/16/10 mueckenh@rz.fh-augsburg.de
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6/15/10 Jesse F. Hughes
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6/15/10 Mike Terry
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6/16/10 Virgil
6/16/10 mueckenh@rz.fh-augsburg.de
6/16/10 Virgil
6/15/10 lwalke3@lausd.net
6/15/10 george
6/15/10 george
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6/15/10 Virgil
6/15/10 Aatu Koskensilta
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6/16/10 Herman Jurjus
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6/17/10 Peter Webb
6/17/10 Virgil
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
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6/17/10 Tim Little
6/17/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/17/10 ross.finlayson@gmail.com
6/17/10 ross.finlayson@gmail.com
6/18/10 Tim Little
6/18/10 Marshall
6/20/10 ross.finlayson@gmail.com
7/12/10 ross.finlayson@gmail.com
6/17/10 Tim Little
6/17/10 Peter Webb
6/17/10 Virgil
6/17/10 Tim Little
6/17/10 Peter Webb
6/18/10 Virgil
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
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6/18/10 Virgil
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
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6/19/10 Peter Webb
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6/19/10 Peter Webb
6/19/10 Tim Little
6/20/10 Peter Webb
6/20/10 Mike Terry
6/20/10 Peter Webb
6/20/10 Tim Little
6/20/10 Peter Webb
6/20/10 Tim Little
6/21/10 Peter Webb
6/21/10 Daryl McCullough
6/21/10 mueckenh@rz.fh-augsburg.de
6/21/10 Virgil
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 Virgil
6/21/10 Mike Terry
6/21/10 Mike Terry
6/20/10 Tim Little
6/20/10 Tim Little
6/20/10 Peter Webb
6/20/10 Tim Little
6/19/10 Tim Little
6/19/10 Peter Webb
6/19/10 Tim Little
6/20/10 Peter Webb
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6/20/10 Peter Webb
6/20/10 Tim Little
6/20/10 Peter Webb
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6/21/10 Peter Webb
6/21/10 Virgil
6/21/10 Tim Little
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 mueckenh@rz.fh-augsburg.de
6/23/10 Virgil
6/21/10 Mike Terry
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6/20/10 Tim Little
6/21/10 Peter Webb
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6/19/10 Peter Webb
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6/18/10 Peter Webb
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6/19/10 mueckenh@rz.fh-augsburg.de
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6/20/10 Virgil
6/20/10 Peter Webb
6/20/10 Tim Little
6/21/10 mueckenh@rz.fh-augsburg.de
6/21/10 Virgil
6/21/10 mueckenh@rz.fh-augsburg.de
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6/22/10 mueckenh@rz.fh-augsburg.de
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6/19/10 Tim Little
6/19/10 Peter Webb
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6/20/10 Peter Webb
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6/20/10 Peter Webb
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6/19/10 Peter Webb
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6/20/10 Tim Little
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6/17/10 Virgil
6/17/10 mueckenh@rz.fh-augsburg.de
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6/17/10 Virgil
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2/17/13
2/17/13
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6/17/10 Virgil
6/17/10 Peter Webb
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6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 K_h
6/18/10 Peter Webb
6/19/10 K_h
6/19/10 Peter Webb
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6/19/10 Virgil
6/19/10 K_h
6/19/10 Peter Webb
6/20/10 K_h
6/21/10 Owen Jacobson
6/21/10 Peter Webb
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6/18/10 Peter Webb
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6/18/10 Peter Webb
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6/18/10 Tim Little
6/18/10 Peter Webb
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6/19/10 Peter Webb
6/19/10 Tim Little
6/19/10 Peter Webb
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6/19/10 Daryl McCullough
6/19/10 Peter Webb
6/20/10 Tim Little
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6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
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6/19/10 Peter Webb
6/19/10 Tim Little
6/19/10 Peter Webb
6/20/10 Tim Little
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6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/19/10 Tim Little
6/19/10 Peter Webb
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6/19/10 Peter Webb
6/19/10 Tim Little
6/20/10 Peter Webb
6/20/10 Tim Little
6/20/10 Peter Webb
6/19/10 Daryl McCullough
6/19/10 mueckenh@rz.fh-augsburg.de
6/19/10 Virgil
6/19/10 Tim Little
6/19/10 Peter Webb
6/19/10 Tim Little
6/20/10 Peter Webb
6/20/10 Tim Little
6/20/10 Peter Webb
6/20/10 Tim Little
6/21/10 Peter Webb
6/21/10 Mike Terry
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 Tim Little
6/18/10 Peter Webb
6/18/10 mueckenh@rz.fh-augsburg.de
6/18/10 Virgil
6/16/10 Jack Markan
6/16/10 Jack Markan
6/16/10 Aatu Koskensilta
6/16/10 Aatu Koskensilta
6/16/10 Jack Markan
6/16/10 mueckenh@rz.fh-augsburg.de
6/16/10 Virgil
6/16/10 Jack Markan
6/17/10 Jack Markan
6/17/10 mueckenh@rz.fh-augsburg.de
6/18/10 mueckenh@rz.fh-augsburg.de
6/18/10 Virgil
6/16/10 Jack Markan
6/16/10 mueckenh@rz.fh-augsburg.de
6/16/10 Virgil
6/16/10 Jack Markan
6/16/10 Jack Markan
6/16/10 mueckenh@rz.fh-augsburg.de
6/16/10 Virgil
6/17/10 mueckenh@rz.fh-augsburg.de
6/17/10 Virgil
6/17/10 Jack Markan
6/17/10 mueckenh@rz.fh-augsburg.de
6/17/10 Virgil
6/17/10 Jack Markan
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6/27/10 Frederick Williams | 8,556 | 20,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-30 | longest | en | 0.888195 |
https://help.syscad.net/Boiling_Point_Elevation_Discussion | 1,722,865,180,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00788.warc.gz | 248,299,286 | 6,425 | # Boiling Point Elevation Discussion
Navigation: User Guide -> Boiling Point Elevation -> Boiling Point Elevation Discussion
If we plot the curves for both pure water (dashed) and a liquid containing impurities (such as salt water or Bayer Liquor), we find the latter has moved up at any point. The difference between these two curves is the Boiling Point Elevation (BPE) at that particular pressure.
Note that according to this definition, $\displaystyle{ \beta }$ is intrinsically a function of pressure. Formally
• $\displaystyle{ \beta = T_L^{s}(p) - T_w^{s}(p) }$
where:
$\displaystyle{ T_w^{s}(p) }$ is the boiling point of water expressed as a function of pressure,
$\displaystyle{ T_{L}^{s}(p) }$ is the boiling point for any liquor.
We have a number of possible ways of interpreting BPE:
## BPE at Stream Pressure
The BPE at stream pressure is difference between the temperature at which the liquor would boil and that which water would boil at, at the stream pressure.
## BPE at Stream Temperature
Care must be taken if the boiling point elevation is specified as a function of temperature - Does the correlation refer to the liquor boiling temperature or water boiling temperature?
See the graph below:
The BPE at stream temperature can be interpreted in two ways:
• We can choose the liquor vapour pressure at that temperature as the basis for saturation (point 1 in the above diagram); or
• We can choose the water vapour pressure at that temperature as the basis for saturation (point 2 in the above diagram).
Please see Boiling Point Elevation for a description of the Boiling Point Elevation equation used in the Standard Species model in SysCAD. | 379 | 1,677 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.883616 |
https://web.williams.edu/Mathematics/sjmiller/public_html/math/talks/talks.html | 1,716,973,832,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00534.warc.gz | 510,671,652 | 54,177 | Steven J. Miller Professor of Mathematics, Williams College 18 Hoxsey St, Williamstown, MA (sjm1 AT williams.edu) NOTE: At some point my pages will be migrating to https://sites.williams.edu/sjm1/ [Click here for general advice] curriculum vita (click here for annotated bibliography) teaching statement Takeaways (all classes) Possible thesis / colloquium projects Opportunities in Mathematics (REUs, gradschool) photos of Cam and Kayla
If there are several versions of a talk, the later version is usually the more 'complete' (ie, correct) version.
1. From Pythagoras to Pi: Part I: https://youtu.be/JHvmP-1KXYc (slides here). Given to 7th grade son who is in Algebra I. 41 minutes
2. From Pythagoras to Pi: Part II: https://youtu.be/ISo8kXDP-6U (slides here). Given to 7th grade son who is in Algebra I. 27 minutes
3. What do you MEAN? https://youtu.be/jBKZaCxpgSE (word file here, pdf here) (3/19/2020): Comfort with Algebra sufficient: 40 minutes
4. Shoe size and age: https://youtu.be/aDxdDifvlKM (word file here, pdf here) (3/20/2020): Just need to be able to multiply and add, say grade 1 or 2 and up: 20 minutes
5. Mono-variants: https://youtu.be/PbZhVLXyatY (powerpoint here, pdf here) (3/23/2020): Grade 5 and up: 28 minutes
6. I Love Rectangles Game: https://youtu.be/JHtrzARHwHU (powerpoint here, pdf here) (3/24/2020): Aimed for K, should be good for all ages. 13 minutes
7. From Pascal to Calculus: Part I: https://youtu.be/dv15VTyEWyQ (powerpoint here, pdf here) (3/25/2020): For those knowing Algebra I (equations of lines): 52 minutes
8. From Pascal to Calculus: Part II: https://youtu.be/D6OnleQJ1XM (powerpoint here, pdf here) (3/26/2020): For those knowing Algebra I (equations of lines): 40 minutes
9. Games: Tic Tac Toe, Chocolate Bar, Coins, Devil: https://Mathyoutu.be/4Kvx-JHBvs0 (slides here) (3/27/2020): Much of it can be done with K, 2nd and higher to be safe: 38 minutes
10. Induction and Sums: Part I: Induction https://youtu.be/3orVsQECaag (powerpoint here, pdf here) (3/30/2020): Assuming Algebra I: 32 minutes
11. Induction and Sums: Part II: Geometric Series Formula https://youtu.be/CJjglF65x6g (powerpoint here, pdf here): (3/31/2020): Assuming Algebra I: 25 minutes
12. Induction and Sums: Part III: From the Geometric Series Formula to Primes https://youtu.be/UWNM8EtzoMI (powerpoint here, pdf here) (4/4/2020): Assuming Algebra: 22 minutes
13. The Three Hat Problem and Error Correcting Codes:https://youtu.be/oMeKf7AhAa4 (powerpoint here, pdf here) (4/6/2020): Aimed for 3rd Grade and up: 26 minutes
14. How to Attack Problems I: We WILL Cross That Bridge: https://youtu.be/JH6uxmgWoFQ (powerpoint here, pdf here) (4/8/2020): Aimed for 4th Grade and up: 20 minutes
15. How to Attack Problems II: Legal 21: https://youtu.be/dlBVLlt4PPA (powerpoint here, pdf here) (4/9/2020): Aimed for 4th/5th Grade and up: 14 minutes
16. Polynomials and Applications: Part I: Lecture 1: Lines: https://youtu.be/NNxBhUTzexA (powerpoint here, pdf here) (4/10/2020): Aimed at 6th/7th grade and up: 9 minutes
17. Polynomials and Applications: Part 2: Lecture 2: Introduction to Quadratics: Plotting, Simple Examples and Roots: https://youtu.be/bkIWsNcw-bU (powerpoint here, pdf here) (4/13/2020): Aimed at 7th grade and up: 17 minutes
18. Polynomials and Applications: Part 2: Lecture 3: Introduction to Quadratics: Plotting, Simple Examples and Roots: https://youtu.be/u8Xx7r0VrJ4 (powerpoint here, pdf here) (4/14/2020): Aimed at 7th grade and up: 18 minutes.
19. Polynomials and Applications: Part 2: Lecture 4: Application 1: Trajectories: https://youtu.be/hAw9vSNMzhg (powerpoint here, pdf here) (4/15/2020): Aimed at 7th grade and up: 24 minutes.
20. Polynomials and Applications: Part 3: Lecture 5: Application 2: Fibonacci Numbers: https://youtu.be/WB9gLTASXCw (powerpoint here, pdf here) (4/16/2020): Aimed at 7th grade and up: 24 minutes.
21. Polynomials and Applications: Part 4: Lecture 6: Application 3: Recurrence Relations and Gambling! https://youtu.be/Esa2TYwDmwA (4/17/2020): Aimed at 7th grade and up: 7 minutes
22. Polynomials and Applications: Part 5: Lecture 6: Application 4: Finding Trajectories (revisited) https://youtu.be/7bfeNy4XW5I (powerpoint here, pdf here) (4/21/2020): Aimed at 7th grade and up: 10 minutes
23. Polynomials and Applications: Part 6: Lecture 7: Application 5: Codes: https://youtu.be/wSEXv5PXxu0 (powerpoint here, pdf here) (4/24/2020): Aimed at 7th grade and up: 11 minutes
24. Introduction to Probability: Part 1: The Factorial Function: https://youtu.be/Un3tWIX-kDY (powerpoint here, pdf here) (4/27/2020): Aimed at 7th grade and up: 13 minutes
25. Introduction to Probability: Part 2: Permutations: https://youtu.be/hhYtkpcQQAY (powerpoint here, pdf here) (4/29/2020): Aimed at 7th grade and up: 11 minutes
26. Introduction to Probability: Part 3: Combinations: https://youtu.be/8w8HWvhWM34 (powerpoint here, pdf here) (4/29/2020): Aimed at 7th grade and up: 19 minutes
27. Introduction to Probability: Part 4: Darth Vader Problem: https://youtu.be/qsUYmXGgngE (powerpoint here, pdf here) (5/4/2020): Aimed at 7th grade and up: 19 minutes
28. Introduction to Probability: Part 5: Double Sixes Problem: https://youtu.be/zJTaXORiH9o (powerpoint here, pdf here) (5/6/2020): Aimed at 7th grade and up: 15 minutes (re-recorded; no TAs)
29. Introduction to Probability: Part 6: Long Suits: https://youtu.be/-iWWLLlCRy4 (powerpoint here, pdf here) (5/8/2020): Aimed at 7th grade and up: 18 minutes
30. Introduction to Probability: Part 7: Long Suits: https://youtu.be/ke5RLxIW6v4 (powerpoint here, pdf here) (5/11/2020): Aimed at 7th grade and up: 18 minutes
31. Introduction to Probability: Part 8: All trump, Poker: https://youtu.be/InhvzliUb4I (powerpoint here, pdf here) (5/13/2020): Aimed at 7th grade and up: 15 minutes
32. Introduction to Probability: Part 9: Advanced Combinatorics: https://youtu.be/WFcEQeCoG5k (powerpoint here, pdf here) (5/19/2020): Aimed at 7th grade and up: 16 minutes
• Talks Online
• Random Matrix Theory and $$L$$-Functions
• Random Matrix Theory and $$L$$-Functions I: Analytic Number Theory Seminar, Ohio State (10/20/03): pdf
• Random Matrix Theory and $$L$$-Functions II: Analytic Number Theory Seminar, Ohio State (10/27/03): pdf
• From Random Matrix Theory to $$L$$-Functions: Special Number Theory Seminar, Tel Aviv (12/23/04): pdf (see also the expanded talk given in Utah in 2009 below)
• Identifying Symmetry Groups of Zeros of Families of $$L$$-Functions: Number Theory and Random Matrix Theory Workshop, CMS Summer 2005, Waterloo (6/1/05): ps pdf; tabbed version ps dvi
• Identifying and breaking the symmetry group of zeros of families of $$L$$-functions: Number Theory Seminar, CUNY Lehman (10/20/06): pdf
• A Symplectic Test of the $$L$$-Functions Ratios Conjecture: Algebra Seminar, Brown University (9/17/07): paper.pdf AMS Special Session on $$L$$-functions and automorphic forms, Courant (20 minute version, 3/16/08): pdf tabbed version pdf Johns Hopkins University (50 minute version, 4/4/08): pdf (untabbed version: pdf); Cornell University (50 minute version, 6/5/08): pdf (untabbed version: pdf).
• From Random Matrix Theory to $$L$$-Functions: Graduate Workshop on Zeta Functions, $$L$$-Functions and their Applications, Utah Valley University, June 2, 2009: pdf I've created a webpage with links to talks and papers related to this conference; click here for these papers and links.
• Random Matrix Theory and Number Theory: Progress report from 2009 SMALL REU at Williams College (final presentation of my students John Goes, Steven Jackson, David Montague, Eve Ninsuwan, Ryan Peckner and Vincent Pham): Williams College (8/11/09): pdf
• Low Lying Zeros of Number Field $$L$$-Functions (presented by Ryan Peckner). Young Mathematicians Conference, Ohio State (8/29/09): pdf
• Tests of the $$L$$-Functions Ratios Conjecture. Maine - Quebec Number Theory Conference (10/3/09): pdf Rutgers University (expanded version, 3/2/10): pdf
• The n-level density of zeros of quadratic Dirichlet $$L$$-functions (presented by Jake Levinson), Young Mathematicians Conference, August 19, 2011. pdf
• Low-lying zeros of cuspidal Maass forms (presented by Oleg Lazarev and Liyang Zhang), Young Mathematicians Conference, August 20, 2011. pdf Maine-Quebec Number Theory Conference, October 2, 2011. pdf
• Determinantal Expansions in Random Matrix Theory and Number Theory (with Nicholas Triantafillou), Maine-Quebec Number Theory Conference, September 29, 2012. pdf (trimmed pdf)
• Low-lying zeros of cuspidal Maass forms (blackboard talk given by Levent Alpoge), Maine-Quebec Number Theory Conference, September 29, 2012.
• Low-lying zeros of GL(2) $$L$$-functions, University of Michigan, October 22, 2012. part 1 part 2
• Low-lying zeros of GL(2) $$L$$-functions, AMS Special Session on Arithmetic Statistics, I, Joint Meetings SD, January 10, 2013 (includes excised orthogonal ensemble for elliptic curves and results for holomorphic cuspidal newforms and Maass forms). pdf Expanded Version: Quebec-Vermont Number Theory Seminar, Concordia University, March 21, 2013pdf
• Newman's Conjecture for Automorphic and Function Field $$L$$-functions (presented by Alan Chang), Maine-Quebec Number Theory Conference (10/5/13). pdf; CANT May 28, 2014 pdf (video online here).
• The $$n$$-Level Density of Dirichlet $$L$$-Functions (presented by Kyle Pratt and Minh-Tam Trinh), Maine-Quebec Number Theory Conference (10/5/13). pdf
• Problems in the theory of low-lying zeros, Simons Symposium on Families of Automorphic Forms and the Trace Formula, Puerto Rico, January 27, 2014. pdf
Results in the theory of low-lying zeros, Simons Symposium on Families of Automorphic Forms and the Trace Formula, Puerto Rico, January 28, 2014. pdf
• Biases in Moments of Satake Parameters and Models for L-function Zeros (with Kevin Yang), Maine-Quebec Number Theory Conference, October 3, 2015pdf
• Biases in Moments of Satake Parameters and in Zeros near the Central Point in Families of L-Functions, Computational Aspects of L-functions, ICERM, Providence, RI (11/10/15). pdf
• Gaps Between Zeros of GL(2) L-functions, Southern New England Conference on Quadratic Forms and Modular Forms, June 2, 2016. pdf
• Extending Agreement in the Katz-Sarnak Density Conjecture (joint with Peter Cohen and Roger Van Peski), Quebec-Maine Number Theory Conference, October 8, 2016. pdf
• One-level density for holomorphic cusp forms of arbitrary level, 31st Annual Workshop on Automorphic Forms, ETSU, Mar 6, 2017. pdf
• Lower Order Biases in Fourier Coefficients of Elliptic Curve and Cuspidal Newform families (with Jared Lichtman, Eric Winsor and Jianing Yang), Maine-Qu\'ebec Number Theory Conference, October 14, 2017. pdf MASON VI, March 17, 2023. pdf
• Variance of Gaussian Primes Across Sectors and The Hecke L-Function Ratios Conjecture (with Yujin Kim and Shannon Sweitzer), Maine-Qu\'ebec Number Theory Conference, October 14, 2017. pdf
• Lower Order Terms for the Variance of Gaussian Primes across Sectors (given by Ezra Waxman), 32nd Autormorphic Forms Workshop, Tufts University, May 22, 2018. pptx pdf
• Optimal Test Functions for n-Level Densities and Applications to Central Point Vanishing (with Charles Devlin VI), Maine-Qu\'ebec Number Theory Conference, October 5, 2019. pdf (video: https://youtu.be/I4j374CQgfg)
• How Low Can We Go? Understanding Zeros of L-Functions Near The Central Point, New York Number Theory Seminar, February 18, 2021. pdf
• The Katz-Sarnak Density Conjecture and Bounding Central Point Vanishing of L-Functions, Second Int'l Webinar: Recent Developments in Number Theory, School of Applied Sciences (Mathematics), Kalinga Institute of Industrial Technology University, Bhubaneswar, India, October 3, 2021. pdf (video: https://youtu.be/dsdMFl0yPfw) University of Rochester, November 5, 2021 pdf (video: https://youtu.be/zDJasLIaXhY)
• The Katz-Sarnak Density Conjecture and Bounding Central Point Vanishing of L-Functions (expanded version with improved bounds), Upstate New York Number Theory Conference, Rochester, NY April 2, 2023. pdf (video: https://youtu.be/Be3zNr4kKVw) (20 minute version) 35th Automorphic Forms Workshop, LSU May 23, 2023. pdf
• Combinatorics in Analyzing L-Function Coefficients and Applications to Low-Lying Zeros, Special Session in Number Theory in Celebration of the 70th Birthday of Ram Murty, CMS Summer Meeting, 4 June 2023. pdf (video: )
• Upper Bounds for the Lowest First Zero in Families of Cuspidal Newforms, 36th Automorphic Forms Workshop, Oklahoma State University, May 21, 2024. pdf (talk: https://youtu.be/imMo7_yUDxs)
• Group Theory in Compound Families of L-Functions, Zassenhaus Groups and Friends Conference, Texas State University, June 2, 2024. pdf
• Random Matrix Theory and Elliptic Curves
• 1 and 2 Level Density Functions for Families of Elliptic Curves: Evidence for the Underlying Group Symmetries.
(Thesis defense) Princeton (5/26/02), Number Theory Seminar Ohio State (5/30/02): pdf Boston University (2/10/03): pdf
• Evidence for a Spectral Interpretation of the Zeros of Families of Elliptic Curves. Joint meeting of the AMS and UMI Pisa (6/13/02): pdf title Shorter version (20 minutes): AMS Sectional, Salt Lake City (10/27/02): dvi pdf
• Random Matrix Theory and Elliptic Curves: Evidence for the Underlying Group Symmetries (with three appendices). Johns Hopkins University (3/3/04): pdf An alternate version: Five College Number Theory Seminar, Amherst, MA, (4/20/04): pdf
• Random Matrix Theory models for zeros near the central point (and applications to elliptic curves) (45 minutes) Workshop on Spectral Theory and Automorphic Forms, Montreal, May 2004. ps. An alternate version (60 minutes): Boston University (2/10/03): ps Brown University (9/13/04 - expanded): dvi pdf Another alternate version: (120 minutes) Fellowship of the Ring Seminar, Brandeis University (4/1/05): ps pdf; tabbed version ps dviAnother alternate version (30 minutes), Advances in Number Theory and Random Matrix Theory, Rochester, NY (6/7/06): pdf; tabbed version pdf paper the talk is based on (to appear in Experimental Mathematics)
• How the Manhattan Project helps us understand primes and elliptic curves (with appendices on Dirichlet $$L$$-Fns, Random Graphs and a bibliography). Colloquium, University of Connecticut (3/24/05): ps pdf; tabbed version ps dvi.
• From the Manhattan Project to Number Theory: How nuclear physics helped us understand primes: Theoretical Physics Seminar, Brown University (4/12/06): pdf; relevant papers: Investigations of Zeros Near the Central Point of Elliptic Curve $$L$$-Functions (to appear in Experimental Mathematics pdf (data available online).
• Finite conductor models for zeros near the central point of elliptic curve $$L$$-functions (with Eduardo Duenez, Duc Khiem Huynh and Jon P. Keating). $$L$$-functions, ranks of elliptic curves and random matrix theory workshop, Banff (7/12/07): slides: pdf (my part); pdf (Duenez); ppt (Keating); pdf (Huynh) pdf (combined). AMS Special Session on Number Theory, Wesleyan University, Middletown, CT (10/11/08): pdf Conference on Modular Forms and Related Topics, Beirut, May 28, 2018. pdf CNTA XV, Universite Laval, July 9, 2018. pdf University of Maine, March 29, 2023. pdf University of Maryland, March 29, 2023 pdf (talk here: https://youtu.be/cNsSR5PloTY)
• Finite conductor models for zeros near the central point of elliptic curve $$L$$-functions (with Eduardo Duenez, Duc Khiem Huynh and Jon P. Keating). (This is an expanded version, and includes results towards an average version of the Birch and Swinnerton-Dyer Conjecture, as well as the results from Duc Khiem's thesis on modeling the first zero of quadratic twists by Jacobi ensembles with discretization and lower order terms included.) University of Rochester: pdf Five College Number Theory Seminar: pdf Williams College (abridged 40 minute version): pdf Maine - Quebec Number Theory Conference (20 minute version, full theory, 10/1/11): pdf Brown University (10/24/11): pdf
• Closed-form moments in elliptic curve families and low-lying zeros, Simons Symposium on Families of Automorphic Forms and the Trace Formula, Puerto Rico, January 31, 2014.
• From Sato-Tate Distributions to Low-Lying Zeros, Frobenius distributions of curves, CIRM, February 2014 pdf (and alsothe SouthEastern Regional Meeting on Numbers (SERMON XXVII), Wofford College, April 26, 2014 pdf): YouTube version of SERMON talk here: http://youtu.be/VBzVAvZ6k6A
• From the Manhattan Project to Elliptic Curves: The Ohio State University, March 24, 2014. pdf UMass Boston, February 4, 2014. pdf Washinigton State University, October 12, 2015. pdf MASON IV (3/7/20). pdf (video here: https://youtu.be/p15X3ERNvLs)
• Finite conductor models for zeros near the central point of elliptic curve L-functions: Yale University, April 15, 2014. pdf Boston College, April 30, 2015. pdf (includes Bias Conjecture) Duke University (9/7/16) pdf
• Results on GL(2) $$L$$-Functions: Biases in Coefficients and Gaps Between Zeros, Families of Automorphic Forms and the Trace Formula, Banff International Research Station, Dec 1, 2014: video online here (slides here). Brown University, Number Theory Seminar, April 2, 2018. pdf
• Large Gaps Between Zeros of GL(2) L-Functions (with Owen Barrett and Karl Winsor), 29th Automorphic Forms Workshop, University of Michigan, March 2, 2015. pdf
• Generalizing repulsion of elliptic curve zeros near the central point to other GL(2) forms (with Owen Barrett), 29th Automorphic Forms Workshop, University of Michigan, March 2, 2015. pdf
• Biases in the second moments of Fourier coefficients in one-parameter families of elliptic curves (with Blake Mackall and Karl Winsor), 29th Automorphic Forms Workshop, University of Michigan, March 3, 2015. pdf AMS Special Session on Analytic Number Theory and Automorphic Forms, Washington State University, April 23, 2017. pdf
• Biases in Second Moments of Elliptic Curves (Aditya Jambhale and Akash L. Narayanan), Maine-Quebec Number Theory Conference, September 30, 2023. pdf
• Machine Learning in Elliptic Curves and Beyond: From Conjectures to Theorems to Conjectures, Hybrid Conference on AI-Math, Institute of Mathematics and Statistics, State University of Rio de Janeiro, RJ, Brazil, February 28, 2024. pdf AMS Special Session on Artificial Intelligence in Mathematics, Milwaukee, April 20, 2024. pdf
• Elliptic Curves
• Ranks of One-Parameter Families of Elliptic Curves Over Q(T) and Thoughts on the Excess Rank Question (with five appendices). Boston College (3/10/03): dvi pdf
• Constructing 1-Parameter Families of Elliptic Curves over Q(T) with Moderate Rank. AMS Sectional, Boulder, CO (10/4/03): pdf
• The effect of zeros of elliptic curve $$L$$-functions at the central point on nearby zeros. AMS Sectional, Lawrenceville, NJ (4/18/04): pdf Expanded Version: Algebra Seminar, Brown University (10/24/05): pdf tabbed version dvi
• Towards an Average'' Version of the Birch and Swinnerton-Dyer Conjecture (presented by John Goes). Young Mathematicians Conference, Ohio State (8/29/09): pdf
• Biases in Moments of Elliptic Curve, 30th Automorphic Forms Workshop, Wake Forest University, March 8, 2016 pdf
• Rank and Bias in Families of Hyperelliptic Curves, Quebec-Maine Number Theory Conference, October 7, 2018. pdf
• Rank and Bias in Families of Curves via Nagao's Conjecture, AMS Special Session on A Showcase of Number Theory at Undergraduate Institutions, JMM, Baltimore (1/17/19): pdf 33rd Automorphic Forms Workshop, Duquesne University (3/8/19): pdf
• Applications of Moments of Dirichlet Coefficients in Elliptic Curve Families, Murmurations in Arithmetic, ICERM, July 7, 2023: pdf
• Random Matrix Ensembles
• Eigenvalue Statistics for Ensembles of Random Matrices (especially Toeplitz Ensembles & Diophantine Obstructions). Probability and Ergodic Theory Seminar, Ohio State (10/30/03): pdf Boston University (6/7/04 expanded): pdf Brown University (9/15/04 expanded): pdf
• On the probability that random graphs are Ramanujan. Expanders and Ramanujan Graphs: Construction and Applications, AMS National Meeting, San Diego (1/9/08): pdf
• Eigenvalue statistics for Toeplitz ensembles. ICM Satellite Meeting in Probability & Stochastic Processes, Bangalore, India (August, 2010): pdf
• The Limiting Eigenvalue Density for the Ensemble of m-Circulant Matrices (presented by Gene Kopp and Murat Kologlu, SMALL progress report (8/4/10): pdf Expanded Version: Young Mathematicians Conference, Ohio State (8/28/10): pdf
• Eigenvalue Statistics of Toeplitz and Block m-Circulant Ensembles (presented with Murat Kologlu), AMS Special Session on Random Processes, Worcester, MA (4/9/11): pdf
• Painleve VI and Tracy-Widom Distributions in Random Graphs, Random Matrix Theory and Number Theory, AMS Special Session on Random Processes, Worcester, MA (4/9/11): pdf
• Distributions of Eigenvalues of Weighted, Structured Matrix Ensembles (presented by Olivia Beckwith and Karen Shen), Young Mathematicians Conference, August 19, 2011. pdf
• Eigenvalue Statistics for Toeplitz and Circulant Ensembles, Analysis and Probability Seminar, UConn, March 2, 2012. pdf Expanded version July 3, 2012 (with Gene Kopp, Murat Kologlu and Karen Shen) at the second Institute of Mathematical Sciences Asia Pacific Rim Meeting, Tsukuba, Japan. pdf (zipped files for slides here)
• Random Matrix Ensembles with Split Limiting Behavior} (joint with Peter Cohen, Yen Nhi Truong Vu and Roger Van Peski), International Conference of The Indian Mathematics Consortium in cooperation with American Mathematical Society, Banaras Hindu University, December 15, 2016. pdf Joint Meetings of the AMS/MAA, Atlanta, January 5, 2017. pdf
• Benford's Law (see the compilation page I made for my Bentley talk for links to papers and other resources) (see also colloquium talks)
• Benford's Law, Values of $$L$$-functions and the 3x+1 Problem. Boston College (10/19/04), University of Michigan (11/15/04), University of Arizona (1/11/06), Brown University (3/20/06): slides paper.pdf. Number Theory Special Session, AMS Sectional Meeting, WPI, Worcester, MA (4/25/09, 20min version): pdf Mathematica programs: Just the 3x+1 Map Benford Investigations Some Numerical Simulations: Data
• Benford's Law and Order Statistics. Brown University (2/1/06) paper.pdf
• Poisson Summation and Benford's Law: From values of $$L$$-functions to the 3x+1 problem to products of random variables. University of Bristol (12/12/07); Workshop on the Theory and Applications of Benford's Law, Santa Fe, NM (12/18/07): pdf
• The logarithmic link between economic, hydrologic, and seismic statistics (with Mark Nigrini), Workshop on the Theory and Applications of Benford's Law, Santa Fe, NM (12/17/07):
• Chains of distributions and Benford's Law (with Dennis Jang, Jung Uk Kang, Alex Kruckman and Jun Kudo), Workshop on the Theory and Applications of Benford's Law, Santa Fe, NM (12/17/07): pdf
• Benford's Law: Why the IRS might care about the 3x+1 problem and ζ(s) (general introduction to the subject), Western New England College (2/11/08): pdf (untabbed version pdf); Smith College (10/7/08): pdf; Williams College, Bronfman Science Lunch (10/21/08, 20min version): pdf Central Connecticut State University (9/25/09): pdf; Bentley University (2/1/10): pdf (untabbed version: pdf; click here for links to papers and other resources) Five Colleges Number Theory Seminar, Amherst (1/31/12, 50min version, more number theoretic): pdf
• Theory and applications of Benford's law to fraud detection, or: Why the IRS should care about number theory! IRS (Boston Offices, 3/28/08): pdf (untabbed version pdf). Williams College, SMALL Summer Colloquium (6/23/10, 40min version): pdf. Hampshire College Summer Program (7/14/2011) pdf. Smith College Sigma Xi talk (11/15/11) pdf. Mount Holyoke College (2/1/12) pdf Math 161, Brown University (11/12/12): pdf (video of the talk is available here)
• Theory and Applications of Benford's Law (Iranian Elections, Climate Data and the Weibull) (Tory Cuff and Allie Lewis presenters), SMALL report, August 3, 2010: pdf
• Benford's law and dependent random variables (presented by Thealexa Becker), WIMIN Conference, Smith College (9/24/11) pdf
• Benford's Law, Values of $$L$$-Functions and the $$3x+1$$ Problem, or: Why the IRS cares about Number and Ergodic Theory, Ergodic Theory Seminar, University of Illinois, March 26, 2013. pdf
• Benford's Law: Why the IRS cares about Algebra and Number Theory (and why you should too!), SMALL Faculty Seminar, July 22, 2015. pdf SACNAS, Washington, DC, October 2015. pdf PME Induction Ceremony, Holy Cross, April 20, 2016. pdf Duke University, September 7, 2016. pdf Washington State University, April 21, 2017. pdf
• Can math detect fraud? CSI: Mathematics: The natural behavior of numbers. Science Cafe, Northampton, September 26, 2016. pdf (Video: https://youtu.be/42vYLnY8tnA)
• Benford's Law and the 3x+1 Problem, or: Why the IRS cares about Discrete Dynamical Systems. AMS Northeastern Sectional, Discrete Dynamical Systems, April 22, 2018. pdf
• Benford's Law: Why the IRS and others care about Number Theory (and you should too!), Cross-domain Conference on Benford's Law application, Stresa, Italy, July 10, 2019. pdf PIMS Distinguished Lecture, UNBC, November 14, 2019 (changed to 11/4/2020: video here: https://youtu.be/NSuXJRIFcXc). pdf Vassar College pdf (video: https://youtu.be/fm3s-xjcnAE). UConn Math Club (9/16/20) pdf. New York Number Theory Seminar (6/16/22) pdf. Texas State REU (7/1/22) pdf University of Pittsburgh (11/11/22) pdf
• Benford's Law and Random Integer Decomposition with Congruence Stopping Condition, MAA Southeastern Sectional Meeting, Special Session on the Theory of Integer Sequences, University of Tennessee, Saturday, March 16, 2024 (video here: https://youtu.be/FIjIFsmDk0c). pdf.
• See also some of the talks in the Additive Number Theory section and the colloquium section.
• Additive Number Theory: More Sum Than Difference Sets, Ramsey Theory, Ramanujan Primes, ...
• When almost all sets are difference dominated. Analysis Seminar, Brown University (9/12/07) paper.pdf; Williams College (9/12/08): pdf Wesleyan University (11/20/08): pdf Workshop on Combinatorial and Additive Number Theory (CANT 2009), CUNY Graduate Center, New York, May 2009: pdf
• Explicit constructions of infinite families of MSTD sets (presented by Dan Scheinerman), Number Theory Session, Joint Mathematical Meetings, Washington, DC (1/05/09): pdf
• The Circle Method and Class Groups of Quadratic Fields (Carlos Dominguez, presenter), SMALL report, August 4, 2010. pdf Expanded Version: Young Mathematicians Conference, Ohio State (8/27/10): pdf
• Constructing explicit families of generalized families of MSTD sets (with Sean Pegado and Luc Robinson), May 26, 2011 at CANT 2011: pdf
• Constructing Generalized Sum-Dominant Sets (presented by Geoff Iyer and Liyang Zhang), Young Mathematicians Conference, August 19, 2011. pdf; Maine-Quebec Number Theory Conference, October 2, 2011. pdf; May 23, 2012 at CANT 2012: pdf
• The Distribution of Generalized Ramanujan Primes (presented by Nadine Amersi and Ryan Ronan), Young Mathematicians Conference, August 20, 2011. pdf; May 24, 2012 at CANT 2012: pdf
• The Distribution of the Number of Missing Sums in Sumsets (presented by Oleg Lazarev), Young Mathematicians Conference, August 21, 2011. pdf. May 23, 2012 at CANT 2012: pdf
• Most Sets are Balanced in Finite Groups (presented by Kevin Vissuet), Special Session on Additive and Combinatorial Number Theory, AMS Sectional, Akron, Ohio, October 21, 2012 pdf
• When Almost All Sets Are Difference Dominated, Number Theory Seminar, University of Illinois, March 25, 2013. pdf
• When Almost All Generalized Sumsets Are Difference-Dominated (given by Ginny Hogan), CANT, May 21, 2013. pdf
• Coordinate Sum and Difference Sets of $$d$$-dimensional Modular Hyperbolas (given by Amanda Bower), CANT, May 21, 2013. pdf
• Sums and Differences of Correlated Random Sets (given by Thao Do), Garden State Undergraduate Math Conference, April 5, 2014. pdf
• Continued Fraction Digit Averages and Maclaurin's Inequalities, CANT, May 28, 2014. pdf (talk online here), Quebec-Maine Number Theory Conference, Sept 28, 2014 pdf AMS Special Session on Continued Fractions, JMM 2015, January 12, 2015. pdf
• MSTD Subsets and Properties of Divots in the Distribution of Missing Sums (with Victor Xu and Xiaorong Zhang), CANT May 26, 2016 pdf
• When almost all sets are difference dominated in Z/nZ (given by Adam Lott), Integers Conference, University of West Georgia, October 7, 2016. pdf
• A geometric perspective on the MSTD question (given by Carsten Peterson), Integers Conference, University of West Georgia, October 7, 2016. pdf
• Ramsey Theory over Number Fields, Finite Fields and Quaternions, CANT, May 26, 2017 pdf
• Within-perfect & near-perfect numbers (with Kevin Kwan), CANT May 22, 2018 pdf
• Phase Transitions in the Distribution of Missing Sums and a Powerful Family of MSTD Sets, Integers 2018, Augusta, Georgia, October 4, 2018. pdf
• Recent progress in sumsets, generalized Schreier sets, generalized Zeckendorf decompositions, and prime and square-free walks (with Joshua Ackerman, Tudor Popescu, Zimu Xiang), CANT May 23, 2019
• Distribution of missing differences in diffsets (given by Fei Peng), CANT June 5, 2020 pdf
• Distribution of missing sums in correlated sumsets (given by Dylan King), CANT June 5, 2020 pdf
• Large sets are sumsets (given by Ben Baily and Henry Fleischmann), Combinatorial and Additive Number Theory, May 2022. pdf
• Schreier multisets and the s-step Fibonacci sequences, Integers Conference, University of Georgia, May 18, 2023.
• Additive Number Theory: Zeckendorf Decompositions (images for Zeckendorf / gaps)
• Cookie Monster Meets the Fibonacci Numbers. Mmmmmm -- Theorems! Workshop on Combinatorial and Additive Number Theory (CANT 2010), CUNY Graduate Center, New York, May 2010: pdf (untabbed version pdf); Hampshire College Summer Program (expanded version, 7/15/10): pdf; Smith College (1/28/11, 30 minute version): pdf Special Session on Undergraduate Research, Holy Cross, 4/9/10 (presented by Murat Kologlu) pdf; expanded version from May 26, 2011 for CANT 2011: pdf expanded version with gaps for Algebra Seminar, UConn (2/7/12) pdf AMS Special Session on Difference Equations, Washington, DC (4/27/12, 20 minutes) pdf University of Michigan (11/30/17) pdf Williams and Texas State REUs (7/8/20) pdf
• From the Fibonacci Numbers to a Central Limit Type Theorem (presented by Yinghui Wang). SMALL report, August 4, 2010. pdf Expanded Version: Young Mathematicians Conference, Ohio State (8/28/10): pdf Even more expanded version: Williams College Seminar October 1, 2010. pdf
• Gaps between summands in generalized Zeckendorf decompositions (presented by Olivia Beckwith), Young Mathematicians Conference, August 19, 2011. pdf
• To Infinity and Beyond: Gaps Between Summands in Zeckendorf Decompositions, May 22, 2012 at CANT 2012: pdf
• Distribution of Summands in Generalized Zeckendorf Decompositions (presented by Rachel Insoft and Philip Tosteson), AMS Sectional Meeting, Rochester, September 22, 2012: pdf AMS Sectional Meeting, Rochester, September 22, 2012: pdf Special Session on Additive and Combinatorial Number Theory, AMS Sectional, Akron, Ohio, October 21, 2012 pdf
• Distribution of summands in generalized Zeckendorf decompositions, AMS Session on Number Theory, I, Joint Meetings San Diego, January 9, 2013. pdf
• Mind the Gap: Distribution of Gaps in Generalized Zeckendorf Decompositions, CANT, May 21, 2013. pdf Williams Math/Stats Faculty Seminar (9/12/13), Maine - Quebec Number Theory Conference (10/5/13), AMS Special Session on Difference Equations (10/12/13). pdf Expanded version dealing with Kentucky Sequence: 16th International Fibonacci Conference, Rochester, NY (7/25/2014): pdf
• Coin Flips, Fibonacci Numbers and Gaps! (given by Phil Tosteson), AMS Special Session on Difference Equations, March 29, 2014. pdf
• Introduction to Zeckendorf Decompositions, AIM, July 7, 2014; Progress Report Images July 9, 2014 (pdf); Final report July 11, 2014 (pdf).
• From Fibonacci Quilts to Benford's Law through Zeckendorf Decompositions, AMS Special Session on Difference Equations, JMM 2015, January 10, 2015. pdf CANT May 19, 2015. pdf
• From the Kentucky Sequence to Benford's Law through Zeckendorf Decompositions, AMS Special Session on Difference Equations, Washington, DC, March 7, 2015. pdf
• The Fibonacci Quilt Sequence (presented by Pari Ford), AMS Special Session on Difference Equations, Washington, DC, March 7, 2015. pdf
• Convergence rates in generalized Zeckendorf decomposition problems (with Ray Li, Zhao Pan and Huanzhong Xu), CANT May 26, 2016 pdf
• Convergence Rates in Generalized Zeckendorf Decompositions (with Zhao Pan and Huanzhong Xu), 17th International Fibonacci Confernece, Caen, France (6/28/2016): pdf
• On Summand Minimality of Generalized Zeckendorf Decompositions (given by Chi Huynh, Carsten Peterson and Nhi Truong), AMS Contributed Paper Session on Undergraduate Research, JMM Atlanta 2017. pdf
• Generalizations of Zeckendorf's Theorem to Two-Dimensional Sequences (with Joshua Siktar), AMS Special Session on Discrete Neural Networking and Applications, I, JMM, San Diego, January 11, 2018. pdf
• Distributions in Generalized Zeckendorf Decompositions (with Yujin Kim, Shannon Sweitzer, Eric Winsor and Jianing Yang), AMS Special Session on Discrete Neural Networking and Applications, I, JMM, San Diego, January 11, 2018. pdf.
• Monovariants to Zeckendorf Decompositions and Games, The Eighteenth International Conference on Fibonacci Numbers and Their Applications, Dalhousie University, July 2, 2018. pdf
• Gaps Of Summands of The Zeckendorf Lattice, presented by SMALL 2019 and Eureka 2018, 19th International Fibonacci Conference, July 21, 2020.
• Completeness of Positive Linear Recurrence Sequences, presented by SMALL 2020, 19th International Fibonacci Conference, July 21, 2020.
• Generalizing Zeckendorf's Theorem to non-constant coefficient recurrences, presented by Eureka 2020 and Polymath REU (pdf), 19th International Fibonacci Conference, July 22, 2020.
• Asymptotic Analysis For Lattice Walks Derived From Zeckendorf Decompositions, presented by CMU students and Eureka 2019, 19th International Fibonacci Conference, July 23, 2020. pdf
• Generalizing Zeckendorf's Theorem to Homogeneous Linear Recurrences, presented by Thomas Martinez, 19th International Fibonacci Conference, July 23, 2020.
• The Generalized Bergman game (given by Faye Jackson and Luke Reifenberg), Combinatorial and Additive Number Theory, May 2022pdf
• Why I love Monovariants: From Zombies to Conway's Soldiers via the Golden Mean, 7th Number Theory Meeting - Torino, September 28, 2023. pdf Kansas State University, November 8, 2023: pdf Western New England PME Induction Ceremony, April 26, 2024: pdf
• The theory of normalization constants and Zeckendorf decompositions, Combinatorial and Additive Number Theory 2024, Cuny Graduate Center, May 22, 2024.
• Additive Number Theory: Erdos Distance Problems (images for Zeckendorf / gaps)
• Classification of Crescent Configurations on Four and Five Points (given by Chi Huynh), Integers Conference, University of West Georgia, October 7, 2016. pdf AMS-MAA-SIAM Special Session on Research in Mathematics by Undergraduates and Students in Post-Baccalaureate Programs, III, Joint Math Meetings, Atlanta, January 2017 (given by Becky Durst, Chi Huynh, Max Hlavacek) pdf
• Crescent Configurations In Non-Euclidean Norms, (given by Dylan King and Cathy Wahlenmayer), Maine-Qu\'ebec Number Theory Conference, October 5, 2019.
• Angle variants of the Erdos distinct distance problem (given by Henry Fleischmann and Ethan Pesikof), Combinatorial and Additive Number Theory, May 2022.
• Erdos Distinct Angle Problems (given by Henry Fleischmann and Ethan Pesikoff), Combinatorial and Additive Number Theory, May 2022. pdf
• Computers, Dynamical Systems and Mathematics Education
• Computers in Undergraduate Education and Zeros of Elliptic Curves. Minnesota (8/10/02): pdf Expanded Version: pdf
• Statistical Investigations as a Tool in Undergraduate Mathematics Research: Poster and Workshop Sessions
International Conference on Statistics Honolulu, Hawaii (6/5/03 and 6/7/03): pdf
• NSF Workshop on Computation in Algebra, Number Theory and Combinatorics. Washington DC (9/21/02): ps and ps
• The Pythagorean Won-Loss Theorem: An introduction to modeling. Great Activities for an Introductory Statistics Class, AMS National Meeting, San Diego, (1/7/08): pdf
• Workshop on Mathematicians in Mathematics Education, participant, Institute for Mathematics Education, Tucson, Arizona, March 20 - 22, 2008.
• Progress Report: Statistics in Function Fields, Sage Days 21, Seattle, Washington, May 27, 2010. pdf
• Introduction to Matlab: IAS Women in Mathematics: May 18, 2011. Click here for files.
• Virus dynamics on star graphs (presented by Thealexa Becker), WIMIN Conference, Smith College (9/24/11) pdf; expanded version at AMS Special Session on Difference Equations, University of Maryland, March 29, 2014. pdf
• Why more is better: the power of multiple proofs, ATMIM Spring Conference, Assabet Valley Regional Technical High School, Grade 9 - 12 Session (3/23/13) pdf
• From M&Ms to Mathematics, or, How I learned to answer questions and help my kids love math, ATMIM Spring Conference, Assabet Valley Regional Technical High School, Keynote Address (3/23/13) pdf HCSSiM, July 17++, 2013. pdf; Dr. Philip O. Coakley Middle School, Norwood, MA, June 9, 2014. pdf. Maine-Quebec Number Theory Conference, October 3, 2015pdf Math League, College of New Jersey, July 25, 2016 and July 20, 2017. pdf Prospective Days, Williams College, August 11, 2017. pdf
• Why Cookies And M&Ms Are Good For You (Mathematically), Stuyvesant High School, May 9, 2014. pdf (talk online here)
• YouTube University: The Benefits of Recording Lectures, Blended Learning Conference, May 22, 2014. pdf (talk online here)
• Challenges and Opportunities for Graduate School Bound Liberal Arts Students, AMS Committee on Education, Washington, DC, October 30, 2015. slides pdf handout pdf
• Building YouTube University Brick by Brick, AMS-MAA Special Session on Innovative Ideas in Enhancing Success in Mathematics Classes, Joint Meetings of the AMS-MAA, Seattle, January 6, 2015. pdf
• Balancing responsibilities in academia, panelist at the 30th Automorphic Forms Workshop, Wake Forest University, March 7, 2016.
• Results and Research in REUs: Cookie Monster Meets the Fibonacci Numbers. Mmmmmm -- Theorems! Florida Institute of Technology, May 20, 2016. pdf
• Springboards to Mathematics: From Zombies to Zeckendorf, Math League Summer Program, College of New Jersey, July 25, 2016. pdf
• Portable Lecture Capture, LACOL Teaching with Tech 2017 Lightning Round, Vassar College, June 15, 2017. pdf (powerpoint slides here)
• Where's my remote? Shared upper level math courses across schools, LACOL 2017 Session 7, Vassar College, June 16, 2017. pdf (video here: http://lacol.net/shared-math/)
• YouTube University: Shared upper level math courses across schools, AMS-MAA Special Session: Novel Methods of Enhancing Success in Mathematics Classes, Joint Mathematics Meeting, San Diego, January 2018. pdf
• Where's the Remote? Upper-Level Math/Stats Hybrid Course Sharing for the Liberal Arts (with Jingchen (Monika) Hu), EDUCAUSE Learning Initiative (ELI) Annual Meeting, New Orleans, January 30, 2018. pdf
• Exploration into a Framework for Digitally Shared Courses (with Liz Evans, Lioba Gerhardi and Monika Hu), LACOL Workshop, Carlton College, June 1, 2018.
• From M&Ms to Modern Mathematics: Introduction to Plotting, Logarithms and Patterns, Williamstown Elementary School, October 1, 2018. pdf
• The Explicit Sato-Tate Conjecture in Arithmetic Progressions, Quebec-Maine Number Theory Conference, October 7, 2018. pdf
• Summer Data Science (online): By and for the liberal arts, panelist in discussion on an online class co-teaching, LACOL Data Science Workshop, June 6, 2019.
• YouTube University II: Shared upper level math courses across schools, Joint Math Meetings, AMS Special Session on Pedagogical Innovations That Lead to Successful Mathematics, Denver, Jan 16, 2020. pdf (online: https://youtu.be/rjn1sn3WLZg)
• Extending P y t h a g o r a s, Williams College 3/9/20: pdf
• From Zombies to Fibonaccis: An Introduction to the Theory of Games, Michigan Math Circle, April 23 and 30, 2020. pdf (part I) pdf (part II) NJ Math Camp, July 17, 2023. pdf
• Engaging students in research in the covid pandemic, Cross Atlantic Representation Theory and Other Topics ONline (CARTOON), May 30, 2020. pdf
• All I need to know about probability I learned from Darth Vader and James Bond, Hampshire College Institute for Summer Studies in Mathematics, August 3, 2020. (powerpoint here, pdf here)
• Analyzing Virus Dynamics on k-level Starlike Graphs (with Akihiro Takigawa), Spring 2021 AMS Eastern Sectional Meeting: Special Session on Applications and Asymptotic Properties of Discrete Dynamical Systems: A Session in Honor of the Retirement of Orlando Merino. pdf
• CUR-Goldwater Faculty Mentor Award, May 12, 2021, Expanded Version. https://youtu.be/NGS5mHJcnlQ (powerpoint here, pdf here)
• Computational Thinking Modules: From Data to Results (through chocolate!), STANYS (Science Teachers of New York State) Conference: November 7, 2021: https://youtu.be/xYy_LAy-gHo (powerpoint here, pdf here).
• The Fibonacci Sequence and Math Outreach (joint with Cameron and Kayla Miller, Anna Mello and her classes, and SMALL 2022), the 20th International Fibonacci Conference, Sarajevo, July 29, 2022: (powerpoint here, pdf here).
• Modeling Beyond the Classroom: Linking Students and Industry, Integrating Math Modeling and Interdisciplinarity into Your Classroom, MathFest, August 5, 2022: https://youtu.be/G1ttDfUzaqA (powerpoint here, pdf here). COMAP Contributed Paper Session: Integrating Modeling into Established Courses: Joint Math Meetings: Boston: Jan 7, 2023 (powerpoint here, pdf here).
• From Zombies to Sports: An Introduction to the Theory of Games, Michigan Math Circle, April 23 and 30, 2020. pdf (part I) pdf (part II) NJ Math Camp, July 17, 2023. pdf
• AMS Expects that every Mathematician will do their ODEs: From the Battle of Trafalgar to Calculus (or Nelson to Newton), AMS Special Session on Modeling to Motivate the Teaching of Mathematics of Differential Equations JMM, San Francisco, January 3, 2024. pdf (video here: https://youtu.be/rRBWahHdMS4 SIMIODE EXPO 2024: February 11, 2024. pdf
• Colloquium / General Talks (mostly Random Matrix Theory and Number Theory)
• How the Manhattan Project helped us understand primes. Colloquium, University of Cincinatti (10/16/03) and Ohio State (10/1/03): pdf. An alternate version (with appendices on Dirichlet $$L$$-Fns, Elliptic Curves and a bibliography): Colloquium, Colby College (3/8/05): ps pdf; tabbed version ps dvi. Another alternate version (with appendices on Dirichlet $$L$$-Fns, Random Graphs and a bibliography). Colloquium, University of Connecticut (3/24/05): ps pdf; tabbed version ps dvi
• From Nuclear Physics to Number Theory: How the Manhattan Project helped us understand Primes: SUMS Conference, Brown University (2/12/05): ps pdf; tabbed version ps dvi This is similar to the above colloquia, but aimed more at undergraduates.
• $$L$$-functions and Random Matrix Theory. Brown University (10/6/06). See pdf for a more extensive version.
• From the Manhattan Project to Number Theory: How Nuclear Physics Helps Us Understand Primes (20 minutes). Bronfman Science Lunch, Williams College (6/23/09): pdf (20 minute version) pdf (expanded version). Wellesley College (2/2/10): pdf (25 minute version)
• Heuristics and Ballpark Estimates: From the 3x+1 problem to counting primes and birthdays: PROMYS, Boston University (7/28/09)
• Pythagoras at the Bat: an Introduction to Statistics and Mathematical Modeling: Wellesley (9/21/09): pdf
• How low can we go? Lower order terms in CLTs from Benford's Law to Elliptic Curves (25 minute version). NES MAA Fall 2009 Meeting (11/21/09): pdf (no tab pdf)
• Great Expectations, or: Expect More, Work Less. Wellesley (2/3/10): pdf
• The Riemann Hypothesis at 150: From Primes to Nuclei and Many Things Between. Williams College (11/18/09) Blackboard talk, but see the following page for relevant papers on the subject.
• From the Manhattan Project to Elliptic Curves. Smith College (1/28/11) pdf Dartmouth College, January 28, 2013. pdf (no tab version: pdf)
• From Cookie Monster to the IRS: Some Fruitful Interactions between Probability, Combinatorics and Number Theory. UNC Charlotte (2/1/11) pdf
• Cookie Monster Meets the Fibonacci Numbers. Mmmmmm -- Theorems! Pi Mu Epsilon Induction Ceremony, College of the Holy Cross (5/5/11) pdf Smith College (9/13/11) pdf (click here for the talk) Amherst College (9/21/11) and Mount Holyoke College (10/5/11) pdf SUMS, Brown University (3/10/12; presented by Louis Gaudet, alternate 18 minute version emphasizing gaps) pdf Colby College (3/28/12) pdf Summer Science Lecture Series, Williams College (6/19/12, 20 minutes) pdf; Williams College (9/14/12), Middlebury College (9/28/12), Wesleyan College (10/19/12) and Brown University (11/12/12) pdf Expanded Version for Yale University with data for longest gap part (4/14/14): pdf (video here: http://youtu.be/5e6HsfxqVSE), Yale University (with rectangle game) (2/23/24) pdf (video here)
• Biases: From Benford's Law to Additive Number Theory via the IRS and Physics. SMALL Summer REU, Williams College (6/22/11) pdf
• Number Theory and Random Matrix Theory Progress Report (Nadine Amersi, Thealexa Becker, Olivia Beckwith, Geoffrey Iyer, Oleg Lazarev, Karen Shen, Alec Greaves-Tunnell, Ryan Ronan, Liyang Zhang). SMALL End of Summer Progress Report, Williams College (8/2/11) pdf (zipped file here). This work was also presented at a poster session. Poster 1 (RMT, Maass forms, Viruses), Poster 2 (Benford, MSTD, c-Ramanujan).
• From Random Matrix Theory to Number Theory, SMALL colloquium, July 25, 2012 (similar to Utah graduate talk). pdf
• Number Theory and Probability: SMALL '13 Projects, Williams College, June 19, 2013. pdf
• The Latke-Hamantaschen Debate: The Primacy of Three, Williams College, March 19, 2014. pptx pdf (YouTube clip: http://www.youtube.com/watch?v=tt4_4YajqRM
• , Williams College, March 20, 2014 (supplemental lecture for multivariable calculus).
