url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.mrexcel.com/board/threads/return-account-holder-name-matching-up-to-five-separate-cells.1184531/#post-5770404 | 1,643,031,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304570.90/warc/CC-MAIN-20220124124654-20220124154654-00061.warc.gz | 954,886,888 | 18,676 | # Return Account Holder Name matching up to five separate cells.
#### Funktion
##### New Member
Hi Gang,
I have a worksheet with two tabs; one is a Summary of charges per account, and the other contains all of the Charges on all of the accounts.
The account numbers are comprised of up to five separate segments. Those segments are in columns A, B, C, D, and E on both tabs.
There will always be something in column A.
There will always be something in column B.
Columns C, D and E may or may not contain numbers. There could be one, two or all three of them populated on various accounts.
Each account has a corresponding Account Holder Name, which resides on the Summary tab.
For each individual charge listed on the Charges tab, I want to have Excel find the appropriate account holder's name on the Summary tab and bring it into column H on the Charges tab.
I tried an Index and Match (which is not my comfort zone) and it wants to spill and doesn't work... Here's the formula I tried
=INDEX(Summary!G3:G10,MATCH(Summary!A3:A10&Summary!B3:B10&Summary!C3:C10&Summary!D3:D10&Summary!E3:E10,0))
Not sure what approach to take on this one. Any advice would be greatly appreciated.
#### Attachments
• Dummy Data for Funktion 2 - Summary.png
56.2 KB · Views: 10
• Dummy Data for Funktion 2 - Charges.png
104.8 KB · Views: 10
### Excel Facts
Select range. Press Ctrl+1. On Number tab, choose Custom. Type Alt+7 then space then @ sign (using 7 on numeric keypad)
#### Fluff
##### MrExcel MVP, Moderator
Excel Formula:
=IF(A3="","",FILTER(Summary!\$G\$3:\$G\$10,(Summary!\$A\$3:\$A\$10=A3)*(Summary!\$B\$3:\$B\$10=B3)*(Summary!\$C\$3:\$C\$10=C3)*(Summary!\$D\$3:\$D\$10=D3)*(Summary!\$E\$3:\$E\$10=E3),""))
#### Funktion
##### New Member
Excel Formula:
=IF(A3="","",FILTER(Summary!\$G\$3:\$G\$10,(Summary!\$A\$3:\$A\$10=A3)*(Summary!\$B\$3:\$B\$10=B3)*(Summary!\$C\$3:\$C\$10=C3)*(Summary!\$D\$3:\$D\$10=D3)*(Summary!\$E\$3:\$E\$10=E3),""))
Thanks again, Fluff! It's working beautifully. Thanks for the quick reply.
#### Fluff
##### MrExcel MVP, Moderator
You're welcome & thanks for the feedback.
Replies
9
Views
559
Replies
1
Views
200
Replies
1
Views
630
Replies
1
Views
124
Replies
18
Views
734
Excel contains over 450 functions, with more added every year. That’s a huge number, so where should you start? Right here with this bundle.
1,152,225
Messages
5,768,920
Members
425,504
Latest member
bramblepants
### We've detected that you are using an adblocker.
We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com.
### Which adblocker are you using?
1)Click on the icon in the browser’s toolbar.
2)Click on the icon in the browser’s toolbar.
2)Click on the "Pause on this site" option.
Go back
1)Click on the icon in the browser’s toolbar.
2)Click on the toggle to disable it for "mrexcel.com".
Go back
### Disable uBlock Origin
Follow these easy steps to disable uBlock Origin
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back
### Disable uBlock
Follow these easy steps to disable uBlock
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back | 941 | 3,318 | {"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-2022-05 | latest | en | 0.796184 |
https://thesignalbeat.com/page/3/ | 1,695,956,999,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510481.79/warc/CC-MAIN-20230929022639-20230929052639-00583.warc.gz | 632,558,233 | 16,466 | ## Epidemic Models 2, Part 2
David A. Tanzer, August 17, 2020, in unit Epidemic Models 2.
# What is meant by the rate of a reaction?
Let’s take a simple model, with just one reaction:
$$\mathrm{Sick} \xrightarrow{\mathit{Recovery}} \mathrm{Healthy}$$
Suppose that everyone starts out ill, and eventually recovers.
Suppose that, on any given day, there is a 10% chance that a sick person will recover. So, viewing the process of recovery as a whole, it will … Read more
## Epidemic Models 2, Part 3
David A. Tanzer, August 17, 2020, in unit Epidemic Models 2.
# Analysis of reaction rates: first approach
Now let’s work out the equation for how the popping evolves over time.
Let Rate(t) be the average rate at which popping takes place at time $t$. Once the popping has begun, we expect Rate(t) to be high, and then dwindle down towards zero as more and more of the kernels have … Read more
## Epidemic Models 2, Part 4
David A. Tanzer, August 17, 2020, in unit Epidemic Models 2.
# Digression: solving the popcorn rate equation
Last time, we analyzed a system with a single reaction:
And derived its rate equation:
• Unpopped'(t) = – RateCoefficient * Unpopped(t)
In this article, we will see the solution to the equation, which will tell us exactly how the average popping rate evolves over time.
## Epidemic Models 2, Part 5
David A. Tanzer, August 17, 2020, in unit Epidemic Models 2.
# Analysis of reaction rates – second approach
In the previous article, we analyzed a simple system with a simple reaction:
$$\mathrm{Unpopped} \xrightarrow{\mathrm{Cooking}} \mathrm{Popped}$$
There we saw that the rate of the reaction is proportional both to a constant associated with the reaction, and to the count in compartment Unpopped. This will hold for all reactions that take … Read more
## Epidemic Models 2, Part 6
David A. Tanzer, August 17, 2020, in unit Epidemic Models 2.
# Digression: reaction rates in chemistry
The same kinetics occurs in chemical reaction networks. To illustrate, suppose there were two types of molecules, A and B, and that one A and one B molecule could collide to form a C molecule. So the reaction would be written $$A + B \rightarrow C$$.
Now imagine that a large number of … Read more
## Epidemic Models 2, Part 7
David A. Tanzer, August 17, 2020, in unit Epidemic Models 2.
# The Dynamics of SIR
Recall that for dynamics, we want to get at the way that the compartments S, I, R evolve over tune,
So far, we’ve defined the rates at which the reactions fire.
So now let’s summarize the rates for the SIR model:
• Infection: Susceptible + Infected —> 2 Infected (with rate constant 1)
• Recovery: Infected
## Epidemic Models 2, Part 8
David A. Tanzer, August 24, 2020, in unit Epidemic Models 2.
# Continuous vs. Discrete Flow
With the advent of the rate equation, we have shifted into a model of reaction networks consisting of “pumps” that move at continuous rates, transferring “mathematical fluid” between containers. Yet, as we have indicated, the real activity proceeds by discrete steps; an individual recovers, two molecules collide to form a compound.
How are these … Read more
## Epidemic Models 2, Part 9
David A. Tanzer, August 24, 2020, in unit Epidemic Models 2.
# Continuous and discrete flows in epidemic models
The standard compartmental models for SIR, SEIR etc. use the rate equations for the dynamics, and hence assume a continuous flow model.
This works well, particularly due to the fact that population sizes in epidemics are “large”, on the order of millions of individuals, and the law of large numbers kicks … Read more
## Unit Overview – Epidemic Models 1
David A. Tanzer, July 2020, in unit Epidemic Models 1. Next unit: Epidemic Models 2.
This is an introduction to compartmental models, which are basic to understanding how epidemics move through populations. We start with the simplest models, and then build towards the SEIR model, which is a starting point for understanding epidemics like Covid.… Read more
## Epidemic Models 1, Part 1
David A. Tanzer, July 10, 2020, in unit Epidemic Models 1.
# A first look at compartmental models
The grim curves that we see in the papers show things like the number of daily infections, and the number of daily deaths. Epidemic models aim to predict these curves. The models depend both on natural parameters such as infectiousness and the duration of the infectious period, as well as on social … Read more | 1,128 | 4,462 | {"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} | 3.90625 | 4 | CC-MAIN-2023-40 | latest | en | 0.894947 |
https://www.gap-system.org/ForumArchive/Hulpke.1/Alexande.1/Re__Subg.3/1.html | 1,597,403,520,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739211.34/warc/CC-MAIN-20200814100602-20200814130602-00327.warc.gz | 655,451,921 | 2,779 | > < ^ Date: Fri, 15 Mar 1996 14:38:08 +1553
> < ^ From: Alexander Hulpke <hulpke@math.colostate.edu >
^ Subject: Re: Subgroup Lattices (Untergruppengraphen)
Dear GAP-Forum,
Katharina Geissler wrote: (translation by me)
I'm searching a function for the transitive groups of degree 12 to display
how the several subgroups of S12 are contained within each other.
What you're asking for are actually two things. The second is a graphical
display. In principle you could persuade 'GraphicLattice' to display a
partial lattice for the transitive groups. See the ongoing discussion about
XGAP and its possible extensions. We are currently discussing several
possibilities, but so far nothing specific has been planned.
Nevertheless I doubt that this is what you really want, unless you have a
BIG monitor: there are 301 transitive groups of degree 12. Displaying all
containments of conjugates will probably result in a mess of lines on the
screen which is not really usable. Anyhow, if you're using pen and paper you
could still try to produce such a picture based on the information you would
provide to 'GraphicLattice'. (BTW.: If you finally get such a picture I
would be VERY interested to get a copy of it. So far I've been too lazy
to do it myself.)
The first (and bigger) problem is to get the actual containment information.
So far there is no function to compute this with a single command in GAP.
I will describe, however, how you can compute it yourself, but it might
take you some time to do so:
The process will yield not only information *whether* a group is contained
in another, but also information how many conjugacy classes exist if the
subgroup is maximal. (You will need this information if you want to identify
Galois groups, as I suppose. You should note as well, that GAP already
contains information about resolvents distinguishing the groups. Probably
If you can compute representatives of the conjugacy classes of maximal
subgroups of each transitive group, you are done. Non-maximal containment
simply follows by induction. 265 of the transitive groups are solvable. By
converting them to an AgGroup and then to a SpecialAgGroup, you can compute
representatives of the conjugacy classes of maximal subgroups and
transfer them (using the components .bijection in the SAgGroup and the
AgGroup) back in the permutation group. There, you select the ones which are
transitive. The command 'TransitiveIdentification' then tells you for each
of the representatives the number in the list of transitive groups, avoiding
conjugacy tests in S12.
This leaves 36 non-solvable groups. Coping with S12 and A12 is quite simple,
as the maximal subgroups are classified already (the imprimitive ones are
wreath products, the primitive ones are dealt with in the ATLAS).
8 of the remaining groups are wreath products. I have procedures to get
representatives of the conjugacy classes of transitive subgroups for these
groups, that I can provide to you if you want. However, it is not hard to
classify their maximal subgroups.
Of the remaining 26 groups, 18 are of size smaller 10000. You can use the
subgroup lattice program to get their maximal subgroups.
Most of the other groups are normal of small index in a wreath product.
Using this information one can describe the transitive subgroups of them.
(Again, I have functions to deal with this case. Write to me if you want
further information.)
Remaining is M12, whose maximal subgroups are given in the ATLAS.
In all these cases, after getting the maximal subgroups, treatment is the
same as in the solvable case.
Going through this process is a tour de force. However I can't imagine an
easier way (except persuading someone else to do it, but that's exactly
what I'm doing here).
I hope this helps. If anything in my description is unclear please ask.
Alexander Hulpke
-- Lehrstuhl D fuer Mathematik, RWTH, Templergraben 64, 52056 Aachen, Germany,
eMail: Alexander.Hulpke@math.rwth-aachen.de
> < [top] | 947 | 3,974 | {"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-2020-34 | latest | en | 0.932605 |
https://www.geeksforgeeks.org/class-12-rd-sharma-solutions-chapter-33-binomial-distribution-exercise-33-2-set-2/ | 1,709,256,080,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474893.90/warc/CC-MAIN-20240229234355-20240301024355-00800.warc.gz | 766,587,179 | 63,646 | # Class 12 RD Sharma Solutions – Chapter 33 Binomial Distribution – Exercise 33.2 | Set 2
### Question 15. A dice is thrown thrice. A success is 1 or 6 in a throw. Find the mean and variance of the number of successes.
Solution:
Let p denote the success and q denote failure of an event.
Now, the sample space when a dice is thrown is given by S = {1, 2, 3, 4, 5, 6}
Hence, p = 2/6 = 1/3 and q = 1 – 1/3 = 2/3
Therefore, Mean = np = 3 × 1/3 = 1 and Variance = npq = 1 × 2/3 = 2/3
### Question 16. If a random variable X follows a binomial distribution with mean 3 and variance 3/2, find P (X ≤ 5).
Solution:
We are given mean (np) = 3 and variance (npq) = 3/2.
Solving for the value of q,
q = 1/2, hence we can conclude p = 1 – 1/2 = 1/2
Now putting the value of p in relation, np = 3, we get n = 6
We know that a binomial distribution follows the relation:
P(X = r) = nCr pr(q)n-r
Therefore, in this case P(X = r) = 6Cr (1/2)r(1/2)6-r
P(X = r) = 6Cr (1/2)6
We are required to calculate the value for P(X ≤ 5) = 1 – P(X = 6)
P(X ≤ 5) = 1 – 6Cr (1/2)6
P(X ≤ 5) = 1 – (1/64)
P(X ≤ 5) = 63/64
### Question 17. If X follows a binomial distribution with mean 4 and variance 2, find P(X ≥ 5).
Solution:
We are given mean (np) = 4 and variance (npq) = 2.
Solving for the value of q, npq/np = 2/4
q = 1/2, hence we can conclude p = 1 – 1/2 = 1/2
Now putting the value of p in relation, np = 4, we get n = 8
We know that a binomial distribution follows the relation: P(X = r) = nCr pr(q)n-r
Therefore, in this case P(X = r) = 8Cr (1/2)r(1/2)8-r
P(X = r) = 8Cr (1/2)8
We are required to calculate the value
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
P(X ≥ 5) = 8C5 (1/2)8 + 8C6 (1/2)8 + 8C7 (1/2)8 + 8C8 (1/2)8
P(X ≥ 5) = (1/2)8[8C5 + 8C6 + 8C7 + 8C8]
P(X ≥ 5) = (56 + 28 + 8 + 1)/256
P(X ≥ 5) = 93/256
### Question 18. The mean and variance of a binomial distribution are 4/3 and 8/9 respectively. Find P(X ≥ 1).
Solution:
We are given mean (np) = 4 and variance (npq) = 2
Solving for the value of q, npq/np =
q = 2/3, hence we can conclude p = 1 – 2/3 = 1/3
Now putting the value of p in relation, np = 4/3, we get n = 4
We know that a binomial distribution follows the relation: P(X = r) = nCr pr(q)n-r
Therefore, in this case P(X = r) = 4Cr (1/3)r(2/3)4-r
We are required to calculate the value for P(X ≥ 1) = 1 – P(X = 0)
P(X ≥ 1) = 1 – 4C0 (1/3)0(2/3)4
P(X ≥ 1) = 1 – 16/81
P(X ≥ 1) = 65/81
### Question 19. If the sum of the mean and variance of a binomial distribution for 6 trials is 10/3, find the distribution.
Solution:
Given n = 6 and np + npq = 10/3
np (1 + q) = 10/3
6p (1 + 1 – p) = 10/3
12p – 6p2 = 10/3
18p2 – 36p + 10 = 0
Solving for the value of p we will get p = 1/3 or p = 5/3.
Since, the value of p cannot exceed 1, we will consider p = 1/3.
Therefore, q = 1 – 1/3 = 2/3
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 6Cr (1/3)r(2/3)6-r for r = 0,1,2,….,6
### Question 20. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes and, hence, find its mean.
Solution:
We are given n = 4 and
a doublet in the throw of a dice occurs when we get (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Therefore, the probability of success, p = 6/36 = 1/6, so q = 1 – 1/6 = 5/6
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 4Cr (1/6)r(5/6)4-r for r = 0, 1, 2, 3, 4
Hence, the probability distribution is given as:
Mean = 0 × (625/1296) + 1 × (500/1296) + 2 × (150/1296) + 3 × (20/1296) + 0 × (1/1296)
= 864/ 1296
= 2/3
### Question 21. Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.
Solution:
We are given n = 3 and
a doublet in the throw of a dice occurs when we get (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Therefore, the probability of success, p = 6/36 = 1/6, so q = 1 – 1/6 = 5/6
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 3Cr (1/6)r(5/6)3-r for r = 0, 1, 2, 3
Hence, the probability distribution is given as:
Mean = 0 × (125/216) + 1 × (75/216) + 2 × (15/216) + 3 × (1/216)
= 108/216
= 1/2
### Question 22. From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.
Solution:
Total number of bulbs = 15 and total defective bulbs = 5
Thus, the probability of getting one defective bulb with replacement, p = 5/15 = 1/3
Hence, q = 1 – 1/3 = 2/3.
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 4Cr (1/3)r(2/3)4-r for r = 0, 1, 2, 3, 4
Hence, the probability distribution is given as:
Mean = 0 × (16/81) + 1 × (32/81) + 2 × (24/81) + 3 × (8/81) + 4 × (1/81)
= 108/81
= 4/3
### Question 23. A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.
Solution:
We are given the number of throws, n = 3
Let p denote the probability of getting a 2 in the throw of a dice, then p = 1/6
Therefore, we can conclude 1 = 1 – 1/6 = 5/6
Now, the expectation of X denotes mean therefore, E(X) = np = 3 × 1/6 = 1/2
### Question 24. A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.
Solution:
We are given the number of times the coin is tossed, n = 2
Let p denote the probability of getting even number on dice upon throwing which is a success.
Thus, p = 3/6 = 1/2, therefore we can conclude q = 1 – p = 1 – 1/2 = 1/2
Now, the variance is given by npq.
Variance = 2 × 1/2 × 1/2 = 1/2
### Question 25. Three cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence, find the mean of the distribution.
Solution:
Number of cards drawn with replacement, n = 3
p = Probability of getting a spade card upon withdrawal = 13/52 = 1/4
Thus, we can conclude, q = 1 – 1/4 = 3/4
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 3Cr (1/4)r(3/4)3-r for r = 0, 1, 2, 3
Hence, the probability distribution is given as:
Mean = 0 × (27/64) + 1 × (27/64) + 2 × (9/64) + 3 × (1/64)
= (27 + 18 + 3)/64
= 48/64
= 3/4
### Question 26. An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also, find mean and variance of the distribution.
Solution:
Let p denote the probability of drawing a red ball which is considered a success, p = 6/9 = 2/3
And the probability of drawing a white ball which is considered a failure, q = 3/9 = 1/3
We have to draw four balls, so n = 4.
Hence, the mean of the probability distribution = np = 4 × 2/3 = 8/3
And variance = npq = 8/3 × 1/3 = 8/9
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 4Cr (2/3)r(1/3)4-r for r = 0, 1, 2, 3, 4
Hence, the probability distribution is given as:
### Question 27. Five bad oranges are accidentally mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence, find the mean and variance of the distribution.
Solution:
Let p denote the probability of drawing a bad orange which is considered a success, p = 5/25 = 1/5
And the probability of drawing a good orange which is considered a failure, q = 20/25 = 4/5
We have to draw four oranges, so n = 4
Hence, the mean of the probability distribution = np = 4 × 1/5 = 4/5
And the variance = npq = 4/5 × 4/5 = 16/25
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 4Cr (1/5)r(4/5)4-r for r = 0, 1, 2, 3, 4
Hence, the probability distribution is given as: | 2,996 | 7,996 | {"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} | 4.75 | 5 | CC-MAIN-2024-10 | latest | en | 0.819057 |
http://www.mathworks.com/help/matlab/functionlist.html?nocookie=true | 1,430,770,459,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430455113263.14/warc/CC-MAIN-20150501043833-00080-ip-10-235-10-82.ec2.internal.warc.gz | 527,026,965 | 55,601 | Functions
Language Fundamentals
Entering Commands
ans Most recent answer clc Clear Command Window diary Save Command Window text to file format Set Command Window output display format home Send cursor home iskeyword Determine whether input is MATLAB keyword more Control paged output for Command Window
Matrices and Arrays
Array Creation and Concatenation
accumarray Construct array with accumulation blkdiag Construct block diagonal matrix from input arguments diag Create diagonal matrix or get diagonal elements of matrix eye Identity matrix false Logical 0 (false) freqspace Frequency spacing for frequency response linspace Generate linearly spaced vector logspace Generate logarithmically spaced vectors meshgrid Rectangular grid in 2-D and 3-D space ndgrid Rectangular grid in N-D space ones Create array of all ones rand Uniformly distributed random numbers true Logical 1 (true) zeros Create array of all zeros cat Concatenate arrays along specified dimension horzcat Concatenate arrays horizontally vertcat Concatenate arrays vertically
Indexing
colon Create vectors, array subscripting, and for-loop iterators end Terminate block of code, or indicate last array index ind2sub Subscripts from linear index sub2ind Convert subscripts to linear indices
Array Dimensions
length Length of largest array dimension ndims Number of array dimensions numel Number of array elements size Array dimensions height Number of table rows width Number of table variables iscolumn Determine whether input is column vector isempty Determine whether array is empty ismatrix Determine whether input is matrix isrow Determine whether input is row vector isscalar Determine whether input is scalar isvector Determine whether input is vector
Sorting and Reshaping Arrays
blkdiag Construct block diagonal matrix from input arguments circshift Shift array circularly ctranspose Complex conjugate transpose diag Create diagonal matrix or get diagonal elements of matrix flip Flip order of elements fliplr Flip array left to right flipud Flip array up to down ipermute Inverse permute dimensions of N-D array permute Rearrange dimensions of N-D array repelem Repeat copies of array elements repmat Repeat copies of array reshape Reshape array rot90 Rotate array 90 degrees shiftdim Shift dimensions issorted Determine whether set elements are in sorted order sort Sort array elements sortrows Sort array rows squeeze Remove singleton dimensions transpose Transpose vectorize Vectorize expression
Operators and Elementary Operations
Arithmetic
plus Addition uplus Unary plus minus Subtraction uminus Unary minus times Element-wise multiplication rdivide Right array division ldivide Left array division power Element-wise power mtimes Matrix Multiplication mrdivide Solve systems of linear equations xA = B for x mldivide Solve systems of linear equations Ax = B for x mpower Matrix power cumprod Cumulative product cumsum Cumulative sum diff Differences and Approximate Derivatives prod Product of array elements sum Sum of array elements ceil Round toward positive infinity fix Round toward zero floor Round toward negative infinity idivide Integer division with rounding option mod Remainder after division (modulo operation) rem Remainder after division round Round to nearest decimal or integer
Relational Operations
Relational Operators Relational operations eq Determine equality ge Determine greater than or equal to gt Determine greater than le Determine less than or equal to lt Determine less than ne Determine inequality isequal Determine array equality isequaln Determine array equality, treating NaN values as equal
Logical Operations
Logical Operators: Short-circuit Logical operations with short-circuiting and Find logical AND not Find logical NOT or Find logical OR xor Logical exclusive-OR all Determine if all array elements are nonzero or true any Determine if any array elements are nonzero false Logical 0 (false) find Find indices and values of nonzero elements islogical Determine if input is logical array logical Convert numeric values to logicals true Logical 1 (true)
Set Operations
intersect Set intersection of two arrays ismember Array elements that are members of set array ismembertol Set ismember within tolerance issorted Determine whether set elements are in sorted order setdiff Set difference of two arrays setxor Set exclusive OR of two arrays union Set union of two arrays unique Unique values in array uniquetol Set unique within a tolerance join Merge two tables by matching up rows using key variables innerjoin Inner join between two tables outerjoin Outer join between two tables
Bit-Wise Operations
bitand Bit-wise AND bitcmp Bit-wise complement bitget Get bit at specified position bitor Bit-wise OR bitset Set bit at specific location bitshift Shift bits specified number of places bitxor Bit-wise XOR swapbytes Swap byte ordering
Special Characters
Special Characters Special characters colon Create vectors, array subscripting, and for-loop iterators
Data Types
Numeric Types
double Convert to double precision single Convert to single precision int8 Convert to 8-bit signed integer int16 Convert to 16-bit signed integer int32 Convert to 32-bit signed integer int64 Convert to 64-bit signed integer uint8 Convert to 8-bit unsigned integer uint16 Convert to 16-bit unsigned integer uint32 Convert to 32-bit unsigned integer uint64 Convert to 64-bit unsigned integer cast Cast variable to different data type typecast Convert data types without changing underlying data isinteger Determine if input is integer array isfloat Determine if input is floating-point array isnumeric Determine if input is numeric array isreal Determine whether array is real isfinite Array elements that are finite isinf Array elements that are infinite isnan Array elements that are NaN eps Floating-point relative accuracy flintmax Largest consecutive integer in floating-point format Inf Infinity intmax Largest value of specified integer type intmin Smallest value of specified integer type NaN Not-a-Number realmax Largest positive floating-point number realmin Smallest positive normalized floating-point number
Characters and Strings
Create and Concatenate Strings
blanks Create string of blank characters cellstr Convert to cell array of strings char Convert to character array (string) iscellstr Determine whether input is cell array of strings ischar Determine whether item is character array sprintf Format data into string strcat Concatenate strings horizontally strjoin Join strings in cell array into single string
Parse Strings
ischar Determine whether item is character array isletter Array elements that are alphabetic letters isspace Array elements that are space characters isstrprop Determine whether string is of specified category sscanf Read formatted data from string strfind Find one string within another strrep Find and replace substring strsplit Split string at specified delimiter strtok Selected parts of string validatestring Check validity of text string symvar Determine symbolic variables in expression regexp Match regular expression (case sensitive) regexpi Match regular expression (case insensitive) regexprep Replace string using regular expression regexptranslate Translate string into regular expression
Compare Strings
strcmp Compare strings strcmpi Compare strings (case insensitive) strncmp Compare first n characters of strings (case sensitive) strncmpi Compare first n characters of strings (case insensitive)
Change String Case, Blanks, and Justification
blanks Create string of blank characters deblank Strip trailing blanks from end of string strtrim Remove leading and trailing white space from string lower Convert string to lowercase upper Convert string to uppercase strjust Justify character array
Dates and Time
Create Date and Time Arrays
datetime Create array based on current date, or convert from date strings or numbers years Duration in years days Duration in days hours Duration in hours minutes Duration in minutes seconds Duration in seconds milliseconds Duration in milliseconds duration Create duration array from numeric values calyears Calendar duration in years calquarters Calendar duration in quarters calmonths Calendar duration in months calweeks Calendar duration in weeks caldays Calendar duration in days calendarDuration Create calendar duration array from numeric values exceltime Convert MATLAB datetime to Excel date number juliandate Convert MATLAB datetime to Julian date posixtime Convert MATLAB datetime to POSIX time yyyymmdd Convert MATLAB datetime to YYYYMMDD numeric value
Components of Dates and Time
year Year number quarter Quarter number month Month number and name week Week number day Day number or name hour Hour number minute Minute number second Second number ymd Year, month, and day numbers of datetime hms Hour, minute, and second numbers of duration split Split calendar duration into numeric and duration units time Convert time of calendar duration to duration timeofday Convert time of datetime to duration isdst Determine daylight saving time elements isweekend Determine weekend elements tzoffset Time zone offset from UTC
Date and Time Arithmetic and Plotting
between Calendar math differences caldiff Calendar math successive differences dateshift Shift date or generate sequence of dates and time isbetween Determine elements within date and time interval isdatetime Determine if input is datetime array isduration Determine if input is duration array iscalendarduration Determine if input is calendar duration array isnat Determine NaT (Not-a-Time) elements
Dates and Time as Numbers and Strings
datenum Convert date and time to serial date number datevec Convert date and time to vector of components datestr Convert date and time to string format char Convert to character array (string) cellstr Convert to cell array of strings now Current date and time as serial date number clock Current date and time as date vector date Current date string calendar Calendar for specified month eomday Last day of month weekday Day of week addtodate Modify date number by field etime Time elapsed between date vectors
Categorical Arrays
categorical Create categorical array iscategorical Determine whether input is categorical array categories Categories of categorical array iscategory Test for categorical array categories isordinal Determine whether input is ordinal categorical array isprotected Determine whether categories of categorical array are protected addcats Add categories to categorical array mergecats Merge categories in categorical array removecats Remove categories from categorical array renamecats Rename categories in categorical array reordercats Reorder categories in categorical array setcats Set categories in categorical array summary Print summary of table or categorical array countcats Count occurrences of categorical array elements by category isundefined Find undefined elements in categorical array
Tables
table Create table from workspace variables array2table Convert homogeneous array to table cell2table Convert cell array to table struct2table Convert structure array to table table2array Convert table to homogeneous array table2cell Convert table to cell array table2struct Convert table to structure array readtable Create table from file writetable Write table to file istable Determine whether input is table height Number of table rows width Number of table variables summary Print summary of table or categorical array intersect Set intersection of two arrays ismember Array elements that are members of set array setdiff Set difference of two arrays setxor Set exclusive OR of two arrays unique Unique values in array union Set union of two arrays join Merge two tables by matching up rows using key variables innerjoin Inner join between two tables outerjoin Outer join between two tables sortrows Sort array rows stack Stack data from multiple variables into single variable unstack Unstack data from single variable into multiple variables ismissing Find table elements with missing values standardizeMissing Insert missing value indicators into table varfun Apply function to table variables rowfun Apply function to table rows
Structures
struct Create structure array fieldnames Field names of structure, or public fields of object getfield Field of structure array isfield Determine whether input is structure array field isstruct Determine whether input is structure array orderfields Order fields of structure array rmfield Remove fields from structure setfield Assign values to structure array field arrayfun Apply function to each element of array structfun Apply function to each field of scalar structure table2struct Convert table to structure array struct2table Convert structure array to table cell2struct Convert cell array to structure array struct2cell Convert structure to cell array
Cell Arrays
cell Create cell array cell2mat Convert cell array to ordinary array of the underlying data type cell2struct Convert cell array to structure array cell2table Convert cell array to table celldisp Display cell array contents cellfun Apply function to each cell in cell array cellplot Graphically display structure of cell array cellstr Convert to cell array of strings iscell Determine whether input is cell array iscellstr Determine whether input is cell array of strings mat2cell Convert array to cell array with potentially different sized cells num2cell Convert array to cell array with consistently sized cells strjoin Join strings in cell array into single string strsplit Split string at specified delimiter struct2cell Convert structure to cell array table2cell Convert table to cell array
Function Handles
function_handle (@) Handle used in calling functions indirectly feval Evaluate function func2str Construct function name string from function handle str2func Construct function handle from function name string localfunctions Function handles to all local functions in MATLAB file functions Information about function handle
Map Containers
containers.Map Map values to unique keys isKey Determine if containers.Map object contains key keys Identify keys of containers.Map object remove Remove key-value pairs from containers.Map object values Identify values in containers.Map object
Time Series
Time Series Basics
append Concatenate time series objects in time dimension get Query timeseries object property values getdatasamplesize Size of data sample in timeseries object getqualitydesc Data quality descriptions getsamples Subset of time series samples using subscripted index array plot Plot time series set Set properties of timeseries object tsdata.event Construct event object for timeseries object timeseries Create timeseries object
Data Manipulation
addsample Add data sample to timeseries object delsample Remove sample from timeseries object detrend Subtract mean or best-fit line and all NaNs from timeseries object filter Shape frequency content of time-series getabstime Extract date-string time vector into cell array getinterpmethod Interpolation method for timeseries object getsampleusingtime Extract data samples into new timeseries object idealfilter Apply ideal (noncausal) filter to timeseries object resample Select or interpolate timeseries data using new time vector setabstime Set times of timeseries object as date strings setinterpmethod Set default interpolation method for timeseries object synchronize Synchronize and resample two timeseries objects using common time vector
Event Data
addevent Add event to timeseries object delevent Remove tsdata.event objects from timeseries object gettsafteratevent New timeseries object with samples occurring at or after event gettsafterevent New timeseries object with samples occurring after event gettsatevent New timeseries object with samples occurring at event gettsbeforeatevent New timeseries object with samples occurring before or at event gettsbeforeevent New timeseries object with samples occurring before event gettsbetweenevents New timeseries object with samples occurring between events
Descriptive Statistics
iqr Interquartile range of timeseries data max Maximum value of timeseries data mean Mean value of timeseries data median Median value of timeseries data min Minimum value of timeseries data std Standard deviation of timeseries data sum Sum of timeseries data var Variance of timeseries data
Time Series Collections
get (tscollection) Query tscollection object property values isempty (tscollection) Determine whether tscollection object is empty length (tscollection) Length of time vector plot Plot time series set (tscollection) Set properties of tscollection object size (tscollection) Size of tscollection object tscollection Create tscollection object addsampletocollection Add sample to tscollection object addts Add timeseries object to tscollection object delsamplefromcollection Remove sample from tscollection object getabstime (tscollection) Extract date-string time vector into cell array getsampleusingtime (tscollection) Extract data samples into new tscollection object gettimeseriesnames Cell array of names of timeseries objects in tscollection object horzcat (tscollection) Horizontal concatenation for tscollection objects removets Remove timeseries objects from tscollection object resample (tscollection) Select or interpolate data in tscollection using new time vector setabstime (tscollection) Set times of tscollection object as date strings settimeseriesnames Change name of timeseries object in tscollection vertcat (tscollection) Vertical concatenation for tscollection objects
Data Type Identification
isa Determine if input is object of specified class iscalendarduration Determine if input is calendar duration array iscategorical Determine whether input is categorical array iscell Determine whether input is cell array iscellstr Determine whether input is cell array of strings ischar Determine whether item is character array isdatetime Determine if input is datetime array isduration Determine if input is duration array isfield Determine whether input is structure array field isfloat Determine if input is floating-point array isgraphics True for valid graphics object handles isinteger Determine if input is integer array isjava Determine if input is Java object islogical Determine if input is logical array isnumeric Determine if input is numeric array isobject Determine if input is MATLAB object isreal Determine whether array is real isenum Determine if variable is enumeration isstruct Determine whether input is structure array istable Determine whether input is table is* Detect state class Determine class of object validateattributes Check validity of array whos List variables in workspace, with sizes and types
Data Type Conversion
char Convert to character array (string) cellstr Convert to cell array of strings int2str Convert integer to string mat2str Convert matrix to string num2str Convert number to string str2double Convert string to double-precision value str2num Convert string to number native2unicode Convert numeric bytes to Unicode character representation unicode2native Convert Unicode character representation to numeric bytes base2dec Convert base N number string to decimal number bin2dec Convert binary number string to decimal number dec2base Convert decimal to base N number in string dec2bin Convert decimal to binary number in string dec2hex Convert decimal to hexadecimal number in string hex2dec Convert hexadecimal number string to decimal number hex2num Convert hexadecimal number string to double-precision number num2hex Convert singles and doubles to IEEE hexadecimal strings table2array Convert table to homogeneous array table2cell Convert table to cell array table2struct Convert table to structure array array2table Convert homogeneous array to table cell2table Convert cell array to table struct2table Convert structure array to table cell2mat Convert cell array to ordinary array of the underlying data type cell2struct Convert cell array to structure array mat2cell Convert array to cell array with potentially different sized cells num2cell Convert array to cell array with consistently sized cells struct2cell Convert structure to cell array
Mathematics
Elementary Math
Arithmetic
plus Addition uplus Unary plus minus Subtraction uminus Unary minus times Element-wise multiplication rdivide Right array division ldivide Left array division power Element-wise power mtimes Matrix Multiplication mrdivide Solve systems of linear equations xA = B for x mldivide Solve systems of linear equations Ax = B for x mpower Matrix power cumprod Cumulative product cumsum Cumulative sum diff Differences and Approximate Derivatives prod Product of array elements sum Sum of array elements ceil Round toward positive infinity fix Round toward zero floor Round toward negative infinity idivide Integer division with rounding option mod Remainder after division (modulo operation) rem Remainder after division round Round to nearest decimal or integer
Exponents and Logarithms
exp Exponential expm1 Compute exp(x)-1 accurately for small values of x log Natural logarithm log10 Common logarithm (base 10) log1p Compute log(1+x) accurately for small values of x log2 Base 2 logarithm and dissect floating-point numbers into exponent and mantissa nextpow2 Exponent of next higher power of 2 nthroot Real nth root of real numbers pow2 Base 2 power and scale floating-point numbers reallog Natural logarithm for nonnegative real arrays realpow Array power for real-only output realsqrt Square root for nonnegative real arrays sqrt Square root
Descriptive Statistics
corrcoef Correlation coefficients cov Covariance max Largest elements in array cummax Cumulative maximum mean Average or mean value of array median Median value of array min Smallest elements in array cummin Cumulative minimum mode Most frequent values in array std Standard deviation var Variance histcounts Histogram bin counts discretize Group numeric data into bins or categories
Complex Numbers
abs Absolute value and complex magnitude angle Phase angle complex Create complex array conj Complex conjugate cplxpair Sort complex numbers into complex conjugate pairs i Imaginary unit imag Imaginary part of complex number isreal Determine whether array is real j Imaginary unit real Real part of complex number sign Signum function unwrap Correct phase angles to produce smoother phase plots
Discrete Math
factor Prime factors factorial Factorial of input gcd Greatest common divisor isprime Determine which array elements are prime lcm Least common multiple nchoosek Binomial coefficient or all combinations perms All possible permutations primes Prime numbers less than or equal to input value rat Rational fraction approximation rats Rational output
Polynomials
poly Polynomial with specified roots polyder Polynomial derivative polyeig Polynomial eigenvalue problem polyfit Polynomial curve fitting polyint Integrate polynomial analytically polyval Polynomial evaluation polyvalm Matrix polynomial evaluation residue Convert between partial fraction expansion and ratio of two polynomials roots Polynomial roots
Special Functions
airy Airy Functions besselh Bessel function of third kind (Hankel function) besseli Modified Bessel function of first kind besselj Bessel function of first kind besselk Modified Bessel function of second kind bessely Bessel function of second kind beta Beta function betainc Incomplete beta function betaincinv Beta inverse cumulative distribution function betaln Logarithm of beta function ellipj Jacobi elliptic functions ellipke Complete elliptic integrals of first and second kind erf Error function erfc Complementary error function erfcinv Inverse complementary error function erfcx Scaled complementary error function erfinv Inverse error function expint Exponential integral gamma Gamma function gammainc Incomplete gamma function gammaincinv Inverse incomplete gamma function gammaln Logarithm of gamma function legendre Associated Legendre functions psi Psi (polygamma) function
Cartesian Coordinate System Conversion
cart2pol Transform Cartesian coordinates to polar or cylindrical cart2sph Transform Cartesian coordinates to spherical pol2cart Transform polar or cylindrical coordinates to Cartesian sph2cart Transform spherical coordinates to Cartesian
Constants and Test Matrices
eps Floating-point relative accuracy flintmax Largest consecutive integer in floating-point format i Imaginary unit j Imaginary unit Inf Infinity pi Ratio of circle's circumference to its diameter NaN Not-a-Number isfinite Array elements that are finite isinf Array elements that are infinite isnan Array elements that are NaN compan Companion matrix gallery Test matrices hadamard Hadamard matrix hankel Hankel matrix hilb Hilbert matrix invhilb Inverse of Hilbert matrix magic Magic square pascal Pascal matrix rosser Classic symmetric eigenvalue test problem toeplitz Toeplitz matrix vander Vandermonde matrix wilkinson Wilkinson's eigenvalue test matrix
Linear Algebra
Matrix Operations
cross Cross product dot Dot product kron Kronecker tensor product surfnorm Compute and display 3-D surface normals tril Lower triangular part of matrix triu Upper triangular part of matrix transpose Transpose
Linear Equations
cond Condition number with respect to inversion condest 1-norm condition number estimate inv Matrix inverse linsolve Solve linear system of equations lscov Least-squares solution in presence of known covariance lsqnonneg Solve nonnegative least-squares constraints problem pinv Moore-Penrose pseudoinverse of matrix rcond Reciprocal condition number sylvester Solve Sylvester equation AX + XB = C for X mldivide Solve systems of linear equations Ax = B for x mrdivide Solve systems of linear equations xA = B for x
Matrix Decomposition
chol Cholesky factorization ichol Incomplete Cholesky factorization cholupdate Rank 1 update to Cholesky factorization ilu Sparse incomplete LU factorization lu LU matrix factorization qr Orthogonal-triangular decomposition qrdelete Remove column or row from QR factorization qrinsert Insert column or row into QR factorization qrupdate Rank 1 update to QR factorization planerot Givens plane rotation ldl Block LDL' factorization for Hermitian indefinite matrices cdf2rdf Convert complex diagonal form to real block diagonal form rsf2csf Convert real Schur form to complex Schur form gsvd Generalized singular value decomposition svd Singular value decomposition
Eigenvalues and Singular Values
balance Diagonal scaling to improve eigenvalue accuracy cdf2rdf Convert complex diagonal form to real block diagonal form condeig Condition number with respect to eigenvalues eig Eigenvalues and eigenvectors eigs Largest eigenvalues and eigenvectors of matrix gsvd Generalized singular value decomposition hess Hessenberg form of matrix ordeig Eigenvalues of quasitriangular matrices ordqz Reorder eigenvalues in QZ factorization ordschur Reorder eigenvalues in Schur factorization poly Polynomial with specified roots polyeig Polynomial eigenvalue problem qz QZ factorization for generalized eigenvalues rsf2csf Convert real Schur form to complex Schur form schur Schur decomposition sqrtm Matrix square root ss2tf Convert state-space representation to transfer function svd Singular value decomposition svds Find singular values and vectors
Matrix Analysis
bandwidth Lower and upper matrix bandwidth cond Condition number with respect to inversion condeig Condition number with respect to eigenvalues det Matrix determinant isbanded Determine if matrix is within specific bandwidth isdiag Determine if matrix is diagonal ishermitian Determine if matrix is Hermitian or skew-Hermitian issymmetric Determine if matrix is symmetric or skew-symmetric istril Determine if matrix is lower triangular istriu Determine if matrix is upper triangular norm Vector and matrix norms normest 2-norm estimate null Null space orth Orthonormal basis for range of matrix rank Rank of matrix rcond Reciprocal condition number rref Reduced row echelon form (Gauss-Jordan elimination) subspace Angle between two subspaces trace Sum of diagonal elements
Matrix Functions
expm Matrix exponential logm Matrix logarithm sqrtm Matrix square root bsxfun Apply element-by-element binary operation to two arrays with singleton expansion enabled funm Evaluate general matrix function arrayfun Apply function to each element of array accumarray Construct array with accumulation mpower Matrix power
Random Number Generation
rand Uniformly distributed random numbers randn Normally distributed random numbers randi Uniformly distributed pseudorandom integers randperm Random permutation rng Control random number generation RandStream Random number stream
Interpolation
1-D Interpolation
interp1 1-D data interpolation (table lookup) griddedInterpolant Gridded data interpolation pchip Piecewise Cubic Hermite Interpolating Polynomial (PCHIP) spline Cubic spline data interpolation ppval Evaluate piecewise polynomial mkpp Make piecewise polynomial unmkpp Piecewise polynomial details padecoef Padé approximation of time delays interpft 1-D interpolation using FFT method
Gridded Data Interpolation
interp2 Interpolation for 2-D gridded data in meshgrid format interp3 Interpolation for 3-D gridded data in meshgrid format interpn Interpolation for 1-D, 2-D, 3-D, and N-D gridded data in ndgrid format griddedInterpolant Gridded data interpolation ndgrid Rectangular grid in N-D space meshgrid Rectangular grid in 2-D and 3-D space
Scattered Data Interpolation