• Some Results on Low-Lying Zeros of L-Functions, Williams College, July 16, 2014 (SMALL colloquium). pdf
• Why more is better: the power of multiple proofs, Hampshire College, July 31, 2014. pdf
• He's just going through a phase: Miller's SMALL students and Phase Transitions, Williams College, October 10, 2014. pdf
• From Fibonacci Quilts to Benford's Law through Zeckendorf Decompositions, Williams College, Science Talk, November 11, 2014. pdf
• Why the IRS cares about the Riemann Zeta Function and Number Theory (and you should too!), Williams College talk to the faculty February 12, 2015 (http://youtu.be/JmBXZCbbwas, slides here), Carnegie Mellon 3/25/15 and Winona State University 3/30/15: pdf
• Extending the Pythagorean Formula, Williams College (4/8/15): http://youtu.be/idIHcgapMG4 (slides here). Modified version (all slides, no blackboard: 20 minutes Williams Summer Science Talk): http://youtu.be/8HtABCG6ACg (slides here). Expanded version (all slides, 50 minutes: Hampshire College Summer Studies in Mathematics, 7/28/2015): (slides here) Washington State University (10/12/15) (see Hampshire College slides), Window on Williams (10/16/15). pdf Hampshire College (8-3-16). pdf Williams SMALL REU, July 26, 2017 and TCNJ Math Camp, July 29, 2017. pdf Williams March 9, 2020 https://www.youtube.com/watch?v=EH6PUS2OwUY&t=931s
• Success and/or Significance, A musician and mathematician discuss beauty, perfection, and faith in the liberal arts, Veritas forum, Williams College, April 14, 2016. pdf pptx (with videos)
• From the Manhattan Project to Elliptic Curves: Introduction to Random Matrix Theory, SMALL Faculty Talk, Williams College, July 27, 2016 pdf SUMS, Brown University, March 18, 2017. pdf. SMALL Faculty Talk, Williams College, July 5, 2017. pdf Analysis Seminar, Virginia Tech, April 6, 2018. pdf. SMALL Faculty Talk, Williams College, June 13, 2018. pdf. Yale University, October 8, 2018. pdf Math Club at U(M), January 21, 2021. pdf (video here: https://youtu.be/j4imqkJ7Eg8)
• Secrets of the Tax System with Steve Miller: Keep More of Your MOOLA: Williams College 9/16/2016: (slides here)
• Egg Drop Mathematics: It's all it's cracked up to be: Math League Summer Program, July 17 and 29, 2017; July 14, 2019. University of Michigan, November 30, 2017. REU in Computational Statistics, UNC Greenesboro, July 10, 2023, TCNJ Math Camp, July 18 and 25, 2023. pdf
• From C to Shining C: Complex Dynamics from Combinatorics to Coastlines, Maritime Studies Program of Williams College and Mystic Seaport, Mystic, CT, October 20, 2017. pdf (video here: https://youtu.be/TMILk79N_Bs) Michigan Math Circle, April 30, 2020. pdf Hampshire College 8/8/2022 (video: https://youtu.be/GO3EbWUesBQpdf
• SNAP DECISIONS: Management Lessons from Legos, Lever, North Adams, January 25, 2018. ppt pdf
• From Zombies to Fibonaccis: An Introduction to the Theory of Games. TCNJ Math Camp, July 16 and 29 (Part I) and 17 and 30 (Part II), 2018. pdf Hampshire College, July 24, 2018. pdf
• The German Tank Problem: Math/Stats At War, University of Connecticut Math Department Awards Dinner, April 26, 2019. pdf Hampshire College, July 8, 2019. pdf (Hampshire College talk: https://youtu.be/IOcI2c3QQbw) Yale University, September 27, 2019. pdf (video: https://youtu.be/nZJUbwC_bTw). The Christie Lecture, NES/MAA Fall 2019 Meeting, 11/22/2019. pdf (video: https://youtu.be/I3ngtIYjw3w). PROMYS August 7, 2020 pdf UNC Greensboro REU, June 9, 2022 pdf
• Inrtoduwtion to Erorr Dwtetcion and Czorrectmon, TCNJ Math Camp, July 14, 2019. MathFest, August 1, 2019. pdf (video: https://youtu.be/RPWMAqPhv6w); 50 minute video from class: https://youtu.be/YOs8QYU3Wq8 Hampshire College, June 30, 2021. University of Michigan, October 1, 2021. pdf (video: https://youtu.be/tuL0K1XtGBE)
• VCTAL CRYPTOGRAPHY LECTURES: June 19-20, 2019: Burlington, VT: See VCTAL Lectures Burlington 2019: Introduction to cryptography, error detection / correction.
• Of the Origin and Design of Eigenvalues in General, with Concise Remarks on the English Constitution. Dearborn REU (6/24/19) and SMALL REU (7/3/19). pdf
• Why I love Monovariants: From Zombies to Conway's Soldiers via the Golden Mean. Math Club, University of Michigan, September 10, 2020. pdf (video: https://youtu.be/LWWc4q3e-RY) Williams College, October 13, 2021. pdf (video: https://youtu.be/b6CrLgHawm8). Baruch College, August 4, 2022. pdf
• Lessons I've learned (many the hard way), Baruch REU July 6, 2021 and Polymath Jr REU July 8, 2021. Williams College (10/13/21) pdf (video: )
• From Monovariants to Zeckendorf Decompositions and Games, and Random Matrix Theory, Williams College (7/14/21) and Texas Tech (7/29/21). pdf (video: https://youtu.be/Kayru_V75V8)
• Classifying GOATS (like Brady, Russell and Ruth) by their tails (joint with Rick Cleary), MathFest August 5, 2021 pdf
• Connections of Class Numbers to the Group Structure of Generalized Pythagorean Triples (with Thomas Jaklitsch, Thomas Martinez, and Sagnik Mukherjee), International Conference on Class Groups of Number Fields 2021 pdf https://youtu.be/BbeFny-C5sU
• Why I love Monovariants: From Zombies to Conway's Soldiers to Fibonacci Games, AMS Eastern Sectional, UMass Amherst 10/1/22 pdf (video from version given in Math 409: 9/26/33: https://youtu.be/p2-QfjgitQ8University of Pittsburgh 11/10/22 pdf University of Maryland 3/29/23 pdf (talk online: https://youtu.be/n5I39r_CyuA) YPMD III, 17-1/17 mod 10, 17*119, Math Camp (New Jersey) 7/25/23, Williams College 7/26/23 pdf
• Problem Child: Using Problems as a Springboard to Research, MathFest: Problem Creation and Problem Solving Session, August 5, 2023 pdf
• Math/Stat Thesis Showcase, Williams College, April 12, 2024 pdf
• Baseball / Sabermetrics Talks
• The Pythagorean Won-Loss Formula in Baseball (An Introduction to Statistics and Modeling) (60 minute version) Brown University (9/28/05) and (12/7/05): slides(tab) slide(no tab) paper_pdf. Shorter 15 minute version: 2006 Hudson River Undergraduate Mathematics Conference (4/8/06): slides(tab) slides(no tab)Pythagoras at the Bat: An Introduction to Statistics and Modeling (20 minute version): Society For American
• Baseball Research, Boston Meeting, Boston, MA (5/20/06): slides(tab) slides(no tab) paper_pdf. Williams College (40 minute version, 1/15/08): pdf. Holy Cross (50 minute version, 2/7/08): pdf. Western New England College (40 minute version, 1/15/08): pdf. Connecticut Smoky Joe Wood SABR Chapter, Hamden, CT (30 minute version, 2/16/08) pdf. PROMYS Program, Boston University (40 minute version, 7/25/08): pdf. Bennington College (2/27/09): pdf Hampshire College (7/22/09): pdf Wellesley (9/21/09): pdf Awards Night, University of Connecticut (4/12/10): pdf Virginia Tech (3/28/11): pdf UMass Amherst (10/19/11) and Fitchburg State University (11/3/11): pdf Boston College (3/29/12) pdf Science Days talk to Prospective Williams Students (8/15/14) pdf and 8/12/16 pdf and 8/10/18 pdf University of Vermont (12/2/16, with an intro on undergraduate research) pdf Developer Thursdays at Cloud85 (6/8/17): pdf University of Michigan (11/29/17): pdf San Francisco (to Williams Northern California Alumni Society, expanded version with xSLG) (1/4/24) pdf Siena College (head-to-head added) (2/8/24): pdf Western New England (head-to-head expanded): (4/26/24): pdf
• Pythagoras on the Ice: Hockey Conference, Babson (10/1/16): pdf
• Webpages with talks / posters by students and colleagues at conferences
• Joint Meeting of the AMS/MAA, 2012: Boston, January 4 to 7, 2012: click here for a webpage with all the talks, or click here for notes on the Panel on Undergraduate Research.
• Young Mathematicians Conference: Ohio State, July 27-29, 2012: click here for a webpage with all the talks.
• SMALL 2012 posters: Poster 1 (Gaps, Phase Transitions) Poster 2 (RMT, Maass)
• SMALL 2012: end of summer talk, Williams College, August 7, 2012.
• Joint Meetings of the AMS/MAA, 2013: San Diego, January 9 to 12, 2013: click here for a webpage with all the talks.
• SMALL 2013: UMass Amherst 7/30/13 Williams 7/31/13 YMC August 2013: Click here for talks and posters.
• AMS Special Session on Difference Equations: Baltimore 2014: homepage for session, including links to talks available on YouTube.
• 2014 SMALL PANTHers Group: (Projects supervised with Caroline Turnage-Butterbaugh, Nathan McNew and Ben Weiss)
• Biases in Elliptic Curve Families (Blake Mackall, Christina Rapti, Karl Winsor): Yale REU Conference, July 25, 2014: pdf
• A Benford Walk Down Wall Street (Xixi Edelsbrunner, Karen Huan, Blake Mackall, Jasmine Powell, Madeleine Weinstein): Yale REU Conference, July 25, 2014: pdf
• Complex Ramsey Theory (Andrew Best, Karen Huan, Nathan McNew, Jasmine Powell, Kimsy Tor, Madeleine Weinstein): Yale REU Conference, July 25, 2014: pdf
• Large gaps between zeros of GL(2) L-functions (Owen Barrett, Brian McDonald, Patrick Ryan, Karl Winsor): Yale REU Conference, July 25, 2014: pdf (talk), pdf (poster)
• Benfordness of Zeckendorf Decompositions (Andrew Best, Patrick Dynes, Xixi Edelsbunner, Brian McDonald, Kimsy Tor and Madeleine Weinstein): 16th International Conference on Fibonacci Numbers and Their Applications, July 25, 2014: pdf Fall Southeastern Sectional Meeting of the AMS, Special Session on Difference Equations, March 7, 2014: pdf
• Cookie Monster Meets the Fibonacci Numbers --- Mmmm, Theorems! (Andrew Best, Patrick Dynes, Xixi Edelsbunner, Brian McDonald, Kimsy Tor and Madeleine Weinstein): 16th International Conference on Fibonacci Numbers and Their Applications, July 25, 2014: pdf
• 2014 Quebec-Maine Number Theory Conference
• SMALL 2015
• UConn REU Conference: A Ramsey Theoretic Approach to Function Fields and Quaternions (Megumi Asada, Sarah Manski), July 28, 2015. pdf
• Large Gaps Between Zeros of L-Functions Associated to GL(2) Cusp Forms (given by David Burt & Blaine Talbut), Maine-Quebec Number Theory Conference, October 3, 2015. pdf
• A Ramsey Theoretic Approach to Finite Fields and Quaternions (given by Sarah Manski), Maine-Quebec Number Theory Conference, October 3, 2015. pdf
• Biases in Moments of Satake Parameters and Models for $L$-function Zeros (with Kevin Yang), Maine-Quebec Number Theory Conference, October 3, 2015. pdf
• From M\&Ms to Mathematics, or, how I learned to answer questions and help my kids love math (joint with Cam and Kayla Miller), Maine-Quebec Number Theory Conference, October 3, 2015. pdf (see also https://www.youtube.com/watch?v=BXVRxLk8PKs&feature=youtu.be )
• SMALL 2016 and Carnegie Mellon 2016
• Williams College Summer Science Poster Session (if presented at Ohio State as well, just listed there) and/or Williams Summer REU Conferencce
• RMT Poster: Checkerboard Matrices and Limiting Spectral Distributions: Peter Cohen, Max Hlavacek, Carsten Sprunger, Paula Burkhardt, Kevin Yang,
Jonathan Dewitt, Nhi Truong, and Roger Van Peski
• Zeckendorf Poster: Summand Minimality of Generalized Zeckendorf Representations of Non-Negative Linear Recurrence Relations: Katherine Cordwell, Max Hlavacek, Ngoc Yen Chi Huynh, Carsten Peterson, and Yen Nhi Truong Vu
• Distance Talk: Optimal point sets determining few distinct triangles: Alyssa Epstein and Adam Lott: Williams College REU Conference: July 29, 2016
• Distance Poster: Classification of All Crescent Configurations on Four and Five Points: Rebecca Durst and Magda Hlavacek
• Young Mathematicians Conference: Ohio State August 19-21, 2016 (the posters were also presented at Williams College)
• Integer Complexity Poster: Some Bounds on Integer Complexity: Katherine Cordwell and Aaditya Sharma
• MSTD Poster: When almost all sets are difference dominated in $$\mathbb{Z}/n\mathbb{Z}$$: Adam Lott
• $$L$$-Functions Poster: Second Moments of Satake Parameters of Convolutions of Families of $$L$$-functions: Andrew Kwon
• Zeckendorf Talk: A Collection of Central Limit Type Theorems in Generalized Zeckendorf Decompositions: Ray Li
• Distance Talk: Classification of All Crescent Configurations on Four and Five Points: Rebecca Durst and Magda Hlavacek
• RMT Talk: Deviations from large eigenvalues of a special matrix ensemble: Carsten Sprunger and Peter Cohen
• $$L$$-Functions and RMT Talk: Extending agreement in the Katz-Sarnak Density Conjecture: Peter Cohen and Carsten Sprunger
• $$L$$-Functions Talk: The Langlands Program: Beyond Endoscopy: Oscar Gonzalez and Chung Hang Kwan (paper here)
• Zeckendorf Talk: On Summand Minimality of Generalized Zeckendorf Decompositions: Katherine Cordwell and Magda Hlavacek
• Number Theory Talk: Near-perfect, within-perfect, and order-a-abundant numbers: Adam Lott and Chung Hang Kwan
• SMALL 2017
• SMALL 2018 (plus CMU)
• (Maria R Ross and Casimir M Kothari), Young Mathematicians Conference, Columbus, Ohio, August 10, 2018. pdf
• (Madeleine L Farris), Young Mathematicians Conference, Columbus, Ohio, August 10, 2018. pdf
• (Mengxi Wang and Lily Shao), Young Mathematicians Conference, Columbus, Ohio, August 10, 2018. pdf
• (Noah Luntzlara and Hunter Wieman), Young Mathematicians Conference, Columbus, Ohio, August 11, 2018. pdf
• (Ben C Logsdon and Trajan A Hammonds), Young Mathematicians Conference, Columbus, Ohio, August 11, 2018. pdf
• (Hung Viet Chu and Lily Shao), Young Mathematicians Conference, Columbus, Ohio, August 11, 2018. pdf
• (Joshua Siktar), Young Mathematicians Conference, Columbus, Ohio, August 12, 2018.
• (poster; Mengxi Wang and Noah Luntzlara), Young Mathematicians Conference, Columbus, Ohio, August, 2018. pdf
• (poster; Trajan A Hammonds and Casimir M Kothari), Young Mathematicians Conference, Columbus, Ohio, August, 2018. pdf
• SMALL 2019
• Erdos Distance Problem
• Crescent Configurations In Non-Euclidean Norms Speakers: Sara Fish, Dylan King, Catherine Wahlenmayer, Conference for New England REU. 7/23/19. pdf Young Mathematicians Conference, August 9, 2019 pdf Maine-Qu\'ebec Number Theory Conference, October 5, 2019.
• Convolutions
• Extensions of Autocorrelation Inequalities with Applications to Additive Combinatorics, Young Mathematicians Conference, August 10, 2019 pdf
• L-functions
• Zeros of L-Functions near the Central Point and Optimal Test Functions Speakers: Charles Devlin, Conference for New England REU. 7/23/19. pdf Young Mathematicians Conference, August 9, 2019 pdf Maine-Qu\'ebec Number Theory Conference, October 5, 2019. pdf
• MSTD
• Distribution of Missing Sums in Correlated Sumsets Speakers: Thomas Martinez, Dylan King and Chenyang Sun, Conference for New England REU. 7/23/19. pdf Young Mathematicians Conference, August 10, 2019. pdf Poster at Harvy Mudd September 2019. pdf Shenandoah Undergraduate Mathematics and Statistics Conference, September 21, 2019 pdf
• Random Matrix Theory
• ABBA and the Random Matrix Discotheque Speakers: Keller Blackwell and Wanqiao Xu (Joint work with Neelima Borade, Charles Devlin VI, Renyuan Ma, and Leticia Mattos da Silva), Conference for New England REU. 7/23/19. pdf Joint Math Meetings, Denver, January 15, 2020. pdf
• Spies and Traitors: Random Matrix Kaleidoscopes and their Turncoat Eigenvalues Speakers: Neelima Borade and Renyuan Ma, Conference for New England REU. 7/23/19. pdf Young Mathematicians Conference, August 10, 2019. pdf Poster at the Joint Mathematics Meetings, Denver January 17, 2020 pdf
• When Bands Play in Random Matrix Theory, Young Mathematicians Conference, August 10, 2019. pdf (poster) pdf (talk)
• Zeckendorf
• Generalizing Zeckendorf's Theorem to Homogeneous Linear Recurrences Speakers: Clayton Mizgerd, Thomas Martinez and Chenyang Sun, Conference for New England REU. 7/23/19. pdf; 20th International Fibonacci Conference, July (pre-recorded), 2022.
• Fibonacci Quilt Game Speakers: Alexandra Newlon, Neelima Borade, Annie Xu, Catherine Wahlenmayer, Conference for New England REU. 7/23/19. pdf. Poster at YMC at Ohio State: Aug 2010: pdf (zipped files here Women in Mathematics in New England, September 21, 2019. Women in Mathematics in New England, September 21, 2019. pdf Undergraduate Mathematics Symposium, University of Illinois (11/2/19). pdf JMM Poster, January 2019. pdf
• Distribution of gaps in Zeckendorf Decompositions from d-dimensional Lattices: Neelima Borade and Annie Xu, Young Mathematicians Conference, August 10, 2019. pdf
• SMALL / POLYMATH 2020
• Erdos Distance Problem
• Tinkering with Lattices: A New Take on the Erdos Distance Problem (Elena Kim, Fernando Trejos Suarez, Jason Zhao), Young Mathematicians Conference, August 2020. https://youtu.be/vP9cqGTwQcU. Also Quebec-Maine Number Theory Conference, September 26, 2020.
• MSTD
• Constructions of Generalized MSTD Sets in Higher Dimensions (Elena Kim, John Lentfer, John Haviland, Phuc Lam, Fernando Trejos Suarez), Young Mathematicians Conference, August 2020. Video: https://youtu.be/znYNJYFgE48
• More Sums Than Differences Sets in Finite Non-Abelian Groups (John Haviland, John Lentfer, Elena Kim, Phuc Lam, Fernando Trejos Suarez), Young Mathematicians Conference, August 2020. https://youtu.be/_K9KoDVlv1Q
• Optimal Test Functions
• Determining optimal test functions for 2-level densities (Ela Boldyriew, Fangu Chen, Jason Zhao), Young Mathematicians Conference, August 2020. https://youtu.be/2sqlbz06J1U Quebec-Maine Number Theory Conference, September 26, 2020 (Jason Zhao):
• Prime Walks
• Prime Walks to Infinity in Z[sqrt(2)] (Daniel Sarnecki and Bencheng Li), Northeast REU Math Conference and Young Mathematicians Conference, August 2020. Video: https://youtu.be/cXnM0E-11qg; also PAJAMAS September 20, 2020: pdf (video here)
• Walking to Infinity Along Some Number Theory Sequences (Tudor Popescu, Saam Rasool, Ploy Wattanawanichkul). PAJAMAS September 19, 2020: pdf (video here)
• RMT
• Split limiting behavior of random matrices with prescribed discrete spectra (Yuxin Lin, Jiahui Yu, Fangu Chen), Young Mathematicians Conference, August 2020. https://youtu.be/7fg5ub5j7uI
• Virus Propagation
• Extending Virus Dynamics to k-level Star-Like Graphs (Akihiro Takigawa, Jack Murphy, Rodrigo Bravo), Young Mathematicians Conference, August 2020. https://youtu.be/vyuYEgQcchY
• Zeckendorf
• Juggling Coefficients in Complete Recurrent Sequences (Ela Boldyriew, Phuc Lam, John Lentfer), Young Mathematicians Conference, August 2020. https://www.youtube.com/watch?v=Au4lKJlthZ4&feature=youtu.be
• Analytic Approaches to Completeness of Generalized Fibonacci Sequences (Fernando Trejos, John Lentfer, John Haviland, Ela Boldyriew, Phuc Lam), Young Mathematicians Conference, August 2020. https://www.youtube.com/watch?v=R_PYpJpb9wU&feature=emb_logo
• Bounding the Zeroing Algorithm: A Tool for Investigating Linear Recurrence Relations (Jack Murphy), Young Mathematicians Conference, August 2020.s://youtu.be/j0npXDk4KGI
• Zeckendorf Games (Kevin Ke, Carl Ye, Vashisth Tiwari), Northeast REU Math Conference and Young Mathematicians Conference, August 2020. https://www.youtube.com/watch?v=usOXOlY0tlE
• Juggling Coeffcients in Complete Recurrent Sequences (given by Elzbieta Boldyriew, John Haviland, Phuc Lam, John Lentfer and Fernando Trejos Suarez), CANT 2021, 5/26/21. pdf (video here)
• SMALL / POLYMATH 2021
• Benford's Law
• Benfordness of the Riemann Mapping Function for the Reciprocal of the Mandelbrot Set (given by Filippo Beretta, Jesse Dimino, Weike Fang), Young Mathematicians Conference, Ohio State, August 2021.
• Erdos Distance Problems
• Thresholds for the Existence of Similar Area Configurations (given by Ben Bailey and Arian Nadjimzadah), Young Mathematicians Conference, Ohio State, August 2021.
• Extending Erdos Distinct Distance Problems to Angles (given by Henry Fleischmann Hongyi Hu), Young Mathematicians Conference, Ohio State, August 2021. pdf
• Erdos Distinct Distance Problem (given by Henry Fleischmann and Ethan Pesikoff), AMS Special Session on Distance Problems in Continuous, Discrete and Finite Field Settings, Joint Math Meetings, Jan 7, 2023.
• Fibonacci/Zeckendorf
• Complexity of the Zeckendorf Graph Game (given by Ben Baily, Faye Jackson, Ethan Pesikoff), Young Mathematicians Conference, Ohio State, August 2021.
• A Stochastic Central Limit Theorem and Applications to Integer Decompositions (given by Ben Bailey, Alicia Smith-Reina, Yingzi Yang), Young Mathematicians Conference, Ohio State, August 2021.
• The Bergman Game (given by Faye Jackson, Ethan Pesikoff, Luke Reifenberg), Young Mathematicians Conference, Ohio State, August 2021. 20th International Fibonacci Conference, July 26, 2022.
• Winning Strategy of Two-Player, Multiplayer and Multialliance Generalized Zeckendorf Game (given by ), Young Mathematicians Conference, Ohio State, August 2021.
• Sums of Reciprocals of Recurrence Relations (given by Sophia Davis, Alicia Reina Smith and Eliel Sosis), 20th International Fibonacci Conference, July 28, 2022. https://youtu.be/jfaS-3pqi_k
• General Number Theory
• Reducibility of Sets in Generalized Settings (given by Justine Dell, Henry Fleischmann, Faye Jackson), Young Mathematicians Conference, Ohio State, August 2021. pdf
• Extending Support in calculating the n-level density of low lying zeros of families of L-functions (given by Justine Dell, Simran Khunger, Alexander Shashkov, Stephen Willis), Young Mathematicians Conference, Ohio State, August 2021. Maine-Quebec Number Theory October 5, 2021. pdf Upstate Number Theory Conference, Union College, October 24, 2021. pdf
• One-level density of a family of L-functions over function fields (given by Dang Dang, Hari Iyer, Sanford Lu), Young Mathematicians Conference, Ohio State, August 2021.
• Prime walk (Walking to Infinity Along Some Number Theory Sequences) (given by Fei Peng, Tudor Popescu), Young Mathematicians Conference, Ohio State, August 2021.
• A Novel Generalization of Diophantine $m$-tuples in Finite Fields (given by Arjun Nigam, Kyle Onghai, Dishant Saikia, Lalit Sharma), Young Mathematicians Conference, Ohio State, August 2021.
• Random Matrix Theory
• Closed form densities for the limiting spectral measure of random circulant Hankel matrices (given by Teresa Dunn, Henry Fleischmann, Alexander Shashkov), Young Mathematicians Conference, Ohio State, August 2021. pdf
• SMALL / POLYMATH / OTHER 2022
• Benford's Law
• Benfordness of lower-dimensional spaces resulting from hyper-box fragmentation (given by Livia Betti and Zoe McDonald), Young Mathematicians Conference August 14, 2022. pdf
• Benfordness of measurements resulting from box fragmentation (given by Livia Betti and Zoe McDonald), The International Conference on Advances in Interdisciplinary Statistics
and Combinatorics, October 8, 2022. pdf
• A New Benford Test for Clustered Data with Applications to American Elections (given by Katie Anderson), International Conference on Advances in Interdisciplinary Statistics and Combinatorics, October 8, 2022. pdf
• CGDS
• Distinct Angles and Angle Chains in Three Dimensions (given by Ruben Ascoli and Jacob Lehmann Duke), Young Mathematicians Conference August 14, 2022. pdf (poster at williams: pdf)
• Learning Spheres and Chains in F_q^d (given by Wyatt Milgrim and Ryan Jeong), Young Mathematicians Conference August 14, 2022. pdf Symposium for Undergraduate Mathematics Research (SUMR), SUNY New Paltz, September 10, 2022. pdf
• Distinct Angles and Angle Chains in $\textbf{R}^3$} (given by Ruben Ascoli), AMS Special Session on Distance Problems in Continuous, Discrete and Finite Field Settings, Joint Math Meetings, Jan 7, 2023. pdf
• VC-Dimension and Distance Chains in F_q^d (given by Wyatt Milgrim), AMS Special Session on Distance Problems in Continuous, Discrete and Finite Field Settings, Joint Math Meetings, Jan 7, 2023. pdf
• Generalized German Tank Problem
• Generalizing the German Tank Problem: Math/Stats at War! (given by Anthony Lee), AISC Conference at UNC Greensboro, October 8, 2022. pdf
• Fibonacci/Zeckendorf
• Continuing Analysis of the Zeckendorf Game (given by SMALL 2022), 20th International Fibonacci Conference, July 28, 2022. pdf
• Generalized (c-k)-nacci Zeckendorf Game (given by Carl Ye), 20th International Fibonacci Conference, July (pre-recorded), 2022. pdf
• Generalizing Minimality Properties of Far-Difference Fibonacci Decompositions (given by Andrew Keisling and Jacob Lehmann), Young Mathematicians Conference August 13, 2022.
• On Zeckendorf Related Partitions Using the Lucas Sequence (given by David Luo), 20th International Fibonacci Conference, July 28, 2022. pdf
• Random and Maximal Lengths of Zeckendorf Games (given by Guilherme Dantas and Prakod Ngamlamai), Young Mathematicians Conference August 14, 2022. pdf
• S-Legal Index Difference Decompositions (given by SMALL 2022), 20th International Fibonacci Conference, July 28, 2022. pdf and Joint Math Meetings January 2023 pdf (and MASON VI, March 18, 2023).
• Short-Range and Random Differences in the Number of Summands of Zeckendorf Decompositions (given by SMALL 2022), 20th International Fibonacci Conference, July 25, 2022.
• Short-Range Differences of the Number of Summands of Zeckendorf Decompositions (given by Guilherme Dantas, Xuyan Liu and Jack Miller, Young Mathematicians Conference August 14, 2022. pdf
• The Far Difference Game (given by SMALL 2022), 20th International Fibonacci Conference, July 25, 2022. pdf
• The Far Difference Game (given by SMALL 2022), Young Mathematicians Conference, August 14. 2022, pdf
• Walking to Infinity on the Fibonacci Sequence (given by Nawapan Wattanawanichkul), 20th International Fibonacci Conference, July (pre-recorded), 2022. pdf
• The Accelerated Zeckendorf Game (given by Ajmain Yamin), Hunter College Math Colloquium, March 16, 2023. pdf
• L-Functions
• An alternative method for calculating Bessel integrals appearing in L-function zero statistics (given by Astrid Lilly and Santiago Velazquez), Young Mathematicians Conference, August 13, 2022. pdf
• Bounding ranks of cuspidal newforms through excised orthogonal ensembles (given by Astrid Lilly and Xuyan Liu), Young Mathematicians Conference, August 13, 2022. pdf
• An excised orthogonal model for families of cusp forms (Andrew Keisling, Xuyan Lu, Annika Mauro Zoe McDonald, Santiago Miguel Jack Miller), Quebec-Maine Number Theory Conference, October 15, 2022. pdf ( video here: https://youtu.be/d_FOSmou9mk )
• Generalized harmonic estimates for the n-level density of L-functions (given by Annika Mauro and Jack Miller), Young Mathematicians Conference August 13, 2022. pdf
• Extending the support of 1- and 2-level densities for cusp form L-functions under square-root cancellation hypotheses (Annika Mauro and Jack Miller), Quebec-Maine Number Theory Conference, October 15, 2022. pdf (video here: https://youtu.be/CDwHaLM7umE )
• MSTD
• Binomial Sets Under Z-Linear Forms (Ryan Jeong), Young Mathematicians Conference, August 16, 2023.
• Sum and Difference Sets in Semidirect Products of Groups (given by Matthew Phang and Andrew Keisling), Young Mathematicians Conference August 13, 2022. pdf
• Sum and Difference Sets in Generalized Dihedral Groups (given by Ruben Ascoli), Midwest Conference on Combinatorics and Combinatorial Computing, October 22, 2022. pdf CANT 2023 pdf
• On the Relative Sizes of Complements of Generalized Sumsets (given by Ryan Jeong), The 21st International Conference on Random Structures and Algorithms, June 16, 2023. pdf
• Polymath Jr REU
• Avoiding 3-term Geometric Progressions in Non-Commutative Settings (Polymath Ramsey Group), Polymath REU Conference, August 13, 2022. pdf
• Linear Recurrences of Order at Most Two in Nontrivial Divisors (given by Liyang Shen), CANT 2023, May 23, 2023. pdf
• On Some Observations about a prime factorization sequence called F-palindromes (given by Chris Bispels and Garam Choi), AMS Special Session on Polymath Jr: Mentoring and Learning, Joint Math Meetings, Jan 7, 2023. pdf
• A Characterization of Prime $v$-palindromes (given by Muhammet Boran and Daniel Tsai), CANT 2023, May 26, 2023. pdf
• SMALL / POLYMATH / OTHER 2023
• Benford's Law (SMALL)
• Generalized stick fragmentation and Benford's Law (Xinyu Fang and Maxwell Sun), Young Mathematicians Conference, August 16, 2023. pdf
• Diophantine Equations and Number Theory (Polymath Jr)
• On a pair of Diophantine Equations (S. U. , H. V. Chu, J. Liu, G. Luan, R. Marasinghe and S. J. Miller), Polymath Jr Conference, August 13, 2023. pdf Joint Math Meetings, AMS Special Session on Polymath Jr, January 6, 2024. pdf Combinatorial and Additive Number Theory: May 22, 2024: pFIXdf
• Power Sums of Primes in Arithmetic Progression (Muhammet Boran, Yildiz, John Byun, Zhangze Li, Steven Joel Miller and Stephanie Reyes), Joint Math Meetings, AMS Special Session on Polymath Jr, January 6, 2024. pdf
• Sum of Consecutive Terms of Pell and Related Sequences (Alexander Gong), Young Mathematicians Conference: August 2023.
• Fibonacci Numbers
• Fixed Points in the Fibonacci Sequence (given by Amanda Verga), Poster Session, Joint Mathematics Meetings, San Francisco, January 2024. pdf
• Dynamics of the Fibonacci Order of Appearance Map (given by Amanda Verga), Contributed Talk, Joint Mathematics Meetings, San Francisco, January 2024. pdf
• L-Functions (SMALL)
• Biases in Second Moments of Elliptic Curves (Zoe Batterman and Aditya Jambhale), Young Mathematicians Conference, August 16, 2023. pdf
• Biases in Second Moments of Elliptic Curves (Aditya Jambhale and Akash L. Narayanan), Maine-Quebec Number Theory Conference, September 30, 2023. pdf
• Modeling the Vanishing of L-functions at the Central Point (Zoe Batterman and Akash L. Narayanan), Young Mathematicians Conference, August 16, 2023.
• Modeling the Vanishing of L-functions at the Central Point (Zoe Batterman and Chris Yao), Maine-Quebec Number Theory Conference, October 1, 2023. pdf
• MSTD (SMALL)
• Limiting Behavior in Missing Sums of Sumsets (poster: Aditya Jambhale, Rauan Kaldybayev, Chris Yao), Williams College Science Poster Session: August 4, 2023: pdf
• Limiting Behavior in Missing Sums of Sumsets (Rauan Kaldybayev), Young Mathematicians Conference: August 16, 2023: pdf Combinatorial and Additive Number Theory: May 22, 2024: pdf
RESEARCH INTERESTS: Analytic Number Theory, Random Matrix Theory, Probability (zeros and n-level statistics for families of $$L$$-functions, especially families of elliptic curves with rank over Q(T), classical random matrix theory, random graphs, computational number theory, Benford's Law, cryptography, linear programming, sabermetrics, ...).
COURSES: Possible thesis / colloquium projects Green Chicken contest Takeaways (all classes) Advice
RECENT CONFERENCES: Undergrad Research Panel (Boston 2012) Undergrad Research Session (HC 2011 and BC 2013)
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# L03_Eigenvalues___Eigenvectors - CE507 Lecture 3 More Index...
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CE507 Lecture 3 More Index Notations ; Eigenvalues & Eigenvectors I. More Index Notation Index Notation for Differentiation Given: ) , , ( ) ~ ( 3 2 1 x x x f x f = is a scalar field (scalar function) of space. Definition: 1 1 , f f x = Thus = = = k j i ijk j i ij i i x x x f f x x f f x f f 3 2 , , , Therefore 222 2 11 22 33 123 ,,,, , ii fff f f xxx ∂∂∂ =++= + + = the Laplacian of f Vector Field: i i e x f x f x f x f x f ˆ ) ~ ( )) ~ ( ), ~ ( ), ~ ( ( ) ~ ( ~ 3 2 1 = = Gradient of a Scalar Field: i i e f f f f x f x f x f f grad f ˆ , ) , , , , , ( ) , , ( . ~ 3 2 1 3 2 1 = = = = Note: k k f e f , ˆ . ~ = Del Operator:
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123 ˆ (, ,) , kk e xx x ∂∂ ∇= = ± Divergence of a Vector Field: k k f x f x f x f f f f x x x f , ) , , , , , ( * ) , , ( ~ ~ 3 3 2 2 1 1 3 2 1 3 2 1 = + + = = Curl of a Vector Field: 3 2 1 1 2 2 1 3 3 1 1 3 2 2 3 3 2 1 3 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ~ ~ ~ e x f x f e x f x f e x f x f f f f x x x e e e f Curl f + + = = = × , ? ( ???) jj i ijk k ijk k i fe f e x εε == Divergence Theorem: Given a closed surface S enclosing a volume “V”. At each infinitesimal area “ds” on the surface, let “ n ˆ ” be its outward normal, then for a given vector field,” f ~ ” : ˆ VS f dV f n dS ∇⋅ = ∫∫∫ ∫∫ ±± ± w x f ~ is assumed to be continuous and differentiable. Physically, f ~ can be considered as a current density (volume of flow / time) of a hypothetical fluid. L.H.S is then the net Rate at which the “fluid” passes out of the volume V.The R.H.S is the rate at which
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## This note was uploaded on 02/28/2008 for the course CE 507 taught by Professor Lee during the Fall '07 term at USC.
### Page1 / 7
L03_Eigenvalues___Eigenvectors - CE507 Lecture 3 More Index...
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Ask a homework question - tutors are online | 792 | 2,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-26 | longest | en | 0.717377 |
https://thegrandparadise.com/essay-tips/what-is-bell-curve-mentality/ | 1,708,503,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00035.warc.gz | 585,723,253 | 12,138 | # What is bell curve mentality?
## What is bell curve mentality?
What do you mean by bell curve mentality? It is a social theory designed to sort people out and identify a limited number of winners. It suggests there’s no need to even try to get some kids to learn because they were’nt born with the right stuff, that some students have the ability to achieve while some do not.Aban 13, 1393 AP
## How do you curve a grade with a square root?
To curve you take the square root of the student’s grade and multiply it by 10. Looking at the example below, let’s say a student scored a 75 on their test. We take the square root of 75, which is about 8.666, and multiply it by 10 giving them an 86.6% curved grade.Azar 26, 1397 AP
## How do you do relative grading?
The other kind of grading system is called relative grading. In this system, grades are given based on the student’s score compared to the others in the class. This system is used in some universities and colleges and even in some advanced high school classes.Aban 12, 1394 AP
A – is the highest grade you can receive on an assignment, and it’s between 90% and 100% B – is still a pretty good grade! D – this is still a passing grade, and it’s between 59% and 69% F – this is a failing grade.Esfand 6, 1399 AP
## How do teachers compute grades?
Get the total score for each component. Divide the total raw score by the highest possible score then multiply the quotient by 100%. Convert Percentage Scores to Weighted Scores.
## When should I curve an exam?
If the class does significantly lower than I think they should have, I will consider curving the exam. Also, courses have certain historical distributions. For example, in an entry-level course I may want an average (mean) of 80-82% with several A’s. In classes like that, failing grades are not unusual.Dey 2, 1387 AP
## What grade is Square Root taught?
Ch 7: 5th Grade Math: Exponents & Square Roots.
4 60%
5 50%
6 40%
7 30%
## Is a 60 a passing grade?
In primary and secondary schools, a D is usually the lowest passing grade. However, there are some schools that consider a C the lowest passing grade, so the general standard is that anything below a 60 or 70 is failing, depending on the grading scale.
## How do you simplify the square root of 90?
1. Thus we can now write √90=3√5⋅2=3√10.
2. √20=2√5.
3. We proceed the same way and write √56=2√2⋅7=2√14.
## What is AP curve?
The p-curve can be used to detect p-hacking. The authors define this curve as “the distribution of statistically significant p values for a set of independent findings. Its shape is a diagnostic of the evidential value of that set of findings.”
## What is a grading method?
The grading method defines how the grade for a single attempt of the activity is determined. There are 4 grading methods: Learning objects – The number of completed/passed learning objects. Highest grade – The highest score obtained in all passed learning objects. Average grade – The mean of all the scores.
## What is bell curve used for?
The term “bell curve” is used to describe a graphical depiction of a normal probability distribution, whose underlying standard deviations from the mean create the curved bell shape. A standard deviation is a measurement used to quantify the variability of data dispersion, in a set of given values around the mean.Bahman 27, 1399 AP
## How are Bell Curve grades calculated?
A simple method for curving grades is to add the same amount of points to each student’s score. A common method: Find the difference between the highest grade in the class and the highest possible score and add that many points. If the highest percentage grade in the class was 88%, the difference is 12%.
## How many methods of grading are used?
Methods of Grading There are three basic methods of pattern grading. There is not a superior method; they are all equally capable of producing a correct garment grade. These include: Cut and spread: This is the easiest method, which acts as the basis of the other two methods.
## What are the types of grading?
Types of Grading Systems
• Percentage Grading – From 0 to 100 Percent.
• Letter grading and variations – From A Grade to F Grade.
• Norm-referenced grading – Comparing students to each other usually letter grades.
• Mastery grading – Grading students as “masters” or “passers” when their attainment reaches a prespecified level. | 1,031 | 4,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-10 | latest | en | 0.953269 |
http://auscblacks.com/multiplication-table-worksheets/multiplication-tables-to-49-one-per-page-a/ | 1,534,603,659,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213689.68/warc/CC-MAIN-20180818134554-20180818154554-00055.warc.gz | 40,338,059 | 9,374 | ## Auscblacks
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By Tina C. Kral on July 03 2018 09:55:40
Relate the numbers and multiplication tables to everyday life. Try to find ways to incorporate the math tables into daily life. This will help you truly understand the numbers instead of simply memorizing the material. If you know that a special holiday is only 8 weeks away, you can use your multiplication facts to figure out how many days away it is. There are 7 days in a week, so 7 times 8 is 56. Your holiday is only 56 days away!
I have seen a lot of kids quickly pass off their 2’s, 5’s, and 10’s. These tables have an obvious pattern and are much easier to learn. Then there is a serious slow down as kids hit the 3’s, 4’s, and 6’s. By the time they get to the 7’s, 8’s, and 9’s they’ve decided that multiplication is way too hard, and math isn’t their thing.
Use flashcards. Make multiplication cards for each number set. Although this may seem tedious, the process of making the cards will actually help you to learn them. Once you’ve made them, spend some time each day studying until you know them all. Focus on one number set at a time. When you go through the cards, put the ones you get wrong back into the pile so you see them multiple times.
When it comes to multiplication, finger tricks are very useful tools, and they are fun! This trick is for multiplication tables of 6, 7, 8, and 9. When you need to find an answer quickly, finger tricks can save the day. | 363 | 1,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-34 | latest | en | 0.95077 |
http://mathhelpforum.com/calculus/75007-confused-u-subs-problem.html | 1,506,205,269,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689779.81/warc/CC-MAIN-20170923213057-20170923233057-00196.warc.gz | 212,208,207 | 10,699 | # Thread: confused with u-subs problem
1. ## confused with u-subs problem
integral of 1/(9 + 4x^2)
The book says u should be 2/3x. Can someone enlighten me on why this is? I am dumbfounded
2. Originally Posted by TYTY
integral of 1/(9 + 4x^2)
The book says u should be 2/3x. Can someone enlighten me on why this is? I am dumbfounded
Have you tried it ....?
Note that $\frac{1}{9 + 4x^2} = \frac{1}{9} \cdot \frac{1}{1 + \left(\frac{2x}{3}\right)^2}$.
3. Originally Posted by mr fantastic
Have you tried it ....?
Note that $\frac{1}{9 + 4x^2} = \frac{1}{9} \cdot \frac{1}{1 + \left(\frac{2x}{3}\right)^2}$.
please be patient with me here as I still don't understand this. (2x/3)^2 x 9 would have to equal 4x^2 again here right? but wouldn't it be (18x/3)^2 = 6x^2?
edit: I guess because of the square, you simplify using square root or something. The calculator agrees with your result but somehow it still goes over my head
4. Originally Posted by TYTY
please be patient with me here as I still don't understand this. (2x/3)^2 x 9 would have to equal 4x^2 again here right? but wouldn't it be (18x/3)^2 = 6x^2?
$9 \cdot \left( \frac{2x}{3} \right)^2 = 9 \cdot \frac{2x}{3} \cdot \frac{2x}{3} = 9 \cdot \frac{4x^2}{9} = 4x^2$.
If you're studying calculus it's expected you can do this. And do it easily.
I suggest you extensively revise basic algebra, index laws etc. if you hope to successfully tackle the sort of calculus questions you're going to get given n the coming days and weeks.
5. Originally Posted by mr fantastic
$9 \cdot \left( \frac{2x}{3} \right)^2 = 9 \cdot \frac{2x}{3} \cdot \frac{2x}{3} = 9 \cdot \frac{4x^2}{9} = 4x^2$.
If you're studying calculus it's expected you can do this. And do it easily.
I suggest you extensively revise basic algebra, index laws etc. if you hope to successfully tackle the sort of calculus questions you're going to get given n the coming days and weeks.
I imagine you're right about needing to review. I haven't taken any classes in 5 years now I find myself in this nightmare situation. Thank you for your explanation though - it did clear it up for me and now in retrospect it does seem kind of obvious. | 677 | 2,168 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-39 | longest | en | 0.915312 |
http://mathhelpforum.com/algebra/210896-solve-proportions-using-cross-products.html | 1,527,091,111,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00240.warc.gz | 185,000,172 | 10,008 | # Thread: Solve Proportions using cross products
1. ## Solve Proportions using cross products
My sister thinks these problems are unsolvable for x. I can't find any examples in my book. I know how to solve basic proportion problems, but these two are hard.
1. x : x-3 = x+4 : x
2. x+1 : 6 = x-1 : x
Thanks!
2. ## Re: Solve Proportions using cross products
Hey BrokenRitual.
Try getting rid of the denominators and collect the terms.
For the first one we have x^2 + 4x =x^2 - 3x or 7x = 0 => x = 0 but this is a contradiction since x is on the denominator and can't be zero. So no solution exists.
You try the same kind of technique for the second one.
3. ## Re: Solve Proportions using cross products
Hello, BrokenRitual!
Why do you say these are "hard"?
$\displaystyle [1]\;\;x : x-3 \:=\: x+4 : x$
We have: .$\displaystyle \frac{x}{x-3} \:=\:\frac{x+4}{x}$
Cross product: .$\displaystyle (x)(x) \:=\:(x-3)(x+4)$
n . . . . . . . . . . . . .$\displaystyle x^2 \:=\:x^2 + x - 12$
. . . . . . . . . . . . . . . $\displaystyle 0 \:=\:x - 12$
. . . . . . . . . . . . . . . $\displaystyle x \:=\:12$
$\displaystyle [2]\;\; x+1 : 6 \:=\: x-1 : x$
We have: .$\displaystyle \frac{x+1}{6} \:=\:\frac{x-1}{x}$
Cross product: .$\displaystyle (x+1)(x) \:=\:6(x-1)$
n . . . . . . . . . . . . . $\displaystyle x^2 + x \:=\:6x-6$
. . . . . . . . . . .$\displaystyle x^2 - 5x + 6 \:=\:0$
. . . . . . . . .$\displaystyle (x-2)(x-3) \:=\:0$
. . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:2,\:3$ | 563 | 1,513 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-22 | latest | en | 0.571693 |
http://jabsto.com/Tutorial/topic-64/Microsoft-Excel-2010-Formulas-322.html | 1,552,933,789,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201521.60/warc/CC-MAIN-20190318172016-20190318194016-00229.warc.gz | 104,900,659 | 5,129 | Microsoft Office Tutorials and References
In Depth Information
The Basic Excel Financial Functions
Financial function arguments
The five basic Excel financial functions have many common arguments. The arguments and their
meanings are listed here:
rate: The interest rate, expressed as a percentage, that is paid on a loan or used to
discount future cash flows.
The period that the interest rate covers must be the same period used for nper and pmt.
nper: The number of periods. This could be the number of payments on a loan or the
number of periods that money is kept in a savings account.
The number of periods must be expressed in the same terms as rate and pmt. A 30-year
mortgage with monthly payments, for instance, would have an nper of 360.
pmt: The amount of each payment. For these financial functions, the payments must be
the same amount and made at regular intervals. The payment amount is normally made up
of both principal and interest.
fv: The future value. This is the last cash flow that settles the transaction. In many cases,
the payments settle the transaction (for example, pay off the loan), so there is no future
value.
pv: The present value. This is the first cash flow that starts the transaction, such as
borrowing money on a loan or putting money into a savings account.
If the transaction is made up of just payments, there may not be a present value.
type: Whether the payments are made in arrears (0 or default) or in advance (1).
guess: An approximation of the result. When computing an interest rate, Excel must
perform many iterations to get the answer. You can help Excel by specifying a guess
argument that you expect to be close to the actual result.
The example in this section computes the present value of a series of future receipts, sometimes
called an annuity. You get one payment of \$1,200 each year for ten years. The value of those
payments right now is \$6,780.27.
=PV(12%,10,1200,0,0)
In other words, if the payer offered you more than \$6,800 right now (so he wouldn’t have to
make the payments to you in the future), you would take it. If he offered you less, you would
pass and wait for the regular payments.
The file basic financial formulas.xlsx on the companion CD-ROM contains all
the examples in this section.
Search JabSto ::
Custom Search | 520 | 2,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-13 | latest | en | 0.938263 |
https://civilservicereview.com/tag/pcsr-review-guide/ | 1,620,531,170,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988955.89/warc/CC-MAIN-20210509032519-20210509062519-00404.warc.gz | 186,303,812 | 11,764 | ## 2018 PCSR Civil Service Review Guide 15
2018 PCSR Civil Service Review Guide 15
Week 15: January 8 – 14, 2018
PART I: MATHEMATICS
A. MOTION PROBLEM
Youtube Tutorial Videos (Taglish)
Lesson 1: Introduction to Motion Problems
Lesson 2: How to Solve Motion Problems Part 1
Lesson 3: How to Solve Motion Problems Part 2
Lesson 4: How to Solve Motion Problems Part 3
Lesson 5: How to Solve Motion Problems Part 4
Lesson 6: How to Solve Motion Problems Part 5
B. Articles
Article 1: Introduction to Motion Problems
Article 2: How to Solve Motion Problems Part 1
Article 3: How to Solve Motion Problems Part 2
Article 4: How to Solve Motion Problems Part 3
Article 5: How to Solve Motion Problems Part 4
PART II: ENGLISH
Part III: CLERICAL (Sub-Professional)
A. Spelling
PART IV: OTHERS
4 Things to do 3 Months Before the Exam
MORE REVIEWERS
1. Math English Tutorials and more at the PCSR Civil Service Review website:
http://civilservicereview.com
2. More than 700 Taglish Math Tutorials at the Sipnayan Youtube channel:
3. Sipnayan Website
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## 2018 PCSR Civil Service Review Guide 1
Week 1: September 24 – October 1, 2017
This week, we are going to learn about Least Common Multiple (LCM) and Greatest Common Divisor (GCD). LCM is needed in adding and subtracting dissimilar fractions, while GCD is needed in reducing fractions to lowest terms. In addition to LCM and GCD, we will also learn about converting improper fractions to mixed number and vice versa.
PART I: MATHEMATICS
A. GCD
Youtube Tutorial Videos (Taglish)
Lesson 1: Understanding Common Divisors
Lesson 2: Common Divisors and Greatest Common Divisor
Lesson 3: Exercises on Getting the GCD
Lesson 4: Getting the GCD by Prime Factorization
Lesson 5: GCD and Reducing Fractions to Lowest Terms
Tutorial Articles
B. LCM
Youtube Tutorial Videos (Taglish)
Lesson 1: Least Common Multiple Part 1
Lesson 2: Least Common Multiple Part 2
Lesson 3: Least Common Multiple Part 3
Tutorial Article
How to Get the Least Common Multiple of Numbers
C. Lowest Terms, Improper Fractions and Mixed Form
Youtube Tutorial Videos (Taglish)
Tutorial Articles
D. More about LCM and GCD
Difference between LCM and GCD
PART II: ENGLISH
A. Vocabulary
Civil Service Exam Vocabulary Review Part 1
B. Grammar
Lesson 1: Verb and Tenses: An Introduction
Lesson 2: The Subject-Verb Agreement Rules Part 1
Lesson 3: The Subject-Verb Agreement Rules Part 2
Lesson 4: Subject-Verb Agreement Exercise
C. Spelling
Civil Service Review Spelling Quiz 1 Set 2
Part III: CLERICAL (Subprofessional)
PART III: TIPS AND TRICKS
7 Tips on Answering Reading Comprehension Questions
MORE REVIEWERS
1. Math English Tutorials and more at the PCSR Civil Service Review website:
http://civilservicereview.com
2. More than 700 Taglish Math Tutorials at the Sipnayan Youtube channel:
3. Sipnayan Website
http://sipnayan.com
Like our Facebook Pages and be updated:
## August 2017 PCSR CIVIL SERVICE REVIEW GUIDE 16
After learning work problems, we now learn how to answers sequences and series.
PART I: MATH
B. Articles
Part II: ENGLISH
A. Spelling
Spelling Quiz 16
B. Vocabulary
Civil Service Exam Vocabulary Review 16
## August 2017 PCSR CIVIL SERVICE REVIEW GUIDE 15
After learning work problems, we learn about rate, base, and percentage. If you want to further your knowledge about them, you may also want to study about their applications which are discount and interest.
PART I: MATH
A. RATE BASE AND PERCENTAGE
Part 1 Introducing Rate, Base, and Percentage
Part 2 Calculating for Percentage
Part 3 Calculating for the Rate
Part 4 Calculating for the Base
Part 5 Calculating Rate, Base, and Percentage
Part 6 Calculating Discounts and Interests
B. DISCOUNT
Part 1 Discount Sample Problem 1
Part 2 Discount Sample Problem 2
Part 3 Discount Sample Problem 3
Part 4 Discount Sample Problem 4
Part 5 Discount Sample Problem 5
Part 6 Discount Problems and Rate, Base, and Percentage
Part II: ENGLISH
A. Vocabulary
Civil Service Vocabulary Quiz
B. Spelling
Spelling Quiz 15
Enjoy!
## August 2017 PCSR Civil Service Review Guide 12
After learning how to solve number problems, let’s learn how to solve age problems. Watch the videos below and read the articles. Exercises will be posted on Wednesday.
PART I: MATH
B. Articles
Part II: ENGLISH
B. Vocabulary
Civil Service Exam Vocabulary 12
C. Spelling
Civil Service Exam Spelling Quiz 12
Enjoy!
## August 2017 PCSR Civil Service Review Guide 11
Here’s the Review Guide 11 for the Civil Service Exam.
PART I: MATH
A. Videos
— Series 1 —
How to Solve Number Problems Mentally 1
How to Solve Number Problems Mentally 2
How to Solve Number Problems Mentally 3
B. Articles
Part II ENGLISH
A. Vocabulary
Civil Service Exam Vocabulary Review 11
B. Grammar and Correct Usage
How to Use the Articles A, An, and The
When to use the Pronouns I and me
C. Spelling
Civil Service Review Spelling Quiz 11
Enjoy! | 1,313 | 4,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-21 | latest | en | 0.779031 |
https://math.stackexchange.com/questions/915340/who-has-a-winning-strategy-in-the-hamilton-circle-game | 1,701,225,363,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00815.warc.gz | 443,801,900 | 35,910 | # Who has a winning strategy in the hamilton-circle-game?
The game starts with a graph with $n$ vertices and no edges. The players alternately add edges until the graph contains a hamilton-circle. The player who made the last move loses.
Who has a winning strategy in this game depending on the number of vertices ? For $n = 3$, the game is boring because player $2$ always wins, no matter how the play is. For $n=4$, player $2$ has a winning strategy.
Computation over all labeled graphs shows that for $n = 5$ and $n = 6$ the first player has a winning strategy, for $n = 7$ and $n = 8$ hasn't. My actual conjecture is the following: first player has a winnning strategy if and only if $n = 4k + 1$ or $n = 4k + 2$. The reason for my conjecture is that complete graph on $n - 1$ vertices has even number of edges in these cases. Also I have checked supposition that avoiding dummy moves (i. e., moves that end game instantly and undesirably) is good enough strategy, but no, it isn't even for $n = 5$.
• The next subresult for $n = 9$ is that the first player has a winning strategy and it confirms conjecture. Feb 1, 2017 at 16:08 | 308 | 1,136 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-50 | latest | en | 0.960684 |
http://www-cgi.cs.cmu.edu/afs/cs.cmu.edu/user/ajw/www/doc/gcl/html/Geometry.cc-source.html | 1,558,480,357,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256586.62/warc/CC-MAIN-20190521222812-20190522004812-00157.warc.gz | 224,903,495 | 2,070 | Main Page Class Hierarchy Compound List File List Compound Members File Members
# Geometry.cc
Go to the documentation of this file.
```00001 /*
00002 File: Geometry.cc
00003
00004 Function: Implements Geometry.h
00005
00006 Author(s): Andrew Willmott
00007
00008 Copyright: (c) 1995-2000, Andrew Willmott
00009
00010 Notes:
00011
00012 */
00013
00014 #include "gcl/Geometry.h"
00015 #include "gcl/VecUtil.h"
00016
00017 #include <stdarg.h>
00018
00019 IndexList Indexes(Int first, ...)
00020 {
00021 IndexList result;
00022 va_list ap;
00023
00024 va_start(ap, first);
00025
00026 while (first != IDX_END)
00027 {
00028 result.Append(first);
00029 first = va_arg(ap, int);
00030 }
00031
00032 va_end(ap);
00033
00034 return(result);
00035 }
00036
00037 Transform Align(const Vector &xAxis, const Vector &yAxis, const Vector &zAxis)
00038 {
00039 Transform align;
00040
00041 // Create a matrix which will align scene with the xAxis, yAxis, zAxis
00042 align[0] = MVector4(xAxis[0], yAxis[0], zAxis[0], 0.0);
00043 align[1] = MVector4(xAxis[1], yAxis[1], zAxis[1], 0.0);
00044 align[2] = MVector4(xAxis[2], yAxis[2], zAxis[2], 0.0);
00045 align[3] = vl_w;
00046 #ifdef VL_ROW_ORIENT
00047 align = trans(align);
00048 #endif
00049
00050 return(align);
00051 }
00052
00053 Transform AlignToDir(const Vector &dir)
00054 {
00055 Vector x, y, z;
00056
00057 if (sqrlen(dir) == 0.0)
00058 return(vl_I);
00059 y = norm(dir);
00060 x = norm(FindOrthoVector(y));
00061 z = cross(x, y);
00062
00063 return(Align(x, y, z));
00064 }
00065
00066
00067 #ifdef CL_TMPL_INST
00068 template class Array<Vector>;
00069 #endif
```
Generated at Sat Aug 5 00:16:59 2000 for Graphics Class Library by 1.1.0 written by Dimitri van Heesch, © 1997-2000 | 619 | 1,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-22 | latest | en | 0.204485 |
https://essaysglobe.com/project-2-2/ | 1,656,219,158,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00609.warc.gz | 294,129,159 | 14,472 | PROJECT 2
Project 2 Instructions
Please also pay close attention to any additional specifications provided by your professor. Professors often will clarify their expectations regarding the format and presentation of your submission.
Ever wonder how your electricity bill is computed? The following table gives the watts of power that are typically required by several major (and not so major) appliances. (Source: http://www.wholesalesolar.com/solar-information/how-to-save-energy/power-table, with some numbers rounded off for computational convenience.)
Table 1:
Appliance Central Air Conditioner Water Heater Clothes Dryer Oven Dishwasher Fridge Ceiling Fan LCD TV Watts 9000 5000 4500 4000 2400 2200 220 250
Assume that the number of hours that the aforementioned appliances run during an average day is given in the table below:
Table 2:
Appliance Central Air Conditioner Water Heater Clothes Dryer Oven Dishwasher Fridge Ceiling Fan LCD TV Hours 5 3 3.5 2 2.5 8 13 6
Add columns to the Project 2 Data spreadsheet that show the following for each appliance:
1. The number of kilowatts of power required by each of these appliances (1 kilowatt = 1000 watts).
2. If the appliance runs the number of hours specified in Table 2, compute the kilowatt-hours (kWh) of energy that each appliance consumes during an average day. Note: kWh are found by multiplying the power of an appliance by the number of hours that it is in use.
3. The number of kilowatt-hours used by each appliance in the entire YEAR.
4. Assuming that electricity costs \$0.13 per kWh, find the electricity cost of running each appliance for the entire year.
5. According to the US Energy Information Administration (see “How much coal, natural gas or petroleum is used to generate a kilowatt-hour of electricity” on https://www.eia.gov/tools/faqs/), it requires 0.00052 tons of coal to generate 1 kWh of electricity. Use this conversion to find how much coal would be required to run each of these appliances for a year.
6. The same source also states that 0.00173 barrels of oil will generate 1 kWh of electricity. Like question 5, find how much oil each appliance would consume on an annual basis.