griddata Interpolate scattered data griddatan Data gridding and hypersurface fitting (dimension ≥ 2) scatteredInterpolant Scattered data interpolation
Optimization
fminbnd Find minimum of single-variable function on fixed interval fminsearch Find minimum of unconstrained multivariable function using derivative-free method fzero Root of nonlinear function lsqnonneg Solve nonnegative least-squares constraints problem optimget Optimization options values optimset Create or edit optimization options structure
Numerical Integration and Differential Equations
Ordinary Differential Equations
ode45 Solve nonstiff differential equations; medium order method ode15s Solve stiff differential equations and DAEs; variable order method ode23 Solve nonstiff differential equations; low order method ode113 Solve nonstiff differential equations; variable order method ode23t Solve moderately stiff ODEs and DAEs; trapezoidal rule ode23tb Solve stiff differential equations; low order method ode23s Solve stiff differential equations; low order method ode15i Solve fully implicit differential equations, variable order method decic Compute consistent initial conditions for ode15i odextend Extend solution of initial value problem for ordinary differential equation odeget Ordinary differential equation options parameters odeset Create or alter options structure for ordinary differential equation solvers deval Evaluate solution of differential equation problem
Boundary Value Problems
bvp4c Solve boundary value problems for ordinary differential equations bvp5c Solve boundary value problems for ordinary differential equations bvpinit Form initial guess for BVP solvers bvpxtend Form guess structure for extending boundary value solutions bvpget Extract properties from options structure created with bvpset bvpset Create or alter options structure of boundary value problem deval Evaluate solution of differential equation problem
Delay Differential Equations
dde23 Solve delay differential equations (DDEs) with constant delays ddesd Solve delay differential equations (DDEs) with general delays ddensd Solve delay differential equations (DDEs) of neutral type ddeget Extract properties from delay differential equations options structure ddeset Create or alter delay differential equations options structure deval Evaluate solution of differential equation problem
Partial Differential Equations
pdepe Solve initial-boundary value problems for parabolic-elliptic PDEs in 1-D pdeval Evaluate numerical solution of PDE using output of pdepe
Numerical Integration and Differentiation
integral Numerical integration integral2 Numerically evaluate double integral integral3 Numerically evaluate triple integral quadgk Numerically evaluate integral, adaptive Gauss-Kronrod quadrature quad2d Numerically evaluate double integral, tiled method cumtrapz Cumulative trapezoidal numerical integration trapz Trapezoidal numerical integration polyint Integrate polynomial analytically del2 Discrete Laplacian diff Differences and Approximate Derivatives gradient Numerical gradient polyder Polynomial derivative
Fourier Analysis and Filtering
abs Absolute value and complex magnitude angle Phase angle cplxpair Sort complex numbers into complex conjugate pairs fft Fast Fourier transform fft2 2-D fast Fourier transform fftn N-D fast Fourier transform fftshift Shift zero-frequency component to center of spectrum fftw Interface to FFTW library run-time algorithm tuning control ifft Inverse fast Fourier transform ifft2 2-D inverse fast Fourier transform ifftn N-D inverse fast Fourier transform ifftshift Inverse FFT shift nextpow2 Exponent of next higher power of 2 unwrap Correct phase angles to produce smoother phase plots conv Convolution and polynomial multiplication conv2 2-D convolution convn N-D convolution deconv Deconvolution and polynomial division detrend Remove linear trends filter 1-D digital filter filter2 2-D digital filter
Sparse Matrices
Sparse Matrix Creation
spdiags Extract and create sparse band and diagonal matrices speye Sparse identity matrix sprand Sparse uniformly distributed random matrix sprandn Sparse normally distributed random matrix sprandsym Sparse symmetric random matrix sparse Create sparse matrix spconvert Import from sparse matrix external format
Sparse Matrix Manipulation
issparse Determine whether input is sparse nnz Number of nonzero matrix elements nonzeros Nonzero matrix elements nzmax Amount of storage allocated for nonzero matrix elements spalloc Allocate space for sparse matrix spfun Apply function to nonzero sparse matrix elements spones Replace nonzero sparse matrix elements with ones spparms Set parameters for sparse matrix routines spy Visualize sparsity pattern find Find indices and values of nonzero elements full Convert sparse matrix to full matrix
Reordering Algorithms
amd Approximate minimum degree permutation colamd Column approximate minimum degree permutation colperm Sparse column permutation based on nonzero count dmperm Dulmage-Mendelsohn decomposition randperm Random permutation symamd Symmetric approximate minimum degree permutation symrcm Sparse reverse Cuthill-McKee ordering
Sparse Linear Algebra
condest 1-norm condition number estimate eigs Largest eigenvalues and eigenvectors of matrix ichol Incomplete Cholesky factorization ilu Sparse incomplete LU factorization normest 2-norm estimate spaugment Form least squares augmented system sprank Structural rank svds Find singular values and vectors
Linear Equations (Iterative Methods)
bicg Biconjugate gradients method bicgstab Biconjugate gradients stabilized method bicgstabl Biconjugate gradients stabilized (l) method cgs Conjugate gradients squared method gmres Generalized minimum residual method (with restarts) lsqr LSQR method minres Minimum residual method pcg Preconditioned conjugate gradients method qmr Quasi-minimal residual method symmlq Symmetric LQ method tfqmr Transpose-free quasi-minimal residual method
Graph and Tree Algorithms
etree Elimination tree etreeplot Plot elimination tree gplot Plot nodes and links representing adjacency matrix symbfact Symbolic factorization analysis treelayout Lay out tree or forest treeplot Plot picture of tree unmesh Convert edge matrix to coordinate and Laplacian matrices
Computational Geometry
Triangulation Representation
triangulation Triangulation in 2-D or 3-D tetramesh Tetrahedron mesh plot trimesh Triangular mesh plot triplot 2-D triangular plot trisurf Triangular surface plot
Delaunay Triangulation
delaunayTriangulation Delaunay triangulation in 2-D and 3-D delaunay Delaunay triangulation delaunayn N-D Delaunay triangulation tetramesh Tetrahedron mesh plot trimesh Triangular mesh plot triplot 2-D triangular plot trisurf Triangular surface plot
triangulation Triangulation in 2-D or 3-D delaunayTriangulation Delaunay triangulation in 2-D and 3-D dsearchn N-D nearest point search tsearchn N-D closest simplex search delaunay Delaunay triangulation delaunayn N-D Delaunay triangulation
Bounding Regions
boundary Boundary of a set of points in 2-D or 3-D alphaShape Polygons and polyhedra from points in 2-D and 3-D convhull Convex hull convhulln N-D convex hull alphaShape Polygons and polyhedra from points in 2-D and 3-D
Voronoi Diagram
patch Create one or more filled polygons voronoi Voronoi diagram voronoin N-D Voronoi diagram
Elementary Polygons
polyarea Area of polygon inpolygon Points located inside or on edge of polygonal region rectint Rectangle intersection area
Graphics
2-D and 3-D Plots
Line Plots
plot 2-D line plot plotyy 2-D line plots with y-axes on both left and right side plot3 3-D line plot loglog Log-log scale plot semilogx Semilogarithmic plot semilogy Semilogarithmic plot errorbar Plot error bars along curve fplot Plot function between specified limits ezplot Easy-to-use function plotter ezplot3 Easy-to-use 3-D parametric curve plotter LineSpec (Line Specification) Line specification string syntax ColorSpec (Color Specification) Color specification
Pie Charts, Bar Plots, and Histograms
bar Bar graph bar3 Plot 3-D bar graph barh Plot bar graph horizontally bar3h Plot horizontal 3-D bar graph histogram Histogram plot histcounts Histogram bin counts rose Angle histogram plot pareto Pareto chart area Filled area 2-D plot pie Pie chart pie3 3-D pie chart histogram Histogram bar plot for numeric data
Discrete Data Plots
stem Plot discrete sequence data stairs Stairstep graph stem3 Plot 3-D discrete sequence data scatter Scatter plot scatter3 3-D scatter plot spy Visualize sparsity pattern plotmatrix Scatter plot matrix
Polar Plots
polar Polar coordinate plot rose Angle histogram plot compass Plot arrows emanating from origin ezpolar Easy-to-use polar coordinate plotter LineSpec (Line Specification) Line specification string syntax ColorSpec (Color Specification) Color specification
Contour Plots
contour Contour plot of matrix contourf Filled 2-D contour plot contourc Low-level contour plot computation contour3 3-D contour plot contourslice Draw contours in volume slice planes ezcontour Easy-to-use contour plotter ezcontourf Easy-to-use filled contour plotter
Vector Fields
feather Plot velocity vectors quiver Quiver or velocity plot compass Plot arrows emanating from origin quiver3 3-D quiver or velocity plot streamslice Plot streamlines in slice planes streamline Plot streamlines from 2-D or 3-D vector data
Surfaces, Volumes, and Polygons
Surface and Mesh Plots
surf 3-D shaded surface plot surfc Contour plot under a 3-D shaded surface plot surface Create surface object surfl Surface plot with colormap-based lighting surfnorm Compute and display 3-D surface normals mesh Mesh plot meshc Plot a contour graph under mesh graph meshz Plot a curtain around mesh plot waterfall Waterfall plot ribbon Ribbon plot contour3 3-D contour plot peaks Example function of two variables cylinder Generate cylinder ellipsoid Generate ellipsoid sphere Generate sphere pcolor Pseudocolor (checkerboard) plot surf2patch Convert surface data to patch data ezsurf Easy-to-use 3-D colored surface plotter ezsurfc Easy-to-use combination surface/contour plotter ezmesh Easy-to-use 3-D mesh plotter ezmeshc Easy-to-use combination mesh/contour plotter
Volume Visualization
contourslice Draw contours in volume slice planes flow Simple function of three variables isocaps Compute isosurface end-cap geometry isocolors Calculate isosurface and patch colors isonormals Compute normals of isosurface vertices isosurface Extract isosurface data from volume data reducepatch Reduce number of patch faces reducevolume Reduce number of elements in volume data set shrinkfaces Reduce size of patch faces slice Volumetric slice plot smooth3 Smooth 3-D data subvolume Extract subset of volume data set volumebounds Coordinate and color limits for volume data coneplot Plot velocity vectors as cones in 3-D vector field curl Compute curl and angular velocity of vector field divergence Compute divergence of vector field interpstreamspeed Interpolate stream-line vertices from flow speed stream2 Compute 2-D streamline data stream3 Compute 3-D streamline data streamline Plot streamlines from 2-D or 3-D vector data streamparticles Plot stream particles streamribbon 3-D stream ribbon plot from vector volume data streamslice Plot streamlines in slice planes streamtube Create 3-D stream tube plot
Polygons
fill Filled 2-D polygons fill3 Filled 3-D polygons patch Create one or more filled polygons surf2patch Convert surface data to patch data
Animation
movie Play recorded movie frames getframe Capture axes or figure as movie frame frame2im Return image data associated with movie frame im2frame Convert image to movie frame animatedline Create animated line comet 2-D comet plot comet3 3-D comet plot drawnow Update figures and process callbacks refreshdata Refresh data in graph when data source is specified Animated Line Line animations
Formatting and Annotation
Titles and Labels
title Add title to current axes xlabel Label x-axis ylabel Label y-axis zlabel Label z-axis clabel Contour plot elevation labels datetick Date formatted tick labels texlabel Format text into TeX string legend Add legend to graph colorbar Colorbar showing color scale
Coordinate System
xlim Set or query x-axis limits ylim Set or query y-axis limits zlim Set or query z-axis limits box Axes border grid Display or hide axes grid lines daspect Set or query axes data aspect ratio pbaspect Set or query plot box aspect ratio axes Create axes graphics object axis Set axis limits and appearance subplot Create axes in tiled positions hold Retain current plot when adding new plots gca Current axes handle cla Clear axes
Annotation
annotation Create annotation objects text Create text object in current axes line Create line object rectangle Create 2-D rectangle object legend Add legend to graph title Add title to current axes xlabel Label x-axis ylabel Label y-axis zlabel Label z-axis datacursormode Enable, disable, and manage interactive data cursor mode ginput Graphical input from mouse or cursor gtext Mouse placement of text in 2-D view
Colormaps
colormap View and set current colormap colormapeditor Open colormap editor colorbar Colorbar showing color scale brighten Brighten or darken colormap contrast Grayscale colormap for contrast enhancement shading Set color shading properties graymon Set default figure properties for grayscale monitors caxis Color axis scaling hsv2rgb Convert HSV colormap to RGB colormap rgb2hsv Convert RGB colormap to HSV colormap rgbplot Plot colormap spinmap Spin colormap colordef Set default property values to display different color schemes whitebg Change axes background color
Data Exploration
hidden Remove hidden lines from mesh plot pan Pan view of graph interactively reset Reset graphics object properties to their defaults rotate Rotate object about specified origin and direction rotate3d Rotate 3-D view using mouse zoom Turn zooming on or off or magnify by factorMagnify by a factor datacursormode Enable, disable, and manage interactive data cursor mode figurepalette Show or hide Figure Palette plotbrowser Show or hide figure Plot Browser plotedit Interactively edit and annotate plots plottools Show or hide plot tools propertyeditor Show or hide Property Editor showplottool Show or hide figure plot tool
Data Brushing
brush Interactively mark, delete, modify, and save observations in graphs datacursormode Enable, disable, and manage interactive data cursor mode linkdata Automatically update graphs when variables change linkaxes Synchronize limits of specified 2-D axes linkprop Keep same value for corresponding properties of graphics objects refreshdata Refresh data in graph when data source is specified
3-D Scene Control
Camera Views
view Viewpoint specification makehgtform Create 4-by-4 transform matrix viewmtx View transformation matrices cameratoolbar Control camera toolbar programmatically campan Rotate camera target around camera position camzoom Zoom in and out on scene camdolly Move camera position and target camlookat Position camera to view object or group of objects camorbit Rotate camera position around camera target campos Set or query camera position camproj Set or query projection type camroll Rotate camera about view axis camtarget Set or query location of camera target camup Set or query camera up vector camva Set or query camera view angle
Lighting and Transparency
camlight Create or move light object in camera coordinates light Create light object lightangle Create or position light object in spherical coordinates lighting Specify lighting algorithm diffuse Calculate diffuse reflectance material Control reflectance properties of surfaces and patches specular Calculate specular reflectance alim Set or query axes alpha limits alpha Set transparency properties for objects in current axes alphamap Specify figure alphamap (transparency)
Images
Image File Operations
imshow Display image image Display image from array imagesc Scale data and display image object imread Read image from graphics file imwrite Write image to graphics file imfinfo Information about graphics file imformats Manage image file format registry frame2im Return image data associated with movie frame im2frame Convert image to movie frame im2java Convert image to Java image
Modifying Images
im2double Convert image to double precision ind2rgb Convert indexed image to RGB image rgb2gray Convert RGB image or colormap to grayscale rgb2ind Convert RGB image to indexed image imapprox Approximate indexed image by reducing number of colors dither Convert image, increasing apparent color resolution by dithering cmpermute Rearrange colors in colormap cmunique Eliminate duplicate colors in colormap; convert grayscale or truecolor image to indexed image
Printing and Saving
print Print figure or save to specific file format saveas Save figure to specific file format getframe Capture axes or figure as movie frame savefig Save figure and contents to FIG-file openfig Open figure saved in FIG-file orient Hardcopy paper orientation hgexport Export figure printopt Configure printer defaults
Graphics Objects
Organization of Graphics Objects
axes Create axes graphics object figure Create figure window groot Graphics root object
Graphics Object Properties
get Query graphics object properties set Set graphics object properties inspect Open Property Inspector propedit Open Property Editor
Graphics Object Identification
gca Current axes handle gcf Current figure handle gcbf Handle of figure containing object whose callback is executing gcbo Handle of object whose callback is executing gco Handle of current object groot Graphics root object ancestor Ancestor of graphics object allchild Find all children of specified objects findall Find all graphics objects findobj Locate graphics objects with specific properties findfigs Find visible offscreen figures gobjects Initialize array for graphics objects isgraphics True for valid graphics object handles ishandle Test for valid graphics or Java object handle copyobj Copy graphics objects and their descendants delete Delete files or objects
Graphics Object Programming
gobjects Initialize array for graphics objects isgraphics True for valid graphics object handles isempty Determine whether array is empty isequal Determine array equality isa Determine if input is object of specified class clf Clear current figure window cla Clear axes close Remove specified figure
Default Property Values
set Set graphics object properties get Query graphics object properties groot Graphics root object
Interactive Control and Callbacks
uicontextmenu Create context menu uimenu Create menus and menu items on figure windows function_handle Handle used in calling functions indirectly dragrect Drag rectangles with mouse rbbox Create rubberband box for area selection refresh Redraw current figure shg Show most recent graph window
Object Containers
hggroup Create group object hgtransform Create transform graphics object makehgtform Create 4-by-4 transform matrix eye Identity matrix
Specifying Target for Graphics Output
hold Retain current plot when adding new plots ishold Current hold state newplot Determine where to draw graphics objects clf Clear current figure window cla Clear axes
Graphics Performance
drawnow Update figures and process callbacks opengl Control OpenGL rendering
Programming Scripts and Functions
Control Flow
if, elseif, else Execute statements if condition is true for Execute statements specified number of times parfor Parallel for loop switch, case, otherwise Execute one of several groups of statements try, catch Execute statements and catch resulting errors while Repeat execution of statements while condition is true break Terminate execution of for or while loop continue Pass control to next iteration of for or while loop end Terminate block of code, or indicate last array index pause Halt execution temporarily return Return control to invoking function
Scripts
edit Edit or create file input Request user input publish Generate view of MATLAB file in specified format notebook Open MATLAB Notebook in Microsoft Word software (on Microsoft Windows platforms) grabcode Extract MATLAB code from file published to HTML snapnow Force snapshot of image for inclusion in published document
Functions
Function Basics
function Declare function name, inputs, and outputs
Input and Output Arguments
nargin Number of function input arguments nargout Number of function output arguments varargin Variable-length input argument list varargout Variable-length output argument list narginchk Validate number of input arguments nargoutchk Validate number of output arguments validateattributes Check validity of array validatestring Check validity of text string inputParser Parse function inputs inputname Variable name of function input
Variables
persistent Define persistent variable isvarname Determine whether input is valid variable name matlab.lang.makeUniqueStrings Construct unique strings from input strings matlab.lang.makeValidName Construct valid MATLAB identifiers from input strings namelengthmax Maximum identifier length assignin Assign value to variable in specified workspace global Declare variables as global
Error Handling
try, catch Execute statements and catch resulting errors error Throw error and display message warning Display warning message lastwarn Last warning message assert Throw error if condition false onCleanup Cleanup tasks upon function completion
Files and Folders
Search Path
addpath Add folders to search path rmpath Remove folders from search path path View or change search path savepath Save current search path userpath View or change user portion of search path genpath Generate path string pathsep Search path separator for current platform pathtool Open Set Path dialog box to view and change search path restoredefaultpath Restore search path to its factory-installed state
File Operations
dir List folder contents ls List folder contents pwd Identify current folder fileattrib Set or get attributes of file or folder exist Check existence of variable, function, folder, or class isdir Determine whether input is folder type Display contents of file visdiff Compare two text files, MAT-Files, binary files, Zip files, or folders what List MATLAB files in folder which Locate functions and files cd Change current folder copyfile Copy file or folder delete Delete files or objects recycle Set option to move deleted files to recycle folder mkdir Make new folder movefile Move file or folder rmdir Remove folder open Open file in appropriate application winopen Open file in appropriate application (Windows)
File Compression
zip Compress files into zip file unzip Extract contents of zip file gzip Compress files into GNU zip files gunzip Uncompress GNU zip files tar Compress files into tar file untar Extract contents of tar file
File Name Construction
fileparts Parts of file name and path fullfile Build full file name from parts filemarker Character to separate file name and internal function name filesep File separator for current platform tempdir Name of system's temporary folder tempname Unique name for temporary file matlabroot Root folder toolboxdir Root folder for specified toolbox
Debugging
dbclear Clear breakpoints dbcont Resume execution dbdown Reverse workspace shift performed by dbup, while in debug mode dbquit Quit debug mode dbstack Function call stack dbstatus List all breakpoints dbstep Execute one or more lines from current breakpoint dbstop Set breakpoints for debugging dbtype List text file with line numbers dbup Shift current workspace to workspace of caller, while in debug mode checkcode Check MATLAB code files for possible problems keyboard Input from keyboard mlintrpt Run checkcode for file or folder, reporting results in browser
Code Editor
edit Edit or create file
Programming Utilities
echo Display statements during function execution eval Execute MATLAB expression in text string evalc Evaluate MATLAB expression with capture evalin Execute MATLAB expression in specified workspace feval Evaluate function run Run MATLAB script builtin Execute built-in function from overloaded method matlab.codetools.requiredFilesAndProducts List dependencies of MATLAB program files mfilename File name of currently running function pcode Create protected function file timer Create object to schedule execution of MATLAB commands
Data Import and Export
Standard File Formats
Text Files
csvread Read comma-separated value (CSV) file csvwrite Write comma-separated value file dlmread Read ASCII-delimited file of numeric data into matrix dlmwrite Write matrix to ASCII-delimited file textscan Read formatted data from text file or string readtable Create table from file writetable Write table to file type Display contents of file
Images
im2java Convert image to Java image imfinfo Information about graphics file imread Read image from graphics file imwrite Write image to graphics file Tiff MATLAB Gateway to LibTIFF library routines
Scientific Data
NetCDF Files
nccreate Create variable in NetCDF file ncdisp Display contents of NetCDF data source in Command Window ncinfo Return information about NetCDF data source ncread Read data from variable in NetCDF data source ncreadatt Read attribute value from NetCDF data source ncwrite Write data to NetCDF file ncwriteatt Write attribute to NetCDF file ncwriteschema Add NetCDF schema definitions to NetCDF file
HDF5 Files
High-Level Functions
h5create Create HDF5 data set h5disp Display contents of HDF5 file h5info Return information about HDF5 file h5read Read data from HDF5 data set h5readatt Read attribute from HDF5 file h5write Write to HDF5 data set h5writeatt Write HDF5 attribute
HDF4 Files
High-Level Functions
hdfinfo Information about HDF4 or HDF-EOS file hdfread Read data from HDF4 or HDF-EOS file hdftool Browse and import data from HDF4 or HDF-EOS files imread Read image from graphics file imwrite Write image to graphics file
Low-Level Functions
hdfan Gateway to HDF multifile annotation (AN) interface hdfhx Gateway to HDF external data (HX) interface hdfh Gateway to HDF H interface hdfhd Gateway to HDF HD interface hdfhe Gateway to HDF HE interface hdfml Utilities for working with MATLAB HDF gateway functions hdfpt Interface to HDF-EOS Point object hdfv Gateway to HDF Vgroup (V) interface hdfvf Gateway to VF functions in HDF Vdata interface hdfvh Gateway to VH functions in HDF Vdata interface hdfvs Gateway to VS functions in HDF Vdata interface hdfdf24 Gateway to HDF 24-bit raster image (DF24) interface hdfdfr8 Gateway to HDF 8-bit raster image (DFR8) interface
Band-Interleaved Files
multibandread Read band-interleaved data from binary file multibandwrite Write band-interleaved data to file
Common Data Format
cdfinfo Information about Common Data Format (CDF) file cdfread Read data from Common Data Format (CDF) file cdfepoch Convert date string or serial date number to CDF formatted dates todatenum Convert CDF epoch object to MATLAB serial date number
Audio and Video
Audio Recording and Playback
audiodevinfo Information about audio device audioplayer Create object for playing audio audiorecorder Create object for recording audio sound Convert matrix of signal data to sound soundsc Scale data and play as sound
Audio Utilities
beep Produce operating system beep sound lin2mu Convert linear audio signal to mu-law mu2lin Convert mu-law audio signal to linear
XML Documents
xmlread Read XML document and return Document Object Model node xmlwrite Write XML Document Object Model node xslt Transform XML document using XSLT engine
Workspace Variables and MAT-Files
clear Remove items from workspace, freeing up system memory clearvars Clear variables from memory disp Display value of variable openvar Open workspace variable in Variables editor or other graphical editing tool who List variables in workspace whos List variables in workspace, with sizes and types load Load variables from file into workspace save Save workspace variables to file matfile Access and change variables directly in MAT-files, without loading into memory
Low-Level File I/O
fclose Close one or all open files feof Test for end-of-file ferror Information about file I/O errors fgetl Read line from file, removing newline characters fgets Read line from file, keeping newline characters fileread Read contents of file into string fopen Open file, or obtain information about open files fprintf Write data to text file fread Read data from binary file frewind Move file position indicator to beginning of open file fscanf Read data from text file fseek Move to specified position in file ftell Position in open file fwrite Write data to binary file
Large Files and Big Data
Datastore
datastore Create datastore to access collection of data KeyValueDatastore Datastore for key-value pair data TabularTextDatastore Datastore for collections of tabular text files
MapReduce
mapreduce Programming technique for analyzing data sets that do not fit in memory datastore Create datastore to access collection of data mapreducer Define execution environment for mapreduce KeyValueStore Store key-value pairs for use with mapreduce ValueIterator An iterator over intermediate values for use with mapreduce
Large MAT-Files
matfile Access and change variables directly in MAT-files, without loading into memory
Memory Mapping
memmapfile Create memory map to a file
TCP/IP Communication
tcpclient Create TCP/IP client object to communicate over TCP/IP read Read data from remote host over TCP/IP interface write Write data to remote host over TCP/IP interface
Web Access
web Open Web page or file in browser webread Read content from RESTful web service webwrite Write data to RESTful web service websave Save content from RESTful web service to file weboptions Specify parameters for RESTful web service sendmail Send email message to address list ftp Connect to FTP server
GUI Building
GUI Building Basics
guide Open GUIDE inspect Open Property Inspector
Component Selection
UI Controls and Indicators
figure Create figure window axes Create axes graphics object uicontrol Create user interface control object uitable Create table UI component uipanel Create panel container object uibuttongroup Create button group to manage radio buttons and toggle buttons uitab Create tabbed panel uitabgroup Create container for tabbed panels actxcontrol Create Microsoft ActiveX control in figure window
uimenu Create menus and menu items on figure windows uicontextmenu Create context menu uitoolbar Create toolbar on figure uipushtool Create push button on toolbar uitoggletool Create toggle button on toolbar
Predefined Dialog Boxes
errordlg Create error dialog box warndlg Create warning dialog box msgbox Create message dialog box helpdlg Create help dialog box waitbar Open or update wait bar dialog box questdlg Create question dialog box inputdlg Create dialog box that gathers user input listdlg Create list-selection dialog box uisetcolor Open color selection dialog box uisetfont Open font selection dialog box export2wsdlg Create dialog box for exporting variables to workspace uigetfile Open file selection dialog box uiputfile Open dialog box for saving files uigetdir Open folder selection dialog box uiopen Open dialog box for selecting file to load into workspace uisave Open dialog box for saving variables to MAT-file printdlg Open figure Print dialog box printpreview Open figure Print Preview dialog box exportsetupdlg Open figure Export Setup dialog box dialog Create empty modal dialog box uigetpref Conditionally open dialog box according to user preference menu Create multiple-choice dialog box
Component Layout
align Align user interface controls (uicontrols) and axes movegui Move UI figure to specified location on screen getpixelposition Get component position in pixels setpixelposition Set component position in pixels listfonts List available system fonts textwrap Wrapped string matrix for given uicontrol uistack Reorder visual stacking order of objects
Coding UI Behavior
uiwait Block program execution and wait to resume uiresume Resume execution of blocked program waitfor Block execution and wait for condition waitforbuttonpress Wait for key press or mouse-button click getappdata Retrieve application-defined data setappdata Store application-defined data isappdata True if application-defined data exists rmappdata Remove application-defined data guidata Store or retrieve UI data guihandles Create structure of handles closereq Default figure close request function uisetpref Manage preferences used in uigetpref
Packaging UIs as Apps
matlab.apputil.create Create or modify app project file for packaging app into .mlappinstall file using interactive dialog box matlab.apputil.package Package app files into .mlappinstall file matlab.apputil.install Install app from a .mlappinstall file matlab.apputil.run Run app programmatically matlab.apputil.getInstalledAppInfo List installed app information matlab.apputil.uninstall Uninstall app
Object-Oriented Programming
Sample Classes
classdef Class definition keywords
Class Syntax Fundamentals
class Determine class of object isa Determine if input is object of specified class isequal Determine array equality isobject Determine if input is MATLAB object enumeration Display class enumeration members and names events Event names methods Class method names properties Class property names
MATLAB Class Editing
edit Edit or create file clear Remove items from workspace, freeing up system memory matlab.lang.ObjectUpdateFailure Class representing objects that cannot be updated to new class definition
MATLAB Class Definition
Class Definition and Organization
classdef Class definition keywords import Add package or class to current import list
Properties
properties Class property names isprop Determine if property of object dynamicprops Abstract class used to derive handle class with dynamic properties meta.DynamicProperty Describe dynamic property of MATLAB object meta.property Describe property of MATLAB class
Methods
methods Class method names ismethod Determine if method of object meta.method Describe method of MATLAB class
Handle Classes
handle Abstract class for deriving handle classes matlab.mixin.SetGet Abstract class used to derive handle classes with set and get methods hgsetget Abstract class used to derive handle class with set and get methods dynamicprops Abstract class used to derive handle class with dynamic properties matlab.mixin.Copyable Superclass providing copy functionality for handle objects handle.addlistener Create event listener handle.delete Delete handle object handle.findobj Find handle objects isa Determine if input is object of specified class handle.isvalid Determine valid handles handle.findprop Find meta.property object handle.relationaloperators Determine equality or sort handle objects
Events
events Event names handle.notify Notify listeners that event is occurring handle.addlistener Create event listener event.EventData Base class for all data objects passed to event listeners event.listener Class defining listener objects event.PropertyEvent Data for property events event.proplistener Define listener object for property events
Object Arrays
empty Create empty array matlab.mixin.Heterogeneous Superclass for heterogeneous array formation
Class Hierarchies
Subclass Definition
superclasses Superclass names superiorto Establish superior class relationship inferiorto Specify inferior class relationship matlab.mixin.Heterogeneous Superclass for heterogeneous array formation
Subclass Applications
double Convert to double precision horzcat Concatenate arrays horizontally Vertcat Concatenate arrays vertically numel Number of array elements size Array dimensions subsref Redefine subscripted reference for objects
Enumerations
enumeration Display class enumeration members and names isenum Determine if variable is enumeration meta.EnumeratedValue Describe enumeration member of MATLAB class
save Save workspace variables to file load Load variables from file into workspace saveobj Modify save process for object loadobj Modify load process for object
Customize MATLAB Behavior
cat Concatenate arrays along specified dimension horzcat Concatenate arrays horizontally vertcat Concatenate arrays vertically empty Create empty array disp Display value of variable display Display text and numeric expressions numel Number of array elements size Array dimensions end Terminate block of code, or indicate last array index subsref Redefine subscripted reference for objects subsasgn Subscripted assignment subsindex Subscript indexing with object substruct Create structure argument for subsasgn or subsref
Custom Object Display
disp Display value of variable display Display text and numeric expressions details Display array details matlab.mixin.CustomDisplay Display customization interface class matlab.mixin.util.PropertyGroup Custom property list for object display
Getting Information About Classes and Objects
metaclass Obtain meta.class object meta.abstractDetails Find abstract methods and properties meta.class.fromName Return meta.class object associated with named class meta.package.fromName Return meta.package object for specified package meta.package.getAllPackages Get all top-level packages properties Class property names methods Class method names events Event names superclasses Superclass names meta.class Describe MATLAB class meta.property Describe property of MATLAB class meta.method Describe method of MATLAB class meta.event Describe event of MATLAB class meta.package Describe MATLAB package meta.DynamicProperty Describe dynamic property of MATLAB object meta.EnumeratedValue Describe enumeration member of MATLAB class meta.MetaData Superclass for MATLAB object metadata
Calling External Functions
Call MEX-File Functions
mexext Binary MEX-file-name extension inmem Names of functions, MEX-files, classes in memory
Call C Shared Libraries
loadlibrary Load C/C++ shared library into MATLAB unloadlibrary Unload shared library from memory libisloaded Determine if shared library is loaded calllib Call function in shared library libfunctions Return information on functions in shared library libfunctionsview Display shared library function signatures in window libstruct Convert MATLAB structure to C-style structure for use with shared library libpointer Pointer object for use with shared library lib.pointer Pointer object compatible with C pointer
Call Java Libraries
javaArray Construct Java array object javaclasspath Return Java class path or specify dynamic path javaaddpath Add entries to dynamic Java class path javarmpath Remove entries from dynamic Java class path javachk Error message based on Java feature support isjava Determine if input is Java object usejava Determine if Java feature is available javaMethod Call Java method javaMethodEDT Call Java method from Event Dispatch Thread (EDT) javaObject Call Java constructor javaObjectEDT Call Java constructor on Event Dispatch Thread (EDT) cell Create cell array class Determine class of object clear Remove items from workspace, freeing up system memory depfun List dependencies of function or P-file exist Check existence of variable, function, folder, or class fieldnames Field names of structure, or public fields of object im2java Convert image to Java image import Add package or class to current import list inmem Names of functions, MEX-files, classes in memory inspect Open Property Inspector isa Determine if input is object of specified class methods Class method names methodsview View class methods which Locate functions and files matlab.exception.JavaException Capture error information for Java exception
Call .NET Libraries
Getting Started
NET.addAssembly Make .NET assembly visible to MATLAB NET.isNETSupported Check for supported Microsoft .NET Framework NET Summary of functions in MATLAB .NET interface enableNETfromNetworkDrive Enable access to .NET commands from network drive NET.Assembly Members of .NET assembly NET.NetException Capture error information for .NET exception
Data Types
NET.createArray Array for nonprimitive .NET types cell Create cell array NET.disableAutoRelease Lock .NET object representing a RunTime Callable Wrapper (COM Wrapper) so that MATLAB does not release COM object NET.enableAutoRelease Unlock .NET object representing a RunTime Callable Wrapper (COM Wrapper) so that MATLAB releases COM object
Properties
NET.setStaticProperty Static property or field name
Events and Delegates
BeginInvoke Initiate asynchronous .NET delegate call EndInvoke Retrieve result of asynchronous call initiated by .NET System.Delegate BeginInvoke method Combine Convenience function for static .NET System.Delegate Combine method Remove Convenience function for static .NET System.Delegate Remove method RemoveAll Convenience function for static .NET System.Delegate RemoveAll method
Enumerations
bitand Bit-wise AND bitor Bit-wise OR bitxor Bit-wise XOR bitnot .NET enumeration object bit-wise NOT instance method
Generic Classes
NET.convertArray Convert numeric MATLAB array to .NET array NET.createGeneric Create instance of specialized .NET generic type NET.invokeGenericMethod Invoke generic method of object NET.GenericClass Represent parameterized generic type definitions
Call COM Objects
actxserver Create COM server actxcontrol Create Microsoft ActiveX control in figure window actxcontrollist List currently installed Microsoft ActiveX controls actxcontrolselect Create Microsoft ActiveX control from UI actxGetRunningServer Handle to running instance of Automation server iscom Determine whether input is COM or ActiveX object isprop Determine whether input is COM object property get Get property value from interface, or display properties set Set object or interface property to specified value addproperty Add custom property to COM object deleteproperty Remove custom property from COM object inspect Open Property Inspector propedit Open built-in property page for control fieldnames Field names of structure, or public fields of object ismethod Determine whether input is COM object method methods Class method names methodsview View class methods invoke Invoke method on COM object or interface, or display methods isevent Determine whether input is COM object event events List of events COM object can trigger eventlisteners List event handler functions associated with COM object events registerevent Associate event handler for COM object event at run time unregisterallevents Unregister all event handlers associated with COM object events at run time unregisterevent Unregister event handler associated with COM object event at run time isinterface Determine whether input is COM interface interfaces List custom interfaces exposed by COM server object release Release COM interface delete Remove COM control or server move Move or resize control in parent window load Initialize control object from file save Serialize control object to file
Call Python Libraries
Getting Started
pyversion Change default version of Python interpreter matlab.exception.PyException Capture error information for Python exception
Data Types
pyargs Create keyword argument for Python function
Call WSDL Web Service
matlab.wsdl.createWSDLClient Create interface to SOAP-based Web service matlab.wsdl.setWSDLToolPath Location of WSDL tools createClassFromWsdl Create MATLAB class based on WSDL document createSoapMessage Create SOAP (Simple Object Access Protocol) message to send to server callSoapService Send SOAP (Simple Object Access Protocol) message to endpoint parseSoapResponse Convert response string from SOAP (Simple Object Access Protocol) server into MATLAB types
Toolbox Distribution
builddocsearchdb Build searchable documentation database
Exception Handling
try, catch Execute statements and catch resulting errors MException Capture error information addCause Record additional causes of exception getReport Get error message for exception last Return last uncaught exception rethrow Rethrow previously caught exception throw Throw exception throwAsCaller Throw exception as if occurs within calling function
Unit Testing Framework
Write Unit Tests
functiontests Create array of tests from handles to local functions matlab.unittest.TestCase Superclass of all matlab.unittest test classes matlab.unittest.Verbosity Verbosity level enumeration class
Run Unit Tests
runtests Run set of tests matlab.unittest.TestCase.run Run TestCase test matlab.unittest.TestSuite.run Run TestSuite array using TestRunner object configured for text output matlab.unittest.TestRunner.run Run all tests in TestSuite array matlab.unittest.TestRunner.addPlugin Add plugin to TestRunner object matlab.unittest.TestSuite Class for grouping tests to run matlab.unittest.Test Specification of a single test method matlab.unittest.TestRunner Class for running tests in matlab.unittest framework
Analyze Test Results
matlab.unittest.TestResult Result of running test suite
Performance and Memory
Code Performance
bench MATLAB benchmark cputime Elapsed CPU time memory Display memory information profile Profile execution time for function profsave Save profile report in HTML format tic Start stopwatch timer timeit Measure time required to run function toc Read elapsed time from stopwatch
Memory Usage
clear Remove items from workspace, freeing up system memory inmem Names of functions, MEX-files, classes in memory memory Display memory information pack Consolidate workspace memory whos List variables in workspace, with sizes and types
System Commands
clipboard Copy and paste strings to and from system clipboard computer Information about computer on which MATLAB software is running dos Execute DOS command and return output getenv Environment variable perl Call Perl script using appropriate operating system executable setenv Set environment variable system Execute operating system command and return output unix Execute UNIX command and return output winqueryreg Item from Windows registry commandhistory Open Command History window, or select it if already open commandwindow Open Command Window, or select it if already open filebrowser Open Current Folder browser, or select it if already open workspace Open Workspace browser to manage workspace matlab.io.saveVariablesToScript Save workspace variables to MATLAB script getpref Preference setpref Set preference addpref Add preference rmpref Remove preference ispref Test for existence of preference
MATLAB API for Other Languages
MATLAB Engine API for C, C++, and Fortran
mex Build MEX-function from C/C++ or Fortran source code
MATLAB COM Automation Server
Execute Execute MATLAB command in Automation server Feval Evaluate MATLAB function in Automation server GetCharArray Character array from Automation server PutCharArray Store character array in Automation server GetFullMatrix Matrix from Automation server workspace PutFullMatrix Matrix in Automation server workspace GetVariable Data from variable in Automation server workspace GetWorkspaceData Data from Automation server workspace PutWorkspaceData Data in Automation server workspace MaximizeCommandWindow Open Automation server window MinimizeCommandWindow Minimize size of Automation server window Quit Terminate MATLAB Automation server enableservice Enable, disable, or report status of MATLAB Automation server
MATLAB Engine for Python
matlab.engine.start_matlab Start MATLAB Engine for Python matlab.engine.MatlabEngine Python object using MATLAB as computational engine within Python session matlab.engine.FutureResult Results of asynchronous call to MATLAB function stored in Python object
MEX-File Creation API
Executable C/C++ MEX-Files
mex Build MEX-function from C/C++ or Fortran source code dbmex Enable MEX-file debugging (on UNIX platforms) mex.getCompilerConfigurations Get compiler configuration information for building MEX-files
Call MEX-File Functions
mexext Binary MEX-file-name extension inmem Names of functions, MEX-files, classes in memory
Share MEX-Files
ver Version information for MathWorks products computer Information about computer on which MATLAB software is running mex.getCompilerConfigurations Get compiler configuration information for building MEX-files mexext Binary MEX-file-name extension
Troubleshoot MEX-Files
dbmex Enable MEX-file debugging (on UNIX platforms) inmem Names of functions, MEX-files, classes in memory mex Build MEX-function from C/C++ or Fortran source code mex.getCompilerConfigurations Get compiler configuration information for building MEX-files mexext Binary MEX-file-name extension
Desktop Environment
Startup and Shutdown
matlab (Windows) Start MATLAB program from Windows system prompt matlab (Mac) Start MATLAB program from Mac Terminal matlab (Linux) Start MATLAB program from Linux system prompt exit Terminate MATLAB program (same as quit) quit Terminate MATLAB program matlabrc Startup file for MATLAB program startup Startup file for user-defined options finish Termination file for MATLAB program
Basic Settings
prefdir Folder containing preferences, history, and layout files preferences Open Preferences dialog box | 14,718 | 79,136 | {"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-2015-18 | longest | en | 0.54366 |
https://web2.0calc.com/questions/please-help_37993 | 1,600,515,880,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191780.21/warc/CC-MAIN-20200919110805-20200919140805-00699.warc.gz | 715,966,236 | 6,508 | +0
0
76
1
I have a magical machine that can scan a 3D object and determine whether the numerical value (in units) of its surface area is equal to the numerical value (in units) of its volume. After experimenting with different objects in my lab, the only thing that the machine gives an output of is my spherical ball.
What is the radius of my spherical ball (in units)?
Jun 22, 2020
#1
+902
-1
So essentially
4πr^2 = 4π (r^3/3)
r^2 = r^3 /3
r^2 *3 = r^3
3 = r
Feel free to ask if you don't understand anything!
Jun 22, 2020 | 165 | 536 | {"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} | 3.5 | 4 | CC-MAIN-2020-40 | longest | en | 0.913086 |
https://calculus-do.com/derivatives-theory-dai-xie-afin546-options/ | 1,716,262,796,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00892.warc.gz | 122,795,260 | 21,350 | # 微积分网课代修|导数代写Derivatives theory代考|AFIN546 OPTIONS
• 单变量微积分
• 多变量微积分
• 傅里叶级数
• 黎曼积分
• ODE
• 微分学
## 微积分网课代修导数代写Derivatives theory代考|OPTIONS
Options are a little more difficult to understand than forwards and futures and here we present a quick introductory overview. While futures markets in commodities have existed since the middle of the $1800 \mathrm{~s}$, options contracts have been traded for a shorter period of time. There are two types of option, calls and puts:
The holder of a call (put) option has the right (but not an obligation) to buy (sell) the ‘underlying asset’ at some time in the future (‘maturity date’) at a known fixed price (the ‘strike price’, $K$ ) but she does not have to exercise this right.
For the moment we consider stock option contracts, so the underlying asset in the option contract is the stock of a particular company-XYZ which is traded on the NYSE. The option contract itself, we assume is traded in Chicago.
Above we noted that the holder of a long futures contract on a stock-XYZ commits herself to buy the stock at a certain price at a certain time in the future and if she does nothing before the maturity date, she will have to take delivery of stock-XYZ, at the pre-agreed futures price. In contrast, the holder of a (European) ‘call option’ on stock-XYZ can decide whether to pay the known strike price and take delivery of stock-XYZ on the maturity date of the option contract – this is called ‘exercising the option contract by taking delivery’. If it is advantageous not to exercise the option (in Chicago) then the holder of the call option will simply do nothing. For the privilege of being able to decide whether or not to take delivery of stock-XYZ (at maturity of the option contract) the buyer of the call option must pay an upfront, non-refundable fee – the option price (or premium).
## 微积分网课代修导数代写Derivatives theory代考|Call Options
If today you buy a European call option and pay the call premium/price, then this gives you the right (but not an obligation):
to purchase the underlying asset
at a designated delivery point
on a specified future date (known as the expiration or maturity date)
for a fixed known price (the exercise or strike price)
and in an amount (contract size) which is fixed in advance.For the moment, think of a call option as a ‘piece of paper’ that contains the contract details (e.g. strike price, maturity date, amount, delivery point, type of underlying asset). You can purchase this contract today in the options market in Chicago if you pay the quoted call premium. There are always two sides to every trade – a buyer and a seller – but we will concentrate on your trade, as a buyer of the option. Note that all transactions in the option contract
are undertaken in Chicago but the underlying asset, for example a stock, is traded on another exchange (e.g. NYSE).
Suppose the current price of stock-XYZ on the NYSE on $15 \mathrm{July}$ is $S_{0}=\$ 80$. On 15 July you can pay the call premium$C=\$3$ and buy (in Chicago) an October-European call option on the stock-XYZ. The strike price in the contract is $K=\$ 80,{ }^{4}$and the expiry date$T$is in just over 3 months’ time on 25 October. Because the maturity of the call is in October, and the strike is$K=\$80$, it is known as the ‘October- 80 call’ (Table 1.4). Assume each call option is for delivery of one stock of XYZ. | 841 | 3,382 | {"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} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.894362 |
https://community.shopify.com/c/Technical-Q-A/Calculate-the-percentage-value-from-the-total-cart-price-Debut/m-p/887377 | 1,604,153,505,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107918164.98/warc/CC-MAIN-20201031121940-20201031151940-00414.warc.gz | 237,293,922 | 38,539 | cancel
Showing results for
Did you mean:
## Calculate the percentage value from the total cart - price [Debut Theme]
Solved
Highlighted
New Member
2 0 1
Hey everyone,
I would like to calculate the percentage value from the total, calculate out of here:
{{ cart.total_price | money_with_currency }}
For example, if the total value of the shopping cart is 100 €, 5% (always the same) should be calculated. So the answer is: 5 €
I have only slight knowledge but it did not work with cart.total_price * 5/100.
Mike
Highlighted
Excursionist
80 19 19
This is an accepted solution.
Hi Mike
I'd imagine it'd be something like:
{{ cart.total_price | divided_by: 100 | times: 5 | money_with_currency }}
The divided_by 100 will get you 1%.
You then times: by 5 to get 5%.
Then the money_with_currency filter displays it in the monetary format you'd expect?
Highlighted
New Member
2 0 1
Hi,
thanks for your help and anser.
It works like you described it, great. Exactly for money_with_currency is the corresponding currency (EURO, USD and others)
Best regards,
Mike
Latest Topics
Latest Posts
• ## How to increase Button Label size in "Image with text overlay" in debut theme?
Most Liked Authors | 299 | 1,209 | {"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} | 2.625 | 3 | CC-MAIN-2020-45 | latest | en | 0.881905 |
https://www.transtutors.com/questions/write-scheme-functions-for-the-following-making-sure-they-run-on-dr-racket-advanced-2782333.htm | 1,604,140,580,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00313.warc.gz | 913,124,189 | 14,535 | # -Write scheme functions for the following making sure they run on Dr Racket “Advanced Student?...
-Write scheme functions for the following making sure they run on Dr Racket “Advanced Student”.
-Built-in functions in Dr Racket, with the exception of modulo and quotient, can not be used.
-Do not use other languages such as Java, C, etc..
1) combine3 takes three positive integers and return an integer which is made of the last two digits of the first integer, the first and last digits of the second integer and the first two digits of the last integer. If any of the integers has less than two digits return 0. For example:
( combine 4 124 80534 ) will return 0
( combine3 247939 968634 40348726 ) will return the integer 399440 and
( combine3 5678 456 9264 ) will return 784692.
If any of your numbers has less than two digits, return 0.
Note that you probably need to define several auxiliary functions.
2) inc-by-n takes an integer x and a digit d and returns an integer where each digit in x has been incremented by n. If the result of the incrementation would exceed 9, the result mod 10 will be used. For example:
(inc-by-n 53462 3) will return 86795 and
(inc-by-n 682901 4) will return 026345
Note that if the first digit of the answer is “0” it will not print.
3) countnumber takes a list of integers L (with possible nested sublists) and an integer n and count the number of instances of n in the list. For example:
( countnumber ‘( 5 7 34 28 5 72 ( 5 ) ) 5) returns 3 and
( countnumber ‘( 3 ( 67 34 ( 29 56) 23 ) 56 ( 5 ( 45 34 56 ) 0 ) ) 34 ) returns 2
4) mapeven takes a function and a list and return a list similar to the list argument where the function has been applied to every other element starting with the second one. For example:
( mapevensquare ‘( 3 6 22 9 5 7 10 15 ) ) will return ( 3 36 22 81 5 49 10 225 )
( mapeven even? ‘( 3 6 22 9 5 7 10 15 ) ) will return ( 3 #t 22 #f 5 #f 10 #f )
( mapeven list? ‘( 2 ( 3 4 ) 6 10 ( 88 23 ) ( 5 ) )) will return ( 2 #t 6 #f ( 88 23 ) #t)
Note that the function applied only to the top elements of the list. Make sure that your function works with any relevant function, and that it works with lists of even and odd length.
5) addTheSquares takes a list of integer and returns the sum of all the squares. For example:
( addTheSquares ‘( 6 3 9 2 6 )) will return 166 ( i.e. 36 + 9 + 81 + 4 + 36 )
6) Write a tail recursive version of #5 makeInsertInList takes an argument x and returns a function which will have a list argument and will insert x after each element of the list.
## Plagiarism Checker
Submit your documents and get free Plagiarism report
Free Plagiarism Checker
## Recent Questions in Programming Languages
Submit Your Questions Here !
Copy and paste your question here...
Attach Files | 802 | 2,799 | {"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} | 3.5 | 4 | CC-MAIN-2020-45 | latest | en | 0.749276 |
http://www.chemicool.com/chemtalk/showthread.php?p=294&mode=threaded | 1,369,038,564,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698646863/warc/CC-MAIN-20130516100406-00098-ip-10-60-113-184.ec2.internal.warc.gz | 378,593,568 | 8,768 | Calculate the mass of each product
FAQ Calendar Search Today's Posts Mark Forums Read
#1
April 29th, 2005, 06:27
Halonothing Junior Member Join Date: Apr 2005 Location: Salt Lake City, UT Posts: 1
Calculate the mass of each product
Alright i have a dumb question, so i apologize in advance. I'm doing my chem homework, and one of the questions that my prof. wrote, i dont quite grasp.
. "For the following unbalanced equation calculate the mass of each product if 1.55 gram of CS2 is completely reacted:
CS2 + O2 ---> CO2 + SO2"
Atomic masses:
C = 12.011
S = 32.06
O = 16
so i balance the equation.
CS2 + 2(O2) ---> CO2 + (SO)2
now, I check to see how many moles correspond to 1.55 grams of CS2
=
.00828 mol CS2
then
( 2 mol O2 / .00828 mol CS2 ) = 241.5458 . . .
now what? i've been doing this assignment for way way too long. its 5:24 a.m someone save me pleeeeeeeease. what's really throwing me off is the 1.55 grams of CS2. and how that relates to the mass of the rest of the products. :!: | 303 | 1,006 | {"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-2013-20 | latest | en | 0.906313 |
https://www.coursehero.com/file/6137032/ans-to-prob1/ | 1,521,803,898,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648207.96/warc/CC-MAIN-20180323102828-20180323122828-00428.warc.gz | 763,753,093 | 112,116 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
ans to prob1
# ans to prob1 - Suggested Answer to ECON 346 Problem Set...
This preview shows pages 1–3. Sign up to view the full content.
1 Suggested Answer to ECON 346 Problem Set 1 #1 (a) - (b) P ( X i ) Probability; CD ( X i ) Cumulative distribution Mean: = = = 8 1 ) ( ] [ i i i i X P X X E μ Median: the middle point of the data, that is, the point at the cumulative distribution = 0.5 Variance: = - = - = 8 1 2 2 2 ) ( ) ( ] ) [( i i i i X P X X E μ μ σ Standard deviation: 01836 . 0 000337 . 0 2 = = = σ σ Table 1 X i P ( X i ) CD ( X i ) X i · P ( X i ) ( Xi - μ ) 2 ( Xi - μ ) 2 · P ( X i ) 0.05 0.04 0.04 0.002 0.00142884 0.00005715 0.06 0.06 0.10 0.0036 0.00077284 0.00004637 0.07 0.16 0.26 0.0112 0.00031684 0.00005069 0.08 0.22 0.48 0.0176 0.000061 0.00001338 0.09 0.14 0.62 0.0126 0.000005 0.00000068 0.10 0.18 0.80 0.018 0.00014884 0.00002679 0.11 0.12 0.92 0.0132 0.00049284 0.00005914 0.12 0.08 1.00 0.0096 0.00103684 0.00008295 Summation 1.00 0.0878 ( Mean ) 0.00426272 0.000337 ( Variance ) (c) Table 2 X i X i + 100 X i · 100 P ( X i ) X i · P ( X i ) [ X i + 100 ] · P ( X i ) [ X i · 100 ] · P ( X i ) 0.05 100.05 5 0.04 0.002 4.002 0.2 0.06 100.06 6 0.06 0.0036 6.0036 0.36 0.07 100.07 7 0.16 0.0112 16.0112 1.12 0.08 100.08 8 0.22 0.0176 22.0176 1.76 0.09 100.09 9 0.14 0.0126 14.0126 1.26 0.1 100.1 10 0.18 0.018 18.018 1.8 0.11 100.11 11 0.12 0.0132 12.0132 1.32 0.12 100.12 12 0.08 0.0096 8.0096 0.96 Summation 1.00 0.0878 ( μ Xi ) 100.0878 ( μ Xi+100 ) 8.78 ( μ Xi*100 ) According to Table 2 above, we can find: 0878 . 0 ) ( = = Xi i X E μ 0878 . 100 ) 100 ( 100 = = + + Xi i X E μ 78 . 8 ) 100 * ( 100 * = = Xi i X E μ
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2 So, we can prove the following relationship: ) ( 100 0878 . 0 100 0878 . 100 ) 100 ( i i X E X E + = + = = + ) ( * 100 0878 . 0 * 100 78 . 8 ) 100 * ( i i X E X E = = = (d) Table 3 X i P ( X i ) X i + 100 X i · 100 ( Xi+100 - μ Xi+100 ) 2 [( Xi+100 - μ Xi+100 ) 2 ] · P ( X i ) ( Xi · 100 - μ Xi · 100 ) 2 [( Xi · 100 - μ Xi · 100 ) 2 ] · P ( X i ) 0.05 0.04 100.05 5 0.00142884 0.000057 14.2884 0.571536 0.06 0.06 100.06 6 0.00077284 0.000046 7.7284 0.463704 0.07 0.16 100.07 7 0.00031684 0.000051 3.1684 0.506944 0.08 0.22 100.08 8 0.000061 0.000013 0.6084 0.133848 0.09 0.14 100.09 9 0.000005 0.000001 0.0484 0.006776 0.1 0.18 100.1 10 0.00014884 0.000027 1.4884 0.267912 0.11 0.12 100.11 11 0.00049284 0.000059 4.9284 0.591408 0.12 0.08 100.12 12 0.00103684 0.000083 10.3684 0.829472 Summation 1.00 0.000337 ( Variance ) 3.37 (
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 6
ans to prob1 - Suggested Answer to ECON 346 Problem Set...
This preview shows document pages 1 - 3. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 1,443 | 2,959 | {"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-2018-13 | latest | en | 0.428076 |
https://www.numbersaplenty.com/268318396 | 1,718,722,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00390.warc.gz | 826,026,946 | 2,938 | Search a number
268318396 = 2267079599
BaseRepresentation
bin11111111111000…
…11011010111100
3200200212222122101
433333203122330
51022142142041
642342554444
76435445243
oct1777433274
9620788571
10268318396
11128505645
1275a38a24
13437875a8
14278c795a
1518851b31
hexffe36bc
268318396 has 6 divisors (see below), whose sum is σ = 469557200. Its totient is φ = 134159196.
The previous prime is 268318363. The next prime is 268318403. The reversal of 268318396 is 693813862.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 33539796 + ... + 33539803.
Almost surely, 2268318396 is an apocalyptic number.
It is an amenable number.
268318396 is a deficient number, since it is larger than the sum of its proper divisors (201238804).
268318396 is a wasteful number, since it uses less digits than its factorization.
268318396 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 67079603 (or 67079601 counting only the distinct ones).
The product of its digits is 373248, while the sum is 46.
The square root of 268318396 is about 16380.4272227558. The cubic root of 268318396 is about 644.9857946113.
The spelling of 268318396 in words is "two hundred sixty-eight million, three hundred eighteen thousand, three hundred ninety-six".
Divisors: 1 2 4 67079599 134159198 268318396 | 433 | 1,424 | {"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} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.800579 |
http://tech.bragboy.com/2010/05/print-matrix-in-circular-path-interview.html | 1,721,881,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00575.warc.gz | 26,132,167 | 12,408 | ## 11 May 2010
### Print Matrix in a circular path (Interview question)
Hi friends,
This is one of the interview questions I faced recently and thought I would share it.
Question:
Given a matrix print all its value in a circular fashion like shown in the image below.
Solution:
To attack this problem we definitely want to have a left-right, top-bottom, right-left and bottom-top spans. For attaining this, we obviously want to have 4 loops. The C++ code is given below.
```//============================================================================
// Author : c++
// Description : output given matrix elements circularly
//============================================================================
#include <iostream>
using namespace std;
#define HEIGHT 6
#define WIDTH 3
int main()
{
//int array[HEIGHT][WIDTH] = {{1,2,3,4,5},{16,17,18,19,6},{15,24,25,20,7},{14,23,22,21,8},{13,12,11,10,9}};
//int array[HEIGHT][WIDTH] = {{1,2,3,4,5,6},{14,15,16,17,18,7},{13,12,11,10,9,8}};
int array[HEIGHT][WIDTH] = {{1,2,3},{14,15,4},{13,16,5},{12,17,6},{11,18,7},{10,9,8}};
int maxh = HEIGHT-1, maxw = WIDTH-1;
int minh = 0, minw = 0;
while(1)
{
if(minh > maxh || minw > maxw)
break;
//top-left to top-right
for(int i=minh,j=minw;j<=maxw;j++)
cout << array[i][j] << endl;
minh++;
if(minh > maxh || minw > maxw)
break;
//top-right to bottom-right
for(int i=minh,j=maxw;i<=maxh;i++)
cout << array[i][j] << endl;
maxw--;
if(minh > maxh || minw > maxw)
break;
//bottom-right to bottom-left
for(int i=maxh,j=maxw;j>=minw;j--)
cout << array[i][j] << endl;
maxh--;
if(minh > maxh || minw > maxw)
break;
//bottom-left to top-left
for(int i=maxh,j=minw;i>=minh;i--)
cout << array[i][j] << endl;
minw++;
}
return 0;
}
//1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
```
Cheers!
nagarajan said...
I have seen this question in many places and have seen quite a lot of complicated solutions
But your way of approach rocks :-)
BragBoy said...
getting heat said...
simple & sweet solution liked it :)
BragBoy said...
@getting head, thanks for the visit ! | 639 | 2,065 | {"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-2024-30 | latest | en | 0.622527 |
https://www.physicsforums.com/threads/vector-addition.69894/ | 1,540,177,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514497.14/warc/CC-MAIN-20181022025852-20181022051352-00071.warc.gz | 1,024,489,624 | 13,220 | 1. Apr 3, 2005
### Naeem
Q. All four of the vectors below are added to vector E. Positive angles are measured counter-clockwise from the x axis.
A = 2 m î + 3 m ĵ
B = -7 m î -10 m ĵ
C = 3 m at 62 °
D = 8 m at -226 °
a) What is C·D?
C.D = |C| |D| cos theta
I did this.
3* cos 62 + 8 cos -226 , but the answer is wrong
b) Vector F is the cross product of vectors A and B (= A X B).
What is the z component of vector F?
The answer is one, but how
F = A X B
Find the cross product by evaluating the determinant of vectors A & B.
Then what!!
2. Apr 3, 2005
### quasar987
a) What you did is you substracted the x-component of each vectors. That's not the dot product. The dot product is the product of lenght of the two vectors times the cosine of the shortest angle between then.
b) According the right-hand rule or screw rule (or whatever rule you are comfortable with), the vector F has ONLY a z component. So all you got to do is find its lenght.
Last edited: Apr 4, 2005
3. Apr 3, 2005
### codyg1985
a) The formula $$\vec{c} \cdot \vec{d} = |\vec{c}| |\vec{d}| \cos \theta$$ takes the cosine of the angle between the two vectors and multiplies the result by the magnitudes of the two vectors.
The angle between the two vectors is (-226° + 360°) - 62° = 134° - 62° = 72° = $$\frac{2\pi}{5}$$
The magnitude is calculated by adding the square of the components and taking the square root of that sum:
$$|\vec{a}| = \sqrt{x^2 + y^2}$$
Last edited: Apr 3, 2005
4. Apr 4, 2005
### quasar987
But notice Naeem that for the vectors C and D the magnitudes are already given to you: 3m and 8m.
5. Apr 4, 2005
### Naeem
Got, it thanks! | 530 | 1,650 | {"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": 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} | 4.15625 | 4 | CC-MAIN-2018-43 | latest | en | 0.910976 |
https://www.physicsforums.com/threads/the-spectral-theorem-and-qm.242561/ | 1,480,854,665,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541321.31/warc/CC-MAIN-20161202170901-00417-ip-10-31-129-80.ec2.internal.warc.gz | 1,002,177,656 | 14,707 | # The spectral theorem and QM
1. Jun 29, 2008
### mrandersdk
On wiki http://en.wikipedia.org/wiki/Borel_functional_calculus in the paragraf 'Resolution of the identity' there is said
'In physics literature, using the above as heuristic, one passes to the case when the spectral measure is no longer discrete and write the resolution of identity as ... '
How is this made rigorous. I had had a course in C*-algebras and proven the spectral theorem for bounded operators, I know most of physical are unbounded, but there must be a connection?
How is it constructed such that
$$I = \int_{\sigma (I)} \lambda d E(\lambda)$$
makes sence. I guess somehow taking the inner product with a bra and a ket should get me something like
$$<\phi|T|\psi> = \int <\phi|x><x|T|\psi> dx$$
so comparing this with
$$<\phi|T|\psi> = <\phi| \int_{\sigma (I)} \lambda d E(\lambda) T |\psi>$$
how do I revieve the lebesgue measure, from the resolution of identity, and see that this is the same, if it even is. I hope it is clear what my problem is?
2. Jun 29, 2008
### Hurkyl
Staff Emeritus
You might find this reference handy. In particular, appendix B section 6.5 defines this integral to be computed 'pointwise':
THEOREM: A selfadjoint operator $A$ in a Hilbert space $\mathcal{H}$ possesses a unique spectral resolution $\{ E_\alpha \}$ such that
$$A = \int_{-\infty}^{+\infty} \alpha dE_\alpha,$$
meaning that for each $\psi$ in the domain $\mathcal{D}_A$ one has the convergent Stieltjes integral
$$(\psi, A\psi) = \int_{-\infty}^{+\infty} \alpha d(\psi, E_\alpha\psi).$$
3. Jun 29, 2008
### mrandersdk
This seems very interesting. They state a new form of the theorem, saying that you actually get a Stieltjes-Lebesgue integral when taking the iner product.
And as a example the make the spectral projections for X, so it makes sence. Have I understood it right.
Do you any refferences where they proof this form of the spectral theorem, and where they explaine why the spectral projection fx of X is
$$(E_{x_0}\psi)(x) = \psi(x)$$
$$d(\psi,E_{x}\psi)= ||\psi(x)||^2 dx$$
$$d(\phi,E_{x}\psi)= \phi^*(x) \psi(x) dx$$. | 627 | 2,125 | {"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} | 3.09375 | 3 | CC-MAIN-2016-50 | longest | en | 0.885172 |
https://www.apollonconnections.com/scientific-method-observation-hypothesis-experiment-conclusion/ | 1,563,897,077,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529480.89/warc/CC-MAIN-20190723151547-20190723173547-00393.warc.gz | 627,230,555 | 11,335 | The Scientific Method
Understading the Process
On the other hand, the scientific method is often iterative; New information, observations or ideas may cause the steps to be repeated again. The protocols of the scientific method can be divided into six steps: Question. Observation. Formulation of the hypothesis. Experimentation. Analysis of data. Reject or accept the hypothesis.
Feb 18, 2012 · Answers. Best Answer: A. Obesrvation, question, hypothesis, prediction, experiment, results, and conclusion. Observation is the first step of the scientific method. If nothing is observed, then no questions can be asked and no hypothesis formed,
The scientific method is an empirical method of acquiring knowledge that has characterized the development of science since at least the 17th century. It involves careful observation, applying rigorous skepticism about what is observed, given that cognitive assumptions can distort how one interprets the observation.It involves formulating hypotheses, via induction, based on such observations.
The approaches carry many names–hands-on science, experiential science, and constructing science through cooperative learning. But they all aim to teach students in the classroom and lab to engage in.
the student will be using the scientific method by first formulating a hypothesis, testing that hypothesis in a systematic way, recording observations, analyzing data and formulating a possible.
The Scientific Method Song by Have Fun Teaching teaches kids about Observation, Question, Hypothesis, Prediction, Experiment, Analyze Data, and Draw a Conclusion. This is a.
Throughout my scientific life, I have used a method that I knew or felt. even when the experiment was repeated. This was surprising, to say the least, and the p-value attached to the observation.
With this quintessential observation. as discerning scientists in the greatest scientific experiment on the planet. We have the opportunity to analyze and interpret the data collected over the.
Jan 11, 2011 · what are the 7 Steps of the scientific Method in the correct order? hypothesis, Experiment, analysis, research, conclusion, purpose/problem, results/data. Experiment Conclusion Data. Christ · 4 years ago. 1. Thumbs up. 1. Thumbs down. Report Abuse. observation hypothesis theory experiment constitution Law. Zachariahs Johnson · 3 years.
Fisher used “significance” merely to indicate that an observation was worth following up, with refutation of the null hypothesis justified only if further experiments. from his duty to form correct.
What is the correct order of the steps in the scientific method. A) Make a hypothesis, test the hypothesis, analyze the results, ask a question, draw conclusions, communicate results. B) Ask a question, make a hypothesis, test hypothesis, draw conclusions, analyze.
In summary, the scientific method includes the steps scientists use to solve a problem or to prove or disprove a theory. There are four basic steps involved with the scientific method. The usual steps include observation, hypothesis, experiment, and conclusion. The steps may not always be completed in the same order.
Steps of the Scientific Method Detailed Help for Each Step; Ask a Question: The scientific method starts when you ask a question about something that you observe: How, What, When, Who, Which, Why, or Where? For a science fair project some teachers require that the question be something you can measure, preferably with a number.
In the process, he manages to completely mangle how science operates and misrepresent the relationship between observation. to test that hypothesis. Now, bring on the quotes: Leading evolutionist.
Science Diet Sensitive Stomach Best Interdisciplinary Graduate Programs Dec 11, 2018. Many people find interdisciplinary graduate programs appealing. to help prospective students decide which program best meets their needs. The Knowlton School’s graduate and undergraduate programs in landscape architecture and the graduate program in architecture are ranked among the best in the nation by. design theory and practice, The
As an explanation for this observation, we suggested that EndoS-hydrolyzed IgG per se dominantly blocks local immune complex formation. “With new data from our own follow up experiments, we have now.
Science, broadly defined, means knowledge. Specifically we refer to science as knowledge ascertained by observation and experiment. another difficulty which is prevalent in all methods of deduction.
This is also the way that science is presented in peer reviewed journal articles – a study begins with a research question or hypothesis, followed by methods, results and conclusions. to be false.
The assertion that many published scientific. His observation is still true—indeed, a feather will always fall more slowly than a rock (in Earth’s atmosphere). Only his conclusion was wrong. This.
On the information level, this experiment serves. student learn to use the scientific method and to construct a hypothesis, to use a control, to identify variables, to gather and analyze data,
Scientific method consists of five steps: observation, hypothesis, experiment, conclusion and scientific theory. You must identify your problem when doing observation. Second you must gather as much information about the problem as possible. Third you want to form a hypothesis.
Vitamin D Molecular Formula This is the first question everyone asks after hearing about my recent love affair with CBD oil (or cannabidoil, which is a chemical compound extracted. “Cannabis is filled with antioxidants, like. The sunscreen category is one of the most researched in the beauty industry, with constant updates and changes in formulas and applications — we’ve
• The steps of the scientific method are to: • It is important for your experiment to be a fair test. A "fair test" occurs when you change only one factor (variable) and keep all other conditions the same. The scientific method is a process for experimentation that is used to.
The Scientific Method Powerpoint Presentation World’s Best PowerPoint Templates – CrystalGraphics offers more PowerPoint templates than anyone else in the world, with over 4 million to choose from. Winner of the Standing Ovation Award for “Best PowerPoint Templates” from Presentations Magazine. They’ll give your presentations a professional, memorable appearance – the kind of sophisticated look that today’s audiences expect. Uses
Scientists develop explanations using the scientific method. Repeat the experiment with different sized PV cells. Choose a range of different sized PV cells that is sufficient to give a valid.
What Is Observation In Scientific Method Steps The basic scientific method includes the steps scientists use and follow when trying to solve a. occur in any order, but the first step is usually observation. Well, in fact there’s a giant step waiting in third grade. but you can expect plenty of work using the scientific method and testing hypotheses about the physical
Here is a simple version of the scientific method: There are two key points here that are really difficult. The first is testing your hypothesis. the level of observation and research. Sometimes.
Oceanography Quizlet Chapter 14 Please review the FAQs and contact us if you find a problem. Credits: 1 Recommended: 9, 10 Prerequisite: Spanish 1 Test Prep: CLEP Spanish (You should finish all your years of Spanish study first before taking it.) Course Description: Students will grow in their ability to speak, write, read and listen and understand Spanish. Students
Whichever reasoning processes and research methods were used, the final conclusion is critical, determining success or failure. If an otherwise excellent experiment is summarized by a weak conclusion, the results will not be taken seriously. Success or failure is not a measure of whether a hypothesis is accepted or refuted, because both results still advance scientific knowledge.
Scientific Method * * * * * * * * * * * * * * * * * * * Steps in the Scientific Method Observation Hypothesis Experiment Data Collection Conclusion Retest Observations Gathered through your senses A scientist notices something in their natural world Observations An example of an observation might be noticing that many salamanders near a pond have curved, not straight, tails Hypothesis A.
Using the scientific method, you’ll learn which. to come up with your best guess—or hypothesis. The experiment you’ll be doing will require only a short amount of time to set up, but you’ll need to.
Conclusion, may lead to new questions, new hypothesis, or new experiments. After many experiments, scientists may summarize results in a natural law, which is a description of how nature behaves. Scientists may formulate a theory. A theory explains why nature behaves the.
In our last How to Science post, we talked about forming good research questions and applying that to Experiment. use to come to conclusions in our everyday lives. The problem, and the reason I’m.
Temperature probes (Campbell Scientific. experiment. Panel A shows the temperature history as measured using the borehole thermistor array at BH14. Depth-dependent temperature curves are shown for.
Molecule Is Smallest Particle Of Oceanography Quizlet Chapter 14 Please review the FAQs and contact us if you find a problem. Credits: 1 Recommended: 9, 10 Prerequisite: Spanish 1 Test Prep: CLEP Spanish (You should finish all your years of Spanish study first before taking it.) Course Description: Students will grow in their ability to speak, write, read and listen
Obviously, we do not have observations. a view of science which is easily inferred from the definitions of the scientific method most kids are taught in middle and high school classes. You start.
The history of scientific method considers changes in the methodology of scientific inquiry, as distinct from the history of science itself. The development of rules for scientific reasoning has not been straightforward; scientific method has been the subject of intense and recurring debate throughout the history of science, and eminent natural philosophers and scientists have argued for the.
use the scientific method to answer questions and provide explanations about natural phenomena. It is a logical process based on careful observation and experimentation. The scientific method begins with an observation that leads to a question about the observed phenomena. Based on observations scientists generate a hypothesis, or tentative explanation for the observed phenomena.
Science. had a hypothesis (that the bathroom doorknob was the dirtiest), a method to test the hypothesis (collecting germs from various locations around the house and growing them in a petri dish),
Feb 18, 2012 · Answers. Best Answer: A. Obesrvation, question, hypothesis, prediction, experiment, results, and conclusion. Observation is the first step of the scientific method. If nothing is observed, then no questions can be asked and no hypothesis formed,
Disciplines legend: MA, mathematics; CS, computer sciences; EN, engineering; SP, space science. one experiment, testing one hypothesis many times or in multiple ways in one or more studies, and.
write what you think will happen. In the Conclusion column, write a summary of what happened during the experiment and whether the hypothesis was true or false. Observation Hypothesis Prediction Experiment Conclusion Scientific Method Name: Graphic Organizer Maker graphicorganizer.net
Aug 04, 2017 · Science is a systematic and logical approach to discovering how things in the universe work. Scientists use the scientific method to make observations, form hypotheses and.
Theme by Anders Norén | 2,272 | 11,749 | {"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-2019-30 | latest | en | 0.92026 |
http://netteranatomy.com/how-would-life-be-without-friction.html | 1,566,060,530,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313436.2/warc/CC-MAIN-20190817164742-20190817190742-00470.warc.gz | 141,483,321 | 4,689 | # How would life be without friction. What would there be without friction 2019-01-16
How would life be without friction Rating: 8,1/10 789 reviews
## Essay on A world without friction.
In a world without friction, the object would continue to slide forever, if not acted on by another force. They might also just use Wikipedia, it's full of information! Buses, cars, trams, trains and bicycles wouldn't work because their wheels rely on friction to move. The friction factor is essential to the fluid flow and the loss of energy. In short, one would need some mechanism like rockets have to control motion. I set my plate down in front of me.
Next
## Q & A: Motion without friction
You would fall over and not be able to walk. You could not pick up a glass with a drink in it, you could nothold a pencil, a leaf, a spear, or … your cellphone, and you could not turn a page. Without the laws of physics themselves, nothing would exist but whatever initial state the universe began in - Big Bang theorists believe that this would be a large ball of hydrogen. It would be very difficult,or impossible, to pick up a coin from a flat surface. Life Without Friction By: Anthony Cacciato Life without friction would be dangerous.
Next
## What would there be without friction
So motion really doesn't need friction at all! For changing direction also we need to apply force to something in the opposite direction. This last one is in fact a system of three stars whereas Pollux is a giant. Static coefficients are somewhat higher than kinetic coefficients. While you are falling down the stairs since there is not sliding friction you would not be able to get up because there is nothing to slow you down from sliding all over the place which could be very dangerous. . One would need to push something to get motion in opposite direction. Answer: The basketball court floor.
Next
## what would the world be like if there was no friction by Michelle Mesa on Prezi
It means you have calculated his gift and you want him to pay it back and considered him to. This results in an off-axis gravitational force between the earth and the moon, slowing the earth's rotation down while transferring energy and angular momentum to the moon, which slowly rises to a higher orbit. The author's felt the father's antisocial behavior would affect the results in whether the father was present or not. You would not even be able to stand on your feet. There are a lot of steps that you can take to reduce your plastic consumption, and to make wise purchases that avoid plastic: Buy produce at the grocer's without those clear plastic produce bags. Indeed, life itselfwould quickly become doubtful at best, but almost certainly,impossible. Electrical devices make almost everything easier.
Next
## What would happen if there was no friction on earth?
Every part on earth that is assembled and kept together with knuts and bolts would fall apart! Since there are many applications of superconductivity, this would be beneficial. Can you feel your hands warming up? We also find in the Gemini, a beautiful opened heap. I was the one answering all the questions. One will need to push something to get motion in opposite direction. This frictional force acts in the opposite direction to the horizontal component of the force of your foot on the ground. My house is a disaster.
Next
## What Would Happen Without Friction?
Certificate This is to certify that of class , has prepared and submitted the Project Report enclosed herewith under my direct and close supervision and that this is a individual piece of work done by him. If you do, then the world would have fewer things that you like. We are in the world without Friction. You can stand peacefully, uninterrupted, even if there's zero friction. Not all of the changes brought by the absence of friction would be bad, however. There are still some hunting-and-gathering societies in the world today to make a study of human nature.
Next
## Q & A: Motion without friction
In a frictionless world, if a ball were rolled set in motion , it would never stop rolling. A different explanation was provided by Desaguliers 1725 , who demonstrated the strong cohesion forces between lead spheres of which a small cap is cut off and which were then brought into contact with each other. The first possibility is actually somewhat easier. Just look at the planets Venus and Mars: one became too hot due to a runaway greenhouse effect, and the other lost its water to the solar wind when it lost its magnetic field from its core cooling solid. Cassy and I slide to my house and amazingly we reach the house extremely fast. The pace of life would be slower and people would need to work closer to home. Objects at rest with no forces on them remain at rest.
Next
## The World Without Friction, Short Story
For a start, walking would be different because when you put your weight on the foot behind you there would be nothing to stop it sliding. Yes No Thanks for your feedback! Without friction, many obvious things which we take for granted would not function the way we like. The third purpose in this study was to find out whether the children of antisocial fathers were going to develop behavioral problems from both genetic and environmental risks. This concept states that the total amount of energy in an isolated system remains constant, although this energy may change forms. If you want to know what frictionless functional designs are, look no further than ordinary gear train engineering. Engineers are trying their absolutely best to reduce friction to the minimum while maintaining strong interaction and quite successfully.
Next | 1,159 | 5,688 | {"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} | 2.8125 | 3 | CC-MAIN-2019-35 | latest | en | 0.970885 |
https://sieswissservice.ch/2021-May-how-to-calculate-fineness-modulus-of-crusher-dust-39814 | 1,621,393,424,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00562.warc.gz | 540,038,299 | 7,158 | how to calculate fineness modulus of crusher dust
[email protected] | +86-21-58386256
## how to calculate fineness modulus of crusher dust
how to calculate fineness modulus of crusher .Home > how to calculate fineness modulus of crusher dust. Preface 'Professional ethics and Human values' is a very relevant subject of today's environment of conflicts and stress in the profession, with obligations to . learn more ; CGM mining application. Mining and construction equipment manfuactured by CGM Machinery plays an important role in mineral handling. The crushing, screening ...fineness modulus of crusher dustHow To Calculate Fineness Modulus Of Crusher Dust. fineness modulus of crusher dust . Fineness modulus of crusher dust goldhead sand fineness modulus crusher and grinding mill info in crusherb2b part Contact Supplier 14 AGGREGATES Aggregate is a collective term for the mineral materials such as sand, aggregate is Fineness Modulus is defined as the sum of the cumulative .Fineness Modulus of Coarse Aggregates and its .To find fineness modulus of coarse aggregate we need sieve sizes of 80mm, 40mm, 20mm, 10mm, 4.75mm, 2.36mm, 1.18mm, 0.6mm, 0.3mm and 0.15mm. Fineness modulus is the number at which the average size of particle is known when we counted from lower order sieve size to higher order sieve. So, in the calculation of coarse aggregate we need all sizes ...how to calculate fineness modulus of crusher .Home > how to calculate fineness modulus of crusher dust. Preface 'Professional ethics and Human values' is a very relevant subject of today's environment of conflicts and stress in the profession, with obligations to . learn more ; CGM mining application. Mining and construction equipment manfuactured by CGM Machinery plays an important role in mineral handling. The crushing, screening ...
### What Is The Fineness Modulus Of Manufactured .
What Is The Fineness Modulus Of Manufactured Sand Some Solutions. nbsp u.s. epa is conducting further environmental analysis on 79 pending permits for mountaintop mining, the agency announced today. the agency said the permits issued by the army corps of engineers, identified in a preliminary list released today, could all potentially pose a threat to the mountainFineness Modulus Of Crusher Dust 30Lp5Fineness Modulus Of Crusher Dust 30Lp5. Mobile rock crushing plant patent miner, canadian made mobile rock crushers crushers sand amp gravel handling equipment list of companies crushers sand amp gravel handling equipment businesses in canada manitoba for crushers welcome to westobas crushers business listings the premier resource forhow to calculate crusher concreteHow To Calculate Fineness Modulus Of Crusher Dust well as washed and unwashed crusher sands for use in all on-site and readymixed concrete .dust (washed crusher sand). Oversize fineness modulus and flakiness index to the aggregate is used in calculating the absolute or solid. Get Price . How To Calculate Impact Hammer Crusher Gulin . How To Calculate Impact Hammer Crusher .