7. Add a “Total” row to the spreadsheet that shows how much electricity is consumed by all 8 appliances combined in the course of a day and of a year. Also, find the cost, the amount of coal that would be required, and the barrels of oil that would be required in order to run all 8 appliances for a year.
Submit Project 2 by 11:59 p.m. (ET) on Monday of Module/Week 7. | 589 | 2,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-27 | longest | en | 0.880934 |
http://www.cfd-online.com/Forums/cfx/95304-numerical-oscillations-cfx-water-boiling-problem-print.html | 1,440,998,116,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065534.46/warc/CC-MAIN-20150827025425-00126-ip-10-171-96-226.ec2.internal.warc.gz | 349,796,049 | 3,005 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- CFX (http://www.cfd-online.com/Forums/cfx/)
- - Numerical oscillations in CFX with water boiling problem. (http://www.cfd-online.com/Forums/cfx/95304-numerical-oscillations-cfx-water-boiling-problem.html)
michujo December 13, 2011 04:59
Numerical oscillations in CFX with water boiling problem.
3 Attachment(s)
Hi everyone, this is my first post here so nice to meet you.
I'm a CFX user and I'm trying to calculate a water boiling problem, but I'm getting weird results. The problem looks like this:
- Water enters an annular tube (space between two coaxial cylinders) where it is heated up through the inner cylinder wall until it starts to evaporate (see attached picture).
- The cylinder is 340 mm long, diameters of inner and outer cylinders are 12.7 mm and 25.4 mm. Water enters 31 K below the saturation temperature at a velocity of 0.4 m/s. Inlet Reynolds number is Re_inlet = 5071.
- A heat flux is applied through the inner cylinder wall.
- I believe the computational mesh is pretty good, what do you think? (see attached picture)
- Computational parameters: I'm using upwind scheme for the advection terms. I did not set the Wall Boling model, phase change is expected to occur when the liquid temperature reaches the saturation temperature.
THE PROBLEM: when water starts to evaporate I get spurious oscillations! I tried different things to try and suppress them but they're still there! (attached picture shows water vapour volume fraction on several cut planes, saturated at volume_fraction_vapour = 0.1)
THE QUESTION: Is there anyone among you who got this same problem in the past? Can you give me any hint as to what might be the reason of this problem?
I thank you guys for your help, this is driving me crazy...
Cheers.
ghorrocks December 13, 2011 17:15
Why do you say this is spurious? Boiling is a very lumpy process and often ends up in slug flows and things like that.
Upwinding for the advection terms will cause excessive damping. Use a second order scheme.
Are you doing this steady state or transient?
michujo December 15, 2011 04:42
Dear Glenn, I'm running it as a steady state-calculation.
I also tried the high order scheme available in CFX with similar results ...
I was trying to perform this simulation in order to validate this boiling model, so I took this case from the literature. In the experiment (and their simulations with CFX!) they get smooth solutions where vapour volume fraction grows along the tube so I was hoping to get similar results.
Any other hint you might think of?
Thanks!
Cheers.
ghorrocks December 15, 2011 17:18
This FAQ covers some issues of relevance here:
http://www.cfd-online.com/Wiki/Ansys...gence_criteria
michujo December 16, 2011 10:00
Dear Glenn,
thanks for your reply. I will have a look at the document.
Cheers.
All times are GMT -4. The time now is 01:15. | 718 | 2,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2015-35 | longest | en | 0.915837 |
https://artofproblemsolving.com/wiki/index.php/Bolzano-Weierstrass_theorem | 1,702,031,398,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00358.warc.gz | 133,220,178 | 10,348 | # Bolzano-Weierstrass theorem
The Bolzano-Weierstrass theorem is a theorem which states that every infinite set in a closed bounded region must have an infinite convergent subsequence. (In fact, we may even ensure that this infinite convergent subsequence converges arbitrarily quickly, but this is beside the point.)
## Proof
Suppose that the set $X$ lies in the closed bounded region $K_0$. We can divide it into two equally-sized closed bounded regions, $K'_0$ and $K''_0$, which can be bounded by smaller neighborhoods. Then by the infinitary pigeonhole principle, at least one of these regions must contain an infinite number of elements of $X$; choose one of these and designate it $K_1$. Repeat this construction to get an even smaller closed bounded region containing infinitely many elements of $X$, called $K_2$. By iterating this construction countably infinitely many times, we obtain an infinite sequence of closed bounded regions $K_n$, each of which is half as small as its predecessor and which also and each of which also contains infinitely many elements of $X$.
Finally, we choose elements $x_n \in K_n \cap X \forall n$. By construction, the sequence $x_n$ approaches $\cap_n K_n$, which is simply a point, and we are done. | 284 | 1,247 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-50 | latest | en | 0.966626 |
https://ham.stackexchange.com/questions/16691/how-to-connect-ic-271-dc-power-connector-to-12-v-battery | 1,716,468,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00153.warc.gz | 244,145,921 | 37,274 | # How to connect IC-271 DC power connector to 12 V battery
What I am trying to do
Radio is an IC-271, ~ 20 years old. I want to run it with a 12 V lead acid battery. I need to connect the battery to the radio. The radio's DC power connector is a 6-pin molex(?). The radio is designed to be used with power supply IC-PS15. On page 4 of the power supply manual, it shows what the power connector pins are.
Pins 2, 5 are ground. <- I understand this.
Pin 6 is for 110 or 240 V?? <- I don't understand this.
Pins 1, 3, 4 <- I don't know what these are by looking at the diagram.
Question
How can I connect a 12 V battery to the power connector? What should I do with each pin?
My level
I am getting started with ham radio and have basic electronics knowledge.
• Hello and welcome to ham.stackexchange.com! May 18, 2020 at 23:50
I assume you're looking at this schematic:
Pins 2 and 5 are ground, as indicated by the chassis-ground symbol they're wired to. Since 1 and 4 also come from the power supply's low-voltage side, they must be the +12V output.
Now, Pin 6 is going to the AC power cord, which is odd. But, if we follow the wire on pin 3 that is drawn parallel to it, we see that it goes to the transformer. So, there's a simple explanation for this: this is a circuit that allows the power supply to be turned on or off by a switch on the radio. If pins 3 and 6 are not connected, the power supply gets no power, and if they are it does.
So, there is no need to worry about those two pins if you are supplying battery power — you could wire a switch across them to be able to use the switch that is somewhere on the radio, or you can also ignore them entirely.
(Incidentally, the "110 or 240 V" description is referring to selecting the right current rating for the fuse given the supply voltage; it's not intended as a description of the function of the wire, but it did give you a hint!)
That was how to understand the pins, but if I identify this right, there's a much simpler solution to your problem: this looks like the de-facto-standard 6-pin power connector used on many HF radios. You can get a premade adapter from amateur radio accessory suppliers such as Powerwerx HF6-PP or MFJ MFJ-5535M. That simplifies things down to the two pins of an Anderson PowerPole connector, a very common connector to use for attaching batteries to amateur radio gear.
Of course, this requires a matching Anderson PowerPole plug on the battery, but you can also get such premade cables, or make your own — if you're at all interested in getting into portable or backup power, solar power systems, or just having several pieces of radio gear, it's quite useful to be able to make your own power cables to adapt various things to a common standard and to interoperate with other hams' gear and other equipment built for PowerPole connectors (e.g. including power meters and distribution boards), so getting a crimping tool for installing PowerPole connectors is a much more worthwhile investment than getting one for assembling one of those 6-pin plugs.
• With your explanation, the schematic makes more sense now. And the power connector you linked to is just the one on my radio. I couldn't find it at my neighborhood parts store, but it looks easy to get online. Thank you! May 17, 2020 at 4:01
From the IC-271A/E manual, page 3:
Pins 1 and 4 are both positive (12V or 13.8V or whatever), 2 and 5 are both ground, and 3 and 6 are left unconnected. This is the common six-pin power connector found on many HF rigs (but more recently replaced by a four-pin connector in Icom's case). | 883 | 3,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.959654 |
https://in-theory.blogspot.com/2007_01_21_archive.html | 1,718,248,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00888.warc.gz | 306,519,943 | 12,063 | ## Thursday, January 25, 2007
### The Multi-dimensional Polynomial van der Waerden Theorem
[This is the second part of Bill Gasarch's guest post on "polynomial" versions of van der Waerden's theorem. -- Luca]
This is a the second of two Guest Postings on the Polynomial VDW Theorem. This one is on the Multi-dimensional Poly VDW Theorem.
Our starting point is VDW's theorem, which is the same starting point Luca had in his postings on Szemeredi's theorem (here, here, here, here, and here) and I had in my first posting.
VDW:
For all $c,k\in N$, for any $c$-coloring $COL:Z \rightarrow [c]$ there exists $a,d\in Z$, $d\ne 0$, such that
$COL(a) = COL(a+d) = COL(a+2d) = \cdots = COL(a+(k-1)d).$
What would a multidimensional version of VDW look like? What would a multidimensional version of Poly VDW look like? Before stating that, we give corollaries to both to provide the flavor.
Corollary of Two-dimensional VDW:
For all $c$, for any $c$-coloring $COL: Z\times Z\rightarrow [c]$, there exists a square with all corners the same color. (Formally there exists $a,b,d$,$d\ne 0$, such that
$COL(a,b)=COL(a+d,b)=COL(a,b+d)=COL(a+d,b+d).$
Think of this as
$COL((a,b)+d(0,0))=COL((a,b)+d(1,0))=$
$COL((a,b)+d(0,1))=COL((a,b)+d(1,1)).$)
Corollary of Two-dimensional Poly VDW:
For all $c$, for any $c$-coloring $COL: Z\times Z\rightarrow [c]$, there exists a rectangle with all corners the same color where the side is the square of its length. (Formally there exists $a,b,d$, $d\ne 0$, such that
$COL(a,b)=COL(a+d,b)=COL(a,b+d^2)=COL(a+d,b+d^2).$
Think of this as
$COL((a,b)+(0,0))=COL((a,b)+d(1,0))=$
$COL((a,b)+d^2(0,1))=COL((a,b)+d(1,0)+d^2(0,1)).$)
In the full Multidimensional theorems the set $\{ (0,0), (0,1), (1,0), (1,1) \}$ will be replaced by an arbitrary finite set.
Multidimensional VDW:
Let $e\in N$. Let $F=\{\overline f_1, \overline f_2, \ldots, \overline f_k\}$ be a finite set of points in $Z^e$ (e.g., $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$ for the square case in the corollary to Multidim VDW). For all $c$, for any $c$-coloring $COL: Z^e\rightarrow [c]$, there exists $\overline a\in Z^e$, $d\in N$, $d\ne0$, such that
$COL(\overline a + d\overline f_1)= COL(\overline a + d\overline f_2) = \cdots =COL(\overline a + d\overline f_k).$
This was proven by Furstenberg and Weiss (Topological dynamics and combinatorial number theory, Journal d'Analyse Mathematique, Vol. 34, 61--85, 1978); however, it was later observed that it follows from the (ordinary) Hales-Jewitt Theorem.
Furstenberg and Katznelson also proved a density version of the Multidimensional VDW Theorem (An ergodic Szemeredi's theorem for commuting transformations, Journal d'Analyse Mathematique, Vol. 34, 275--291, 1978). Roughly speaking, if $A\subseteq Z^d$ is dense enough then there exists $\overline a\in Z^e$, $d\in N$, $d\ne 0$ such that $\{ \overline a+d\overline f \mid f\in F \}\subseteq A$.
Consider the functions
$\overline a + d\overline f_1, \overline a + d\overline f_2, \cdots, \overline a + d\overline f_k.$
One may consider replacing these functions with a more complicated function of $d$ and $\overline f_1, \ldots, \overline f_k$. This leads to the following theorem.
Multidimensional Poly VDW:
Let $e,r\in n$. Let $p: Z^r \rightarrow Z^e$. Let $F=\{\overline f_1, \overline f_2, \ldots, \overline f_s\}$ be a finite set of points in $Z^r$. Let $f_i = (f_{i1},\ldots,f_{ir}).$ For all $c$, for any $c$-coloring $COL: Z^e\rightarrow [c]$, there exists $\overline a\in Z^e$, $d\in Z^r$, $d=(d_1,\ldots,d_r)$, $d\ne \overline 0$, such that
$COL(\overline a + P(d_1f_{11},d_2f_{12},\ldots,d_rf_{1r}))= COL(\overline a + P(d_1f_{21},d_2f_{22},\ldots,d_rf_{2r}))$
$=COL(\overline a + P(d_1f_{31},d_2f_{r2},\ldots,d_rf_{3r}))$
$\vdots$
$COL(\overline a + P(d_1f_{s1},d_2f_{r2},\ldots,d_rf_{sr}))$
This was first proven by Bergelson and Leibman (Polynomial extensions of van der Waerden's and Szemeredi's theorems, Journal of the American Math Society 1996, Vol. 9, 725--753.
They actually proved a density version.
There is an alternative proof by Bergelson and Liebman (Set-polynomials and Polynomial extension of the Hales-Jewett Theorem, Annals of Math, Vol. 150, 1999, 33-75.}
The proof is in two steps
• Prove the Poly Hales-Jewett Theorem (henceforth Poly HJ). (This proof is not elementary.)
• Prove Multi-dimensional Poly VDW Theorem from Poly HJ. (This part was elementary.)
There is a purely combinatorial proof of the Multi-dimensional Poly VDW Theorem, though it is not stated in the literature:
• Walters has a combinatorial proof of Poly HJ in the paper of his mentioned in my last post.
• As noted above, Bergelson and Leibman showed how to get from the Poly HJ theorem to the Multi-dimensional Poly VDW theorem.
Putting all this together there is an elementary proof of the Multi-dimensional Poly VDW theorem. There is a general version as well, similar to the Gen Poly VDW from
my last post, but I won't state it here.
These types of theorems have been generalized in Polynomial Szemeredi theorems for countable modules over integral domains and finite fields by Bergelson, Leibman, and McCutcheon, Journal d'Analyse Mathematique, Vol 95, 243--296, 2005.
There has also been some work on restricting what $d$ or $\overline d$ could be in the above theorems. We state the easiest of such theorems. It is derivable from the (ordinary) Hales-Jewitt theorem.
VDW's with $d$ restricted:
Let $C\subseteq N$. Let $D$ be the set of all sums of distinct elements of $C$. For all $c,k\in N$, for any $c$-coloring $COL:Z \rightarrow [c]$ there exists $a\in Z$, $d\in D$, $d\ne 0$, such that
$COL(a) = COL(a+d) = COL(a+2d) = \cdots = COL(a+(k-1)d).$
More complicated versions of this for multidimensional polynomials were proven by Bergelson and McCutcheon (An Ergodic IP Polynomial Szemeredi Theorem by Bergelson and McCutcheon. Memoirs of the American Math Society, Vol 46, 2000.
Open Problems:
1. Consider the one-dimensional VDW theorem over the reals. Is there an easy analytic proof of this, or of some subcases of it?
2. If we allow polynomials with constant term what happens. More concretely: For all finite sets $F\subseteq Z[x]$, determine $c$ such that:
• For any $c$-coloring $COL:Z\rightarrow [c]$ there exists $a,d\in Z$, $d\ne 0$, such that $\{ a+p(d) \mid p\in F\}$ is monochromatic.
• There exists a $c+1$-coloring $COL:Z\rightarrow [c]$ such that for all $a,d\in Z$, $d\ne 0$, $\{ a+p(d) \mid p\in F\}$ is not monochromatic.
3. Everything I've mentioned above except the Poly HJ theorem has a density analoge that is difficult to proof. Find easier proofs. This is not well defined since it depends on how you define `easier'. Combinatorial may be one definition, but those can be rough too.
I would like to thank Alexander Leibman for his help in preparing this post.
## Wednesday, January 24, 2007
### But most disturbingly ...
My new hero is the student in a class of mine who sent me the following email:
There is always a student who shows up 10 minutes late to each of the (...) lectures, carrying a box of takeout food. He proceeds to noisily eat it while you lecture. This is highly distracting [as well as disturbing; this guy doesn't know how to use chopsticks], especially for those of us who chose to follow the rules. Please take some action as to inform students that eating in class is disrespectful not only to you, but distracting to other students who are trying to take notes. Thanks.
See, eating noisily during class is annoying enough. But not knowing how to use chopsticks? That's disturbing! | 2,377 | 7,577 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-26 | longest | en | 0.783497 |
https://kinerjamall.com/help-78 | 1,674,944,720,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00873.warc.gz | 371,792,640 | 3,935 | # Pre calculus problem solver with steps free
Math can be a challenging subject for many students. But there is help available in the form of Pre calculus problem solver with steps free. Keep reading to learn more!
## The Best Pre calculus problem solver with steps free
There is Pre calculus problem solver with steps free that can make the technique much easier. I To solve an inequality equation, you need to find the values of the variable that make the equation true. In other words, you need to find the values of the variable that make the left side of the equation equal to the right side of the equation. To do this, you can use the same methods that you would use to solve a regular equation.
Math can be difficult for a lot of people, but with practice and understanding, it can become much easier. Calculus is a math topic that a lot of people struggle with, but with the help of a good teacher or tutor, it can be conquered. There are a lot of concepts and formulas that need to be learned in order to be successful in calculus, but once they are understood, the math problems become much easier to solve.
This is because it involves a lot of higher level math concepts. However, there are some ways to make solving calculus problems easier. One way is to use a graphing calculator. This can help you visualize the problem and see what is going on. There are also online resources that can help you with calculus problems. These can provide step-by-step instructions on how to solve the problem. Finally, if you are really struggling
Math app that solves any problem If you're looking for a math app that can solve any problem, then you'll want to check out Math Solver. This app is designed to help you solve any math problem, no matter how difficult it may be. Just enter the problem into the app, and it will provide you with a step-by-step solution. Plus, if you're struggling with a particular concept, you can watch video lessons within the app to help you understand it better
equations are mathematical problems that can be solved using algebraic methods. Word problems are mathematical problems that can be solved by reading and understanding the problem, and using algebraic methods to solve. | 450 | 2,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-06 | latest | en | 0.956761 |
https://www.ademcetinkaya.com/2023/01/fssiu-fortistar-sustainable-solutions.html | 1,696,006,730,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00738.warc.gz | 667,395,900 | 56,769 | Outlook: Fortistar Sustainable Solutions Corp. Unit is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Wait until speculative trend diminishes
Time series to forecast n: 29 Jan 2023 for (n+8 weeks)
Methodology : Active Learning (ML)
## Abstract
Fortistar Sustainable Solutions Corp. Unit prediction model is evaluated with Active Learning (ML) and Pearson Correlation1,2,3,4 and it is concluded that the FSSIU stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
## Key Points
1. Reaction Function
2. Should I buy stocks now or wait amid such uncertainty?
3. What is prediction in deep learning?
## FSSIU Target Price Prediction Modeling Methodology
We consider Fortistar Sustainable Solutions Corp. Unit Decision Process with Active Learning (ML) where A is the set of discrete actions of FSSIU stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Pearson Correlation)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Active Learning (ML)) X S(n):→ (n+8 weeks) $∑ i = 1 n a i$
n:Time series to forecast
p:Price signals of FSSIU stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## FSSIU Stock Forecast (Buy or Sell) for (n+8 weeks)
Sample Set: Neural Network
Stock/Index: FSSIU Fortistar Sustainable Solutions Corp. Unit
Time series to forecast n: 29 Jan 2023 for (n+8 weeks)
According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for Fortistar Sustainable Solutions Corp. Unit
1. If an entity previously accounted for a derivative liability that is linked to, and must be settled by, delivery of an equity instrument that does not have a quoted price in an active market for an identical instrument (ie a Level 1 input) at cost in accordance with IAS 39, it shall measure that derivative liability at fair value at the date of initial application. Any difference between the previous carrying amount and the fair value shall be recognised in the opening retained earnings of the reporting period that includes the date of initial application.
2. An entity is not required to incorporate forecasts of future conditions over the entire expected life of a financial instrument. The degree of judgement that is required to estimate expected credit losses depends on the availability of detailed information. As the forecast horizon increases, the availability of detailed information decreases and the degree of judgement required to estimate expected credit losses increases. The estimate of expected credit losses does not require a detailed estimate for periods that are far in the future—for such periods, an entity may extrapolate projections from available, detailed information.
3. The accounting for the time value of options in accordance with paragraph 6.5.15 applies only to the extent that the time value relates to the hedged item (aligned time value). The time value of an option relates to the hedged item if the critical terms of the option (such as the nominal amount, life and underlying) are aligned with the hedged item. Hence, if the critical terms of the option and the hedged item are not fully aligned, an entity shall determine the aligned time value, ie how much of the time value included in the premium (actual time value) relates to the hedged item (and therefore should be treated in accordance with paragraph 6.5.15). An entity determines the aligned time value using the valuation of the option that would have critical terms that perfectly match the hedged item.
4. When an entity designates a financial liability as at fair value through profit or loss, it must determine whether presenting in other comprehensive income the effects of changes in the liability's credit risk would create or enlarge an accounting mismatch in profit or loss. An accounting mismatch would be created or enlarged if presenting the effects of changes in the liability's credit risk in other comprehensive income would result in a greater mismatch in profit or loss than if those amounts were presented in profit or loss
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
Fortistar Sustainable Solutions Corp. Unit is assigned short-term Ba1 & long-term Ba1 estimated rating. Fortistar Sustainable Solutions Corp. Unit prediction model is evaluated with Active Learning (ML) and Pearson Correlation1,2,3,4 and it is concluded that the FSSIU stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
### FSSIU Fortistar Sustainable Solutions Corp. Unit Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBaa2Baa2
Balance SheetBa1Caa2
Leverage RatiosCaa2B2
Cash FlowBaa2Baa2
Rates of Return and ProfitabilityBaa2B3
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 90 out of 100 with 756 signals.
## References
1. Farrell MH, Liang T, Misra S. 2018. Deep neural networks for estimation and inference: application to causal effects and other semiparametric estimands. arXiv:1809.09953 [econ.EM]
2. Chen X. 2007. Large sample sieve estimation of semi-nonparametric models. In Handbook of Econometrics, Vol. 6B, ed. JJ Heckman, EE Learner, pp. 5549–632. Amsterdam: Elsevier
3. J. Baxter and P. Bartlett. Infinite-horizon policy-gradient estimation. Journal of Artificial Intelligence Re- search, 15:319–350, 2001.
4. E. Altman. Constrained Markov decision processes, volume 7. CRC Press, 1999
5. N. B ̈auerle and J. Ott. Markov decision processes with average-value-at-risk criteria. Mathematical Methods of Operations Research, 74(3):361–379, 2011
6. Semenova V, Goldman M, Chernozhukov V, Taddy M. 2018. Orthogonal ML for demand estimation: high dimensional causal inference in dynamic panels. arXiv:1712.09988 [stat.ML]
7. D. Bertsekas. Min common/max crossing duality: A geometric view of conjugacy in convex optimization. Lab. for Information and Decision Systems, MIT, Tech. Rep. Report LIDS-P-2796, 2009
Frequently Asked QuestionsQ: What is the prediction methodology for FSSIU stock?
A: FSSIU stock prediction methodology: We evaluate the prediction models Active Learning (ML) and Pearson Correlation
Q: Is FSSIU stock a buy or sell?
A: The dominant strategy among neural network is to Wait until speculative trend diminishes FSSIU Stock.
Q: Is Fortistar Sustainable Solutions Corp. Unit stock a good investment?
A: The consensus rating for Fortistar Sustainable Solutions Corp. Unit is Wait until speculative trend diminishes and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of FSSIU stock?
A: The consensus rating for FSSIU is Wait until speculative trend diminishes.
Q: What is the prediction period for FSSIU stock?
A: The prediction period for FSSIU is (n+8 weeks) | 1,969 | 8,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-40 | latest | en | 0.841692 |
https://talibiddeenjr.wordpress.com/2008/05/31/solving-algebra-word-problems/ | 1,532,201,043,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592654.99/warc/CC-MAIN-20180721184238-20180721204238-00354.warc.gz | 778,883,250 | 18,086 | ## Solving Algebra Word Problems
31 May
I wrote out the following guide to help my older kids solve simple algebraic word problems.
Algebra Word
Problem Getting Started Quick Guide
After reading the problem all the way through:
1. 1. Set your variable to be what you are looking for (what you are
asked to solve for in some cases ). (WRITE IT OUT)
a. Example: The question asks you to find the number of red marbles
so you write:
x = number of red marbles
a. Look for keywords
The following words may signal or signals is, was, are, will be, yields, sold for = increased by, more than, combined, together, total of, sum, added to addition decreased by, minus, less, difference between/of, less than, fewer than subtraction of, times, multiplied by product of, increased/decreased by a factor of multiplication per, a, out of, ratio of, quotient of percent (divide by 100) division
Some
words taken from www.purplemath.com
b. Take phrases at a time and write down what they translate into
3. 3. Simplify/collect like terms
4. 4. Solve for your variable (get it by itself)
a. Plug your answer back into the original equation, if the sides
are equal, then the solution is correct, insha Allah.
Purple Math.com has tips on solving algebra word problems | 305 | 1,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-30 | latest | en | 0.907353 |
https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_11&oldid=55949 | 1,607,106,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00415.warc.gz | 193,024,413 | 10,554 | # 2005 AMC 8 Problems/Problem 11
## Problem
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up$90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up \$90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?
$\textbf{(A)}\ \textdollar 1.06\qquad\textbf{(B)}\ \textdollar 0.53 \qquad\textbf{(C)}\ \textdollar 0\qquad\textbf{(D)}\ \textdollar 0.53\qquad\textbf{(E)}\ \textdollar 1.06$
## Solution
The price Jacks rings up is $(90.00)(1.06)(0.80)$. The price Jill rings up is $(90.00)(0.80)(1.06)$. By the commutative property of multiplication, these quantities are the same, and the difference is $\boxed{\textbf{(C)}\ \textdollar 0}$.
## See Also
2005 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Login to AoPS | 447 | 1,318 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-50 | latest | en | 0.676518 |
https://www.owlgen.in/what-is-investment-function-according-to-economics/ | 1,675,291,212,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499953.47/warc/CC-MAIN-20230201211725-20230202001725-00788.warc.gz | 960,751,597 | 25,877 | # What is Investment Function According to Economics?
## Investment Function:
The classical economists opined that saving as well as investment both are a function of rate of interest. Saving is a direct function of rate of interest while investment is an indirect function of rate of interest. An adjustment in rate of interest brings about equilibrium in money market.
But J. M. Keynes did not agree to it. He opined that investment depends on the expected returns on it. Investment takes place in an economy because it gives certain returns. The return to investment can be known by estimating marginal efficiency of investment. It refers to addition in total output when capital stock increases by one unit.
Rate of interest is the opportunity cost of investment. If invested funds are borrowed one, then it is explicit cost and even if the entrepreneur invests his own funds, it is implicit cost. So, whether to undertake an investment or not, an investor compares rate of interest with return on investment i.e. marginal efficiency of capital.
• If Marginal efficiency of capital > Rate of Interest, it is advisable to undertake investment.
• If Marginal efficiency of capital < Rate of Interest, it is advisable not to undertake investment.
• If Marginal efficiency of capital = Rate of Interest, consider other factors like risk involved, future expectations etc.
The relationship between investment and MEI is negative i.e. as we increase investment, the additional returns expected on it keeps on decreasing. It is shown with the help of following diagram.
Investment is undertaken not only for the present but for the future and continues for a long time, therefore, uncertainty also lays a vital role in taking investment decisions.
There are many econometric models of investment behaviour and two of these are explained below:
The Accelerator Model: it states that rate of investment depends on the level of AD. In other words, it states that the desirable level of capital stock in an economy is a constant fraction of the level of output.
K = h. Q
Where K is capital stock h is constant and Q is level of output. Therefore, more will be the level of output more will be the rate of investment. During boom, when entrepreneurs are certain about future returns, they tend to invest more. On the other hand, during recession, the rate of investment is less due to uncertainty regarding returns. Another notable point is that investment is not only influenced by level of output, but it also influences the level of output. (Multiplier concept). | 495 | 2,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-06 | latest | en | 0.950526 |
https://www.unitsconverters.com/en/Dam/S2-To-Fm/S2/Utu-3422-3427 | 1,675,009,668,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00235.warc.gz | 1,046,942,907 | 39,929 | Formula Used
1 Meter per Square Second = 0.1 Decameter per Square Second
1 Meter per Square Second = 1E+15 Femtometer per Square Second
1 Decameter per Square Second = 1E+16 Femtometer per Square Second
## dam/s² to fm/s² Conversion
The abbreviation for dam/s² and fm/s² is decameter per square second and femtometer per square second respectively. 1 dam/s² is 1E+16 times bigger than a fm/s². To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including dam/s² to fm/s² conversion.
## Decameter per Square Second to fm/s²
Check our Decameter per Square Second to fm/s² converter and click on formula to get the conversion factor. When you are converting acceleration from Decameter per Square Second to fm/s², you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
## dam/s² to Femtometer per Square Second
The formula used to convert dam/s² to Femtometer per Square Second is 1 Decameter per Square Second = 1E+16 Femtometer per Square Second. Measurement is one of the most fundamental concepts. Note that we have Fahrenheit as the biggest unit for length while Yottaampere is the smallest one.
## Convert dam/s² to fm/s²
How to convert dam/s² to fm/s²? Now you can do dam/s² to fm/s² conversion with the help of this tool. In the length measurement, first choose dam/s² from the left dropdown and fm/s² from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from fm/s² to dam/s²? You can check our fm/s² to dam/s² converter.
## dam/s² to fm/s² Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like acceleration finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like dam/s² to fm/s² through multiplicative conversion factors. When you are converting acceleration, you need a Decameter per Square Second to Femtometer per Square Second converter that is elaborate and still easy to use. Converting dam/s² to Femtometer per Square Second is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Decameter per Square Second to fm/s², this tool is the answer that gives you the exact conversion of units. You can also get the formula used in dam/s² to fm/s² conversion along with a table representing the entire conversion.
Let Others Know | 658 | 2,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-06 | latest | en | 0.867782 |
http://apmonitor.com/me575/index.php/Main/TwoBarTruss?action=print | 1,643,296,947,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00242.warc.gz | 4,423,711 | 5,693 | ### Two Bar Truss Design
A design of the truss is specified by a unique set of values for the analysis variables: height (H), diameter, (d), thickness (t), separation distance (B), modulus of elasticity (E), and material density (rho). Suppose we are interested in designing a truss that has a minimum weight, will not yield, will not buckle, and does not deflect "excessively,” and so we decide our model should calculate weight, stress, buckling stress and deflection.
This introductory assignment is designed as a means of demonstrating the optimization capabilities of a number of software packages. Below are tutorials for solving this problem with a number of software tools. Below are a few step-by-step tutorials.
#### Python (GEKKO) Solution
GEKKO is optimization software for mixed-integer and differential algebraic equations. It is coupled with large-scale solvers for linear, quadratic, nonlinear, and mixed integer programming (LP, QP, NLP, MILP, MINLP).
import numpy as np
# import gekko, pip install if needed
from gekko import GEKKO
# create new model
m = GEKKO()
# declare model parameters
width = m.Param(value=60)
thickness = m.Param(value=0.15)
density = m.Param(value=0.3)
modulus = m.Param(value=30000)
# declare variables and initial guesses
height = m.Var(value=30.00,lb=10.0,ub=50.0)
diameter = m.Var(value=3.00,lb=1.0,ub=4.0)
weight = m.Var()
# intermediate variables with explicit equations
leng = m.Intermediate(m.sqrt((width/2)**2 + height**2))
area = m.Intermediate(np.pi * diameter * thickness)
iovera = m.Intermediate((diameter**2 + thickness**2)/8)
stress = m.Intermediate(load * leng / (2*area*height))
buckling = m.Intermediate(np.pi**2 * modulus \
* iovera / (leng**2))
deflection = m.Intermediate(load * leng**3 \
/ (2 * modulus * area * height**2))
# implicit equations
m.Equation(weight==2*density*area*leng)
m.Equation(weight < 24)
m.Equation(stress < 100)
m.Equation(stress < buckling)
m.Equation(deflection < 0.25)
# minimize weight
m.Minimize(weight)
# solve optimization
m.solve() # remote=False for local solve
print ('')
print ('--- Results of the Optimization Problem ---')
print ('Height: ' + str(height.value))
print ('Diameter: ' + str(diameter.value))
print ('Weight: ' + str(weight.value))
#### Solution Contour Plot (Python)
## Generate a contour plot
# Import some other libraries that we'll need
# matplotlib and numpy packages must also be installed
import matplotlib
import numpy as np
import matplotlib.pyplot as plt
# Constants
pi = 3.14159
dens = 0.3
modu = 30000.0
# Analysis variables
wdth = 60.0
thik = 0.15
# Design variables at mesh points
x = np.arange(10.0, 30.0, 2.0)
y = np.arange(1.0, 3.0, 0.3)
hght, diam = np.meshgrid(x, y)
# Equations and Constraints
leng = ((wdth/2.0)**2.0 + hght**2)**0.5
area = pi * diam * thik
iovera = (diam**2.0 + thik**2.0)/8.0
wght = 2.0 * dens * leng * area
strs = load * leng / (2.0 * area * hght)
buck = pi**2.0 * modu * iovera / (leng**2.0)
defl = load * leng**3.0 / (2.0*modu * area * hght**2.0)
# Create a contour plot
# Visit https://matplotlib.org/examples/pylab_examples/contour_demo.html
# for more examples and options for contour plots
plt.figure()
# Weight contours
CS = plt.contour(hght, diam, wght)
plt.clabel(CS, inline=1, fontsize=10)
# Stress<100
CS = plt.contour(hght, diam, strs,[100.0],colors='k',linewidths=[4.0])
plt.clabel(CS, inline=1, fontsize=10)
# Deflection<0.25
CS = plt.contour(hght, diam, defl,[0.25],colors='b',linewidths=[4.0])
plt.clabel(CS, inline=1, fontsize=10)
# Stress-Buckling<0
CS = plt.contour(hght, diam, strs-buck,[0.0],colors='r',linewidths=[4.0])
plt.clabel(CS, inline=1, fontsize=10)
plt.title('Two Bar Optimization Problem')
plt.xlabel('Height')
plt.ylabel('Diameter')
# Save the figure as a PNG
plt.savefig('contour1.png')
# Create a new figure to see more detail
plt.figure()
# Weight contours
CS = plt.contour(hght, diam, wght)
plt.clabel(CS, inline=1, fontsize=10)
# Stress<100
CS = plt.contour(hght, diam, strs,[90.0,100.0],colors='k',linewidths=[0.5, 4.0])
plt.clabel(CS, inline=1, fontsize=10)
# Deflection<0.25
CS = plt.contour(hght, diam, defl,[0.22,0.25],colors='b',linewidths=[0.5, 4.0])
plt.clabel(CS, inline=1, fontsize=10)
# Stress-Buckling<0
CS = plt.contour(hght, diam, strs-buck,[-5.0,0.0],colors='r',linewidths=[0.5, 4.0])
plt.clabel(CS, inline=1, fontsize=10)
plt.title('Two Bar Optimization Problem')
plt.xlabel('Height')
plt.ylabel('Diameter')
# Save the figure as a PNG
plt.savefig('contour2.png')
# Show the plots
plt.show()
#### Objective Function Plot
One part of the assignment asks you to select width and load as variables for a 3d optimal surface plot and plot the solution of the optimization problem to minimize deflection at each of the width / load combinations. This tutorial example shows how to do this same activity but for the alternative problem of minimizing weight.
This assignment can be completed in collaboration with others. Additional guidelines on individual, collaborative, and group assignments are provided under the Expectations link. | 1,498 | 5,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | latest | en | 0.715295 |
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What brought the American semiconductor market back from the
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24 Oct 2008, 02:52
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What brought the American semiconductor market back from the verge of bankruptcy shortly after the tech slump was special government funded research.
(A) What brought
(B) The thing that brought
(C) That which brought
(D) Bringing
(E) What has brought
for this problem, I chose option D because I thought that it was concise. However, the OA is A. Why is A better than D? Would anyone be so kind to explain it to me? I'll really appreciate it.
Thanks
If you have any questions
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24 Oct 2008, 04:38
"what" clearly references "special government funded research".
Try re-writing the sentence as:
special government funded research brought the American semiconductor market......
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24 Oct 2008, 08:23
Why is E incorrect here?
Rest of the sentence is in past tense . Hence we don't need present perfect with has in E
if the sentence were has been government funded research, then E would have been correct.
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24 Oct 2008, 11:04
here is my take on A :
(A) What brought (short and precise )
(B) The thing that brought ("that thing" --- > awkward)
(C) That which brought ?(this confused me for a while .. and i think in some other post i saw this as the answer)
(D) Bringing (present continous tense not required .. rest of the sentence is in past tense)
(E) What has brought (present perfect is not required as the portion of the sentence ,which is not underlined, talks about past tense "was" )
i will definitely go with A but is it a right/preferred way to start a sentence from "What"
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26 Oct 2008, 01:22
Why is E incorrect here?
the reason E is wrong is that the whole sentence is talking about the past. When you use the present perfect in option E, you're rather saying that an action started in the past and it is still going on today. That is not what you want to say. You want to talk about an action that happened in the past and is also completed in the past, period.
Re: SC: Complicated question [#permalink] 26 Oct 2008, 01:22
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Display posts from previous: Sort by | 1,073 | 4,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-04 | latest | en | 0.917319 |
http://www.chegg.com/homework-help/questions-and-answers/the-magnitude-of-a-velocity-vector-is-called-speed-suppose-that-a-wind-is-blowing-from-the-q3513222 | 1,369,376,306,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704234586/warc/CC-MAIN-20130516113714-00062-ip-10-60-113-184.ec2.internal.warc.gz | 378,842,265 | 8,363 | ## Vectors
The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 30 km/h. (This means that the direction from which the wind blows is 45° west of the northerly direction.) A pilot is steering a plane in the direction N60°E at an airspeed (speed in still air) of 100 km/h. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. Give your answers correct to one decimal place.
N °E (true course)
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https://www.nationalreview.com/the-agenda/gabriel-rossman-future-la-carte-pricing-reihan-salam/ | 1,638,075,583,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00227.warc.gz | 1,004,442,976 | 72,644 | # Gabriel Rossman on the Future of A La Carte Pricing
Gabriel Rossman, a sociologist at UCLA, is one of the sharpest observers of the media landscape I’ve come across, and he has a very thought-provoking post on how Steve Jobs and Jeff Bezos might succeed where Kevin Martin’s FCC failed, i.e., they might give television consumers more choice about the kind of content they purchase than the traditional cable business model allows. But the result, interestingly, might be a sharp reduction in the quality of the television programming we consume:
A switch to a la carte will probably result in an increase in consumer surplus per unit demanded but a drastic decrease in quantity supplied. (“Consumer surplus” is econ jargon for the subjective experience of a “bargain”).
Suppose that my household values watching True Blood and Mad Men at \$5 an episode, Top Chef at \$2 an episode, and Mythbusters and Toddlers and Tiaras at 50 cents an episode. Now suppose that my next door neighbors have the exact opposite set of preferences. In both cases there is a total of \$13 of demand for television per household per week. If the cable company charges \$12.99 per week, both my neighbor and I will write the check, but do so reluctantly as we’re just barely this side of the indifference curve.
Now suppose that someone (say, Apple or Amazon) starts selling shows a la carte. If the price point is 50 cents both my neighbor and I will still watch all five shows. However we’ll only be paying \$2.50 a week and will be getting \$10.50 in consumer surplus. If the price point is \$2, each of us will get three shows and pay \$6, for \$6 in consumer surplus. If the price point is \$4.99 we’ll each buy two shows, pay almost \$10, and get two cents of consumer surplus. There is no price point where we both pay \$12.99 like we used to. At any one a la carte price point, both my neighbor and I will pay less than we used to, watch the same or less amount of tv, and get the same or higher consumer surplus.
Thus a switch to an a la carte model implies much lower costs to the consumer. Because revenues would fall, so would production by some combination of reduced numbers of shows and reduced production values. Basically, we’re looking at an end to the television renaissance we’ve enjoyed since the late 1990s as people like me decide that we’d rather pay \$10 or \$20 a month for the few shows we love and do without the rest than pay \$50 a month for a bunch of stuff, most of which we don’t even really like.
Of course, there is another structural shift at work, namely the rise of technologies that are sharply reducing the cost of digital effects that can mask skimpy production budgets. In conclusion, Gabriel writes:
That is to say, television may not be economically viable when priced on an a la carte basis and this could lead to a decline in volume and possibly quality of original programming. This will probably involve a slow decline but could be catastrophic. The most likely scenario for a catastrophic collapse is if the studios forecast that a la carte means declining revenue and try to pare back their cost structure in anticipation. This would probably lead to a militant slate getting elected at both WGA and SAG and an even worse strike / soft strike than we had on the last contract cycle.
But this could mean that U.S. viewers will consume more entertainment that is peer-produced, produced overseas, or by institutions that otherwise work outside of the WGA and SAG structure. Quality writing is of course very important, but perhaps quality writers will attach themselves to hyper-realistic animation projects, etc. No more actors! Or rather a somewhat smaller cadre of actors who focus primarily on screen-capture work, and who make middle-class wages rather than outsized sums. There are many, many ways this phenomenon could play out.
All that said, I think Gabriel’s main point is beyond dispute: a la carte pricing has the potential to seriously disrupt the current structure of the television industry. Expect WGA and SAG to wield their influence to press for even more onerous regulation of the entertainment industries. As I’ve been saying for a really long time, this creates an opportunity for the right and the so-called copyleft to form an alliance built on libertarian principle and mutual interest.
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posted by on .
Use the trig Identities to find the other 5 trig functions.
Problem
7.)Tan(90-x)=-3/8
8.)Csc x=-13/5
9.)Cot x=square root of 3
10.)Sin(90-x)=-.4563
11.)Sec(-x)=4
12.)Cos x=-.2351
I need HELP!
• Trigonometry - ,
I will do one for you.
You try some of the others and let me know what you got.
#8
csc x = -13/5, then
sin x = -5/13
so by CAST, angle x is in either III or IV
construct a right-angled triangle with the side opposite angle x as 5 and hypotenuse 13 (sin ß = opp/hyp)
By Pythagoras the adjacent side is 12
so in III, (x is appr. 202.62º)
sin x = -5/13
cos x = -12/13
sec x = -13/12
tan x = 5/12
cot x = 12/5
in IV, (x is appr. 337.38º)
sin x = -5/13
cos x = 12/13
sec x = 13/12
tan x = -5/12
cot x = -12/5
• Trigonometry - ,
Would #7 be Sin- 3 over square root of 73 Cos- 8 over square root of 73 Tan- -8/3 Csc- square root of 73 over 3 Cot- -3/8
• Trigonometry - ,
sin x =4/5 and cos y = 5/3
sin x = 4/5
cos x = 3/5
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February 2008, 2(1): 121-131. doi: 10.3934/ipi.2008.2.121
An inverse boundary value problem for a nonlinear wave equation
1 Department of Mathematics, Hokkaido University, Kita 10, Nishi 8, Kita-Ku, Sapporo, Hokkaido, 060-0810, Japan 2 Department of Mathematics, Faculty of Science and Technology, Tokyo University of Science, Yamazaki 2641, Noda, Chiba, 278-8510, Japan
Received June 2007 Published January 2008
An inverse boundary value problem for nonlinear wave equation of divergence form in one space dimension is considered. By assuming the nonlinear term is unknown, we show the linear and quadratic part of this term can be identified from the Dirichlet to Neumann map. Here, the nonlinearity is only in terms of the first derivative with respect to the space variable, and the linear and quadratic parts are defined in terms of this derivative. The identification not only gives the uniqueness but also the reconstruction.
Citation: Gen Nakamura, Michiyuki Watanabe. An inverse boundary value problem for a nonlinear wave equation. Inverse Problems & Imaging, 2008, 2 (1) : 121-131. doi: 10.3934/ipi.2008.2.121
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2018 Impact Factor: 1.469 | 1,732 | 5,692 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.730979 |
https://www.i-programmer.info/programming/python/14276-programmers-python-parameters.html?start=1 | 1,611,230,016,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00749.warc.gz | 825,667,562 | 11,244 | Programmer's Python - Parameters
Written by Mike James
Monday, 11 January 2021
Article Index
Programmer's Python - Parameters
Variable Parameters
## Variable Parameters
Suppose you want to define a function which can accept any number of parameters. The solution to this is to specify a parameter of the form *name. In this case all of the arguments, as many as supplied, are packed into a tuple ready for you to process.
For example:
```def sum(*nums):
total=0
for n in nums:
total=total+n
Notice that the parameter when used within the function doesn’t have the * prefix. You can now call sum with as many numbers to add up as you require:
`print(sum(1,2,3,4,5,6))`
You can also specify a variable set of parameters after a number of positional parameters.
For example:
```def sum(a,b,*nums):
total=a+b
for n in nums:
total=total+n
print(sum(1,2,3,4,5))```
In this case 1 and 2 are stored in a and b, and the remainder of the arguments are stored in nums as a tuple.
You obviously can’t have any positional parameters beyond a catch all *name but you can have keyword only parameters. For example:
```def sum(*nums,a,b):
total=a+b
for n in nums:
total=total+n
If you just call this version of sum with positional arguments you will see an error message. You have to call this with named parameters:
`print(sum(1,2,3,4,5,a=6,b=7))`
It doesn’t matter where you specify the named parameters but you have to do it – hence the term “keyword only” parameter.
You can also set defaults for keyword only parameters.
For example:
```def sum(*nums,a=1,b):
total=a+b
for n in nums:
total=total+n
Now a is still a keyword only parameter but if you don’t specify a value for it when you call the function, the default will be used. Notice that if you want keyword only parameters to be optional you have to supply a default:
`print(sum(1,2,3,4,5,b=7))`
and you cannot omit b because it has no default.
Keyword only parameters are stored in the __kwdefaults__ attribute of the function as a dictionary object.
For example:
`print(sum.__kwdefaults__)`
prints:
`{'a': 1}`
What about a variable number of arbitrary keyword parameters?
In the same way that *name stores any number of positional parameters in a tuple **name will store any number of keyword parameters in a dictionary.
For example:
```def sum(**namedNums):
total=0
for name,value in namedNums.items():
total=total+value
print(name,value)
creates a function that can only be called using keyword parameters and you can make up the keywords when you call the function.
For example:
`print(sum(one=1,two=2,three=3))`
prints one 1, two 2, three 3 and the sum of the values, 6.
You can also define a function that has any number of positional parameters and any number of keyword parameters by using *name and **name together. Notice that in this case *name has to come first and you can also have some positional parameters as long as they come before *name.
For example:
```def sum(a,b,*nums,**namedNums):
total=a+b
for value in nums:
total=total+value
for name,value in namedNums.items():
total=total+value
print(name,value)
Now you have to call sum with two positional parameters, but after that you can have as many positional or keyword parameters you want to use.
For example:
`print(sum(1,2,3,4,one=1,two=2,three=3))`
in this case 1 is stored in a, 2 in b, (3,4) in nums, and the keyword parameters are stored in namedNums.
## Unpacking Into Parameters
You can think of *name and **name as being packing operations. They pack the positional and keyword parameters into tuples and dictionaries respectively.
It sometimes happens that what you have is already packed into a tuple or a dictionary and you want to pass it to a function as a parameter list. The easiest solution is to use the unpacking operators * and ** which unpack a tuple or a dictionary respectively.
For example:
```def sum(a,b):
total=a+b
tuple=(1,2)
print(sum(*tuple))```
In this case the tuple (1,2) is unpacked into the parameters a and b.
If there are too many parameters after unpacking you will generate an error:
```tuple=(1,2,3,4)
print(sum(*tuple))```
generates:
`TypeError: sum() takes 2 positional arguments but 4 were given`
You can of course avoid this by using the * operator in the function definition:
```def sum(a,b,*nums):
total=a+b
Any unused parameters that result from the unpacking are now packed into a tuple referenced by nums.
The same idea works with dictionaries and keyword parameters:
```dict={"a":1,"b":2}
print(sum(**dict))```
In this case dict is unpacked into a and b with values 1 and 2 respectively.
As with tuples the parameters have to exist but you can repack the dictionary into a parameter using **name.
#### Not in this extract:
• Return Values
• Docstrings and Annotations
• Decorators
• Making the Wrap Transparent – functools
• Multiple Decorators
• Decorator Factories
## Summary
• Python allows defaults for parameters to be set.
• You can call functions using parameters as keyword parameters.
• Specifying *name is a way of using a variable number of positional parameters – any arguments supplied that are not used by positional parameters are stored in *name as a tuple.
• Specifying **kwname works in the same way as *name but any unused keyword arguments are stored in it as a dictionary.
• Docstrings can be used to create rudimentary documentation for a function.
• Annotations can be applied to parameters and return values – usually to indicate the type.
• Decorators are functions that take a single function as an argument and return a single function – they are used to transform functions.
• A decorator is written as @dec in front of a function definition and this is equivalent to f=dec(f) where f is the name of the function.
• Multiple decorators are applied starting with the decorator closest to the function working up the page.
• You can also use a decorator factory @decfac(parameters) with parameters. It is evaluated at once and has to return a decorator which is applied to the function in the usual way.
## Is now available as a print book: Amazon
#### Contents
1. Hello Python World
2. Variables, Objects and Attributes
3. The Function Object
Extract - Function Objects
Extract - Local and Global
Extract - Parameters ***NEW
Extract - Decorators
6. Class Methods and Constructors
Extract - Objects Become Classes
7. Inside Class
8. Metaclass
Extract - Properties
10. Custom Attribute Access
Extract - Custom Attributes
Extract - Default Methods
11. Single Inheritance
12. Multiple Inheritance
13. Class and Type
Extract - Class
14. Type Annotation
Extract - Type Annotation
## Related Articles
Creating The Python UI With Tkinter
Creating The Python UI With Tkinter - The Canvas Widget
The Python Dictionary
Arrays in Python
Advanced Python Arrays - Introducing NumPy | 1,606 | 6,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-04 | latest | en | 0.741355 |
https://programmingpraxis.com/2013/02/12/binary-search-tree-in-order-predecessor-and-successor/ | 1,680,218,875,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949506.62/warc/CC-MAIN-20230330225648-20230331015648-00020.warc.gz | 538,642,370 | 28,698 | ## Binary Search Tree: In-Order Predecessor And Successor
### February 12, 2013
We have seen several flavors of binary search trees, including classic bsts, treaps, red/black trees, and avl trees. For all of them we provided a `lookup` function that returned a requested key/value pair. In today’s exercise we write functions that return the predecessor and successor of a requested key, allowing a program to walk the tree in order. There are two common variants of these functions, one using parent pointers and one without; our exercise will use the variant without parent pointers, which is generally more useful.
We give the basic logic for the `successor` function; the `predecessor` function is similar, but reversed. A recursive function descends the tree searching for the requested key, keeping track of the current tree as the possible successor every time if follows the left sub-tree. If it finds the key, its successor is the minimum element of the right sub-tree, found by chasing left sub-trees until reaching a null node. If the search reaches a null node and hence fails to find the key, the next largest key in the tree is the successor, found in the possible successor that was saved at each recursive call.
Your task is to write `predecessor` and `successor` functions for binary search trees. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
### 3 Responses to “Binary Search Tree: In-Order Predecessor And Successor”
1. jpverkamp said
Here’s a version in Racket: Predecessor and successor in a binary search tree
I went a slightly different direction than the solution here with a shared function to return the node containing the value we were searching for and the last left and right branch all at once (the ability to return multiple `value`s is always nice). That made the actual `predecessor` and `successor` functions somewhat simpler:
```; get the predecessor to a value
(define (predecessor tr < val)
(define-values (l v r) (find-nodes tr < val))
(cond
[(tree-left? v)
(maximum (tree-left v))]
[(and (not (void? r)) (not (empty? r)))
(tree-value r)]))
```
I also included code for using this to implement a sorting algorithm. Dump everything into a tree, find the `minimum`, then repeatedly call `successor` until you run out of elements. Am I incorrect in thinking that this is actually still `O(n log n)`, albeit with a higher constant than most? It sorts a million element list only about 3x slower than a naive implementation of `quicksort` did, but insertion sort (known `O(n^2)`) didn’t even finish before I got bored.
2. I finally got around to this challenge. Here’s the code for successor in python.
```class Node:
def __init__(self, val=None, l=None, r=None):
self.l = l
self.r = r
self.val = val
def insert(node, val):
if node == None:
return
elif node.val == val:
return
elif node.val > val:
if node.l is None:
node.l = Node(val)
else:
insert(node.l, val)
else:
if node.r is None:
node.r = Node(val)
else:
insert(node.r, val)
def suc(node, val, ancestor=Node(None)):
'''
>>> mynode = Node(5)
>>> for v in "6 12 8 3 2 5 3 4 11 7 1".split():
... insert(mynode,int(v))
>>> suc(mynode,5)
6
>>> suc(mynode,8)
11
>>> suc(mynode,11)
12
>>> suc(mynode,12)
>>> suc(mynode,2)
3
>>> suc(mynode,9)
11
>>> suc(mynode,15)
'''
if node is None:
return ancestor.val
if node.val == val:
if node.r is None:
return ancestor.val
else: # Find leftmost of right node
node = node.r
while node.l is not None:
node = node.l
return node.val
elif node.val > val:
return suc(node.l, val, node)
else:
return suc(node.r, val, ancestor)
``` | 944 | 3,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-14 | latest | en | 0.9058 |
https://mapleprimes.com/products/maple?page=1575 | 1,568,532,749,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514570830.42/warc/CC-MAIN-20190915072355-20190915094355-00366.warc.gz | 580,548,479 | 28,616 | ## Hide output in for loop...
Stupid question maybe, but how can you hide output of a for loop? The usual : does not work here... Thanks for your help.
## Maximization of x1+x2+...+xn...
Imagine you have to solve the following problem. xi*(a-(x1+x2+...+xi+...+xn)-b) However, we can not see ex ante that all the x's are the same. So, I would like to differentiate to xi and equal the derivative to zero. So, I should have xi in function of the other x's. I hope anyone can see what I mean and I beg you to answer me because this is very important for me. Thanks a lot. Greets Jan
Hello, suppose you have the following function which gives 1 if both its arguments are even: G := (i,j) -> if is(i,even) and is(j,even) then 1; else 0; end if; Then what is the difference between these two statements: sum( sum( G(i,j), i=0..5 ), j=0..5 ); add( add( G(i,j), i=0..5 ), j=0..5 ); The first one gives 0 and no sum is actually computed (you can see this by adding print(i) and print(j) statements in the body of G). The second one gives the correct result, namely 9. How come i is not replaced in sum? -- Thanks for any reply, Franky.
## plotting multiple 3d data sets on a single plot...
How do i plot a bunch of 3D data sets. If it were just one set i could use pointplot3d i suppose. with(plots): > l1 := [[1,2,2],[3,4,2],[1,5,6],[7,8,9]]; > l2 := [[1,2,2.5],[3,4,2],[1,5,6],[7,8,9]]; > plots[pointplot3d]([l1,l2], axes=boxed, labels=[x, y, z], > connect=true); #DOESNT WORK More exactly i want to plot the data sets (11 rows each) pasted below. 1.000000000000 1.000000000000 .571428571400 2.000000000000 1.000000000000 .139941691000 3.000000000000 1.000000000000 .068542869040 4.000000000000 1.000000000000 .041965021860 5.000000000000 1.000000000000 .028776014990
## Embedding Information...
I was wondering if anyone was familiar with an easy way to embed or hide information into a Maple worksheet (Classic worksheet or Maple 9/9.5 standard worksheet, NOT a Maple 10 standard worksheet)? I want to be able to be able to have something equivalent to a checksum. We are producing Maple worksheets as homework for students, and I want to have something that the sudent cannot easily change to make sure that he or she is not submitting a friend's homework.
## Graphing differential equation...
Hey guys, I'm having a bit of trouble graphing the following: f'''(x) + f(x)*f''(x) = 0 where f(0)=f'(0)=0 and f'(infinity)=1 My specific problem is that I can't set f'(0)=0 and f'(infinity)=1 thanks, Rebecca
I am trying to solve (numerically) the following system of differential equations using dsolve (Maple 9): DGLS:=seq(seq(diff(B[i,k](t),t)=add(add(W(j,l,i,k)*B[j,l](t)-W(i,k,j,l)*B[i,k](t),l=1..N),j=1..N),k=1..N),i=1..N); The problem is posed by the coefficients W(i,k,j,l). I want to make the W(i,j,k,l) dependent on the unknown functions B[i,k](t). For that purpose I wrote a Maple program W := proc(i::integer, k::integer, j::integer, l::integer) which calculates a "transition probability" Wtrans. If Wtrans does not depend on the B[j,l] (e.g. Wtrans:=1/N^2;), everything works nice. However, if I try to use the B[i,k] somehow (e.g. if NKrit_jl > B[j,l] then Wtrans := 0.5 fi;), then error messages from the procedure W result like "Error, invalid terms in product".
## Differentiating with respect to structured-symbols...
by: Maple
When you try to differentiate a Maple expression with respect to one of the constituents (of expression) which is not simple atomic symbol, Maple could not perform operation, e.g. ``` > diff( sin(x(t)), x(t)); Error, invalid input: diff received x(t), which is not valid for its 2nd argument ```. One solution to handle this problem is described in book (Maple book Chap.12)of Walter Ganz (Walter Ganz). In his trick structured symbol is temporarily replaced (via substitution) with a local variable, differentiation is performed with respect to local variable and finally local variable is replaced with original structured symbol.
## Sorting with Indices
by: Maple
Suppose you want to sort a list L ( of numbers ) and also determine order of elements as a list of indices corresponding to elements of original list L, that is, you want such a integer list "I" that `"seq(L[I[j],j=1..nops(L))"` to be equivalent to sorted list. This functionality is present in MATLAB in `"[B,IX] = sort(...)"` syntax and i come up with this problem while trying to convert a MATLAB function (GaussQuadratureWeights) to maple. The procedure described resolves problem using a few MAPLE commands including MAPLE's built-in sort function.
## Extracting an integer coefficient from a term...
What is an efficient, robust, way to extract an integer coefficient from a single term? My first thought was using lcoeff, however, it doesn't work if the term contains constants (say Pi) or floats. Currently I'm using patmatch,
```icoeff := proc(t)
local k,x,kx;
return `if`(patmatch(t, k::'nonunit'(integer)*x::anything,kx)
,eval(k,kx)
,1);
end proc:
map(icoeff, [0, 1, -3.0, -3, -0., 3.0*Pi, 4*Pi, -12/5*I]);
[0, 1, 1, -3, 1, 1, 4, 1]
```
## Can anyone tell me the difference of maple used in...
One of my friends told me that Maple used in Linux(SUSE) is unstable,^_^，what he said may be wrong.I need more information about these,Anyone has even used maple in Linux(SUSE)? Can you tell me the difference of maple using in between Linux(SUSE) and Linux(RedHat)? I want to install maple V10 in SuperComputer Center for parallel computing.
by: Maple
See here. | 1,556 | 5,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-39 | longest | en | 0.863367 |
https://firstrays.com/growing-conditions/artificial-lighting-intensity/ | 1,656,702,652,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103945490.54/warc/CC-MAIN-20220701185955-20220701215955-00368.warc.gz | 301,812,692 | 24,825 | # Artificial Lighting Intensity
When discussing artificial light, the questions “How far from the lights should the plants be?” and “How large of an area can they cover?” come up frequently. Those are, unfortunately, not really simple questions to answer.