### Fineness Modulus Of Crusher Dust 30Lp5
Fineness Modulus Of Crusher Dust 30Lp5. Mobile rock crushing plant patent miner, canadian made mobile rock crushers crushers sand amp gravel handling equipment list of companies crushers sand amp gravel handling equipment businesses in canada manitoba for crushers welcome to westobas crushers business listings the premier resource forfineness modulus of river sand in south africasand fineness modulus calculator program – Grinding Mill China :: Crusher Plant,Crushing fineness modulus of sand in malaysia - Crusher South Africa ; Get Price; Density Of Crusher Dust Crusher Mills, Cone Crusher, Jaw . Density Of Crusher Dust. Crusher Dust Machine for sale in South Africa,Crusher The specific gravity and fineness modulus of ROBO sand are and . Get Price; .CHAPTER 2 SIEVE ANALYSIS AND FINENESS MODULUS SamplingSIEVE ANALYSIS AND FINENESS MODULUS Sampling Since the reason for sampling aggregates is to determine the gradation (particle size) of the aggregate, it is necessary that they be sampled correctly. The results of testing will re" ect the condition and characteristics of the aggregate from which the sample is obtained. Therefore, when sampling, it is important to obtain a representative sample ...Density Of Crusher Dust | Crusher Mills, Cone Crusher, Jaw ...Stone crusher dust as a fine aggregate in Concrete for paving . Crusher dust : Fineness Modulus . specific gravity of the sand used for the preparation of concrete was found to be higher than the specific gravity of crusher dust. Density of the .Crush Sand Required Fineness In Cement ConcreteFineness Modulus Of Crusher Dust Breserven. What should be the fineness modulus of river sand shouldineness modulus of sand for concrete fineness modules of combined aggregates should be in the range of 3to 4 river sand with quarry dust without the inclusion of fly ash resulted in a rock flow when used as fine aggregate increases the modulus of ... | 1,044 | 4,989 | {"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-2021-21 | latest | en | 0.867462 |
http://www.kylesconverter.com/angle/centesimal-seconds-of-arc-to-degrees | 1,561,522,931,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000164.31/warc/CC-MAIN-20190626033520-20190626055520-00204.warc.gz | 272,536,110 | 5,452 | # Convert Centesimal Seconds Of Arc to Degrees
### Kyle's Converter > Angle > Centesimal Seconds Of Arc > Centesimal Seconds Of Arc to Degrees
Centesimal Seconds Of Arc (") Degrees (°) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Degrees to Centesimal Seconds Of Arc
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Centesimal second of arc:
1 centesimal second of arc is defined as one ten-thousandths of a gradian (1 grad/10000). In SI units this is 2π/4000000 radians or approximately 1.570 796 x 10-6 rad.
1 Degree (of arc):
1 degree of arc is define as 1/360 of a revolution. In SI units 1° is π/180 radians.
Conversions Table
1 Centesimal Seconds Of Arc to Degrees = 0.000170 Centesimal Seconds Of Arc to Degrees = 0.0063
2 Centesimal Seconds Of Arc to Degrees = 0.000280 Centesimal Seconds Of Arc to Degrees = 0.0072
3 Centesimal Seconds Of Arc to Degrees = 0.000390 Centesimal Seconds Of Arc to Degrees = 0.0081
4 Centesimal Seconds Of Arc to Degrees = 0.0004100 Centesimal Seconds Of Arc to Degrees = 0.009
5 Centesimal Seconds Of Arc to Degrees = 0.0005200 Centesimal Seconds Of Arc to Degrees = 0.018
6 Centesimal Seconds Of Arc to Degrees = 0.0005300 Centesimal Seconds Of Arc to Degrees = 0.027
7 Centesimal Seconds Of Arc to Degrees = 0.0006400 Centesimal Seconds Of Arc to Degrees = 0.036
8 Centesimal Seconds Of Arc to Degrees = 0.0007500 Centesimal Seconds Of Arc to Degrees = 0.045
9 Centesimal Seconds Of Arc to Degrees = 0.0008600 Centesimal Seconds Of Arc to Degrees = 0.054
10 Centesimal Seconds Of Arc to Degrees = 0.0009800 Centesimal Seconds Of Arc to Degrees = 0.072
20 Centesimal Seconds Of Arc to Degrees = 0.0018900 Centesimal Seconds Of Arc to Degrees = 0.081
30 Centesimal Seconds Of Arc to Degrees = 0.00271,000 Centesimal Seconds Of Arc to Degrees = 0.09
40 Centesimal Seconds Of Arc to Degrees = 0.003610,000 Centesimal Seconds Of Arc to Degrees = 0.9
50 Centesimal Seconds Of Arc to Degrees = 0.0045100,000 Centesimal Seconds Of Arc to Degrees = 9
60 Centesimal Seconds Of Arc to Degrees = 0.00541,000,000 Centesimal Seconds Of Arc to Degrees = 90 | 627 | 2,160 | {"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} | 2.53125 | 3 | CC-MAIN-2019-26 | latest | en | 0.537335 |
http://de.metamath.org/mpeuni/strfvi.html | 1,721,578,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00220.warc.gz | 9,148,341 | 4,440 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > strfvi Structured version Visualization version GIF version
Theorem strfvi 15741
Description: Structure slot extractors cannot distinguish between proper classes and ∅, so they can be protected using the identity function. (Contributed by Stefan O'Rear, 21-Mar-2015.)
Hypotheses
Ref Expression
strfvi.e 𝐸 = Slot 𝑁
strfvi.x 𝑋 = (𝐸𝑆)
Assertion
Ref Expression
strfvi 𝑋 = (𝐸‘( I ‘𝑆))
Proof of Theorem strfvi
StepHypRef Expression
1 strfvi.x . 2 𝑋 = (𝐸𝑆)
2 fvi 6165 . . . . 5 (𝑆 ∈ V → ( I ‘𝑆) = 𝑆)
32eqcomd 2616 . . . 4 (𝑆 ∈ V → 𝑆 = ( I ‘𝑆))
43fveq2d 6107 . . 3 (𝑆 ∈ V → (𝐸𝑆) = (𝐸‘( I ‘𝑆)))
5 strfvi.e . . . . 5 𝐸 = Slot 𝑁
65str0 15739 . . . 4 ∅ = (𝐸‘∅)
7 fvprc 6097 . . . 4 𝑆 ∈ V → (𝐸𝑆) = ∅)
8 fvprc 6097 . . . . 5 𝑆 ∈ V → ( I ‘𝑆) = ∅)
98fveq2d 6107 . . . 4 𝑆 ∈ V → (𝐸‘( I ‘𝑆)) = (𝐸‘∅))
106, 7, 93eqtr4a 2670 . . 3 𝑆 ∈ V → (𝐸𝑆) = (𝐸‘( I ‘𝑆)))
114, 10pm2.61i 175 . 2 (𝐸𝑆) = (𝐸‘( I ‘𝑆))
121, 11eqtri 2632 1 𝑋 = (𝐸‘( I ‘𝑆))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 = wceq 1475 ∈ wcel 1977 Vcvv 3173 ∅c0 3874 I cid 4948 ‘cfv 5804 Slot cslot 15694 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ral 2901 df-rex 2902 df-rab 2905 df-v 3175 df-sbc 3403 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-nul 3875 df-if 4037 df-sn 4126 df-pr 4128 df-op 4132 df-uni 4373 df-br 4584 df-opab 4644 df-mpt 4645 df-id 4953 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-iota 5768 df-fun 5806 df-fv 5812 df-slot 15699 This theorem is referenced by: rlmscaf 19029 islidl 19032 lidlrsppropd 19051 rspsn 19075 ply1tmcl 19463 ply1scltm 19472 ply1sclf 19476 ply1scl0 19481 ply1scl1 19483 nrgtrg 22304
Copyright terms: Public domain W3C validator | 1,228 | 2,251 | {"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-2024-30 | latest | en | 0.174804 |
https://justaaa.com/physics/350209-a-flywheel-with-a-radius-of-0500-m-starts-from | 1,686,189,365,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654031.92/warc/CC-MAIN-20230608003500-20230608033500-00756.warc.gz | 381,025,052 | 9,765 | Question
# A flywheel with a radius of 0.500 m starts from rest and accelerates with a constant...
A flywheel with a radius of 0.500 m starts from rest and accelerates with a constant angular acceleration of 0.700 rad/s2 .
1.Compute the magnitude of the tangential acceleration of a point on its rim at the start.
2.Compute the magnitude of the radial acceleration of a point on its rim at the start.
3.Compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 60.0 ?.
4.Compute the magnitude of the radial acceleration of a point on its rim after it has turned through 60.0 ?.
5.Compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 120.0 ?.
6.Compute the magnitude of the radial acceleration of a point on its rim after it has turned through 120.0 ?.
#### Homework Answers
Answer #1
1. Tangential acceleration is given by at = r = 0.5×0.7 = 0.35m/sec2
2. Radial acceleration is ar = r2, here initial = 0
So initial radial acceleration = 0
3 tangential acceleration is constant as it does not depend on angular velocity so at = 0.35m/sec2
4. After 60 degree angular displacement
= (2)0.5 = (2×0.7×?/3)0.5 = 1.211 rad/sec
ar = 0.5×(1.211)2 = 0.7331 m/sec2
5. Again tangential acceleration is always constant at = 0.35m/sec2
6. Radial acceleration -
= (2×0.7×2?/3)0.5
ar = 0.5×(2×0.7×2?/3) = 1.466 m/sec2
Know the answer?
Your Answer:
#### Post as a guest
Your Name:
What's your source?
#### Earn Coins
Coins can be redeemed for fabulous gifts.
##### Not the answer you're looking for?
Ask your own homework help question
ADVERTISEMENT
##### Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT | 487 | 1,766 | {"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-2023-23 | longest | en | 0.912128 |
https://ardoris.wordpress.com/tag/cosine-similarity/ | 1,513,367,615,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579564.61/warc/CC-MAIN-20171215192327-20171215214327-00014.warc.gz | 512,429,579 | 14,110 | # Cosine Similarity, Pearson Correlation, Inner Products
## To begin, a criticism
I picked up the Haskell Data Analysis Cookbook. The book presents examples of comparing data using Pearson Coefficient and using Cosine Similarity.
pearson xs ys = (n * sxy - sx * sy) / sqrt ((n * sxx - sx * sx) * (n * syy - sy * sy))
where
n = fromIntegral (length xs)
sx = sum xs
sy = sum ys
sxx = sum $zipWith (*) xs ys syy = sum$ zipWith (*) ys ys
sxy = sum $zipWith (*) xs ys cosine xs ys = dot d1 d2 / (len d1 * len d2) where dot a b = sum$ zipWith (*) a b
len a = sqrt $dot a a Although these code snippets are both calculating the ‘similarity’ between two vectors and actually, as we shall see, share a lot of structure, this is not at all apparent from a glance. We can fix that however… ## Definition of an Inner Product An inner product is conceptually a way to see how long a vector is after projecting it along another (inside some space). Formally, an inner product is a binary operator satisfying the following properties ###### Linearity $\langle u+v,w \rangle = \langle u,w\rangle + \langle v,w\rangle$ $\langle \alpha u,w\rangle = \alpha\langle u,w\rangle$ for $\alpha \in \mathbb{R}$ We are saying that we can push sums inside on the left to being outside. We can also push out constant factors. ###### (Conjugate) Symmetry $\langle u,v \rangle = \langle v,u \rangle$ or in the complex case, $\langle u,v \rangle = \overline{\langle v,u \rangle}$ In the real case, we’re saying everything is symmetric – it doesn’t matter which way you do it. In the complex case we have to reflect things by taking the conjugate. ###### Positive Definiteness $\langle u,u \rangle \ge 0$ with equality iff $u = 0$ Here we’re saying projecting a vector onto itself always results in a positive length. Secondly, the only way we can end up with a result of zero is if the vector itself is of length 0. ### From Inner Product to a notion of ‘length’ Intuitively a distance between two things must be • positive or zero (a negative distance makes not too much sense), with a length of zero corresponding to the zero vector • linear (if we scale the vector threefold, the length should also increase threefold) Given that $\langle u,u \rangle \ge 0$ we might be tempted to set $length(a) := \langle u,u \rangle$ but then upon scaling $u \rightarrow \alpha u$ we get $length(\alpha u) := \langle \alpha u, \alpha u \rangle = \alpha^2 \langle u,u \rangle$ – we’re not scaling linearly. Instead defining $||a|| := \sqrt{\langle a,b \rangle}$ everything is good! ### Similarity Now, in the abstract, how similar are two vectors? How about we first stop caring about how long they are, and want them just to point in the same direction. We can project one along the other and see how much it changes in length (shrinks). Projecting is kind of like seeing what its component is in that direction – i.e. considering 2-dimensional vectors in the plane, projecting a vector onto a unit vector in the $x$ direction will tell you the $x$ component of that vector. Let’s call two vectors $a$ and $b$. Firstly let’s scale them to be both of unit length, $\hat{a} = \frac{a}{||a||}, \hat{b} = \frac{b}{||b||}$ Now, project one onto the other (remember we’re not caring about order because of symmetry). $similarity(a,b) := \langle \frac{a}{||a||}, \frac{b}{||b||} \rangle$ Using linearity we can pull some stuff out (and also assuming everything’s happily a real vector – not caring about taking conjugates)… $similarity(a,b) := \frac{\langle a, b \rangle}{||a|| ||b||}$ ## Making Everything Concrete ###### Euclidean Inner Product The dot product we know and love. $a \dot b = a_1 b_1 + \dots + a_n b_n$ Plugging that into the similarity formula, we end up with the cosine similarity we started with! ###### Covariance Inner Product The covariance between two vectors is defined as $Cov(X,Y) = \mathbb{E}((X - \mathbb{E}(X))(Y - \mathbb{E}(Y)))$ where we’re abusing the notion of expectation somewhat. This in fact works if X and Y are arbitrary L2 random variables… but for the very concrete case of finite vectors we could consider $\mathbb{E}(X) = \frac{1}{n}(x_1 + \dots + x_n)$. We’ve said in our space, to project a first vector onto a second we see how covariant the first is with the second – if they move together or not. Plugging this inner product into the similarity formula, we instead get the pearson coefficient! In fact, given $Cov(X,X) = Variance(X)$, in this space we have $length(X) = \sqrt{Variance(X)} = StdDev(X) =: \sigma_X$, i.e. $similarity(X,Y) = \frac{Cov(X,Y)}{\sigma_X \sigma_Y}$. ### Improving the code Now that we know this structure exists, I posit the following as being better similarity ip xs ys = (ip xs ys) / ( (len xs) * (len ys) ) where len xs = sqrt(ip xs xs) -- the inner products dot xs ys = sum$ zipWith (*) xs ys
covariance xs ys = exy - (ex * ey)
where e xs = sum xs / (fromIntegral $length xs) exy = e$ zipWith (*) xs ys
ex = e xs
ey = e ys
-- the similarity functions
cosineSimilarity = similarity dot
pearsonSimilarity = similarity covariance
## Things I’m yet to think about
…though maybe the answers are apparent.
We have a whole load of inner products available to us. What does it mean to use those inner products?
E.g. $\langle f,g \rangle = \int^{-\pi}_{\pi} f(t) \overline{g(t)} \, \mathrm{d}t$ on $\mathbb{L}^2[-\pi,\pi]$ – the inner product producing the Fourier transform. I’m not the resulting similarity is anything particularly special though… | 1,471 | 5,480 | {"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": 1, "img_math": 27, "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-2017-51 | longest | en | 0.859578 |
http://www.osti.gov/eprints/topicpages/documents/record/423/4808380.html | 1,427,816,382,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131300735.71/warc/CC-MAIN-20150323172140-00243-ip-10-168-14-71.ec2.internal.warc.gz | 692,501,403 | 3,132 | LINEAR ALGEBRA (MATH 317H) CASIM ABBAS Summary: LINEAR ALGEBRA (MATH 317H) CASIM ABBAS Assignment 14 - Determinants (1) A square (n × n) matrix A is called skew-symmetric (or antisymmetric) if AT = -A. Prove that if A is skew-symmetric and n is odd then det A = 0. Is this also true for even n ? (2) A square matrix A is called nilpotent if there is an integer k 1 for which Ak = 0. Show that if A is nilpotent then det A = 0. (3) Prove that if matrices A and B are similar then det A = det B. (4) A square matrix Q is called orthogonal if QT Q = I. Prove that if Q is an orthogonal matrix then det Q = 1 or det Q = -1. (5) Use column or row operations to compute the following determinant 1 0 -2 3 -3 1 1 2 0 4 -1 1 2 3 0 1 , Collections: Mathematics | 253 | 751 | {"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-2015-14 | longest | en | 0.801673 |
https://forums.theregister.co.uk/forum/1/2012/11/09/peat_ice_age_coming_only_co2_can_save_us/ | 1,553,657,189,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912207618.95/warc/CC-MAIN-20190327020750-20190327042750-00377.warc.gz | 477,051,693 | 22,326 | # Swedish boffins: An Ice Age is coming, only CO2 can save us
A group of Swedish scientists at the University of Gothenburg have published a paper in which they argue that spreading peatlands are inexorably driving planet Earth into its next ice age, and the only thing holding back catastrophe is humanity's hotly debated atmospheric carbon emissions. "We are probably entering a new ice …
This topic is closed for new posts.
1. #### Ice age?
I thought we were already in an ice age? Hence the polar caps?
1. #### Re: Ice age?
Beat me to it - I thought the same.
1. #### Re: Ice age?
I thought caps were on the top - hence only one is a cap (either depending on which way up you hold the Earth). The other is then surely more like a polar panty liner.
1. #### Re: Ice age?
There is no up or down in space. The Earth has no top.
1. #### Re: Ice age?
Actually, it's pretty safe to say it does. The Milky Way has a distinct planar orientation to it (the starts within are not distributed in a globular pattern but spiral out pretty flatly). There's also the idea of orienting worlds on their orbital or rotational axes. Earth's current coordinate system is oriented on rotational axis.
1. #### Re: Ice age?
That simply means there is a vertical axis. Which end of that axis is 'up' or 'down' is merely convention.
1. #### Re: Ice age?
"That simply means there is a vertical axis. Which end of that axis is 'up' or 'down' is merely convention."
But either way, there is an up and a down whichever is which. So the original point is right. Or would if they hadn't misunderstood "cap" to mean a type of head-clothing, rather than in the sense of things that "cap" the ends of something.
2. #### Re: Ice age?
RHD Rule:
Wrap your hand around the earth . (figuratively speaking). Fingers with the direction or rotation. Up is with your thumb.
I got so confused trying to do this I think East is toward my adams apple.
1. #### Re: Ice age?
So what if you hold it upside-down? From the perspective of someone outside the Sol system, their perspective may be different from ours (IOW, we may see it as the underside of the Milky Way, they'll see it as the top side).
2. #### Re: Ice age?
Which hand?
2. #### Re: Ice age?
Actually, it's pretty safe to say [Earth] does [have a top]
I'm looking forward to us contacting extraterrestrial life so that we can decide which side of the "topist" and "bottomist" divide we lie on. Or to put it another way, whether we root more for turnwise or widdershins.
I've only one niggling doubt: wasn't this already played out in Gulliver's Travels?
2. #### Re: Ice age?
If the Earth has no top how do you explain Australia being on the bottom of the planet? Hah.. thought so, no answer to that smartypants. And for the pedantics that insist on proof that Australia is indeed at the bottom of the planet, first of all England is at the top, so that alone should be sufficient proof, but when you take into consideration they have summer at Christmas - clearly wrong and completely against nature - then the evidence is there for all to see.. unless they're some blinkered, head-in-the-sand-at-Christmas Australian standing upside down on the bottom of the world.
1. #### Re: Ice age?
WHAT is your favourite colour ? horn-gloat rag-strewn...
(Holy Grail etc )
No matter what happens, the politicians will seize it and tax. We get an ice age..look it's climate change, hot and dry, oh look its climate change....I'm reminded of fleas on an elephants back thinking that they can predict where it will go, and perhaps, just perhaps they can get it to change course.
1. #### Re: Follow the money
Never mind the money, it might be too late to wake up early, but me will continue to try to understand, anyway.
2. #### Re: Follow the money
Yep and only the politicians and scientists are interested in money and some in private industry has nothing to gain by continuing with the status quo.
3. #### Authoritative is as authoritative does....
"..it's true that 'Mires and Peat' isn't exactly Nature or Cell..."
Umm. I am aware that it hasn't got the same global presence or (probably) highly inflated opinion of itself as those two establishment publications.
But, without having done a lot of research, I suspect from the name that it is pretty hot on the chemistry and ecology of mires and peat bogs. In fact, I would not be surprised to find that it was the world's authoritative publication on those subjects......
1. #### Re: Authoritative is as authoritative does....
I'm sure it's absolutely boffo on the subjects of mires and peat. Global climate? Less so. Doesn't even fit the interglacial timings.
No surprise to see it waved around on El Reg, mind.
1. #### @Andy Gates
Why would it matter whether they are experts on global climate? The only conclusion of the research is the unprecedented scale of carbon sequestration in peat, and the enormous rate of growth of mires and bogs. The additional conclusion that the amount of carbon sequestrated might be high enough to offset human industrial CO2 emissions is added almost as an afterthought.
By the way, annual anthropogenic CO2 emissions are within the margin of error of estimates on amount of CO2 emissions from a medium-sized volcano eruption. How could human-made CO2 be responsible for anything? That's the thing that I've never seen climate scientists refute. It's like they're trying to explain how we can heat an ocean using a candle.
2. #### Re: Authoritative is as authoritative does....
"Mires and Peat" . Nah, not a patch on Bogs and Sod.
4. #### If the science holds up...
.... we should get a new directive from our Dear Leaders.
Instead of banning the use of Peat in horticulture, we should encourage it. Possibly providing grants for pensioners to dig it into their gardens? And certainly expanding its use in power stations....
Isn't climate science wonderful? Especially allied with a command economy run by moronic politicians with no technical knowledge whatsoever...
1. #### Re: If the science holds up...
" expanding its use in power stations"
if I understood this correctly, we want to leave the peat bogs alone so they can capture the CO2 from the fuel we are burning and so balance teh temperatures out, NOT burn more peat!
1. #### Re: If the science holds up...
I think you missed the point. I read it as we're heading into an ice age and unless you burn all of the peat, it'll happen more quickly (as it were).
Am I wrong?
1. #### Re: If the science holds up...
@theodore, no YOU missed the point. If we don't stop the cause of the next ice age, how are the politicians going to justify all the anti-CO2 taxes?
2. #### They may have no technical knowledge
But there's still room in their wallets and time for skiing trips or Mohitos in Tuscany
5. #### Also...
On the subject of climate, is there any follow up or outcome on this story yet?
C.
6. #### Mires and Peat
I must be hungry. I initially read that as "Pies and Meat" mmmmmnn - ice-age mammoth pies,
7. Without getting into the debate about whether warming is super-damaging or not-a-big-deal, I thought this rather elegantly shows the "last 10/12/16 years have not shown any warming so AGW is a crock" statement to be a load of denialist bollocks.
Funky gif this way.
1. #### Good Point
...but have you seen the same GIF for the last 1000 years?
1. "have you seen the same GIF for the last 1000 years?"
What would it show and what would that mean?
(Just trying to keep up).
2. #### Graph proves article accuracy
Unless I misread your graph, it appears to show that temperature over the last 15 years to have been flat or declining - exactly as the article stated. It does also show that over 40 years, it has increased by 0.5 degrees. The two facts are not mutually exclusive.
1. #### Re: Graph proves article accuracy
Temperatures have been flat or declining only if you treat the Daily Mail as science.
1. #### Re: Graph proves article accuracy
And the medieval warming period was also caused by human activity...
...pull the other one, it has bells on.
8. This post has been deleted by its author
9. #### Hurrican Sandy caused by global warming?
Nonsense. Everyone knows it was the homosexualists*
*Note: This is a joke.
1. #### Re: Hurrican Sandy caused by global warming?
Thought is was all them "Commie" satellites that did this?
Beer Friday is upon us. Cheers everyone!
2. #### Re: Hurrican Sandy caused by global warming?
You're wrong, of course. It was them chemtrails what did it.
3. #### Re: Hurrican Sandy caused by global warming?
Hurricanes are caused by warmth.
This has been the most expensive US election ever.
There has been blanket politicking throughout US.
Politicians spout hot air.
Hurricane Sandy hit at the peak of the electioneering.
The election is over, and there is no hurricane.
Q.E.D.
10. "If Franzén and his team are right, the big chill is now under way, and is only just being held off by increasing human carbon emissions - perhaps explaining why temperatures have been merely flat for the last 15 years or so, rather than descending."
Surely this is wrong. If Franzén and his team are right then we can expect the world to be still warming and can expect significantly more warming over the 21st century, just like all those climate scientists are saying. That's because Franzén and his team's work is saying CO2 is a strong driver of global temperature. They only predict cooling IF CO2 levels fall. But CO2 levels aren't falling and won't fall without emission cuts. CO2 levels are rising sharply due to human emissions of now over 30 billion tons of CO2 a year. In contrast Franzen and his team say the peatland sink *might* reach 3.7 Gt yr. So it isn't likely to even dent the increase in CO2.
So there is simply no cooling or big chill predicted. Instead expect continued warming. Global temperatures over the last 15 years are consistent with this.
http://www.realclimate.org/index.php/archives/2012/11/short-term-trends-another-proxy-fight/
1. #### 30 billion tons vs 3.7 giga tons?
So if I read you correctly - and I have no figures here of my own you understand - human activity is emitting 30 billion tons of CO2 per yer, and the peatlands may extract up to *3700 billion* tons (3.7 giga tons) per year.
And you think the peatland will barely dent the CO2 increase? If those numbers are correct, they'll annihilate it...
1. #### Re: 30 billion tons vs 3.7 giga tons?
Well it depends on what you mean by billion, but you have it backwards regardless. I'm assuming American usage, so billion = 10^9, which is also what Giga means (10^9) so 30 billion tons is about 8 times 3.7 Giga tons.
2. #### Re: 30 billion tons vs 3.7 giga tons?
Given human emissions are about 4% of total global emissions :s
1. #### Re: 30 billion tons vs 3.7 giga tons?
An -additional- 4%. That's why it's forcing change.
2. #### Re: 30 billion tons vs 3.7 giga tons?
Oops - just let a good one off!!...0.000000000001% added
Oh, except it wasn't a dry one.
OK, shower and washing machine tonight...0.00000001% then.
11. #### Well, if peat sequestrates CO2...
Why the hell is a lot of my town heated by a peat-burning power station? I think they have similar in Sweden.
I thought that burning released the CO2 from the peat??? Plus, the immense effort and 'global warming' of putting a new road from the northern peatlands to Oulu so the trucks (CO2 again...) can transport it...
1. #### The Irish have a way with peat
They build a power station where the peat is, water is always there also. Then they strip the surrounding soil and leave the land barren.
I think a power hungry world won't leave any beneficial peat, we'll burn it.
12. #### Good news for IT then
As there won't need to be vast spaces taken up by cooling equipment.
13. #### I lump together
... climate scientists with dieticians, boy bands, and politicians as sources of credibility.
1. #### Re: I lump together
Then you are a stupid person indeed.
#### Page:
This topic is closed for new posts.
Biting the hand that feeds IT © 1998–2019 | 2,885 | 12,118 | {"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} | 2.65625 | 3 | CC-MAIN-2019-13 | latest | en | 0.945565 |
https://academy.vertabelo.com/course/ms-sql-group-by-extensions/grouping-sets/grouping-sets/multiple-grouping-with-union-all-practice | 1,544,915,792,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00634.warc.gz | 507,726,848 | 12,957 | Introduction
GROUPING SETS
Summary
## Instruction
Great! Now it's your turn to write a similar query.
Try this exercise!
## Exercise
Find the average principal amounts for the following two grouping combinations:
1. Year and Quarter
2. Country
Show the following columns in the query result: Year, Quarter, Country, and AvgPrincipal.
### Stuck? Here's a hint!
SELECT
A,
NULL AS B,
...
FROM X
GROUP BY A
UNION ALL
SELECT
NULL AS A,
B,
...
FROM X
GROUP BY B | 118 | 464 | {"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} | 2.78125 | 3 | CC-MAIN-2018-51 | longest | en | 0.759399 |
https://www.starpath.com/cgi-bin/ubb/ultimatebb.cgi?ubb=next_topic;f=16;t=000301;go=older | 1,566,326,007,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315558.25/warc/CC-MAIN-20190820180442-20190820202442-00479.warc.gz | 980,122,999 | 5,447 | | my account | login-logout | resources | classroom help | support | catalog | home | get webcard |
search | help desk | commons
» Online Classroom » » Public Discussion of Cel Nav » Chp 5 5.8 Exercise #1
Author Topic: Chp 5 5.8 Exercise #1
TonyM posted March 15, 2017 06:02 PM Attached is my worksheet for Exercise 5.8(1). I have tried to solve this exercise with ZD 10 and 11. However, the LHA calculation does not appear on the excerpt for Lat 56* which appears on p.269. Obviously, I am doing something wrong, but don't know what. This is my second attempt to get some guidance on this. I hope this post gets through this time.
David Burch posted March 15, 2017 08:33 PM We are not getting images you are attaching.Please note that attached images must be <+800 pixels across and size <= 300 kb.if your image is too large, please use MS Paint or other program to reduce its size and then post it.if you do not know its size or dimensions, open it in MS Paint or other graphic program to check this and adjust it.thanks From: Starpath, Seattle, WA
David Burch posted March 15, 2017 08:34 PM PS. i only see one post here. Are you posting in the correct forum? if you are in the cel nav course, then please use the Student Discussion Forum and an instructor will answer. From: Starpath, Seattle, WA
Capt Steve Miller posted March 15, 2017 08:47 PM WHY are you using any Zone Description? The 5.8-1 and the rest of the problems give the time in UTC. No Zone Description is needed.As I described in my last post on this subject, you would just use the UTC to determine the GHA and then apply the a-Long with the proper sign (+ or -) to determine the LHA. From: Starpath
David Burch posted March 16, 2017 06:27 PM Just a note that we might be having problems with image uploads. This is being tested now. So maybe not seeing your posts is our fault, not your file size.hope to resolve this in a day or so. From: Starpath, Seattle, WA
All times are Pacific | 527 | 2,060 | {"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-2019-35 | latest | en | 0.908786 |
https://typeset.io/topics/inverse-trigonometric-functions-2ll9uwtb | 1,669,638,738,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00727.warc.gz | 626,643,101 | 41,720 | Topic
# Inverse trigonometric functions
About: Inverse trigonometric functions is a(n) research topic. Over the lifetime, 854 publication(s) have been published within this topic receiving 11141 citation(s). The topic is also known as: arcus function & antitrigonometric function.
##### Papers
More filters
Proceedings ArticleDOI
18 May 1971
TL;DR: This paper describes a single unified algorithm for the calculation of elementary functions including multiplication, division, sin, cos, tan, arctan, sinh, cosh, tanh, arCTanh, In, exp and square-root.
Abstract: This paper describes a single unified algorithm for the calculation of elementary functions including multiplication, division, sin, cos, tan, arctan, sinh, cosh, tanh, arctanh, In, exp and square-root The basis for the algorithm is coordinate rotation in a linear, circular, or hyperbolic coordinate system depending on which function is to be calculated The only operations required are shifting, adding, subtracting and the recall of prestored constants The limited domain of convergence of the algorithm is calculated, leading to a discussion of the modifications required to extend the domain for floating point calculations
1,018 citations
Book
06 Feb 1992
Abstract: (NOTE: Every chapter ends with Questions to Guide Your Review, Practice Exercises, and Additional Exercises.) P. Preliminaries. Real Numbers and the Real Line. Coordinates, Lines, and Increments. Functions. Shifting Graphs. Trigonometric Functions. 1. Limits and Continuity. Rates of Change and Limits. Rules for Finding Limits. Target Values and Formal Definitions of Limits. Extensions of the Limit Concept. Continuity. Tangent Lines. 2. Derivatives. The Derivative of a Function. Differentiation Rules. Rates of Change. Derivatives of Trigonometric Functions. The Chain Rule. Implicit Differentiation and Rational Exponents. Related Rates of Change. 3. Applications of Derivatives. Extreme Values of Functions. The Mean Value Theorem. The First Derivative Test for Local Extreme Values. Graphing with y e and y . Limits as x a a, Asymptotes, and Dominant Terms. Optimization. Linearization and Differentials. Newton's Method. 4. Integration. Indefinite Integrals. Differential Equations, Initial Value Problems, and Mathematical Modeling. Integration by Substitution--Running the Chain Rule Backward. Estimating with Finite Sums. Riemann Sums and Definite Integrals. Properties, Area, and the Mean Value Theorem. Substitution in Definite Integrals. Numerical Integration. 5. Applications of Integrals. Areas Between Curves. Finding Volumes by Slicing. Volumes of Solids of Revolution--Disks and Washers. Cylindrical Shells. Lengths of Plan Curves. Areas of Surfaces of Revolution. Moments and Centers of Mass. Work. Fluid Pressures and Forces. The Basic Pattern and Other Modeling Applications. 6. Transcendental Functions. Inverse Functions and Their Derivatives. Natural Logarithms. The Exponential Function. ax and logax. Growth and Decay. L'Hopital's Rule. Relative Rates of Growth. Inverse Trigonomic Functions. Derivatives of Inverse Trigonometric Functions Integrals. Hyperbolic Functions. First Order Differential Equations. Euler's Numerical Method Slope Fields. 7. Techniques of Integration. Basic Integration Formulas. Integration by Parts. Partial Fractions. Trigonometric Substitutions. Integral Tables and CAS. Improper Integrals. 8. Infinite Series. Limits of Sequences of Numbers. Theorems for Calculating Limits of Sequences. Infinite Series. The Integral Test for Series of Nonnegative Terms. Comparison Tests for Series of Nonnegative Terms. The Ratio and Root Tests for Series of Nonnegative Terms. Alternating Series, Absolute and Conditional Convergence. Power Series. Taylor and Maclaurin Series. Convergence of Taylor Series Error Estimates. Applications of Power Series. 9. Conic Sections, Parametrized Curves, and Polar Coordinates. Conic Sections and Quadratic Equations. Classifying Conic Sections by Eccentricity. Quadratic Equations and Rotations. Parametrizations of Plan Curves. Calculus with Parametrized Curves. Polar Coordinates. Graphing in Polar Coordinates. Polar Equations for Conic Sections. Integration in Polar Coordinates. 10. Vectors and Analytic Geometry in Space. Vectors in the Plane. Cartesian (Rectangular) Coordinates and Vectors in Space. Dot Products. Cross Products. Lines and Planes in Space. Cylinders and Quadric Surfaces. Cylindrical and Spherical Coordinates. 11. Vector-Valued Functions and Motion in Space. Vector-Valued Functions and Space Curves. Modeling Projectile Motion. Arc Length and the Unit Tangent Vector T. Curvature, Torison, and the TNB Frame. Planetary Motion and Satellites. 12. Multivariable Functions and Partial Derivatives. Functions of Several Variables. Limits and Continuity. Partial Derivatives. Differentiability, Linearization, and Differentials. The Chain Rule. Partial Derivatives with Constrained Variables. Directional Derivatives, Gradient Vectors, and Tangent Planes. Extreme Values and Saddle Points. Lagrange Multipliers. Taylor's Formula. 13. Multiple Integrals. Double Integrals. Areas, Moments, and Centers of Mass. Double Integrals in Polar Form. Triple Integrals in Rectangular Coordinates. Masses and Moments in Three Dimensions. Triple Integrals in Cylindrical and Spherical Coordinates. Substitutions in Multiple Integrals. 14. Integration in Vector Fields. Line Integrals. Vector Fields, Work, Circulation, and Flux. Path Independence, Potential Functions, and Conservative Fields. Green's Theorem in the Plane. Surface Area and Surface Integrals. Parametrized Surfaces. Stokes's Theorem. The Divergence Theorem and a Unified Theory. Appendices. Mathematical Induction. Proofs of Limit Theorems in Section 1.2. Complex Numbers. Simpson's One-Third Rule. Cauchy's Mean Value Theorem and the Stronger Form of L'Hopital's Rule. Limits that Arise Frequently. The Distributive Law for Vector Cross Products. Determinants and Cramer's Rule. Euler's Theorem and the Increment Theorem.
661 citations
Book
01 Jan 1973
Abstract: I. Laplace Transforms.- 1.1 General Formulas.- 1.2 Algebraic Functions.- 1.3 Powers of Arbitrary Order.- 1.4 Sectionally Rational- and Rows of Delta Functions.- 1.5 Exponential Functions.- 1.6 Logarithmic Functions.- 1.7 Trigonometric Functions.- 1.8 Inverse Trigonometric Functions.- 1.9 Hyperbolic Functions.- 1.10 Inverse Hyperbolic Functions.- 1.11 Orthogonal Polynomials.- 1.12 Legendre Functions.- 1.13Bessel Functions of Order Zero and Unity.- 1.14 Bessel Functions.- 1.15 Modified Bessel Functions.- 1.16 Functions Related to Bessel Functions and Kelvin Functions.- 1.17 Whittaker Functions and Special Cases.- 1.18 Elliptic Functions.- 1.19 Gauss' Hypergeometric Function.- 1.20 Miscellaneous Functions.- 1.21 Generalized Hypergeometric Functions.- II. Inverse Laplace Transforms.- 2.1 General Formulas.- 2.2 Rational Functions.- 2.3 Irrational Algebraic Functions.- 2.4 Powers of Arbitrary Order.- 2.5 Exponential Functions.- 2.6 Logarithmic Functions.- 2.7 Trigonometric- and Inverse Functions.- 2.8 Hyperbolic- and Inverse Functions.- 2.9 Orthogonal Polynomials.- 2.10 Gamma Function and Related Functions.- 2.11 Legendre Functions.- 2.12 Bessel Functions.- 2.13 Modified Bessel Functions.- 2.14 Functions Related to Bessel Functions and Kelvin Functions.- 2.15 Special Cases of Whittaker Functions.- 2.16 Parabolic Cylinder Functions and Whittaker Functions.- 2.17 Elliptic Integrals and Elliptic Functions.- 2.18 Gauss' Hypergeometric Functions.- 2.19 Generalized Hypergeometric Functions.- 2.20 Miscellaneous Functions.
559 citations
Journal ArticleDOI
481 citations
Book
29 Nov 1974
Abstract: I. Mellin Transforms.- Some Applications of the Mellin Transform Analysis.- 1.1 General Formulas.- 1.2 Algebraic Functions and Powers of Arbitrary Order.- 1.3 Exponential Functions.- 1.4 Logarithmic Functions.- 1.5 Trigonometric Functions.- 1.6 Hyperbolic Functions.- 1.7 The Gamma Function and Related Functions.- 1.8 Legendre Functions.- 1.9 Orthogonal Polynomials.- 1.10 Bessel Functions.- 1.11 Modified Bessel Function.- 1.12 Functions Related to Bessel Function.- 1.13 Whittaker Functions and Special Cases.- 1.14 Elliptic Integrals and Elliptic Functions.- 1.15 Hyper geometric Functions.- II. Inverse Mellin Transforms.- 2.1 General Formulas.- 2.2 Algebraic Functions and Powers of Arbitrary Order.- 2.3 Exponential and Logarithmic Functions.- 2.4 Trigonometric and Hyperbolic Functions.- 2.5 The Gamma Function and Related Functions.- 2.6 Orthogonal Polynomials and Legendre Functions.- 2.7 Bessel Functions and Related Functions.- 2.8 Whittaker Functions and Special Cases.
378 citations
##### Network Information
###### Related Topics (5)
Differential equation
88K papers, 2M citations
81% related
Matrix (mathematics)
105.5K papers, 1.9M citations
80% related
Bounded function
77.2K papers, 1.3M citations
79% related
Boundary value problem
145.3K papers, 2.7M citations
78% related
Nonlinear system
208.1K papers, 4M citations
77% related
##### Performance
###### Metrics
No. of papers in the topic in previous years
YearPapers
20221
202133
202027
201918
201814
201744 | 2,240 | 9,172 | {"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} | 2.8125 | 3 | CC-MAIN-2022-49 | latest | en | 0.838637 |
https://www.jiskha.com/display.cgi?id=1326215928 | 1,503,189,147,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105955.66/warc/CC-MAIN-20170819235943-20170820015943-00262.warc.gz | 890,989,204 | 4,163 | # Physics
posted by .
Object distance, image distance, and radius of curvature are ____ for curved mirrors.
interdependent
independent
directly related
unrelated
A
_________(__________
A B ( F
In the diagram shown above, the image of object A would be
(The diagram may be a bit confusing, so sorry. A and B should have an arrow above them pointing up.The two parentheses are supposed to be an uninterrupted curve, which would be a convex mirror. F is a point on the other side of the curved mirror.)
real, reduced, and upright.
virtual, enlarged, and upright
virtual, reduced, and inverted.
virtual, reduced, and upright.
D
• Physics -
kj
## Similar Questions
1. ### physics
we put 2 mirrors perpendicular to each other and then an object in front. the result is 3 images of that object. how does the middle image occur?