First of all, we must remember that light intensity drops off as the inverse square of the distance. In other words, doubling the distance from the light source decreases the light intensity to one-fourth of what it was, and tripling the distance drops it to only one-ninth, and so-on, as depicted below.
As the inverse-square law applies to relative distances and intensities, we could just as easily start our measurement at the arbitrary “3” in that diagram, meaning that the intensity at “9”, being three times the distance, would have 1/9th the intensity as at the “3”. The implication to plant lighting, therefore, is that by using light sources of different intensities, you can match the light levels to your growing area. For example, if light source “A” has the same intensity at one foot distance as light source “B” has at three feet, then the light intensity for “B” at nine feet is equal to that from “A” at three.
Another aspect of the inverse-square law is that it applies to light radiating in all directions, so the points of equal intensity form a sphere with the light source at its center. That is important to us because the floors and table-tops we grow our plants on are not curved like the surface of a sphere, so if our light source is positioned directly over a round table, for example, the light intensity would decrease as you move from the center to the edge of the table.
That is all fine and well, but most bulbs do not emit light from a single point, and it and the diagrams above do not take into account the effect of a reflector.
In the case of a single fluorescent tube, the “point” light source becomes, in essence, a “line” of light, so it takes that circular light distribution and “stretches” it into an elongated shape with rounded ends. One might also argue that the longer length of a fluorescent tube might “reinforce” or “strengthen” the light intensity under the bulb, as the light emitted from the middle of the bulb is reinforced by the light spreading from the points on either side of it, plus that coming from a little farther out, plus a little farther out from that, etc., etc. (We’ll return to this in a moment.)
Reflectors on lights are both limiting and intensifying. Limiting, in that they block the rays from escaping from all directions for the light source. That’s actually not all bad, as often a well-designed reflector cuts off the light that would be of practical use to the grower. Referring back to the diagram above, a decent reflector might limit the rays to the area within the legs of the table.
But a reflector can also be a HUGE addition, as it can reflect and redirect the light that would be lost upward and to the sides, down toward the plants. The graph below shows what would be the calculated inverse-square drop-off in light intensity from a single-bulb 24-watt, two-foot T5 light fixture, plus the actual light meter measurements, that clearly show the effect off the reflector coupled with the “line of light” effect mentioned earlier.
To show that the improvement was not entirely from the reflected light, the graph below shows the same fixture with the reflector removed. The blue line in this graph – “with reflector” – is the same data as the red line above, and it is clear that the “no reflector” curve still shows an improvement over the calculated light intensity curve, above.
We have not addressed the use of multiple bulbs, but that’s simply a multiplier to the information above. | 803 | 3,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-27 | latest | en | 0.953782 |
https://discussions.unity.com/t/force-calculation-formulae/895038 | 1,726,077,193,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00706.warc.gz | 174,851,349 | 7,988 | Force Calculation Formulae
Hello all!
I am trying to figure out how the ground contact forces are calculated when using the wheel collider component. Like I understand that slip is computed followed by a series of calculations. But surprisingly, I couldn’t find a single reference on what’s happen in the engine. Worst part is I don’t have a premium tech support packages and I can’t reach out to the support itself. Does someone knows how and where I can find this information?
A little context - I am a mechanical engineer trying to use unity to simulate robots in environments developed based on high fidelity point cloud and texture data rendered from images. I want to make a realistic simulation and so looking for understanding how unity physics engine is put together. I know that Nvidia PhysX drives unity but is it safe to assume that their source code is exactly how calculations in unity are done? Any inputs on this matter would be highly appreciated, TIA!
1 Like
Hello.
WheelCollider only expose Normal force, in WheelCollider.force.
Longitudinal and Lateral friction forces are not exposed. I also can’t find any thing regarding tire force in PhysX either.
Based on what you said, you don’t need to simulate high speed movement. What you can do is to ignore wheelcollider tire force altogether (set stiffness to 0), and make your own tire friction force using Normal force exposed from wheelcollider, using the formula F = Fz * mu, with mu is the friction coefficient, Fz is the normal force
Most racing game also went this route, using F = Fz * mu until the rigidbody reaches a high enough speed to use slip-based friction
I’m going to reply to the background of this question:
As far as I am aware, Unity 3D physics is PhysX, Unity wraps the PhysX functions in its own calls, so not all parameters are exposed to you and some defaults may be set without you realising (to values that may be hard to find), but the actual physics engine is PhysX.
I’m a mechanical engineer who worked for many years building physics based models and I would not recommend relying on anything in a game engine to model real world engineering problems. The number one requirement of game physics is that it be as efficient as possible; accuracy is not a priority, it only needs to produce a believable approximation.
Here are some trivial examples to think about:
Friction (Unity - Manual: Physic Material asset reference)
“Please note that the friction model used by the Nvidia PhysX engine is tuned for performance and stability of simulation, and does not necessarily present a close approximation of real-world physics. In particular, contact surfaces which are larger than a single point (such as two boxes resting on each other) will be calculated as having two contact points, and will have friction forces twice as big as they would in real world physics. You may want to multiply your friction coefficients by 0.5 to get more realistic results in such a case.”
Integration Scheme (Integration Basics | Gaffer On Games)
PhysX uses semi-implicit Euler integration. You can see we had some fun with this a while back ( Not reaching exact jump height using kinematic equation? (following Seb Lague's kinematic tutorials) ). The positional error will depend on the magnitude of the acceleration, but will of course be cumulative.
Rigidbody Constraints ( Object rotates despite axis is frozen )
These are unreliable at best and would need to be wrapped up in some code to have any confidence in.
You could look at implementing a different physics engine, PyBullet would seem a promising choice for robotics (although I have never used it). You can find an early port here (Bullet Physics For Unity | Physics | Unity Asset Store), but if this is a serious project it would probably be better to do some research and re-port the latest version.
However, I think Unity offers good potential to build a tool for creating a GUI environment and to pre- and post-process data for technical applications. Good luck!
We don’t modify the underlying source apart from a potential fix or compatibility related thing. Most of what you select goes directly through to PhysX and certainly the simulation, when running, is all PhysX.
The same goes for 2D in the case of Box2D. | 868 | 4,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.9341 |
https://nzetc.victoria.ac.nz/tm/scholarly/tei-Stout80a-t10-body-d1-d4.html | 1,718,648,935,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00816.warc.gz | 375,397,293 | 6,328 | # Mortality from Five to Seventy-Five Years of Age
### Mortality from Five to Seventy-Five Years of Age.
After considering the various methods avail-able for passing from the estimated mean population and the recorded deaths to the probability of living at each age, and being desirous of preserving all the well-marked features of the observations, the method employed by Mr. J, Milne in the construction of the well-known Carlisle table, and commonly known as the graphic method, was adopted. A description of the process, illustrated by a diagram of the population curve, is given in his treatise, vol. I., p. 101. In describing it we shall, as far as possible, use Mr. Milne's own words. Taking a sheet of sectional ruled paper we draw thereon a straight line of indefinite length, A 2 as a base for future operations.
A B C D E F G Z
And for the first group of ages assume A B at pleasure; then let B C, C D, D E, &c., be to A B in the same ratio of the 2nd, 3rd, 4th, &c., groups of ages respectively to the first interval A B. At each of the points A, B, C, &c, erect perpendiculars to A Z; then in the perpendicular that passes through B, assume Bbat pleasure, connect with the perpendicular on A and form parallelogram. In those that [unclear: pa] through C, D, E, &c, take the points c, d, e, [unclear: &c] so that the parallelograms Ab, Bc, Cd, &c, [unclear: be] completed, each of the others may be to [unclear: fi] first in the ratio of the numbers living in [unclear: f] corresponding group of ages to the number [unclear: b] the first interval. In this way the area of [unclear: th] parallelograms are made to represent the [unclear: num] her living in each group of ages.
Next let a line (as little curved as the [unclear: other] conditions will admit of) be described [unclear: through] these parallelograms, so that the point [unclear: describ] it, in its motion from the first ordinate Aa, [unclear: m] continually approach towards Z, and may [unclear: ne] change its direction abruptly, so as to form, [unclear: b] angle in its path. (In other words it must [unclear: be] a flowing line). Also, let the line thus [unclear: described] so cut each of the parallelograms, ahove-[unclear: me] tioned, that the area comprehended by the [unclear: bas] the two sides of the parallelogram [unclear: perpendicula] thereto and the portion of the line [unclear: which] intercepted between those sides, may be [unclear: equ] to the area of the same parallelogram. [unclear: Th] is, the curve in cutting the parallelograms [unclear: m] add an area to each equal to the area cut [unclear: of] so that the area of the parallelogram shall [unclear: be] exactly the same as before.
So shall the number of the living in [unclear: an] assigned year of age be to the given [unclear: number] the interval including that year, in the [unclear: rates] the area insisting upon the portion of the [unclear: bea] corresponding with the year assigned, to [unclear: be] area of the parallelogram in which it is [unclear: for] Similar parallelograms were set out for [unclear: f] deaths in the same groups of ages, and a [unclear: cen] drawn through them on the same condition.
We thus get two curves representing [unclear: res] tively the population and the deaths at [unclear: ea] nge, and dividing the deaths by the [unclear: popul] we obtain the function, called by Dr. [unclear: Farr] the "rate of mortality," and by Dr. [unclear: Sprag] the "central death rate," the symbol for [unclear: which] is mx. Then, in accordance with a [unclear: suggestion] by Mr. G. King (Journal Institute [unclear: Actual]vol. xxiv, p. 203), the values of mso [unclear: form] were plotted out on cross-ruled paper, [unclear: aa] carefully adjusted by drawing fresh [unclear: curre] to remove irregularities introduced by [unclear: famil] drawing of the first two curves, but [unclear: with] removing any of the characteristics of the [unclear: table]
Then in order to test the effect of the [unclear: grand] tion, the number living at each age [unclear: war] multiplied by mx, and the result gave the pected deaths. The sum of these should be [unclear: the] total number of deaths, according to the [unclear: org] nal facts. If there is any considerable [unclear: diff] ence the m's must be re-adjusted so [unclear: as] remove it. The results as finally adjusted [unclear: we] arranged in quinquennial groups, and will [unclear: be] found compared with the actual death Table D, hereunder :
Males Females. Ages, Expected. Actual. Expected. 5-9 135.02 135.85 111-29 112.38 [unclear: 5-10] 10-14 77.81 76.92 77.07 75.08 [unclear: 10-15] 16-19 105.82 105.62 98.50 102.08 [unclear: 15-20] 20-24 134.04 135.77 120.74 150.09 [unclear: 20-25] 26-29 130.60 129.31 117.27 112.62 [unclear: 25-30] 30-34 135.09 136.02 112.45 107.85 [unclear: 30-35] 35-30 150.44 149.92 111.09 115.31 [unclear: 35-40] 40-44 175.67 175.38 104.07 103.46 [unclear: 40-45] 45-49 196.13 l96.00 93.18 89.46 [unclear: 45-50] 50-54 204.67 204.54 87.04 90.15 [unclear: 50-55] 55-59 173.05 174.15 74 07 75.92 [unclear: 55-60] 60-64 177.44 171.08 79.18 77.62 [unclear: 60-65] 65-69 134.32 133.15 72.93 73.62 [unclear: 65-70] 70-74 1114.07 111.31 72.93 72.23 [unclear: 70-75] Totals. 2040.97 2040.92 1332.16 1323.47 Total
page 7
From the m's we derive px, the probability of living a year by the formula.
and then starting from an arbitrary radix (in the present case 100,000 for each sex) by continuous multiplication we construct the column lxin each table, and this gives the number living at each age. | 1,694 | 5,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.939625 |
https://www.airmilescalculator.com/distance/skg-to-jsh/ | 1,718,525,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00013.warc.gz | 586,448,297 | 35,323 | # How far is Sitia from Thessaloniki?
The distance between Thessaloniki (Thessaloniki Airport) and Sitia (Sitia Public Airport) is 404 miles / 650 kilometers / 351 nautical miles.
The driving distance from Thessaloniki (SKG) to Sitia (JSH) is 609 miles / 980 kilometers, and travel time by car is about 16 hours 40 minutes.
404
Miles
650
Kilometers
351
Nautical miles
1 h 15 min
## Distance from Thessaloniki to Sitia
There are several ways to calculate the distance from Thessaloniki to Sitia. Here are two standard methods:
Vincenty's formula (applied above)
• 403.767 miles
• 649.800 kilometers
• 350.864 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 404.195 miles
• 650.489 kilometers
• 351.236 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Thessaloniki to Sitia?
The estimated flight time from Thessaloniki Airport to Sitia Public Airport is 1 hour and 15 minutes.
## Flight carbon footprint between Thessaloniki Airport (SKG) and Sitia Public Airport (JSH)
On average, flying from Thessaloniki to Sitia generates about 84 kg of CO2 per passenger, and 84 kilograms equals 186 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Thessaloniki to Sitia
See the map of the shortest flight path between Thessaloniki Airport (SKG) and Sitia Public Airport (JSH).
## Airport information
Origin Thessaloniki Airport
City: Thessaloniki
Country: Greece
IATA Code: SKG
ICAO Code: LGTS
Coordinates: 40°31′10″N, 22°58′15″E
Destination Sitia Public Airport
City: Sitia
Country: Greece
IATA Code: JSH
ICAO Code: LGST
Coordinates: 35°12′57″N, 26°6′4″E | 526 | 1,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.791631 |
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# Get customized homework help now!
Skewness
Skewness is defined as asymmetry in the distribution of the sample data values. Values on one side of the distribution tend to be further from the 'middle' than values on the other side. For skewed data, the usual measures of location will give different values, for example, mode < median < mean would indicate positive (or right) skewness.
Positive (or right) skewness is more common than negative (or left) skewness.
For univariate data Y1, Y2, ..., YN, the formula for skewness is:
Where is the mean, is the standard deviation, and N is the number of data points. The skewness for a normal distribution is zero, and any symmetric data should have skewness near zero. Negative values for the skewness indicate data that are skewed left and positive values for the skewness indicate data that are skewed right. By skewed left, we mean that the left tail is long relative to the right tail. Similarly, skewed right means that the right tail is long relative to the left tail. Some measurements have a lower bound and are skewed right. For example, in reliability studies, failure times cannot be negative
Properties of skewness
Skewness can be infinite, as when
Pr [ X > x ] = x -3 for x > 1 , Pr [ X < 1 ] = 0
or undefined, as when
Pr [ X < x ] = (1 - x) -3/ 2 for negative x
Pr [ X > x ] = (1 + x) -3/ 2 for positive x.
In this latter example, the third cumulant is undefined. One can also have distributions such as
Pr [ X > x ] = x -2 for x > 1 , Pr [ X < 1 ] = 0
where both the second and third cumulants are infinite, so the skewness is again undefined. If Y is the sum of n indepent random variables, all with the same distribution as X, then the third cumulant of Y is n times that of X and the second cumulant of Y is n times that of X
so Skew [ Y ] = Skew [ X ] / √n . This shows that the skewness of the sum is smaller, as it approaches a Gaussian distribution in accordance with the central limit theorem.
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https://bookdown.org/jkang37/stochastic-process-lecture-notes/lecture02.html | 1,623,652,000,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611445.13/warc/CC-MAIN-20210614043833-20210614073833-00168.warc.gz | 154,539,864 | 7,039 | # Chapter 2 Stationarity, Spectral Theorem, Ergodic Theorem(Lecture on 01/07/2021)
Definition 2.1 (Uncorrelated increaments) We say that a stochastic process $$X$$ has uncorrelated (orthogonal) increaments if for any $$t_i<t_j<t_k<t_l\in T$$, $$Cov(t_j-t_i,t_l-t_k)=0$$.
Definition 2.2 (Cross-covariance Function) A useful function for the study of coevolution of two stochastic processes, say $$X$$ and $$Y$$, defined on the same probability space and with the same index set, is the cross-covariance function, defined as $$$C_{X,Y}(t_i,t_j)=Cov(X_{t_i},Y_{t_j})=E(X_{t_i}Y_{t_j})-E(X_{t_i})E(Y_{t_j}) \tag{2.1}$$$ for $$t_i,t_j\in T$$. The cross-covariance function measures how correlated are the two processes.
If $$X$$ and $$Y$$ are completely uncorrelated, then $$C_{X,Y}(t_i,t_j)=0,\forall t_i,t_j\in T$$.
Why is f.d.d.s. important?
One of the major usage of the stochastic process is to serve as the prior for the unknown function $$f(\mathbf{x})$$ in a regression problem. In real application, we only have finite number of data. Therefore, what we really need is a prior on the finite dimensional random vector $$(f(\mathbf{x}_1),\cdots,f(\mathbf{x}_n))$$, which is just the f.d.d.s. of the stochastic process.
Definition 2.3 (Strongly Stationarity) A stochastic process $$X$$ is called strong stationary if the f.d.d.s. are invariant under index shift. That is, for any (finite) $$n$$, any $$t_0$$ and for all $$t_1,\cdots,t_n\in T$$, $$(X_{t_1},\cdots,X_{t_n})$$ has the same distribution as $$(X_{t_1+t_0},\cdots,X_{t_n+t_0})$$.
Definition 2.4 (Weakly Stationarity) A stochastic process $$X$$ is called weak stationary if its mean function is constant and its covariance function is invariant under index shift. That is, $$\forall t\in T$$, $$E(X_t)=\mu$$ and for all $$t_i,t_j\in T$$, $$Cov(X_{t_i},X_{t_j})=c(t_i-t_j)$$. It is a function of $$t_i-t_j$$ only.
1. Stationarity is a simplification on f.d.d.s.. The theory and method for a stochastic process are considerably simplified under the assumption of (either strong or weak) stationarity, which imposes certain structures on the finite dimensional distributions.
2. Strong stationarity implies weak stationarity. The converse is not true in general. While for the Gaussian process, the opposite direction is also true. For Gaussian process the f.d.d.s. are all multivariate normal and we can characterize f.d.d.s. completely by the mean function and covariance function.
From the theory of Fourier analysis, any function $$f:\mathbb{R}\to\mathbb{R}$$ with certain properties (including periodicity and continuity) has a unique Fourier representation. $$$f(x)=0.5a_0+\sum_{n=1}^{\infty}(a_n\cos(nx)+b_n\sin(nx)) \tag{2.2}$$$ which express $$f$$ as a sum of varying proportions of regular oscillation.
Fourier representation transforms the uncountably infinite dimensional problem of estimating unknown function $$f(x)$$ to a countably infinite dimensional problem of estimating $$a_n$$ and $$b_n$$.
We will express the covariance function of a stationary process using the Fourier transformation. Before doing that, let us fix some notations. By weak stationarity we have $$E(X_t)=\mu,\forall t$$ and $$Cov(X_{t_i},X_{t_j})=c(t)$$, where $$t=|t_i-t_j|$$. Then $$Var(X_{t_i})=c(0)=\sigma^2$$ for all $$t\in\mathbb{R}$$. With out loss of generality, we assume $$\mu=0$$ and $$\sigma^2=1$$. Under this assumption, the autocorrelation function and the autocovariance function coincide, i.e. $$Corr(X_{t_i},X_{t_j})=c(t)$$.
We are going to express (using the idea of Fourier transformation) the covariance function as a characteristic function of some random variable. That is $$$c(t)=\int \exp(itx)dF(x)=\int \exp(itx)f(x)dx=E(\exp(itx)) \tag{2.3}$$$ for some distribution function $$F$$. This distribution is called the spectral distribution of the stochastic process. If $$F$$ has a density, then that is called the spectral density.
The spectral distribution gives us another paradigm of studying stochastic process. Instead of following the usually paradigm to study the mean and covariance function, one can also go to the spectral domain and model the spectral density by density estimation methods. Sometimes working with spectral density is much more easier (especially in time series analysis), but when the dimension is higher, working in complex domain becomes harder.
Theorem 2.1 (Bochner Theorem in Stochastic Process) A function $$\phi$$ is a characteristic function of a random variable if
1. $$\phi(0)=1$$, $$|\phi(t)|\leq 1$$ for all $$t$$.
2. $$\phi$$ is uniformly continuous on $$\mathbb{R}$$.
3. $$\phi$$ is a non-negative function. i.e. for all $$t_1,\cdots,t_n$$, and constant $$z_1,\cdots,z_n$$, $$\sum_{i=1}^k\sum_{j=1}^kz_iz_j\phi(t_i-t_j)\geq 0$$.
Thus, only (2) needs to be assumed for $$c(t)$$ to have a spectral density. From (2.3), using the result from Fourier analysis, we can express the spectral density $$f$$ as $$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(-itx)c(t)dt \tag{2.4}$$$
The spectral density and the covariance function has a one-to-one correspondence.
The above representation has been mainly defined when $$T=\mathbb{R}$$. There are some special case when the index is not $$\mathbb{R}$$.
1. If the index point is multivariate, i.e. $$T=\mathbb{R}^d$$, then $$$c(\mathbf{t})=\int_{x_1}\cdots\int_{x_t}\exp(-i\mathbf{t}\mathbf{x})f(\mathbf{x})d\mathbf{x} \tag{2.5}$$$
If the indexing set of the stochastic process is $$\mathbb{Z}$$, i.e. $$X=\{X(n,\omega):n\in\mathbb{Z},\omega\in\Omega\}$$. In that case the above representation is not unique since $$\exp(inx)=\exp(i(n+2\pi)x)$$ when $$x$$ is an integer. However, we can constrain it on the region $$(-\pi,\pi)$$ and define $$$c(n)=\int_{-\pi}^{\pi}\exp(-inu)dF(u)\Longrightarrow f(u)=\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\exp(-inu)c(u), \tag{2.6}$$$ for $$u\in(-\pi,\pi)$$. Then the correspondence is still one-to-one.
When we are concerned about a stochastic process with a discrete index set, that is $$X=\{X(n,\omega):n\in\mathbb{Z},\omega\in\Omega\}$$. We are interested in $$S_n=\sum_{j=1}^nX_j$$. When $$n$$ is large and $$X_j$$ are i.i.d. it is simple to study $$S_n$$, we then have CLT or LLN to give the asymptotic behavior of $$S_n$$. However, for a stochastic process, these assumptions do not hold. Still we have some results for stochastic process, these are known as the ergodicity theorem for stochastic process.
Theorem 2.2 (Ergodic Theorem for Weakly Stationary Stochastic Process) If $$X=\{X(n,\omega):n\in\mathbb{Z},\omega\in\Omega\}$$ is a weakly stationary process. Then there exist a random variable $$Y$$ such that $$E(y)=E(X_1)$$ and $$\frac{S_n}{n}\to Y$$ in mean square, meaning that $$\lim_{n\to\infty}E[(\frac{S_n}{n}-Y)^2]=0$$
Theorem 2.3 (Ergodic Theorem for Strongly Stationary Stochastic Process) If $$X=\{X(n,\omega):n\in\mathbb{Z},\omega\in\Omega\}$$ is a strongly stationary process such that $$E(|X_1|)<\infty$$. Then there exist a random variable $$Y$$ such that $$E(y)=E(X_1)$$ and $$\frac{S_n}{n}\to Y$$ almost surely and in mean square, meaning that $$\lim_{n\to\infty}P[|\frac{S_n}{n}-Y|>\epsilon]=0$$ for any $$\epsilon>0$$.
Definition 2.5 (Brownian Motion) A stochastic process $$B=\{B(t,\omega):t\geq 0,\omega\in\Omega\}$$ is called a Brownian motion, if it satisfies
1. $$B_0=0$$;
2. $$B_t-B_s\sim N(0,t-s)$$, $$\forall t>s\geq 0$$.
3. $$B_t-B_s$$ is independent of $$B_s$$.
4. The function $$t\to B_t$$ is continuous.
Brownian motion is not a stationary stochastic process, because from (2) and (3) in Definition 2.5, $$Cov(B_t,B_s)=\min(t,s)$$ | 2,320 | 7,591 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 6, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-25 | latest | en | 0.81511 |
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72 + 73 = 49 + 343= 392 What do you do to the exponents when the base is the same and your x ? You add the exponents 83 x 84 = = 87 What is Bedmas ? Brackets , exponents , divide , multiply , add , subtract. When you are divide exponents with the same base you …….? Subtract the exponents. 0.25 ÷ 0.24 = = 0.29 What do you do when you add exponents with the same base you…? You do the exponents then add the base numbers. 74 – 73 = 2401 – 343 = 2058 How do you change per cent to decimal ? You move the decimal two places to the right. 88% to 0.88 | 316 | 1,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-26 | latest | en | 0.677101 |
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Best of Donald Catlin
# Gaming Guru
5 July 2003
I'm sure you've all heard that old admonition "Be careful what you say, your words may come back to haunt you." Well, my words did come back to haunt me.
Back in March of 2001 I wrote an article entitled Four Card Poker? I Don't Think So. In that article I argued that the frequencies of four-card Poker hands are such that I thought players would be uncomfortable with the natural hand rankings. In case you missed, the article here is a table showing the frequencies of four card Poker hands.
Hand Frequency Royal Flush 4 Four of a Kind 13 Straight Flush 40 Three of a Kind 2,495 Straight 2,772 Two Pair 2,808 Flush 2,816 Pair 82,368 High Card 177,408 Total - 270,725
Figure 1
Four Card Poker Frequencies
You see that the usual Poker hierarchy is drastically changed when four-card hands are used. Three of a Kind beats a Straight and Two Pair beats a Flush. What is more, the frequencies for Flush through Three of a Kind are very close. It was my contention that Poker players would be uncomfortable with these rankings. I still believe that. Nevertheless, in the past couple of months two different four-card Poker games have come to my attention, one of which, I understand, has been playing for over a year now. So do I have to eat crow on this?
Well, I came close to eating crow but fortunately I did mention in my article that one could produce different numbers by having the player choose the best four-card hand from 5, 6, or 7 cards (although I tempered the remark by indicating that a player would not be happy turning a Full House into Two Pair). So, crow is not on my bill of fare since both of the aforementioned games deal five-card hands and have the player (or dealer) select the best four-card Poker hand from the five. Let's see what this does to the frequencies.
How many Straight Flushes (including Royals) are there? Well there are four suits and the straights in each suit occur as A-J down to 4-A so there are eleven of them. The A-J can be paired with any of the 48 remaining cards; the other 10 can be paired with only 47 of the remaining 48 since, for example, putting a suited J with a 10-7 Straight Flush would produce a J-8 Straight Flush rather than the desired 10-7. Hence there are 4 x 48 + 4 x 10 x 47 or 2072 Straight Flushes. There are 13 four-card Four of a Kinds and any one of the remaining 48 cards can be paired with each to make a five-card hand, so there are 13 x 48 or 624 of these. There are 13 choices of ranks for a Three of a Kind and for each such choice there are four ways to pick the three from the four. Picking 2 of the remaining 48 (1128 ways) we have the number of five-card hands containing a Three of a Kind is 13 x 4 x 1128 or 58,656. Now here is where things get interesting. The two games mentioned above rank hands as follows:
Game #1: Straight Flush Four of a Kind Flush Straight Three of a Kind Two Pair One Pair High Card Game #2: Four of a Kind Straight Flush Three of a Kind Flush Straight Two Pair One Pair High Card
Clearly the inventor of Game #1 wanted to keep the hand rankings the same as they are in regular Poker even though, as we will see, the natural rankings by frequency are different. Game #2, on the other hand, has the top three hands in the correct order according to frequency. Both games rank the Flush above the Straight. Is this correct? Well, note first that if the Flush is ranked above the Straight and we are faced with a situation wherein our five-card hand contains hands of both types, we should opt to pick the Flush rather than the Straight. The calculation is a bit tricky so I'm going to skip it - write to me if you would like details. It turns out that with this strategy there are 116,688 five-card hands that contain four-card Flushes. Of these 2,072 are Straight Flushes so subtracting these we end up with 114,616 five-card hands that contain ordinary Flushes. This leaves 101,808 five-card hands that contain ordinary Straights. So both of the above rankings appear to be in the wrong order. Here are all of the frequencies with the Flush listed above the Straight:
Type Hand Frequency Four of a Kind 624 Straight Flush 2,072 Three of a Kind 58,656 Flush 114,616 Straight 101,808 Two Pair 123,552 One Pair 1,047,552 High Card 1,150,080 Total - 2,598,960
Figure 2
Hand Frequencies when Flush > Straight
Okay, so it looks like we have to rank the Straight higher than the Flush. This means that when faced with a five-card hand that contains both a four-card Flush and a four-card Straight that we should pick the straight rather than the Flush. The following table shows what happens when we do this:
Type Hand Frequency Four of a Kind 624 Straight Flush 2072 Three of a Kind 58,656 Straight 110,464 Flush 105,960 Two Pair 123,552 One Pair 1,047,552 High Card 1,150,080 Total - 2,598,960
Figure 3
Hand Frequencies when Straight > Flush
Oh no! Now there are fewer Flushes than Straights. So there you have it. When four-card Poker hands are selected from fiver- card hands, it is impossible to rank the Straights and Flushes in an order than reflects their natural frequencies. Fascinating!
As a practical matter there is no harm in this paradoxical situation, however, I think that this same phenomenon will occur in other games wherein a hand is selected as being the best hand chosen from a larger hand. There may be some surprises laying in wait there and gaming developers should keep this in mind. See you next month.
Four Card Poker Paradox is republished from Online.CasinoCity.com.
Articles in this Series
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Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.
#### Books by Donald Catlin:
Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.
#### Books by Donald Catlin:
Lottery Book: The Truth Behind the Numbers | 1,524 | 6,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-04 | latest | en | 0.96971 |
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# What is 4c equals 3d plus 3 and c equals d-1?
Updated: 9/19/2023
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c = 6 and d = 7
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千次阅读
2020-05-21 19:19:56
假设原本数据集是如下的 csv 格式，行代表时间，列数代表变量数。
用它来构造机器学习的数据集，也就是有监督标签的样本。
# 代码一
import torch
import torch.utils.data
import os
import numpy as np
import pandas as pd
class MTSDataset(torch.utils.data.Dataset):
"""Multi-variate Time-Series Dataset for *.txt file
Returns:
[sample, label]
"""
def __init__(self,
window,
horizon,
data_name='electricity',
set_type='train', # 'train'/'validation'/'test'
data_dir='./data'):
assert type(set_type) == type('str')
self.window = window
self.horizon = horizon
self.data_dir = data_dir
self.set_type = set_type
file_path = os.path.join(data_dir, data_name, '{}_{}.txt'.format(data_name, set_type))
self.len, self.var_num = rawdata.shape
self.sample_num = max(self.len - self.window - self.horizon + 1, 0)
self.samples, self.labels = self.__getsamples(rawdata)
def __getsamples(self, data):
X = torch.zeros((self.sample_num, self.window, self.var_num))
Y = torch.zeros((self.sample_num, 1, self.var_num))
for i in range(self.sample_num):
start = i
end = i + self.window
X[i, :, :] = torch.from_numpy(data[start:end, :])
Y[i, :, :] = torch.from_numpy(data[end+self.horizon-1, :])
return (X, Y)
def __len__(self):
return self.sample_num
def __getitem__(self, idx):
sample = [self.samples[idx, :, :], self.labels[idx, :, :]]
return sample
dataset = MTSDataset(
window=16,
horizon=3,
data_name='electricity',
set_type='train',
data_dir='.'
)
i = 0
sample = dataset[i]
print(sample[0].shape) # torch.Size([16, 321])
print(sample[1].shape) # torch.Size([1, 321])
# 代码二
import torch
import numpy as np
def normal_std(x):
return x.std() * np.sqrt((len(x) - 1.)/(len(x))) # 样本标准差转化成总体标准差
class Data_utility(object):
# train and valid is the ratio of training set and validation set. test = 1 - train - valid
def __init__(self, file_name, train, valid, cuda, horizon, window, normalize = 2):
self.cuda = cuda;
self.P = window;
self.h = horizon
fin = open(file_name);
self.dat = np.zeros(self.rawdat.shape);
self.n, self.m = self.dat.shape;
self.normalize = 2
self.scale = np.ones(self.m) # 维数等于数据列数
self._normalized(normalize) # 当 normalize = 2, 修改 scale 为每一列的最大绝对值, 使每列数据绝对值最大值为 1
self._split(int(train * self.n), int((train+valid) * self.n), self.n)
self.scale = torch.from_numpy(self.scale).float()
tmp = self.test[1] * self.scale.expand(self.test[1].size(0), self.m) # shape: [batch, n_var]
if self.cuda:
self.scale = self.scale.cuda();
self.scale = Variable(self.scale);
self.rse = normal_std(tmp);
self.rae = torch.mean(torch.abs(tmp - torch.mean(tmp)));
def _normalized(self, normalize):
if (normalize == 0):
self.dat = self.rawdat
#normalized by the maximum value of entire matrix.
if (normalize == 1):
self.dat = self.rawdat / np.max(self.rawdat);
#normlized by the maximum value of each column(sensor).
if (normalize == 2):
for i in range(self.m):
self.scale[i] = np.max(np.abs(self.rawdat[:,i]))
self.dat[:,i] = self.rawdat[:,i] / self.scale[i]
def _split(self, train, valid, test):
train_set = range(self.P+self.h-1, train);
valid_set = range(train, valid)
test_set = range(valid, self.n)
self.train = self._batchify(train_set, self.h) # [batch, p, n_multiv], [batch, n_multiv]
self.valid = self._batchify(valid_set, self.h)
self.test = self._batchify(test_set, self.h)
def _batchify(self, idx_set, horizon):
n = len(idx_set);
X = torch.zeros((n,self.P,self.m));
Y = torch.zeros((n,self.m));
for i in range(n):
end = idx_set[i] - self.h + 1;
start = end - self.P;
X[i,:,:] = torch.from_numpy(self.dat[start:end, :]);
Y[i,:] = torch.from_numpy(self.dat[idx_set[i], :]);
return [X, Y];
def get_batches(self, inputs, targets, batch_size, shuffle=True):
length = len(inputs)
if shuffle:
index = torch.randperm(length)
else:
index = torch.LongTensor(range(length))
start_idx = 0
while (start_idx < length):
end_idx = min(length, start_idx + batch_size)
excerpt = index[start_idx:end_idx]
X = inputs[excerpt]; Y = targets[excerpt];
if (self.cuda):
X = X.cuda();
Y = Y.cuda();
yield Variable(X), Variable(Y);
start_idx += batch_size
data = Data_utility(file_name='solar_AL.txt', train=0.6, valid=0.2, cuda=False, horizon=12, window=24, normalize = 2)
print(data.rawdat.shape) # (52560, 137)
print(data.train[0].shape, data.train[1].shape) # torch.Size([31501, 24, 137]) torch.Size([31501, 137])
def train(data, X, Y, model, criterion, optim, batch_size):
model.train();
total_loss = 0;
n_samples = 0;
for X, Y in data.get_batches(X, Y, batch_size, True): # <<< attention!
output = model(X);
scale = data.scale.expand(output.size(0), data.m) # <<< attention!
loss = criterion(output * scale, Y * scale);
loss.backward();
total_loss += loss.item();
n_samples += (output.size(0) * data.m);
# 参考文献
1. DSANet: Dual Self-Attention Network for Multivariate Time Series Forecasting
2. Modeling Long- and Short-Term Temporal Patterns with Deep Neural Networks
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ts_0.
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展开全文
• 数据是驱动科技发展的源泉，我们平常科研中也常常需要在各种开源数据上验证自己模型的效果。时间序列目前可以使用的开源数据集有哪些呢？本期为大家做一次梳理。
关注微信公众号“时序人”获取更好的阅读体验
#### 时间序列学术前沿
数据是驱动科技发展的源泉，我们平常科研中也常常需要在各种开源数据上验证自己模型的效果。时间序列目前可以使用的开源数据集有哪些呢？本期为大家做一次梳理。
### UCR Time Series
UCR时间序列数据集是时序领域的“Imagnet”，涵盖医疗/电力/地理 等诸多领域，目前全量数据有128种。涉及时间序列预测，回归，聚类等诸多任务，可以说是发Paper必跑数据集，由加州大学河滨分校计算机系的教授 Eamonn Keogh 所在的课题组维护
http://www.cs.ucr.edu/~eamonn/time_series_data/
### FigShare
这是一个研究成果共享平台，这里向全世界开放免费的研究成果及科学数据。
https://figshare.com/
### Awesome Public Datasets
该项目提供了一个非常全面的数据获取渠道，包含各个细分领域的数据库资源，自然科学和社会科学的覆盖都很全面，适合做研究和数据分析的人员。
https://github.com/awesomedata/awesome-public-datasets
### 服务监控数据集
该数据集是由人工神经网络公司Numenta所提供的，包含互联网服务场景下的各种流式数据与评测脚本。 NAB是用于评估数据流实时应用中异常检测算法的新颖基准，它由50多个带有标签的真实世界和人工时间序列数据文件以及为实时应用程序设计的新颖评分机制组成。
https://github.com/numenta/NAB
### 音乐数据库
这是数据集包含了海量的公开音乐数据库，适用于包含音乐推荐、分类在内的各种任务
http://millionsongdataset.com/
### 国家经济数据
国家统计局经常会统计涉及经济民生等多个方面的指标，提供了非常丰富的开源时间序列数据。这里简单为大家列举一些可以获取这些数据的渠道：
##### 国家数据
数据来源中华人民共和国国家统计局，包含了我国经济民生等多个方面的数据，并且在月度、季度、年度都有覆盖，较为全面。
https://data.stats.gov.cn/
##### CEIC
涵盖超过195个国家400多万个时间序列的数据源，最完整的一套超过128个国家的经济数据，能够精确查找GDP、CPI、进口、出口、外资直接投资、零售、销售以及国际利率等深度数据。
https://www.ceicdata.com/zh-hans
##### 万得
被誉为中国的Bloomberg，在金融业有着全面的数据覆盖，金融数据的类目更新非常快，据说很受国内的商业分析者和投资人的亲睐。
https://www.wind.com.cn/
##### 中国统计信息网
国家统计局的官方网站，汇集了全国各级政府各年度的国民经济和社会发展统计信息，建立了以统计公报为主，统计年鉴、阶段发展数据、统计分析、经济新闻、主要统计指标排行等。
http://www.tjcn.org/
### 数据竞赛网站
除了上面整理的常用的开源时间序列数据以外，如果我们想获取针对特定任务的时间序列数据，我们常常可以通过各种竞赛平台获取数据。这些数据集通常干净且科研性非常高。包括：
更多原创内容与系列分享，欢迎关注微信公众号“时序人”获取。
展开全文
• 截止到目前最新整理的的UCR数据集（UCRArchive）。这是以一个完整的压缩包，共有129个数据集。压缩包没有解压码。
• 非线性时间序列 建模 预测>>书的数据集以及代码
• 通过数学模型介绍了，非常热门，应用广泛，高大上的一些机器学习模型
• 第1章 数据集 1.1 数据集概述 1.2 从文件中读取数据 1.3 提取有效数据 第2章 全连接网络的拟合 2.1 定义全连接网络 2.2 定义loss与优化器 2.3 训练前准备 2.4 开始训练 2.5 loss显示 2.6 结果分析 第3章...
目录
第1章 概述
1.1 业务需求概述
1.2 业务分析
1.3 导入库
第1章 数据集
1.1 数据集概述
1.2 从文件中读取数据
1.3 提取有效数据
第2章 全连接网络的拟合
2.1 定义全连接网络
2.2 定义loss与优化器
2.3 训练前准备
2.4 开始训练
2.5 loss显示
2.6 结果分析
第3章 RNN网络的拟合
3.1 定义RNN网络
3.2 生成网络
3.3 定义loss与优化器
3.4 训练前的准备
3.5 开始训练
3.6 显示loss
3.7 分析
## 第1章 概述
### 1.1 业务需求概述
有一批最近52周的销售数据，希望能够根据这些数据，预测未来几周销售数据。
### 1.2 业务分析
这是一个线性拟合的问题，那么每周的数据之间是独立的？还是每周的数据之间有时间的相关性？
本文就采用层数的全连接网络和RNN网络对数据进行拟合与比较。
如果全连接网络的拟合度比RNN网络好，则证明每周的数据是独立无关的。
如果全连接网络的拟合度没有RNN网络好，则证明每周的数据是具有时间相关性。
### 1.3 导入库
%matplotlib inline
import torch
import torch.nn as nn
from torch.nn import functional as F
from torch import optim
import numpy as np
import math, random
import matplotlib.pyplot as plt
import pandas as pd
from torch.utils.data import Dataset
## 第1章 数据集
### 1.1 数据集概述
销售数据是通过Excel存放的：包含多个样本数据，每个样本包含52周的数据，如下图所示：
横轴：不同的样本
纵轴：每个样本包含的序列52周的销售数据。
### 1.2 从文件中读取数据
# 从excel文件中数据
sales_data.head()
### 1.3 提取有效数据
# 提取有效数据
source_data = sales_data
source_data.values.shape
source_data.values
array([[0.00000000e+00, 2.29157429e+00, 2.88283833e+00, ...,
2.52078778e+00, 2.63721001e+00, 2.74892268e+00],
[1.00000000e+00, 6.09664112e-01, 2.52360427e+00, ...,
1.66285479e+00, 2.48192522e+00, 2.41120153e+00],
[2.00000000e+00, 3.42279275e+00, 1.81265116e+00, ...,
1.72005177e+00, 2.16993942e+00, 2.16170153e+00],
...,
[1.50490000e+04, 3.83370716e+00, 3.54807088e+00, ...,
3.03134190e+00, 1.85208859e+00, 3.00563077e+00],
[1.50500000e+04, 3.18643937e+00, 1.70323164e+00, ...,
1.19605309e+00, 3.34099104e+00, 2.92155740e+00],
[1.50510000e+04, 3.10427003e+00, 2.82717086e+00, ...,
1.33781619e+00, 2.71468770e+00, 2.34349413e+00]])
# 显示数据集数据
source_data
Unnamed: 0week_1week_2week_3week_4week_5week_6week_7week_8week_9...week_43week_44week_45week_46week_47week_48week_49week_50week_51week_52
002.2915742.8828383.5818054.0138751.1099242.8342472.6250452.1968844.208887...1.2422502.7710723.7781433.1616343.7926452.0154812.8261842.5207882.6372102.748923
110.6096642.5236042.6095373.6959973.3992722.6166121.6519351.3482953.862523...2.4775052.2845133.4617833.1010523.4424201.9157762.4261681.6628552.4819252.411202
223.4227931.812651-0.0979661.7240643.0936601.7812782.8499462.9497073.560835...2.2196672.3598341.9320292.7319471.8362451.2199331.2227401.7200522.1699392.161702
333.3722832.5772543.8543083.4496792.5176762.6356792.3527062.8562161.948434...1.6709120.9091133.2166521.7753463.2704842.3997093.0320711.7036661.7505851.821736
442.9977271.7994153.6480902.3915672.3767781.8647170.4080622.3467263.260303...2.3040583.0389052.0387332.8251862.2329372.5090502.881796-1.7121312.1403661.926818
..................................................................
15047150471.9527261.5476371.9027890.8132612.6009792.9106382.8783960.5942163.526187...0.4860733.3362523.3072703.0268351.4721163.2207922.6640441.5461533.0269482.611774
15048150480.6528243.5891913.2577072.8212762.1859372.534801-0.7743753.8356953.776809...2.6841531.3849123.1845702.8329412.0920332.6061980.7531933.1605993.0858002.814394
15049150493.8337073.5480713.1941601.9934373.0135471.8250472.3051960.5224753.126647...2.9289371.6337543.0785982.4665630.4893803.5187253.4064663.0313421.8520893.005631
15050150503.1864391.7032323.1965912.407803-0.4743703.8799433.7624083.4156691.790529...1.6426122.3877702.1498930.6884633.1508053.2422092.9727281.1960533.3409912.921557
15051150513.1042702.8271711.1764283.2749672.9346552.3491882.4129351.2190542.843297...2.5512590.8192033.4165323.2420212.8101772.0374872.6816941.3378162.7146882.343494
15052 rows × 53 columns
## 第2章 全连接网络的拟合
### 2.1 定义全连接网络
class FullyConnected(nn.Module):
def __init__(self, x_size, hidden_size, output_size):
super(FullyConnected, self).__init__()
self.hidden_size = hidden_size
self.linear_with_tanh = nn.Sequential(
nn.Linear(10, self.hidden_size),
nn.Tanh(),
nn.Linear(self.hidden_size, self.hidden_size),
nn.Tanh(),
nn.Linear(self.hidden_size, output_size)
)
def forward(self, x):
yhat = self.linear_with_tanh(x)
return yhat
# 输入样本的size
x_size = 1
# 隐藏层的size：try to change this parameters
hidden_size = 2
#数据特征的size
output_size = 10
fc_model = FullyConnected(x_size = x_size, hidden_size = hidden_size, output_size = output_size)
print(fc_model)
fc_model = fc_model.double()
print(fc_model)
FullyConnected(
(linear_with_tanh): Sequential(
(0): Linear(in_features=10, out_features=2, bias=True)
(1): Tanh()
(2): Linear(in_features=2, out_features=2, bias=True)
(3): Tanh()
(4): Linear(in_features=2, out_features=10, bias=True)
)
)
fc_model.state_dict()['linear_with_tanh.0.weight']
tensor([[ 0.2028, 0.0721, 0.0101, 0.2866, -0.3106, 0.1612, 0.0082, -0.1394,
0.2630, -0.2903],
[ 0.0596, -0.2646, 0.2036, -0.2676, -0.0198, 0.2969, 0.2596, 0.0965,
0.0746, -0.0195]], dtype=torch.float64)
### 2.2 定义loss与优化器
#定义loss函数
criterion = nn.MSELoss()
# 使用 Adam 优化器 比课上使用的 SGD 优化器更加稳定
optimizer = optim.AdamW(fc_model.parameters(), lr=0.01)
### 2.3 训练前准备
n_epochs = 30
n_layers = 1
batch_size = seq_length
seq_length = 10
fc_losses = np.zeros(n_epochs)
print(data_loader.batch_size )
### 2.4 开始训练
# 开始训练
for epoch in range(n_epochs):
# 定义用于每个epoch的平均loss
epoch_losses = []
# 读取batch数据
#当读到数据不足batch size时，跳过batch size
if data.shape[0] != batch_size:
continue
#随机的获取长度=seq_length的数据
random_index = random.randint(0, data.shape[-1] - seq_length - 1)
train_x = data[:, random_index: random_index+seq_length]
#train_y = data[:, random_index + 1: random_index + seq_length + 1]
train_y = train_x
#进行前向运算
outputs = fc_model(train_x.double())
#复位梯度
#求此次前向运算的loss
loss = criterion(outputs, train_y)
#反向求导
loss.backward()
#梯度迭代
optimizer.step()
#
epoch_losses.append(loss.detach())
if iter_ == 0:
plt.clf();
plt.ion()
plt.title("Epoch {}, iter {}".format(epoch, iter_))
plt.plot(torch.flatten(train_y),'c-',linewidth=1,label='Label')
#plt.plot(torch.flatten(train_x),'g-',linewidth=1,label='Input')
plt.plot(torch.flatten(outputs.detach()),'r-',linewidth=1,label='Output')
plt.draw();
plt.pause(0.05);
fc_losses[epoch] = np.mean(epoch_losses)
................................................................
### 2.5 loss显示
plt.plot(fc_losses)
### 2.6 结果分析
全连接网络应对这种时序序列问题，拟合性不是很好。
## 第3章 RNN网络的拟合
### 3.1 定义RNN网络
class SimpleRNN(nn.Module):
def __init__(self, x_size, hidden_size, n_layers, batch_size, output_size):
super(SimpleRNN, self).__init__()
self.hidden_size = hidden_size
self.n_layers = n_layers
self.batch_size = batch_size
#self.inp = nn.Linear(1, hidden_size)
self.rnn = nn.RNN(x_size, hidden_size, n_layers, batch_first=True)
self.out = nn.Linear(hidden_size, output_size) # 10 in and 10 out
def forward(self, inputs, hidden=None):
hidden = self.__init__hidden()
#print("Forward hidden {}".format(hidden.shape))
#print("Forward inps {}".format(inputs.shape))
output, hidden = self.rnn(inputs.float(), hidden.float())
#print("Out1 {}".format(output.shape))
output = self.out(output.float());
#print("Forward outputs {}".format(output.shape))
return output, hidden
def __init__hidden(self):
hidden = torch.zeros(self.n_layers, self.batch_size, self.hidden_size, dtype=torch.float64)
return hidden
### 3.2 生成网络
x_size = 1
hidden_size = 2 # try to change this parameters
n_layers = 1
seq_length = 10
output_size = 1
#rnn_model = SimpleRNN(x_size, hidden_size, n_layers, seq_length, output_size)
rnn_model = SimpleRNN(x_size, hidden_size, n_layers, seq_length, output_size)
print(rnn_model)
SimpleRNN(
(rnn): RNN(1, 2, batch_first=True)
(out): Linear(in_features=2, out_features=1, bias=True)
)
### 3.3 定义loss与优化器
# 定义loss
criterion = nn.MSELoss()
#定义优化器
optimizer = optim.AdamW(rnn_model.parameters(), lr=0.01) # 使用 Adam 优化器 比课上使用的 SGD 优化器更加稳定
### 3.4 训练前的准备
# 训练前的准备
n_epochs = 30
rnn_losses = np.zeros(n_epochs)
data_loader = torch.utils.data.DataLoader(source_data.values, batch_size=seq_length, shuffle=True)
### 3.5 开始训练
# 开始训练
for epoch in range(n_epochs):
if t.shape[0] != seq_length: continue
random_index = random.randint(0, t.shape[-1] - seq_length - 1)
train_x = t[:, random_index: random_index+seq_length]
#train_y = t[:, random_index + 1: random_index + seq_length + 1]
train_y = train_x
#获取输出
outputs, hidden = rnn_model(train_x.double().unsqueeze(2), hidden_size)
#梯度复位
#定义损失函数
loss = criterion(outputs.double(), train_y.double().unsqueeze(2))
# 反向求导
loss.backward()
#梯度迭代
optimizer.step()
#记录loss
epoch_losses.append(loss.detach())
#显示拟合图像
if iter_ == 0:
plt.clf();
plt.ion()
plt.title("Epoch {}, iter {}".format(epoch, iter_))
plt.plot(torch.flatten(train_y),'c-',linewidth=1,label='Label')
plt.plot(torch.flatten(train_x),'g-',linewidth=1,label='Input')
plt.plot(torch.flatten(outputs.detach()),'r-',linewidth=1,label='Output')
plt.draw();
plt.pause(0.05);
rnn_losses[epoch] = np.mean(epoch_losses)
### 3.7 分析
RNN网络在第一个epoch完成后，拟合度就非常好。
展开全文
• 时间序列有高度的自相关性，所以我不能随机打乱。 如果1-6月训练，7、8验证，9、10测试的话，实验结果表明测试精度远差于验证逻辑上很好解释，中间差了两个月，市场行情发生了很大的变化，模型没见过跟不...
• 下载地址：laiguokun/multivariate-time-series-data. electricity (行，列) = (26304, 321) 每日的用电量有一定的季节性，图中只画出 10 列 exchange_rate (行，列) = (7588, 8) solar-energy ...
• ## 时间序列数据简介
千次阅读 2021-08-11 19:14:18
前言 1. 引言 1.1 时间序列定义 1.2 应用场景 1.3研究方法的递进 2. 时间序列数据 2.1 数据的获取 2.2 数据时间轴的确定 2.3 时间序列遇到问题
• 时间序列数据集有一个额外的维度——时间。 我们可以将其视为 3D 数据集。 比如说，我们有一个包含 5 个特征和 5 个输入实例的数据集。 那么时间序列数据基本上是该表在第 3 维的扩展，其中每个新表只是新时间步长...
• 时间序列数据必须经过变换才能用来拟合有监督的学习模型。在这种形式下，数据可以立即用于拟合有监督的机器学习算法，甚至多层感知器神经...本文介绍了如何将时间序列数据集转换为三维结构，以便适合CNN或LSTM模型。
• 用Pandas和Streamlit对时间序列数据集进行可视化过滤 介绍 我们每天处理的数据最多的类型可能是时间序列数据。基本上，使用日期，时间或两者同时索引的任何内容都可以视为时间序列数据集。在我们工作中，可能经常...
• MATLAB实现ARIMA时间序列预测数据集
• 可用于进行序列预测的一维数据集，取自某支股票的某时间段内的价格。
• MATLAB时间序列回归Data_TSReg2数据集
• 为了在接下来的程序中对这些时间序列进行聚类以及评估聚类效果，需要读取这些数据，并且将状态（分类）作为标签附在每一段数据上。鉴于标签和时间序列的格式不一致，本人采用了字典格式。 具体代码如下： def ...
• MATLAB实现RNN(循环神经网络)时间序列预测数据集
• data = pd.read_excel(r'你的数据位置',index_col=0)#读取数据 data data1 = data.values[:,:]#截取所需要的数据 data1.shape import numpy as np X = []#开一个空的东西用来存数据 for i in range(0,n，m):#n是你所...
• 提出的TSkmeans算法可以有效地利用时间序列数据集的固有子空间信息来增强聚类性能。 更具体地说，平滑子空间由加权时间戳表示，加权时间戳指示这些时间戳对于聚类对象的相对判别力。 我们工作的主要贡献包括设计一...
• 时间序列数据在许多不同的行业中都非常重要。它在研究、金融行业、制药、社交媒体、网络服务等领域尤为重要。对时间序列数据的分析也变得越来越重要。在分析中有什么比一些好的可视化效果更好呢?没有一些视觉效果，...
• 时间序列预测可以被构造为一个监督学习问题。通过对时间序列数据的重构，可以针对不同问题使用相关的机器学习算法。本文介绍了如何将时间序列问题重新构造为机器学习中的监督学习问题。
... | 8,381 | 19,402 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.268441 |
https://www.tutorela.com/math/area-of-a-regular-hexagon | 1,723,250,461,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782288.54/warc/CC-MAIN-20240809235615-20240810025615-00647.warc.gz | 813,169,236 | 42,106 | # Area of a Regular Hexagon
🏆Practice regular hexagon
## Calculation of the Area of a Regular Hexagon - Let's Calculate It This Way!
The regular hexagon belongs to the family of regular polygons. It is a polygon in which all sides, and all angles, are equal to each other. By its name, we can understand that it is a geometric figure with $6$ different sides. The sum of its internal angles equals $720^o$ degrees. Therefore:
external angle $=60^o$ ; internal angle $=120^o$
## Test yourself on regular hexagon!
Given the hexagon in the drawing:
What is the area?
Three diagonals divide the regular hexagon into six congruent isosceles triangles. Therefore, the calculation of the area of a hexagon, whose side is a:
$6\times\frac{a^2\sqrt{3}}{4}$
Assuming that the length of the side of the regular hexagon is $4$ cm, its area will be:
#### $6\frac{a^2\sqrt{3}}{4}=41.57$
How many sides does a regular hexagon have?
The hexagon is a geometric figure with 6 equal sides
## Exercises with Hexagons
### Exercise 1
Given a regular hexagon
with a perimeter of $72$ cm
Calculate the area of the hexagon
Calculate the area of the hexagon
Solution
Since the hexagon has 6 equal sides we will divide $72$ by $6$
$\frac{72}{6}=12$
Each side is equal to $12$ cm
And then we will place the side in the formula to find the area of the hexagon
$A=6\cdot\frac{(12)^2\cdot\sqrt{3}}{4}=374.12$
The correct answer is $374.12$ cm²
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### Exercise 2
Given that the area of the hexagon is equal to $6$ cm²
What is the value of the sides of the hexagon?
Solution
We place the side of the hexagon as $X$ in the formula
Given that the area is equal to $6$ cm²
$A=\frac{6\cdot(X)^2\cdot\sqrt{3}}{4}$
Multiply by 4
$24=6\cdot X^2\cdot\sqrt{3}~~~~$
Now divide by 6
$4= X^2\cdot\sqrt{3}$
Divide by $~~~~\sqrt{3}$
$X^2=\frac{4}{\sqrt{3}}$
$X=1.526$ cm
$X=1.526$ cm
### Exercise 3
Given the regular hexagon with an area of $(\sqrt{3})^3$
Calculate the value of the sides of the hexagon
Solution
We use the formula to find the area of the hexagon:
Given that $A=(\sqrt{3})^3$ we represent the side as $X$.
Area of the hexagon =
$A=(\sqrt{3})^3$
We represent the side with $X$.
$A=6\cdot\frac{(X)^2\cdot\sqrt{3}}{4}=(\sqrt{3})^3$
We multiply both sides by $4$
$4\cdot(\sqrt{3})^3=6\cdot X^2\cdot\sqrt{3} ~~~~~$
We divide by $:√3$
$4\cdot(\sqrt{3})^2=6\cdot X^2$
$4\cdot3=6X^2 ~~~~~~$
Now we divide by $:6$
$2=X^2$
$X=\sqrt{2}$
The answer is $X=\sqrt{2}$
Do you know what the answer is?
### Exercise 4
Regular hexagon with an area of $12$ cm².
How much is each side of the hexagon worth?
Solution
We place on the side $X$ the formula
(Given: the area of the hexagon is $12$ cm²)
$A=6\cdot\frac{(X)^2\cdot\sqrt{3}}{4}=\frac{12}{1}$
We multiply the whole expression by 4
$6X^2\cdot\sqrt{3}=12\cdot4~~~~~$
We divide by $:6$
$X^2\cdot\sqrt{3}=8$
We divide by $\sqrt{3}$
$X^2=\frac{8}{\sqrt{3}}$
$X=\sqrt{\frac{8}{\sqrt{3}}}=2.149$
The side of the hexagon is worth $2.149$ cm.
### Exercise 5
Given that the area of the regular hexagon is equal to $8$ cm²
Calculate the value of the sides of the hexagon.
Solution
We place in the formula to find the area of the hexagon: $A=8$ and we will represent the side as $X$
$A=6\cdot\frac{(X^2)\sqrt{3}}{4}=\frac{8}{1}$
Multiply by $4$
$8\cdot4=6\cdot X^2\sqrt{3}=\frac{8}{1}$
$32=6\sqrt{3}\cdot X^2 ~~~~~~$
Divide the expression by $:6\sqrt{3}$
$X^2=3.07$
$X^=\sqrt{3.07}$
The correct answer is $X^=\sqrt{3.07}$
### Exercise 6
Given that the area of the regular hexagon has a value of $49$ cm²
Calculate the value of the sides of the hexagon.
Solution
We place in the formula to find the area of the hexagon $A=49$ cm² and the side will be represented as $X$.
$A=6\cdot\frac{X^{2\cdot}\sqrt{3}}{4}=\frac{49}{1}$
Multiply by $4$
$6\sqrt{3}X^2=4\cdot49~~~~~$
Divide by $:6\sqrt{3}$
$X^2=18.86$
Finally, we take the square root $√$
$X=4.34$
$X=4.34$
## Take private lessons online, from the comfort of your home!
Calculating the area of a regular hexagon can only be done if we understand its shape and properties. If you feel that you do not understand your teacher's explanations in the classroom, you can get in touch with one of our private tutors. Even if what you need is to complete part of the material that you have not understood in the classroom, an online private tutor can help you with just a few classes, in which you can study exactly what you are missing.
You can also coordinate an online private math class, right from your personal computer! How do you have an online private class?
• First, you will have a small talk with the teacher.
• Explain to the teacher what topic you want to study.
• Discuss what difficulties you have in your studies.
• The teacher will explain what you do not understand.
• Practice during the class, the material you are studying. For example, calculating the area of a regular hexagon.
Do you think you will be able to solve it?
Well, as you should already know, the more you practice, the better you will understand. First, when you start to apply a new formula, many doubts usually arise. After having done about 5 exercises, you will feel that you are beginning to better understand the new geometric shape you are studying, and the calculation of its area. Therefore, what we recommend is that you practice as much as possible. What do you achieve by exercising?
• It is not necessary to start by memorizing the formula, but you will gradually remember it as you use it.
• When you study regular hexagons, you will come across a variety of exercises, which include different data.
• Each exercise you do will help you become familiar with this geometric shape.
• Gradually, it will take you less time to solve the different problems.
• And you will feel more confident about yourself.
## "What suits me best, a private lesson just for you, or together with a study partner?"
A private class along with another classmate is only recommended in cases where both are interested in studying the same topics. Since when you have a private class, it is more than important to use the time effectively. The ultimate goal is for you to understand the material being taught. If two students take a class together, and both are interested in different topics, surely the class cannot be taken full advantage of.
## Study math throughout the year, and not just before exams.
Mathematics is a subject whose topics are learned gradually. Almost any formula, topic, or model you learn will also apply to the following topics. Therefore, it is important to avoid falling behind. If you feel that you need to reinforce a specific topic, get in touch with one of our teachers. The best way to learn mathematics is by dedicating time and study throughout the year, and not by cramming right before each exam. Good luck!
Do you know what the answer is?
## examples with solutions for regular hexagon
### Exercise #1
Given the hexagon in the drawing:
What is the area?
64.95
### Exercise #2
Given the hexagon in the drawing:
What is the area?
166.27
### Exercise #3
Given the hexagon in the drawing:
What is the area?
41.56
### Exercise #4
A hexagon has an area of 8 cm².
How long are the sides of the hexagon?
### Video Solution
$\sqrt{3}$ cm
### Exercise #5
A hexagon has sides
measuring 6 cm long.
What is the area of the hexagon? | 2,140 | 7,514 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 73, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-33 | latest | en | 0.834525 |
https://www.tensorflow.org/api_docs/python/tf/linalg/matrix_transpose?hl=de | 1,568,997,626,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574050.69/warc/CC-MAIN-20190920155311-20190920181311-00300.warc.gz | 1,064,999,862 | 63,584 | Join us at TensorFlow World, Oct 28-31. Use code TF20 for 20% off select passes.
# tf.linalg.matrix_transpose
Transposes last two dimensions of tensor `a`.
### Aliases:
• `tf.compat.v1.linalg.matrix_transpose`
• `tf.compat.v1.linalg.transpose`
• `tf.compat.v1.matrix_transpose`
• `tf.compat.v2.linalg.matrix_transpose`
• `tf.linalg.transpose`
• `tf.matrix_transpose`
``````tf.linalg.matrix_transpose(
a,
name='matrix_transpose',
conjugate=False
)
``````
#### For example:
``````x = tf.constant([[1, 2, 3], [4, 5, 6]])
tf.linalg.matrix_transpose(x) # [[1, 4],
# [2, 5],
# [3, 6]]
x = tf.constant([[1 + 1j, 2 + 2j, 3 + 3j],
[4 + 4j, 5 + 5j, 6 + 6j]])
tf.linalg.matrix_transpose(x, conjugate=True) # [[1 - 1j, 4 - 4j],
# [2 - 2j, 5 - 5j],
# [3 - 3j, 6 - 6j]]
# Matrix with two batch dimensions.
# x.shape is [1, 2, 3, 4]
# tf.linalg.matrix_transpose(x) is shape [1, 2, 4, 3]
``````
Note that `tf.matmul` provides kwargs allowing for transpose of arguments. This is done with minimal cost, and is preferable to using this function. E.g.
``````# Good! Transpose is taken at minimal additional cost.
tf.matmul(matrix, b, transpose_b=True)
# Inefficient!
tf.matmul(matrix, tf.linalg.matrix_transpose(b))
``````
#### Args:
• `a`: A `Tensor` with `rank >= 2`.
• `name`: A name for the operation (optional).
• `conjugate`: Optional bool. Setting it to `True` is mathematically equivalent to tf.math.conj(tf.linalg.matrix_transpose(input)).
#### Returns:
A transposed batch matrix `Tensor`.
#### Raises:
• `ValueError`: If `a` is determined statically to have `rank < 2`.
#### Numpy Compatibility
In `numpy` transposes are memory-efficient constant time operations as they simply return a new view of the same data with adjusted `strides`.
TensorFlow does not support strides, `linalg.matrix_transpose` returns a new tensor with the items permuted. | 573 | 1,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-39 | longest | en | 0.586649 |
https://isabelle.in.tum.de/repos/isabelle/annotate/ddd97d9dfbfb/src/HOL/Fun.thy | 1,579,461,552,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00346.warc.gz | 516,205,269 | 21,452 | src/HOL/Fun.thy
author haftmann Thu Oct 29 11:41:36 2009 +0100 (2009-10-29) changeset 33318 ddd97d9dfbfb parent 33057 764547b68538 child 34101 d689f0b33047 permissions -rw-r--r--
moved Nat_Transfer before Divides; distributed Nat_Transfer setup accordingly
clasohm@1475 ` 1` ```(* Title: HOL/Fun.thy ``` clasohm@1475 ` 2` ``` Author: Tobias Nipkow, Cambridge University Computer Laboratory ``` clasohm@923 ` 3` ``` Copyright 1994 University of Cambridge ``` huffman@18154 ` 4` ```*) ``` clasohm@923 ` 5` huffman@18154 ` 6` ```header {* Notions about functions *} ``` clasohm@923 ` 7` paulson@15510 ` 8` ```theory Fun ``` haftmann@32139 ` 9` ```imports Complete_Lattice ``` nipkow@15131 ` 10` ```begin ``` nipkow@2912 ` 11` haftmann@26147 ` 12` ```text{*As a simplification rule, it replaces all function equalities by ``` haftmann@26147 ` 13` ``` first-order equalities.*} ``` haftmann@26147 ` 14` ```lemma expand_fun_eq: "f = g \ (\x. f x = g x)" ``` haftmann@26147 ` 15` ```apply (rule iffI) ``` haftmann@26147 ` 16` ```apply (simp (no_asm_simp)) ``` haftmann@26147 ` 17` ```apply (rule ext) ``` haftmann@26147 ` 18` ```apply (simp (no_asm_simp)) ``` haftmann@26147 ` 19` ```done ``` oheimb@5305 ` 20` haftmann@26147 ` 21` ```lemma apply_inverse: ``` haftmann@26357 ` 22` ``` "f x = u \ (\x. P x \ g (f x) = x) \ P x \ x = g u" ``` haftmann@26147 ` 23` ``` by auto ``` nipkow@2912 ` 24` wenzelm@12258 ` 25` haftmann@26147 ` 26` ```subsection {* The Identity Function @{text id} *} ``` paulson@6171 ` 27` haftmann@22744 ` 28` ```definition ``` haftmann@22744 ` 29` ``` id :: "'a \ 'a" ``` haftmann@22744 ` 30` ```where ``` haftmann@22744 ` 31` ``` "id = (\x. x)" ``` nipkow@13910 ` 32` haftmann@26147 ` 33` ```lemma id_apply [simp]: "id x = x" ``` haftmann@26147 ` 34` ``` by (simp add: id_def) ``` haftmann@26147 ` 35` haftmann@26147 ` 36` ```lemma image_ident [simp]: "(%x. x) ` Y = Y" ``` haftmann@26147 ` 37` ```by blast ``` haftmann@26147 ` 38` haftmann@26147 ` 39` ```lemma image_id [simp]: "id ` Y = Y" ``` haftmann@26147 ` 40` ```by (simp add: id_def) ``` haftmann@26147 ` 41` haftmann@26147 ` 42` ```lemma vimage_ident [simp]: "(%x. x) -` Y = Y" ``` haftmann@26147 ` 43` ```by blast ``` haftmann@26147 ` 44` haftmann@26147 ` 45` ```lemma vimage_id [simp]: "id -` A = A" ``` haftmann@26147 ` 46` ```by (simp add: id_def) ``` haftmann@26147 ` 47` haftmann@26147 ` 48` haftmann@26147 ` 49` ```subsection {* The Composition Operator @{text "f \ g"} *} ``` haftmann@26147 ` 50` haftmann@22744 ` 51` ```definition ``` haftmann@22744 ` 52` ``` comp :: "('b \ 'c) \ ('a \ 'b) \ 'a \ 'c" (infixl "o" 55) ``` haftmann@22744 ` 53` ```where ``` haftmann@22744 ` 54` ``` "f o g = (\x. f (g x))" ``` oheimb@11123 ` 55` wenzelm@21210 ` 56` ```notation (xsymbols) ``` wenzelm@19656 ` 57` ``` comp (infixl "\" 55) ``` wenzelm@19656 ` 58` wenzelm@21210 ` 59` ```notation (HTML output) ``` wenzelm@19656 ` 60` ``` comp (infixl "\" 55) ``` wenzelm@19656 ` 61` paulson@13585 ` 62` ```text{*compatibility*} ``` paulson@13585 ` 63` ```lemmas o_def = comp_def ``` nipkow@2912 ` 64` paulson@13585 ` 65` ```lemma o_apply [simp]: "(f o g) x = f (g x)" ``` paulson@13585 ` 66` ```by (simp add: comp_def) ``` paulson@13585 ` 67` paulson@13585 ` 68` ```lemma o_assoc: "f o (g o h) = f o g o h" ``` paulson@13585 ` 69` ```by (simp add: comp_def) ``` paulson@13585 ` 70` paulson@13585 ` 71` ```lemma id_o [simp]: "id o g = g" ``` paulson@13585 ` 72` ```by (simp add: comp_def) ``` paulson@13585 ` 73` paulson@13585 ` 74` ```lemma o_id [simp]: "f o id = f" ``` paulson@13585 ` 75` ```by (simp add: comp_def) ``` paulson@13585 ` 76` paulson@13585 ` 77` ```lemma image_compose: "(f o g) ` r = f`(g`r)" ``` paulson@13585 ` 78` ```by (simp add: comp_def, blast) ``` paulson@13585 ` 79` paulson@33044 ` 80` ```lemma vimage_compose: "(g \ f) -` x = f -` (g -` x)" ``` paulson@33044 ` 81` ``` by auto ``` paulson@33044 ` 82` paulson@13585 ` 83` ```lemma UN_o: "UNION A (g o f) = UNION (f`A) g" ``` paulson@13585 ` 84` ```by (unfold comp_def, blast) ``` paulson@13585 ` 85` paulson@13585 ` 86` haftmann@26588 ` 87` ```subsection {* The Forward Composition Operator @{text fcomp} *} ``` haftmann@26357 ` 88` haftmann@26357 ` 89` ```definition ``` haftmann@26357 ` 90` ``` fcomp :: "('a \ 'b) \ ('b \ 'c) \ 'a \ 'c" (infixl "o>" 60) ``` haftmann@26357 ` 91` ```where ``` haftmann@26357 ` 92` ``` "f o> g = (\x. g (f x))" ``` haftmann@26357 ` 93` haftmann@26357 ` 94` ```lemma fcomp_apply: "(f o> g) x = g (f x)" ``` haftmann@26357 ` 95` ``` by (simp add: fcomp_def) ``` haftmann@26357 ` 96` haftmann@26357 ` 97` ```lemma fcomp_assoc: "(f o> g) o> h = f o> (g o> h)" ``` haftmann@26357 ` 98` ``` by (simp add: fcomp_def) ``` haftmann@26357 ` 99` haftmann@26357 ` 100` ```lemma id_fcomp [simp]: "id o> g = g" ``` haftmann@26357 ` 101` ``` by (simp add: fcomp_def) ``` haftmann@26357 ` 102` haftmann@26357 ` 103` ```lemma fcomp_id [simp]: "f o> id = f" ``` haftmann@26357 ` 104` ``` by (simp add: fcomp_def) ``` haftmann@26357 ` 105` haftmann@31202 ` 106` ```code_const fcomp ``` haftmann@31202 ` 107` ``` (Eval infixl 1 "#>") ``` haftmann@31202 ` 108` haftmann@26588 ` 109` ```no_notation fcomp (infixl "o>" 60) ``` haftmann@26588 ` 110` haftmann@26357 ` 111` haftmann@26147 ` 112` ```subsection {* Injectivity and Surjectivity *} ``` haftmann@26147 ` 113` haftmann@26147 ` 114` ```constdefs ``` haftmann@26147 ` 115` ``` inj_on :: "['a => 'b, 'a set] => bool" -- "injective" ``` haftmann@26147 ` 116` ``` "inj_on f A == ! x:A. ! y:A. f(x)=f(y) --> x=y" ``` haftmann@26147 ` 117` haftmann@26147 ` 118` ```text{*A common special case: functions injective over the entire domain type.*} ``` haftmann@26147 ` 119` haftmann@26147 ` 120` ```abbreviation ``` haftmann@26147 ` 121` ``` "inj f == inj_on f UNIV" ``` paulson@13585 ` 122` haftmann@26147 ` 123` ```definition ``` haftmann@26147 ` 124` ``` bij_betw :: "('a => 'b) => 'a set => 'b set => bool" where -- "bijective" ``` haftmann@28562 ` 125` ``` [code del]: "bij_betw f A B \ inj_on f A & f ` A = B" ``` haftmann@26147 ` 126` haftmann@26147 ` 127` ```constdefs ``` haftmann@26147 ` 128` ``` surj :: "('a => 'b) => bool" (*surjective*) ``` haftmann@26147 ` 129` ``` "surj f == ! y. ? x. y=f(x)" ``` paulson@13585 ` 130` haftmann@26147 ` 131` ``` bij :: "('a => 'b) => bool" (*bijective*) ``` haftmann@26147 ` 132` ``` "bij f == inj f & surj f" ``` haftmann@26147 ` 133` haftmann@26147 ` 134` ```lemma injI: ``` haftmann@26147 ` 135` ``` assumes "\x y. f x = f y \ x = y" ``` haftmann@26147 ` 136` ``` shows "inj f" ``` haftmann@26147 ` 137` ``` using assms unfolding inj_on_def by auto ``` paulson@13585 ` 138` haftmann@31775 ` 139` ```text{*For Proofs in @{text "Tools/Datatype/datatype_rep_proofs"}*} ``` paulson@13585 ` 140` ```lemma datatype_injI: ``` paulson@13585 ` 141` ``` "(!! x. ALL y. f(x) = f(y) --> x=y) ==> inj(f)" ``` paulson@13585 ` 142` ```by (simp add: inj_on_def) ``` paulson@13585 ` 143` berghofe@13637 ` 144` ```theorem range_ex1_eq: "inj f \ b : range f = (EX! x. b = f x)" ``` berghofe@13637 ` 145` ``` by (unfold inj_on_def, blast) ``` berghofe@13637 ` 146` paulson@13585 ` 147` ```lemma injD: "[| inj(f); f(x) = f(y) |] ==> x=y" ``` paulson@13585 ` 148` ```by (simp add: inj_on_def) ``` paulson@13585 ` 149` nipkow@32988 ` 150` ```lemma inj_on_eq_iff: "inj_on f A ==> x:A ==> y:A ==> (f(x) = f(y)) = (x=y)" ``` paulson@13585 ` 151` ```by (force simp add: inj_on_def) ``` paulson@13585 ` 152` nipkow@32988 ` 153` ```lemma inj_eq: "inj f ==> (f(x) = f(y)) = (x=y)" ``` nipkow@32988 ` 154` ```by (simp add: inj_on_eq_iff) ``` nipkow@32988 ` 155` haftmann@26147 ` 156` ```lemma inj_on_id[simp]: "inj_on id A" ``` haftmann@26147 ` 157` ``` by (simp add: inj_on_def) ``` paulson@13585 ` 158` haftmann@26147 ` 159` ```lemma inj_on_id2[simp]: "inj_on (%x. x) A" ``` haftmann@26147 ` 160` ```by (simp add: inj_on_def) ``` haftmann@26147 ` 161` haftmann@26147 ` 162` ```lemma surj_id[simp]: "surj id" ``` haftmann@26147 ` 163` ```by (simp add: surj_def) ``` haftmann@26147 ` 164` haftmann@26147 ` 165` ```lemma bij_id[simp]: "bij id" ``` haftmann@26147 ` 166` ```by (simp add: bij_def inj_on_id surj_id) ``` paulson@13585 ` 167` paulson@13585 ` 168` ```lemma inj_onI: ``` paulson@13585 ` 169` ``` "(!! x y. [| x:A; y:A; f(x) = f(y) |] ==> x=y) ==> inj_on f A" ``` paulson@13585 ` 170` ```by (simp add: inj_on_def) ``` paulson@13585 ` 171` paulson@13585 ` 172` ```lemma inj_on_inverseI: "(!!x. x:A ==> g(f(x)) = x) ==> inj_on f A" ``` paulson@13585 ` 173` ```by (auto dest: arg_cong [of concl: g] simp add: inj_on_def) ``` paulson@13585 ` 174` paulson@13585 ` 175` ```lemma inj_onD: "[| inj_on f A; f(x)=f(y); x:A; y:A |] ==> x=y" ``` paulson@13585 ` 176` ```by (unfold inj_on_def, blast) ``` paulson@13585 ` 177` paulson@13585 ` 178` ```lemma inj_on_iff: "[| inj_on f A; x:A; y:A |] ==> (f(x)=f(y)) = (x=y)" ``` paulson@13585 ` 179` ```by (blast dest!: inj_onD) ``` paulson@13585 ` 180` paulson@13585 ` 181` ```lemma comp_inj_on: ``` paulson@13585 ` 182` ``` "[| inj_on f A; inj_on g (f`A) |] ==> inj_on (g o f) A" ``` paulson@13585 ` 183` ```by (simp add: comp_def inj_on_def) ``` paulson@13585 ` 184` nipkow@15303 ` 185` ```lemma inj_on_imageI: "inj_on (g o f) A \ inj_on g (f ` A)" ``` nipkow@15303 ` 186` ```apply(simp add:inj_on_def image_def) ``` nipkow@15303 ` 187` ```apply blast ``` nipkow@15303 ` 188` ```done ``` nipkow@15303 ` 189` nipkow@15439 ` 190` ```lemma inj_on_image_iff: "\ ALL x:A. ALL y:A. (g(f x) = g(f y)) = (g x = g y); ``` nipkow@15439 ` 191` ``` inj_on f A \ \ inj_on g (f ` A) = inj_on g A" ``` nipkow@15439 ` 192` ```apply(unfold inj_on_def) ``` nipkow@15439 ` 193` ```apply blast ``` nipkow@15439 ` 194` ```done ``` nipkow@15439 ` 195` paulson@13585 ` 196` ```lemma inj_on_contraD: "[| inj_on f A; ~x=y; x:A; y:A |] ==> ~ f(x)=f(y)" ``` paulson@13585 ` 197` ```by (unfold inj_on_def, blast) ``` wenzelm@12258 ` 198` paulson@13585 ` 199` ```lemma inj_singleton: "inj (%s. {s})" ``` paulson@13585 ` 200` ```by (simp add: inj_on_def) ``` paulson@13585 ` 201` nipkow@15111 ` 202` ```lemma inj_on_empty[iff]: "inj_on f {}" ``` nipkow@15111 ` 203` ```by(simp add: inj_on_def) ``` nipkow@15111 ` 204` nipkow@15303 ` 205` ```lemma subset_inj_on: "[| inj_on f B; A <= B |] ==> inj_on f A" ``` paulson@13585 ` 206` ```by (unfold inj_on_def, blast) ``` paulson@13585 ` 207` nipkow@15111 ` 208` ```lemma inj_on_Un: ``` nipkow@15111 ` 209` ``` "inj_on f (A Un B) = ``` nipkow@15111 ` 210` ``` (inj_on f A & inj_on f B & f`(A-B) Int f`(B-A) = {})" ``` nipkow@15111 ` 211` ```apply(unfold inj_on_def) ``` nipkow@15111 ` 212` ```apply (blast intro:sym) ``` nipkow@15111 ` 213` ```done ``` nipkow@15111 ` 214` nipkow@15111 ` 215` ```lemma inj_on_insert[iff]: ``` nipkow@15111 ` 216` ``` "inj_on f (insert a A) = (inj_on f A & f a ~: f`(A-{a}))" ``` nipkow@15111 ` 217` ```apply(unfold inj_on_def) ``` nipkow@15111 ` 218` ```apply (blast intro:sym) ``` nipkow@15111 ` 219` ```done ``` nipkow@15111 ` 220` nipkow@15111 ` 221` ```lemma inj_on_diff: "inj_on f A ==> inj_on f (A-B)" ``` nipkow@15111 ` 222` ```apply(unfold inj_on_def) ``` nipkow@15111 ` 223` ```apply (blast) ``` nipkow@15111 ` 224` ```done ``` nipkow@15111 ` 225` paulson@13585 ` 226` ```lemma surjI: "(!! x. g(f x) = x) ==> surj g" ``` paulson@13585 ` 227` ```apply (simp add: surj_def) ``` paulson@13585 ` 228` ```apply (blast intro: sym) ``` paulson@13585 ` 229` ```done ``` paulson@13585 ` 230` paulson@13585 ` 231` ```lemma surj_range: "surj f ==> range f = UNIV" ``` paulson@13585 ` 232` ```by (auto simp add: surj_def) ``` paulson@13585 ` 233` paulson@13585 ` 234` ```lemma surjD: "surj f ==> EX x. y = f x" ``` paulson@13585 ` 235` ```by (simp add: surj_def) ``` paulson@13585 ` 236` paulson@13585 ` 237` ```lemma surjE: "surj f ==> (!!x. y = f x ==> C) ==> C" ``` paulson@13585 ` 238` ```by (simp add: surj_def, blast) ``` paulson@13585 ` 239` paulson@13585 ` 240` ```lemma comp_surj: "[| surj f; surj g |] ==> surj (g o f)" ``` paulson@13585 ` 241` ```apply (simp add: comp_def surj_def, clarify) ``` paulson@13585 ` 242` ```apply (drule_tac x = y in spec, clarify) ``` paulson@13585 ` 243` ```apply (drule_tac x = x in spec, blast) ``` paulson@13585 ` 244` ```done ``` paulson@13585 ` 245` paulson@13585 ` 246` ```lemma bijI: "[| inj f; surj f |] ==> bij f" ``` paulson@13585 ` 247` ```by (simp add: bij_def) ``` paulson@13585 ` 248` paulson@13585 ` 249` ```lemma bij_is_inj: "bij f ==> inj f" ``` paulson@13585 ` 250` ```by (simp add: bij_def) ``` paulson@13585 ` 251` paulson@13585 ` 252` ```lemma bij_is_surj: "bij f ==> surj f" ``` paulson@13585 ` 253` ```by (simp add: bij_def) ``` paulson@13585 ` 254` nipkow@26105 ` 255` ```lemma bij_betw_imp_inj_on: "bij_betw f A B \ inj_on f A" ``` nipkow@26105 ` 256` ```by (simp add: bij_betw_def) ``` nipkow@26105 ` 257` nipkow@32337 ` 258` ```lemma bij_comp: "bij f \ bij g \ bij (g o f)" ``` nipkow@32337 ` 259` ```by(fastsimp intro: comp_inj_on comp_surj simp: bij_def surj_range) ``` nipkow@32337 ` 260` nipkow@31438 ` 261` ```lemma bij_betw_trans: ``` nipkow@31438 ` 262` ``` "bij_betw f A B \ bij_betw g B C \ bij_betw (g o f) A C" ``` nipkow@31438 ` 263` ```by(auto simp add:bij_betw_def comp_inj_on) ``` nipkow@31438 ` 264` nipkow@26105 ` 265` ```lemma bij_betw_inv: assumes "bij_betw f A B" shows "EX g. bij_betw g B A" ``` nipkow@26105 ` 266` ```proof - ``` nipkow@26105 ` 267` ``` have i: "inj_on f A" and s: "f ` A = B" ``` nipkow@26105 ` 268` ``` using assms by(auto simp:bij_betw_def) ``` nipkow@26105 ` 269` ``` let ?P = "%b a. a:A \ f a = b" let ?g = "%b. The (?P b)" ``` nipkow@26105 ` 270` ``` { fix a b assume P: "?P b a" ``` nipkow@26105 ` 271` ``` hence ex1: "\a. ?P b a" using s unfolding image_def by blast ``` nipkow@26105 ` 272` ``` hence uex1: "\!a. ?P b a" by(blast dest:inj_onD[OF i]) ``` nipkow@26105 ` 273` ``` hence " ?g b = a" using the1_equality[OF uex1, OF P] P by simp ``` nipkow@26105 ` 274` ``` } note g = this ``` nipkow@26105 ` 275` ``` have "inj_on ?g B" ``` nipkow@26105 ` 276` ``` proof(rule inj_onI) ``` nipkow@26105 ` 277` ``` fix x y assume "x:B" "y:B" "?g x = ?g y" ``` nipkow@26105 ` 278` ``` from s `x:B` obtain a1 where a1: "?P x a1" unfolding image_def by blast ``` nipkow@26105 ` 279` ``` from s `y:B` obtain a2 where a2: "?P y a2" unfolding image_def by blast ``` nipkow@26105 ` 280` ``` from g[OF a1] a1 g[OF a2] a2 `?g x = ?g y` show "x=y" by simp ``` nipkow@26105 ` 281` ``` qed ``` nipkow@26105 ` 282` ``` moreover have "?g ` B = A" ``` nipkow@26105 ` 283` ``` proof(auto simp:image_def) ``` nipkow@26105 ` 284` ``` fix b assume "b:B" ``` nipkow@26105 ` 285` ``` with s obtain a where P: "?P b a" unfolding image_def by blast ``` nipkow@26105 ` 286` ``` thus "?g b \ A" using g[OF P] by auto ``` nipkow@26105 ` 287` ``` next ``` nipkow@26105 ` 288` ``` fix a assume "a:A" ``` nipkow@26105 ` 289` ``` then obtain b where P: "?P b a" using s unfolding image_def by blast ``` nipkow@26105 ` 290` ``` then have "b:B" using s unfolding image_def by blast ``` nipkow@26105 ` 291` ``` with g[OF P] show "\b\B. a = ?g b" by blast ``` nipkow@26105 ` 292` ``` qed ``` nipkow@26105 ` 293` ``` ultimately show ?thesis by(auto simp:bij_betw_def) ``` nipkow@26105 ` 294` ```qed ``` nipkow@26105 ` 295` paulson@13585 ` 296` ```lemma surj_image_vimage_eq: "surj f ==> f ` (f -` A) = A" ``` paulson@13585 ` 297` ```by (simp add: surj_range) ``` paulson@13585 ` 298` paulson@13585 ` 299` ```lemma inj_vimage_image_eq: "inj f ==> f -` (f ` A) = A" ``` paulson@13585 ` 300` ```by (simp add: inj_on_def, blast) ``` paulson@13585 ` 301` paulson@13585 ` 302` ```lemma vimage_subsetD: "surj f ==> f -` B <= A ==> B <= f ` A" ``` paulson@13585 ` 303` ```apply (unfold surj_def) ``` paulson@13585 ` 304` ```apply (blast intro: sym) ``` paulson@13585 ` 305` ```done ``` paulson@13585 ` 306` paulson@13585 ` 307` ```lemma vimage_subsetI: "inj f ==> B <= f ` A ==> f -` B <= A" ``` paulson@13585 ` 308` ```by (unfold inj_on_def, blast) ``` paulson@13585 ` 309` paulson@13585 ` 310` ```lemma vimage_subset_eq: "bij f ==> (f -` B <= A) = (B <= f ` A)" ``` paulson@13585 ` 311` ```apply (unfold bij_def) ``` paulson@13585 ` 312` ```apply (blast del: subsetI intro: vimage_subsetI vimage_subsetD) ``` paulson@13585 ` 313` ```done ``` paulson@13585 ` 314` nipkow@31438 ` 315` ```lemma inj_on_Un_image_eq_iff: "inj_on f (A \ B) \ f ` A = f ` B \ A = B" ``` nipkow@31438 ` 316` ```by(blast dest: inj_onD) ``` nipkow@31438 ` 317` paulson@13585 ` 318` ```lemma inj_on_image_Int: ``` paulson@13585 ` 319` ``` "[| inj_on f C; A<=C; B<=C |] ==> f`(A Int B) = f`A Int f`B" ``` paulson@13585 ` 320` ```apply (simp add: inj_on_def, blast) ``` paulson@13585 ` 321` ```done ``` paulson@13585 ` 322` paulson@13585 ` 323` ```lemma inj_on_image_set_diff: ``` paulson@13585 ` 324` ``` "[| inj_on f C; A<=C; B<=C |] ==> f`(A-B) = f`A - f`B" ``` paulson@13585 ` 325` ```apply (simp add: inj_on_def, blast) ``` paulson@13585 ` 326` ```done ``` paulson@13585 ` 327` paulson@13585 ` 328` ```lemma image_Int: "inj f ==> f`(A Int B) = f`A Int f`B" ``` paulson@13585 ` 329` ```by (simp add: inj_on_def, blast) ``` paulson@13585 ` 330` paulson@13585 ` 331` ```lemma image_set_diff: "inj f ==> f`(A-B) = f`A - f`B" ``` paulson@13585 ` 332` ```by (simp add: inj_on_def, blast) ``` paulson@13585 ` 333` paulson@13585 ` 334` ```lemma inj_image_mem_iff: "inj f ==> (f a : f`A) = (a : A)" ``` paulson@13585 ` 335` ```by (blast dest: injD) ``` paulson@13585 ` 336` paulson@13585 ` 337` ```lemma inj_image_subset_iff: "inj f ==> (f`A <= f`B) = (A<=B)" ``` paulson@13585 ` 338` ```by (simp add: inj_on_def, blast) ``` paulson@13585 ` 339` paulson@13585 ` 340` ```lemma inj_image_eq_iff: "inj f ==> (f`A = f`B) = (A = B)" ``` paulson@13585 ` 341` ```by (blast dest: injD) ``` paulson@13585 ` 342` paulson@13585 ` 343` ```(*injectivity's required. Left-to-right inclusion holds even if A is empty*) ``` paulson@13585 ` 344` ```lemma image_INT: ``` paulson@13585 ` 345` ``` "[| inj_on f C; ALL x:A. B x <= C; j:A |] ``` paulson@13585 ` 346` ``` ==> f ` (INTER A B) = (INT x:A. f ` B x)" ``` paulson@13585 ` 347` ```apply (simp add: inj_on_def, blast) ``` paulson@13585 ` 348` ```done ``` paulson@13585 ` 349` paulson@13585 ` 350` ```(*Compare with image_INT: no use of inj_on, and if f is surjective then ``` paulson@13585 ` 351` ``` it doesn't matter whether A is empty*) ``` paulson@13585 ` 352` ```lemma bij_image_INT: "bij f ==> f ` (INTER A B) = (INT x:A. f ` B x)" ``` paulson@13585 ` 353` ```apply (simp add: bij_def) ``` paulson@13585 ` 354` ```apply (simp add: inj_on_def surj_def, blast) ``` paulson@13585 ` 355` ```done ``` paulson@13585 ` 356` paulson@13585 ` 357` ```lemma surj_Compl_image_subset: "surj f ==> -(f`A) <= f`(-A)" ``` paulson@13585 ` 358` ```by (auto simp add: surj_def) ``` paulson@13585 ` 359` paulson@13585 ` 360` ```lemma inj_image_Compl_subset: "inj f ==> f`(-A) <= -(f`A)" ``` paulson@13585 ` 361` ```by (auto simp add: inj_on_def) ``` paulson@5852 ` 362` paulson@13585 ` 363` ```lemma bij_image_Compl_eq: "bij f ==> f`(-A) = -(f`A)" ``` paulson@13585 ` 364` ```apply (simp add: bij_def) ``` paulson@13585 ` 365` ```apply (rule equalityI) ``` paulson@13585 ` 366` ```apply (simp_all (no_asm_simp) add: inj_image_Compl_subset surj_Compl_image_subset) ``` paulson@13585 ` 367` ```done ``` paulson@13585 ` 368` paulson@13585 ` 369` paulson@13585 ` 370` ```subsection{*Function Updating*} ``` paulson@13585 ` 371` haftmann@26147 ` 372` ```constdefs ``` haftmann@26147 ` 373` ``` fun_upd :: "('a => 'b) => 'a => 'b => ('a => 'b)" ``` haftmann@26147 ` 374` ``` "fun_upd f a b == % x. if x=a then b else f x" ``` haftmann@26147 ` 375` haftmann@26147 ` 376` ```nonterminals ``` haftmann@26147 ` 377` ``` updbinds updbind ``` haftmann@26147 ` 378` ```syntax ``` haftmann@26147 ` 379` ``` "_updbind" :: "['a, 'a] => updbind" ("(2_ :=/ _)") ``` haftmann@26147 ` 380` ``` "" :: "updbind => updbinds" ("_") ``` haftmann@26147 ` 381` ``` "_updbinds":: "[updbind, updbinds] => updbinds" ("_,/ _") ``` haftmann@26147 ` 382` ``` "_Update" :: "['a, updbinds] => 'a" ("_/'((_)')" [1000,0] 900) ``` haftmann@26147 ` 383` haftmann@26147 ` 384` ```translations ``` haftmann@26147 ` 385` ``` "_Update f (_updbinds b bs)" == "_Update (_Update f b) bs" ``` haftmann@26147 ` 386` ``` "f(x:=y)" == "fun_upd f x y" ``` haftmann@26147 ` 387` haftmann@26147 ` 388` ```(* Hint: to define the sum of two functions (or maps), use sum_case. ``` haftmann@26147 ` 389` ``` A nice infix syntax could be defined (in Datatype.thy or below) by ``` haftmann@26147 ` 390` ```consts ``` haftmann@26147 ` 391` ``` fun_sum :: "('a => 'c) => ('b => 'c) => (('a+'b) => 'c)" (infixr "'(+')"80) ``` haftmann@26147 ` 392` ```translations ``` haftmann@26147 ` 393` ``` "fun_sum" == sum_case ``` haftmann@26147 ` 394` ```*) ``` haftmann@26147 ` 395` paulson@13585 ` 396` ```lemma fun_upd_idem_iff: "(f(x:=y) = f) = (f x = y)" ``` paulson@13585 ` 397` ```apply (simp add: fun_upd_def, safe) ``` paulson@13585 ` 398` ```apply (erule subst) ``` paulson@13585 ` 399` ```apply (rule_tac [2] ext, auto) ``` paulson@13585 ` 400` ```done ``` paulson@13585 ` 401` paulson@13585 ` 402` ```(* f x = y ==> f(x:=y) = f *) ``` paulson@13585 ` 403` ```lemmas fun_upd_idem = fun_upd_idem_iff [THEN iffD2, standard] ``` paulson@13585 ` 404` paulson@13585 ` 405` ```(* f(x := f x) = f *) ``` paulson@17084 ` 406` ```lemmas fun_upd_triv = refl [THEN fun_upd_idem] ``` paulson@17084 ` 407` ```declare fun_upd_triv [iff] ``` paulson@13585 ` 408` paulson@13585 ` 409` ```lemma fun_upd_apply [simp]: "(f(x:=y))z = (if z=x then y else f z)" ``` paulson@17084 ` 410` ```by (simp add: fun_upd_def) ``` paulson@13585 ` 411` paulson@13585 ` 412` ```(* fun_upd_apply supersedes these two, but they are useful ``` paulson@13585 ` 413` ``` if fun_upd_apply is intentionally removed from the simpset *) ``` paulson@13585 ` 414` ```lemma fun_upd_same: "(f(x:=y)) x = y" ``` paulson@13585 ` 415` ```by simp ``` paulson@13585 ` 416` paulson@13585 ` 417` ```lemma fun_upd_other: "z~=x ==> (f(x:=y)) z = f z" ``` paulson@13585 ` 418` ```by simp ``` paulson@13585 ` 419` paulson@13585 ` 420` ```lemma fun_upd_upd [simp]: "f(x:=y,x:=z) = f(x:=z)" ``` paulson@13585 ` 421` ```by (simp add: expand_fun_eq) ``` paulson@13585 ` 422` paulson@13585 ` 423` ```lemma fun_upd_twist: "a ~= c ==> (m(a:=b))(c:=d) = (m(c:=d))(a:=b)" ``` paulson@13585 ` 424` ```by (rule ext, auto) ``` paulson@13585 ` 425` nipkow@15303 ` 426` ```lemma inj_on_fun_updI: "\ inj_on f A; y \ f`A \ \ inj_on (f(x:=y)) A" ``` nipkow@15303 ` 427` ```by(fastsimp simp:inj_on_def image_def) ``` nipkow@15303 ` 428` paulson@15510 ` 429` ```lemma fun_upd_image: ``` paulson@15510 ` 430` ``` "f(x:=y) ` A = (if x \ A then insert y (f ` (A-{x})) else f ` A)" ``` paulson@15510 ` 431` ```by auto ``` paulson@15510 ` 432` nipkow@31080 ` 433` ```lemma fun_upd_comp: "f \ (g(x := y)) = (f \ g)(x := f y)" ``` nipkow@31080 ` 434` ```by(auto intro: ext) ``` nipkow@31080 ` 435` haftmann@26147 ` 436` haftmann@26147 ` 437` ```subsection {* @{text override_on} *} ``` haftmann@26147 ` 438` haftmann@26147 ` 439` ```definition ``` haftmann@26147 ` 440` ``` override_on :: "('a \ 'b) \ ('a \ 'b) \ 'a set \ 'a \ 'b" ``` haftmann@26147 ` 441` ```where ``` haftmann@26147 ` 442` ``` "override_on f g A = (\a. if a \ A then g a else f a)" ``` nipkow@13910 ` 443` nipkow@15691 ` 444` ```lemma override_on_emptyset[simp]: "override_on f g {} = f" ``` nipkow@15691 ` 445` ```by(simp add:override_on_def) ``` nipkow@13910 ` 446` nipkow@15691 ` 447` ```lemma override_on_apply_notin[simp]: "a ~: A ==> (override_on f g A) a = f a" ``` nipkow@15691 ` 448` ```by(simp add:override_on_def) ``` nipkow@13910 ` 449` nipkow@15691 ` 450` ```lemma override_on_apply_in[simp]: "a : A ==> (override_on f g A) a = g a" ``` nipkow@15691 ` 451` ```by(simp add:override_on_def) ``` nipkow@13910 ` 452` haftmann@26147 ` 453` haftmann@26147 ` 454` ```subsection {* @{text swap} *} ``` paulson@15510 ` 455` haftmann@22744 ` 456` ```definition ``` haftmann@22744 ` 457` ``` swap :: "'a \ 'a \ ('a \ 'b) \ ('a \ 'b)" ``` haftmann@22744 ` 458` ```where ``` haftmann@22744 ` 459` ``` "swap a b f = f (a := f b, b:= f a)" ``` paulson@15510 ` 460` paulson@15510 ` 461` ```lemma swap_self: "swap a a f = f" ``` nipkow@15691 ` 462` ```by (simp add: swap_def) ``` paulson@15510 ` 463` paulson@15510 ` 464` ```lemma swap_commute: "swap a b f = swap b a f" ``` paulson@15510 ` 465` ```by (rule ext, simp add: fun_upd_def swap_def) ``` paulson@15510 ` 466` paulson@15510 ` 467` ```lemma swap_nilpotent [simp]: "swap a b (swap a b f) = f" ``` paulson@15510 ` 468` ```by (rule ext, simp add: fun_upd_def swap_def) ``` paulson@15510 ` 469` paulson@15510 ` 470` ```lemma inj_on_imp_inj_on_swap: ``` haftmann@22744 ` 471` ``` "[|inj_on f A; a \ A; b \ A|] ==> inj_on (swap a b f) A" ``` paulson@15510 ` 472` ```by (simp add: inj_on_def swap_def, blast) ``` paulson@15510 ` 473` paulson@15510 ` 474` ```lemma inj_on_swap_iff [simp]: ``` paulson@15510 ` 475` ``` assumes A: "a \ A" "b \ A" shows "inj_on (swap a b f) A = inj_on f A" ``` paulson@15510 ` 476` ```proof ``` paulson@15510 ` 477` ``` assume "inj_on (swap a b f) A" ``` paulson@15510 ` 478` ``` with A have "inj_on (swap a b (swap a b f)) A" ``` nipkow@17589 ` 479` ``` by (iprover intro: inj_on_imp_inj_on_swap) ``` paulson@15510 ` 480` ``` thus "inj_on f A" by simp ``` paulson@15510 ` 481` ```next ``` paulson@15510 ` 482` ``` assume "inj_on f A" ``` nipkow@27165 ` 483` ``` with A show "inj_on (swap a b f) A" by(iprover intro: inj_on_imp_inj_on_swap) ``` paulson@15510 ` 484` ```qed ``` paulson@15510 ` 485` paulson@15510 ` 486` ```lemma surj_imp_surj_swap: "surj f ==> surj (swap a b f)" ``` paulson@15510 ` 487` ```apply (simp add: surj_def swap_def, clarify) ``` wenzelm@27125 ` 488` ```apply (case_tac "y = f b", blast) ``` wenzelm@27125 ` 489` ```apply (case_tac "y = f a", auto) ``` paulson@15510 ` 490` ```done ``` paulson@15510 ` 491` paulson@15510 ` 492` ```lemma surj_swap_iff [simp]: "surj (swap a b f) = surj f" ``` paulson@15510 ` 493` ```proof ``` paulson@15510 ` 494` ``` assume "surj (swap a b f)" ``` paulson@15510 ` 495` ``` hence "surj (swap a b (swap a b f))" by (rule surj_imp_surj_swap) ``` paulson@15510 ` 496` ``` thus "surj f" by simp ``` paulson@15510 ` 497` ```next ``` paulson@15510 ` 498` ``` assume "surj f" ``` paulson@15510 ` 499` ``` thus "surj (swap a b f)" by (rule surj_imp_surj_swap) ``` paulson@15510 ` 500` ```qed ``` paulson@15510 ` 501` paulson@15510 ` 502` ```lemma bij_swap_iff: "bij (swap a b f) = bij f" ``` paulson@15510 ` 503` ```by (simp add: bij_def) ``` haftmann@21547 ` 504` nipkow@27188 ` 505` ```hide (open) const swap ``` haftmann@21547 ` 506` haftmann@31949 ` 507` haftmann@31949 ` 508` ```subsection {* Inversion of injective functions *} ``` haftmann@31949 ` 509` nipkow@33057 ` 510` ```definition the_inv_into :: "'a set => ('a => 'b) => ('b => 'a)" where ``` nipkow@33057 ` 511` ```"the_inv_into A f == %x. THE y. y : A & f y = x" ``` nipkow@32961 ` 512` nipkow@33057 ` 513` ```lemma the_inv_into_f_f: ``` nipkow@33057 ` 514` ``` "[| inj_on f A; x : A |] ==> the_inv_into A f (f x) = x" ``` nipkow@33057 ` 515` ```apply (simp add: the_inv_into_def inj_on_def) ``` nipkow@32961 ` 516` ```apply (blast intro: the_equality) ``` nipkow@32961 ` 517` ```done ``` nipkow@32961 ` 518` nipkow@33057 ` 519` ```lemma f_the_inv_into_f: ``` nipkow@33057 ` 520` ``` "inj_on f A ==> y : f`A ==> f (the_inv_into A f y) = y" ``` nipkow@33057 ` 521` ```apply (simp add: the_inv_into_def) ``` nipkow@32961 ` 522` ```apply (rule the1I2) ``` nipkow@32961 ` 523` ``` apply(blast dest: inj_onD) ``` nipkow@32961 ` 524` ```apply blast ``` nipkow@32961 ` 525` ```done ``` nipkow@32961 ` 526` nipkow@33057 ` 527` ```lemma the_inv_into_into: ``` nipkow@33057 ` 528` ``` "[| inj_on f A; x : f ` A; A <= B |] ==> the_inv_into A f x : B" ``` nipkow@33057 ` 529` ```apply (simp add: the_inv_into_def) ``` nipkow@32961 ` 530` ```apply (rule the1I2) ``` nipkow@32961 ` 531` ``` apply(blast dest: inj_onD) ``` nipkow@32961 ` 532` ```apply blast ``` nipkow@32961 ` 533` ```done ``` nipkow@32961 ` 534` nipkow@33057 ` 535` ```lemma the_inv_into_onto[simp]: ``` nipkow@33057 ` 536` ``` "inj_on f A ==> the_inv_into A f ` (f ` A) = A" ``` nipkow@33057 ` 537` ```by (fast intro:the_inv_into_into the_inv_into_f_f[symmetric]) ``` nipkow@32961 ` 538` nipkow@33057 ` 539` ```lemma the_inv_into_f_eq: ``` nipkow@33057 ` 540` ``` "[| inj_on f A; f x = y; x : A |] ==> the_inv_into A f y = x" ``` nipkow@32961 ` 541` ``` apply (erule subst) ``` nipkow@33057 ` 542` ``` apply (erule the_inv_into_f_f, assumption) ``` nipkow@32961 ` 543` ``` done ``` nipkow@32961 ` 544` nipkow@33057 ` 545` ```lemma the_inv_into_comp: ``` nipkow@32961 ` 546` ``` "[| inj_on f (g ` A); inj_on g A; x : f ` g ` A |] ==> ``` nipkow@33057 ` 547` ``` the_inv_into A (f o g) x = (the_inv_into A g o the_inv_into (g ` A) f) x" ``` nipkow@33057 ` 548` ```apply (rule the_inv_into_f_eq) ``` nipkow@32961 ` 549` ``` apply (fast intro: comp_inj_on) ``` nipkow@33057 ` 550` ``` apply (simp add: f_the_inv_into_f the_inv_into_into) ``` nipkow@33057 ` 551` ```apply (simp add: the_inv_into_into) ``` nipkow@32961 ` 552` ```done ``` nipkow@32961 ` 553` nipkow@33057 ` 554` ```lemma inj_on_the_inv_into: ``` nipkow@33057 ` 555` ``` "inj_on f A \ inj_on (the_inv_into A f) (f ` A)" ``` nipkow@33057 ` 556` ```by (auto intro: inj_onI simp: image_def the_inv_into_f_f) ``` nipkow@32961 ` 557` nipkow@33057 ` 558` ```lemma bij_betw_the_inv_into: ``` nipkow@33057 ` 559` ``` "bij_betw f A B \ bij_betw (the_inv_into A f) B A" ``` nipkow@33057 ` 560` ```by (auto simp add: bij_betw_def inj_on_the_inv_into the_inv_into_into) ``` nipkow@32961 ` 561` berghofe@32998 ` 562` ```abbreviation the_inv :: "('a \ 'b) \ ('b \ 'a)" where ``` nipkow@33057 ` 563` ``` "the_inv f \ the_inv_into UNIV f" ``` berghofe@32998 ` 564` berghofe@32998 ` 565` ```lemma the_inv_f_f: ``` berghofe@32998 ` 566` ``` assumes "inj f" ``` berghofe@32998 ` 567` ``` shows "the_inv f (f x) = x" using assms UNIV_I ``` nipkow@33057 ` 568` ``` by (rule the_inv_into_f_f) ``` berghofe@32998 ` 569` haftmann@31949 ` 570` haftmann@22845 ` 571` ```subsection {* Proof tool setup *} ``` haftmann@22845 ` 572` haftmann@22845 ` 573` ```text {* simplifies terms of the form ``` haftmann@22845 ` 574` ``` f(...,x:=y,...,x:=z,...) to f(...,x:=z,...) *} ``` haftmann@22845 ` 575` wenzelm@24017 ` 576` ```simproc_setup fun_upd2 ("f(v := w, x := y)") = {* fn _ => ``` haftmann@22845 ` 577` ```let ``` haftmann@22845 ` 578` ``` fun gen_fun_upd NONE T _ _ = NONE ``` wenzelm@24017 ` 579` ``` | gen_fun_upd (SOME f) T x y = SOME (Const (@{const_name fun_upd}, T) \$ f \$ x \$ y) ``` haftmann@22845 ` 580` ``` fun dest_fun_T1 (Type (_, T :: Ts)) = T ``` haftmann@22845 ` 581` ``` fun find_double (t as Const (@{const_name fun_upd},T) \$ f \$ x \$ y) = ``` haftmann@22845 ` 582` ``` let ``` haftmann@22845 ` 583` ``` fun find (Const (@{const_name fun_upd},T) \$ g \$ v \$ w) = ``` haftmann@22845 ` 584` ``` if v aconv x then SOME g else gen_fun_upd (find g) T v w ``` haftmann@22845 ` 585` ``` | find t = NONE ``` haftmann@22845 ` 586` ``` in (dest_fun_T1 T, gen_fun_upd (find f) T x y) end ``` wenzelm@24017 ` 587` wenzelm@24017 ` 588` ``` fun proc ss ct = ``` wenzelm@24017 ` 589` ``` let ``` wenzelm@24017 ` 590` ``` val ctxt = Simplifier.the_context ss ``` wenzelm@24017 ` 591` ``` val t = Thm.term_of ct ``` wenzelm@24017 ` 592` ``` in ``` wenzelm@24017 ` 593` ``` case find_double t of ``` wenzelm@24017 ` 594` ``` (T, NONE) => NONE ``` wenzelm@24017 ` 595` ``` | (T, SOME rhs) => ``` wenzelm@27330 ` 596` ``` SOME (Goal.prove ctxt [] [] (Logic.mk_equals (t, rhs)) ``` wenzelm@24017 ` 597` ``` (fn _ => ``` wenzelm@24017 ` 598` ``` rtac eq_reflection 1 THEN ``` wenzelm@24017 ` 599` ``` rtac ext 1 THEN ``` wenzelm@24017 ` 600` ``` simp_tac (Simplifier.inherit_context ss @{simpset}) 1)) ``` wenzelm@24017 ` 601` ``` end ``` wenzelm@24017 ` 602` ```in proc end ``` haftmann@22845 ` 603` ```*} ``` haftmann@22845 ` 604` haftmann@22845 ` 605` haftmann@21870 ` 606` ```subsection {* Code generator setup *} ``` haftmann@21870 ` 607` berghofe@25886 ` 608` ```types_code ``` berghofe@25886 ` 609` ``` "fun" ("(_ ->/ _)") ``` berghofe@25886 ` 610` ```attach (term_of) {* ``` berghofe@25886 ` 611` ```fun term_of_fun_type _ aT _ bT _ = Free ("", aT --> bT); ``` berghofe@25886 ` 612` ```*} ``` berghofe@25886 ` 613` ```attach (test) {* ``` berghofe@25886 ` 614` ```fun gen_fun_type aF aT bG bT i = ``` berghofe@25886 ` 615` ``` let ``` wenzelm@32740 ` 616` ``` val tab = Unsynchronized.ref []; ``` berghofe@25886 ` 617` ``` fun mk_upd (x, (_, y)) t = Const ("Fun.fun_upd", ``` berghofe@25886 ` 618` ``` (aT --> bT) --> aT --> bT --> aT --> bT) \$ t \$ aF x \$ y () ``` berghofe@25886 ` 619` ``` in ``` berghofe@25886 ` 620` ``` (fn x => ``` berghofe@25886 ` 621` ``` case AList.lookup op = (!tab) x of ``` berghofe@25886 ` 622` ``` NONE => ``` berghofe@25886 ` 623` ``` let val p as (y, _) = bG i ``` berghofe@25886 ` 624` ``` in (tab := (x, p) :: !tab; y) end ``` berghofe@25886 ` 625` ``` | SOME (y, _) => y, ``` berghofe@28711 ` 626` ``` fn () => Basics.fold mk_upd (!tab) (Const ("HOL.undefined", aT --> bT))) ``` berghofe@25886 ` 627` ``` end; ``` berghofe@25886 ` 628` ```*} ``` berghofe@25886 ` 629` haftmann@21870 ` 630` ```code_const "op \" ``` haftmann@21870 ` 631` ``` (SML infixl 5 "o") ``` haftmann@21870 ` 632` ``` (Haskell infixr 9 ".") ``` haftmann@21870 ` 633` haftmann@21906 ` 634` ```code_const "id" ``` haftmann@21906 ` 635` ``` (Haskell "id") ``` haftmann@21906 ` 636` nipkow@2912 ` 637` ```end ``` | 12,939 | 33,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-05 | latest | en | 0.395008 |
https://mikesmathpage.wordpress.com/2020/10/10/revisiting-futility-closets-paradox-of-the-2nd-ace-with-my-younger-son/ | 1,680,382,173,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00320.warc.gz | 445,237,637 | 25,612 | # Revisiting Futility Closet’s “Paradox of the 2nd Ace” with my younger son
This week my younger son told me he wanted to learn a bit more about statistics. By lucky coincidence I happened to stumble on one of our old projects while trying to answer a question on twitter:
This project was about a fun probability problem I learned in this tweet from Jon Cook:
I stared by introducing the problem to my son and asked what he thought the answer was going to be:
Then we started in on the calculations. Finding the probability that someone having “at least one ace” had more than one was a little challenging, but we found the right approach after a few tries:
Next up was calculating the probability that someone who had the ace of spades would have a second ace. After the work we did in the last part, this calculation was easier:
Finally, we went to the Futility Closet page to see the numbers and then I asked my son why he thought the surprising result was true.
Definitely a fun project and a neat probability surprise for kids to see.
## One thought on “Revisiting Futility Closet’s “Paradox of the 2nd Ace” with my younger son”
1. Hmm, it feels like there should be a nice way to simplify those binomial sums, but I don’t see it off-hand. | 279 | 1,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-14 | latest | en | 0.972072 |
https://www.tutorialspoint.com/python-reform-k-digit-elements | 1,680,328,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00760.warc.gz | 1,129,691,521 | 9,658 | # Python – Reform K digit elements
When it is required to reform K digit elements, a list comprehension and the ‘append’ method are used.
## Example
Below is a demonstration of the same
my_list = [231, 67, 232, 1, 238, 31, 793]
print("The list is :")
print(my_list)
K = 3
print("The value of K is ")
print(K)
temp = ''.join([str(ele) for ele in my_list])
my_result = []
for index in range(0, len(temp), K):
my_result.append(int(temp[index: index + K]))
print("The resultant list is :")
print(my_result)
## Output
The list is :
[231, 67, 232, 1, 238, 31, 793]
The value of K is
3
The resultant list is :
[231, 672, 321, 238, 317, 93]
## Explanation
• A list is defined and is displayed on the console.
• The value for K is initialized and is displayed on the console.
• A list comprehension is used to iterate over the elements in the list and convert it to a string type, and join it by spaces.
• This is assigned to a variable.
• An empty list is defined.
• The value up to K is iterated over and the elements from index 0 to K is appended to the empty list.
• This is the output which is displayed on the console. | 316 | 1,137 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-14 | latest | en | 0.803513 |
https://healthyincandyland.com/qa/question-what-is-single-tone-signal.html | 1,603,695,366,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890586.57/warc/CC-MAIN-20201026061044-20201026091044-00491.warc.gz | 357,593,943 | 8,497 | # Question: What Is Single Tone Signal?
## What is the frequency of a signal?
Frequency is the number of occurrences of a repeating event per unit of time.
It is also referred to as temporal frequency, which emphasizes the contrast to spatial frequency and angular frequency.
Frequency is measured in units of hertz (Hz) which is equal to one occurrence of a repeating event per second..
## What is the disadvantage of FM over AM?
Disadvantages of FM over AM are FM systems have a much wider bandwidth than AM systems and therefore more prone to selective fading. … FM transmitter and receiver require booster circuits as they have poorer signal to noise ratio at high audio frequencies.
AM transmissions are much more susceptible than FM or digital signals are to interference, and often have lower audio fidelity. Thus, AM broadcasters tend to specialise in spoken-word formats, such as talk radio, all news and sports, leaving the broadcasting of music mainly to FM and digital stations.
## What is the best frequency?
Furthermore, 432 Hz resonates with 8 Hz (the Schumann Resonance), the documented fundamental electromagnetic “beat” of Earth. It just feels better. Research says that music tuned from this frequency is easier to listen to, brighter, clearer, and contains more inherent dynamic range.
## What is single tone frequency modulation?
Single Tone Frequency Modulation. For the single tone frequency modulation,i.e the modulating signal x(t) be a sinusoidal signal of amplitude Em and frequency fm . Therefore, x(t) = Em cos (2πfmt) The unmodulated carrier is represented by the expression : ec = Ec sin (ωct + φ)
## What is the difference between carrier signal and message signal?
The sinusoidal signal that is used in the modulation is known as the carrier signal, or simply “the carrier”. The signal that is used in modulating the carrier signal(or sinusoidal signal) is known as the “data signal” or the “message signal”.
## Which is better AM or FM?
FM is less prone to interference than AM. However, FM signals are impacted by physical barriers. FM has better sound quality due to higher bandwidth. … In AM radio broadcasting, the modulating signal has bandwidth of 15kHz, and hence the bandwidth of an amplitude-modulated signal is 30kHz.
## What are the advantages of frequency modulation?
Advantage of frequency modulation :Amplitude of frequency modulation signal is remain constant.Less susceptible to noise.Provides good sound quality.More efficient use of power.Operate in very high frequency.
## Why does AM radio exist?
One reason AM broadcast is still around it is used mostly for talk radio. There are a few FM talk radio stations, most are on AM. Talk radio does not require the quality of audio that FM can provide with music, so many AM stations went with talk radio. … So many AM stations remain on because of this.
## What is the frequency of space?
Exactly 47.71 MHz. The entire universe would begin to contract and the world as we know it would end.
## How do you calculate frequency?
Divide the wavelength into the velocity to calculate the frequency, expressed as described above as the number of cycles per second, or Hertz – written “Hz.” For example, a water wave with a wavelength of 1 foot traveling at a speed of 4 inches per second has a frequency of 1/3 feet/second divided by 1 foot = . 33 Hz.