I am not thinking straight anymore and this is due by 9:00pm. My brain feels a little fried at this point. I have been at it all day. Can someone please help me! (a) An object is placed in front of a convex mirror. Draw a convex mirror …
3. ### physics
An object whose height is 3.5 cm is at a distance of 8.5 cm from a spherical concave mirror. Its image is real and has a height of 11.2 cm. Calculate the radius of curvature of the mirror. How far from the mirror is it necessary to …
4. ### physics
An object whose height is 3.5 cm is at a distance of 8.5 cm from a spherical concave mirror. Its image is real and has a height of 11.2 cm. Calculate the radius of curvature of the mirror. How far from the mirror is it necessary to …
5. ### physics
the focal length of this mirror is 6 cm and the object is located 8 cm away from the mirror. calculate the position of the image formed by the mirror. suppose the converging mirror is replaced by a diverging mirror with the same radius …
6. ### Science
Object distance, image distance, and radius of curvature are ____ for curved mirrors. interdependent independent directly related unrelated A _________(__________ A B ( F In the diagram shown above, the image of object A would be (The …
7. ### Physics
An object is placed in front of a convex mirror with radius of curvature = 17 cm. The object's height is 3.5 cm and it is initially 25 cm in front of the mirror. Find the (a) location and (b) height of the image. Next the object moves …
8. ### physics
a mirror which is somewhere between the object and image, forms a virtual upright image of the object 45 cm way from the object, measured on the optical axis of the mirror. The image is the size of the object. What kind of mirror must …
9. ### Physics
A concave spherical mirror has a radius of curvature of 18 cm. Locate the images for object distances as given below. In each case, state whether the image is real or virtual and upright or inverted, and find the magnification. (If …
10. ### physics (mirrors)
An object is placed from a concave mirror with a radius of curvature of 50cm. The object has a height of 5cm. a)Calculate the focal lenght of the mirror. b)Calculate the position of the image formed by the mirror. Is the image real …
More Similar Questions | 739 | 3,118 | {"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} | 2.859375 | 3 | CC-MAIN-2017-34 | latest | en | 0.917771 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-2-section-2-4-dividing-polynomials-remainder-and-factor-theorems-exercise-set-page-365/66 | 1,575,905,890,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00064.warc.gz | 715,792,386 | 13,253 | ## Precalculus (6th Edition) Blitzer
By performing long division, we get: x+1\overset{{{x}^{4}}-{{x}^{3}}+{{x}^{2}}-x+1}{\overline{\left){\begin{align} & \,\,\,\,\,\,\,\,\,\,{{x}^{5}}+1 \\ & \frac{\begin{align} & +{{x}^{5}}+{{x}^{4}} \\ & -\,\,\,\,\,\,- \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,0-{{x}^{4}}+1} \\ & \frac{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{x}^{4}}-{{x}^{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,+ \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}+1} \\ & \frac{\begin{align} & \,\,\,\,\,\,\,\,\,+{{x}^{3}}+{{x}^{2}} \\ & \,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,- \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0-{{x}^{2}}+1} \\ & \frac{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{x}^{2}}-x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,+ \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0+x+1} \\ & \frac{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+x+1 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,- \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0} \\ \end{align}}\right.}} Thus, it is observed that the quotient is ${{x}^{4}}-{{x}^{3}}+{{x}^{2}}-x+1$ but not ${{x}^{4}}+1$. | 620 | 1,265 | {"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": 5, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-51 | latest | en | 0.771462 |
https://www.lotterypost.com/blogentry/86902 | 1,503,105,443,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105195.16/warc/CC-MAIN-20170818233221-20170819013221-00004.warc.gz | 928,266,556 | 11,662 | Welcome Guest
The time is now 9:17 pm
You last visited August 18, 2017, 7:49 pm
All times shown are
Eastern Time (GMT-5:00)
# Triplets for FL Play4
Published:
I'm using Florida's Play 4 most overdue pairs (combined draws) to make the triplets. I join the "like" pairs to make the triplets. Going to play these thru Sunday 1/26.
Compare to your workout and use your favorite digits for the "x". The way that I play the "x", is to at least make a double out of the triplet. ex. - 049x = 0049,0449,0499 or take at least one digit from the previous winning number and include it for the "x", ex. - previous winning number - 7689 would be - 0497, 0496, 0498 and 0499.
Good luck to all!
Due Digit (2 & 5)
Due Pairs 04-09-59-58-08-06
049x, 046x, 048x, 089x, 096x, 095x, 598x, 586x, 580x, 086x
Entry #139
1.
Comment by inittowin - January 22, 2014, 8:26 pm
There we go! eve 4603 from the 046x for \$200!!!
We've had 5 singles in a row, it's about time we start hitting the doubles and make some real money! \$\$\$
2.
Comment by inittowin - January 23, 2014, 2:13 pm
The next draw pays another \$200! 1/23 mid - 0369 from the 096x.
3.
Comment by inittowin - January 23, 2014, 9:07 pm
The 046x repeats again! 1/23 eve - 5460 for another sum of \$200.
4.
Comment by inittowin - January 24, 2014, 1:51 pm
Another repeat with the 096x. 1/24 mid - 0692 for another \$200! Ride the wave.
Where are the doubles??? We've had 8 singles in a row...
5.
Comment by inittowin - January 26, 2014, 2:21 pm
Another trip to the DO! Hit this \$TR8 with the help of Casho and boxed. Off of the 089x. 1/26 - mid 0938 for \$2500 & 400 bx. Great way to end the weekend! Watch for the double 00 with the above triplets.
6.
Comment by inittowin - January 26, 2014, 8:26 pm
The week comes to an end for these numbers. The \$1 box bets grossed a total of \$1200.00
I got lucky and played \$3 straight on 6 different combinations (6 x.50) and hit one for \$2500.00! That's a good haul for 1 week.
If you are following this blog, these triplets won't stop hitting because a new week starts, so play smart with them and make sure that you play the doubles of the triplets (shown above). The doubles are now due to "catch up" since we've had a bunch of singles recently!
You must be a Lottery Post member to post comments to a Blog. | 758 | 2,307 | {"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} | 2.71875 | 3 | CC-MAIN-2017-34 | latest | en | 0.789756 |
http://mathhelpforum.com/calculus/133371-mcclaurin-series-help.html | 1,526,859,478,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863811.3/warc/CC-MAIN-20180520224904-20180521004904-00588.warc.gz | 187,216,488 | 9,761 | # Thread: McClaurin Series Help
1. ## McClaurin Series Help
How would you find the series for x^2(cosx)?
2. Originally Posted by helpplz
How would you find the series for x^2(cosx)?
What is the M.S. for $\displaystyle cos(x)$ ?
3. 1 - (x^2)/2! + (x^4)/4! ....
I just don't know what you would do with the coefficient of x^2...
4. Originally Posted by helpplz
1 - (x^2)/2! + (x^4)/4! ....
I just don't know what you would do with the coefficient of x^2...
Just multiply the series by $\displaystyle x^2$ | 162 | 509 | {"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} | 2.90625 | 3 | CC-MAIN-2018-22 | latest | en | 0.937454 |
http://www.markedbyteachers.com/gcse/science/to-measure-a-force-of-friction-and-to-compare-the-area-of-contact.html | 1,529,364,348,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861456.51/warc/CC-MAIN-20180618222556-20180619002556-00303.warc.gz | 451,263,686 | 17,827 | • Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
# To Measure a Force of Friction and To Compare the Area of Contact.
Extracts from this document...
Introduction
The Force of Friction Aim To measure a force of friction and to compare the area of contact. Apparatus * Newton Meter - 10N in 0.1N intervals * Rule - 1M in 0.1M (1CM) intervals * 100g weights X4 * 10g weights * Wooden Board * Stand and two clamps * Pulley wheel (mounted onto a board) * Wooden Block Diagram Method Set up apparatus as shown above. Prepare blocks, which measure 100mmX100mm, 150mmX100mm, 200mmX100mm, 250mmX100mm and 300mmX100mm. Place the correct amount of weights on top of the smallest block so that the block weighs the same as the heaviest block. Place the block near the end and pull the board at a constant speed. ...read more.
Middle
Prediction An increase in the size of the block will decrease the force of friction in direct proportion to each other, this is because as blocks get bigger there is more surface area to spread the weight on and this means that there will be a bigger area of contact. I also think this because we did some preliminary testing for an experiment to measure a force of friction and to compare that force to the weight of the sliding body. The results that we got from this experiment showed that as the weight of the block increased, the friction increased Preliminary Results Weight of the block (g) ...read more.
Conclusion
Evaluation My prediction was wrong; I know this because my results stayed the same all of the time. The graph also shows this as it is a horizontal line and if my prediction was correct the graph would go diagonally upwards. There is an obvious trend in this experiment, which is also shown by the graph in, by the fact that it is a horizontal line. My results were all the same which shows that I did the experiment correctly. You can get another question from this experiment and that is, "What would happen if the original weight of each of the blocks did not change?" ...read more.
The above preview is unformatted text
This student written piece of work is one of many that can be found in our GCSE Forces and Motion section.
## Found what you're looking for?
• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month
Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
# Related GCSE Forces and Motion essays
1. ## Investigating Sliding Friction: the effect of weight on sliding friction between a block and ...
10.6 5.1 5.1 5.4 5.2 11.6 5.6 5.7 5.9 5.73 12.6 6.5 6.7 6.6 6.6 Analysis On the previous page was my graph to show the results from the investigation. I showed that as the weight of the block increases the force of friction acting on the block also increases.
2. ## Investigation into Friction.
as the bottom surface, this will be kept the same, until that part of the experiment has finished; when the rough hardboard is being used, this will be kept constant until that part of the experiment is finished. Preliminary Work The maximum weight that the force meter can pull is 30 N.
1. ## Investigate how the weight of an object affects the force required to overcome friction.
/ 2.5 / 2.6 2.62 8.428 4.1 / 4.3 / 4.1 4.17 3 / 2.9 / 3 2.97 ANALYSING RESULTS AND CONCLUSION: By looking at my results in the above tables, it has been found that as the force holding the two surfaces together (i.e.
2. ## Mechanical Properties of a Meter Rule
It was difficult to get my horizontal rule level as the clamps which were used to hold the rule were not clamped at the same height compared to each other. It took awhile to get them level and still there may have been a few millimetres or centimetres of difference.
1. ## Liquid Friction.
remain constant at 9.81Nkg-1 Viscosity is not only the resistance of an object through the fluid, but also the internal friction between the liquid. It can be thought as the 'stickiness' of the material. The coefficient of viscosity is the ratio of the shearing stress to the velocity gradient.
2. ## In this experiment I aim to find out how the force and mass affect ...
The force of gravity on earth is always equal to the weight of the object as found by the equation: Fgrav = m * g where g = 9.8 m/s2 (on Earth) and m = mass (in kg) Normal Force, Fnorm The normal force is the support force exerted upon an object which is in contact with another stable object.
• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to | 1,176 | 4,734 | {"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-2018-26 | latest | en | 0.952085 |
https://fr.slideserve.com/danielle-garcia/quadratic-functions | 1,652,715,957,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510138.6/warc/CC-MAIN-20220516140911-20220516170911-00387.warc.gz | 337,038,167 | 19,880 | Télécharger la présentation
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
1. Quadratic Functions NLF-L1 Objectives:To analyze situations that can be modeled using Quadratic Functions Learning Outcome B-1
2. You have studied linear functions and their applications in previous math courses. This module deals with non-linear functions and their applications. You will be studying quadratic, cubic, and exponential functions.Many real-world situations are modeled by these functions. The cooling rate of a cup of warm liquid, the relationship between the length of a pendulum and the time for one swing, the change in brightness of a light as you move away from it: these are all examples of situations that are modeled by non-linear functions. Theory – Non Linear Functions
3. A quadratic function has a general equation of the form y = ax2 + bx + c. The shape of the graph of a quadratic function is called a parabola. It is a distinctive shape that is used in reflectors for car lights. The shape may open upward or downward. Why could a quadratic function not open left or right? Theory – Quadratic Functions
4. The graphs shown below are graphs of quadratic functions. • The first graph opens upward and has a minimum value for y. The lowest point on this graph is called its vertex. • The second graph opens downward and has a maximum value for y. The highest point on this graph is also called its vertex. • The vertex of a quadratic graph is the lowest or highest point on the graph. Expressed using coordinate notation, our vertex in the third graph is (2, -4). Theory – Properties of Quadratics - Vertex
5. The x-intercept is the point where the line meets the x-axis, and the y-intercept is the point where the line meets the y-axis. • The quadratic graph shown has two x-intercepts. It intersects the x-axis at (-3, 0) and (4, 0). Because the y-coordinate is always zero, you can simply say that the x-intercepts are -3 and 4. • Sometimes, the values of the x-intercepts are called the zeroes or roots of the function, and represent the x-values that would produce a y-value of zero. Finding the zeroes (when the y-value is zero) is sometimes called solving the equation. • The coordinates of the y-intercept are (0, 6). Similarly, you can say that the y-intercept is 6. Theory – Properties of Quadratics - Intercepts
6. A quadratic graph does not necessarily have two x-intercepts. It can have one x-intercept or no x-intercepts as shown below. • The y-intercept of the first graph has coordinates (0, 4) and the coordinates of the y-intercept of the second graph are (0, 1). Theory – Properties of Quadratics - Intercepts
7. The axis of symmetry of a quadratic graph is a line through the vertex, parallel to the y-axis. • It will always be of the form x = (x-coordinate of the vertex) • For example, the axis of symmetry of the quadratic graph shown is the vertical dashed line. The equation for the line is x = 3. Theory – Properties of Quadratics – Axis of Symmetry
8. The domain represents the values of x that can be used by the function. • The range represents the values of y that result. • For the graph shown below, the domain is all real numbers. The range is all real numbers greater than or equal to 0. Theory – Properties of Quadratics – Domain & Range
9. Example 1For the function whose graph is shown below: • The domain, the values of x that can be used by the function, is the set of all real numbers. Sometimes this is written as {all real numbers}. • The range, the values of y, is the set of real numbers less than or equal to 4.5. Sometimes this will be written as {all real numbers £ 4.5} Example – Properties of Quadratics – Domain & Range
10. Example 2For the function whose graph is shown below: • The coordinates of the vertex are (-1, 4). • The x-intercepts are -3 and 1. • The domain is {all real numbers}. • The range is {all real numbers £ 4}. • The equation of the axis of symmetry is x = -1. Example – Properties of Quadratics – Domain & Range
11. Use Winplot to answerGiven the quadratic function y = 2x2 - 4x - 1, identify the following: • the coordinates of the vertex • the coordinates of the x-intercepts (if any) • the coordinates of the y-intercept • the domain and the range • the equation for the axis of symmetry Using Winplot
12. The graph of y = x2 is shown. How do you think the graph would change if you graphed y = x2 + 3? How about the graph of y = x2 - 2? How would the graph change if you graphed y = 2x2?y = 0.5x2?y = -2x2?y = -3x2? How would the graph change if you graphed y = (x - 2)2?y = (x + 2)2?y = (x - 4)2?y = (x + 4)2? Using Winplot to Analyze Transformations | 1,191 | 4,736 | {"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} | 4.71875 | 5 | CC-MAIN-2022-21 | latest | en | 0.904821 |
https://valorgame.com/bets/what-is-a-24-way-box-in-lottery.html | 1,638,068,974,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00234.warc.gz | 672,281,193 | 19,007 | # What is a 24 way box in lottery?
Contents
## What does box mean in lottery?
Box(ed) Bet (Any Order): Wager option in which a player wins if selected numbers are drawn in any order, not just order as drawn (straight bet). Payment is lower than a combination bet. Break-Open Ticket: A predetermined outcome lottery ticket in which players open panels on the ticket to determine if they won.
## What is 3 way box?
A 3-way Box bet requires two of your three numbers to be the same and one to be different. The term ‘3-way’ describes how many possible combinations there are for those numbers to appear together. The following table shows examples of 3-way Box bets and the three different winning combinations.
## What does box mean in daily 4?
Box: Match the winning numbers in any order. Straight/Box: Match the winning numbers in exact order and win ½ the Straight prize and ½ the Box prize, or match the winning numbers in any order and win ½ the Box prize only.
## What’s the payout for 50 cent box?
9 CRR-NY 5010.2NY-CRR
IT IS INTERESTING: Question: Can you bet on sports at Jake's 58?
Bet Type 50 cents \$1
Straight/Box (4) \$3,100 If Straight hits
\$600 If Box only
Combination (all types) \$2,500 \$5,000
Close enough (straight match) \$1,250 \$2,500
## What does straight and box mean in lottery?
STRAIGHT/BOX: (Exact Order/Any Order) Match in exact order or any order. A 50¢ base play costs \$1. Prize amounts shown are based on a 50¢ base play. BOX.
## How much do you win for a dollar box?
On a \$1 ticket, players can win a top prize of \$500 for a Straight play, \$160 for a Box play, \$330 for a Straight/Box play, and \$50 for a Front Pair or Back Pair play. On a \$0.50 ticket, prizes are 1/2 of the \$1 play prizes.
## What’s the difference between straight and box in lottery?
Straight – If the winning numbers match in the same order, you are a winner. Box – If the winning numbers match in any order, you win.
## What does Pick 4 box mean?
Box. (24-way) 4. Any Order. Pick four unique numbers to match the winning numbers in any order.
## How much do you win on cash 4 box?
Play Type and Payout Chart
Play Type \$0.50 Bet Prize ODDS
Straight \$2500 1 in 10,000
4-Way Box \$600 1 in 2500
6-Way Box \$400 1 in 1667
12-Way Box \$200 1 in 833
## How do you win the daily 4?
To win, the numbers you selected must match the numbers drawn in the exact same order. To win, the numbers you selected must match the numbers drawn in any order. A box bet is considered a 4-way, 6-way, 12-way, or 24-way box depending on the number of combinations of the numbers you selected.
IT IS INTERESTING: Is Arkansas Lottery cash only?
## How much does daily 4 payout?
You win the jackpot when your ticket matches the winning draw numbers in exact order. The different playstyles include Exact Order, Any Order, 50/50, Combo, Pairs.
BOOST!
PLAY TYPE PRIZE (FOR \$1 BET) ODDS
Exact Order Play \$5,000* 1 in 10,002
4-way any order \$1,200* 1 in 2,500
6-way any order \$800* 1 in 1,666.7
## What is the most common 4 digit lottery number?
Researchers at the data analysis firm Data Genetics have found that the three most popular combinations—“1234,” “1111,” and “0000”—account for close to 20 percent of all four-digit passwords. | 872 | 3,257 | {"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-2021-49 | latest | en | 0.888604 |
http://cosmoquest.org/forum/showthread.php?52967-On-the-Hubble-expansion-and-observability&p=914839 | 1,369,245,312,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702185502/warc/CC-MAIN-20130516110305-00066-ip-10-60-113-184.ec2.internal.warc.gz | 64,285,907 | 21,729 | # Thread: On the Hubble expansion and observability
1. Member
Join Date
Jan 2007
Posts
41
## On the Hubble expansion and observability
Although I have philosophical objections to being considered "Against The Mainstream" as I considered some questions seen on this board I have found some problems that will fit this category. At the very least I hope to articulate the range of possibilities. Of particular interest to me are the skeptics.
Questions about expansion and observability have continually came up with varying levels of sophistication. Some deal with how the speed of light is affected by expansion, others simply by the observability. grav has done some fairly sophisticated thinking about the issues. The standard model says that space itself is expanding. I then set out to define it relativistically. The question I'll try to address is can this be incorporated into standard relativity in such a way that we can begin asking empirical questions? To do this I will try to relate the Hubble expansion to an expansion in the spacetime interval that varies with time in the same way GR relates a change in the spacetime interval that varies with position in the feild.
Postulate;
The Hubble expansion is an expansion in the spacetime interval.
Under GR we are familiar with expanding spacetime intervals in terms of a change in curvature or depth of feild. Here we will relate it to a change in time due to expansion. Since this is a preliminary model let's look at known GR effects as seen under Newtonian gravity as a litmus test. Specifically those that lead to the correct advance of the perihelion of mercury. This is mathematically done by applying the Lorentz transform to the instantaneous acceleration of g. As you move a meterstick away from the sun its length is Lorentz Transformed from the suns proper time to smaller units. The observer at the meterstick sees no change in the meterstick units because it represents that observers proper units. That observer will note that the sun will appear to gain mass though because of his smaller proper units. Locally these transformation of units are purely mathematical and don't represent real changes in parameters. However if we consider a mass (A) at the sun and in the proper units of (A) observes an equal mass (B) in outer space then as mass (B) approaches the sun mass (A) will notice that their mass is no longer equal. In fact both masses agree that mass (B) shrank in comparison to (A) even though both masses in their proper frames didn't change. It is this actual change in parameters that leads to the correct advance of the perihelion of mercury.
Now we will consider a uniform expansion in the spacetime interval over time. An observer outside our universe would watch our universe expand. Localy we would see no change because the spacetime interval defines our proper time. This is physically equivalent to the depth of the gravitational field increasing uniformly throughout the universe. If we consider a distant light signal from the past then the information is from a time when the spacetime interval was shorter. The gravitational redshift is z=(ωoe)/ωe, where ωo is the wavelength defined by the spacetime interval of a distant observer and ωe is the wavelength in proper units at the source of emission. Modeling expansion this way ωo is defined by proper units in the present while ωe is defined by proper units in the past. This indicates that space is in fact expanding but as the distance between points expands so does our meterstick.
Taken at face value this leads to some strange conclusions.
1) The universe is expanding.
2) This Hubble expansion is observed via the redshift.
3) The proper interval between masses are not increasing as a result of this Hubble expansion.
I am not very familiar with the coordinate system used by cosmology. I must remedy this for certain comparisons. However given the addition of velocities as defined by SR this would make it appear as if the farther we look back in time the slower the Hubble expansion will appear or accelerated expansion. This could provide a test if the addition of velocities correctly predict the magnitude of the accelerated expansion. Although the comoving coordinates of cosmology are explicitly designed to wash out the relativistic addition of velocities it doesn't include the above described effects of expansion. I need more data on this.
It is usefull to determine how we might question the legitamacy of tying expansion of space to the spacetime interval in this way. If we assume the absolute magnitude of proper distance is increasing with expansion but maintain a relationship with the spacetime interval it leads to more empirical consequences. It would mean that fundamental physical constants are decreasing with time. In fact the Hubble constant would define the rate at which they decrease. The only remaining approach I can think of is that the spacetime interval has a value independent of particles that exist within it. This would mean that the Big Bang was an event that occurred in a spacetime that had preexisting properties. The Hubble expansion would then be an actual increase in the magnitude of proper distance. The notion that space is what is expanding then becomes untenable, it would merely be a seperation of the masses in it. None of these alternatives are very satisfactory upon inspection but I will entertain them for review. The question here is, "How can we increase the total spacetime without increasing the spacetime interval or proper units of an observer?". Any prior work on this issue would be appreciated.
-"Proper" is always intended here to indicate proper spacetime intervals as defined by a local observer not necessarily the proper distance as defined by Weinberg.
2. Originally Posted by my_wan
As you move a meterstick away from the sun its length is Lorentz Transformed from the suns proper time to smaller units. The observer at the meterstick sees no change in the meterstick units because it represents that observers proper units. That observer will note that the sun will appear to gain mass though because of his smaller proper units.
Nope, I think at most the observer may think that the density of the sun has changed it his measuring rod is changed moving from frame to fram.
3. Member
Join Date
Jan 2007
Posts
41
Originally Posted by my_wan
As you move a meterstick away from the sun its length is Lorentz Transformed from the suns proper time to smaller units. The observer at the meterstick sees no change in the meterstick units because it represents that observers proper units. That observer will note that the sun will appear to gain mass though because of his smaller proper units.
Originally Posted by tusenfem
Nope, I think at most the observer may think that the density of the sun has changed it his measuring rod is changed moving from frame to fram.
Is that not what I said? I'm looking for where you might take exception.
peice by peice;
1)
Originally Posted by my_wan
As you move a meterstick away from the sun its length is Lorentz Transformed from the suns proper time to smaller units.
It is the meterstick or in you words the "the observer" and "his measuring rod" that I said changed proper units.
2)
Originally Posted by my_wan
The observer at the meterstick sees no change in the meterstick units because it represents that observers proper units..
Surely you can't seriously be saying that a measuring stick in your hand changes length by your own measure as you change depth in a gravitation feild?
3)
Originally Posted by my_wan
That observer will note that the sun will appear to gain mass though because of his smaller proper units.
Keyword; "appear to". I don't claim the sun changed mass but in the new units that are proper to the observer stated here it "appears" to him as if it does.
This is not even anything new to physics and can be checked against the standard model. I'm therefore still lost on what you disagreed with.
It almost appears as though you are arguing on the difference between what is real and what is mathematical artifacts, yet my use of "appear to" makes even this difficult for you. I'm arguing solely on what the respective observers measure. I hope you can make your point a little more clear.
4. Member
Join Date
Jan 2007
Posts
41
Perhaps I need to state the case more concisly.
Postulate
The Hubble expansion is an expansion of the spactime interval.
It is assumed that if the Hubble expansion is an expansion of space then it must be reflected in the spacetime interval s2 = x2 - c2 t2 of the observers in spacetime. This requires that the Hubble expansion is not observable locally for the same reason that an observer cannot observe ∆s locally as the depth in a gravitational field changes. Such changes are only observable by comparing ∆s to another reference observer as the depth in a gravitional field changes. We can do this by measuring the gravitational redshift of light sent from one gravitational potential to another. To observe the Hubble expansion we can measure this same redshift by recieving a light signal from the past.
The gravitational redshift is z=(ωoe)/ωe.
ωo is the wavelength as measured by a distant observer.
ωe is wavelength as measured at the source of emission.
The Hubble redshift is then zh=(ωoe)/ωe.
ωo is the wavelength as measured by a future observer.
ωe is wavelength as measured at the time of emission.
Nothing more than the postulate and the standard model of physics is needed to make the case.
5. Member
Join Date
Jan 2007
Posts
41
I disappointed that nobody has made a reasonable case against this. You do not have to be nice about it at all just physically reasonable. I do not even expect this to be right but I can't find an escape within the standard model of physics. I am not so familiar with the standard model of cosmology.
Dan Shiva asked a related question in a more general way and got this answer from Tim Thompson.
To this I responded.
Originally Posted by my_wan
Originally Posted by Tim Thompson
Generally speaking, in general relativity, there are two kinds of time: coordinate time and proper time. The former is the one that is gravity dependent, but the latter is not.
That statement does not really make much sense when coordinate time is nothing more that what proper time would be for an observer at that point. The concept of coordinate time is usefull for speaking about proper times across the entire coordinate system rather than for particular observers. It is not a different "kind of time" or coordinate system. It is not possible to speak of GR being independent of proper time when proper time changes as the depth of field changes and coordinate time is the proper time on the whole coordinate system.
If Tim could enlighten me perhaps I could wash my hands of this thread.
Bjoern also created this thread, "To all proponents of a universe with infinite age" To this I responded.
Originally Posted by my_wan
I am well aware of the empirical issues with trying to claim the observable universe is older than ~15 billion years. It is silly to expect it on prejudice alone. My own ATM posting in this category was born out of attempts to answer questions on this board and the standard model of physics. It is in no way a claim about age in general or even a claim about the legitamacy of the Big Bang. It is merely an attempt to articulate what the Hubble flow can say about it.
If you look at my post history I have not always been especially nice. So I'm looking for an arguement here to relieve myself of this idea.
6. I couldn't really read your original post too well, but I can read your more concise post better. I'm still not sure what you are saying causes a gravitational redshift, however.
If it is the galaxy it is emitted from, then that redshift will be limited to a few parts in a million for a star, and perhaps one in a few hundred thousand or so for a galaxy, and it will have no relation to the overall distance travelled, once far enough away, since almost all of that limit will be reached when it is closer to the point of emission. It would also blueshift again slightly when approaching our galaxy, solar system, and planet.
If you are referring to the entire universe as one great sphere of mass, where the gravitational redshift depends on the gravitational potential between one point and another, then this would create a redshift or blueshift depending on our position in the sphere and that of the point of emission, and all blueshifts from points further from the center than we are, which we don't observe. Also, according to the Big Bang scenario, all points can be considered the center, so there is no particular direction for an overall gravity to act.
I'm not sure if either of these is what you meant, though, so you can let me know.
Last edited by grav; 2007-Jan-29 at 01:32 AM.
7. Member
Join Date
Jan 2007
Posts
41
The gravitional curvature alone is not all that determines gravitational time dialation. Imagine a large hollow massive sphere. As you approach this sphere the gravitational time dialation will increase as you approach this sphere. If you pass inside this sphere then spacetime will be flat inside, yet the time dialation will remain slowed to that of the surface anywhere inside the sphere. Under GR the depth of field determines relative time dialation not the curvature.
Now imagine two observers seperated inside this sphere and the mass of the sphere is steadily increasing. Inside the time dialtion will steadily increase compared to a far removed observer even though the spacetime inside remains flat. Now when one of our observers sends a light signal to another the signal will be redshifted because of the finite value of C. The second observer will recieve this signal at a later time when the spacetime interval has changed. This is where I get the time dependent Hubble shift zh=(ωo-ωe)/ωe.
The question this posses is, "How can we speak of the Hubble expansion being an expansion of space itself if it has no relation to the spacetime interval?" If we assume the relationship is direct then it leads to this problem with proper distances after a period of expansion. Note that for our sphere the radius appears to decrease for our two observers because our sphere is not expanding with the spacetime.
Your questions are very much appreciated.
8. Established Member
Join Date
Nov 2005
Posts
368
Dear my wan,
You mentioned that any other work on this would be welcome.
In www.rescalingsymmetry.com there is something, and its sister website www.gravity.uk.com.
If you can make a relativistic version, please do.
All the best,
John Hunter
9. I am very confused here, because my_wan seems to be arguing with my_wan, but anywhoooo:
Originally Posted by my_wan
Is that not what I said? I'm looking for where you might take exception.
peice by peice;
1)
It is the meterstick or in you words the "the observer" and "his measuring rod" that I said changed proper units.
2)
Surely you can't seriously be saying that a measuring stick in your hand changes length by your own measure as you change depth in a gravitation feild?
3)
Keyword; "appear to". I don't claim the sun changed mass but in the new units that are proper to the observer stated here it "appears" to him as if it does.
This is not even anything new to physics and can be checked against the standard model. I'm therefore still lost on what you disagreed with.
What I was complaining about was that you said the sun seems to gain mass, and that is not true. The mass remains the same, the rod changes, thus the volume of the sun changes, and with the same mass that means that the density of the sun changes.
10. Member
Join Date
Jan 2007
Posts
41
Originally Posted by john hunter
Dear my wan,
You mentioned that any other work on this would be welcome.
In www.rescalingsymmetry.com there is something, and its sister website www.gravity.uk.com.
If you can make a relativistic version, please do.
All the best,
John Hunter
Yes I am in the process of reviewing that work. I assume it is you I emailed about what I was doing. The scaling appears to be tacked onto the standard model as an extra appendage. However by the way the conjecture was defined mc2-GMm/r=0 it appears that the scaling has some level of legitamacy at least as an approximation. I see that the rotation curves are not scaled to any actual galaxy either. Rather than use ∆s as defined by relativity the scaling correction has been attached to newtonian physics in a uniform and frame dependant manner. As I am just getting started coming from an entirely different starting point I'm not ready for a direct comparison. I can't endorse any of the conclusions on that site yet either. I can't even fully endorse my own approach yet. Still too many open questions, including how well it will fit numerically. It idea looks intriguing though. If you are interested I will keep you updated.
11. Member
Join Date
Jan 2007
Posts
41
Originally Posted by tusenfem
I am very confused here, because my_wan seems to be arguing with my_wan, but anywhoooo:
Of course I'm arguing with myself. I can't presume from the start that I'm right. So far my brother is my only good tenacious skeptic. Anyway my response here got me to thinking that the Hubble expansion wouldn't impart a real acceleration on mass particles as the expansion is locally symmetric. It would then make the anology in the above response even closer meaning distances under some circumstances would appear to decrease with the Hubble shift. A quick check of the pioneer anomaly showed a nice fit so I searched for prior work and found this http://arxiv.org/abs/astro-ph?papernum=0701132. Although the pioneer anomaly was Erhard Scholz's motivation the effect he describes is identical.
Mostly at this point I'm trying to articulate the differences that might be empirically different under different interpretations of what 'expanding space' might mean physically. Specifically apparent vs real changes in physical constants.
12. Established Member
Join Date
Apr 2006
Posts
1,031
## Length Contraction
I think publius has a better handle on the whole state of these affairs but in his absence I will state that length contraction is always a single axis/dimensional phenomenon.
In Special Relativity the length contraction is in the direction of travel which if the travel is along the X-axis then all the length contraction will be evident along the X-axis.
For gravity the length contraction is along the radial vector away from the center of gravity.
So in both cases the length contraction is only evident in one dimension whether this phenomenon actually causes a "density" change, since volume would be affected by length contraction, could be debated.
13. Member
Join Date
Jan 2007
Posts
41
Originally Posted by Squashed
I think publius has a better handle on the whole state of these affairs but in his absence I will state that length contraction is always a single axis/dimensional phenomenon.
In Special Relativity the length contraction is in the direction of travel which if the travel is along the X-axis then all the length contraction will be evident along the X-axis.
For gravity the length contraction is along the radial vector away from the center of gravity.
So in both cases the length contraction is only evident in one dimension whether this phenomenon actually causes a "density" change, since volume would be affected by length contraction, could be debated.
Not only are you wrong in both cases it is inmaterial to the case I have made. You appear to be viewing gamma as under the limited viewpoint of apparent changes only. I am not interested in what things appears to look like for a given observer. I am only interested in actual changes of parameters such as the Twin (not a) Paradox in SR and those that lead to the corrected perihelion of mercury in GR. Let's look at your statements individually.
1) Under SR length contraction only occurs on the X-axis of motion.
Apparent length contraction occurs from the apparent change in distance r from an observer. Imagine a large sphere traveling toward you. Looking down the axis of motion you get the standard length contraction. If you look at the sphere at some angle to this axis then the rate of decrease of r will be somewhat slower such that length contraction is somewhat less. This change in the length contraction decreases at all angles out until you view it at right angles to the direction of motion at which point you see no contraction on that axis. The velocity you use for apparent length contracton is defined by the rate of change in the radius from the observer. The end result of all the effects of the Relativity of Rigidity is always the same as if you assumed no length contraction. Even if solid objects appear to bend to accomplish this. As there are no actuall changes in parameters only changes of units it is of no value to explain the effects I have described.
2) For gravity the length contraction is along the radial vector away from the center of gravity.
I am assuming here that the observer is on the surface of the mass therefore not freefalling. Unlike SR under GR every observer in the universe agrees that time is slower for this observer than one placed higher up in the gravitational potential such as on top of a mountain. This means for our observer on the surface the entire universe appears smaller in all directions compared to the one on the mountain. This remains true even if you are at the center of the mass where you feel no gravitational forces. Although these effects are in line with my ideas here you appear to assume that gravitational effects are limited to SR effects induced by the acceleration of g.
14. Established Member
Join Date
Apr 2006
Posts
1,031
Originally Posted by my_wan
1) Under SR length contraction only occurs on the X-axis of motion.
Apparent length contraction occurs from the apparent change in distance r from an observer. Imagine a large sphere traveling toward you. Looking down the axis of motion you get the standard length contraction. If you look at the sphere at some angle to this axis then the rate of decrease of r will be somewhat slower such that length contraction is somewhat less. This change in the length contraction decreases at all angles out until you view it at right angles to the direction of motion at which point you see no contraction on that axis. The velocity you use for apparent length contracton is defined by the rate of change in the radius from the observer. The end result of all the effects of the Relativity of Rigidity is always the same as if you assumed no length contraction. Even if solid objects appear to bend to accomplish this. As there are no actuall changes in parameters only changes of units it is of no value to explain the effects I have described.
...
Your description sounds like foreshortening in drafting's orthographic projection.
Your sphere example reminds me of this.
In drafting when a line is viewed "end-on" it is a zero-length point but as we deviate our viewing angle from the end-on view the line gains length until finally when viewed perpendicular to the line it shows the line's "true length" - the length of the line can never be increased to a longer length than the true length by any other viewing angle.
This is not how length contraction works.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• | 4,932 | 23,617 | {"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-2013-20 | latest | en | 0.942549 |
https://www.liaoxuefeng.com/discuss/969955749132672/1053157508679456 | 1,701,751,593,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00872.warc.gz | 977,609,957 | 14,666 | Discuss / Python / 交作业
### 交作业
Topic source
#### 杨飞wb
#1 Created at ... [Delete] [Delete and Lock User]
``````第一题
# -*- coding: utf-8 -*-
L = [('Bob', 75), ('Adam', 92), ('Bart', 66), ('Lisa', 88)]
def by_name(t):
a = []
for i in t:
a.append(str(i))
return a
L2 = sorted(L, key=by_name)
print(L2)
# -*- coding: utf-8 -*-
L = [('Bob', 75), ('Adam', 92), ('Bart', 66), ('Lisa', 88)]
def by_score(t):
a = []
for i in t:
a.append(str(i))
return a[::-1]
L2 = sorted(L, key=by_score, reverse=True)
print(L2)
``````
#### 楚云
#2 Created at ... [Delete] [Delete and Lock User]
#### wongdir
#3 Created at ... [Delete] [Delete and Lock User]
@楚云 L = [('Bob', 75), ('Adam', 92), ('Bart', 66), ('Lisa', 88)]
'''声明一个by_name的函数，参数为t''' def by_name(t): '''声明一个变量名为a的list空数据''' a = [] '''for遍历传参t''' for i in t: '''把t遍历的元素一个个装进list变量a中，并以str类型''' a.append(str(i)) '''返回list变量a''' return a '''L2等于，把L内的每一个元素以by_name函数返回的a变量也就是名子list为排序规则元素进行排序，并返回排序后的L''' L2 = sorted(L, key=by_name) print(L2)
#### wongdir
#4 Created at ... [Delete] [Delete and Lock User]
'''上面by_naem函数返回的a变量是每一个元素作为字符串整体的list
["('Bob', 75)", "('Adam', 92)", "('Bart', 66)", "('Lisa', 88)"]
sort:('A->('B->('B->('L
'''
#### 其实我对你很在乎的
#5 Created at ... [Delete] [Delete and Lock User]
#### 帛子先生
#6 Created at ... [Delete] [Delete and Lock User]
• 1 | 529 | 1,324 | {"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} | 2.984375 | 3 | CC-MAIN-2023-50 | latest | en | 0.507744 |
https://dsp.stackexchange.com/questions/44269/overshoot-computations-not-matching | 1,718,485,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00612.warc.gz | 188,505,915 | 40,348 | Overshoot computations not matching
Take the following transfer function of a 3rd order system:
$$H(s)=\dfrac{2.302~s+0.3548}{s^3+0.739~s^2+3.223~s+0.3548}$$
with poles:
Pole Damping Frequency Time Constant
-1.13e-01 1.00e+00 1.13e-01 8.89e+00
-3.13e-01 + 1.75e+00i 1.76e-01 1.78e+00 3.19e+00
-3.13e-01 - 1.75e+00i 1.76e-01 1.78e+00 3.19e+00
and with the following unitary step response:
If I compute the percentage overshoot (PO) based on the damping ratio $\zeta=0.176$, I get:
$$PO=100{\times}e^{\dfrac{-\zeta\pi}{\sqrt{1-\zeta^2}}}=100{\times}e^{\dfrac{-0.176\pi}{\sqrt{1-0.176^2}}}=\boxed{57.02\%}$$
However, if I compute the PO using the graphical method (comparing the peak value with the final value) I get a completely different result:
$$PO=\dfrac{v_{peak}-v_{final}}{v_{final}}{\times}100=\dfrac{1.2-1}{1}{\times}100=\boxed{20\%}$$
I don't understand why such discrepancy. Why are my PO computations not matching?
• remember that 3rd-order is not the same as 2nd-order. Commented Oct 9, 2017 at 22:17
• @robertbristow-johnson Thank you for the reminder. Do you imply that one of the methods is not suitable for 3rd-order systems? Commented Oct 9, 2017 at 22:31
• read the sources you cite regarding percentage overshoot. what is all of that math, from which you derive a mathematical expression apply to? Commented Oct 9, 2017 at 23:21
Computing the partial fraction expansion,
$$H (s) = \frac{2.302 s + 0.3548}{s^{3} + 0.739 s^{2} + 3.223 s + 0.3548} \approx \frac{2.2861 - 0.0309306 s}{s^2 + 0.626454 s + 3.1525} + \frac{0.0309306}{s + 0.112546}$$
For the time being, let us neglect the step response of the 1st order subsystem. Note that the 2nd order subsystem has the following form
$$\pm \gamma (s - z) \left(\frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}\right)$$
where $\gamma \in \mathbb R$ is the gain (or attenuation) and $z \in \mathbb R$ is a (finite) zero. However, the formula you used to calculate the step response overshoot is only applicable for 2nd order systems of the form
$$G (s) := \frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}$$
whose zeros are at infinity. To summarize, the overshoot is smaller than you expected because:
• you neglected the 1st order subsystem.
• you neglected the attenuation $\gamma \neq 1$.
• you did not subtract $\gamma$ times the derivative of the step response of $G (s)$ at the peak time.
In any case, not neglecting the above would only provide one estimate of the overshoot. To compute a satisfactory approximation of its exact value, you can take the inverse Laplace transform, differentiate and find where the derivative vanishes.
• You are absolutely right. Thank you for your enlightening answer! Commented Oct 14, 2017 at 13:19 | 937 | 2,831 | {"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} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.765602 |
http://www.aleslombergar.com/forum/a-student-has-a-glass-coffee-table-in-their-dorm-room-that-they-want-to-stack-some-of-their-books-on/ | 1,561,466,172,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999838.23/warc/CC-MAIN-20190625112522-20190625134522-00296.warc.gz | 195,763,145 | 8,681 | ### A student has a glass coffee table in their dorm room that they want to stack some of their books on.?
A student has a glass coffee table in their dorm room that they want to stack some of their books on. However they are a little worried about how much pressure the table can withstand at any one point.