## What is tone modulation?
Tone modulation is “a modulation in which the modulation is carried out by a single frequency (tone) signal”. The toned (single frequency) modulating signal consists of only one frequency component and this signal is modulated with a carrier signal. … Consider the general equation for amplitude modulation signal. | 767 | 3,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-45 | latest | en | 0.952718 |
pseudoprofound.wordpress.com | 1,500,935,762,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424931.1/warc/CC-MAIN-20170724222306-20170725002306-00683.warc.gz | 718,452,246 | 37,578 | # Recursive (not recurrent!) Neural Nets in TensorFlow
For the past few days I’ve been working on how to implement recursive neural networks in TensorFlow. Recursive neural networks (which I’ll call TreeNets from now on to avoid confusion with recurrent neural nets) can be used for learning tree-like structures (more generally, directed acyclic graph structures). They are highly useful for parsing natural scenes and language; see the work of Richard Socher (2011) for examples. More recently, in 2014, Ozan İrsoy used a deep variant of TreeNets to obtain some interesting NLP results.
The best way to explain TreeNet architecture is, I think, to compare with other kinds of architectures, for example with RNNs:
In RNNs, at each time step the network takes as input its previous state s(t-1) and its current input x(t) and produces an output y(t) and a new hidden state s(t). TreeNets, on the other hand, don’t have a simple linear structure like that. With RNNs, you can ‘unroll’ the net and think of it as a large feedforward net with inputs x(0), x(1), …, x(T), initial state s(0), and outputs y(0),y(1),…,y(T), with T varying depending on the input data stream, and the weights in each of the cells tied with each other. You can also think of TreeNets by unrolling them – the weights in each branch node are tied with each other, and the weights in each leaf node are tied with each other. The TreeNet illustrated above has different numbers of inputs in the branch nodes. Usually, we just restrict the TreeNet to be a binary tree – each node either has one or two input nodes. There may be different types of branch nodes, but branch nodes of the same type have tied weights.
The advantage of TreeNets is that they can be very powerful in learning hierarchical, tree-like structure. The disadvantages are, firstly, that the tree structure of every input sample must be known at training time. We will represent the tree structure like this (lisp-like notation):
`(S (NP that movie) (VP was) (ADJP cool))`
In each sub-expression, the type of the sub-expression must be given – in this case, we are parsing a sentence, and the type of the sub-expression is simply the part-of-speech (POS) tag. You can see that expressions with three elements (one head and two tail elements) correspond to binary operations, whereas those with four elements (one head and three tail elements) correspond to trinary operations, etc.
The second disadvantage of TreeNets is that training is hard because the tree structure changes for each training sample and it’s not easy to map training to mini-batches and so on.
### Implementation in TensorFlow
There are a few methods for training TreeNets. The method we’re going to be using is a method that is probably the simplest, conceptually. It consists of simply assigning a tensor to every single intermediate form. So, for instance, imagine that we want to train on simple mathematical expressions, and our input expressions are the following (in lisp-like notation):
```1
(+ 1 2)
(* (+ 2 1) 2)
(+ (* 1 2) (+ 2 1))```
Now our full list of intermediate forms is:
```a = 1
b = 2
c = (+ a b)
d = (+ b a)
e = (* d b)
f = (* a b)
g = (+ f d)```
For example, `f = (* 1 2)`, and `g = (+ (* 1 2) (+ 2 1))`. We can see that all of our intermediate forms are simple expressions of other intermediate forms (or inputs). Each of these corresponds to a separate sub-graph in our tensorflow graph. So, for instance, for `*`, we would have two matrices `W_times_l` and `W_times_r`, and one bias vector `bias_times`. And for computing `f`, we would have:
`f = relu(W_times_l * a + W_times_r * b + bias_times)`
Similarly, for computing d we would have:
`d = relu(W_plus_l * b + W_plus_r * a + bias_plus)`
The full intermediate graph (excluding input and loss calculation) looks like:
```a = W_input * [1, 0]
b = W_input * [0, 1]
c = relu(W_plus_l * a + W_plus_r * b + bias_plus)
d = relu(W_plus_l * b + W_plus_r * a + bias_plus)
e = relu(W_times_l * d + W_times_r * b + bias_times)
f = relu(W_times_l * a + W_times_r * b + bias_times)
g = relu(W_plus_l * f + W_plus_r * d + bias_plus)
output1 = sigmoid(W_output * a)
output2 = sigmoid(W_output * c)
output3 = sigmoid(W_output * e)
output4 = sigmoid(W_output * g)```
For training, we simply initialize our inputs and outputs as one-hot vectors (here, we’ve set the symbol `1` to `[1, 0]` and the symbol `2` to `[0, 1]`), and perform gradient descent over all W and bias matrices in our graph. The advantage of this method is that, as I said, it’s straightforward and easy to implement. The disadvantage is that our graph complexity grows as a function of the input size. This isn’t as bad as it seems at first, because no matter how big our data set becomes, there will only ever be one training example (since the entire data set is trained simultaneously) and so even though the size of the graph grows, we only need a single pass through the graph per training epoch. However, it seems likely that if our graph grows to very large size (millions of data points) then we need to look at batch training.
Batch training actually isn’t that hard to implement; it just makes it a bit harder to see the flow of information. We can represent a ‘batch’ as a list of variables: `[a, b, c]`. So, in our previous example, we could replace the operations with two batch operations:
```[a, b] = W_input * [[1, 0], [0, 1]]
[c, d, g] = relu(W_plus_l * [a, b, f] + W_plus_r * [b, a, d] + bias_plus)
[e, f] = relu(W_times_l * [d, a] + W_times_r * [b, b] + bias_times)
output = sigmoid(W_output * [a, c, e, g])```
You’ll immediately notice that even though we’ve rewritten it in a batch way, the order of variables inside the batches is totally random and inconsistent. This is the problem with batch training in this model: the batches need to be constructed separately for each pass through the network. If we think of the input as being a huge matrix where each row (or column) of the matrix is the vector corresponding to each intermediate form (so `[a, b, c, d, e, f, g]`) then we can pick out the variables corresponding to each batch using tensorflow’s `tf.gather` function. So for instance, gathering the indices `[1, 0, 3]` from `[a, b, c, d, e, f, g]` would give `[b, a, d]`, which is one of the sub-batches we need. The total number of sub-batches we need is two for every binary operation and one for every unary operation in the model.
For the sake of simplicity, I’ve only implemented the first (non-batch) version in TensorFlow, and my early experiments show that it works. For example, consider predicting the parity (even or odd-ness) of a number given as an expression. So `1` would have parity 1, `(+ 1 1)` (which is equal to 2) would have parity 0, `(+ 1 (* (+ 1 1) (+ 1 1)))` (which is equal to 5) would have parity 1, and so on. Training a TreeNet on the following small set of training examples:
```1
[+,1,1]
[*,1,1]
[*,[+,1,1],[+,1,1]]
[+,[+,1,1],[+,1,1]]
[+,[+,1,1],1 ]
[+,1,[+,1,1]]```
Seems to be enough for it to ‘get the point’ of parity, and it is capable of correctly predicting the parity of much more complicated inputs, for instance:
`[+,[+,[+,1,1],[+,[+,1,1],[+,1,1]]],1]`
Correctly, with very high accuracy (>99.9%), with accuracy only diminishing once the size of the inputs becomes very large. The code is just a single python file which you can download and run here.
## 6 thoughts on “Recursive (not recurrent!) Neural Nets in TensorFlow”
1. Christopher Snyder says:
Interesting. I had not seen this example before. There is a standing issue in tensorflow for implementing recursive networks, and I’m not sure what the status is. I think that the main roadblock is that recursive networks like the Socher paper you reference need to dynamically create the computation graph from intermediate results.
In other cases where there are few (in your example there is only one possible) computation graph to consider, I think it currently works well.
Also, I would mention to the interested reader that a lot of the time that training examples may have multiple correct tree-structures.
Like
1. In my example there is more than one possible computation graph.
The computation graphs are combined together and trained simultaneously.
Like
2. Have you considered using TensorFlow’s iteration constructs for this? If the graph is in topologically sorted order (for a tree this is just postorder) you can sweep through the entire graph in a linear fashion, both forward and back. You can then accommodate graphs of any size, up to the maximum number of nodes that you preallocate. Iteration is done entirely within TensorFlow rather than Python.
Like | 2,202 | 8,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-30 | longest | en | 0.928523 |
http://openstudy.com/updates/56461541e4b01c719e703eeb | 1,519,599,848,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00148.warc.gz | 265,798,466 | 8,252 | • anonymous
Betty earns $52.65 per week working as a lifeguard. She worked as a lifeguard for 5 weeks. She also earned$46.25 painting faces at a carnival. She spent $16.25 of her total earnings to buy a book. Betty put of the remaining money in her savings account. Betty used the following steps to find the total money she put in her saving account: Step 1: [($52.65 ⋅ 5) + $46.25 −$16.25] ⋅ 1/6 Step 2: [$263.25 +$46.25 − $16.25] ⋅ 1/6 Step 3: 293.25 ⋅ Step 4:$1,759.50 In which step did Betty first make an error? Use words to explain how Betty can correct her error.
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Looking for something else?
Not the answer you are looking for? Search for more explanations. | 429 | 1,501 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-09 | latest | en | 0.659329 |
https://innerself.com/content/living/science-a-technology/16083-4-easy-steps-can-make-you-a-math-whiz.html?tmpl=component&print=1&layout=default&page= | 1,516,094,406,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886397.2/warc/CC-MAIN-20180116090056-20180116110056-00626.warc.gz | 725,494,044 | 6,536 | # 4 Easy Steps Can Make You A Math Whiz
Many people find mathematics daunting. If true, this piece is for you. If not, this piece is still for you.
What do you think of when you think about mathematics? Perhaps you think about x’s and y’s, intractable fractions, or nonsensical word problems. The cartoonist Gary Larson once depicted hell’s library as containing only giant tomes of word problems. You know, “If a train leaves New York…”
I was trained as a mathematician, and I will let you in on a trade secret: That is not what mathematics is, nor where it lives. It’s true that learning mathematics often involves solving problems, but it should focus on the joy of solving puzzles, rather than memorizing rules.
I invite you to see yourself as a problem solver and mathematician. And I’d like to introduce you to the man who once invited me to the study of problem solving: George Pólya.
## Math Pólya’s way
For many reasons, not the least of which is that Pólya died in 1985, you will meet him as I did – through his wildly successful “How to Solve It.” Penned in 1945, this book went on to sell over one million copies and was translated into 17 languages.
As a mathematician, Pólya worked on a wide range of problems, including the study of heuristics, or how to solve problems. When you read “How to Solve It,” it feels like you’re taking a guided tour of Pólya’s mind. This is because his writing is metacognitive – he writes about how he thinks about thinking. And metacognition is often the heart of problem solving.
Pólya’s problem solving plan breaks down to four simple steps:
1. Make sure you understand the problem.
2. Make a plan to solve the problem.
3. Carry out the plan.
There it is. Problem solving in the palm of your hand – math reduced to four steps.
Here’s a classic problem from research on mathematics education done by Jean Lave. A man, let’s call him John, is making ¾ of a recipe that calls for 2/3 cup of cottage cheese. What do you think John did? What would you do?
If you’re like me, you might immediately dive into calculations, perhaps struggling with what the fractions mean, working to remember the rules for arithmetic. That’s what John seemed to do, at first. But then he had a Eureka! moment.
John measured 2/3 cup of cottage cheese, then dumped it onto a cutting board. He patted the cheese into a circle and drew lines into it, one vertical, one horizontal, dividing the cheese patty into quarters. He then carefully pushed one quarter of the cottage cheese back into its container. Voilá! Three-quarters of 2/3 cup of cottage cheese remained.
John is a mathematician and problem solver. First, he understood the problem: He needed ¾ of what the recipe called for, which was 2/3 cup. Then, he made a plan, most likely visualizing in his head how he would measure and divide the cottage cheese. Finally, he carried out the plan.
Did he check his answer? That remains unclear, but we can check the validity of his work for him. Did he indeed end up with ¾ of 2/3 cup of cottage cheese? Yes, because the full amount was reduced by one-quarter, leaving three-quarters.
## Another approach
Would this solution work with different foods or serving sizes? So long as a person could divide that serving into quarters, yes, the plan would work.
Could we solve the problem another way with the same result? Sure — there are many ways to solve this problem, and they should all result in the same half-cup answer. Here is one.
How to find 3/4 of 2/3.
Jennifer Ruef, CC BY
Notice that this solution uses pictures. New brain research validates what mathematics educators have been saying for decades: Pictures help us think. Drawing pictures also happens to be another of Pólya’s suggestions.
John probably made use of one of Pólya’s most important suggestions: Can you think of a related problem?
Of course, this is a cheesy problem – sorry, I really didn’t even try to fight that pun – which is a common complaint about story problems. I chose it because it has delighted math researchers for years, and because John is quite clever in his solution. He is also extremely mathematical.
I’ve taught mathematics, and how to teach mathematics, for nearly 30 years. For over a decade, it was my job to convince high school freshmen not only that algebra was meaningful, but that it was meant for them, and they for it. In my work, I’ve met many people who love mathematics and many who find it overwhelming and nonsensical. And so it’s an important part of my work to help people see the beauty and wonder of mathematics, and think of themselves as mathematicians.
These messages are especially important for parents helping children learn mathematics. If you understand the problem you’re trying to solve, you’re well on your way to solving it. And you, yes you, are a problem solver.
We all know it’s not always so simple to solve problems. Pólya did too. That’s the glory of it – the messy, wonderful, powerful adventure.
Jennifer Ruef, Assistant Professor of Education Studies, University of Oregon
### Related Books:
##### How to Solve It: A New Aspect of Mathematical Method (Princeton Science Library)
Author: G. Polya
Binding: Paperback
Publisher: Princeton University Press
List Price: \$19.95
##### Mathematics and Plausible Reasoning [Two Volumes in One]
Author: George Polya
Binding: Paperback
Publisher: Martino Fine Books
List Price: \$25.00 | 1,221 | 5,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-05 | latest | en | 0.967401 |
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# Brilliant Ideas Of Decimal Place Value Resources Teaching On Games 5th Grade Math Worksheets Fraction To
By Fleurette Auger at December 03 2018 21:44:59
It's helpful having printable worksheets for something like this, because parents often go through quite a few of these before the child masters writing the numbers or letters correctly. What is my goal? It must be specific, challenging and attainable. Many people set but don't achieve their goals because the goals are simply too vague, too small or too big. For instance, "I'm going to get in shape this year" is a very poor goal.
At the grassroots level, teachers in schools are given a packed curriculum for the year. Schools try to teach the students a number of procedures without delving much into its finer details. Hence, the student is left in a confounding position as to when a particular procedure must be used. The key ingredient to understanding math is constant practice and math assignment help. Unfortunately, this is not a common scenario among the popular math classes.Connect The Letter To The Correct Sound/Word: These are activities where you draw a line between a letter and the picture items that start with that letter. For instance, you'd draw a line from the letter A to the word "Apple" and the letter L to the word "Lemon". This activity is good, but takes a lot of monitoring to make sure that students are correctly connecting the letters. It's best as a homework activity, where parents can help to make sure their children are correctly connecting the letters to the words.
## Gallery of Decimal Games 5th Grade
In math bingo, each student is given a bingo card (also known as a "bingo worksheet" or "bingo board") printed with numbers. These aren't necessarily the standard bingo numbers, but rather are the answers to a number of different math problems. The game is then played exactly like a normal game of bingo, with the teacher playing the part of the bingo caller, but instead of the teacher calling out the numbers printed on the cards, the teacher instead calls out math problems (the teacher may also write the problem on the blackboard). The students' task is to solve each problem, and then look for the number on their bingo card. If you are looking for an article that describes the basics of Excel and introduces the interface and concepts for beginners, you have come to the right place. Microsoft Excel is a powerful business application that is organized into a structural hierarchy of Workbooks, Worksheets, and Cells.
Parents can go onto the various online portals to find suitable option to help their children learn math solving problem skills. Thorough research over the internet will give a wide range of websites that offers right set of math worksheets for children. 3rd grade math worksheets problems are experienced by many kids and their parents are usually frustrated. Not any more, there is information on how to solve the problem of your kids math homework. So what does the research show? The scientifically-based research shows that there are specific strategies that enhance reading for comprehension. Reading strategies are detailed steps students implement before reading the text, while reading the text and after reading the text. The strategies are purposeful in enhancing comprehension. | 660 | 3,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-18 | latest | en | 0.958268 |
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167 - The Sultan's Successors
Why does this code get WA? I just use formula to get the 8 positions and rotate the board in 4 directions to get the max sum. Where are the bugs?
[c]
#include<stdio.h>
void main(void)
{
int board[8][8],k,x,y,z,ans[4],max;
scanf("%d",&k);
for(x=0;x<k;x++)
{
for(y=0;y<8;y++)
for(z=0;z<8;z++)
scanf("%d",&board[y][z]);
ans[0]=board[0][5]+board[1][0]+board[2][4]+board[3][1]+board[4][7]+board[5][2]+board[6][6]+board[7][3];
ans[1]=board[0][6]+board[1][4]+board[2][2]+board[3][0]+board[4][5]+board[5][7]+board[6][1]+board[7][3];
ans[2]=board[0][4]+board[1][1]+board[2][5]+board[3][0]+board[4][6]+board[5][3]+board[6][7]+board[7][2];
ans[3]=board[0][4]+board[1][6]+board[2][0]+board[3][2]+board[4][7]+board[5][5]+board[6][3]+board[7][1];
for(y=0,max=0;y<4;y++)
if(ans[y]>max)
max=ans[y];
printf("%5d\n",max);
}
}
[/c]
visser
New poster
Posts: 8
Joined: Mon Jun 10, 2002 11:13 am
Location: Netherlands
There are a bit more solutions than 4 for the 8-queen problem.
htl
Experienced poster
Posts: 185
Joined: Fri Jun 28, 2002 12:05 pm
Location: Taipei, Taiwan
[c]
#include<stdio.h>
void main(void)
{
int board[8][8],k,x,y,z,ans[8],max;
scanf("%d",&k);
for(x=0;x<k;x++)
{
for(y=0;y<8;y++)
for(z=0;z<8;z++)
scanf("%d",&board[y][z]);
ans[0]=board[0][5]+board[1][0]+board[2][4]+board[3][1]+board[4][7]+board[5][2]+board[6][6]+board[7][3];
ans[1]=board[0][6]+board[1][4]+board[2][2]+board[3][0]+board[4][5]+board[5][7]+board[6][1]+board[7][3];
ans[2]=board[0][4]+board[1][1]+board[2][5]+board[3][0]+board[4][6]+board[5][3]+board[6][7]+board[7][2];
ans[3]=board[0][4]+board[1][6]+board[2][0]+board[3][2]+board[4][7]+board[5][5]+board[6][3]+board[7][1];
ans[4]=board[0][2]+board[1][7]+board[2][3]+board[3][6]+board[4][0]+board[5][5]+board[6][1]+board[7][4];
ans[5]=board[0][3]+board[1][1]+board[2][7]+board[3][5]+board[4][0]+board[5][2]+board[6][4]+board[7][6];
ans[6]=board[0][3]+board[1][6]+board[2][2]+board[3][7]+board[4][1]+board[5][4]+board[6][0]+board[7][5];
ans[7]=board[0][1]+board[1][3]+board[2][5]+board[3][7]+board[4][2]+board[5][0]+board[6][6]+board[7][4];
for(y=0,max=0;y<8;y++)
if(ans[y]>max)
max=ans[y];
printf("%5d\n",max);
}
}
[/c]
xenon
Learning poster
Posts: 100
Joined: Fri May 24, 2002 10:35 am
Location: Scheveningen, Holland
Do some research. There are quite a bit more solutions to the 8queen
htl
Experienced poster
Posts: 185
Joined: Fri Jun 28, 2002 12:05 pm
Location: Taipei, Taiwan
I found 12 fundamental solutions. Then I rotate one of them in 4 directions and turn over to rotate it in 4 directions again. Then I got 8*12=96 solutions. Am I right?
xenon
Learning poster
Posts: 100
Joined: Fri May 24, 2002 10:35 am
Location: Scheveningen, Holland
close, but no sigar
You're allmost there
Some of the fundamental solutions will have symmetry, so they will remain unchanged after rotation or reflection. The actual number is...
[SPOILER] XCII [/SPOILER]
Happy hunting,
-xenon
htl
Experienced poster
Posts: 185
Joined: Fri Jun 28, 2002 12:05 pm
Location: Taipei, Taiwan
So I have to calculate the solutions instead of list all of them?
xenon
Learning poster
Posts: 100
Joined: Fri May 24, 2002 10:35 am
Location: Scheveningen, Holland
Well, I think it is for you to decide how you implement the problem.
Basically, anything is alowed as long as it gives the correct answer in the correct format within 30 seconds
soyoja
Experienced poster
Posts: 106
Joined: Sun Feb 17, 2002 2:00 am
Location: Seoul, South Korea
Contact:
I have a question.
I try to solve this problem using general backtracking algorithm.
So I get 22 solution.
But what is the difference between general 8 - queen solution and this problem's solution?
Only pivoting and rotating my 8 - queen solution is this problem's answer?
If it is true, then why general 8 - queen solution can't catch all possible
case? ( Xenon said that there are total 92 solutions )
My program is wrong or other explaination is exist?
Dominik Michniewski
Guru
Posts: 834
Joined: Wed May 29, 2002 4:11 pm
Location: Wroclaw, Poland
Contact:
If I correct remember, after rotating and mirroring all solutions generating from backtrack we have 192 possible solutions ...
Maybe this help us
DM
If you really want to get Accepted, try to think about possible, and after that - about impossible ... and you'll get, what you want ....
Born from ashes - restarting counter of problems (800+ solved problems)
soyoja
Experienced poster
Posts: 106
Joined: Sun Feb 17, 2002 2:00 am
Location: Seoul, South Korea
Contact:
Oh... I'm sorry. My program was wrong.
In the classic n - queens problem, 8*8 chess board has 92 solutions.
( From the description of problem 10401. )
Dominik Michniewski
Guru
Posts: 834
Joined: Wed May 29, 2002 4:11 pm
Location: Wroclaw, Poland
Contact:
Sorry, (
I check my program(Acc) and I saw that I use only 92 solutions for 8-queen problem ....
That must be other problem with your code ...
After generating all possibilities you have only to find maximum sum of fields, in which queen are placed .... maybe your search algorithm isn't good ?
DM
If you really want to get Accepted, try to think about possible, and after that - about impossible ... and you'll get, what you want ....
Born from ashes - restarting counter of problems (800+ solved problems)
soyoja
Experienced poster
Posts: 106
Joined: Sun Feb 17, 2002 2:00 am
Location: Seoul, South Korea
Contact:
At last, I accepted my code.
It's only my mistake using backtracking algorithm.
Using general N - Queen algorithm, this problem can solve easily.
Pavl0
New poster
Posts: 16
Joined: Sun Apr 18, 2004 2:57 pm
my prog find all 92 positions at start prog end use it far all cases
J get AC
J use iterratin methot for nqp for 8 is fast
Ali Arman Tamal
Learning poster
Posts: 76
Joined: Sat Jan 15, 2005 5:04 pm
Location: Dhaka
Contact:
HINT FOR 167 [Sultan's Successor]
There will be 92 non-attacking combinations.
You will have to use Backtracking to find them all.
Hope it helps !! | 2,058 | 6,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-40 | latest | en | 0.392353 |
https://byjusexamprep.com/tips-to-solve-reasoning-puzzles-rbi-assistant-mains-i | 1,660,908,420,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573667.83/warc/CC-MAIN-20220819100644-20220819130644-00274.warc.gz | 166,772,099 | 70,250 | # Tips & Tricks to Solve Reasoning Puzzles Fast for RBI Assistant Mains
By Gaurav Singh Bhadouria|Updated : May 1st, 2020
Tips & Tricks to Solve Reasoning Puzzles Fast in RBI Assistant Mains! We all are very much aware of the present situation across the globe. The situation has led to the cancellation and delay in the conduct of many competitive exams. One such exam is RBI Assistant Mains 2020. After a gap of 3 years, RBI had conducted it, so students had a great expectation from this exam. In such a condition, students should not lose focus on their exam preparation and should continue their preparation with full efforts.
Talking of any Mains exam, all the sections play an important role in clearing and scoring good marks. In this article, we will discuss how to approach and effective tips to solve the reasoning puzzles fast in RBI Assistant Mains 2020 exam. Puzzles, as we all know are asked in a good number. In the main exam, puzzles might be asked of around 20-25 marks out of a total of 40 marks. Once you have done enough practice of puzzles, you can easily score well in the reasoning ability section.
Get Banking Exams Important Updates, Study-Notes, Free PDF's & more, Join BYJU'S Exam Prep Banking Telegram Group Join Now
## Tips to Solve Reasoning Puzzles Fast for RBI Assistant Mains
### Types of Puzzles:
• Linear Puzzle
• Circular Puzzle
• Floor/Bbox based Puzzles
• Scheduling Puzzles
• An uncertain number of persons based Puzzles
• Square/Rectangle Arrangement based Puzzle
• Combination of above Puzzles
The candidates should go through the complete puzzle at once before attempting it. Focus on the following important points
• Number of parameters given
• Number of positive statements
• The number of independent statements/dependent statements.
• Check the direction of sitting
• Take care of words like adjacent, just above or below, not more than, exactly mid, floors above, etc
### Right Approach to attempt a puzzle in the examination:
1. Analyze and interpret all the direct statements given in the puzzle
2. Also, jot down all the negative statements given in the puzzle
3. Write them down on a piece of paper in a simple form such that you do not have to look at the given puzzle time and again. E.g. if the statement is “A sits 4th to the right of B” then you can simply write it as “A →4R B”. This will save your time spent in writing down information.
4. In case of multiple possibilities, draw each one possibility and rule out them one by one till you are left with just one final answer
5. In the end, you will be left with only one case that will be in sync with all the positive as well as the negative statements
It is always better to understand how to solve a puzzle with the help of an example. So, let us now solve one question in detail along with its solution.
Example:
A certain number of people are seating around a circular table facing the center. At most eight persons are sitting between Z and I. Three persons are sitting between O and L. Three persons are sitting between Z and N. N sits to the immediate left of L. From the right of J, six people are sitting between J and K. I sits third to the left of J. L sits fifth to the right of K. M sits fourth to the right of O. M sits fourth to the left of I.
Solution:
1) I sits third to the left of J.
2) From the right of J, six people are sitting between J and K.
3) L sits fifth to the right of K.
4) N sits immediate left of L.
5) M sits fourth to the right of O
6) M sits fourth to the left of I.
7) Three persons are sitting between Z and N.
8) Three persons are sitting between O and L.
9) At most eight persons are sitting between Z and I.
Therefore, option D is the correct answer.
We hope it will be clear enough now as how to approach a puzzle. Hope you will find this article helpful and relevant. | 890 | 3,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-33 | longest | en | 0.922302 |
https://hyblunt.com/qa/question-what-is-a-formula-of-percentage.html | 1,620,624,318,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00522.warc.gz | 333,413,300 | 8,755 | # Question: What Is A Formula Of Percentage?
## How do I find the percentage of two numbers without a calculator?
Divide the number by 10 to find 10%.
In this case, 10% is 24.
Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72.
## What is percentage of a number?
In mathematics, a percentage is a number or ratio that represents a fraction of 100. It is often denoted by the symbol “%” or simply as “percent” or “pct.” For example, 35% is equivalent to the decimal 0.35, or the fraction.
## How do you find 70 percent of a number?
Example 1. Find 70% of 80. Following the shortcut, we write this as 0.7 × 80. Remember that in decimal multiplication, you multiply as if there were no decimal points, and the answer will have as many “decimal digits” to the right of the decimal point as the total number of decimal digits of all of the factors.
## What grade is a 56 out of 80?
70%Convert fraction (ratio) 56 / 80 Answer: 70%
## How do you find 20% of a number?
If you know what the whole number is and you know what percent of that number you are looking for, you multiply. For example, if you are looking for 20% of 100, you multiply 100 by 0.2. If you want to find what percent of 100 is equal to 20, you would divide 100 by 20.
## How do I figure out a percentage of two numbers?
If you want to know what percent A is of B, you simple divide A by B, then take that number and move the decimal place two spaces to the right. That’s your percentage! To use the calculator, enter two numbers to calculate the percentage the first is of the second by clicking Calculate Percentage.
## How do I calculate percentage of a total?
Example: how to calculate percent of total: Find percent of total for each of the following numbers: 100, 400 and 600. First, find the total. Add up 100 + 400 + 600 = 1,100. Next, let’s figure out what percent of our 1,100 total is 100.
## What is discount formula?
Find the original price (for example \$90 ) Get the the discount percentage (for example 20% ) Calculate the savings: 20% of \$90 = \$18. Subtract the savings from the original price to get the sale price: \$90 – \$18 = \$72.
## How do I calculate percentage on calculator?
How to Calculate Percentages with a CalculatorIf your calculator has a “%” button. Let’s say you wanted to find 19 percent of 20. Press these buttons: 1 9 % * 2 0 = … If your calculator does not have a “%” button. Step 1: Remove the percent sign and add a couple of zeros after the decimal point. 19% becomes 19.00.Nov 29, 2014
## How do I calculate mean?
The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.
## What is algebra formula?
Algebra includes both numbers and letters. Numbers are fixed, i.e. their value is known. Letters or alphabets are used to represent the unknown quantities in the algebra formula.
## How do you find 15% of a number?
15% is 10% + 5% (or 0.15 = 0.1 + 0.05, dividing each percent by 100). Thinking about it this way is useful for two reasons. First, it’s easy to multiply any number by 0.1; just move the decimal point left one digit. For example, 75.00 x 0.1 = 7.50, or 346.43 x 0.1 = 34.64 (close enough).
## How do I calculate a percentage?
How To Calculate PercentDetermine the total or whole amount.Divide the number to be expressed as a percent by the total. In most cases, you’ll divide the smaller number by the larger number.Multiple the resulting value by 100.Feb 6, 2020
## What is percentage and its formula?
If we have to calculate percent of a number, divide the number by whole and multiply by 100. Hence, the percentage means, a part per hundred. The word per cent means per 100. It represented by the symbol “%”….Percentage Chart.FractionsPercentage1/250%1/333.33%1/425%1/520%10 more rows
## What’s the easiest way to calculate percentages?
Generally, the way to figure out any percentage is to multiply the number of items in question, or X, by the decimal form of the percent. To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1. | 1,124 | 4,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-21 | latest | en | 0.92623 |
https://id.scribd.com/doc/315057507/modul-tropikal-kimia-garam | 1,618,817,137,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00134.warc.gz | 423,188,187 | 102,672 | Anda di halaman 1dari 20
# Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
PROGRAM KECEMERLANGAN
PANITIA KIMIA
2016
## Modul Topikal : GARAM
Aktiviti 1 : Tentukan keterlarutan garam (GL : garam larut ; GTL : Garam tak larut
No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Formula of Salt
PbCO3
NaCl
CaSO4
AgNO3
K2CO3
FeCl3
Na2SO4
NH4NO3
CuSO4
PbCl2
ZnCO3
Ca(NO3)2
Na2CO3
AgCl
PbSO4
Pb(NO3)2
(NH4)2CO3
Mg(NO3)2
Na2SO4
Solubility ( , X )
No
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
Solubility ( , X )
Formula of Salt
MgCO3
KCl
(NH4)2SO4
Cu(NO3)2
NaNO3
CaCl2
BaSO4
KNO3
Ag2CO3
MgCl2
ZnSO4
Ba(NO3)2
FeCO3
NH4Cl
Fe(NO3)3
MgSO4
BaCO3
ZnCl2
FeSO4
Activity 2 :
Classify the following salts into soluble salt and insoluble salt in water.
Kelaskan garam berikut kepada garam terlarut dan garam tak terlarutkan dalam air
Calcium carbonate
Zinc chloride
Silver chloride
Magnesium chloride, MgCl2
Potassium carbonate
Lead (II) sulphate, PbSO4
## Lead (II) nitrate
Ammonium chloride
Answer ;
Soluble salt / Garam terlarut
## Insoluble salt / Garam tak terlarut
Activity 3 : tuliskan ion-ion yang hadir. Tuliskan persamaan ion yang terlibat
Insoluble salt
Ions present
Ag+ ,
## Magnesium carbonate, MgCO3
Ionic equation
Cl-
.. , ..
.. + ..
..
.. + ..
..
Activity 4 :
a) name solution T / nama larutan T
add dilute
Zinc, Zn
metal
nitric acid
HNO3
Colourless
solution, T
Gas W
..
b) write a balanced chemical equation for the
reaction.
Tulis persamaan kimia seimbang bagi tindak balas
Diagram 1
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
## c) name gas W / nama gas W
..
Solid K
add dilute, Z
Pepejal K
tambah Z cair
a) i) name solid K .: ..
nama pepejal K :
Colourless solution,MgCl2
Larutan tak berwarna
ii) name Z .
Nama Z
: ..
reaction.
## Colourless gas produces,M
Gas tak berwarna terbebas,M
Water, H2O
## c) state the confirmatory test the present of M gas.
Nyatakan ujian pengesahan kehadiran gas M.
Air, H2O
..
Activity 5 :
1. The following equation represents the reaction between
silver nitrate solution and hydrochloric acid.
AgNO3 (aq) + NaCl
(aq)
AgCl
(s)
+ NaNO3 (aq)
## (i) name the type of reaction.
.
(ii) write the ionic equation for the reaction.
tuliskan persamaan ion bagi tindak balas.
2.
add dilute
Lead (II) nitrate
Pb(NO3)2
White precipitate
T
sulphuric acid,
H2SO4
reaction.
..
## c) write an ionic equation for the reaction
tuliskan persamaan ion bagi tindak balas.
Activity 6 :
Diagram 1 shows a series of reaction for zinc compound
Rajah 1 menunjukkan satu siri tindak balas bagi sebatian
zink
Zinc nitrate
Zink nitrat
Sodium carbonate
Natriumk karbonat
## i) What is the colour of zinc carbonate ?
Apakah warna zink karbonat?
Zinc carbonate
Zink karbonat
## Zinc nitrate solution reacts with sodium carbonate solution
to form zinc carbonate precipitate.
Larutan zink nitrat bertindak balas dengan larutan natrium
karbonat untuk membentuk mnedakan zink karbonat.
## iii) write a balanced chemical equation for the
reaction.
Tuliskan persamaan kimia seimbang bagi tindak balas
.....................................................................................
Aktiviti 7 :
1. ZnCO3
2. MgCO3
3. CaCO3
4. PbCO3
5. CuCO3
6. ZnCO3
## Pemanasan garam nitrat
Heating of nitrate metal.
## Pemanasan garam karbonat
Heating of carbonate metal
. + CO2
. + .
.. + CO2
. +
. +
. + CO2
7.
8.
9.
10.
11.
12.
Cu(NO3)2
. + O2 + ....
....
KNO2 + ....
AgNO3
. + +.
.
MgO + O2 + .
Pb(NO3)2
.. + .. +
Zn(NO3)2 .... + .. +....
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
Activity 8 :
Zinc carbonate
ZnCO3
## i) How you can convert zinc carbonate to zinc oxide ?
Zinc oxide
ZnO
..........................................................................................
ii) write a balanced chemical equation : ...........................................................
Activity 9 :
.
1. Draw a labelled diagram for the heating of zinc carbonate to produce zinc oxide and carbon dioxide gas. In your
diagram show how carbon dioxide is tested.
Lukis gambar rajah berlabel bagi pemanasan zink karbonat untuk menghasilkan zink oksida dan gas karbon dioksida. Dalam
gambar rajah anda, tunjukkan bagaimana gas karbon dioksida itu diuji.
Heated up
Lead (II)
carbonate
2.
Solid K
Mix of gases R
and Q
Pepejal K
a)
what is solid K. :
b)
c)
## write a balance chemical equation to represent the effect of heat on Pb(NO3) 2
..
Aktiviti 10 :
1.
Zinc carbonate
ZnCO3
Heated up
Solid U
Gas V
Solid U is yellow in colour when hot and white in colour when cooled.
a)
What is solid U : ..
b)
## How to convert zinc carbonate to solid U. .
b)
i) Name gas V : ..
ii) What is observed when gas V is passed through limewater? : ..
iii) Write the chemical equation for reaction between gas V with limewater. ..
c)
2.
## Name the cation in solid K. : ..
Salt P
Garam P
Metal oxide X
Metal oksida X
Heat
Gas Y
Gas Y
Colour of metal oxide X is yellow when hot and white when cold. Gas Y turns lime water milky.
a)
Name gas Y
: .
b)
: .
c)
Name salt P
: .
d)
## Write an equation to represent the action of heat on salt P .
Activity 11 :
White precipitate Q
Dissolved in excess
sodium hydroxide
solution
Reaction I
Colourless K
solution
+ NaOH (aq)
Reaction II
+ Na2SO4 (aq)
White
precipitate R
@azemi_chemistry panel smssi
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
## a) In reaction I, white precipitate Q is formed which is
soluble in excess sodium hydroxide solution.
Name all the ions which are probably presence in
solution K.
## d) i) Name two cations which produced white
precipitate that will not dissolve in excess
sodium hydroxide solution.
...
## ii) state one confirmatory test for one cation
presence in d) i)
..
## b) Referring to reaction I and II, name the cation
presence in solution K.
.......................
Activity 12 :
Step 1
1.
Step 2
Zinc oxide
Zink oksida
Salt solution P
Larutan garam P
## add nitric acid
Zinc carbonate
add solution Q
Zink karbonat
Figure 1
Figure 1 show the steps involved in the preparation of zinc carbonate.
a)
## write a balanced equation for the information of salt solution P. .........................................................
b)
explain briefly how you can obtain a solution of salt solution P. ..........................................................
c) (i) name solution Q that is required to be added to salt solution P in step 2 to produce zinc carbonate.
...............................................................................................................................................
(ii) name the type of reaction involved in step 2.......................................................................................
SOALAN 13 :
Jadual menunjukkan ion positif dan ion negatif dalam tiga larutan garam
Table shows the positive and negative ions in three salt solutions.
Nama garam
Ion positif
Name of salt
Ion negatif
Positive Ion
Negative Ion
Kuprum(II) sulfat
Cu2+
SO42-
Natrium sulfat
Na+
SO42-
Plumbum(II) nitrat
Pb2+
NO3-
Sodium sulphate
## Gunakan maklumat dalam jadual 4 untuk menjawab soalan berikut
Use the information in table 4 to answer the following questions.
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
a)
## Apakah nama lain bagi ion bercas positif
b)
..........................................................................................................
Namakan ion dalam larutan kuprum(II) sulfat
c)
## Name the ions in copper (II) sulphate solution.
..........................................................................................................
Write the formula for lead(II) nitrat.
..........................................................................................................
SOALAN 14 :
Diagram 1 shows a flow chart of the qualitative analysis of substance X.
Rajah 1 menunjukkan carta alir analisis kualitatif bagi sebatian X.
a) Based on diagram 1, identify the
Black powder X
## Berdasarkan rajah 1, kenal pasti
Serbuk hitam X
i) Black powder X .
Serbuk hitam X
solution
## Tambah larutan asid hidroklorik, HCl
Larutan biru Y
Blue solution Y
iii) Cation and anion of Y solution.
Larutan biru Y
Solution Y + Sodium
hydroxide, NaOH solution
Larutan Y + larutan natrium
hidroksida, NaOH
Blue
precipitate
## Kation dan anion larutan Y
Solution Y + Silver
nitrate, AgNO3 solution
Larutan Y + Larutan
argentum nitrat, AgNO3
...
[4 marks]
White precipitate
Mendakan putih
Mendakan biru
SOALAN 15 : AT F4 2012
Diagram 6 shows the reaction and observation for salt X.
Rajah 6 menunjukkan tindak balas dan pemerhatian untuk garam X.
Salt X
Garam X
Add ammonia solution
Blue precipitate
Mendakan biru
## Add excess ammonia solution
Tambahkan larutan ammonia berlebihan
5
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
Z
(a) Refer to Diagram 6,
Merujuk kepada Rajah 6,
(i) What is gas Y?
Apakah gas Y?
[1 mark]
(ii) Write a chemical equation for the reaction between salt X and dilute hydrochloric acid.
Tuliskan persamaan kimia untuk tindak balas antara garam X dan asid hidroklorik cair.
[2 marks]
(iii) What is the observation in Z?
Apakah pemerhatian dalam Z?
## (iv) State the ions present in salt X.
Nyatakan ion-ion yang hadir dalam garam X.
[1 mark]
[1 mark]
(b) A simple experiment can be conducted in the laboratory to produce copper(II) sulphate solution.
Satu eksperimen ringkas boleh dijalankan dalam makmal untuk menyediakan larutan kuprum(II) sulfat.
(i) Describe briefly how the experiment can be conducted.
Huraikan dengan ringkas bagaimana eksperimen ini boleh dijalankan.
[3 marks]
(ii) Describe a test to identify sulphate ion in copper(II) sulphate solution.
Huraikan satu ujian untuk mengenal pasti ion sulfat dalam larutan kuprum(II) sulfat.
.
[3 marks]
SOALAN 16 :
Diagram 21 shows a series of reactions of lead(II) carbonate.
Rajah 21 menunjukkan satu siri tindak balas bagi plumbum(II) karbonat.
Lead(II) carbonate
Plumbum(II) karbonat
Heat
Panaskan
Solid P
Pepejal P
Colourless gas Q
Gas tidak berwarna Q
I
II
asid nitrik cair
Solution R
Larutan R
III
## Potassium iodide solution
Larutan kalium iodida
Precipitate X
Mendakan X
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
a)
i)
## Name the solid P/ Namakan pepejal P.
..
ii)
iii)
[1 mark ]
Draw a labelled diagram of the set up of apparatus can be used to produce solid P and to
identify gas Q in step I.
Lukiskan gambar rajah berlabel bagi susunan radas yang boleh digunakan untuk menghasilkan
pepejal P dan mengenalpasti gas Q dalam langkah I
[2 marks ]
Write the chemical equation for the decomposition of lead(II) carbonate when heated.
Tuliskan persamaan kimia bagi penguraian plumbum(II) karbonat apabila dipanaskan.
..........
b)
i)
[1 mark ]
.
ii)
[1 mark ]
iii)
[1 mark ]
## Write the ionic equation for the formation of precipitate X.
Tuliskan persamaan ion bagi pembentukan mendakan X.
.
iv)
[1 mark ]
## How to obtain precipitate X from the mixture.
Bagaimanakah mendakan X diasingkan daripada campuran.
...
[1 mark ]
SOALAN 17:
Diagram 4.1 shows the steps of preparation of salt G. Excess lead(II) oxide powder is dissolved in
50 cm3 of 1.0 mol dm-3 nitric acid.
Rajah 4.1 menunjukkan langkah-langkah bagi penyediaaan garam G. Serbuk plumbum(II) oksida
berlebihan dilarutkan dalam 50 cm3 asid nitrik 1.0 mol dm-3.
## Lead(II) oxide powder
Serbuk plumbum(II) oksida
50 cm3 of 1.0 moldm-3 nitric acid
50 cm3 asid nitrik 1.0 moldm-3
Salt solution
Larutan garam
7
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
## Diagram 4.1 / Rajah 4.1
(a) What is the meaning of salt?
Apakah maksud bagi garam?
[1 mark]
## (b) Write the chemical formula of salt G formed.
Tuliskan formula kimia bagi garam G yang terbentuk.
......
(c) Why is excess lead(II) oxide powder added to nitric acid?
Mengapakah serbuk plumbum(II) oksida berlebihan ditambahkan kepada asid nitrik ?
......
(d) Write the ionic equation for the reaction between lead(II) oxide and nitric acid.
Tuliskan persamaan ion bagi tindak balas antara plumbum (II) oksida dan asid nitrik.
[1 mark]
[1 mark
[2 marks]
(e) Salt G formed contains nitrate ion. Describe a chemical test to verify the ion.
Garam G yang terbentuk mengandungi ion nitrat.Huraikan satu ujian kimia untuk mengesahkan
ion itu.
....
....
...
[2 marks]
SOALAN 18:
Diagram 5 shows a flow chart for the qualitative analysis of salt W. The green colour of carbonate salt W is
heated strongly to produce black colour of solid X and colourless gas Z.
Rajah 5 menunjukkan carta alir analisis kualitatif bagi garam W. Garam karbonat W yang berwarna hijau
Salt W dengan kuat
Solid X
Heat
Colourless
gas Z Z .
dipanaskan
menghasilkan pepejal
X berwarna +
hitam dan gas
tak berwarna
Garam W
Pepejal X
Gas tak berwarna Z
Panaskan
Process I
Proses I
Process II
Proses II
+ Hydrochloric acid
+ Asid hidroklorik
Process III
Proses III
Blue solution Y
Larutan biru Y
## + Sodium hydroxide solution
+ larutan natrium hidroksida
## + silver nitrate solution
+ larutan argentum nitrat
@azemi_chemistry panel smssi
8
Blue precipitate
Mendakan biru
White precipitate
Mendakan putih
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
## (a) Based on Diagram 5,
Berdasarkan Rajah 5,
(i)
Diagram 5/Rajah 5
## State the name of salt W and solid X.
Nyatakan nama bagi garam W dan pepejal X.
Salt W :.....................................................................................................................................
Garam W
Solid X :.............................................................................................................................................
Pepejal X
[2 marks]
(ii)
## describe a chemical test to identify gas Z.
huraikan satu ujian kimia untuk mengenal pasti gas Z.
............................................................................................................................................................
............................................................................................................................................................
(iii)
## what is the name of reaction in Process I?
apakah nama tindak balas dalam Proses I?
........................................................................................................................................
(iv)
## write a balanced chemical equation for the reaction in Process I.
tuliskan persamaan kimia yang seimbang bagi tindak balas dalam Proses I.
...............................................................................................................
[2 marks]
[1 mark]
[2 marks]
(b) Based on the observation in Process II and Process III, state the cation and anion present in solution Y.
Berdasarkan pemerhatian dalam Proses II dan Proses III, nyatakan kation dan anion yang hadir dalam
larutan Y.
Cation..............................................................................................................
Kation
Anion .......................................................................................................
Anion
(c) (i)
Write the ionic equation for the reaction occur in Process III.
Tuliskan persamaan ion bagi tindak balas yang berlaku dalam Proses III.
..........................................................................................................................
(ii)
[2 marks]
## What is the name of reaction occur in Process III?
Apakah nama tindak balas yang berlaku dalam Proses III?
[1 mark]
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
[1 mark]
## Magnesium nitrat adalah garam larut.
Huraikan bagaimanakah penyediaan magnesium nitrat kering
dalam makmal. Jawapan and mestilah mengandungi
## Bahan kimia / substance
Prosedur / procedures
Persamaan kimia yang terlibat dalam tindak balas/ chemical equation
## Persamaan kimia : Mg + HNO3 Mg(NO3)2 + H2
Chemical equation
1.
## Masukkan 25 cm3 asid nitrik 0.1 moldm-3 ke dalam bikar
Pour 25 cm3 of 0.1 moldm-3 nitric acid ke dalam bikar
2.
## Panaskan asid nitrik perlahan-lahan
Heat nitric acid slowly
3.
## Tambahkan serbuk magnesium ke dalam bikar sehingga berlebihan
Add magnesium powder into a beaker until excess.
4.
5.
6.
7.
8.
## Turaskan garam / Filter the salt
9.
Keringkan garam di antara dua kertas turas / Dry the salt between two pieces of filter paper.
## SOALAN 21 : Eksperimen garam terlarutkan
Zink klorida adalah garam larut.
Huraikan bagaimanakah penyediaan zink klorida kering dalam makmal. Jawapan
and mestilah mengandungi
Bahan kimia
Prosedur
Persamaan kimia yang terlibat dalam tindak balas
SOALAN 22 :
## Penyediaan garam tak terlarutkan
Anda telah dibekalkan dengan bahan berikut:
You are given the following substance :
Argentum nitrat
Natrium klorida
10
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
Silver nitrate
Sodium chloride
Terangkan bagaimana anda boleh menyediakan garam dengan menggunakan bahan-bahan di atas.
Describe how to prepare a sample of salt by using the above substances .
Jawapan :
[6 marks]
Prosedur / Procedures;
1. Masukkan 25 cm3 larutan barium nitrat, Ba(NO3)2 0.1 mol dm-3 ke dalam bikar.
Pour 25 cm3 of 0.1 mol dm-3 barium nitrat, Ba(NO3)2 into a beaker.
2. Masukkan 25 cm3 larutan natrium klorida,NaCl 0.1 mol dm-3 ke dalam bikar.
Add 25 cm3 of 0.1 mol dm-3 sodium chloride,NaCl into a beaker.
3. Kacau campuran dengan rod kaca
Stir the mixture using glass rod
4. Turas campuran / Filter the mixture
5. Bilas baki turasan dengan air suling.
Rinse the residue with distilled water
6. Keringkan garam di antara dua kertas turas.
Dry the salt/residue in between sheet of filter papers to dry
## SOALAN 23 : Garam Tak larut Insoluble salt
Plumbum(II) sulfat adalah garam tak larut.
Huraikan bagaimanakah penyediaan plumbum(II) sulfat kering dalam makmal.
Lead(II) iodide an insoluble salt. Describe how the preparation the dry lead(II) sulphate in the laborotory
[6 marks]
SOALAN 24 :
Ujian pengesahan kimia
Confirmory test
Huraikan ujian-ujian kimia yang boleh digunakan untuk mengenal kation dan anion dalam larutan zink klorida
Describe the chemical tests that can be used to verify the cation and anion in zinc chloride
[5 marks]
Ujian pengesahan ion zink
Ujian pengesahan ion klorida
1. Masukkan larutan ke dalam tabung uji.
1. Masukkan larutan ke dalam tabung uji.
2. Masukkan 2 cm3 ammonia akueus ke dalam
tabung uji sehingga berlebihan.
3. Mendakan putih terbentuk
2.
3.
4.
## 4. dan larut dalam berlebihan Mengesahkan
kehadiran ion zink
tabung uji .
5.
6.
11
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
SOALAN 25 :
Ujian kimia
Huraikan ujian-ujian kimia yang boleh digunakan untuk mengenal kation dan anion dalam larutan plumbum(II) nitrat
Describe the chemical tests that can be used to verify the cation and anion in lead(II) nitrate
## SOALAN 26 : Pnyediaan garam tak terlarutkan
Anda telah dibekalkan dengan bahan berikut:
You are given the following substance :
Barium nitrat
Barium nitrate
magnesium sulfat
magnesium sulfat
Terangkan bagaimana anda boleh menyediakan garam dengan menggunakan bahan-bahan di atas.
Describe how to prepare a sample of salt by using the above substances .
[6 marks]
SOALAN 27 :
Diagram 4 shows Experiments I and II in the preparation of a salt.
Rajah 4 menunjukkan Eksperimen 1 dan II dalam penyediaan garam.
Experiment
Eksperimen
Method
Kaedah
0.1 mol dm-3 hydrochloric acid
0.1 mol dm-3 asid hidroklorik
## 25.0 cm3 of 0.2 mol dm-3 sodium hydroxide + phenolphthalein
25.0 cm3 natrium hidroksida 0.2 mol dm-3 + fenolftalein
12
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
## 10 cm3 of 1.0 mol
dm-3 lead(II) nitrate
solution
10 cm3 larutan
plumbum(II) nitrat
1.0 mol dm-3
II
## Excess potassium iodide solution
Larutan kalium iodida
berlebihan
Precipitate X
Mendakan X
Diagram 4
Rajah 4
(a) Based on Experiment I:
Berdasarkan Eksperimen I :
## (i) State the name for the reaction.
Nyatakan nama bagi tindak balas itu.
................................................................
[1 mark]
(ii) Write the chemical equation for the reaction that occurs in the conical flask.
Tuliskan persamaan kimia bagi tindakbalas yang berlaku di dalam kelalang kon.
....................................................................................................................................
[1 mark]
(iii) State the colour change in the conical flask at the end point. .............................................................
Nyatakan perubahan warna di dalam kelalang kon pada takat akhir.
[1 mark]
(iv) Calculate the volume of hydrochloric acid used to neutralise the sodium hydroxide solution.
Hitungkan isi padu asid hidroklorik yang digunakan untuk meneutralkan larutan natrium hidroksida.
[2 marks]
(b) Based on Experiment II:
Berdasarkan Eksperimen II :
(i) State the name of the reaction. ..............................................................................................................
Nyatakan nama bagi tindak balas itu.
[1 mark]
(ii) State the name of precipitate X. ...............................................................................................................
Nyatakan nama bagi mendakan X.
[1 mark]
(iii) Write the ionic equation for the reaction. ...........................................................................................................
Tuliskan persamaan ion bagi tindak balas itu.
[1 mark]
(iv) Calculate the maximum mass of precipitate X formed. [Relative atomic mass ; Pb=207, I=127]
Hitungkan jisim maksimum mendakan X yang terbentuk. [Jisim atom relatif ; Pb=207, I=127]
[2 marks]
SOALAN 28 :
5
10.0 cm of 1.0 mol dm-3 sodium iodide is poured to 8 different test tubes. Different volume of 1.0 moldm-3
lead(II) nitrate are added to each test tube. The height of precipitate formed in each test tube is measured.
The graph below is obtained when the height of precipitate is plotted against the volume of lead (II) nitrate
solution.
Height of
precipitate/
cm
13
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
1 2
(a)
## Name the precipitate formed....................................................................................................
[ 1 mark ]
3
10 cm of 1.0 mol dm-3 lead (II) nitrate is added to
Estimate the height of the precipitate formed.
(b)
## sodium iodide solution.
..........................................................................................................................................
[ 1 mark ]
(d)
Based on the above graph, what is the volume of lead (II) nitrate needed to completely react with
sodium iodide solution?
...............................................................................................................................................
[ 1 mark ]
Calculate
(i) the number of moles of potassium iodide in the reaction.
(e)
[ 1 mark ]
(ii) the number of moles of lead (II) nitrate that has completely reacted with potassium iodide solution.
[ 1 mark ]
(f)
## Write the ionic equation for the formation of the precipitate.
................................................................................................................................................
[ 1 mark ]
The concentration of lead (II) nitrate is changed from 1.0 mol dm-3 to 2.0 mol dm-3 .
Sketch the graph obtained on the above graph.
(g)
[ 1 mark]
(h)
The mixture in test tube 8 is filtered. State the observation when the sodium hydroxide solution is
added to the filtrate until in excess .
............................................................................................................................
...................
[ 2 marks ]
SOALAN 29 :
Diagram 5 shows a flow chart for the qualitative analysis of salt W. The green colour of carbonate salt W is heated
strongly to produce black colour of solid X and colourless gas Z.
Rajah 5 menunjukkan carta alir analisis kualitatif bagi garam W. Garam karbonat W yang berwarna hijau dipanaskan dengan kuat
menghasilkan pepejal X berwarna hitam dan gas tak berwarna Z .
Salt W
Garam W
Solid X
Pepejal X
Heat
Panaskan
Process I
Proses I
Process II
Proses II
+ Hydrochloric acid
+ Asid hidroklorik
Blue solution Y
Larutan biru Y
## + Sodium hydroxide solution
+ larutan natrium hidroksida
Process III
Proses III
## + silver nitrate solution
+ larutan argentum nitrat
14
Blue precipitate
Mendakan biru
Colourless gas Z
Gas tak berwarna Z
## @azemi_chemistry panel smssi
White precipitate
Mendakan putih
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
Diagram 5/Rajah 5
(d) Based on Diagram 5,
Berdasarkan Rajah 5,
(v)
## State the name of salt W and solid X.
Nyatakan nama bagi garam W dan pepejal X.
Salt W :...............................................................................Solid X : ............................................................
Garam W
Pepejal X
[2 marks]
(vi)
(vii)
## describe a chemical test to identify gas Z.
huraikan satu ujian kimia untuk mengenal pasti gas Z.
............................................................................................................................................................
............................................................................................................................................................
[2 marks]
what is the name of reaction in Process I?
apakah nama tindak balas dalam Proses I?
........................................................................................................................................
[1 mark]
(viii)
## write a balanced chemical equation for the reaction in Process I.
tuliskan persamaan kimia yang seimbang bagi tindak balas dalam Proses I.
............................................................................................................................................................
[2 marks]
(e) Based on the observation in Process II and Process III, state the cation and anion present in solution Y.
Berdasarkan pemerhatian dalam Proses II dan Proses III, nyatakan kation dan anion yang hadir dalam larutan
Y.
Cation : ................................................................................. Anion :.......................................................................
Kation
Anion
[2 marks]
(f)
(i)
(ii)
Write the ionic equation for the reaction occur in Process III.
Tuliskan persamaan ion bagi tindak balas yang berlaku dalam Proses III.
............................................................................................................................................................
[1 mark]
What is the name of reaction occur in Process III?/
Apakah nama tindak balas yang berlaku dalam Proses III?
[1 mark]
SOALAN 30 :
## Diagram 6.1 shows a series of reaction of copper compound.
Rajah 6.1 menunjukkan satu siri tindak balas bagi sebatian kuprum
Copper (II)
nitrate
Kuprum (II)
nitrat
Sodium carbonate
Natrium karbonat
Copper(II)
carbonate
Kuprum(II)
karbonat
Heat
Panaskan
Copper(II)
oxide
Kuprum(II)
oksida
Gas X
Acid Y
Asid Y
Copper(II) sulphate
Kuprum(II) sulfat
15
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
## Diagram 6.1 /Rajah 6.1
(a) Copper(II) nitrate solution reacts with sodium carbonate solution to form copper(II) carbonate precipitate.
Larutan kuprum(II) nitrat bertindak balas dengan larutan natrium karbonat membentuk mendakan kuprum(II)
karbonat.
(i)
## What is the colour of copper(II) carbonate? / Apakah warna kuprum(II) karbonat?
.....................
[1 mark]
(ii)
State the name of the reaction / Nyatakan nama bagi tindak balas itu.
............................................................................................................................................................
[1 mark]
## (b) Heating of copper(II) carbonate produces copper(II) oxide and gas X.
Pemanasan kuprum(II) karbonat menghasilkan kuprum(II) oksida dan gas X.
(i)
..
[1 mark]
(ii)
## Write a balanced chemical equation of the reaction.
Tuliskan persamaan kimia seimbang bagi tindak balas itu.
........................................................................
[1 mark]
(i)
Draw a labeled diagram for the heating of copper(II) carbonate to produce copper(II) oxide and
gas X. In your diagram show how gas X is tested.
Lukiskan gambar rajah berlabel bagi pemanasan kuprum(II) karbonat untuk menghasilkan kuprum(II) oksida dan gas
X. Dalam rajah anda tunjukkan bagaimana gas X diuji.
[2 marks]
## (c) Copper(II) carbonate reacts with acid Y to produce copper(II) sulphate.
The chemical equation is shown below.