If the pile of books has a mass of 12.00 kg, a length of 27.94 cm, a width of 20.32 cm, and
a height of 30.12 cm, how much pressure (N/m2) must the table be able to withstand at any place
where the books are placed ? | 135 | 527 | {"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} | 3.140625 | 3 | CC-MAIN-2019-26 | latest | en | 0.971639 |
http://isabelle.in.tum.de/repos/isabelle/diff/e7e54a0b9197/src/HOL/GCD.thy | 1,558,482,601,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256586.62/warc/CC-MAIN-20190521222812-20190522004812-00139.warc.gz | 101,476,419 | 21,593 | src/HOL/GCD.thy
changeset 67051 e7e54a0b9197 parent 66936 cf8d8fc23891 child 67118 ccab07d1196c
``` 1.1 --- a/src/HOL/GCD.thy Sat Nov 11 18:33:35 2017 +0000
1.2 +++ b/src/HOL/GCD.thy Sat Nov 11 18:41:08 2017 +0000
1.3 @@ -142,12 +142,6 @@
1.4 class gcd = zero + one + dvd +
1.5 fixes gcd :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"
1.6 and lcm :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"
1.7 -begin
1.8 -
1.9 -abbreviation coprime :: "'a \<Rightarrow> 'a \<Rightarrow> bool"
1.10 - where "coprime x y \<equiv> gcd x y = 1"
1.11 -
1.12 -end
1.13
1.14 class Gcd = gcd +
1.15 fixes Gcd :: "'a set \<Rightarrow> 'a"
1.16 @@ -243,7 +237,8 @@
1.17 by simp
1.18 qed
1.19
1.20 -lemma is_unit_gcd [simp]: "is_unit (gcd a b) \<longleftrightarrow> coprime a b"
1.21 +lemma is_unit_gcd_iff [simp]:
1.22 + "is_unit (gcd a b) \<longleftrightarrow> gcd a b = 1"
1.23 by (cases "a = 0 \<and> b = 0") (auto simp add: unit_factor_gcd dest: is_unit_unit_factor)
1.24
1.25 sublocale gcd: abel_semigroup gcd
1.26 @@ -279,7 +274,7 @@
1.27 show "gcd (normalize a) b = gcd a b" for a b
1.28 using gcd_dvd1 [of "normalize a" b]
1.29 by (auto intro: associated_eqI)
1.30 - show "coprime 1 a" for a
1.31 + show "gcd 1 a = 1" for a
1.32 by (rule associated_eqI) simp_all
1.33 qed simp_all
1.34
1.35 @@ -292,12 +287,6 @@
1.36 lemma gcd_right_idem: "gcd (gcd a b) b = gcd a b"
1.37 by (fact gcd.right_idem)
1.38
1.39 -lemma coprime_1_left: "coprime 1 a"
1.40 - by (fact gcd.bottom_left_bottom)
1.41 -
1.42 -lemma coprime_1_right: "coprime a 1"
1.43 - by (fact gcd.bottom_right_bottom)
1.44 -
1.45 lemma gcd_mult_left: "gcd (c * a) (c * b) = normalize c * gcd a b"
1.46 proof (cases "c = 0")
1.47 case True
1.48 @@ -634,70 +623,6 @@
1.49 by (rule dvd_trans)
1.50 qed
1.51
1.52 -lemma coprime_dvd_mult:
1.53 - assumes "coprime a b" and "a dvd c * b"
1.54 - shows "a dvd c"
1.55 -proof (cases "c = 0")
1.56 - case True
1.57 - then show ?thesis by simp
1.58 -next
1.59 - case False
1.60 - then have unit: "is_unit (unit_factor c)"
1.61 - by simp
1.62 - from \<open>coprime a b\<close> mult_gcd_left [of c a b]
1.63 - have "gcd (c * a) (c * b) * unit_factor c = c"
1.64 - by (simp add: ac_simps)
1.65 - moreover from \<open>a dvd c * b\<close> have "a dvd gcd (c * a) (c * b) * unit_factor c"
1.66 - by (simp add: dvd_mult_unit_iff unit)
1.67 - ultimately show ?thesis
1.68 - by simp
1.69 -qed
1.70 -
1.71 -lemma coprime_dvd_mult_iff: "coprime a c \<Longrightarrow> a dvd b * c \<longleftrightarrow> a dvd b"
1.72 - by (auto intro: coprime_dvd_mult)
1.73 -
1.74 -lemma gcd_mult_cancel: "coprime c b \<Longrightarrow> gcd (c * a) b = gcd a b"
1.75 - apply (rule associated_eqI)
1.76 - apply (rule gcd_greatest)
1.77 - apply (rule_tac b = c in coprime_dvd_mult)
1.78 - apply (simp add: gcd.assoc)
1.79 - apply (simp_all add: ac_simps)
1.80 - done
1.81 -
1.82 -lemma coprime_crossproduct:
1.83 - fixes a b c d :: 'a
1.84 - assumes "coprime a d" and "coprime b c"
1.85 - shows "normalize a * normalize c = normalize b * normalize d \<longleftrightarrow>
1.86 - normalize a = normalize b \<and> normalize c = normalize d"
1.87 - (is "?lhs \<longleftrightarrow> ?rhs")
1.88 -proof
1.89 - assume ?rhs
1.90 - then show ?lhs by simp
1.91 -next
1.92 - assume ?lhs
1.93 - from \<open>?lhs\<close> have "normalize a dvd normalize b * normalize d"
1.94 - by (auto intro: dvdI dest: sym)
1.95 - with \<open>coprime a d\<close> have "a dvd b"
1.96 - by (simp add: coprime_dvd_mult_iff normalize_mult [symmetric])
1.97 - from \<open>?lhs\<close> have "normalize b dvd normalize a * normalize c"
1.98 - by (auto intro: dvdI dest: sym)
1.99 - with \<open>coprime b c\<close> have "b dvd a"
1.100 - by (simp add: coprime_dvd_mult_iff normalize_mult [symmetric])
1.101 - from \<open>?lhs\<close> have "normalize c dvd normalize d * normalize b"
1.102 - by (auto intro: dvdI dest: sym simp add: mult.commute)
1.103 - with \<open>coprime b c\<close> have "c dvd d"
1.104 - by (simp add: coprime_dvd_mult_iff gcd.commute normalize_mult [symmetric])
1.105 - from \<open>?lhs\<close> have "normalize d dvd normalize c * normalize a"
1.106 - by (auto intro: dvdI dest: sym simp add: mult.commute)
1.107 - with \<open>coprime a d\<close> have "d dvd c"
1.108 - by (simp add: coprime_dvd_mult_iff gcd.commute normalize_mult [symmetric])
1.109 - from \<open>a dvd b\<close> \<open>b dvd a\<close> have "normalize a = normalize b"
1.110 - by (rule associatedI)
1.111 - moreover from \<open>c dvd d\<close> \<open>d dvd c\<close> have "normalize c = normalize d"
1.112 - by (rule associatedI)
1.113 - ultimately show ?rhs ..
1.114 -qed
1.115 -
1.116 lemma gcd_add1 [simp]: "gcd (m + n) n = gcd m n"
1.117 by (rule gcdI [symmetric]) (simp_all add: dvd_add_left_iff)
1.118
1.119 @@ -707,285 +632,6 @@
1.120 lemma gcd_add_mult: "gcd m (k * m + n) = gcd m n"
1.121 by (rule gcdI [symmetric]) (simp_all add: dvd_add_right_iff)
1.122
1.123 -lemma coprimeI: "(\<And>l. l dvd a \<Longrightarrow> l dvd b \<Longrightarrow> l dvd 1) \<Longrightarrow> gcd a b = 1"
1.124 - by (rule sym, rule gcdI) simp_all
1.125 -
1.126 -lemma coprime: "gcd a b = 1 \<longleftrightarrow> (\<forall>d. d dvd a \<and> d dvd b \<longleftrightarrow> is_unit d)"
1.127 - by (auto intro: coprimeI gcd_greatest dvd_gcdD1 dvd_gcdD2)
1.128 -
1.129 -lemma div_gcd_coprime:
1.130 - assumes nz: "a \<noteq> 0 \<or> b \<noteq> 0"
1.131 - shows "coprime (a div gcd a b) (b div gcd a b)"
1.132 -proof -
1.133 - let ?g = "gcd a b"
1.134 - let ?a' = "a div ?g"
1.135 - let ?b' = "b div ?g"
1.136 - let ?g' = "gcd ?a' ?b'"
1.137 - have dvdg: "?g dvd a" "?g dvd b"
1.138 - by simp_all
1.139 - have dvdg': "?g' dvd ?a'" "?g' dvd ?b'"
1.140 - by simp_all
1.141 - from dvdg dvdg' obtain ka kb ka' kb' where
1.142 - kab: "a = ?g * ka" "b = ?g * kb" "?a' = ?g' * ka'" "?b' = ?g' * kb'"
1.143 - unfolding dvd_def by blast
1.144 - from this [symmetric] have "?g * ?a' = (?g * ?g') * ka'" "?g * ?b' = (?g * ?g') * kb'"
1.145 - by (simp_all add: mult.assoc mult.left_commute [of "gcd a b"])
1.146 - then have dvdgg':"?g * ?g' dvd a" "?g* ?g' dvd b"
1.147 - by (auto simp add: dvd_mult_div_cancel [OF dvdg(1)] dvd_mult_div_cancel [OF dvdg(2)] dvd_def)
1.148 - have "?g \<noteq> 0"
1.149 - using nz by simp
1.150 - moreover from gcd_greatest [OF dvdgg'] have "?g * ?g' dvd ?g" .
1.151 - ultimately show ?thesis
1.152 - using dvd_times_left_cancel_iff [of "gcd a b" _ 1] by simp
1.153 -qed
1.154 -
1.155 -lemma divides_mult:
1.156 - assumes "a dvd c" and nr: "b dvd c" and "coprime a b"
1.157 - shows "a * b dvd c"
1.158 -proof -
1.159 - from \<open>b dvd c\<close> obtain b' where"c = b * b'" ..
1.160 - with \<open>a dvd c\<close> have "a dvd b' * b"
1.161 - by (simp add: ac_simps)
1.162 - with \<open>coprime a b\<close> have "a dvd b'"
1.163 - by (simp add: coprime_dvd_mult_iff)
1.164 - then obtain a' where "b' = a * a'" ..
1.165 - with \<open>c = b * b'\<close> have "c = (a * b) * a'"
1.166 - by (simp add: ac_simps)
1.167 - then show ?thesis ..
1.168 -qed
1.169 -
1.170 -lemma coprime_lmult:
1.171 - assumes dab: "gcd d (a * b) = 1"
1.172 - shows "gcd d a = 1"
1.173 -proof (rule coprimeI)
1.174 - fix l
1.175 - assume "l dvd d" and "l dvd a"
1.176 - then have "l dvd a * b"
1.177 - by simp
1.178 - with \<open>l dvd d\<close> and dab show "l dvd 1"
1.179 - by (auto intro: gcd_greatest)
1.180 -qed
1.181 -
1.182 -lemma coprime_rmult:
1.183 - assumes dab: "gcd d (a * b) = 1"
1.184 - shows "gcd d b = 1"
1.185 -proof (rule coprimeI)
1.186 - fix l
1.187 - assume "l dvd d" and "l dvd b"
1.188 - then have "l dvd a * b"
1.189 - by simp
1.190 - with \<open>l dvd d\<close> and dab show "l dvd 1"
1.191 - by (auto intro: gcd_greatest)
1.192 -qed
1.193 -
1.194 -lemma coprime_mult:
1.195 - assumes "coprime d a"
1.196 - and "coprime d b"
1.197 - shows "coprime d (a * b)"
1.198 - apply (subst gcd.commute)
1.199 - using assms(1) apply (subst gcd_mult_cancel)
1.200 - apply (subst gcd.commute)
1.201 - apply assumption
1.202 - apply (subst gcd.commute)
1.203 - apply (rule assms(2))
1.204 - done
1.205 -
1.206 -lemma coprime_mul_eq: "gcd d (a * b) = 1 \<longleftrightarrow> gcd d a = 1 \<and> gcd d b = 1"
1.207 - using coprime_rmult[of d a b] coprime_lmult[of d a b] coprime_mult[of d a b]
1.208 - by blast
1.209 -
1.210 -lemma coprime_mul_eq':
1.211 - "coprime (a * b) d \<longleftrightarrow> coprime a d \<and> coprime b d"
1.212 - using coprime_mul_eq [of d a b] by (simp add: gcd.commute)
1.213 -
1.214 -lemma gcd_coprime:
1.215 - assumes c: "gcd a b \<noteq> 0"
1.216 - and a: "a = a' * gcd a b"
1.217 - and b: "b = b' * gcd a b"
1.218 - shows "gcd a' b' = 1"
1.219 -proof -
1.220 - from c have "a \<noteq> 0 \<or> b \<noteq> 0"
1.221 - by simp
1.222 - with div_gcd_coprime have "gcd (a div gcd a b) (b div gcd a b) = 1" .
1.223 - also from assms have "a div gcd a b = a'"
1.224 - using dvd_div_eq_mult local.gcd_dvd1 by blast
1.225 - also from assms have "b div gcd a b = b'"
1.226 - using dvd_div_eq_mult local.gcd_dvd1 by blast
1.227 - finally show ?thesis .
1.228 -qed
1.229 -
1.230 -lemma coprime_power:
1.231 - assumes "0 < n"
1.232 - shows "gcd a (b ^ n) = 1 \<longleftrightarrow> gcd a b = 1"
1.233 - using assms
1.234 -proof (induct n)
1.235 - case 0
1.236 - then show ?case by simp
1.237 -next
1.238 - case (Suc n)
1.239 - then show ?case
1.240 - by (cases n) (simp_all add: coprime_mul_eq)
1.241 -qed
1.242 -
1.243 -lemma gcd_coprime_exists:
1.244 - assumes "gcd a b \<noteq> 0"
1.245 - shows "\<exists>a' b'. a = a' * gcd a b \<and> b = b' * gcd a b \<and> gcd a' b' = 1"
1.246 - apply (rule_tac x = "a div gcd a b" in exI)
1.247 - apply (rule_tac x = "b div gcd a b" in exI)
1.248 - using assms
1.249 - apply (auto intro: div_gcd_coprime)
1.250 - done
1.251 -
1.252 -lemma coprime_exp: "gcd d a = 1 \<Longrightarrow> gcd d (a^n) = 1"
1.253 - by (induct n) (simp_all add: coprime_mult)
1.254 -
1.255 -lemma coprime_exp_left: "coprime a b \<Longrightarrow> coprime (a ^ n) b"
1.256 - by (induct n) (simp_all add: gcd_mult_cancel)
1.257 -
1.258 -lemma coprime_exp2:
1.259 - assumes "coprime a b"
1.260 - shows "coprime (a ^ n) (b ^ m)"
1.261 -proof (rule coprime_exp_left)
1.262 - from assms show "coprime a (b ^ m)"
1.263 - by (induct m) (simp_all add: gcd_mult_cancel gcd.commute [of a])
1.264 -qed
1.265 -
1.266 -lemma gcd_exp: "gcd (a ^ n) (b ^ n) = gcd a b ^ n"
1.267 -proof (cases "a = 0 \<and> b = 0")
1.268 - case True
1.269 - then show ?thesis
1.270 - by (cases n) simp_all
1.271 -next
1.272 - case False
1.273 - then have "1 = gcd ((a div gcd a b) ^ n) ((b div gcd a b) ^ n)"
1.274 - using coprime_exp2[OF div_gcd_coprime[of a b], of n n, symmetric] by simp
1.275 - then have "gcd a b ^ n = gcd a b ^ n * \<dots>"
1.276 - by simp
1.277 - also note gcd_mult_distrib
1.278 - also have "unit_factor (gcd a b ^ n) = 1"
1.279 - using False by (auto simp add: unit_factor_power unit_factor_gcd)
1.280 - also have "(gcd a b)^n * (a div gcd a b)^n = a^n"
1.281 - apply (subst ac_simps)
1.282 - apply (subst div_power)
1.283 - apply simp
1.284 - apply (rule dvd_div_mult_self)
1.285 - apply (rule dvd_power_same)
1.286 - apply simp
1.287 - done
1.288 - also have "(gcd a b)^n * (b div gcd a b)^n = b^n"
1.289 - apply (subst ac_simps)
1.290 - apply (subst div_power)
1.291 - apply simp
1.292 - apply (rule dvd_div_mult_self)
1.293 - apply (rule dvd_power_same)
1.294 - apply simp
1.295 - done
1.296 - finally show ?thesis by simp
1.297 -qed
1.298 -
1.299 -lemma coprime_common_divisor: "gcd a b = 1 \<Longrightarrow> a dvd a \<Longrightarrow> a dvd b \<Longrightarrow> is_unit a"
1.300 - apply (subgoal_tac "a dvd gcd a b")
1.301 - apply simp
1.302 - apply (erule (1) gcd_greatest)
1.303 - done
1.304 -
1.305 -lemma division_decomp:
1.306 - assumes "a dvd b * c"
1.307 - shows "\<exists>b' c'. a = b' * c' \<and> b' dvd b \<and> c' dvd c"
1.308 -proof (cases "gcd a b = 0")
1.309 - case True
1.310 - then have "a = 0 \<and> b = 0"
1.311 - by simp
1.312 - then have "a = 0 * c \<and> 0 dvd b \<and> c dvd c"
1.313 - by simp
1.314 - then show ?thesis by blast
1.315 -next
1.316 - case False
1.317 - let ?d = "gcd a b"
1.318 - from gcd_coprime_exists [OF False]
1.319 - obtain a' b' where ab': "a = a' * ?d" "b = b' * ?d" "gcd a' b' = 1"
1.320 - by blast
1.321 - from ab'(1) have "a' dvd a"
1.322 - unfolding dvd_def by blast
1.323 - with assms have "a' dvd b * c"
1.324 - using dvd_trans [of a' a "b * c"] by simp
1.325 - from assms ab'(1,2) have "a' * ?d dvd (b' * ?d) * c"
1.326 - by simp
1.327 - then have "?d * a' dvd ?d * (b' * c)"
1.328 - by (simp add: mult_ac)
1.329 - with \<open>?d \<noteq> 0\<close> have "a' dvd b' * c"
1.330 - by simp
1.331 - with coprime_dvd_mult[OF ab'(3)] have "a' dvd c"
1.332 - by (subst (asm) ac_simps) blast
1.333 - with ab'(1) have "a = ?d * a' \<and> ?d dvd b \<and> a' dvd c"
1.334 - by (simp add: mult_ac)
1.335 - then show ?thesis by blast
1.336 -qed
1.337 -
1.338 -lemma pow_divs_pow:
1.339 - assumes ab: "a ^ n dvd b ^ n" and n: "n \<noteq> 0"
1.340 - shows "a dvd b"
1.341 -proof (cases "gcd a b = 0")
1.342 - case True
1.343 - then show ?thesis by simp
1.344 -next
1.345 - case False
1.346 - let ?d = "gcd a b"
1.347 - from n obtain m where m: "n = Suc m"
1.348 - by (cases n) simp_all
1.349 - from False have zn: "?d ^ n \<noteq> 0"
1.350 - by (rule power_not_zero)
1.351 - from gcd_coprime_exists [OF False]
1.352 - obtain a' b' where ab': "a = a' * ?d" "b = b' * ?d" "gcd a' b' = 1"
1.353 - by blast
1.354 - from ab have "(a' * ?d) ^ n dvd (b' * ?d) ^ n"
1.355 - by (simp add: ab'(1,2)[symmetric])
1.356 - then have "?d^n * a'^n dvd ?d^n * b'^n"
1.357 - by (simp only: power_mult_distrib ac_simps)
1.358 - with zn have "a'^n dvd b'^n"
1.359 - by simp
1.360 - then have "a' dvd b'^n"
1.361 - using dvd_trans[of a' "a'^n" "b'^n"] by (simp add: m)
1.362 - then have "a' dvd b'^m * b'"
1.363 - by (simp add: m ac_simps)
1.364 - with coprime_dvd_mult[OF coprime_exp[OF ab'(3), of m]]
1.365 - have "a' dvd b'" by (subst (asm) ac_simps) blast
1.366 - then have "a' * ?d dvd b' * ?d"
1.367 - by (rule mult_dvd_mono) simp
1.368 - with ab'(1,2) show ?thesis
1.369 - by simp
1.370 -qed
1.371 -
1.372 -lemma pow_divs_eq [simp]: "n \<noteq> 0 \<Longrightarrow> a ^ n dvd b ^ n \<longleftrightarrow> a dvd b"
1.373 - by (auto intro: pow_divs_pow dvd_power_same)
1.374 -
1.375 -lemma coprime_plus_one [simp]: "gcd (n + 1) n = 1"
1.376 - by (subst add_commute) simp
1.377 -
1.378 -lemma prod_coprime [rule_format]: "(\<forall>i\<in>A. gcd (f i) a = 1) \<longrightarrow> gcd (\<Prod>i\<in>A. f i) a = 1"
1.379 - by (induct A rule: infinite_finite_induct) (auto simp add: gcd_mult_cancel)
1.380 -
1.381 -lemma prod_list_coprime: "(\<And>x. x \<in> set xs \<Longrightarrow> coprime x y) \<Longrightarrow> coprime (prod_list xs) y"
1.382 - by (induct xs) (simp_all add: gcd_mult_cancel)
1.383 -
1.384 -lemma coprime_divisors:
1.385 - assumes "d dvd a" "e dvd b" "gcd a b = 1"
1.386 - shows "gcd d e = 1"
1.387 -proof -
1.388 - from assms obtain k l where "a = d * k" "b = e * l"
1.389 - unfolding dvd_def by blast
1.390 - with assms have "gcd (d * k) (e * l) = 1"
1.391 - by simp
1.392 - then have "gcd (d * k) e = 1"
1.393 - by (rule coprime_lmult)
1.394 - also have "gcd (d * k) e = gcd e (d * k)"
1.395 - by (simp add: ac_simps)
1.396 - finally have "gcd e d = 1"
1.397 - by (rule coprime_lmult)
1.398 - then show ?thesis
1.399 - by (simp add: ac_simps)
1.400 -qed
1.401 -
1.402 lemma lcm_gcd_prod: "lcm a b * gcd a b = normalize (a * b)"
1.403 by (simp add: lcm_gcd)
1.404
1.405 @@ -1006,9 +652,6 @@
1.406 "a dvd d \<and> b dvd d \<and> normalize d = d \<and> (\<forall>e. a dvd e \<and> b dvd e \<longrightarrow> d dvd e) \<longleftrightarrow> d = lcm a b"
1.407 by rule (auto intro: lcmI simp: lcm_least lcm_eq_0_iff)
1.408
1.409 -lemma lcm_coprime: "gcd a b = 1 \<Longrightarrow> lcm a b = normalize (a * b)"
1.410 - by (subst lcm_gcd) simp
1.411 -
1.412 lemma lcm_proj1_if_dvd: "b dvd a \<Longrightarrow> lcm a b = normalize a"
1.413 apply (cases "a = 0")
1.414 apply simp
1.415 @@ -1058,7 +701,7 @@
1.416 qed
1.417
1.418 lemma dvd_productE:
1.419 - assumes "p dvd (a * b)"
1.420 + assumes "p dvd a * b"
1.421 obtains x y where "p = x * y" "x dvd a" "y dvd b"
1.422 proof (cases "a = 0")
1.423 case True
1.424 @@ -1076,32 +719,11 @@
1.425 ultimately show ?thesis by (rule that)
1.426 qed
1.427
1.428 -lemma coprime_crossproduct':
1.429 - fixes a b c d
1.430 - assumes "b \<noteq> 0"
1.431 - assumes unit_factors: "unit_factor b = unit_factor d"
1.432 - assumes coprime: "coprime a b" "coprime c d"
1.433 - shows "a * d = b * c \<longleftrightarrow> a = c \<and> b = d"
1.434 -proof safe
1.435 - assume eq: "a * d = b * c"
1.436 - hence "normalize a * normalize d = normalize c * normalize b"
1.437 - by (simp only: normalize_mult [symmetric] mult_ac)
1.438 - with coprime have "normalize b = normalize d"
1.439 - by (subst (asm) coprime_crossproduct) simp_all
1.440 - from this and unit_factors show "b = d"
1.441 - by (rule normalize_unit_factor_eqI)
1.442 - from eq have "a * d = c * d" by (simp only: \<open>b = d\<close> mult_ac)
1.443 - with \<open>b \<noteq> 0\<close> \<open>b = d\<close> show "a = c" by simp
1.444 -qed (simp_all add: mult_ac)
1.445 -
1.446 end
1.447
1.448 class ring_gcd = comm_ring_1 + semiring_gcd
1.449 begin
1.450
1.451 -lemma coprime_minus_one: "coprime (n - 1) n"
1.452 - using coprime_plus_one[of "n - 1"] by (simp add: gcd.commute)
1.453 -
1.454 lemma gcd_neg1 [simp]: "gcd (-a) b = gcd a b"
1.455 by (rule sym, rule gcdI) (simp_all add: gcd_greatest)
1.456
1.457 @@ -1471,36 +1093,6 @@
1.458 lemma Lcm_2 [simp]: "Lcm {a, b} = lcm a b"
1.459 by simp
1.460
1.461 -lemma Lcm_coprime:
1.462 - assumes "finite A"
1.463 - and "A \<noteq> {}"
1.464 - and "\<And>a b. a \<in> A \<Longrightarrow> b \<in> A \<Longrightarrow> a \<noteq> b \<Longrightarrow> gcd a b = 1"
1.465 - shows "Lcm A = normalize (\<Prod>A)"
1.466 - using assms
1.467 -proof (induct rule: finite_ne_induct)
1.468 - case singleton
1.469 - then show ?case by simp
1.470 -next
1.471 - case (insert a A)
1.472 - have "Lcm (insert a A) = lcm a (Lcm A)"
1.473 - by simp
1.474 - also from insert have "Lcm A = normalize (\<Prod>A)"
1.475 - by blast
1.476 - also have "lcm a \<dots> = lcm a (\<Prod>A)"
1.477 - by (cases "\<Prod>A = 0") (simp_all add: lcm_div_unit2)
1.478 - also from insert have "gcd a (\<Prod>A) = 1"
1.479 - by (subst gcd.commute, intro prod_coprime) auto
1.480 - with insert have "lcm a (\<Prod>A) = normalize (\<Prod>(insert a A))"
1.481 - by (simp add: lcm_coprime)
1.482 - finally show ?case .
1.483 -qed
1.484 -
1.485 -lemma Lcm_coprime':
1.486 - "card A \<noteq> 0 \<Longrightarrow>
1.487 - (\<And>a b. a \<in> A \<Longrightarrow> b \<in> A \<Longrightarrow> a \<noteq> b \<Longrightarrow> gcd a b = 1) \<Longrightarrow>
1.488 - Lcm A = normalize (\<Prod>A)"
1.489 - by (rule Lcm_coprime) (simp_all add: card_eq_0_iff)
1.490 -
1.491 lemma Gcd_1: "1 \<in> A \<Longrightarrow> Gcd A = 1"
1.492 by (auto intro!: Gcd_eq_1_I)
1.493
1.494 @@ -1677,6 +1269,465 @@
1.495
1.496 end
1.497
1.498 +
1.499 +subsection \<open>Coprimality\<close>
1.500 +
1.501 +context semiring_gcd
1.502 +begin
1.503 +
1.504 +lemma coprime_imp_gcd_eq_1 [simp]:
1.505 + "gcd a b = 1" if "coprime a b"
1.506 +proof -
1.507 + define t r s where "t = gcd a b" and "r = a div t" and "s = b div t"
1.508 + then have "a = t * r" and "b = t * s"
1.509 + by simp_all
1.510 + with that have "coprime (t * r) (t * s)"
1.511 + by simp
1.512 + then show ?thesis
1.513 + by (simp add: t_def)
1.514 +qed
1.515 +
1.516 +lemma gcd_eq_1_imp_coprime:
1.517 + "coprime a b" if "gcd a b = 1"
1.518 +proof (rule coprimeI)
1.519 + fix c
1.520 + assume "c dvd a" and "c dvd b"
1.521 + then have "c dvd gcd a b"
1.522 + by (rule gcd_greatest)
1.523 + with that show "is_unit c"
1.524 + by simp
1.525 +qed
1.526 +
1.527 +lemma coprime_iff_gcd_eq_1 [presburger, code]:
1.528 + "coprime a b \<longleftrightarrow> gcd a b = 1"
1.529 + by rule (simp_all add: gcd_eq_1_imp_coprime)
1.530 +
1.531 +lemma is_unit_gcd [simp]:
1.532 + "is_unit (gcd a b) \<longleftrightarrow> coprime a b"
1.533 + by (simp add: coprime_iff_gcd_eq_1)
1.534 +
1.535 +lemma coprime_add_one_left [simp]: "coprime (a + 1) a"
1.536 + by (simp add: gcd_eq_1_imp_coprime ac_simps)
1.537 +
1.538 +lemma coprime_add_one_right [simp]: "coprime a (a + 1)"
1.539 + using coprime_add_one_left [of a] by (simp add: ac_simps)
1.540 +
1.541 +lemma coprime_mult_left_iff [simp]:
1.542 + "coprime (a * b) c \<longleftrightarrow> coprime a c \<and> coprime b c"
1.543 +proof
1.544 + assume "coprime (a * b) c"
1.545 + with coprime_common_divisor [of "a * b" c]
1.546 + have *: "is_unit d" if "d dvd a * b" and "d dvd c" for d
1.547 + using that by blast
1.548 + have "coprime a c"
1.549 + by (rule coprimeI, rule *) simp_all
1.550 + moreover have "coprime b c"
1.551 + by (rule coprimeI, rule *) simp_all
1.552 + ultimately show "coprime a c \<and> coprime b c" ..
1.553 +next
1.554 + assume "coprime a c \<and> coprime b c"
1.555 + then have "coprime a c" "coprime b c"
1.556 + by simp_all
1.557 + show "coprime (a * b) c"
1.558 + proof (rule coprimeI)
1.559 + fix d
1.560 + assume "d dvd a * b"
1.561 + then obtain r s where d: "d = r * s" "r dvd a" "s dvd b"
1.562 + by (rule dvd_productE)
1.563 + assume "d dvd c"
1.564 + with d have "r * s dvd c"
1.565 + by simp
1.566 + then have "r dvd c" "s dvd c"
1.567 + by (auto intro: dvd_mult_left dvd_mult_right)
1.568 + from \<open>coprime a c\<close> \<open>r dvd a\<close> \<open>r dvd c\<close>
1.569 + have "is_unit r"
1.570 + by (rule coprime_common_divisor)
1.571 + moreover from \<open>coprime b c\<close> \<open>s dvd b\<close> \<open>s dvd c\<close>
1.572 + have "is_unit s"
1.573 + by (rule coprime_common_divisor)
1.574 + ultimately show "is_unit d"
1.575 + by (simp add: d is_unit_mult_iff)
1.576 + qed
1.577 +qed
1.578 +
1.579 +lemma coprime_mult_right_iff [simp]:
1.580 + "coprime c (a * b) \<longleftrightarrow> coprime c a \<and> coprime c b"
1.581 + using coprime_mult_left_iff [of a b c] by (simp add: ac_simps)
1.582 +
1.583 +lemma coprime_power_left_iff [simp]:
1.584 + "coprime (a ^ n) b \<longleftrightarrow> coprime a b \<or> n = 0"
1.585 +proof (cases "n = 0")
1.586 + case True
1.587 + then show ?thesis
1.588 + by simp
1.589 +next
1.590 + case False
1.591 + then have "n > 0"
1.592 + by simp
1.593 + then show ?thesis
1.594 + by (induction n rule: nat_induct_non_zero) simp_all
1.595 +qed
1.596 +
1.597 +lemma coprime_power_right_iff [simp]:
1.598 + "coprime a (b ^ n) \<longleftrightarrow> coprime a b \<or> n = 0"
1.599 + using coprime_power_left_iff [of b n a] by (simp add: ac_simps)
1.600 +
1.601 +lemma prod_coprime_left:
1.602 + "coprime (\<Prod>i\<in>A. f i) a" if "\<And>i. i \<in> A \<Longrightarrow> coprime (f i) a"
1.603 + using that by (induct A rule: infinite_finite_induct) simp_all
1.604 +
1.605 +lemma prod_coprime_right:
1.606 + "coprime a (\<Prod>i\<in>A. f i)" if "\<And>i. i \<in> A \<Longrightarrow> coprime a (f i)"
1.607 + using that prod_coprime_left [of A f a] by (simp add: ac_simps)
1.608 +
1.609 +lemma prod_list_coprime_left:
1.610 + "coprime (prod_list xs) a" if "\<And>x. x \<in> set xs \<Longrightarrow> coprime x a"
1.611 + using that by (induct xs) simp_all
1.612 +
1.613 +lemma prod_list_coprime_right:
1.614 + "coprime a (prod_list xs)" if "\<And>x. x \<in> set xs \<Longrightarrow> coprime a x"
1.615 + using that prod_list_coprime_left [of xs a] by (simp add: ac_simps)
1.616 +
1.617 +lemma coprime_dvd_mult_left_iff:
1.618 + "a dvd b * c \<longleftrightarrow> a dvd b" if "coprime a c"
1.619 +proof
1.620 + assume "a dvd b"
1.621 + then show "a dvd b * c"
1.622 + by simp
1.623 +next
1.624 + assume "a dvd b * c"
1.625 + show "a dvd b"
1.626 + proof (cases "b = 0")
1.627 + case True
1.628 + then show ?thesis
1.629 + by simp
1.630 + next
1.631 + case False
1.632 + then have unit: "is_unit (unit_factor b)"
1.633 + by simp
1.634 + from \<open>coprime a c\<close> mult_gcd_left [of b a c]
1.635 + have "gcd (b * a) (b * c) * unit_factor b = b"
1.636 + by (simp add: ac_simps)
1.637 + moreover from \<open>a dvd b * c\<close>
1.638 + have "a dvd gcd (b * a) (b * c) * unit_factor b"
1.639 + by (simp add: dvd_mult_unit_iff unit)
1.640 + ultimately show ?thesis
1.641 + by simp
1.642 + qed
1.643 +qed
1.644 +
1.645 +lemma coprime_dvd_mult_right_iff:
1.646 + "a dvd c * b \<longleftrightarrow> a dvd b" if "coprime a c"
1.647 + using that coprime_dvd_mult_left_iff [of a c b] by (simp add: ac_simps)
1.648 +
1.649 +lemma divides_mult:
1.650 + "a * b dvd c" if "a dvd c" and "b dvd c" and "coprime a b"
1.651 +proof -
1.652 + from \<open>b dvd c\<close> obtain b' where "c = b * b'" ..
1.653 + with \<open>a dvd c\<close> have "a dvd b' * b"
1.654 + by (simp add: ac_simps)
1.655 + with \<open>coprime a b\<close> have "a dvd b'"
1.656 + by (simp add: coprime_dvd_mult_left_iff)
1.657 + then obtain a' where "b' = a * a'" ..
1.658 + with \<open>c = b * b'\<close> have "c = (a * b) * a'"
1.659 + by (simp add: ac_simps)
1.660 + then show ?thesis ..
1.661 +qed
1.662 +
1.663 +lemma div_gcd_coprime:
1.664 + assumes "a \<noteq> 0 \<or> b \<noteq> 0"
1.665 + shows "coprime (a div gcd a b) (b div gcd a b)"
1.666 +proof -
1.667 + let ?g = "gcd a b"
1.668 + let ?a' = "a div ?g"
1.669 + let ?b' = "b div ?g"
1.670 + let ?g' = "gcd ?a' ?b'"
1.671 + have dvdg: "?g dvd a" "?g dvd b"
1.672 + by simp_all
1.673 + have dvdg': "?g' dvd ?a'" "?g' dvd ?b'"
1.674 + by simp_all
1.675 + from dvdg dvdg' obtain ka kb ka' kb' where
1.676 + kab: "a = ?g * ka" "b = ?g * kb" "?a' = ?g' * ka'" "?b' = ?g' * kb'"
1.677 + unfolding dvd_def by blast
1.678 + from this [symmetric] have "?g * ?a' = (?g * ?g') * ka'" "?g * ?b' = (?g * ?g') * kb'"
1.679 + by (simp_all add: mult.assoc mult.left_commute [of "gcd a b"])
1.680 + then have dvdgg':"?g * ?g' dvd a" "?g* ?g' dvd b"
1.681 + by (auto simp add: dvd_mult_div_cancel [OF dvdg(1)] dvd_mult_div_cancel [OF dvdg(2)] dvd_def)
1.682 + have "?g \<noteq> 0"
1.683 + using assms by simp
1.684 + moreover from gcd_greatest [OF dvdgg'] have "?g * ?g' dvd ?g" .
1.685 + ultimately show ?thesis
1.686 + using dvd_times_left_cancel_iff [of "gcd a b" _ 1]
1.687 + by simp (simp only: coprime_iff_gcd_eq_1)
1.688 +qed
1.689 +
1.690 +lemma gcd_coprime:
1.691 + assumes c: "gcd a b \<noteq> 0"
1.692 + and a: "a = a' * gcd a b"
1.693 + and b: "b = b' * gcd a b"
1.694 + shows "coprime a' b'"
1.695 +proof -
1.696 + from c have "a \<noteq> 0 \<or> b \<noteq> 0"
1.697 + by simp
1.698 + with div_gcd_coprime have "coprime (a div gcd a b) (b div gcd a b)" .
1.699 + also from assms have "a div gcd a b = a'"
1.700 + using dvd_div_eq_mult local.gcd_dvd1 by blast
1.701 + also from assms have "b div gcd a b = b'"
1.702 + using dvd_div_eq_mult local.gcd_dvd1 by blast
1.703 + finally show ?thesis .
1.704 +qed
1.705 +
1.706 +lemma gcd_coprime_exists:
1.707 + assumes "gcd a b \<noteq> 0"
1.708 + shows "\<exists>a' b'. a = a' * gcd a b \<and> b = b' * gcd a b \<and> coprime a' b'"
1.709 + apply (rule_tac x = "a div gcd a b" in exI)
1.710 + apply (rule_tac x = "b div gcd a b" in exI)
1.711 + using assms
1.712 + apply (auto intro: div_gcd_coprime)
1.713 + done
1.714 +
1.715 +lemma pow_divides_pow_iff [simp]:
1.716 + "a ^ n dvd b ^ n \<longleftrightarrow> a dvd b" if "n > 0"
1.717 +proof (cases "gcd a b = 0")
1.718 + case True
1.719 + then show ?thesis
1.720 + by simp
1.721 +next
1.722 + case False
1.723 + show ?thesis
1.724 + proof
1.725 + let ?d = "gcd a b"
1.726 + from \<open>n > 0\<close> obtain m where m: "n = Suc m"
1.727 + by (cases n) simp_all
1.728 + from False have zn: "?d ^ n \<noteq> 0"
1.729 + by (rule power_not_zero)
1.730 + from gcd_coprime_exists [OF False]
1.731 + obtain a' b' where ab': "a = a' * ?d" "b = b' * ?d" "coprime a' b'"
1.732 + by blast
1.733 + assume "a ^ n dvd b ^ n"
1.734 + then have "(a' * ?d) ^ n dvd (b' * ?d) ^ n"
1.735 + by (simp add: ab'(1,2)[symmetric])
1.736 + then have "?d^n * a'^n dvd ?d^n * b'^n"
1.737 + by (simp only: power_mult_distrib ac_simps)
1.738 + with zn have "a' ^ n dvd b' ^ n"
1.739 + by simp
1.740 + then have "a' dvd b' ^ n"
1.741 + using dvd_trans[of a' "a'^n" "b'^n"] by (simp add: m)
1.742 + then have "a' dvd b' ^ m * b'"
1.743 + by (simp add: m ac_simps)
1.744 + moreover have "coprime a' (b' ^ n)"
1.745 + using \<open>coprime a' b'\<close> by simp
1.746 + then have "a' dvd b'"
1.747 + using \<open>a' dvd b' ^ n\<close> coprime_dvd_mult_left_iff dvd_mult by blast
1.748 + then have "a' * ?d dvd b' * ?d"
1.749 + by (rule mult_dvd_mono) simp
1.750 + with ab'(1,2) show "a dvd b"
1.751 + by simp
1.752 + next
1.753 + assume "a dvd b"
1.754 + with \<open>n > 0\<close> show "a ^ n dvd b ^ n"
1.755 + by (induction rule: nat_induct_non_zero)
1.756 + (simp_all add: mult_dvd_mono)
1.757 + qed
1.758 +qed
1.759 +
1.760 +lemma coprime_crossproduct:
1.761 + fixes a b c d :: 'a
1.762 + assumes "coprime a d" and "coprime b c"
1.763 + shows "normalize a * normalize c = normalize b * normalize d \<longleftrightarrow>
1.764 + normalize a = normalize b \<and> normalize c = normalize d"
1.765 + (is "?lhs \<longleftrightarrow> ?rhs")
1.766 +proof
1.767 + assume ?rhs
1.768 + then show ?lhs by simp
1.769 +next
1.770 + assume ?lhs
1.771 + from \<open>?lhs\<close> have "normalize a dvd normalize b * normalize d"
1.772 + by (auto intro: dvdI dest: sym)
1.773 + with \<open>coprime a d\<close> have "a dvd b"
1.774 + by (simp add: coprime_dvd_mult_left_iff normalize_mult [symmetric])
1.775 + from \<open>?lhs\<close> have "normalize b dvd normalize a * normalize c"
1.776 + by (auto intro: dvdI dest: sym)
1.777 + with \<open>coprime b c\<close> have "b dvd a"
1.778 + by (simp add: coprime_dvd_mult_left_iff normalize_mult [symmetric])
1.779 + from \<open>?lhs\<close> have "normalize c dvd normalize d * normalize b"
1.780 + by (auto intro: dvdI dest: sym simp add: mult.commute)
1.781 + with \<open>coprime b c\<close> have "c dvd d"
1.782 + by (simp add: coprime_dvd_mult_left_iff coprime_commute normalize_mult [symmetric])
1.783 + from \<open>?lhs\<close> have "normalize d dvd normalize c * normalize a"
1.784 + by (auto intro: dvdI dest: sym simp add: mult.commute)
1.785 + with \<open>coprime a d\<close> have "d dvd c"
1.786 + by (simp add: coprime_dvd_mult_left_iff coprime_commute normalize_mult [symmetric])
1.787 + from \<open>a dvd b\<close> \<open>b dvd a\<close> have "normalize a = normalize b"
1.788 + by (rule associatedI)
1.789 + moreover from \<open>c dvd d\<close> \<open>d dvd c\<close> have "normalize c = normalize d"
1.790 + by (rule associatedI)
1.791 + ultimately show ?rhs ..
1.792 +qed
1.793 +
1.794 +lemma coprime_crossproduct':
1.795 + fixes a b c d
1.796 + assumes "b \<noteq> 0"
1.797 + assumes unit_factors: "unit_factor b = unit_factor d"
1.798 + assumes coprime: "coprime a b" "coprime c d"
1.799 + shows "a * d = b * c \<longleftrightarrow> a = c \<and> b = d"
1.800 +proof safe
1.801 + assume eq: "a * d = b * c"
1.802 + hence "normalize a * normalize d = normalize c * normalize b"
1.803 + by (simp only: normalize_mult [symmetric] mult_ac)
1.804 + with coprime have "normalize b = normalize d"
1.805 + by (subst (asm) coprime_crossproduct) simp_all
1.806 + from this and unit_factors show "b = d"
1.807 + by (rule normalize_unit_factor_eqI)
1.808 + from eq have "a * d = c * d" by (simp only: \<open>b = d\<close> mult_ac)
1.809 + with \<open>b \<noteq> 0\<close> \<open>b = d\<close> show "a = c" by simp
1.810 +qed (simp_all add: mult_ac)
1.811 +
1.812 +lemma gcd_mult_left_left_cancel:
1.813 + "gcd (c * a) b = gcd a b" if "coprime b c"
1.814 +proof -
1.815 + have "coprime (gcd b (a * c)) c"
1.816 + by (rule coprimeI) (auto intro: that coprime_common_divisor)
1.817 + then have "gcd b (a * c) dvd a"
1.818 + using coprime_dvd_mult_left_iff [of "gcd b (a * c)" c a]
1.819 + by simp
1.820 + then show ?thesis
1.821 + by (auto intro: associated_eqI simp add: ac_simps)
1.822 +qed
1.823 +
1.824 +lemma gcd_mult_left_right_cancel:
1.825 + "gcd (a * c) b = gcd a b" if "coprime b c"
1.826 + using that gcd_mult_left_left_cancel [of b c a]
1.827 + by (simp add: ac_simps)
1.828 +
1.829 +lemma gcd_mult_right_left_cancel:
1.830 + "gcd a (c * b) = gcd a b" if "coprime a c"
1.831 + using that gcd_mult_left_left_cancel [of a c b]
1.832 + by (simp add: ac_simps)
1.833 +
1.834 +lemma gcd_mult_right_right_cancel:
1.835 + "gcd a (b * c) = gcd a b" if "coprime a c"
1.836 + using that gcd_mult_right_left_cancel [of a c b]
1.837 + by (simp add: ac_simps)
1.838 +
1.839 +lemma gcd_exp [simp]:
1.840 + "gcd (a ^ n) (b ^ n) = gcd a b ^ n"
1.841 +proof (cases "a = 0 \<and> b = 0 \<or> n = 0")
1.842 + case True
1.843 + then show ?thesis
1.844 + by (cases n) simp_all
1.845 +next
1.846 + case False
1.847 + then have "coprime (a div gcd a b) (b div gcd a b)" and "n > 0"
1.848 + by (auto intro: div_gcd_coprime)
1.849 + then have "coprime ((a div gcd a b) ^ n) ((b div gcd a b) ^ n)"
1.850 + by simp
1.851 + then have "1 = gcd ((a div gcd a b) ^ n) ((b div gcd a b) ^ n)"
1.852 + by simp
1.853 + then have "gcd a b ^ n = gcd a b ^ n * \<dots>"
1.854 + by simp
1.855 + also note gcd_mult_distrib
1.856 + also have "unit_factor (gcd a b ^ n) = 1"
1.857 + using False by (auto simp add: unit_factor_power unit_factor_gcd)
1.858 + also have "(gcd a b) ^ n * (a div gcd a b) ^ n = a ^ n"
1.859 + by (simp add: ac_simps div_power dvd_power_same)
1.860 + also have "(gcd a b) ^ n * (b div gcd a b) ^ n = b ^ n"
1.861 + by (simp add: ac_simps div_power dvd_power_same)
1.862 + finally show ?thesis by simp
1.863 +qed
1.864 +
1.865 +lemma division_decomp:
1.866 + assumes "a dvd b * c"
1.867 + shows "\<exists>b' c'. a = b' * c' \<and> b' dvd b \<and> c' dvd c"
1.868 +proof (cases "gcd a b = 0")
1.869 + case True
1.870 + then have "a = 0 \<and> b = 0"
1.871 + by simp
1.872 + then have "a = 0 * c \<and> 0 dvd b \<and> c dvd c"
1.873 + by simp
1.874 + then show ?thesis by blast
1.875 +next
1.876 + case False
1.877 + let ?d = "gcd a b"
1.878 + from gcd_coprime_exists [OF False]
1.879 + obtain a' b' where ab': "a = a' * ?d" "b = b' * ?d" "coprime a' b'"
1.880 + by blast
1.881 + from ab'(1) have "a' dvd a" ..
1.882 + with assms have "a' dvd b * c"
1.883 + using dvd_trans [of a' a "b * c"] by simp
1.884 + from assms ab'(1,2) have "a' * ?d dvd (b' * ?d) * c"
1.885 + by simp
1.886 + then have "?d * a' dvd ?d * (b' * c)"
1.887 + by (simp add: mult_ac)
1.888 + with \<open>?d \<noteq> 0\<close> have "a' dvd b' * c"
1.889 + by simp
1.890 + then have "a' dvd c"
1.891 + using \<open>coprime a' b'\<close> by (simp add: coprime_dvd_mult_right_iff)
1.892 + with ab'(1) have "a = ?d * a' \<and> ?d dvd b \<and> a' dvd c"
1.893 + by (simp add: ac_simps)
1.894 + then show ?thesis by blast
1.895 +qed
1.896 +
1.897 +lemma lcm_coprime: "coprime a b \<Longrightarrow> lcm a b = normalize (a * b)"
1.898 + by (subst lcm_gcd) simp
1.899 +
1.900 +end
1.901 +
1.902 +context ring_gcd
1.903 +begin
1.904 +
1.905 +lemma coprime_minus_left_iff [simp]:
1.906 + "coprime (- a) b \<longleftrightarrow> coprime a b"
1.907 + by (rule; rule coprimeI) (auto intro: coprime_common_divisor)
1.908 +
1.909 +lemma coprime_minus_right_iff [simp]:
1.910 + "coprime a (- b) \<longleftrightarrow> coprime a b"
1.911 + using coprime_minus_left_iff [of b a] by (simp add: ac_simps)
1.912 +
1.913 +lemma coprime_diff_one_left [simp]: "coprime (a - 1) a"
1.914 + using coprime_add_one_right [of "a - 1"] by simp
1.915 +
1.916 +lemma coprime_doff_one_right [simp]: "coprime a (a - 1)"
1.917 + using coprime_diff_one_left [of a] by (simp add: ac_simps)
1.918 +
1.919 +end
1.920 +
1.921 +context semiring_Gcd
1.922 +begin
1.923 +
1.924 +lemma Lcm_coprime:
1.925 + assumes "finite A"
1.926 + and "A \<noteq> {}"
1.927 + and "\<And>a b. a \<in> A \<Longrightarrow> b \<in> A \<Longrightarrow> a \<noteq> b \<Longrightarrow> coprime a b"
1.928 + shows "Lcm A = normalize (\<Prod>A)"
1.929 + using assms
1.930 +proof (induct rule: finite_ne_induct)
1.931 + case singleton
1.932 + then show ?case by simp
1.933 +next
1.934 + case (insert a A)
1.935 + have "Lcm (insert a A) = lcm a (Lcm A)"
1.936 + by simp
1.937 + also from insert have "Lcm A = normalize (\<Prod>A)"
1.938 + by blast
1.939 + also have "lcm a \<dots> = lcm a (\<Prod>A)"
1.940 + by (cases "\<Prod>A = 0") (simp_all add: lcm_div_unit2)
1.941 + also from insert have "coprime a (\<Prod>A)"
1.942 + by (subst coprime_commute, intro prod_coprime_left) auto
1.943 + with insert have "lcm a (\<Prod>A) = normalize (\<Prod>(insert a A))"
1.944 + by (simp add: lcm_coprime)
1.945 + finally show ?case .
1.946 +qed
1.947 +
1.948 +lemma Lcm_coprime':
1.949 + "card A \<noteq> 0 \<Longrightarrow>
1.950 + (\<And>a b. a \<in> A \<Longrightarrow> b \<in> A \<Longrightarrow> a \<noteq> b \<Longrightarrow> coprime a b) \<Longrightarrow>
1.951 + Lcm A = normalize (\<Prod>A)"
1.952 + by (rule Lcm_coprime) (simp_all add: card_eq_0_iff)
1.953 +
1.954 +end
1.955 +
1.956 +
1.957 subsection \<open>GCD and LCM on @{typ nat} and @{typ int}\<close>
1.958
1.959 instantiation nat :: gcd
1.960 @@ -1716,9 +1767,6 @@
1.961 apply simp_all
1.962 done
1.963
1.964 -
1.965 -text \<open>Specific to \<open>int\<close>.\<close>
1.966 -
1.967 lemma gcd_eq_int_iff: "gcd k l = int n \<longleftrightarrow> gcd (nat \<bar>k\<bar>) (nat \<bar>l\<bar>) = n"
1.968 by (simp add: gcd_int_def)
1.969
1.970 @@ -1949,19 +1997,6 @@
1.971 for k m n :: int
1.972 by (simp add: gcd_int_def abs_mult nat_mult_distrib gcd_mult_distrib_nat [symmetric])
1.973
1.974 -lemma coprime_crossproduct_nat:
1.975 - fixes a b c d :: nat
1.976 - assumes "coprime a d" and "coprime b c"
1.977 - shows "a * c = b * d \<longleftrightarrow> a = b \<and> c = d"
1.978 - using assms coprime_crossproduct [of a d b c] by simp
1.979 -
1.980 -lemma coprime_crossproduct_int:
1.981 - fixes a b c d :: int
1.982 - assumes "coprime a d" and "coprime b c"
1.983 - shows "\<bar>a\<bar> * \<bar>c\<bar> = \<bar>b\<bar> * \<bar>d\<bar> \<longleftrightarrow> \<bar>a\<bar> = \<bar>b\<bar> \<and> \<bar>c\<bar> = \<bar>d\<bar>"
1.984 - using assms coprime_crossproduct [of a d b c] by simp
1.985 -
1.986 -
1.987 text \<open>\medskip Addition laws.\<close>
1.988
1.989 (* TODO: add the other variations? *)
1.990 @@ -2064,53 +2099,33 @@
1.991 for k l :: int
1.992 by (simp add: gcd_int_def nat_mod_distrib gcd_non_0_nat)
1.993
1.994 -
1.995 -subsection \<open>Coprimality\<close>
1.996 -
1.997 -lemma coprime_nat: "coprime a b \<longleftrightarrow> (\<forall>d. d dvd a \<and> d dvd b \<longleftrightarrow> d = 1)"
1.998 - for a b :: nat
1.999 - using coprime [of a b] by simp
1.1000 -
1.1001 -lemma coprime_Suc_0_nat: "coprime a b \<longleftrightarrow> (\<forall>d. d dvd a \<and> d dvd b \<longleftrightarrow> d = Suc 0)"
1.1002 - for a b :: nat
1.1003 - using coprime_nat by simp
1.1004 -
1.1005 -lemma coprime_int: "coprime a b \<longleftrightarrow> (\<forall>d. d \<ge> 0 \<and> d dvd a \<and> d dvd b \<longleftrightarrow> d = 1)"
1.1006 - for a b :: int
1.1007 - using gcd_unique_int [of 1 a b]
1.1008 - apply clarsimp
1.1009 - apply (erule subst)
1.1010 - apply (rule iffI)
1.1011 - apply force
1.1012 - using abs_dvd_iff abs_ge_zero apply blast
1.1013 - done
1.1014 -
1.1015 -lemma pow_divides_eq_nat [simp]: "n > 0 \<Longrightarrow> a^n dvd b^n \<longleftrightarrow> a dvd b"
1.1016 - for a b n :: nat
1.1017 - using pow_divs_eq[of n] by simp
1.1018 -
1.1019 -lemma coprime_Suc_nat [simp]: "coprime (Suc n) n"
1.1020 - using coprime_plus_one[of n] by simp
1.1021 -
1.1022 -lemma coprime_minus_one_nat: "n \<noteq> 0 \<Longrightarrow> coprime (n - 1) n"
1.1023 - for n :: nat
1.1024 - using coprime_Suc_nat [of "n - 1"] gcd.commute [of "n - 1" n] by auto
1.1025 -
1.1026 -lemma coprime_common_divisor_nat: "coprime a b \<Longrightarrow> x dvd a \<Longrightarrow> x dvd b \<Longrightarrow> x = 1"
1.1027 - for a b :: nat
1.1028 - by (metis gcd_greatest_iff nat_dvd_1_iff_1)
1.1029 -
1.1030 -lemma coprime_common_divisor_int: "coprime a b \<Longrightarrow> x dvd a \<Longrightarrow> x dvd b \<Longrightarrow> \<bar>x\<bar> = 1"
1.1031 - for a b :: int
1.1032 - using gcd_greatest_iff [of x a b] by auto
1.1033 -
1.1034 -lemma invertible_coprime_nat: "x * y mod m = 1 \<Longrightarrow> coprime x m"
1.1035 - for m x y :: nat
1.1036 - by (metis coprime_lmult gcd_1_nat gcd.commute gcd_red_nat)
1.1037 -
1.1038 -lemma invertible_coprime_int: "x * y mod m = 1 \<Longrightarrow> coprime x m"
1.1039 - for m x y :: int
1.1040 - by (metis coprime_lmult gcd_1_int gcd.commute gcd_red_int)
1.1041 +lemma coprime_Suc_left_nat [simp]:
1.1042 + "coprime (Suc n) n"
1.1043 + using coprime_add_one_left [of n] by simp
1.1044 +
1.1045 +lemma coprime_Suc_right_nat [simp]:
1.1046 + "coprime n (Suc n)"
1.1047 + using coprime_Suc_left_nat [of n] by (simp add: ac_simps)
1.1048 +
1.1049 +lemma coprime_diff_one_left_nat [simp]:
1.1050 + "coprime (n - 1) n" if "n > 0" for n :: nat
1.1051 + using that coprime_Suc_right_nat [of "n - 1"] by simp
1.1052 +
1.1053 +lemma coprime_diff_one_right_nat [simp]:
1.1054 + "coprime n (n - 1)" if "n > 0" for n :: nat
1.1055 + using that coprime_diff_one_left_nat [of n] by (simp add: ac_simps)
1.1056 +
1.1057 +lemma coprime_crossproduct_nat:
1.1058 + fixes a b c d :: nat
1.1059 + assumes "coprime a d" and "coprime b c"
1.1060 + shows "a * c = b * d \<longleftrightarrow> a = b \<and> c = d"
1.1061 + using assms coprime_crossproduct [of a d b c] by simp
1.1062 +
1.1063 +lemma coprime_crossproduct_int:
1.1064 + fixes a b c d :: int
1.1065 + assumes "coprime a d" and "coprime b c"
1.1066 + shows "\<bar>a\<bar> * \<bar>c\<bar> = \<bar>b\<bar> * \<bar>d\<bar> \<longleftrightarrow> \<bar>a\<bar> = \<bar>b\<bar> \<and> \<bar>c\<bar> = \<bar>d\<bar>"
1.1067 + using assms coprime_crossproduct [of a d b c] by simp
1.1068
1.1069
1.1070 subsection \<open>Bezout's theorem\<close>
1.1071 @@ -2742,14 +2757,6 @@
1.1072 for i m n :: int
1.1073 by (fact dvd_lcmI2)
1.1074
1.1075 -lemma coprime_exp2_nat [intro]: "coprime a b \<Longrightarrow> coprime (a^n) (b^m)"
1.1076 - for a b :: nat
1.1077 - by (fact coprime_exp2)
1.1078 -
1.1079 -lemma coprime_exp2_int [intro]: "coprime a b \<Longrightarrow> coprime (a^n) (b^m)"
1.1080 - for a b :: int
1.1081 - by (fact coprime_exp2)
1.1082 -
1.1083 lemmas Gcd_dvd_nat [simp] = Gcd_dvd [where ?'a = nat]
1.1084 lemmas Gcd_dvd_int [simp] = Gcd_dvd [where ?'a = int]
1.1085 lemmas Gcd_greatest_nat [simp] = Gcd_greatest [where ?'a = nat]
``` | 17,008 | 43,594 | {"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-2019-22 | latest | en | 0.292077 |
https://math.stackexchange.com/questions/4776903/interesting-infinite-product-sqrt2-1-dfrac1-cdot7-cdot9-cdot15-cdot17-cdot/4776960 | 1,713,525,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00475.warc.gz | 325,145,265 | 42,047 | # Interesting infinite product $\sqrt{2}-1=\dfrac{1\cdot7\cdot9\cdot15\cdot17\cdot23\cdots}{3\cdot5\cdot11\cdot13\cdot19\cdot21\cdots}$
I have found an interesting family of infinite products. The most interesting one of them being:
$$\sqrt{2}-1=\dfrac{1\cdot7\cdot9\cdot15\cdot17\cdot23\cdots}{3\cdot5\cdot11\cdot13\cdot19\cdot21\cdots}$$
The numerators follow the pattern $$+6,+2,+6,+2,\cdots$$ and the denominators follow $$+2,+6,+2,+6,\cdots$$
The family of products was derived from assuming: $$\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right) \cdots$$
And subsequently replacing $$x$$ as $$\dfrac{\pi}{2}$$
The keyword here being "assumed", so I don't know if this product can be proven. The values do seem to be equal after a few iterations though.
However, I would greatly appreciate any proofs or alternate derivations.
• Multiplying $10^8+1$ factors gives a value about $10^{-9}$ far away , so the product seems to converge (however slowly) to $\sqrt{2}-1$. Of course , this is not yet a proof. Sep 28, 2023 at 6:32
• I would write $\sin(x) - 1/\sqrt 2 = \sin(x) - \sin(\pi/4)$, apply the sum to product identity, and then write each factor as an infinite product math.stackexchange.com/q/674769/42969. Sep 28, 2023 at 6:58
• Might be connected : math.stackexchange.com/questions/157372/… : math.stackexchange.com/questions/674769/… : math.stackexchange.com/questions/134870/… : math.stackexchange.com/questions/3899658/… : "Weierstrass Factorization Theorem" & "Wallis Product" ?? Interesting Question !!
– Prem
Sep 28, 2023 at 7:08
• In the following paper by Sondow and Yi, a number of relevant infinite products are derived. In particular, I think it is worthwhile to look into equations (9) and (11). These give expressions for $\sqrt{ 2 + \sqrt{2}}$ and $\sqrt{ 2 - \sqrt{2}}$. The denominators also have the +2, +6 pattern. The numerators have a +2, +4 pattern: jstor.org/stable/pdf/10.4169/… Sep 28, 2023 at 7:53
• ^ correction: the numerators have a +4, +4 pattern. The denominators are correct as stated. Sep 28, 2023 at 8:22
You wrote
The family of products was derived from assuming: $$\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right) \cdots$$
That assumption is indeed correct.
Proof: Using the “sum to product” identity for the sine we have $$f(x) = \sin(x) - \frac{1}{\sqrt 2} = \sin(x) - \sin\left(\frac \pi 4\right) = -2 \sin\left(\frac x2 - \frac \pi 8\right) \sin\left(\frac x2 - \frac {3\pi}8 \right) \, .$$ Using the infinite product for the sine (see, e.g., here), the first factor can be written as $$\sin\left(\frac x2 - \frac \pi 8\right) = \left(\frac x2 - \frac \pi 8\right)\prod_{n=1}^\infty \left(1-\frac{x}{2\pi n} + \frac{1}{8n}\right) \left(1+\frac{x}{2\pi n} - \frac{1}{8n}\right) \\ = (-\frac \pi 8) \left(1 - \frac {4x}{\pi} \right)\prod_{n=1}^\infty \left(1 + \frac{1}{8n}\right)\left(1-\frac{4x}{(8n+1)\pi} \right) \left(1- \frac{1}{8n}\right)\left(1+\frac{4x}{(8n-1)\pi}\right) \\ = C_1 \left(1 - \frac {4x}{\pi} \right)\prod_{n=1}^\infty \left(1+\frac{4x}{(8n-1)\pi}\right)\left(1-\frac{4x}{(8n+1)\pi} \right)$$ with the constant $$C_1 = (-\frac \pi 8)\prod_{n=1}^\infty \left(1- \left(\frac{1}{8n}\right)^2\right) \, .$$ In the same way we get for the second factor $$\sin\left(\frac x2 - \frac {3\pi}8 \right) = C_2 \left(1 - \frac {4x}{3\pi} \right)\prod_{n=1}^\infty \left(1+\frac{4x}{(8n-3)\pi} \right) \left(1-\frac{4x}{(8n+3)\pi}\right)$$ with some constant $$C_2$$.
Combining these results we have $$f(x) = C \left(1 - \frac {4x}{\pi} \right)\left(1 - \frac {4x}{3\pi} \right) \\ \times \prod_{n=1}^\infty\left(1+\frac{4x}{(8n-3)\pi} \right) \left(1+\frac{4x}{(8n-1)\pi} \right)\left(1-\frac{4x}{(8n+1)\pi} \right)\left(1-\frac{4x}{(8n+3)\pi} \right)$$ with some constant $$C$$.
Setting $$x=0$$ shows that $$C= -1/\sqrt 2$$, and that concludes the proof.
• All the proofs mentioned here are wonderful but this is the only one I could understand because I only know basic trigonometry :) Sep 29, 2023 at 5:15
• What is the justification for commuting infinite products? Oct 28, 2023 at 7:37
• @Isomorphism: The absolute convergence of the infinite product. Oct 28, 2023 at 7:56
This is an alternative derivation, not a proof. Noting the pattern of $$+8$$ between alternating multiplicands. Your infinite product can be expressed as two separate products:
$$\prod_{k=0}^{\infty}{\left( \dfrac{8k + 1}{8k + 3}\right)} \cdot \prod_{k=1}^{\infty}{\left( \dfrac{8k - 1}{8k - 3}\right)} \\= \prod_{k=0}^{\infty}{\left( \dfrac{8k + 1}{8k + 3}\right)} \cdot 3\prod_{k=0}^{\infty}{\left( \dfrac{8k - 1}{8k - 3}\right)}$$
Now, consider the more general form:
$$\prod_{k=0}^{n}{\left( \dfrac{8k + 1}{8k + 3}\right)} \cdot 3\prod_{k=0}^{n}{\left( \dfrac{8k - 1}{8k - 3}\right)} \\= 3\cdot\left(\dfrac{\left(n + \frac{1}{8}\right)! \cdot \left(\frac{3}{8}\right)!}{3\cdot\left(n + \frac{3}{8}\right)! \cdot \left(\frac{1}{8}\right)!} \right) \cdot \left( \dfrac{\left(n - \frac{1}{8}\right)! \cdot \left(-\frac{3}{8}\right)!}{3\cdot\left(n - \frac{3}{8}\right)! \cdot \left(-\frac{1}{8}\right)!} \right)$$
As $$n \to \infty$$, the product converges to
$$\dfrac{\left(\frac{3}{8}\right)! \cdot \left(-\frac{3}{8}\right)!}{3\cdot\left(\frac{1}{8}\right)! \cdot \left(-\frac{1}{8}\right)!}$$
Recall that $$(-z)!\cdot z! = \Gamma{(1 - z)}\cdot\Gamma{(1 + z)} = \Gamma{(1 - z)}\cdot z\cdot\Gamma{(z)}$$. By the reflection identity, $$\Gamma{(1 - z)}\cdot z\cdot\Gamma{(z)} = \dfrac{\pi z}{\sin{\pi z}}$$
Hence,
$$\dfrac{\left(\frac{3}{8}\right)! \cdot \left(-\frac{3}{8}\right)!}{3\cdot\left(\frac{1}{8}\right)! \cdot \left(-\frac{1}{8}\right)!} \\= \dfrac{\frac{3\pi/8}{\sin{3\pi/8}}}{3\cdot\frac{\pi/8}{\sin{\pi/8}}} \\= \dfrac{\sin{\pi/8}}{\sin{3\pi/8}} \\= \sqrt{2} - 1$$
You can use the sine expansion formulas in the last step.
• May I ask a silly question? There was a $3$ at the top. Why has it gone to the bottom? Sep 28, 2023 at 10:57
• @BobDobbs You question is not silly :D. Thanks for hinting the error. I forgot to include the $\dfrac{1}{3}$ factors after evaluation of both products. Will correct now! Sep 28, 2023 at 11:48
Consider the partial product $$P_k=\frac{\prod _{n=1}^k a_n}{\prod _{n=1}^k b_n}$$ where $$a_n=4 n+(-1)^n-2\qquad \qquad\text{and} \qquad\qquad b_n=4 n-(-1)^n-2$$ that is to say $$P_k=\prod _{n=1}^k \left(1+\frac{2 (-1)^n}{4 n-(-1)^n-2}\right)$$ which is $$P_k=\frac{ \Gamma \left(\left\lfloor \frac{k-1}{2}\right\rfloor +\frac{9}{8}\right) \Gamma \left(\left\lfloor \frac{k}{2}\right\rfloor +\frac{7}{8}\right)}{\Gamma \left(\left\lfloor \frac{k-1}{2}\right\rfloor +\frac{11}{8}\right) \Gamma \left(\left\lfloor \frac{k}{2}\right\rfloor +\frac{5}{8}\right)}\,\,\tan \left(\frac{\pi }{8}\right)$$
The big front factor oscillates but converge very slowly to $$1$$. Then the result.
Asymptotically $$P_{2p}=\left( 1+\frac{1}{8 p}+\frac{1}{128 p^2}+O\left(\frac{1}{p^3}\right) \right)\,\,\tan \left(\frac{\pi }{8}\right)$$ $$P_{2p+1}=\left(1-\frac{1}{8 p}+\frac{9}{128 p^2}+O\left(\frac{1}{p^3}\right) \right)\,\,\tan \left(\frac{\pi }{8}\right)$$ and $$\frac{P_{2p+2} }{P_{2p} }=1-\frac{1}{8 p^2}+O\left(\frac{1}{p^3}\right)$$ $$\frac{P_{2p+3} }{P_{2p+1} }=1+\frac{1}{8 p^2}+O\left(\frac{1}{p^3}\right)$$
As an (overkill) alternative, consider Hurwitz's zeta function $$\zeta(s,a)=\sum_{n\geq 0}(n+a)^{-s}$$ which satisfies (cf. 25.11.18 in https://dlmf.nist.gov/25, or, for a proof, Section 13.21 in Whittaker and Watson's Course of Modern Analysis) $$\zeta(0,a)=\tfrac 12-a,\qquad \frac d{ds}\zeta(s,a)\big|_{s=0}=\log\Gamma(a)-\frac 12\log(2\pi).\qquad\qquad(\star)$$ Introduce the function \begin{align*} f(s)&=\sum_{n\geq 1}(-1)^n\bigl((4n-1)^{-s}+(4n+1)^{-s}\bigr) \\ &=8^{-s}\bigl[-\zeta (s,\tfrac{3}{8})-\zeta(s,\tfrac{5}{8})+\zeta(s,\tfrac{7}{8})+\zeta(s,\tfrac{9}{8})\bigr] \end{align*} We have $$\sum_{n\geq 1}(-1)^n\bigl(\log(4n-1)+\log(4n+1)\bigr)=\log\biggl(\frac{7\cdot9\cdot15\cdot 17\cdots}{3\cdot 5\cdot 11\cdot 13\cdots}\biggr)=\frac d{ds}f(s)\big|_{s=0}$$ which can be evaluated thanks to $$(\star)$$ and Euler's reflection formula for the Gamma function.
Not an answer by any means but a few interesting observations:
Since it has been proved by the other wonderful answers that:$$\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right) \cdots$$
One could take the logarithm on both sides:
$$\log{(\dfrac{1}{\sqrt{2}}-\sin(x))}=\log{(\dfrac{1}{\sqrt{2}})}+\log\left(1-\dfrac{4x}{\pi}\right)+\log\left(1-\dfrac{4x}{3\pi}\right)+\log\left(1+\dfrac{4x}{5\pi}\right)+\log\left(1+\dfrac{4x}{7\pi}\right) \cdots$$
And differentiate:
$$\dfrac{-\cos(x)}{\frac{1}{\sqrt{2}}-0}=>\sqrt{2}=0+\dfrac{4}{\pi(1-\dfrac{4x}{\pi})}+\dfrac{4}{3\pi(1-\dfrac{4x}{3\pi})}+\dfrac{-4}{5\pi(1-\dfrac{4x}{5\pi})}+\dfrac{-4}{7\pi(1-\dfrac{4x}{7\pi})}+\cdots$$
When $$x=0,$$ the resulting summation becomes:
$$\dfrac{\pi}{2\sqrt{2}}=1+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots$$, So one can derive a subset of the Dirichlet L-summations from the $$\sin(x)-k$$ representation
Additionally replace $$x=-\dfrac{\pi}{4}$$ in: $$\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right) \cdots$$
To get, $$1=(1+\dfrac{1}{3})(1-\dfrac{1}{5})(1-\dfrac{1}{7})(1+\dfrac{1}{9})(1+\dfrac{1}{11})\cdots$$
:) | 3,995 | 9,717 | {"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": 47, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-18 | latest | en | 0.792682 |
https://math.answers.com/Q/You_have_1.85_in_nickels_and_dimes._There_are_five_more_dimes_than_nickels._How_many_nickels_are_there | 1,675,667,977,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00226.warc.gz | 406,700,340 | 50,573 | 0
# You have 1.85 in nickels and dimes. There are five more dimes than nickels. How many nickels are there?
Wiki User
2009-09-22 05:36:36
10(x+5) + 5(x) = 185
15x + 50 = 185
15x = 135
x = 9
there are 9 nickels
Wiki User
2009-09-22 05:36:36
Study guides
20 cards
➡️
See all cards
3.8
2036 Reviews | 128 | 304 | {"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} | 3.46875 | 3 | CC-MAIN-2023-06 | latest | en | 0.867998 |
https://www.lotterypost.com/thread/157224 | 1,480,884,926,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541396.20/warc/CC-MAIN-20161202170901-00236-ip-10-31-129-80.ec2.internal.warc.gz | 976,012,656 | 16,890 | Welcome Guest
You last visited December 4, 2016, 3:15 pm
All times shown are
Eastern Time (GMT-5:00)
# An important question for predictors
Topic closed. 8 replies. Last post 9 years ago by codmander.
Page 1 of 1
United States
Member #10438
January 13, 2005
431 Posts
Offline
Posted: June 8, 2007, 2:32 pm - IP Logged
when playing/predicting pick 5 and 6 games have you figured an average between good hits? In other words the number of draws between 4 out of 5 on any givin game and the next 4 of 5 that you get? really what i was looking for is the average on the slump between hits so if one was to go with a predictor timing could be a part of it
mid-Ohio
United States
Member #9
March 24, 2001
19821 Posts
Offline
Posted: June 8, 2007, 2:41 pm - IP Logged
All that information is available in the prediction statistics of each member, you can check their overall statistics or check them by state and/or game types.
* you don't need to buy more tickets, just buy a winning ticket *
Harbinger
D.C./MD.
United States
Member #44103
July 30, 2006
5583 Posts
Online
Posted: June 8, 2007, 4:36 pm - IP Logged
Without too much scientific examination 4 of 5 happens to me every thirty to ninety days, on the 5/39 games. On the 6 of 39, the 5 of 6 has come once since I started about a year ago, the 4 of 6 every couple of weeks, 3 of 6 almost daily. On each of these games I play on, 5 of 39 average 15 lines daily, and the 6/39 25 lines. If I hit I add lines the next day. Yesterday was a low hit day so I did a replay but the day before was better. Here you go a four of five in the abbrev. wheel (missing the 9 (if six was 9!)) are on there but I got a 2 of 5 w/bonus, a 3 of 5 and a 3 of 5 w/bonus. I picked 12 numbers and did an abbreviated wheel. It would be nice to know when you have four or five because a full wheel on this set would have yielded a good sum.
I learned from the challenges this is the way to go, get a set of numbers you like, and do an abbreviated wheel. It is really time consuming to manually pick lines and it was starting to make my head hurt. But picking a set of 10,11,12, etc. numbers and plugging and wheeling is lot easier and when you are close it pays off. I've shown a couple of tickets from day before yesterday, I've got a lot of losers also but I don't scan those, I bag them for this guy.
.
United States
Member #48046
December 7, 2006
1699 Posts
Offline
Posted: June 8, 2007, 4:50 pm - IP Logged
Does one of Mr. Todd's wheels do that?
9 plays for 11 or 12 numbers?
Mabey I'll get a little yellow in my name.
That looks like a good game nine plays for \$6 bucks
Harbinger
D.C./MD.
United States
Member #44103
July 30, 2006
5583 Posts
Online
Posted: June 8, 2007, 5:31 pm - IP Logged
Yes. They alone are worth the upgrade. There are dozens. Everything you need is in one place. No muss no fuss. copy paste, fill in the boxes. One thing left is for an application that fills out playslips, it would be software that prints the slips or fills them out, I've put some thought into doing that but I don't have the time or resources to devote to doing that. A couple of software companies have professed to have such software, I have tried it, and it doesn't work.
It has to utilize some way to import the image of the slip and OCR it permanently, so that you can copy/paste a set of lines through a utility to fill-in the correct spots. and print it. I am going to scan a filled out slip on a sheet of letter size and try and play it. I'll report back.
LOL jarasan.
Harbinger
D.C./MD.
United States
Member #44103
July 30, 2006
5583 Posts
Online
Posted: June 8, 2007, 9:36 pm - IP Logged
Well the replay on DC Daily 6 resulted in 3 4 of 6's, and 10 3 of 5's. Bonus match 5 yielded o goose egg. I played a mix of pick6 4 if 5 of 14 and a 3 if 5 12 abbreviated wheels. I played 62 lines \$31.00 got back \$50 real dollars net \$19 oh boy, I like 5 of 6 and 6 of 6 \$1000 and \$250k respectively. The 15 and 22 weren't on my 12 or 14 lists. On the prediction board this would have won a lot more money but not real money and you are only allowed 50 lines. Again, this is only for entertainment purposes and you don't think I'm blowin' smoke. LOL jarasan
Harbinger
D.C./MD.
United States
Member #44103
July 30, 2006
5583 Posts
Online
Posted: June 8, 2007, 11:11 pm - IP Logged
correction 10 3 of 6's.
NC
United States
Member #11741
February 23, 2005
1233 Posts
Offline
Posted: June 8, 2007, 11:54 pm - IP Logged
Well the replay on DC Daily 6 resulted in 3 4 of 6's, and 10 3 of 5's. Bonus match 5 yielded o goose egg. I played a mix of pick6 4 if 5 of 14 and a 3 if 5 12 abbreviated wheels. I played 62 lines \$31.00 got back \$50 real dollars net \$19 oh boy, I like 5 of 6 and 6 of 6 \$1000 and \$250k respectively. The 15 and 22 weren't on my 12 or 14 lists. On the prediction board this would have won a lot more money but not real money and you are only allowed 50 lines. Again, this is only for entertainment purposes and you don't think I'm blowin' smoke. LOL jarasan
Hello Jarasan,
Nice advertising here. A picture is worth a thousand words!- I believe you are going to hit the whole thing.
Congratulations.
United States
Member #10438
January 13, 2005
431 Posts
Offline
Posted: June 9, 2007, 6:53 pm - IP Logged
intresting seems ive blown it by since my number pickin has been somewhat good yet not getting them together i need to do those abbr. wheels since ive been in somewhat of slump i think a good time would be to see a few losing draws go by than start wheelin
Page 1 of 1 | 1,704 | 5,587 | {"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-2016-50 | latest | en | 0.927686 |
https://brilliant.org/discussions/thread/prove-or-disprove-define-t_nx-as-the-chebyshev/ | 1,627,920,218,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154321.31/warc/CC-MAIN-20210802141221-20210802171221-00254.warc.gz | 163,918,217 | 18,839 | # Chebyshev Polynomials Proof Problem
Prove (or disprove): Define $T_n(x)$ as the Chebyshev polynomial of the first kind with degree $n$. If $p$ is an odd prime, then $\sqrt{\frac{T_p(x) - 1}{x-1}}$ is an irreducible polynomials over the rational numbers.
By definition $T_n(x) = \cos(n \cos^{-1}(x))$. Listing out the first few odd primes of $n$ shows that (for simplicity sake, we take the positive root only)
$T_3(x) - 1 = (4x^3-3x) - 1 = (x-1)(2x+1)^2 \\ \Rightarrow \sqrt{\frac{T_3(x) - 1}{x-1}} = 2x + 1$
$T_5(x) - 1 = (16x^5-20x^3+5x) - 1 = (x-1)(4x^2+2x-1)^2 \\ \Rightarrow \sqrt{\frac{T_5(x) - 1}{x-1}} = 4x^2+2x-1$
$T_7(x) - 1 = (64x^7-112x^5+56x^3-7x) - 1 = (x-1)(8x^3+4x^2-4x-1)^2 \\ \Rightarrow \sqrt{\frac{T_7(x) - 1}{x-1}} = 8x^3+4x^2-4x-1$
\begin{aligned} T_{11}(x) - 1 &=& (1024 x^{11}-2816 x^9+2816 x^7-1232 x^5+220 x^3-11x) - 1 \\ &=& (x-1) (32 x^5+16 x^4-32 x^3-12 x^2+6 x+1)^2 \\ \Rightarrow \sqrt{\frac{T_{11}(x) - 1}{x-1}} &=& 32 x^5+16 x^4-32 x^3-12 x^2+6 x+1 \end{aligned}
All these polynomials are irreducible polynomials over the rationals. Is it true for all odd prime $p$?
Note by Pi Han Goh
6 years, 1 month ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
The solution is as I expected click here.I hope you don't mind that I copied the problem to math overflow.It uses only the multiplicative property of the degrees of number fields and that $x^{p-1}+x^{p-2}+...+1$, the pth cyclotomic polynomial, is irreducible (can be proven by shifting x to x+1 and using eisenstein).
- 6 years, 1 month ago
Haha, it's a bit of an overkill to use OverFlow! I was about to post this in StackExchange if there's no answer. Thanks!
- 6 years, 1 month ago
I posted it on stackexchange, but no one provided a good solution :D
- 6 years, 1 month ago
If you shift x to x+1 the irreducibility of these can be proven with Eisenstein's criterion.
- 6 years, 1 month ago
Thanks! But how do I prove that it's true for ALL odd primes $p$?
- 6 years, 1 month ago
By the way do you have a solution or just spotted that they are irreducible for small values?Vecause I can't figure out what identity to use to make that square root in a better form.Eisenstein still works though
- 6 years, 1 month ago
I don't have a solution, hence the "or disprove" statement. I got inspired by solving via the Challenge Master notes comment on my solution here.
- 6 years, 1 month ago
So it is perfectly possible that a counterexample exists.Well I'm going to read up on Tschebyscheff Polynomials, any reccomendations?Also, I have found a paper that states that (T_p(x)/x/) is irreducible for all primes p.But I don't know how to deal with that -1.
- 6 years, 1 month ago
Recommendation? Haha, no, I read up a lot before posting this, and I'm still pretty stumped. Link please?
- 6 years, 1 month ago
I think that the identity $T_n(x)=cos(n.arccosx)$ can give us the roots of $T_n(x)-1$
- 6 years, 1 month ago
- 6 years, 1 month ago
Well cosx is 1 only when x is 2.pi.something.So that means that $n.arrcosx$ must be 2.pi.something.So $x=cos(2.\pi.k/n)$ are the roots of T_n(x)-1.Problem is, I don't know when they double roots and how to prove that T is their minimal polynomial.It is interesting that these roots are the real parts of the roots of unity.
- 6 years, 1 month ago
Also the identity $(T_n(x)-1)/(x-1)(T_n(x)+1)/(x+1)=U_{n-1}$ might help, since we know the factorisation of U.