Kuprum(II) karbonat bertindak balas dengan asid Y menghasilkan kuprum(II) sulfat. Persamaan kimia itu
ditunjukkan di bawah:
CuCO3
+
Acid Y
(i)
## What is acid Y? /Apakah asid Y?
..
[1 mark]
(ii)
If 12.4 g copper(II) carbonate reacts completely with excess acid Y, calculate the mass of
copper(II) sulphate formed.
[Relative atomic mass: C = 12, O=16 , S = 32, Cu = 64]
Jika 12.4 g kuprum(II) karbonat bertindak balas lengkap dengan asid Y berlebihan, hitung jisim
kuprum(II) sulfat yang terbentuk.
16
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
## [Jisim atom relatif: C = 12, O=16 , S = 32, Cu = 64]
3 marks]
SOALAN 31 :
(a) Diagram 3.1 shows the names for two type of salts.
Rajah 3.1 menunjukkan nama bagi dua jenis garam.
Copper(II) chloride
Kuprum(II) klorida
Lead(II) chloride
Plumbum(II) klorida
Diagram 3.1/Rajah 3.1
(i)
Based on the salt given in Diagram 3.1, write the formula of an insoluble salt.
State the name of reaction to prepare insoluble salt.
(ii)
(b)
Berdasarkan garam yang diberikan dalam Rajah 3.1, tuliskan formula garam yang tak terlarutkan.
Nyatakan nama bagi tindak balas menyediakan garam tak terlarutkan.
[2 marks]
State the suitable chemicals required to produce copper(II) chloride and lead(II) chloride salts.
Nyatakan bahan-bahan kimia yang sesui untuk menyediakan garam kuprum (II) klorida dan
plumbum(II) klorida.
[4 marks]
Diagram 3.2 shows reactions involving solid S. When heated, solid S decomposes to solid T, brown gas U
and colourless gas W. Gas U relights glowing wooden splinter.
Rajah3.2 menunjukkan tindak balas yang melibatkan pepejal S. Bila dipanaskan, pepejal S terurai kepada
pepejal T, gas perang U dan gas tak berwarna U. Gas W menyalakan semula kayu uji berbara.
White solid S
Pepejal putih S
Solid T
Pepejal T
Brown gas U
Gas perang U
Colourless gas W
Gas W tak berwarna
U
+ HNO3 (aq)
Colourless solution X
Larutan tak berwarna X
+ NaOH (aq)
## White precipitate, soluble
in excess NaOH
Mendakan putih, larut
dalam berlebihan NaOH
+ NH3(aq)
17
## White precipitate, soluble
in excess NH3 @azemi_chemistry panel smssi
Mendakan putih, larut
dalam berlebihan NH3
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
## Diagram 3.2/ Rajah3.2
(i) Identify solid S, solid T, gas U and gas W.
Kenal pasti pepejal S, pepejal T, gas U dan gas W
(ii)
[ 4marks]
## Write the chemical equation for the heating of solid S.
Tuliskan persamaan kimia untuk pemanasan pepejal S.
[2 marks]
(c) A student carried out an experiment to construct an ionic equation for the formation of barium sulphate.
Table 3.3 shows the height of precipitate formed when 5.0 cm 3 of 0.5 mol dm-3 potassium sulphate
solution is added with 1.0 cm3, 2.0 cm3, 3.0 cm3, 4.0 cm3, 5.0 cm3, 6.0 cm3, 7.0 cm3 and 8.0 cm3 of 0.5 mol
dm-3 barium chloride solution respectively in eight test tubes.
Seorang pelajar telah menjalankan satu eksperimen untuk membina persamaan ion bagi pembentukan
barium sulfat. Jadual 3.3 menunjukkan tinggi mendakan yang terbentuk apabila 5.0cm3larutan kalium
sulfat0.5 mol dm-3ditambahkan dengan masing-masing 1.0 cm3, 2.0 cm3, 3.0 cm3, 4.0 cm3, 5.0 cm3, 6.0 cm3,
7.0 cm3dan 8.0 cm3larutan barium klorida dalam lapan tabung uji .
Test tube
Tabung uji
## Volume of 0.5 mol dm-3 potassium
sulphate solution / cm3
Isipadu larutan kalium sulfat
0.5 mol dm-3 / cm3
5.0
5.0
5.0
5.0
5.0
5.0
5.0
5.0
## Volume of 0.5 mol dm-3 barium
chloride solution / cm3
Isipadu larutan barium klorida
0.5 mol dm-3 / cm3
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
1.2
1.4
1.6
1.8
2.0
2.0
2.0
2.0
Height of precipitate/ cm
Tinggi mendakan/cm
## Table 3.3 / Jadual 3.3
(i) Based on Table 3.3, draw a graph of the height of the precipitate against volume of 1.0 mol dm -3
barium chloride solution.
.
Berdasarkan Jadual 3.3, lukiskan graf tinggi mendakan melawan isi padu larutan barium klorida 1 mol dm-3
[3 marks]
(ii) On the graph that you have drawn in a(i) , mark the minimum volume of 1.0 mol dm -3 barium chloride
solution needed to react completely with 5.0 cm3 of 1.0 mol dm-3 potassium sulphate solution.
Pada kertas graf yang telah anda lukiskan di (a) (i), tandakan isi padu minimum larutan barium klorida 1.0
mol dm-3 yang diperlukan untuk bertindak balas lengkap dengan5.0 cm3larutan kalium sulfat 1.0 mol dm-3.[1
mark]
18
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
(iii) Calculate the number of mole of barium ions and number of moles of sulphate ions required for the
formation of barium sulphate. Then calculate the number of moles of sulphate ions that react with 1 mole
of barium ion.
Hitungkan bilangan mol ion barium dan bilangan mol ion sulfat yang diperlukan untuk pembentukan barium
sulfat. Kemudian hitungkan bilangan mol ion sulfat yang bertindak balas dengan 1 mol ion barium
[3 marks]
(iv) Write the ionic equation for the formation of barium sulphate
Tuliskan persamaan ion untuk pembentukan barium klorida
[1 mark]
SOALAN 31 :
BAHAGIAN C
(a) A farmer discovers that his vegetables are not growing well due to soil problems. By using your chemistry
knowledge, state two possible causes and ways to overcome the problems by naming the chemical used.
Seorang petani mendapati sayuran yang ditanamnya tidak subur disebabkan masalah tanah. Dengan
menggunakan pengetahuan kimia anda, nyatakan dua penyebab yang mungkin dan cara untuk mengatasi
masalah ini dengan menamakan bahan kimia yang digunakan .
[4 marks]
(b) Table4 shows the information on action of heat for two lead salts, P and Q.
Jadual4menunjukkan maklumat bagi tindakan haba ke atas dua garam plumbum P dan Q ,
Experiment
Eksperimen
Salt P
Garam P
Products
Hasil
Residue R
Baki R
Observation
Pemerhatian
Brown solid when hot, yellow when cold
Pepejal perang bila panas, kuning bila sejuk
Gas A
Gas A
## Lime water become chalky
Air kapur menjadi keruh
19
## Module Latihan Topikal SPM_2016
Bab 8_Tingkatan 4
Heat
Panas
Lime water
Air kapur
Salt Q
Garam Q
Residue R
Baki R
## Brown solid when hot, yellow when cold
Pepejal perang bila panas, kuning bila sejuk
Gas B
Gas B
Brown gas
Gas perang
Gas C
Gas C
## Rekindles glowing splinder
Menyalakan kayu uji berbara
Heat
Panas
Table 4/ Jadual 4
Based on Table 4, identify residue R, gas A, gas B and gas C.
Write the chemical formulae for salt P and Q.
Berdasarkan Jadual 4, kenal pasti baki R, gas A, gas B dan gas C.
Tuliska formula kimia bagi garam P dan garam Q.
[6marks]
(b) By using all the chemical substances given below and suitable apparatus, describe a laboratory
experiment to prepare dry zinc sulphate salt.
Dengan menggunakan bahan kimia yang diberikan di bawah dan alat radas yang sesuai, huraikan satu
eksperimen di makmal untuk menyediakan garam zink sulfat kering.
## zinc nitrate solution
dilute sulphuric acid
sodium carbonate solution
Larutan zink nitrat
Asid sulfurik cair
Larutan natrium karbonat
## In your description, include chemical equations involved.
Dalam huraian anda sertakan persamaan kimia yang terlibat.
20
[12 marks]
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# 26.27_Lec-12 - Physics 241 Lecture 12 Y. E. Kim September...
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Unformatted text preview: Physics 241 Lecture 12 Y. E. Kim September 30, 2010 Chapter 26, Section 3 Chapter 27, Sections 1 - 2 September 30, 2010 University Physics, Chapter 26 and 27 1 September 30, 2010 University Physics, Chapter 26 2 Chapter 26 Section 3 Multi-Loop Circuits ¡ To analyze a multi-loop circuit, identify all complete loops and all junction points in the circuit and DSSO\ .LUFKKRII·V Rules to these parts of the circuit separately ¡ Analyze the single loops in a multi-loop circuit with .LUFKKRII·V /RRS 5XOH DQG WKH MXQFWLRQV ZLWK .LUFKKRII·V Junction Rule, and obtain a system of coupled equations in several unknown variables ¡ These coupled equations can be solved in several ways ¡ Solution with matrices and determinants ¡ Direct substitution September 30, 2010 University Physics, Chapter 26 3 ([DPSOH¡ .LUFKKRII·V 5XOHV ¢£¤ ¡ The circuit here has three resistors, R 1 , R 2 , and R 3 and two sources of emf, V emf,1 and V emf,2 ¡ This circuit cannot be resolved into simple series or parallel structures ¡ To analyze this circuit, we need to assign currents flowing through the resistors ¡ We can choose the directions of these currents arbitrarily September 30, 2010 University Physics, Chapter 26 4 ¡ At junction b the incoming current must equal the outgoing current ¡ At junction a we again equate the incoming current and the outgoing current ¡ But this equation gives us the same information as the previous equation! ¡ We need more information to determine the three currents ² 2 more independent equations ([DPSOH¡ .LUFKKRII·V /DZV ¢£¤ 2 1 3 i i i ¡ 1 3 2 i i i ¡ September 30, 2010 University Physics, Chapter 26 5 ¡ 7R JHW WKH RWKHU HTXDWLRQV ZH PXVW DSSO\ .LUFKKRII·V Loop Rule. ¡ This circuit has three loops. ¡ Left ¡ R 1 , R 2 , V emf,1 ¡ Right ¡ R 2 , R 3 , V emf,2 ¡ Outer ¡ R 1 , R 3 , V emf,1 , V emf,2 ([DPSOH¡ .LUFKKRII·V /DZV ¢£¤ September 30, 2010 University Physics, Chapter 26 6 ([DPSOH¡ .LUFKKRII·V /DZV ¢£¤ ¡ Going around the left loop counterclockwise starting at point b we get ¡ Going around the right loop clockwise starting at point b we get ¡ Going around the outer loop clockwise starting at point b we get ¡ But this equation gives us no new information! 1 1 ,1 2 2 1 1 ,1 2 2 0 e m f e m f i R V i R i R V i R ¡ ¡ ¡ ¢ £ £ 3 3 ,2 2 2 3 3 ,2 2 2 0 e m f e m f i R V i R i R V i R ¡ ¡ ¡ ¢ £ £ 3 3 ,2 ,1 1 1 e m f e m f i R V V i R ¡ ¡ £ £ September 30, 2010 University Physics, Chapter 26 7 ¡ We now have three equations ¡ And we have three unknowns i 1 , i 2 , and i 3 ¡ We can solve these three equations in a variety of ways ([DPSOH¡ .LUFKKRII·V /DZV ¢£¤ 1 3 2 i i i ¡ 1 1 ,1 2 2 e m f i R V i R ¡ ¡ 3 3 ,2 2 2 e m f i R V i R ¡ ¡ 2 3 , 1 2 , 2 1 1 2 1 3 2 3 3 , 1 1 , 2 2 1 2 1 3 2...
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## This note was uploaded on 09/09/2011 for the course PHYS 241 taught by Professor Wei during the Fall '08 term at Purdue.
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26.27_Lec-12 - Physics 241 Lecture 12 Y. E. Kim September...
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View Full Version : OpenGL Lighting Model Basic Question
Joshua Smith
01-06-2006, 07:43 AM
If I:
- set up a scene with a fixed directional light shining on a sphere, and
- I rotate the eye point around that sphere (so the light is moving around relative to the eye), and
- I have LIGHT_MODEL_LOCAL_VIEWER set to FALSE
would you expect the specular highlight to move around on the sphere? I certainly would, because I would expect that specular lighting would be using blinn halfway vectors dotted against vertex normals to compute specular highlight locations. But what I'm getting is the highlight is glued to a particular place on the sphere. As though it is not using a blinn halfway vector, but rather is just dotting against the light direction.
Does it sound like:
1) I'm overestimating the capabilities of OpenGL lighting, or
2) I set something up wrong?
dorbie
01-06-2006, 07:59 AM
The light is transformed through the current modelview matrix to eyespace and stored there.
To move the light relative to the eye in any way you need to respecify it's position with a new matrix on the modelview stack or with a new position.
Joshua Smith
01-06-2006, 08:06 AM
Judging from the diffuse lighting, I would say the light is pointing the right direction. That is, as I rotate the eye around the sphere, the bright spot stays put on the sphere. But the specular highlight is right dead center in the the bright spot as well, no matter how I rotate things, which isn't what I would expect.
Perhaps I can rephrase this question to be more direct:
When computing specular highlights, does OpenGL dot the vertex normals against the light direction or against a vector halfway between the eye & light directions?
dorbie
01-06-2006, 08:44 AM
It should use Blinn's halfway vector (half way between view vector and light source) and dot product with the normal.
With localviewer true the view vector is computed per vertex.
Hmm, localviewer false will use the eyespace z axis as the view vector for all vertices.
Try setting localviewer to true.
Joshua Smith
01-06-2006, 09:22 AM
My specular highlights disappear with localviewer set to true. This is a bit of a mystery to me. From reading the docs, it looks like with localviewer=false, the to-eye vector is just +Z, whereas with localviewer=true, the to-eye vector is |eyePos - vertexPos|, which would be darn close to +Z in eye space, and should be exactly +Z for a vertex in the middle of the projection plane.
So I wouldn't expect the localviewer setting to make much of any difference with a small sphere test case like mine. And I'm quite mystified why localviewer=true would make the highlight disappear altogether.
dorbie
01-06-2006, 10:37 AM
Hmm... are you loading your viewing matrix on the projection matrix stack? Or doing projection on the modelview?
Distinguishing these correctly is critical to certain fixed function features, like specular which requires the view vector to be computed correctly.
You should be doing projection on the projection matrix and doing viewing on the modelview (or software)
Look at the glMatrixMode call.
Joshua Smith
01-06-2006, 11:21 AM
Originally posted by dorbie:
Hmm... are you loading your viewing matrix on the projection matrix stack?Wow. That's scary. You've been doing this a long time, haven't you. Where do I send the beer?
You nailed it. I was setting my camera matrix in the context of the GL_PROJECTION matrix. Seemed reasonable -- I think of camera positioning as part of projection. But moving it down two lines to be in the GL_MODELVIEW made specular highlights work right.
dorbie
01-06-2006, 01:01 PM
:) <paranoia>... maybe I can see your code</paranoia>
Cool, so you're now doing viewing transformation in hardware too. That means you're using a lot of the fixed function pipeline for hardware acceleration.
Now all you have to do is multiply model matrices with the modelview stack and you'll be fully hardware accelerated.
Joshua Smith
01-06-2006, 03:01 PM
Is that really done by the GPU? We'll be deploying on a high-end game card (something from ATI or nVidia in the \$400 range). Has matrix math really be offloaded from the CPU to those cards? (Last I knew, which was years back, gamer graphics cards didn't accelerate much of the matrix math.)
The question is academic, though. There's no way I'd tear up our software pipeline to get a nominal increase like moving those tranforms off the CPU would get me. Our target application will be showing models with modest polygon counts (say, 20,000), but tons of texture. A lot like the models you see on our site: http://www.kaon.com/services3D.html
For what it's worth, making the changes you suggested had no impact at all on performance. But it did solve my perspective correction issues. (Yippee!)
Next step is implementing "pick" functionality, which looks like such a mess in OpenGL, I might just do it in software.
Then I'll start optimizing, although I suspect the application is going to be mostly fill-rate bound, because our polygon counts are so low.
dorbie
01-06-2006, 09:14 PM
It's almost always done on the GPU, not necessarily the matrix mults but certainly the vertex transformation, moreover it's the same cost as the viewing transformation you're already doing since it's a concatenated matrix.
The performance gained will depend on the content, and the platform, intel on chip graphics for example still transforms in software, but almost all graphics cards made for the past 5 years transform and light on the GPU.
Your claim is just entirely content related, but it's also spurious until you move model matrix transform to the GPU before comparrison. You seem to have some middleware, the observation that your demos are fill limited and so you shouldn't care about geometry performance is a cop out.
As for tearing up a perfectly good software pipeline... sigh, how much work is pushmatrix, multmatrix, popmatrix for each transformation node? Nobody is suggesting you tear anything up, just eliminate redundant software transformation when you're already doing the matrix multiplication in hardware... and that's exactly what you're already doing with the viewing transformation.
You seem to look at these things and shy away from the move to hardware, why should a matrix mult for transform nodes significantly complicate your software or involve tearing anything up?
If you can handle the geometry well content developers will take advantage of it and you will see complex geometry, saying you are geometrically simple is a self fulfilling prophecy in this case.
Further, moving work to the GPU frees up the CPU for processing, threading or other system activities.
Joshua Smith
01-07-2006, 05:07 AM
Originally posted by dorbie:
Your claim is just entirely content related, but it's also spurious until you move model matrix transform to the GPU before comparrison. You seem to have some middleware, the observation that your demos are fill limited and so you shouldn't care about geometry performance is a cop out. It's a pretty well established principle that you shouldn't optimize until everything is working and you have evidence of where your bottlenecks are. I'm not completely ruling out moving bits of the transform hierarchy to the GL pipeline, I'm just extremely skeptical that when I get some stats, we'll find that we're spending more than a few milliseconds doing geometry transforms.
Like a game engine, we focus on getting the best performance possible with a known kind of content -- we're not trying to make a great general-purpose engine, but rather an insanely-great special-purpose engine. So yes, we know exactly what kind of content will go into this pipeline, and one of the constraints is that the SAME content will be deployed on the web, which has very well understood limitations on polygon count.
Originally posted by dorbie:
As for tearing up a perfectly good software pipeline... sigh, how much work is pushmatrix, multmatrix, popmatrix for each transformation node? Nobody is suggesting you tear anything up, just eliminate redundant software transformation when you're already doing the matrix multiplication in hardware... and that's exactly what you're already doing with the viewing transformation.The issue has to do with software architecture and layering. I am working to avoid sticking GL stuff all over an existing, highly optimized pipeline. There are a few key places where it is natural to shunt an exisitng data stream out of our pipeline and over to GL. Transforming geometry isn't one of them. (Note that 90% of the "animation" done in the product visualization space we serve involves simply moving the eye point around, so the geometry transformation you're so concerned about hardly ever runs anyway.)
Also, from what I've read, the MODELVIEW matrix stack is generally of fixed depth. In the general purpose case, that must mean that you have to implement this stuff in both places anyway, in case you run out of room in the stack.
dorbie
01-08-2006, 07:00 PM
I've never seen an app legitimately overflow the modelview matrix stack, usually it's a missing popmatrix call, it's guaranteed to be at least 32 deep. Worst case you could manually multiply the last matrix, still no need for software xform, just software matrix mult.
You're already doing the viewing matrix mult, doing the model on there is free. This is not a case of premature optimization, it's about using a resource you're already using in the correct and standard way for 'free' with very little effort. Anyhoo, I'm done trying to persuade you to do the right thing.
Jedimaster
01-10-2006, 09:12 PM
Read the RedBook. I met this problem before too. Light Position will be transformed in ModelViewMatrix. so you should
draw()
<-------->
glLight()
<--------->
dorbie
01-11-2006, 08:28 AM | 2,164 | 9,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-30 | latest | en | 0.924052 |
https://rememberingsomer.com/factors-of-36-that-add-up-to-13/ | 1,653,337,833,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00475.warc.gz | 526,430,922 | 5,988 | Factors that 36 are the perform of integers that us can break-up evenly into 36. There are overall 9 determinants of 36 amongst which 36 is the greatest factor and its positive determinants are 1, 2, 3, 4, 6, 9, 12, 18, and 36. The Pair determinants of 36 room (1, 36), (2, 18), (3, 12), (4, 9) and (6, 6) and also its Prime components are 1, 2, 3, 4, 6, 9, 12, 18, 36.
You are watching: Factors of 36 that add up to 13
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18 and also 36Negative components of 36: -1, -2, -3, -4, -6, -9, -12, -18 and also -36Prime factors of 36: 2, 3Prime factorization of 36: 2 × 2 × 3 × 3 = 22 × 32Sum of determinants of 36: 91
Let us explore much more about factors of 36 and ways to uncover them.
1 What are the components of 36? 2 How to calculation the components of 36? 3 Factors of 36 in Pairs 4 Important Notes 5 FAQs on determinants of 36
## What room the determinants of 36?
Factors of a number are the numbers that divide the given number precisely without any remainder. According to the an interpretation of factors, components of 36 room those number that exactly divide 36 leaving no remainder. In other words, we have the right to say if 2 numbers are multiplied and the product is 36, then the number are determinants of 36.Let us do a hit and also trial technique to find determinants of 36.
Divide 36 ÷ 2 = 18Quotient = 18, remainder = 0Now main point 18 × 2 = 36Hence, 18 and also 2 are the factors of 18.
## How to calculation the components of 36?
We have the right to use different methods favor the divisibility test, element factorization, and the upside-down department method to calculate the factors of 36.The most common technique we use to find determinants is the prime factorization method.
In element factorization, we express 36 as a product the its prime factors.In the department method, we see which numbers division 36 exactly without a remainder.
Let united state calculate determinants of 36 using the adhering to two methods:
Factors that 36 by element factorization element tree methodFactors that 36 by upside-down department method
### Prime administer of 36 by Upside-Down department Method
Now as debated above, in the prime factorization technique we show factors of 36 as a product that prime determinants of 36. Follow the steps to uncover the prime factors of 36 through the element factorization method.
With the assist of divisibility rules, find out the smallest factor of the given number. Here, 36 is an also number. So the is plainly divisible by 2. In various other words, 2 divides 36 with no remainder. Therefore, 2 is the the smallest prime element of 36.36 ÷ 2 = 18. Now discover the prime factors of the acquired quotient. Repeat step 1 and also Step 2 it rotates we acquire a prime number as the quotient. Here, 18 is the quotient.18 ÷ 2 = 9. Here, 9 is the quotient. Now uncover the prime determinants of the 9.9 ÷ 3 = 3. Here 3 is the prime number. Therefore we have the right to stop the process.Prime factorization of 36 using upside-down division method is 2 × 2 × 3 × 3 = 2² × 3².
### Prime administrate of 36 by aspect Tree Method
The aspect tree technique is another visualization an approach to represent determinants of 36. It likewise helps in identify prime determinants of 36. A factor tree is not unique for a offered number. Instead of express 36 as 2 × 18, we have the right to express 36 as 6 × 6.
Prime factorization of 36 using aspect tree method is 2 × 2 × 3 × 3 = 2² × 3².
Explore components using illustrations and also interactive examples
## Factors that 36 in Pairs
Factor pairs room the 2 numbers that, once multiplied, give the number 36. Find all of the components of the number 36 utilizing the steps below:
Step I: begin with 1 and also the number itself.Step II: count up by people to watch if you have the right to multiply two numbers together to obtain your target number.Step III: Stop once you can’t gain any much more numbers in between.Step IV: connect the variable pairs.
Factor bag of 36 are provided below:
36 = 1 × 3636 = 4 × 936 = 6 × 636 =12 × 336 = 18 × 2
Let united state club the numbers in pairs.
The components of 36 in pairs space (1, 36), (4, 9), (6, 6), (12, 3), and also (18, 2).There space a full of 5 bag of determinants of 36.
We deserve to have an unfavorable factors also for a offered number.
Since the product that two an unfavorable numbers is positive <(-) × (-) = +>.The negative pair determinants of 36 are (-1, -36), (-4, -9), (-6, -6), (-12, -3), and (-18, -2)There room a complete of 5 an unfavorable pairs of determinants of 36.
Important Notes:
36 is a composite number as it has factors other than 1 and itself.Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and also 36.Pair factors of 36 are (1, 36), (4, 9), (6, 6), (12, 3), and (18, 2).1 is a factor that every number.Prime factorization of 36 is 2 × 2 × 3 × 3 = 2² × 3².
Challenging Questions:
Find the factors of 3600 with the aid of techniques used in finding factors of 36.Prove the the factor that a number is always less than or equal to the number itself?
## FAQs on factors of 36
### What room the components of 36?
The determinants of 36 space 1, 2, 3, 4, 6, 9, 12, 18, 36 and its an adverse factors room -1, -2, -3, -4, -6, -9, -12, -18, -36.
### How plenty of Factors of 36 are additionally Factors that 22?
Since, the factors of 36 room 1, 2, 3, 4, 6, 9, 12, 18, 36 and the determinants of 22 are 1, 2, 11, 22.Hence, <1, 2> are the common factors that 36 and 22.
See more: Who Was The Oldest Nba Rookie Of The Year ? Was Kobe Rookie Of The Year
### What is the Greatest usual Factor that 36 and 20?
The factors of 36 and also 20 room 1, 2, 3, 4, 6, 9, 12, 18, 36 and also 1, 2, 4, 5, 10, 20 respectively.Common factors of 36 and 20 room <1, 2, 4>.Hence, the GCF that 36 and 20 is 4.
### What is the amount of all the components of 36?
Sum the all factors of 36 = (22 + 1 - 1)/(2 - 1) × (32 + 1 - 1)/(3 - 1) = 91 | 1,845 | 5,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2022-21 | latest | en | 0.894857 |
https://www.traditionaloven.com/building/refractory/graphite/convert-kilogram-kg-of-graphite-to-block-9x4-5x2-5-in-graphite.html | 1,571,298,299,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986673250.23/warc/CC-MAIN-20191017073050-20191017100550-00462.warc.gz | 1,134,743,529 | 11,771 | Graphite 1 kilogram mass to graphite 9x4.5x2.5 inch converter
# graphite conversion
## Amount: 1 kilogram (kg - kilo) of mass Equals: 0.27 graphite 9x4.5x2.5 inch (solid) in block
Converting kilogram to graphite 9x4.5x2.5 inch value in the graphite units scale.
TOGGLE : from graphite 9x4.5x2.5 inch into kilograms in the other way around.
## graphite from kilogram to graphite 9x4.5x2.5 inch Conversion Results:
### Enter a New kilogram Amount of graphite to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
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CONVERT : between other graphite measuring units - complete list.
Conversion calculator for webmasters.
## Graphite Units Converter
The calculator reflects mass density of graphite @ 20 degree Celsius state which is 2.25g/cm3 (calculated not in cold state example 2.09 - 2.23 grams per cubic centimeter.)
Due to its physical properties it is one excellent refractory material. Graphite is lighter in weight in comparison with alumina and still it retains more heat than heavy dense firebricks do. At the same time natural graphite is well abundant on the earth hence, in year 2010, refractory businesses consumed 12,500 tonnes of it only within US.
Convert graphite measuring units between kilogram (kg - kilo) and graphite 9x4.5x2.5 inch (solid) but in the other reverse direction from graphite 9x4.5x2.5 inch into kilograms.
conversion result for graphite: From Symbol Equals Result To Symbol 1 kilogram kg - kilo = 0.27 graphite 9x4.5x2.5 inch solid
# Converter type: graphite measurements
This online graphite from kg - kilo into solid converter is a handy tool not just for certified or experienced professionals.
First unit: kilogram (kg - kilo) is used for measuring mass.
Second: graphite 9x4.5x2.5 inch (solid) is unit of block.
## graphite per 0.27 solid is equivalent to 1 what?
The graphite 9x4.5x2.5 inch amount 0.27 solid converts into 1 kg - kilo, one kilogram. It is the EQUAL graphite mass value of 1 kilogram but in the graphite 9x4.5x2.5 inch block unit alternative.
How to convert 2 kilograms (kg - kilo) of graphite into graphite 9x4.5x2.5 inch (solid)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.26786828683559 * 2 (or divide it by / 0.5)
QUESTION:
1 kg - kilo of graphite = ? solid
1 kg - kilo = 0.27 solid of graphite
## Other applications for graphite units calculator ...
With the above mentioned two-units calculating service it provides, this graphite converter proved to be useful also as an online tool for:
1. practicing kilograms and graphite 9x4.5x2.5 inch of graphite ( kg - kilo vs. solid ) measuring values exchange.
2. graphite amounts conversion factors - between numerous unit pairs.
3. working with - how heavy is graphite - values and properties.
International unit symbols for these two graphite measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for kilogram is:
kg - kilo
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for graphite 9x4.5x2.5 inch is:
solid
### One kilogram of graphite converted to graphite 9x4.5x2.5 inch equals to 0.27 solid
How many graphite 9x4.5x2.5 inch of graphite are in 1 kilogram? The answer is: The change of 1 kg - kilo ( kilogram ) unit of graphite measure equals = to 0.27 solid ( graphite 9x4.5x2.5 inch ) as the equivalent measure for the same graphite type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in kg - kilo - kilograms for graphite amount, the rule is that the kilogram number gets converted into solid - graphite 9x4.5x2.5 inch or any other graphite unit absolutely exactly.
Conversion for how many graphite 9x4.5x2.5 inch ( solid ) of graphite are contained in a kilogram ( 1 kg - kilo ). Or, how much in graphite 9x4.5x2.5 inch of graphite is in 1 kilogram? To link to this graphite kilogram to graphite 9x4.5x2.5 inch online converter simply cut and paste the following.
The link to this tool will appear as: graphite from kilogram (kg - kilo) to graphite 9x4.5x2.5 inch (solid) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,206 | 4,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.780995 |
https://patioheaterpartsonline.com/interesting-and-useful/quick-answer-what-size-rinnai-direct-vent-heater-do-i-need.html | 1,632,572,485,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057622.15/warc/CC-MAIN-20210925112158-20210925142158-00462.warc.gz | 493,901,611 | 11,608 | # Quick Answer: What Size Rinnai Direct Vent Heater Do I Need?
## How big of a Rinnai tankless water heater do I need?
Rinnai tankless gas water heaters need a demand of at least 0.4 to 0.6 gallons per minute of water flow through the hot side to ignite and 0.26 gallons per minute to stay in operation.
## How big of a gas wall heater do I need?
For example, a 300 square foot room typically requires 7,000 BTUs to maintain a comfortable temperature, while a 1,000 square foot room requires 18,000 BTUs. A simple formula to determine your heating needs is: (desired temperature change) x (cubic feet of space) x. 133 = BTUs needed per hour.
## How do you calculate Rinnai flow rate?
To calculate the flow rate, consider how many outlets might possibly be used at once. Calculate the rough flow rate through each outlet for the total flow. The result is roughly what you will use in litres per minute (L/min) at peak demand times in your facility.
## Are Rinnai heaters efficient?
“Rinnai is known for product quality and reliability. Additionally, the Rinnai Q Series Boilers earned between a 95 and 95.4 percent AFUE rating, ranking them among the most efficient heating-only boiler options available.
You might be interested: Quick Answer: How To Stop Kerosene Heater From Smoking?
## What is the downside of a tankless water heater?
The primary disadvantage of on demand or instant hot water heaters is the upfront cost. The smaller units that you often see won’t produce enough hot water to serve most households. They’ll only serve one faucet at a time—a problem if you want to shower while the dishwasher is running.
## How do I calculate what size heater I need?
What Size Heater
1. Multiply the length times the width times the height to calculate the volume in cubic feet.
2. Multiply the room’s cubic footage by 4 if its walls contain fewer than 3 inches of insulation.
## How many square feet will 40000 Btu heat?
To heat a 2,000 square foot home, you will need approximately 40,000 BTU’s of heating power.
## Can a tankless water heater fill a tub?
Please remember that tankless water heaters also come in different sizes, and you need to get a unit that will deliver a strong flow to the tub. A small tankless water heater can take a long time to fill a large tub. Whatever way you choose, at least now you’ll finally be able to relax in a nice, hot whirlpool tub.
## How many gallons per minute should my tankless water heater be?
The best sizing factor for a tankless water heater is its flow per minute rate. A 3.2 gpm, tankless heater can heat 3.2 gallons of water per minute, which is roughly enough to provide hot water to a shower and a sink. To find the correct capacity, add up all of the fixtures that you expect to use at the same time.
You might be interested: Quick Answer: Knocking Sound When Heater Is On In Car?
## How many gallons should my water heater be?
But if you only need a loose estimate of what size you need (versus an exact calculation), follow these guidelines: For 1 to 2 people: 30-40 gallons. For 2 to 3 people: 40-50 gallons. For 3 to 4 people: 50-60 gallons.
## How do I choose a Rinnai heater?
Multiply the square footage by the BTU per square foot to arrive at the total BTU rating you’ll need from your Rinnai heater. All Rinnai heaters are rated for a particular number of BTU, ranging from 8,000 up to 40,000 or more, giving you a lot of flexibility in the model you can buy.
## How long does a Rinnai wall heater last?
The life expectancy (or useful lifespan) of a Rinnai tankless water heater is up to 20 years. Actual life is impacted by water quality, usage and proper maintenance. For comparison, a tankless water heater lasts about twice as long as a tank water heater, making them a better value in the long run.
## Do Rinnai wall heaters use electricity?
Rinnai has some small direct vent wall furnace. One model has a 58W power draw. | 932 | 3,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-39 | latest | en | 0.90015 |
https://www.whizwriters.com/accounting-1371/ | 1,566,521,732,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317688.48/warc/CC-MAIN-20190822235908-20190823021908-00032.warc.gz | 1,008,876,760 | 9,297 | # ACCOUNTING
PROBLEM 1
Kathleen Vohs of the University of Minnesota and her coworkers carried out several randomized comparative experiments on the effects of thinking about money. Here’s part of one such experiment. Ask student subject to unscramble 30 sets of five words to make a meaningful phrase from four of the five words. The control group unscrambled phrases like “cold it desk outside is” into “it is cold outside.” The treatment group unscrambled phrases that lead to thinking about money, turning “high salary desk paying” into “a high-paying salary.” Then each subject worked a hard puzzle, knowing that they could ask for help. Here are the times in seconds until subjects asked for help:
Treatment group:
609 444 242 198 174 55 251 466 443
531 135 241 476 482 362 69 160
Control group:
118 272 412 290 140 104 55 189 126
400 91 63 87 142 141 373 156
The researcher suspected that money is connected with self-sufficiency, so that the treatment group would ask for help less quickly on the average. Do the data support this idea? Use a 5% level of significance.
EXPECTATIONS
– Draw graphs and charts when appropriate and necessary to demonstrate your reasoning! Label all graphs and charts!
– Display formulas. Write complete sentences to summarize your conclusions.
– If use any table values, clearly state which tables you used (e.g. Table A-2, etc.).
-Attach excel output when appropriate or necessary (e.g. a scatterplot, etc.)
HYPOTHESIS TESTING QUESTIONS
Your work for all statistical hypothesis testing questions should include the following:
1. Established Ho and Ha.
2. Summary statistics (either computed or given in the problem)
3. The name of the test (e.g. 2sampleTtest or T-test about correlation, etc.)
4. A formula to compute a test statistic (e.g. 1Prop-Z test statistic, etc.)
5. A p-value of the test and/or a critical value from a statistical table.
6. Clearly state the decision rule you use the reach a conclusion. (You may have to sketch a graph to show rejection regions.) Do you “Reject Ho” or do you “Fail to Reject Ho”?
7. State your conclusion in plain language. Use complete sentences.
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ORDER NOW »» | 539 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-35 | longest | en | 0.868307 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=10891&CurriculumID=48&Method=Worksheet&NQ=4&Num=7.2&Type=B | 1,555,887,737,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532929.54/warc/CC-MAIN-20190421215917-20190422001917-00251.warc.gz | 97,288,632 | 3,205 | Name: ___________________Date:___________________
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### High School Mathematics7.2 Intersection of Sets
The intersection of sets A and B is the set of elements which are common to A and B. Example: If A = {1, 2, 3, 4} and B = {2, 4, 6}, then find A Ç B. Solution: Given that, A = {1, 2, 3, 4} and B = {2, 4, 6}. A Ç B = {1, 2, 3, 4} Ç {2, 4, 6} The common elements in A and B are 2 and 4. Therefore, A Ç B = {2, 4}. Directions: Choose the correct answer. Also write at least 5 examples of your own.
Name: ___________________Date:___________________
### High School Mathematics7.2 Intersection of Sets
Q 1: If A = {1, 3, 5, 7} and B = {2, 4, 6}, then find A Ç B.{1, 3}Null set{1, 2, 3, 4, 5, 6} Q 2: If A = {1, 2, 3, 4} and B = {1, 3, 5}, then find A Ç B.{1, 2, 3, 4}{1, 3}Null set Question 3: This question is available to subscribers only! Question 4: This question is available to subscribers only!
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For unlimited printable worksheets & more, go to http://www.kwizNET.com. | 521 | 1,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-18 | longest | en | 0.819775 |
https://si.wikipedia.org/wiki/%E0%B7%83%E0%B7%90%E0%B6%9A%E0%B7%92%E0%B6%BD%E0%B7%8A%E0%B6%BD:Convert/%E0%B6%BD%E0%B7%9A%E0%B6%9B%E0%B6%BA | 1,591,356,796,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348500712.83/warc/CC-MAIN-20200605111910-20200605141910-00138.warc.gz | 523,719,462 | 44,935 | # සැකිල්ල:Convert/ලේඛය
Template {{convert}} calculates measurements from one unit (you can enter) to another one, and then presents the results. For example:
`{{convert|2|km|mi}}` → 2 kilometres (1.2 mi)
`{{convert|7|mi|km}}` → 7 miless (11 km)
The available units are listed below. Units are always case-sensitive (kW, not KW or kw). Multiple formatting and calculation options are listed below.
## Unit name or symbol (abbreviation): 1 pound or 1 lb?
By default, the first value shows unit name spelled out, the second the symbol (or abbreviation). Use `|abbr=on/off` to change this behaviour:
• `{{convert|1|lb|kg}}` → 1 pound (0.45 kg)
To abbreviate both:
• `{{convert|1|lb|kg|abbr=on}}` → 1 lb (0.45 kg)
• `{{convert|1|lb|kg|abbr=off}}` → 1 pound (0.45 kilograms)
`|abbr=in` is the reverse behaviour to the default, `|abbr=out`:
• `{{convert|1|lb|kg|abbr=in}}` → 1 lb (0.45 kilograms)
When using scientific notation, the default is for the second (converted) value to display with scientific notation:
• `{{convert|100|e6mi|e6km}}` → 100 million miless (160×106 km)
To display thousands, millions, etc., while abbreviating both input and output units use `|abbr=unit`:
• `{{convert|100|e6mi|e6km|abbr=unit}}` → 100 million mi (160 million km)
### Convenience: {{cvt}} has `|abbr=on` by default
Template {{cvt}} is similar to {{convert}}, but has `|abbr=on` as the default behaviour. In {{cvt}}, all other options are available. So:
`{{cvt|1|lb|kg}}` → 1 lb (0.45 kg)
is equal to:
`{{convert|1|lb|kg|abbr=on}}` → 1 lb (0.45 kg)
Use `|adj=on` to produce the adjective form:
• `A {{convert|10|mi|km|adj=on}} distance` → A 10-miles (16 km) distance.
Default behaviour, for comparison:
• `{{convert|10|mi|km}} to go` → 10 miless (16 km) to go.
This option does not work with `|abbr=on`.
## Rounding: 100 ft is 30 m or 30.5 m or 30.48 m?
By definition, 100 ft equals 30.48 m. In practical use, it is common to round the calculated metric number. With that, there are several possibilities.
### Default rounding
By {{Convert}} default, the conversion result will be rounded either to precision comparable to that of the input value (the number of digits after the decimal point—or the negative of the number of non-significant zeroes before the point—is increased by one if the conversion is a multiplication by a number between 0.02 and 0.2, remains the same if the factor is between 0.2 and 2, is decreased by 1 if it is between 2 and 20, and so on) or to two significant figures, whichever is more precise. An exception to this is rounding temperatures (see below).
Examples of rounding
Input Displays as Note
`{{convert|123|ft|m|-1}}` 123 feet (40 m)
`{{convert|123|ft|m}}` 123 feet (37 m) same output as with 0 (below)
`{{convert|123|ft|m|0}}` 123 feet (37 m)
`{{convert|123|ft|m|1}}` 123 feet (37.5 m)
`{{convert|123|ft|m|2}}` 123 feet (37.49 m) exact value is 37.4904 m
`{{convert|500|ft|m|-1}}` 500 feet (150 m)
`{{convert|500|ft|m}}` 500 feet (150 m) same output as with -1 (above)
`{{convert|500|ft|m|0}}` 500 feet (152 m)
`{{convert|500|ft|m|1}}` 500 feet (152.4 m)
`{{convert|500|ft|m|2}}` 500 feet (152.40 m) exact value is 152.4 m
Convert supports four types of rounding:
### Round to a given precision: use a precision number
Specify the desired precision with the fourth unnamed parameter (or third unnamed parameter if the "convert to" parameter is omitted; or fifth unnamed parameter if a range is specified; or fourth unnamed parameter again if a range is specified and the "convert to" parameter is omitted; needs to be replaced with a "precision" named parameter). The conversion is rounded off to the nearest multiple of 110 to the power of this number. For instance, if the result is 8621 and the round number is "-2", the result will be 8600. If the result is "234.0283043" and the round number is "0", the result will be 234.
### Round to a given number of significant figures: `|sigfig=`
To specify the output number to be with n significant figures use `|sigfig=<number>`:
• `{{convert|1200|ft|m|sigfig=4}}` → 1,200 feet (365.8 m)
• `{{convert|1200|ft|m|sigfig=3}}` → 1,200 feet (366 m)
• `{{convert|1200|ft|m|sigfig=2}}` → 1,200 feet (370 m)
• `{{convert|1200|ft|m|sigfig=1}}` → 1,200 feet (400 m)
Default behaviour, for comparison:
• `{{convert|1200|ft|m}}` → 1,200 feet (370 m)
Setting `|sigfig=` is meaningless
• `{{convert|1200|ft|m|sigfig=0}}` → 1,200 feet (370 m)*
### Round to a multiple of 5: 15, 20, 25, ...
Using `|round=5` rounds the outcome to a multiple of 5.
• `{{convert|10|m|ft}}` → 10 metres (33 ft)
• `{{convert|10|m|ft|round=5}}` → 10 metres (35 ft)
Similar: using `|round=25` rounds the outcome to a multiple of 25.
• `{{convert|10|m|ft}}` → 10 metres (33 ft)
• `{{convert|10|m|ft|round=25}}` → 10 metres (25 ft)
Default behaviour, for comparison:
• `{{convert|10|m|ft|sigfig=4}}` → 10 metres (32.81 ft)
In a range, one can round each value individually to the default. Use `|round=each`:
• `{{convert|10 x 200 x 3000|m|ft}}` → 10 by 200 by 3,000 metres (33 ft × 656 ft × 9,843 ft)
• `{{convert|10 x 200 x 3000|m|ft|round=each}}` → 10 by 200 by 3,000 metres (33 ft × 660 ft × 9,800 ft)
### Round to a multiple of a given fraction: 2 3⁄16 inch
Specify the desired denominator using `|frac=<some positive integer>`. (Denominator is the below-the-slash number, for example the 3 in 13).
• `{{convert|5.56|cm|in|frac=16}}` → 5.56 centimetres (2 316 in)
• `{{convert|8|cm|in|frac=4}}` → 8 centimetres (3 14 in)
The fraction is reduced when possible:
• `{{convert|8|cm|in|frac=100}}` → 8 centimetres (3 320 in)
Default behaviour, for comparison:
• `{{convert|8|cm|in}}` → 8 centimetres (3.150 in)
### In temperatures: rounding °C, °F and K
In temperatures, the conversion will be rounded either to precision comparable to that of the input value or to that which would give three significant figures when expressed in kelvins, whichever is more precise.
1. `{{convert|10,000|C|F K}}` → 10,000 °C (18,000 °F; 10,300 K)
2. `{{convert|10,000.1|C|F K}}` → 10,000.1 °C (18,032.2 °F; 10,273.2 K)
3. `{{convert|-272|C|F K}}` → −272 °C (−457.60 °F; 1.15 K)
4. `{{convert|-272|C}}` → −272 °C (−457.60 °F)
5. `{{convert|-300|C|F K}}` → −300 °C (−508.0 °F; −26.9 K)
6. `{{convert|0|C|F K}}` → 0 °C (32 °F; 273 K)
The precision of the input number in example (1) is one digit, but the precision of its Kelvins expression is three, so the precision of the Fahrenheit conversion is made three (made 180...) . (1) and (2) seem to belie the fact that a 0.1 degrees Celsius change is a 0.18 degrees Fahrenheit change, and make the 32 degrees difference shown in (1) begin to seem off somehow. Result (1) seems off until you set the significant figures yourself with `|sigfig=`:
`{{convert|10000|C|sigfig=5}}` → 10,000 °C (18,032 °F)
or you set the precision positionally, relative to the decimal point (zero being at the decimal point):
`{{convert|10000|C|0}}` → 10,000 °C (18,032 °F)
The precision of the input number in example (2) is six, so the precision of the Fahrenheit output is, whereas before, Kelvins had determined it to be three. Examples (3) and (4) show how this can be hidden and generate questions, but it occurs there because the Kelvins conversion generated two fractional parts. (Before it was the input number that generated the fractional part.) In example (3) the three input digits converted into five significant output digits because of the two digits after the decimal point, generated by the Kelvins conversion. This happened again in (5), but in (6) decimal fractions were neither given as input nor induced by the Kelvins conversion.
## Into multiple units: 10 °C (50 °F; 283 K)
Separate the multiple output units by a space:
• `{{convert|10|C|F K}}` → 10 °C (50 °F; 283 K)
• `{{convert|5|km|mi nmi}}` → 5 kilometres (3.1 mi; 2.7 nmi)
If the output unit names contain spaces, use `+` as the separator.
## Ranges of values
A range converts two values and separates them by your choice of words and punctuation.
### A range: 6 to 17 kg (13 to 37 lb)
Range indicators are entered as the second parameter (between the values). Range separators can be:
Range separators in {{convert}}
Separator Convert Result Notes
`-` `{{convert|3|-|6|ft}}` 3–6 feet (0.91–1.83 m) Input can be hyphen (-) or en dash (–), output uses en dash
`–` `–` (en dash) `{{convert|3|–|6|ft}}` 3–6 feet (0.91–1.83 m)
`and` `{{convert|3|and|6|ft}}` 3 and 6 feet (0.91 and 1.83 m)
`and(-)` `{{convert|3|and(-)|6|ft}}` 3 and 6 feet (0.91–1.83 m) `|abbr=on` abbreviates the first unit
`and(-)|abbr=on` `{{convert|3|and(-)|6|ft|abbr=on}}` 3 and 6 ft (0.91–1.83 m)
`or` `{{convert|3|or|6|ft}}` 3 or 6 feet (0.91 or 1.83 m)
`to` `{{convert|3|to|6|ft}}` 3 to 6 feet (0.91 to 1.83 m)
`to(-)` `{{convert|3|to(-)|6|ft}}` 3 to 6 feet (0.91–1.83 m) `|abbr=on` abbreviates the first unit
`to(-)|abbr=on` `{{convert|3|to(-)|6|ft|abbr=on}}` 3 to 6 ft (0.91–1.83 m)
`to about` `{{convert|3|to about|6|ft}}` 3 to about 6 feet (0.91 to about 1.83 m)
`+/-` `{{convert|3|+/-|6|ft}}` 3 ± 6 feet (0.91 ± 1.83 m)
`±` `±` `{{convert|3|±|6|ft}}` 3 ± 6 feet (0.91 ± 1.83 m)
`+` `{{convert|3|+|6|ft}}` 3 + 6 feet (0.91 + 1.83 m)
`,` `{{convert|3|,|6|ft}}` 3, 6 feet (0.91, 1.83 m)
`, and` `{{convert|3|, and|6|ft}}` 3, and 6 feet (0.91, and 1.83 m)
`, or` `{{convert|3|, or|6|ft}}` 3, or 6 feet (0.91, or 1.83 m)
`by` `{{convert|3|by|6|ft}}` 3 by 6 feet (0.91 by 1.83 m)
`x` `{{convert|3|x|6|ft}}` 3 by 6 feet (0.91 m × 1.83 m) `|abbr=on` abbreviates and repeats the first unit
`×` `×` `{{convert|3|×|6|ft}}` 3 by 6 feet (0.91 m × 1.83 m)
`x|abbr=on` `{{convert|3|x|6|ft|abbr=on}}` 3 ft × 6 ft (0.91 m × 1.83 m)
`xx` |xx| is deprecated. Use |x| instead
`*` |*| is deprecated. Use |x| instead
### Multiple dimensions: 6 m × 12 m (20 ft × 39 ft)
Use `by`:
• `{{convert|6|by|12|ft|m}}` → 6 by 12 feet (1.8 by 3.7 m)
Use `×`, multiplication sign, or `x`, letter:
• `{{convert|6|x|12|m|ft}}` → 6 by 12 metres (20 ft × 39 ft)
In science, the formal way is to set `|x|` and `|abbr=on` (keeping dimensions right, like in area = x km2):
• `{{convert|6|x|12|m|ft|abbr=on}}` → 6 m × 12 m (20 ft × 39 ft)
### About feet, inch in ranges and multiples
While it is possible to enter feet, inch in a simple conversion, this is not possible for ranges:
Default behaviour, for comparison:
• `{{convert|1|ft|3|in|mm}}` → 1 foot 3 inches (380 mm)
## Words
### Spelling of unit name: UK metre or US meter?
Default spelling of units is en-UK. To show en-US spelling, use `|sp=us`:
`{{convert|1|m|ft}}` → 1 metre (3.3 ft)—default
`{{convert|1|m|ft|sp=us}}` → 1 meter (3.3 ft)
### Spell out numbers: ten miles
To write a number in words, use `|spell=in`:
• `{{convert|10|mi|m|spell=in}}` → දහය miless (16,000 m)
To spell out both in and out values, use `|spell=on`:
• `{{convert|10|mi|m|spell=on}}` → දහය miless (දහසය thousand metres)
To make first letter a capital, use `|spell=In`, `|spell=On`
• `{{convert|10|mi|m|spell=In}}` → දහය miless (16,000 m)
• `{{convert|10|mi|m|spell=On}}` → දහය miless (දහසය thousand metres)
Remember that the spelling of the units (ft, m) is independently set by `|abbr=`. To the extreme:
• `{{convert|10|mi|m|spell=on|abbr=off|sp=us}}` → දහය miless (දහසය thousand meters)
### Inserted before units: 4 planted acres
• `{{convert|4|acre||adj=pre|planted}}` → 4 planted acres (1.6 ha)
`disp=preunit` is similar, but has no separator after the specified text, and can have different text for the output value:
• `{{convert|4|acre||disp=preunit|planted }}` → 4 planted acres (1.6 planted ha)
• `{{convert|4|acre||disp=preunit|planted |reforested-}}` → 4 planted acres (1.6 reforested-ha)
### After adjective unit: A 10-foot-long corridor
`{{convert|10|ft|m|adj=mid|-long}}` → 10-foot-long (3.0 m)
### Plurals: 1 inch, 2 inches
The unit symbol is singular always. Depending on the preceding number only, a unit name can be shown plural.
• `{{convert|1|metre}}` → 1 metre (3 ft 3 in)
• `{{convert|2|metre}}` → 2 metres (6 ft 7 in)
• `{{convert|2|metre|abbr=on}}` → 2 m (6 ft 7 in)
Exception
Entering the unit spelled `|foot|` forces singular output "foot", whatever the number is.
• `{{convert|100|foot|abbr=off}}` → 100 foot (30 metres)
### Fractions: one-eighth of an imperial pint
The convert template also supports spelling out fractions.
• `{{convert|1/4|oz|g|spell=in}}` → one-quarter ounce (7.1 g)
Any additional words needed for the fraction can also be added at the end of the template.
• `{{convert|1/8|imppt|mL|spell=in|adj=pre|of an}}` → one-eighth of an imperial pint (71 mL)
## Numbers
### Using an SI prefix: gigameter (Gm), or micrometer (μm)
Metric prefixes
Text Symbol Factor
yotta Y 1024
zetta Z 1021
exa E 1018
peta P 1015
tera T 1012
giga G 109
mega M 106
kilo k 103
hecto h 102
deca da 101
(none) (none) 1
deci d 10−1
centi c 10−2
milli m 10−3
micro μ 10−6
nano n 10−9
pico p 10−12
femto f 10−15
atto a 10−18
zepto z 10−21
yocto y 10−24
Units can have an SI prefix like `G` before the unit: `Gm`, and `giga` before the name: `gigameter`. These are plain multiplication factors.
To illustrate, these are trivial calculations (from meter to meter), showing the multiplication factor:
• 12 Gm (1.2×1010 m)
• 12 μm (1.2×10−5 m)
The prefix can be added before the SI unit (here: unit `m` for meter):
• `{{convert|12|Gm|mi|abbr=on}}` → 12 Gm (7,500,000 mi)
• `Mm`: 12 Mm (7,500 mi)
• `km`: 12 km (39,000 ft)
• `mm`: 12 mm (0.47 in)
• `μm`: 12 μm (0.012 mm)
• `um`: 12 μm (0.012 mm) (letter "u" can be used for "μ" here)
The prefix can be used in the output unit:
• `{{convert|12000|mi|Mm|abbr=on}}` → 12,000 mi (19 Mm)
• `{{convert|12|in|μm|abbr=on}}` → 12 in (300,000 μm)
As an exception, the non-SI unit "inch" can have the "μ" prefix too)
• `{{convert|12|μm|μin|abbr=on}}` → 12 μm (470 μin)
### Engineering notation: 7 × 106 m
#### In the unit: e6m
Engineering number notations like 7E6 (for 7 × 106) can be entered as a "prefix" to the unit:
• `{{convert|7|e6m}}` → 7 million metres (23,000,000 ft)
The same is possible for the output unit:
• `{{convert|23,000,000|ft|e6m}}` → 23,000,000 feet (7.0×106 m)
Any standard unit (not a combination, multiple, or built-in unit) can have such a prefix:
• `e3` (thousand),
• `e6` (million),
• `e9` (billion),
• `e12` (trillion),
• `e15` (quadrillion).
### Scientific notation: 1.23×10−14
In scientific notation, a number is written like 1.23×10−14. The plain number has exactly one digit before the decimal point.
With {{convert}}, the input can be in e-notation such as `12.3e4`. This value is displayed as a power of ten, and the output is displayed in scientific notation, except that an output value satisfying 0.01 <= v < 1000 is shown as a normal number. In addition, if the output value is 1000 and sigfig=4 is used, the value is displayed as a normal number.
• `{{convert|12.3e-15|atm|atm|abbr=on}}` → 12.3×10−15 atm (1.23×10−14 atm)
• `{{convert|0.00000005|atm|atm|abbr=on}}` → 0.00000005 atm (5.0×10−8 atm)
### Input with fractions: 1 1⁄2 inches (38.1 mm)
The number to convert can be written in fractions. Both `/` (keyboard slash) and `⁄` (fraction slash) are accepted:
• `{{convert|1/2|in|mm|1}}`12 inch (12.7 mm)
• `{{convert|1⁄2|in|mm|1}}`12 inch (12.7 mm)
With integers, use a `+` sign
• `{{convert|2+1⁄2|in|mm|1}}`2 12 inches (63.5 mm)
When negative, use a hyphen `-` and repeat it:
• `{{convert|-2-1⁄2|in|mm|1}}`−2 12 inches (−63.5 mm)
• `{{convert|2-1⁄2|in|mm|1}}` → 2–12 inch (50.8–12.7 mm) Note: this is read as a range
• `{{convert|-2+1⁄2|in|mm|1}}`[convert: invalid number] N Should be a number, not an expression (do not require a calculation)
### Output with horizontal fraction bar in: 1/2 inch
Using a double slash (`//`) returns a horizontal bar fraction:
• `{{convert|1//2|in|mm|1}}`1/2 inch (12.7 mm)
• `{{convert|2+1//2|in|mm|1}}`2 1/2 inches (63.5 mm)
### Thousands separator: 1,000 mi or 1000 mi
In input, a comma for thousands separator is accepted but not required; a gap (space) is not accepted. In output, by default, the thousand separator is the comma:
• `{{convert|1234567|m|ft}}` → 1,234,567 metres (4,050,417 ft)
• `{{convert|1,234,567|m|ft}}` → 1,234,567 metres (4,050,417 ft)
• `{{convert|1 234 567|m|ft}}`[convert: invalid number] N
Set `|comma=off` to remove the separator from the output:
• `{{convert|1234567|m|ft|comma=off}}` → 1234567 metres (4050417 ft)
Use `|comma=gaps` to use digit grouping by gap (thin space) as a thousands separator:
• `{{convert|1234567|m|ft|comma=gaps}}`1234567 metres (4050417 ft)
Default behaviour, for comparison:
• `{{convert|1234567|m|ft}}` → 1,234,567 metres (4,050,417 ft)
Setting `|comma=5` or `|comma=gaps5` will only add the separator when the number of digits is 5 or more:
• `{{convert|1234|m|ft|comma=5}}` → 1234 metres (4049 ft)
• `{{convert|1234567|m|ft|comma=5}}` → 1,234,567 metres (4,050,417 ft)
• `{{convert|1234|m|ft|comma=gaps5}}` → 1,234 metres (4,049 ft)*
• `{{convert|1234567|m|ft|comma=gaps5}}` → 1,234,567 metres (4,050,417 ft)*
Default behaviour, for comparison:
• `{{convert|1234|m|ft}}` → 1,234 metres (4,049 ft)
## Output manipulation
### Brackets and separators: 10 m [33 ft]
Punctuation that distinguishes the two measurements is set by `|disp=`.
Options are: `b` (the default), `sqbr`, `comma`, `or`, `br`, `x|…`:
• `{{convert|10|m|ft|disp=sqbr}}` → 10 metres [33 ft]
• `{{convert|10|m|ft|disp=comma}}` → 10 metres, 33 ft
• `{{convert|10|m|ft|disp=or}}` → 10 metres or 33 feet
Default behaviour, for comparison:
• `{{convert|10|m|ft}}` → 10 metres (33 ft)
Setting `|disp=br` will force a new line (`<br/>`)
• `{{convert|10|m|ft|disp=br}}` → 10 metres
33 feet
Also `|disp=br()` will force a new line, and keep the brackets:
• `{{convert|10|m|ft|disp=br()}}` → 10 metres
(33 feet)
Setting `|disp=x|…` allows any text as separator:
• `{{convert|10|m|ft|disp=x|_MyText_}}` → 10 metres_MyText_33 ft—(To display spaces, use ` `)
### Flipping (reordering) the two measurements: 1,609.3 metres (1 miles)
Setting `|order=flip` will flip (swap) the two measurements:
• `{{convert|1|mi|m|order=flip}}` → 1,609.3 metres (1 miles)
Default behaviour, for comparison:
• `{{convert|1|mi|m}}` → 1 miles (1,609.3 metres)
When converting to multiple units, the effect is:
• `{{convert|10|km|mi nmi|order=flip}}` → 6.2137 miless; 5.3996 nautical miles (10 kilometres)
• `{{convert|10|km|nmi mi|order=flip}}` → 5.3996 nautical miles; 6.2137 miless (10 kilometres)
### Displaying parts of the result: 2 (1.5)
It is possible to display only parts of the conversion result:
Convert Output Description
`{{convert|2|cuyd|m3}}` 2 cubic yards (1.5 m3) Default behaviour, for comparison
`{{convert|2|cuyd|m3|abbr=values}}` 2 (1.5) Input and output numbers
`{{convert|2|cuyd|m3|disp=unit}}` cubic yards Input unit
`{{convert|2|cuyd|m3|disp=unit|adj=on}}` cubic-yard Input unit, adjective (hyphenated)
`{{convert|2|cuyd|cuyd|0|disp=out|abbr=off}}` 2 cubic yards Input (workaround)
`{{convert|2|cuyd|m3|abbr=~}}` 2 cubic yards [cu yd] (1.5 m3) Input: both name and symbol
`{{convert|2|cuyd|m3|disp=unit2}}` m3 Output unit (symbol)
`{{convert|2|cuyd|m3|disp=unit2|abbr=off}}` cubic metres Output unit (name)
`{{convert|2|cuyd|m3|disp=number}}` 1.5 Output value
`{{convert|2|cuyd|m3|disp=out}}` 1.5 m3 Output value and unit
`{{convert|2|cuyd|m3|disp=out|abbr=off}}` 1.5 cubic metres Output value and unit
### Display both input name and symbol: 2 kilopascals [kPa]
Setting `|abbr=~` returns both name and symbol of the first (input) unit:
• `{{convert|2|kPa|psi|abbr=~}}` → 2 kilopascals [kPa] (0.29 psi)
• `A {{convert|2|kPa|psi|abbr=~|adj=on}} pressure`A 2-kilopascal [kPa] (0.29 psi) pressure
## Table options
For the wikitable structure, there are three options: add a line-break, split the result over columns and make the table sortable.
### Enforced line break
`|disp=br` adds a line-break and omits brackets.
`|disp=br()` adds a line-break and does add brackets to the converted value. This may be useful in tables:
`|disp=br` `|disp=br()`
100 kilometres
62 miless
100 kilometres
(62 miless)
### Table columns showing numbers only
Using {convert} in a table cell, with `|disp=table` splits the result over two columns:
`{{convert|10|m|ft|disp=table}}`
`style="text-align:right;"|10`
`|style="text-align:right;"|33`
`|disp=tablecen` does the same, and also centers the text:
`{{convert|20|m|ft|disp=tablecen}}`
`style="text-align:center;"|20`
`|style="text-align:center;"|66`
m ft
`|disp=table` 10 33
`|disp=tablecen` 20 66
`|disp=<other>` (default) 30 metres (98 ft)
### Sorting
Use `|sortable=on` to include a hidden numerical sortkey in the output, suitable for use in a table with sortable columns. Technically, this places a hidden string before the actual displayed values:
`{{convert|10|m|ft|sortable=on}}` → <span style="display:none">7001100000000000000</span>10 metres (33 ft)
Use both `|disp=table` and `|sortable=on` together to produce table columns (pipe symbols) for each value in sortable columns:
m ft
A 15 34 52
B 15.5 51
C 16.0 52.5
D 16 52
The generated sortkey is calculated in a consistent way based on both the value and its unit as passed to the convert template. In most cases convert uses the passed value converted to SI base units. It is therefore not necessarily the displayed value or other alternate units and is calculated regardless of output format options. Using different units or different order of units in individual rows should therefore not lead to incorrect sorting, although variations in rounding can give surprising results, since an unrounded number is used for the sortkey.
## Units
### All units
The table below lists units supported by template {{convert}}. The list is not complete—more complete lists are linked for each dimension. A complete list is at the full list of units.
Explanation – unit codes
The {{convert}} template uses unit-codes, which are similar to (but not necessarily exactly the same as) the usual written abbreviation for a given unit. These 'unit-codes' are given in column 3 of the following tables, and are accepted as input by the {{convert}} template as the second and third unnamed parameters. For example, `{{convert|100|kg|lb}}` produces "100 kilograms (220 lb)" and `{{convert|100|lb|kg}}` produces "100 pounds (45 kg)"
Some units have alternative unit-codes; these are shown in brackets in column 3, e.g. °F (F): either may be entered for Fahrenheit (but not lower-case f).
The unit-codes should be treated as case-sensitive, especially in cases like `{{convert|100|Mm|mm}}`, which produces "100 megametres (1.0×1011 mm)".
The output of {{convert}} can display multiple converted units, if further unit-codes are specified after the second unnamed parameter (without the | separator). Typical combination output units are listed below in column 7. For example, `{{convert|55|nmi|km mi}}` produces "55 nautical miles (102 km; 63 mi)", and `{{convert|1|oz|ozt g gr}}` produces "1 ounce (0.91 ozt; 28 g; 440 gr)".
Abridged list of units supported by {{Convert}}
Explanation
system unit unit-
code
symbol or
abbrev.
notes sample default
conversion
combination
output units
the system(s) to which the unit belongs units listed by name unit-code to use in template symbols shown in output other notes about the units sample of the default conversion for the unit output codes for multiple conversions
AREA
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
SI square kilometre km2 km2 US spelling: square kilometer
1.0 km2 (0.39 sq mi)
• km2 sqmi
square metre m2 m2 US spelling: square meter
1.0 m2 (11 sq ft)
• m2 sqft
square centimetre cm2 cm2 US spelling: square centimeter
1.0 cm2 (0.16 sq in)
• cm2 sqin
square millimetre mm2 mm2 US spelling: square millimeter
1.0 mm2 (0.0016 sq in)
• mm2 sqin
non-SI metric hectare ha ha 1.0 ha (2.5 acres)
Imperial
&
US customary
square mile sqmi sq mi 1.0 sq mi (2.6 km2)
• sqmi km2
acre acre (none) 1.0 acre (0.40 ha)
square yard sqyd sq yd 1.0 sq yd (0.84 m2)
square foot sqft
(sqfoot)
sq ft long code "sqfoot" outputs square foot (and never feet) 1.0 sq ft (0.093 m2)
• sqft m2 (sqfoot m2)
square inch sqin sq in 1.0 sq in (6.5 cm2)
• sqin cm2
Other square nautical mile sqnmi sq nmi 1.0 sq nmi (3.4 km2; 1.3 sq mi)
dunam dunam (none) For alternative spellings and definitions see the full list 1.0 dunam (0.0010 km2; 0.00039 sq mi)
tsubo tsubo (none) 1.0 tsubo (3.3 m2)
DENSITY
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
Metric kilogram per cubic metre kg/m3 kg/m3 1.0 kg/m3 (1.7 lb/cu yd)
• kg/m3 lb/ft3 (kg/m3 lb/cuft)
• kg/m3 lb/yd3 (kg/m3 lb/cuyd)
gram per cubic metre g/m3 g/m3 1.0 g/m3 (0.0017 lb/cu yd)
• g/m3 kg/m3
• g/m3 lb/ft3 (g/cm3 lb/cuft)
• g/m3 lb/yd3 (g/cm3 lb/cuyd)
Imperial
&
US customary
pound per cubic foot lb/ft3 lb/cu ft 1.0 lb/cu ft (0.016 g/cm3)
• lb/ft3 kg/m3 (lb/cu ft g/m3)
• lb/ft3 g/m3 (lb/cu ft g/m3)
pound per cubic yard lb/yd3 lb/cu yd 1.0 lb/cu yd (0.59 kg/m3)
• lb/yd3 kg/m3 (lb/cuyd kg/m3)
• lb/yd3 g/m3 (lb/cuyd g/m3)
ENERGY
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
SI gigajoule GJ GJ 1.0 GJ (280 kWh)
megajoule MJ MJ 1.0 MJ (0.28 kWh)
kilojoule kJ kJ 1.0 kJ (240 cal)
hectojoule hJ hJ 1.0 hJ (24 cal)
decajoule daJ daJ 1.0 daJ (2.4 cal)
joule J J 1.0 J (0.24 cal)
decijoule dJ dJ 1.0 dJ (0.024 cal)
centijoule cJ cJ 1.0 cJ (0.0024 cal)
millijoule mJ mJ 1.0 mJ (0.00024 cal)
microjoule μJ (uJ) μJ 1.0 μJ (2.4×10−7 cal)
nanojoule nJ nJ 1.0 nJ (2.4×10−10 cal)
cgs megaerg Merg Merg 1.0 Merg (0.10 J)
kiloerg kerg kerg 1.0 kerg (0.10 mJ)
erg erg (none) 1.0 erg (0.10 μJ)
Watt-hour multiples terawatt-hour TWh TWh 1.0 TWh (3.6 PJ)
TW.h TW⋅h
gigawatt-hour GWh GWh 1.0 GWh (3.6 TJ)
GW.h GW⋅h
megawatt-hour MWh MWh 1.0 MWh (3.6 GJ)
MW.h MW⋅h
kilowatt-hour kWh kWh 1.0 kWh (3.6 MJ)
kW.h kW⋅h
watt-hour Wh Wh 1.0 Wh (3.6 kJ)
W.h W⋅h
Electron-
volt
multiples
gigaelectronvolt GeV GeV 1.0 GeV (0.16 nJ)
megaelectronvolt MeV MeV 1.0 MeV (0.16 pJ)
kiloelectronvolt keV keV 1.0 keV (0.16 fJ)
electronvolt eV eV 1.0 eV (0.16 aJ)
millielectronvolt meV meV 1.0 meV (0.16 zJ)
Calorie multiples calorie Cal Cal The thermo-chemical calorie is the default definition. For others, see the full list. 1.0 Cal (4.2 kJ)
megacalorie Mcal Mcal 1.0 Mcal (4.2 MJ)
kilocalorie kcal kcal 1.0 kcal (4.2 kJ)
calorie cal cal 1.0 cal (4.2 J)
millicalorie mcal mcal 1.0 mcal (4.2 mJ)
pound/
ounce
-foot/
inch-hour-minute-second
foot-poundal ftpdl ft⋅pdl 1.0 ft⋅pdl (0.042 J)
foot-pound force ftlbf ft⋅lbf 1.0 ft⋅lbf (1.4 J)
ftlb-f ft⋅lbf
inch-pound force inlbf in⋅lbf 1.0 in⋅lbf (110 mJ)
inlb-f in⋅lbf
inch-ounce force inozf in⋅ozf 1.0 in⋅ozf (7.1 mJ)
inoz-f in⋅ozf
horsepower-hour hph hp⋅h 1.0 hp⋅h (0.75 kWh)
British thermal unit British thermal unit Btu Btu The International Steam Table British thermal unit is used. For others, see the full list. 1.0 Btu (1.1 kJ)
BTU BTU
TNT-based units gigatonne of TNT GtTNT (none) 1.0 gigatonne of TNT (4.2 EJ)
gigaton of TNT GtonTNT (none)
megatonne of TNT MtTNT (none) 1.0 megatonne of TNT (4.2 PJ)
megaton of TNT MtonTNT Mt
kilotonne of TNT ktTNT (none) 1.0 kilotonne of TNT (4.2 TJ)
kiloton of TNT ktonTNT kt
tonne of TNT tTNT (none) 1.0 tonne of TNT (4.2 GJ)
ton of TNT tonTNT (none)
Other
Hartree Eh Eh 1.0 Eh (27 eV)
rydberg Ry Ry 1.0 Ry (14 eV)
tonne of oil equivalent toe toe 1.0 toe (42 GJ)
barrel of oil equivalent BOE BOE 1.0 BOE (6.1 GJ)
cubic foot of natural gas cuftnaturalgas (cufootnaturalgas) Unit-code cufootnaturalgas will show "cubic foot of natural gas" if plural. 1.0 cubic foot of natural gas (1.1 MJ)
litre-atmosphere latm l⋅atm 1.0 l⋅atm (100 J)
Latm L⋅atm
imperial gallon-atmosphere impgalatm imp gal⋅atm 1.0 imp gal⋅atm (460 J)
US gallon-atmosphere USgalatm
(usgalatm)
US gal⋅atm 1.0 US gal⋅atm (380 J)
U.S.galatm
(usgalatm)
U.S. gal⋅atm
FORCE
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
SI giganewton GN GN Allows triple output units. See: full list. 1.0 GN (220,000,000 lbf)
• GN LT-f
• GN LTf
• GN ST-f
• GN STf
meganewton MN MN Allows triple output units. See: full list. 1.0 MN (220,000 lbf)
• MN LT-f
• MN LTf
• MN ST-f
• MN STf
kilonewton kN kN Allows triple output units. See: full list. 1.0 kN (220 lbf)
• kN LT-f
• kN LTf
• kN ST-f
• kN STf
newton N N 1.0 N (0.22 lbf)
• N lb-f
• N lbf
• N oz-f
• N ozf
millinewton mN mN 1.0 mN (0.0036 ozf)
• mN oz-f
• mN ozf
• mN gr-f
• mN grf
micronewton μN (uN) μN 1.0 μN (0.0016 grf)
• μN gr-f
• μN grf
nanonewton nN nN 1.0 nN (1.6×10−6 grf)
• nN gr-f
• nN grf
cgs megadyne Mdyn Mdyn 1.0 Mdyn (2.2 lbf)
kilodyne kdyn kdyn 1.0 kdyn (0.036 ozf)
dyne dyn
(dyne)
dyn 1.0 dyn (0.016 grf)
millidyne mdyn mdyn 1.0 mdyn (1.6×10−5 grf)
Metric gravitational units tonne-force t-f tf 1.0 tf (9.8 kN; 0.98 LTf; 1.1 STf)
tf tf
kilogram-force kg-f kgf 1.0 kgf (9.8 N; 2.2 lbf)
kgf kgf
gram-force g-f gf 1.0 gf (9.8 mN; 0.035 ozf)
gf gf
milligram-force mg-f mgf 1.0 mgf (9.8 μN; 0.015 grf)
mgf mgf
Avoirdupois-based units poundal pdl pdl 1.0 pdl (0.14 N)
long ton-force LT-f LTf 1.0 LTf (10.0 kN)
• LT-f ST-f
LTf long ton-force
• LTf STf
short ton-force ST-f STf 1.0 STf (8.9 kN)
• ST-f LT-f
STf short ton-force
• STf LTf
pound-force lb-f lbf 1.0 lbf (4.4 N)
lbf lbf
grain-force gr-f grf 1.0 grf (640 μN)
grf grf
LENGTH
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
SI megametre Mm Mm US spelling: megameter
1.0 Mm (620 mi)
kilometre km km US spelling: kilometer
1.0 km (0.62 mi)
• km mi
metre m m US spelling: meter
1.0 m (3 ftin)
• m ft
• m ftin
centimetre cm cm US spelling: centimeter
1.0 cm (0.39 in)
• cm in
millimetre mm mm US spelling: millimeter
1.0 mm (0.039 in)
• mm in
micrometre μm (um) μm US spelling: micrometer
1.0 μm (3.9×10−5 in)
nanometre nm nm US spelling: nanometer
1.0 nm (3.9×10−8 in)
non-SI metric ångström Å
(angstrom)
Å 1.0 Å (3.9×10−9 in)
Imperial
&
US customary
miles mi mi 1.0 mi (1.6 km)
• mi km
furlong furlong (none) 1.0 furlong (660 ft; 200 m)
chain chain (none) 1.0 chain (66 ft; 20 m)
rod rd rd For other names of this unit see the full list. 1.0 rd (17 ft; 5.0 m)
fathom fathom (none) fathom ≡ 6 ft 1.0 fathom (6.0 ft; 1.8 m)
yard yd yd assumes the international definition 1.0 yd (0.91 m)
foot ft (foot) ft long code "foot" outputs foot (and never feet) 1.0 ft (0.30 m)
• ftin (feet and inches)
• ft m (foot m)
inch in in 1.0 in (25 mm)
• in cm
• in mm
Other nautical mile nmi nmi the international standard nautical mile
For other nautical miles see the full list.
1.0 nmi (1.9 km; 1.2 mi)
parsec pc pc 1.0 pc (3.3 ly)
light-year ly ly 1.0 ly (63,000 AU)
astronomical unit AU AU 1.0 AU (150,000,000 km; 93,000,000 mi)
MASS
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
SI kilogram kg kg Allows triple output units. See: full list. 1.0 kg (2.2 lb)
• kg lb
• kg st
gram g g 1.0 g (0.035 oz)
• g oz
milligram mg mg 1.0 mg (0.015 gr)
• mg gr
microgram μg (ug) μg 1.0 μg (1.5×10−5 gr)
non-SI metric tonne t t Allows triple output units. See: full list. 1.0 t (0.98 long tons; 1.1 short tons)
• t LT
• t ST
metric ton MT t
• MT LT
• MT ST
Avoirdupois long ton LT (none) 2,240 lb
used mostly in the British Commonwealth.
Allows triple output units. See: full list.
1.0 long ton (1.0 t)
• LT t
• LT MT
• LT ST
long ton long ton
short ton ST (none) 2,000 lb
used mostly in the US.
Allows triple output units. See: full list.
1.0 short ton (0.91 t)
• ST t
• ST MT
• ST LT
short ton short ton
stone st st 14 lb
used mostly in the British Commonwealth except Canada.
Allows triple output units. See: full list.
1.0 st (14 lb; 6.4 kg)
• st kg
• st lb
pound lb lb Allows triple output units. See: full list. 1.0 lb (0.45 kg)
• lb kg
• lb st
ounce oz oz 1.0 oz (28 g)
• oz g
drachm drachm (none) 1.0 drachm (1.8 g)
drachm dram (none)
grain gr gr equivalent to the troy grain 1.0 gr (0.065 g)
Troy troy ounce ozt ozt 1.0 ozt (1.1 oz; 31 g)
other carat carat (none) 1.0 carat (0.20 g)
SPEED
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
SI metre per second m/s m/s US spelling: meter per second
1.0 m/s (3.3 ft/s)
• m/s ft/s (m/s foot/s)
non-SI metric kilometre per hour km/h km/h US spelling: kilometer per hour
1.0 km/h (0.62 mph)
• km/h mph
Imperial
&
US customary
mile per hour mph mph 1.0 mph (1.6 km/h)
• mph km/h
• mph kn
foot per second ft/s (foot/s) ft/s long code "foot/s" outputs foot per second (and never feet) 1.0 ft/s (0.30 m/s)
• ft/s m/s
Maritime units knot kn (knot) kn 1.0 kn (1.9 km/h; 1.2 mph)
• kn mph
TEMPERATURE
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
SI K K K Allows triple output units. See: full list. 1.0 K (−272.15 °C; −457.87 °F)
• K °C (K C)
• K °R (K R)
• K °F (K F)
°C °C (C) °C 1.0 °C (33.8 °F)
• °C K (C K)
• °C °R (C R)
• °C °F (C F)
Imperial
&
US customary
°R °R (R) °R 1.0 °R (0.556 K; −458.670 °F; −272.594 °C)
• °R K (R K)
• °R °C (R C)
• °R °F (R F)
°F °F (F) °F 1.0 °F (−17.2 °C)
• °F K (F K)
• °F °C (F C)
• °F °R (F R)
Celsius change C-change Used for temperature intervals instead of actual temperatures
Example: {{convert|5|C-change|0}} warmer
Result: 5 °C (9 °F) warmer
Fahrenheit change F-change Used for temperature intervals instead of actual temperatures
Example: {{convert|10|F-change|0}} colder
Result: 10 °F (6 °C) colder
TORQUE
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
Industrial
SI newton metre Nm N⋅m Triple combinations are also possible. See the full list. 1.0 N⋅m (0.74 lbf⋅ft)
• Nm kg.m
• Nm lb.ft
Non-SI metric kilogram metre kg.m kg⋅m 1.0 kg⋅m (9.8 N⋅m; 7.2 lb⋅ft)
• kg.m Nm
• kg.m lb.ft
Imperial
&
US customary
pound force-foot lb.ft lb⋅ft 1.0 lb⋅ft (1.4 N⋅m)
• lb.ft Nm
• lb.ft kg-m
Scientific
SI newton metre N.m N⋅m Triple combinations are also possible. See the full list. 1.0 N⋅m (0.74 lbf⋅ft)
• N.m kgf.m
• N.m lbf.ft
Non-SI metric kilogram force-metre kgf.m kgf⋅m 1.0 kgf⋅m (9.8 N⋅m; 7.2 lbf⋅ft)
• kgf.m N.m
• kgf.m lbf.ft
Imperial
&
US customary
pound force-foot lbf.ft lbf⋅ft 1.0 lbf⋅ft (1.4 N⋅m)
• lbf.ft N.m
• lbf.ft kgf.m
VOLUME
(Full list)
system unit unit-
code
abbrev-
iation
notes sample default
conversion
combination
output units
SI cubic metre m3 m3 US spelling: cubic meter
one kilolitre
1.0 m3 (35 cu ft)
cubic centimetre cm3 cm3 US spelling: cubic centimeter
one millilitre
1.0 cm3 (0.061 cu in)
cc cc
cubic millimetre mm3 mm3 US spelling: cubic millimeter
1.0 mm3 (6.1×10−5 cu in)
non-SI metric kilolitre kl kl US spelling: kiloliter
one cubic metre
1.0 kl (35 cu ft)
kL kL
litre l l US spelling: liter
one cubic decimetre
Allows triple output units. See: full list.
1.0 l (0.22 imp gal; 0.26 US gal)
• l impgal
• l USgal
• l U.S.gal
• l USdrygal
• l U.S.drygal
L L
• L impgal
• L impqt
• L USgal
• L U.S.gal
• L USdrygal
• L U.S.drygal
centilitre cl cl US spelling: centiliter
1.0 cl (0.35 imp fl oz; 0.34 US fl oz)
cL cL
millilitre ml ml US spelling: milliliter
one cubic centimetre
1.0 ml (0.035 imp fl oz; 0.034 US fl oz)
• ml impoz
• ml USoz
• ml U.S.oz
mL mL
• mL impoz
• mL USoz
• mL U.S.oz
Imperial
&
US customary
cubic yard cuyd cu yd 1.0 cu yd (0.76 m3)
cubic foot cuft (cufoot) cu ft long code "cufoot" outputs cubic foot (and never feet) 1.0 cu ft (0.028 m3)
cubic inch cuin cu in 1.0 cu in (16 cm3)
Imperial imperial barrel impbbl imp bbl 36 imp gal 1.0 imp bbl (160 l; 36 imp gal; 43 US gal)
imperial bushel impbsh imp bsh 8 imp gal 1.0 imp bsh (36 l; 8.0 imp gal; 8.3 US dry gal)
impbu imp bu
imperial gallon impgal imp gal 4.54609 litres by definition, also
4 imp qt or 8 imp pt or 160 imp fl oz
Allows triple output units. See: full list.
1.0 imp gal (4.5 l; 1.2 US gal)
• impgal l
• impgal L
• impgal USgal
• impgal U.S.gal
• impgal USdrygal
• impgal U.S.drygal
imperial quart impqt imp qt 1/4 imp gal or 40 imp fl oz 1.0 imp qt (1,100 ml; 38 US fl oz)
imperial pint imppt imp pt 1/8 imp gal or 20 imp fl oz 1.0 imp pt (0.57 l)
imperial fluid ounce impoz (impfloz) imp fl oz 1/160 imp gal 1.0 imp fl oz (28 ml; 0.96 US fl oz)
• impoz USoz
• impoz U.S.oz
• impoz ml
• impoz mL
US customary
liquid measure
US barrel USbbl US bbl 31½ US gal
used for liquids except for oil and beer (see the full list)
1.0 US bbl (120 l; 32 US gal; 26 imp gal)
U.S.bbl U.S. bbl
barrel oilbbl bbl 42 US gal 1.0 bbl (0.16 m3)
US beer barrel USbeerbbl
(usbeerbbl)
US bbl 1.0 US bbl (120 l; 31 US gal; 26 imp gal)
U.S.beerbbl
(usbeerbbl)
U.S. bbl
US gallon USgal US gal 231 cubic inches by definition, also
4 US qt or 8 US pt or 128 US fl oz
Allows triple output units. See: full list.
1.0 US gal (3.8 l; 0.83 imp gal)
• USgal l
• USgal L
• USgal impgal
U.S.gal U.S. gal
• USgal l
• USgal L
• USgal impgal
US quart USqt US qt 1/4 US gal or 32 US fl oz 1.0 US qt (950 ml)
U.S.qt U.S. qt
US pint USpt US pt 1/8 US gal or 16 US fl oz 1.0 US pt (0.47 l; 0.83 imp pt)
U.S.pt U.S. pt
US fluid ounce USoz
(USfloz)
US fl oz 1/128 US gal 1.0 US fl oz (30 ml)
• USoz ml
• USoz mL
• USoz impoz
U.S.oz
(U.S.floz)
U.S. fl oz
• U.S.oz ml
• U.S.oz mL
• U.S.oz impoz
US customary
dry measure
US dry barrel USdrybbl US dry bbl 105/32 US bsh 1.0 US dry bbl (0.12 m3)
U.S.drybbl U.S. dry bbl
US bushel USbsh US bsh 2150.42 cubic inches by definition 1.0 US bsh (35 l; 8.0 US dry gal; 7.8 imp gal)
U.S.bsh U.S. bsh
US bushel USbu US bu 2150.42 cubic inches by definition 1.0 US bu (35 l; 8.0 US dry gal; 7.8 imp gal)
U.S.bu U.S. bu
US dry gallon USdrygal US dry gal 1/8 US bsh
Allows triple output units. See: full list.
1.0 US dry gal (4.4 l)
• USdrygal l
• USdrygal L
• USdrygal impgal
U.S.drygal U.S. dry gal
• U.S.drygal l
• U.S.drygal L
• U.S.drygal impgal
US dry quart USdryqt US dry qt 1/32 US bsh 1.0 US dry qt (1,100 ml)
U.S.dryqt U.S. dry qt
US dry pint USdrypt US dry pt 1/32 US bsh 1.0 US dry pt (550 ml)
U.S.drypt U.S. dry pt
PRESSURE
unit unit-
code
abbrev-
iation
sample default
conversion
gigapascal GPa GPa 1.0 GPa (150,000 psi)
megapascal MPa MPa 1.0 MPa (150 psi)
kilopascal kPa kPa 1.0 kPa (0.15 psi)
hectopascal hPa hPa 1.0 hPa (0.015 psi)
pascal Pa Pa 1.0 Pa (0.00015 psi)
millipascal mPa mPa 1.0 mPa (1.5×10−7 psi)
millibar mbar mbar 1.0 mbar (1.0 hPa)
mb mb
decibar dbar dbar 1.0 dbar (10 kPa)
bar bar (none) 1.0 bar (100 kPa)
kilobarye kBa kBa 1.0 kBa (1.0 hPa)
barye Ba Ba 1.0 Ba (0.10 Pa)
standard atmosphere atm atm 1.0 atm (100 kPa)
torr Torr Torr 1.0 Torr (0.13 kPa)
millimetre of mercury mmHg mmHg 1.0 mmHg (0.13 kPa)
inch of mercury inHg inHg 1.0 inHg (3.4 kPa)
pound per square inch psi psi 1.0 psi (6.9 kPa)
FUEL EFFICIENCY
unit unit-
code
notes combinations
kilometres per litre km/l (km/L) Use `km/L` to get "km/L"
• km/l mpgimp
• km/l mpgus
litres per 100 kilometres l/100 km (L/100 km) Use `L/100 km` to get "L/100 km"
• l/100 km mpgimp
• l/100 km mpgus
litres per kilometre l/km (L/km) Use `L/km` to get "L/km"
• l/km impgal/mi
• l/km usgal/mi
miles per imperial gallon mpgimp
• mpgimp mpgus
miles per US gallon mpgus (mpgUS, mpgU.S.) Use `mpgUS` to get "US"
Use `mpgU.S.` to get "U.S."
`mpgus` will give "U.S." if spelling is
set to US & "US" otherwise
• mpgus mpgimp
imperial gallons per mile impgal/mi
• impgal/mi
US gallons per mile usgal/mi (USgal/mi, U.S.gal/mi) As above with the `us` vs `US` vs `U.S.`
• usgal/mi
The `mpgUS`, `mpgU.S.`, `USgal/mi` vs `U.S.gal/mi`, `km/L`, `L/100 km` and `L/km` variants work within combinations also (making 36 combinations in total).
POWER
Watts, kilowatts, milliwatts, etc. are supported. As is horsepower (English & metric). Use standard abbreviations as input code (lowercase hp for horsepower, for example)
POPULATION DENSITY
unit unit-
code
notes combinations
inhabitants per square kilometre PD/sqkm `PD` stands for population density, i.e. humans (inhabitants)
• PD/sqkm PD/sqmi
per square kilometre /sqkm Used when the word 'inhabitants' would be inappropriate.
• /sqkm /sqmi
inhabitants per hectare PD/ha
• PD/ha PD/acre
per hectare /ha
• /ha /acre
inhabitants per square mile PD/sqmi `PD` stands for population density, i.e. humans (inhabitants)
• PD/sqmi PD/sqkm
per square mile /sqmi Used when the word 'inhabitants' would be inappropriate.
• /sqmi /sqkm
inhabitants per acre PD/acre
• PD/acre PD/ha
per acre /acre
• /acre /ha
COST PER UNIT MASS
unit unit-
code
notes combinations
dollars per pound \$/lb "\$" can mean US\$, HK\$, etc.
• (none)
dollars per kilogram \$/kg
• (none)
dollars per troy ounce \$/ozt 12 troy ounces in 1 troy pound
• (none)
### 'per' units: kg/hl
When using a slash (`/`), a unit like `kg/hl` is recognized as kilograms per hectolitre and will be converted with other mass/volume units.
• `{{convert|1000|kg/hl}}` → 1,000 kilograms per hectolitre (100 lb/imp gal)
### Units of difference: 10 °C higher; how much in °F
When a number is the measurement for a difference, one conversion is done. This may occur in temperatures (°C, °F, K). Compare this to a range, when two conversions are done.
The only available units of difference are: `|C-change=`, `|F-change=` and `|K-change=`.
• `{{convert|10|C}}` → 10 °C (50 °F), regular temperature
• `{{convert|10|C-change}}` → 10 °C (18 °F), difference (e.g., "The temperature changed from 30 into 40 °C")
• `{{convert|10-15|C}}` → 10–15 °C (50–59 °F), range (e.g., "The temperature is between 10 and 15 °C")
Consistently, into multiple units:
• `A rise of {{convert|10|C-change|F-change K-change}}` → A rise of 10 °C (18 °F; 10 K)
### Multiple units: 1 ft 5 in
#### In input
Base document § Input multiples lists options for multiple unit input (like `ft,in`). It can catch predefined sets only (units that can be subdivided; e.g., yd into ft):
• `{{convert|1|yd|2|ft|3|in}}` → 1 yard 2 feet 3 inches (1.60 m)
• `{{convert|2|ft|3|in|cm}}` → 2 feet 3 inches (69 cm)
• `{{convert|1|lb|5|oz|g}}` → 1 pound 5 ounces (600 g)
#### In output
Available multiple-unit output options predefined, like `ftin` and `ydftin`. The full list is at § Output multiples.
• `{{convert|2|m|ftin}}` → 2 metres (6 ft 7 in)
• `{{convert|2|m|ft in}}` → 2 metres (6.6 ft; 79 in), using a space, returns the decimal point
Default behaviour, for comparison:
• `{{convert|2|m}}` → 2 metres (6 ft 7 in)
• {{hands}} a length used to measure horses
• {{Long ton}} a weight in ton, cwt, qr and lb
### Currency per unit: \$/mi → \$/km
Using currency symbols in a \$ per unit value, you can convert the per-unit:
• `{{convert|10|\$/mi|\$/km}}` → \$10 per miles (\$6.2/km)
You can set the currency in both values using `|\$=€`:
• `{{convert|10|\$/mi|\$/km|\$=€}}` → €10 per miles (€6.2/km)
It is not possible to convert the currency. So, this result (mixed currencies) is not possible: \$15 per mile (€8.6/km) N
## Parameter list
Parameter options for {{convert}}
Parameter Value Description Note
`|abbr=in` `in` Use symbol for first (left-hand side) unit Unit display
`|abbr=off` `off` Use name for all units Unit display
`|abbr=none` `none`
`|abbr=on` `on` Use symbol for all units Unit display
`|abbr=out` `out` Use symbol for right-hand side unit (default) Unit display
`|abbr=unit` `unit` Use symbol for all units when using scientific notation Unit display
`|abbr=values` `values` Omit both the input and output units: show only the numbers Unit display
`|abbr=~` `~` Shows both unit name and symbol Unit display
`|adj=mid|…` `mid` User-specified text after the input unit; sets `adj=on` (adjective). Expects 1 unnamed parameter. Word adding, adjective
`|adj=on` `on` Unit name is adjective (singular and hyphenated) Grammar, adjective
`|adj=pre|…` `pre` User-specified text before input unit. Expects 1 unnamed parameter. Word adding
`|adj=ri0` `ri0` Round input with precision 0 Input precision
`|adj=ri1` `ri1` Round input with precision 1 Input precision
`|adj=ri2` `ri2` Round input with precision 2 Input precision
`|adj=ri3` `ri3` Round input with precision 3 Input precision
`|comma=5` `5` Only use comma for thousands separator if 5 or more digits Number format
`|comma=gaps` `gaps` Use gaps (space), not comma, for thousands separator Number format
`|comma=off` `off` No thousands separator Number format
`|disp=b` `b` Join input and output using " (...)" (default) Join values
`|disp=sqbr` `sqbr` Join input and output using " [...]" Join values
`|disp=br` `br` Join input and output using "<br/>" Join values
`|disp=comma` `comma` Join input and output using ", " Join values
`|disp=or` `or` Join input and output using " or " Join values
`|disp=number` `number` Display output number only Parts only
`|disp=output number only` `output number only`
`|disp=out` `out` Display only output number and name/symbol Parts only
`|disp=output only` `output only`
`|disp=preunit|…[|…]` `preunit` Text to be inserted after value and before units, for both input and output, with optionally different text for output. Expects 1 or 2 unnamed parameters. Word adding
`|disp=table` `table` Output is suitable for a table cell with align="right" Table columns
`|disp=tablecen` `tablecen` Output is suitable for a table cell with align="center" Table columns
`|disp=unit` `unit` Display input name/symbol only (not input number, not output) Parts only
`|disp=unit2` `unit2` Display output name/symbol only (not input; not output number) Parts only
`|disp=x|…` `x` Join input and output using user-specified text Word adding
`|frac=N` `N` Show imperial number in fractions, denominator=N Number format, fraction
`|input=P2048` `P2048` (e.g.) Reads and converts Wikidata property Inside template
`|lk=in` `in` Link left-hand side unit name or symbol Unit link
`|lk=on` `on` Link all unit names or symbols (but not twice for the same unit) Unit link
`|lk=out` `out` Link right-hand side unit name or symbol Unit link
`|order=flip` `flip` Inverts order of input, output measurements (conversion first) Order
`|qid=Q1056131` `Q1056131` (e.g.) Reads Wikidata property from Wikidata item Inside template; testing
`|round=5` `5` Rounds calculation to the nearest multiple of 5 Output precision
`|round=25` `25` Rounds calculation to the nearest multiple of 25 Output precision
`|round=each` `each` In a range, each number is rounded by the default rounding Output precision
`|sigfig=N` `N` Round output number to N significant figures (N is a positive integer) Output precision
`|sortable=on` `on` Adds invisible sort key Table sort
`|sp=us` `us` Use U.S. spelling ("meter" instead of default "metre") Spelling U.S. names
`|spell=in` `in` Spell input number in words Spelling numbers
`|spell=In` `In` Spell input number in words with first letter uppercase Spelling numbers
`|spell=on` `on` Spell input and output numbers in words Spelling numbers
`|spell=On` `On` Spell input and output numbers in words with first letter uppercase Spelling numbers
`|\$=€` `€` Replace \$-sign with a currency sign, for example in `\$/acre` (no currency conversion happens) Cost per unit
`|debug=yes` `yes` Debugging only. In a sortable table: show the normally hidden sort key Table sort
`|disp=flip` `flip` Deprecated. Use |order=flip Order
`|sing=` Deprecated. Use |adj= Plurals
## Deprecated options
This list:
Deprecated options should not be used. They may produce incorrect or undesired results and there is no guarantee that they will be supported in the future.
• disp=flip is deprecated. Use order=flip instead
• Range separator |xx| is deprecated (not MOS compliant). Use |x| instead
• Range separator |*| is deprecated (not MOS compliant). Use |x| instead
## TemplateData
This is the TemplateData documentation for this template used by VisualEditor and other tools.
TemplateData for Convert
Converts measurements to other units.
Template parameters
This template prefers inline formatting of parameters.
ParameterDescriptionTypeStatus
value`1`
the value to convert
Numberrequired
from unit`2`
the unit for the provided 'value'
Example
km
Linerequired
to units`3`
the unit to convert into
Example
mi
Stringsuggested
precision or suffix`4`
significant digits after decimal dot or, if negative, exponent of ten
Numberoptional
link units`lk`
“on” all, “in” input, “out” output or “off” no units
Default
off
Example
on
Stringoptional
abbreviation`abbr`
display for the units: “on” unit symbols, “off” all unit names in full words, “in” input unit symbol, “out” abbreviated output units, “unit” both input and output units abbreviated when using scientific notation, “values” no units at all
Default
off
Stringoptional
spelling`sp`
“us” display U.S. spelling of unit names
Example
us
Stringoptional
adjective`adj`
adjective form (singular unit name appended by hyphen) “on” or “mid” to put conversion at end
Default
off
Example
on
Booleanoptional
conversion`disp`
display conversion result: “or” after ‘or’, “x” with custom prefix and suffix, “b” in parentheses, “table”/“tablecen”, “output only” alone, “output number only” alone and without unit, “unit” not at all but input unit; if the value is a number it is used as precision
Example
b
Stringoptional
ordering`order`
“flip” returns converted value first, input value second.
Example
flip
Stringoptional
significant figures`sigfig`
number that sets the number of significant figures
Numberoptional
rounding output`round`
“5” rounds the output number to nearest multiple of 5, “25” to nearest multiple of 25, “each” rounds each number in a range
Numberoptional
thousands separator`comma`
Sets or suppresses thousand separator in the numbers. “off” = no separator; “gaps”: use space not comma; “5” and “gaps5”: only add separator when number > 5 positions (10,000 or more)
Default
on
Example
off
Stringoptional
sort key`sortable`
“on” generates a hidden sort key
Example
on
Stringoptional | 18,292 | 49,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-24 | latest | en | 0.581397 |
http://clay6.com/qa/8508/a-straight-line-is-parallel-to-the-x-axis-and-passes-through-the-intersecti | 1,529,370,987,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00133.warc.gz | 61,034,856 | 25,759 | # A straight line is parallel to the x-axis and passes through the intersection of the lines $x+2y+1=0$ and $y=x+7$.Find the equation of the straight line.
This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com
Let us solve the given equations x+2y =-1 and -x+y=7 We get x =-5 and y =2 Since the line parallel to x axis passes through these points, the equation of the required line is y= 2
answered Sep 2, 2015 | 126 | 466 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-26 | latest | en | 0.954796 |
http://programmers.stackexchange.com/questions/162036/algorithmic-problem-quickly-finding-all-s-where-value-x-is-some-given-value/163137 | 1,469,330,619,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00229-ip-10-185-27-174.ec2.internal.warc.gz | 203,369,759 | 18,410 | # Algorithmic problem - quickly finding all #'s where value %x is some given value
Problem I'm trying to solve, apologies in advance for the length:
Given a large number of stored records, each with a unique (String) field S. I'd like to be able to find through an indexed query all records where Hash(S) % N == K for any arbitrary N, K (e.g. given a million strings, find all strings where HashCode(s) % 17 = 5. Is there some way of memoizing this so that we can quickly answer any question of this form without doing the % on every value?
The motivation for this is a system of N distributed nodes, where each record has to be assigned to at least one node. The nodes are numbered 0 - (K-1) , and each node has to load up all of the records that match it's number:
If we have 3 nodes
• Node 0 loads all records where Hash % 3 ==0
• Node 1 loads all records where Hash % 3 ==1
• Node 2 loads all records where Hash % 3 ==2
adding a 4th node, obviously all the assignments have to be recomputed -
• Node 0 loads all records where Hash % 4 ==0
• ...
• etc
I'd like to easily find these records through an indexed query without having to compute the mod individually.
The best I've been able to come up with so far:
If we take the prime factors of N (p1 * p2 * ... )
if N % M == I then p % M == I % p for all of N's prime factors
e.g. 10 nodes :
N % 10 == 6 then
• N % 2 = 0 == 6 %2
• N % 5 = 1 == 6 %5
so storing an array of the "%" of N for the first "reasonable" number of primes for my data set should be helpful. For example in the above example we store the hash and the primes
HASH PRIMES (array of %2, %3, %5, %7, ... ])
16 [0 1 1 2 .. ]
so looking for N%10 == 6 is equivalent to looking for all values where array[1]==1 and array[2] == 1.
However, this breaks at the first prime larger than the highest number I'm storing in the factor table. Is there a better way?
-
You may get a better luck on math.stackexchange.com. You may need to re-phrase this a bit to bridge the gap between CS and Math. – dasblinkenlight Aug 23 '12 at 16:56
Thanks, good suggestion – Steve B. Aug 23 '12 at 17:13
Agreed - the fact that it's an "algorithm" doesn't necessarily mean CS skills :) This is highly mathematical in nature and I second @dasblinkenlight on migrating this to math.SE – PhD Aug 23 '12 at 17:14
presumably, you're doing this for load balancing of some sort. How often are you re-balancing? How often and how do the Strings change? How often does your range of nodes change? If your problem can accept some constraints (such as only allowing N nodes to range between X and Y) then some pragmatic solutions can be presented. They won't necessarily be the mathematically pure solutions, but they will still work. – GlenH7 Aug 23 '12 at 17:39
Have I misunderstood something, or did `N` change its meaning throughout your question? At first, it seemed to be the number of nodes, later, it seems to represent a hash code? – phant0m Aug 31 '12 at 10:58
How important is it that formula be "Hash % num_machines"? This formula is used for distributed caches, like memcached. It works great until you add/remove nodes. At that point, the advice is to abandon it and use consistent hashing.
-
Thanks for the excellent suggestion, although I didn't see an easy application of consistent hashing in this case. Suggestions for a better distribution than Hash % #machines (i.e. anything that's both fair and stable) would also be welcome. – Steve B. Aug 23 '12 at 19:06
Unless I've misunderstood something, your conjecture is incorrect.
If we take the prime factors of N (p1 * p2 * ... )
if N % M == I then p % M == I % p for all of N's prime factors
How did you come up with that?
Let's say `N = 36` and `M = 6`
The prime factorization of `N = 2 * 2 * 3 * 3`. `36 % 6 = 0`
According to your statement, the following should hold:
``````p % 6 = 0 % p = 0
``````
But clearly, this is not the case: `2 % 6 = 2 != 0`
- | 1,074 | 3,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2016-30 | latest | en | 0.900358 |
https://brilliant.org/practice/evaluating-absolute-value-expressions/ | 1,500,699,827,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423901.32/warc/CC-MAIN-20170722042522-20170722062522-00001.warc.gz | 616,340,364 | 21,699 | ###### Waste less time on Facebook — follow Brilliant.
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# Absolute Value
Absolute value is a mathematician's way of judging numbers by their magnitude rather than their positive/negative value. It is the distance of the number from 0 on a number line.
# Evaluating Absolute Value Expressions
If $$15<x<39$$, what is the value of $|x-15|+|x-39|?$
If $$-12 \leq a < 12$$, what is the value of $\lvert a-14 \rvert + \lvert a-12 \rvert + \lvert a+12 \rvert + \lvert a+14 \rvert ?$
If $$9 \leq a < 29$$, what is the value of $|a-9|+\sqrt{(a-29)^2}?$
When $$-19\leq x\leq y\leq4,$$ which of the following is equivalent to $\lvert x+19\rvert+\lvert x-y\rvert+\lvert y-4\rvert?$
If $$-6\leq x\leq-4,$$ what is the value of $\lvert2x+12\rvert+\lvert x+4\rvert+\lvert x-1\rvert?$
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Set Loading... | 294 | 855 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-30 | longest | en | 0.610009 |
https://codedump.io/share/hDWVaQ1Z5zjn/1/how-to-get-the-239s-complement-value-of-a-biginteger-of-arbitrary-length | 1,529,827,530,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866888.20/warc/CC-MAIN-20180624063553-20180624083553-00517.warc.gz | 587,102,888 | 9,139 | Harish - 1 year ago 59
Android Question
# How to get the 2's complement value of a BigInteger of arbitrary length
Is there a method in BigInteger to get the 2's complement value?
For eg: if there is a BigInteger with a negative value
``````BigInteger a = new BigInteger("-173B8EC504479C3E95DEB0460411962F9EF2ECE0D3AACD749BE39E1006FC87B8", 16);
``````
then I want to get the 2's complement in a BigInteger form
``````BigInteger b = E8C4713AFBB863C16A214FB9FBEE69D0610D131F2C55328B641D61EFF9037848
``````
I can subtract the first BigInteger from 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF to get the second BigInteger but is there a generic method to calculate this for a BigInteger of any length?
To make this value its two's complement, you will have to manipulate the contents. That is of course impossible, so you first get the contents out, manipulate them and then get them into a new `BigInteger`:
``````public static BigInteger twosComplement(BigInteger original)
{
// for negative BigInteger, top byte is negative
byte[] contents = original.toByteArray();
// prepend byte of opposite sign
byte[] result = new byte[contents.length + 1];
System.arraycopy(contents, 0, result, 1, contents.length);
result[0] = (contents[0] < 0) ? 0 : (byte)-1;
// this will be two's complement
return new BigInteger(result);
}
public static void main(String[] args)
{
BigInteger a = new BigInteger("-173B8EC504479C3E95DEB0460411962F9EF2ECE0D3AACD749BE39E1006FC87B8", 16);
BigInteger b = twosComplement(a);
System.out.println(a.toString(16).toUpperCase());
System.out.println(b.toString(16).toUpperCase());
// for comparison, from question:
System.out.println("E8C4713AFBB863C16A214FB9FBEE69D0610D131F2C55328B641D61EFF9037848");
}
``````
Output:
``````-173B8EC504479C3E95DEB0460411962F9EF2ECE0D3AACD749BE39E1006FC87B8
E8C4713AFBB863C16A214FB9FBEE69D0610D131F2C55328B641C61EFF9037848
E8C4713AFBB863C16A214FB9FBEE69D0610D131F2C55328B641D61EFF9037848
``````
And this new `BigInteger` is really the two's complement, not just a re-interpretation of the bits.
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 658 | 2,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-26 | latest | en | 0.57623 |
https://solvedlib.com/exercise-15-18-nash-company-reported-the,88541 | 1,675,016,505,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00188.warc.gz | 542,762,556 | 18,539 | # Exercise 15-18 Nash Company reported the following amounts in the stockholders' equity section of its December...
###### Question:
Exercise 15-18 Nash Company reported the following amounts in the stockholders' equity section of its December 31, 2016, balance sheet. Preferred stock, 10%, $100 par (10,000 shares authorized, 1,900 shares issued)$190,000 Common stock, $5 par (101,000 shares authorized, 20,200 shares issued) 101,000 Additional paid-in capital 125,000 Retained earnings 431,000 Total$847,000 During 2017, Nash took part in the following transactions concerning stockholders' equity. 1. Paid the annual 2016 $10 per share dividend on preferred stock and a$2 per share dividend on common stock. These dividends had been declared on December 31, 2016. 2. Purchased 1,800 shares of its own outstanding common stock for $43 per share. Nash uses the cost method. 3. Reissued 700 treasury shares for land valued at$31,600. 4. Issued 530 shares of preferred stock at $104 per share. 5. Declared a 10% stock dividend on the outstanding common stock when the stock is selling for$47 per share. 6. Issued the stock dividend. 7. Declared the annual 2017 $10 per share dividend on preferred stock and the$2 per share dividend on common stock. These dividends are payable in 2018.
No. Account Titles and Explanation Debit Credit 1. Dividends Payable - Preferred Stock 19000 Dividends Payable - Common Stock 38000 2 Cash 57000 2. Treasury Stock 77400 Cash 77400 - 3. Land 31600 Treasury Stock 29900 Paid-in Capital from Treasury Stock דר Treasury Stock - 1700 4. Cash 120 Preferred Stock Paid-in Capital in Excess of Par - Preferred Stock
5. Retained Earnings Common Stock Dividend Distributable Paid-in Capital in Excess of Par - Common Stock 6. Common Stock Dividend Distributable Common Stock Retained Earnings Dividends Payable - Preferred Stock TIT Dividends Payable - Common Stock
##### How do you convert 0.044 into a percent and fraction?
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##### Two capacitors connectec pardllel urc duce ecuivaleni cacacilance Ga Gnen ccnne:e: ser es Ine ecuivaeni capacitance cnly 5,8Ihe incv Jual capacitance eacn capaciton Enter your answers asconding order; Express your answers Using two significant figures separated by comMaSueDufido reda resbt keyboard shortcuts belp;SubmilRequost AnswierProvide Feedback
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##### Ifty -3y _ t = 0then the appropriate integrating factor is_ 0 p(t) = 820 0 u(t) =e-t 0 u(t) = ext 0 u(t) =e-3t 0 p(t) = 1/+3 0 p(t) =t3 0 u(t) = 20 0 u(t) = e-&
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##### Question 11 pts:A researcher collects data on a variable called Memory: Possible values for this variable are 1 correct answer; 2 correct answers, 3 correct answers and 4 correct answers. At the end of the study the researcher has the following data (each number reflects the numbey of correct answers student gave): 4,2,3,3.3,4,1,2,4,4,3,3,3,3,3,2,4,1,1,1,1 What type of data did the researcher collect? Begin your written justification; 'The variable Memory is then complete the sentence to ex
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##### Show that if a wave function $u(1,2, ldots n)$ is an energy eigenfunction of a symmetric Hamiltonian that corresponds to a nondegenerate eigenvalue, it is either symmetric or antisymmetric. Show this first for $n=2$, then for $n=3$, and then indicate how the proof can be extended to arbitrary $n .$
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Breezeless Company produces doors and window frames. It generates profit at $2.00 per door and$1.00 per window frame. Manufacture of these products requires twowork centers. Work center 1 has 120 labor hours available while work center 2 has 80 hours available. Production of one hundred doors requi... | 2,683 | 9,489 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-06 | latest | en | 0.840786 |
https://manasataramgini.wordpress.com/2008/01/06/the-ara-and-the-arbelos/ | 1,686,245,884,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00634.warc.gz | 428,244,467 | 31,701 | ## The Ara and the Arbelos
When I was young, I was fascinated by the blade held by the clobber and the way in which he plied it in fixing shoes and slippers. This blade (Ara in saMskR^ita) had a special shape and I realized a weapon like that was held by some of our deva-s. There was an Indian arrow that had a head much like this tool. Much later, I learnt of the great yavanAchArya Archimedes having studied a curve derived of 3 circles that he named the Arbelos or the clobber’s knife. Obviously he was meaning the same object apparently still in use in leather-work. I wondered if the proto-Indoeuropeans or at least the common ancestor of the Hindus and Greeks had a common device like that (Arbelos and Ara) used in leather work. The Arbelos fascinates mathematicians and laymen alike, and I was no exception to it. I simply wished to relive my journey through the Arbelos – perhaps relive a fantasy of the halcyon days of my childhood. Many fascinating properties of the Arbelos are apparent even at sight:
The most obvious being that the perimeter of the upper boundary of the curve = perimeter of the lower boundary because: d=d1+d2; pi*d/2=pi*(d1+d2)/2;
The hemichord formed by the shared tangent of the two smaller semi-circles is indicated by T. It is the geometric mean of their respective diameters: T²=d1*d2
Further, the area of the Arbelos (A) can be expressed independently of the 3 radii of the semi-circles using just the value T: A=pi*T²/4; thus, the area of the Arbelos is the same as that of a circle with diameter T. this is obvious from baudhAyana’s (Pythagoras) theorem and that any angle inscribed in a semicircle is a right angle.
There are many other interesting properties of the arbelos that have made it a fertile object of investigation in Euclidean geometry over the ages. Even the great geometer Jakob Steiner worked on its properties, like the issue of Pythagorean triplets in the chain of circles. Even in the past 100 years it has been a source of rich new results. One of the most famous features of the arbelos from ancient Greek mathematics concerned the chain of circles inscribed within it. The first circle (K1) in this chain is a tangent to all the 3 semicircles of the arbelos. The next K2 is a tangent to K1 and two of the other semicircles. In this chain of circles the height of the of the center of Kn from the base line is equal to n*dn, where dn is the diameter of Kn. The centers of K1-Kn lie on an ellipse whose foci are the centers of the two semicircles that bound the chain. If the ratio of the diameters of the two smaller semicircles of the arbelos (d1/d2) is rational then the following is observed:
The right triangle formed by the center of Kn, the center of the largest semicircle of the arbelos and the foot of the perpendicular dropped from the center of Kn to the baseline has sides forming a Pythagorean triplet. This chain of circles can be constructed using the method of reflection of points on a circle. Here the reflecting circle is of radius=d and it reflects the right line as the largest semicircle, the left parallel line as the second largest semicircle and the first semicircle bounded by these two lines as the 3rd semicircle of the arbelos. All other inscribed tangential circles are reflected to form the chain.
The family of circles – Archimedean twin, the Bankoff circle and the quadruplet are shown.
Finally there are the famous Archimedean twin circles of the arbelos, which are the tangential to the largest semicircle, either one of the two other semicircles and the segment T. Archimedes showed that these two circles are congruent. Strikingly, numerous other congruent circles come up all over the arbelos, and their discovery and properties has formed a major area of modern studies on this figure. One of these discovered by Bankoff, a rich dentist who was an amateur mathematician: One takes the following 3 points- 1) & 2) the two touches by the first circle of the chain of inscribed circles on the smaller two semicircles of the arbelos. 3) the point where the two smaller semicircles of the arbelos touch. Then one draws the circumcircle of these points. It is congruent to the Archimedean twin circles.Another more recently discovered quadruplet of the family is described thus: Take the perpendiculars from the centers of the smaller semicircles of the arbelos and let them intersect the semicircles at A and B. Draw the lines from the center of the biggest semicircle of the arbelos to the A and B. Then there are two circles passing through each of these points which are tangential to each other and the lines from the former sentence, which are also tangential to the biggest circle to the arbelos. These two pairs of circles form a quadruplet which is congruent to the Archimedean twin circles.
This entry was posted in Scientific ramblings. Bookmark the permalink. | 1,156 | 4,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-23 | latest | en | 0.967633 |
https://www.mql5.com/en/forum/304914 | 1,553,139,429,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202484.31/warc/CC-MAIN-20190321030925-20190321052925-00033.warc.gz | 820,388,982 | 9,837 | # Discussion of article "Applying Monte Carlo method in reinforcement learning"
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204874
New article Applying Monte Carlo method in reinforcement learning has been published:
In the article, we will apply Reinforcement learning to develop self-learning Expert Advisors. In the previous article, we considered the Random Decision Forest algorithm and wrote a simple self-learning EA based on Reinforcement learning. The main advantages of such an approach (trading algorithm development simplicity and high "training" speed) were outlined. Reinforcement learning (RL) is easily incorporated into any trading EA and speeds up its optimization.
After stopping the optimization, simply enable the single test mode (since the best model is written to the file and only that model is to be uploaded):
Let's scroll the history for two months back and see how the model works for the full four months:
We can see that the resulting model lasted another month (almost the entire September), while breaking down in August.
Author: Maxim Dmitrievsky
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MetaQuotes Software Corp.:
New article Applying Monte Carlo method in reinforcement learning has been published:
Author: Maxim Dmitrievsky
Thank you. Is it possible to make the training using the GPU instead of CPU?
28579
jaffer wilson:
Thank you. Is it possible to make the training using the GPU instead of CPU?
Yes, if you rewrite all logic (RF include) on open cl kernels :) also random forest has worst gpu feasibility and parallelism
2172
Thank you Maxim.
I have been playing with the code and introduced different types of data for the feature input vectors. I have tried Close, Open, High, Low, Price Typical, Tick Volumes and derivatives of these.
I was having problems with the Optimiser complaining about "some error after the pass completed", but I finally managed to track this down: the optimiser errors occur if the input vector data has any zero values.
Where I was building a derivative, e.g. Close[1]-Close[2], sometimes the close values are the same giving a derivative of zero. For these types of input vector values I found the simplest fix was to add a constant, say 1000, to all vector values. This cured the optimiser errors and yet allowed the RDF to function.
I have also noticed the unintended consequence of running the same tests over and over, the amount of curve fitting increases for the period tested. Sometimes it is better to delete the recorded RDF files and run the optimisation again.
I am still experimenting and have more ideas for other types of feature.
28579
Mark Flint:
Thank you Maxim.
I have been playing with the code and introduced different types of data for the feature input vectors. I have tried Close, Open, High, Low, Price Typical, Tick Volumes and derivatives of these.
I was having problems with the Optimiser complaining about "some error after the pass completed", but I finally managed to track this down: the optimiser errors occur if the input vector data has any zero values.
Where I was building a derivative, e.g. Close[1]-Close[2], sometimes the close values are the same giving a derivative of zero. For these types of input vector values I found the simplest fix was to add a constant, say 1000, to all vector values. This cured the optimiser errors and yet allowed the RDF to function.
I have also noticed the unintended consequence of running the same tests over and over, the amount of curve fitting increases for the period tested. Sometimes it is better to delete the recorded RDF files and run the optimisation again.
I am still experimenting and have more ideas for other types of feature.
Hi, Mark. Depends of feature selection algorithm (in case of this article it's 1 feature/another feature (using price returns) in "recursive elimination func"), so if you have "0" in divider this "some error" can occur if you are using cl[1]-cl[2].
Yes, different optimizer runs can differ, because its used a random sampling, also RDF random algorithm. To fix this you can use MathSrand(number_of_passes) in expert OnInint() func, or another fixed number. | 860 | 4,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-13 | latest | en | 0.935687 |
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