- 6 years, 1 month ago
This post needs to be seen by more people so we can actually solve it lol.Also I still cannot understand why Eisenstein works here (what is special about the coeffs)
- 6 years, 1 month ago | 1,707 | 5,238 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "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} | 4.40625 | 4 | CC-MAIN-2021-31 | latest | en | 0.770443 |
https://mathoverflow.net/questions/435941/a-double-sum-with-complex-numbers-having-stochastic-variables/436090 | 1,726,024,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651343.80/warc/CC-MAIN-20240911020451-20240911050451-00080.warc.gz | 351,518,345 | 27,113 | # A double sum with complex numbers having stochastic variables
I am very confused by a sum I have been trying to solve analytically/ numerically for a long time. It comes from the idea of a physical problem where the observation is made that has a combined response of a number of entities. For example, I want to evaluate the mathematical sum at a observation point $$\omega$$ that looks like the following when $$M \to \infty$$ .
$$L(\omega) = \sum_{n = 0}^{N-1}\sum_{m = 1}^{M} \exp\left(i \left( (\omega_m - \omega)n + \beta_m \right) \right)$$
Where all $$\omega_m$$s are random draws from a normal distribution
$$\omega_m \sim \mathcal{N}(\mu, \sigma^2)$$
and the $$\beta_m$$ are the normal draws from a uniform distribution
$$\beta_m \sim \mathcal{U}[-\pi, +\pi]$$
Let's try to solve it first with the sum with respect to $$m$$ and then $$n$$. The sum with respect to $$m$$ can be approximated to an infinite integral when $$M \to \infty$$.
$$\mathbb{E}(L(\omega)) \approx M \sum_{n = 0}^{N-1} \int_{-\infty}^{+\infty} \int_{-\pi}^{+\pi} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\exp(i(x - \omega)n)\frac{1}{2\pi} \exp(i\beta) d\beta dx$$
This integral is $$0$$ because of the integral $$\int_{-\pi}^{+\pi}\exp(i\beta) d\beta$$.
Let's approach this sum first with respect to $$n$$ and then $$m$$. The function has a closed form with respect to the sum with $$n$$.
$$L(\omega) = \sum_{m = 1}^{M} \frac{\sin\left( \frac{N (\omega_m-\omega)}{2} \right)}{\sin\left( \frac{(\omega_m-\omega)}{2} \right)} \exp\left(i \left( (1-N)\frac{\omega_m-\omega}{2} - \beta_m \right) \right)$$
If I take the expectation here with an integral approximation, it also becomes $$0$$. However, I proceeded with finding the expression for the absolute value of $$L(\omega)$$ from the above expression.
$$|L(\omega)|^2 = \sum_{m = 1}^{M} \left| \frac{\sin\left( \frac{N (\omega_m-\omega)}{2} \right)}{\sin\left( \frac{(\omega_m-\omega)}{2} \right)} \right|^2 + \sum_{p = 1}^{P} \sum_{q = 1}^{Q} \frac{\sin\left( \frac{N (\omega_p-\omega)}{2} \right)}{\sin\left( \frac{(\omega_p-\omega)}{2} \right)} \frac{\sin\left( \frac{N (\omega_q-\omega)}{2} \right)}{\sin\left( \frac{(\omega_q-\omega)}{2} \right)} \cos\left( (1 - N) \frac{(\omega_p-\omega_q)}{2} + \beta_q - \beta_p \right)$$. The sum with $$p$$ and $$q$$ are similar to $$m$$. The second term is clearly $$0$$ based on the same type of approximations with expected value when $$M \to \infty$$. So, taking only the first term, the expected value becomes,
$$|L(\omega)|^2 \approx M \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) \left| \frac{\sin\left( \frac{N (x-\omega)}{2} \right)}{\sin\left( \frac{(x-\omega)}{2} \right)} \right|^2 dx$$
Can I further reduce this to a closed form or a form that can only depend on $$N$$ numerically?
Let's take a simpler form by choosing $$\mu = 0$$, $$\sigma = 1$$ and find the integral at $$\omega = 0$$.
$$|L(0)|^2 \approx M \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(x)^2}{2}\right) \left| \frac{\sin\left( \frac{N (x)}{2} \right)}{\sin\left( \frac{(x)}{2} \right)} \right|^2 dx$$
Your final integral can be readily evaluated by expanding the fraction of sines into sums of exponentials $$e^{ikx/2}$$ with integer $$k$$, and integrating term by term with the Gaussian weight, to arrive at $$|L(0)|^2 \equiv\frac{M}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-x^2/2} \frac{\sin^2\left(Nx/2 \right)}{\sin^2\left( x/2 \right)} \, dx=$$ $$=MN+2M e^{-N^2/2} \sum _{k=1}^{N-1} k e^{\frac{1}{2} \left(2 k N-k^2\right)}.$$
• Thank you! However, when I look for the geometric series sum $\exp(ixk/2)$ from $k=0$ to $k=N-1$, I see that it is $\frac{\cos(1/4 (1 + N) x) \sin((N x)/4)}{ \sin(x/4)} + i \frac{sin((N x)/4) sin(1/4 (1 + N) x)} {sin(x/4)}$ in a trigonometric form. It is a complex expression. Did you mean that it can be written in a different way, or did you mean it is this exact sum? Commented Dec 6, 2022 at 9:55
• Also the final term looks like it is a divergent sum, am I correct? I quickly did a simulation with respect to $N$ and it looks like it diverges quickly. Commented Dec 6, 2022 at 10:10
• the expression on the second line is an exact evaluation of the integral on the first line; it increases linearly in $N$, with a slope of $M\times \sqrt{2\pi}$. Commented Dec 6, 2022 at 11:49
• Thank you so much for the answer again. It gave me a lot of perspective. My simulations agree with this. Commented Dec 7, 2022 at 16:39
I tried the way @Carlo suggested in the answer.
First, I tried expanding the sine ratio term.
$$\left( \frac{\sin(Nx/2)}{\sin(x/2)} \right)^2 = \left( \frac{ \exp(i N x/2) - \exp(-iN x/2) }{ \exp(i x/2) - \exp(-ix/2) } \right)^2$$ $$= \exp({-i(N-1)x}) \left( \frac{ \exp(i N x) - 1 }{ \exp(i x) - 1 } \right)^2$$
If I replace $$t = \exp(ix)$$, then this expression becomes,
$$= t^{-(N-1)} \left( \frac{ t^{N} - 1 }{ t - 1 } \right)^2$$
$$= t^{-(N-1)} [1 + t + t^2 + t^3 + ... + t^{N-1}]^2$$
$$= t^{-(N-1)} [1 + 2t + 3t^2 + 4t^3 + ... + N t^{N-1} + (N-1)t^{N+1} + ... + t^{2N-1}]$$
$$= t^{-(N-1)} [1 + \sum_{p=1}^{N-1} (p+1) t^{p} + (N-p) t^{N+p}]$$ $$= [t^{-(N-1)} + \sum_{p=1}^{N-1} (p+1) t^{p-N+1} + (N-p) t^{p+1} ]$$ $$= [\exp({-ix(N-1)}) + \sum_{p=1}^{N-1} (p+1) \exp({ix(p-N+1)}) + (N-p) \exp({ix(p+1)} )]$$
So the original integral is,
$$I = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \exp(-x^2/2) \left[\exp({-ix(N-1)}) + \sum_{p=1}^{N-1} (p+1) \exp({ix(p-N+1)}) + (N-p) \exp({ix(p+1)} )\right] dx$$
We know the integral of
$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \exp(-x^2/2) \exp(iax) dx = \exp{(-a^2/2)}$$
Hence,
$$I = \exp\left({-\frac{(N-1)^2}{2}}\right) + \sum_{p=1}^{N-1} (p+1) \exp\left({-\frac{(p-N+1)^2}{2}}\right) + (N-p) \exp\left({-\frac{(p+1)^2}{2}}\right)$$
So,
$$|L(0)|^2 \approx M \left[ \exp\left({-\frac{(N-1)^2}{2}}\right) + \sum_{p=1}^{N-1} (p+1) \exp\left({-\frac{(p-N+1)^2}{2}}\right) + (N-p) \exp\left({-\frac{(p+1)^2}{2}}\right) \right]$$
Is this what I am supposed to get? Just want to verify it. The solutions don't look exactly the same.
• try to compare with the answer for $N=5$, which is $$|L(0)|^2=M\left[5+\frac{2}{e^8}+\frac{4}{e^{9/2}}+\frac{6}{e^2}+\frac{8}{\sqrt{e}}\right].$$ Your formula does not agree... Commented Dec 7, 2022 at 12:34
• Indeed it is different. Did I do something wrong? Is this the same you suggested when you said it could be expressed in terms of a sum over $\exp(ipx/2)$ ? Commented Dec 7, 2022 at 13:06
• what I would suggest you do is to first try a small value of $N$, like $N=3$; then you should be able to directly check whether each step is correct or not; once you have that under control you can generalise to arbitrary $N$ more reliably. Commented Dec 7, 2022 at 13:08
• Excellent. I did it with $N=3$ and I saw a nice pattern. I arrived at your solution. Thanks! In fact, it can be even written in a divergent series form $|L(0)|^2 = M( N + 2 \sum_{p=1}^{N-1} \frac{p}{e^{(N-p)^2/2}} )$ Commented Dec 7, 2022 at 13:41 | 2,726 | 7,070 | {"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": 49, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-38 | latest | en | 0.768513 |
https://bakingsos.com/zoi1q/41e327-area-of-octagon-inscribed-in-a-circle-formula | 1,627,626,224,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153934.85/warc/CC-MAIN-20210730060435-20210730090435-00384.warc.gz | 135,329,199 | 8,612 | vertex angle is 45º (360º ÷ 8)) A handy formula says the area of a triangle . Here, inscribed means to 'draw inside'. Find formulas for the square's side length, diagonal length, perimeter and area, in terms of r. Strategy. Hi Harry, You can use two of Stephen's responses in the Quandaries and Queries database to find the area. The area of the octagon is approximately 2.828, because the area of a polygon greater than a square is 1/2 the apothum times the perimeter. But first, here are two great tips for this and other problems. and "sandwiching" an angle of 45 deg. the area of one triangle is given by the formula 1/2bh. In order to calculate the area of an octagon, we divide it into small eight isosceles triangles. A square is inscribed in a circle with radius 'r'. Each side of the regular octagon subtends 45^@ at the center. The apothum is the line segment from the center point of the polygon perpendicular to a side. Piece o’ cake. Joined Feb 17, 2004 Messages 1,723. (*****You cannot find this formula in any of the books, since it is my invention. Use \pi \approx 3.14 and \… Math. a circle has 360° so dividing by 8 you get 45° for the apex angle of each isosceles triangle. A regular octagon is inscribed in a circle of radius 15.8 cm. Find the perimeter of the octagon. The area of a circle is given by the formula A = ðr2, where r is the radius. The equal sides of every triangle include angle 45^@. If I know that the sides of my octagon are 8 units, how do I determine the radius of an inscribed circle? The incircle of a regular polygon is the largest circle that will fit inside the polygon and touch each side in just one place (see figure above) and so each of the sides is a tangent to the incircle. Regular octagon ABCDEFGH is inscribed in a circle whose radius is 7 2 2 cm. The perimeter, area, length of diagonals, as well as the radius of an inscribed circle and circumscribed circle will all be available in the blink of an eye. What is the area of the octagon. The key insight to solve this problem is that the diagonal of the square is the diameter of the circle. The word circle is derived from the Greek word kirkos, meaning hoop or ring. Last Updated : 16 Nov, 2018; Given a square of side length ‘a’, the task is to find the side length of the biggest octagon that can be inscribed within it. The diagonals of a square inscribed in a circle intersect at the center of the circle. where a = 10 b =10sin45º = 10*√2/2 First draw the picture of a circle with radius 1, and an octagon inside the circle. So all you have to do to get the area of the octagon is to calculate the area of the square and then subtract the four corner triangles. You now have an isosceles triangle, equal sides r = 6 in. A regular octagon is inscribed in a circle with radius r Find the area enclosed between the circle and the r . In this paper we derive formulas giving the areas of a pentagon or hexagon inscribed in a circle in terms of their side lengths. For problems involving regular octagons, 45°- 45°- 90° triangles can come in handy. Side of Octagon when area is given calculator uses Side=sqrt((sqrt(2)-1)*(Area/2)) to calculate the Side, The Side of Octagon when area is given formula is defined as length of side of the octagon and formula is given by sqrt((sqrt(2)-1)*(area/2)). You can modify this to find the side length of a regular hexagon inscribed in a circle of radius 4 cm. Where n is the number of sides of the regular polygon. \quad Use \quad \pi \approx 3.14 and octagon in … A square inscribed in a circle is one where all the four vertices lie on a common circle. A circle with radius 16 cm is inscribed in a square . math. Please help. Hence the diameter of the inscribed circle is the width of octagon. 1. the sine of the included angle. is half the product of two of its sides and . By formula, area of triangle = absinC, therefore a = 6, b = 6 and C = 45 deg. triangles, whose congruent sides are 5, and . Another way to say it is that the square is 'inscribed' in the circle. Angle of Depression: A Global Positioning System satellite orbits 12,500 miles above Earth's surface. Find the area of the octagon. Draw one side of the octagon(a chord in the circle), and 2 radii connecting the ends to the center. Find the area of a regular octagon inscribed in a circle with radius r. . The trig area rule can be used because #2# sides and the included angle are known:. 1. Formula for Area of an Octagon: Area of an octagon is defined as the region occupied inside the boundary of an octagon. The octagon consists of 8 congruent isosceles. In a regular polygon, there are 8 sides of equal length and equal internal angles – 135 0.An irregular octagon is one which has 8 different sides. Examples: Input: a = 4 Output: 1.65685 Input: a = 5 Output: 2.07107 Recommended: Please try your approach on first, before moving on to the solution. in this article, we cover the important terms related to circles, their properties, and various circle formulas. Diagonals. Area of Octagon: An octagon is a polygon with eight sides; a polygon being a two-dimensional closed figure made of straight line segments with three or more sides.The word octagon comes from the greek oktágōnon, “eight angles”. One side of regular octagon will make 45 degree angle on the center of the circle. The lines joining opposite vertices are diameters. ~~~~~ This octagon is comprised of 8 isosceles triangles, each with two lateral sides of the length r … The area of the circle can be found using the radius given as #18#.. #A = pi r^2# #A = pi(18)^2 = 324 pi# A hexagon can be divided into #6# equilateral triangles with sides of length #18# and angles of #60°#. A regular octagon is inscribed in a circle with radius r . If the number of sides is 3, this is an equilateral triangle and its incircle is exactly the same as the one described in Incircle of a Triangle. What is a regular octagon? Divide the octagon into a total of 8 triangles each with one vertex at the center of the circle and the other vertices on the edge of the circle. D. Denis Senior Member. Program to find the side of the Octagon inscribed within the square. I have two questions that I need help with. Area hexagon = #6 xx 1/2 (18)(18)sin60°# #color(white)(xxxxxxxxx)=cancel6^3 xx 1/cancel2 … These diameters divide the octagon into eight isosceles triangles. Question 35494: find the area of a regular octagon inscribed in a unit cricle Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website! Their lengths are the radius of the circle = 5. Providing instructional and assessment tasks, lesson plans, and other resources for teachers, assessment writers, and curriculum developers since 2011. The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral. now with the use of trig, you can find an expression for the height of the triangle - 10sin45º. Area of triangle = 36(sqrt 2) / … Examples: Input: r = 5 Output: 160.144 Input: r = 8 Output: 409.969 Approach: We know, side of the decagon within the circle, a = r√(2-2cos36)() So, area of the decagon, A regular octagon is inscribed in a circle with a radius of 5 cm. Find the area enclosed between the circle and the octagon in terms of r . Given here is a regular decagon, inscribed within a circle of radius r, the task is to find the area of the decagon.. I don't have anything in my book for octagons, only rectangles. A circle is a closed shape formed by tracing a point that moves in a plane such that its distance from a given point is constant. While the pentagon and hexagon formulas are complicated, we show that each can be written in a surprisingly compact form related to the formula for the discriminant of a cubic polynomial in one variable. The ratio of the area of the incircle to the area of an equilateral triangle, , is larger than that of any non-equilateral triangle. If increasing the radius of a circle by 1 inch gives the resulting circle an area of 100ð square inches, what is the radius of the original circle? So here we. Derivation of Octagon Formulas: Consider a regular octagon with each side “a” units. So, we know the distance from the center to a vertex, which is one, because it is a unti circle. have A = (1/2)(5)(5)(sin 45) = (25√2 ) / 4. : Theorem 4.1. Now, the formula for computing the area of a regular octagon can be given as: Area of an Octagon = $$2a^{2}(1+\sqrt{2})$$ Where ‘a’ is the length of any one side of the octagon. R = 6 in an expression for the circumference of a regular octagon is inscribed in a circle is where... Used because # 2 # sides and the r 8 units, how do I the! Only rectangles inside the boundary of an octagon, we divide it into small eight isosceles triangles is. A geometric shape with 8 equal lengths and 8 equal lengths and equal. Is that the square is inscribed in a circle with radius 16 cm is in... Distance from the Greek word kirkos, meaning hoop or ring - 10sin45º writers,.! Apothum is the radius miles above Earth 's surface and curriculum developers since 2011 calculate! Of their side lengths area, in terms of r 1, and curriculum developers since 2011 and \… regular... Miles above Earth 's surface sides are 5, and other problems a circle of radius cm... Find the side length of a circle with radius ' r ' length of a triangle hexagon inscribed in circle... = n tan ( theta ) R^2 triangle is given by the formula 1/2bh Name: Anna Who are:! Octagon can easily be found by formula, area of the circle tan ( theta ) R^2 to this. Geometric shape with 8 equal angles anything in my book for octagons, 45°- 45°- 90° can... Width of octagon hexagon inscribed in a circle, how could that person use a ’... Area, in terms of their side lengths the Quandaries and Queries database to find the area of triangle absinC..., where r is the diameter of the inscribed circle and various circle formulas a! The area eight isosceles triangles the radius of an octagon inside the boundary an! Providing instructional and assessment tasks, lesson plans, and curriculum developers since 2011 circle, how could person. Of trig, you can find an expression for the height of the circle and octagon... Radius r.: Student regular heptagon inscribed area of octagon inscribed in a circle formula a circle with radius r the area. Octagon in terms of r. Strategy perimeter and area, in terms of their side lengths to... Side lengths one triangle is given by the formula 1/2bh you now have isosceles. ) R^2 pentagon or hexagon inscribed in a square is inscribed in circle. Intersect at the center point of the books, since it is a geometric with... Are 5, and, assessment writers, and curriculum developers since 2011 circumference of a octagon! Kirkos, meaning hoop or ring radius 15.8 cm are 5, and their lengths! Sin 45 ) = ( 25√2 ) / 4 meaning hoop or.. Is inscribed in a square inscribed in a circle in terms of r, properties. The boundary of an octagon is defined as the region occupied inside the boundary of an octagon draw... N is the line segment from the center of the inscribed circle of r 4. Circles, their properties, and other area of octagon inscribed in a circle formula word circle is given by the formula.! The use of trig, you can modify this to find the area of a regular octagon is defined the! Diameters divide the octagon in terms of their side lengths is defined as the region occupied the... The distance from the center of the books, since it is my invention octagon subtends 45^ @ at center. Triangles can come in handy the important terms related to circles, their properties, and an octagon area! If I know that the sides of the inscribed circle use of trig, you can an! Of the inscribed circle for area of a regular octagon is inscribed in a circle with 1... Is a geometric shape with 8 equal angles each side “ a ” units subtends 45^ @ the. Triangle include angle 45^ @ at the center insight to solve this problem is the! Problems involving regular octagons, 45°- 45°- 90° triangles can come in handy first... = ðr2, where r is the number of sides of the triangle - 10sin45º between circle! '' an angle of Depression: a Global Positioning System satellite orbits 12,500 above. And Queries database to find area of octagon inscribed in a circle formula side of the octagon in terms of r. Strategy center of the circle! The areas of a pentagon or hexagon inscribed in a circle of radius 4.. Into small eight isosceles triangles, equal sides of the circle and the octagon can easily be by. A regular octagon will make 45 degree angle on the center of the regular polygon Who you. ÷ 8 ) ) a handy formula says the area enclosed between the circle 's side length of a with... Areas of a circle with radius ' r ' expression for the 's! Terms of their side lengths assessment writers, and an octagon: area of inscribed! Of a regular octagon subtends 45^ @ = 45 deg circle of radius cm! A circle in terms of r number of sides of every triangle include angle 45^ @ at the center,! Vertices lie on a common circle use of trig, you can use two of Stephen 's responses the! Center to a vertex, which is one, because it is that the sides of every triangle angle. The circumference of a square is given by, a = 6 in we cover the important terms related circles... Circle in terms of r. Strategy every triangle include angle 45^ @ at the center of the.. Of two of its sides and an octagon inside the circle related to circles, their properties, and octagon!, which is one, because it is a unti circle two of Stephen 's responses in circle... Boundary of an octagon is inscribed in a circle in terms of r. Strategy equal angles that the of. Octagons, only rectangles way to say it is a unti circle the product of two of its sides.... For octagons, 45°- 45°- 90° triangles can come in handy terms related to circles, properties! Can not find this formula in any of the circle for the square he found the side of... Harry, you can use two of its sides and the included angle are:. Not find this formula in any of the circle circle in terms of.. 15.8 cm ( * * you can find an expression for area of octagon inscribed in a circle formula circumference a... A vertex, which is one where all the four vertices lie on a common circle to calculate the of! Lengths and 8 equal lengths and 8 equal angles the circle but,... The apothum is the width of octagon, assessment writers, and an octagon, we divide into... Book for octagons, only rectangles and Queries database to find the area between! Two questions that I need help with circumference of a area of octagon inscribed in a circle formula octagon is inscribed in a circle of radius cm. Earth 's surface circle in terms of r side lengths intersect at the to! Tasks, lesson plans, and an octagon, we know the distance from the of! Found the side length area of octagon inscribed in a circle formula diagonal length, diagonal length, diagonal length, and. Unti circle giving the areas of a regular octagon is inscribed in a circle with r.! Inscribed in a circle with radius 16 cm is inscribed in a circle of radius 15.8 cm # and... Positioning System satellite orbits 12,500 miles above Earth 's surface anything in my for... Can find an expression for the height of the square is inscribed in a circle intersect at the center side! Of an octagon derivation of octagon formulas: Consider a regular octagon with each of. And an octagon is inscribed in a square is 'inscribed ' in the Quandaries and database! Terms of r database to find the area of a square is diameter. And area, in terms of their side lengths the key insight to solve this problem is the... Of Depression: a Global Positioning System satellite orbits 12,500 miles above Earth 's surface distance from the of! 6, b = 6 in 's surface diameter of the books, since is! Word circle is given by the formula a = 6 and C = 45 deg properties, other! Who are you: Student and C = 45 deg 360º ÷ )... Have anything in my book for octagons, 45°- 45°- 90° triangles can come in handy of my are! In a circle whose radius is 7 area of octagon inscribed in a circle formula 2 cm assessment writers and. Triangle = absinC, therefore a = n tan ( theta ) R^2 = ( 1/2 ) 5. First he found the side of regular octagon inscribed in a circle of radius cm. Defined as the region occupied inside the circle expression for the circumference of a circle with radius r, a. 8 units, how could that person use a calculator ’ s octagon: of. Shape with 8 equal lengths and 8 equal angles the center point of the regular polygon handy! ( 1/2 ) ( 5 ) ( 5 ) ( sin 45 =... Divide it into small eight isosceles triangles did not remember the formula for the height the! Miles above Earth 's surface \pi \approx 3.14 and \… a regular heptagon inscribed in a circle at...: Student way to say it is a geometric shape with 8 equal lengths and 8 equal angles angles. Diagonal of the regular polygon circle whose radius is 7 2 2 cm square is the of. Cm is inscribed in a circle with radius r. angle on the center know. Used because # 2 # sides and which is one, because it is geometric. = absinC, therefore a = n tan ( theta ) R^2 in this article, we know distance. Related to circles, their properties, and other resources for teachers, assessment writers and... | 4,308 | 17,299 | {"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": 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} | 4.125 | 4 | CC-MAIN-2021-31 | latest | en | 0.912719 |
https://web2.0calc.com/questions/can-somebody-please-help | 1,632,742,156,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00275.warc.gz | 642,556,571 | 5,510 | +0
0
153
1
+113
Right now Mr. Rob is three times as old as his oldest son. The sum of their ages is 64. How old are they?
Answer all parts to the question below. Be sure to label each part with: a, b, c, d, or e.
a) Write a let statement to represent his son's age.
b) Write a let statement to represent Mr. Robs age.
c) Write an equation with the given information.
d) What is his son's age?
e) How old is Mr. Rob?
Mar 5, 2021
#1
+34465
+1
s = son age
r = rob age = 3s
3s + s = 64
4s = 64
s = 16 y/o
r = 3 s = 3(16) = 48 y/o
Mar 5, 2021 | 204 | 555 | {"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} | 3.671875 | 4 | CC-MAIN-2021-39 | latest | en | 0.962985 |
https://www.smore.com/g6g8k | 1,558,900,222,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259452.84/warc/CC-MAIN-20190526185417-20190526211417-00416.warc.gz | 934,325,601 | 12,819 | # MRS. FIELDS' 1ST GRADE NEWS
## A PEEK AT THE WEEK
We had a great week! In Reader's Workshop, our focus has been practicing the reading strategies we have learned in our mini lessons. We have learned about Lips The Fish, Eagle Eye, Sammy the Snake and Chunky Monkey just to name a few. These strategies are used when trying to figure out an unknown word when they are reading. I am so excited to see that they are using them. I always ask what reading strategy they used to figure out an unknown word and they can tell me exactly what strategy they used. In math this week, we have concentrated on balancing equations and they have done great understanding that it has to be the same number on both sides to be equal. Next, we learned that the equation is true if it balances and false if it doesn't equal the same number. If the equation is false, then they must make it true by changing one of the numbers to make them balance. This is normally a 4th quarter skill. This has been moved to 1st quarter. This is a spiraling skill that we will be working on all year. They have really done a fantastic job applying the math strategies they have learned to these difficult problems. I am so proud of them! They have turned into junior mathematicians!
## Mrs. Fields' 1st Grade Class
We are 1st graders and we attend South Elementary School. We love school because we love to read, make projects on our i-pads and math station games. Our class works together as a team. Thank you for visiting.
## PUMPKIN MATH AND FALL FAIR
NEXT THURSDAY ID FALL FAIR AT SOUTH. WE HOPE EVERYONE CAN COME AND PLAY SOME GAMES AND SUPPORT THE SCHOOL! DON'T FORGET TO VOTE FOR OUR PUMPKIN IN THE PUMPKIN DECORATING CONTEST. FRIDAY IS PUMPKIN MATH DAY FOR FIRST GRADE. WE COUNT, WEIGH, MEASURE, CUT, SCOOP, WRITE, DRAW AND ESTIMATE ALL IN ONE ACTIVITY. WE ALWAYS HAVE SO MUCH FUN! | 450 | 1,862 | {"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-2019-22 | longest | en | 0.944177 |
http://www.diyaudio.com/forums/subwoofers/164141-spiral-transmission-lineflared-horn-calculator-4.html | 1,529,385,782,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861899.65/warc/CC-MAIN-20180619041206-20180619061206-00027.warc.gz | 402,819,834 | 16,891 | Spiral Transmission Line'Flared Horn Calculator
User Name Stay logged in? Password
Home Forums Rules Articles diyAudio Store Blogs Gallery Wiki Register Donations FAQ Calendar Search Today's Posts Mark Forums Read Search
Please consider donating to help us continue to serve you. Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
Don Hills
diyAudio Member
Join Date: May 2009
Location: New Zealand
Quote:
Originally Posted by heyday So, the question becomes how do we generate a toroidal vortex sound wave so that we may concentrate the power flux of said wave?
We don't. My original post explains why such an attempt is an exercise in futility.
Quote:
Originally Posted by heyday So, if one placed a radii of tubes at the compression area of the radiating driver, with coiled spirals of lengths corresponding likewise, as the permitted reduction above so states, (which, the added spirals would also contribute to the compression ratio of the driver) couldn't one achieve an acoustical induction and at minimum a velocity increase/power amplification of the wave?
One could not. The equations are not applicable to acoustics.
heyday
diyAudio Member
Join Date: Mar 2010
Location: Oklahoma
Quote:
[D]oes introducing a properly designed spiral to a 130 Hz wave create SPL increasing inductance that is not mechanically possible with a 20 Hz wave?
Are you taking the position that Green's Function does not apply to acoustics?
They are discussed rather specifically within the General Solution of [sound] wave theory. Additionally I have posted material showing that Green's Function predicts concentrating the power flux of a sound wave in special circumstances.
Theory aside, it appears from published decriptions also posted by myself that these calculations have been used in designing acoustic instruments that were used to "magnify" soundwaves and thus generate an "acoustic lens," said technology being described as an advance able to reduce the size of "any" acoustical instrument.
So, excuse me if I'm not following your vague generalizations here.
Regards,
Dane
11th April 2010, 10:11 PM #33 Don Hills diyAudio Member Join Date: May 2009 Location: New Zealand The solution is simple. Just answer your own question: "[D]oes introducing a properly designed spiral to a 130 Hz wave create SPL increasing inductance that is not mechanically possible with a 20 Hz wave?" Produce a "properly designed spiral" and the equations that describe its behaviour. Once you learn how to do that, you'll understand my original post.
heyday
diyAudio Member
Join Date: Mar 2010
Location: Oklahoma
Quote:
Originally Posted by Don Hills The solution is simple. Just answer your own question: "[D]oes introducing a properly designed spiral to a 130 Hz wave create SPL increasing inductance that is not mechanically possible with a 20 Hz wave?" Produce a "properly designed spiral" and the equations that describe its behaviour. Once you learn how to do that, you'll understand my original post.
My view of being properly designed would be to design around the equations themselves, which were shown to be (at the compression level) scaled to approximately 10 centimeters to the meter.
So, where's the problem?
heyday
diyAudio Member
Join Date: Mar 2010
Location: Oklahoma
You guys are a great bunch of help. ;-)
That's ok, I have an idea...
Other than that, I have a few interesting experimental observations regarding tapped horns that may be of use to members here.
A while back I ran upon this tapped horn design at speakerplans dot com:
12" Tapped horn - Speakerplans.com Forums - Page 1
Plans are here:
MTH-30
I am in the process of building a house and happened to have some scraps of 7/16ths OSB around so I built a box out of that and coupled it with an old worn out Rockford Fosgate 8 ohm driver that made horrible noises in the sealed and vented boxes I designed for it in accordance with manufacturrrs recommendations.
I was amazed that in the TH it actually sounded half way decent, though most people thought it sounded good. I was pushing it with a Kenwood SW-30HT plate amp I cannabolized after the original driver became useless due to failures of the surrounds.(Running off the LFE output RCA connection from a Kenwood VR-606.) The amp was designed to push 4 ohms at 125 watts. I have seen posts here at diyaudio where it was explained that an 8 ohm driver will always sound better pushed from a 4 ohm amp than a 4 ohm load will, so I won't make any claims regarding the enclosure contribution to the quality of the bass output in yhis instance.
Meanwhile, a buddy gave me a 10" Sony Xploder XSL1037. Though I was able to download a pdf copy of the owners manual for that driver, not one TS parameter was provided, nor have I been able to find that information despite an intensive google search.
Then, a couple of days later my sister gave me a Pioneer SX-316S receiver. I mounted the driver onto a baffle which I screwed over the 12' cutout of the afore-mentioned sealed box and hooked it to the Pioneer out of curiosity. The Pioneer has a regular subwoofer high level speaker output rated as follows:
Quote:
130 W (100 Hz, 10 % THD, 8 Ω)
I barely turned it up and a protection circuit within the amp shut the power down immediately, reminding me that I was using a 4 ohm driver instead of the 8 ohm Rockford I had been connecting to check out the receiever's sound in comparison to the Kenwood.
I immediately recalled reading a post in the diyaudiomobile forum about a triple driver TH design where the drivers were wired in parallel and the poster's statement that
Quote:
Because the impedance peaks of the tapped horn are staggered, we can get away with a wiring scheme which isn't practical with a conventional horn, a sealed box, a vented box, or a bandpass. I'll measure the impedance in the next day or two, and you'll see what I mean. All the impedance "troughs" are staggered, due to pathlength differences for each woofer.
Reference: http://www.diymobileaudio.com/forum/...tml#post840026
This definately called for some experimentation!
So, accordingly, I jury-rigged the Sony into the TH enclosure after removing the 12" from the box. I predicted that the 4 ohm load would not activate the amp's protection circuit because the TH is an impedance matching device with a variable resistance that increases as the amplifier source draws power to push the driver to meet the demands created by high volume low frequency output. (I believe that I read something similar in an explanation posted here at diyaudio somewhere in the multitudes of TH threads available here.)
The result of the experiment greatly exceeded my expectation of the TH to match itself with any impedance condition imposed by the amp so that the impedance mis-match would be invisable to the protection circuit.
Not only was I able to avoid a shut down created by the the low ohm condition of the Sony, but the TH design allowed me to use the Pioneer's speaker setup function to increase the sub output + 10 DB and play music at very high volume levels for several hours without protective shutdown or any discernable heating problems within the amp.
What this all means to me is that a TH enclosure acts in such way that two driver's at the amps rated resistance load may be used within the enclosure in a parallel wiring scheme without adverse effect to the amp.
That being said, I would think that it is most likely probable that at least 3 drivers of the amps rated impedance level may be connected parallel within a TH enclosure without risk of thermal damage to the amplifying equipment.
Though there doesn't seem to be much interest here in my idea's concerning possible technological innovations to improve the Tapped Horn's performance, I plan to continue developing that concept as much as I am able to back over at the Tapped Horn Experiment Box Thread
Regards,
Dane
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Site Site Announcements Forum Problems Amplifiers Solid State Pass Labs Tubes / Valves Chip Amps Class D Power Supplies Headphone Systems Source & Line Analogue Source Analog Line Level Digital Source Digital Line Level PC Based Loudspeakers Multi-Way Full Range Subwoofers Planars & Exotics Live Sound PA Systems Instruments and Amps Design & Build Parts Equipment & Tools Construction Tips Software Tools General Interest Car Audio diyAudio.com Articles Music Everything Else Member Areas Introductions The Lounge Clubs & Events In Memoriam The Moving Image Commercial Sector Swap Meet Group Buys The diyAudio Store Vendor Forums Vendor's Bazaar Sonic Craft Apex Jr Audio Sector Acoustic Fun Chipamp DIY HiFi Supply Elekit Elektor Mains Cables R Us Parts Connexion Planet 10 hifi Quanghao Audio Design Siliconray Online Electronics Store Tubelab Manufacturers AKSA Audio Poutine Musicaltech Holton Precision Audio CSS Dx Classic Amplifiers exaDevices Feastrex GedLee Head 'n' HiFi - Walter Heatsink USA miniDSP SITO Audio Twin Audio Twisted Pear Wild Burro Audio
Similar Threads Thread Thread Starter Forum Replies Last Post testarossa2k Full Range 53 16th August 2017 07:19 PM el`Ol Full Range 0 14th May 2007 08:31 AM Ang Multi-Way 4 9th May 2007 11:49 AM kneadle Multi-Way 45 4th November 2004 12:30 AM
New To Site? Need Help?
All times are GMT. The time now is 05:23 AM. | 2,191 | 10,037 | {"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-2018-26 | latest | en | 0.933842 |
https://www.bartleby.com/solution-answer/chapter-45-problem-11ps-intermediate-algebra-10th-edition/9781285195728/for-problems-11-52-perform-the-indicated-divisionsobjective1-x27x78x6/83b3a42b-78b0-11e9-8385-02ee952b546e | 1,571,194,309,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00118.warc.gz | 817,606,124 | 82,348 | Chapter 4.5, Problem 11PS
### Intermediate Algebra
10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Chapter
Section
### Intermediate Algebra
10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem
# For Problems 11-52, perform the indicated divisions.(Objective1) x 2 − 7 x − 78 x + 6
To determine
To Find:
The value of the polynomials by performing the indicated division by binomials.
Explanation
Division of a polynomial by binomial is done by long-division method.
In this method to find the first quotient term, divide the first dividend term by the first divisor term.
Multiply the divisor with the term obtained by dividing the first dividend term with the first quotient term.
The product obtained is subtracted from the dividend.
Repeat the same process to divide the other terms of the dividend.
Calculation:
Consider the polynomial x2−7x−78x+6.
The following steps are used to solve the problem.
First multiply the divisor by x and write the product x2+6x under the dividend and subtract.
The value obtained is equals to −13x.
Now multiply the divisor by −13 and write the product −13x−78 under the dividend and simplify.
The polynomial x2−7x−78 is divided as follows,
Then,
x+6x−13x2−7x−78
### Still sussing out bartleby?
Check out a sample textbook solution.
See a sample solution
#### The Solution to Your Study Problems
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
Get Started | 394 | 1,551 | {"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} | 4.21875 | 4 | CC-MAIN-2019-43 | latest | en | 0.842853 |
https://mathoverflow.net/questions/74737/where-can-i-find-an-explicit-description-of-the-pseudocolimit-of-a-small-pseudof | 1,631,833,117,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053759.24/warc/CC-MAIN-20210916204111-20210916234111-00416.warc.gz | 433,794,592 | 32,274 | # Where can I find an explicit description of the pseudocolimit of a small pseudofunctor to Cat?
Given a functor from a small category to $Set$, we can describe the colimit set as a quotient of the disjoint union of image sets by an equivalence relation arising from morphisms in the source category (as seen in Wikipedia, or Kashiwara-Schapira's Categories and Sheaves). I am looking for an analogous description on the 2-categorical level:
Is there an explicit description of "the" pseudo-colimit of a pseudo-functor $F$ from a small 2-category to $Cat$, and is it in the literature? In particular, who should I reference for the fact that such pseudo-colimits exist?
Here, by pseudo-colimit, I mean "pseudo-bi-colimit" in the sense of Borceaux's Handbook of Categorical Algebra, i.e., a pair $(L, \pi)$, where $L$ is a small category, and $\pi: F \to \Delta(L)$ is a pseudo-natural transformation to the constant pseudo-functor, such that the functor $Fun(L,B) \to PsNat(F, \Delta(B))$ given by $f \mapsto \Delta(f) \odot \pi$ is an equivalence.
Presumably, this should be a small category whose set of objects is something like a quotient of a disjoint union of object sets, but when I tried to work it out by myself, all of the arrows gave me a headache. I also looked at several sources, e.g., Kelly's Elementary Observations on 2-categorical limits, the elephant, and Borceaux, without success (but I may have missed something).
• If you are content to consider diagrams indexed by 1-categories, you can find a simple construction in SGA 4 Expos\'e VI Section 6: Grothendieck construction applied to F (your "disjoint union") followed by inversion of Cartesian morphisms (your "quotient"). One thorough treatment of the general case (even with weights) is Thomas Fiore's "Pseudo limits, biadjoints, and pseudo algebras." Sep 7 '11 at 14:22
• Both of your suggestions are quite helpful! Thank you very much. Please change this to an answer. Sep 7 '11 at 15:03
• Another nice source is "The stack of microlocal perverse sheaves" by Ingo Waschkies. Sep 15 '11 at 16:39
An answer can more or less be extracted from Kelly's Elementary Observations on 2-categorical limits, at least if you already know that it's there. (-:
First, as Kelly notes in section 6, it would suffice to construct what we may call strict pseudo-colimits, that is pseudo-colimits in your sense for which the functor $f\mapsto \Delta(f)\odot \pi$ is an isomorphism (since any isomorphism of categories is a fortiori also an equivalence).
Second, in Proposition 5.1, Kelly shows how to construct strict pseudo-limits (which he calls merely "pseudo-limits") in any 2-category out of (strict) products, cotensor products, iso-inserters, and iso-equifiers. This is a 2-categorical version of the construction of limits out of products and equalizers. (In the correction to the paper Fibrations in bicategories, Street gives an equivalent construction of non-strict pseudo-limits in terms of products, cotensor products, and descent objects.)
Dually, of course, pseudo-colimits may of course be constructed from coproducts, tensor products, iso-coinserters, and iso-coequifiers. Coproducts and tensor products in $Cat$ are easy — they are disjoint unions and cartesian products — so it suffices to construct iso-coinserters and iso-coequifiers.
At this level, though, I think we do have to descend into writing down strings of composites of arrows modulo equivalence relations. Iso-coinserters and iso-coequifiers, being both particular Cat-weighted colimits, can be constructed in terms of cartesian products in Cat and ordinary unweighted colimits, so it would suffice to understand the latter. We know that Cat is cocomplete (as a 1-category) since it is the models of an essentially algebraic theory, so this settles the existence question. But an explicit description of colimits is going to be kind of messy.
• Thank you. I suspected Kelly had an answer encoded somehow, but I was unable to extract it in full. Sep 8 '11 at 6:55
Here is my comment, rewritten as requested as an (elliptical) answer:
If you are content to consider diagrams indexed by 1-categories, you can find a simple construction in SGA 4 Expos\'e VI Section 6: it is the Grothendieck construction, associating a fibration to F (your "disjoint union"), followed by formal inversion of Cartesian morphisms in the fibration (your "quotient"). One thorough treatment of the general case (even with weights) is in Thomas Fiore's "Pseudo limits, biadjoints, and pseudo algebras."
• By the way, this and its generalization to oo-categories is also corollary 3.3.4.3 in "Higher Topos Theory". Links are collected here: ncatlab.org/nlab/show/2-limit#2ColimitsInCat Sep 13 '11 at 6:20 | 1,160 | 4,734 | {"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-2021-39 | latest | en | 0.893316 |
https://brainmass.com/business/business-policy-and-implementation/relevant-information-decision-making-pecos-printers-407302 | 1,597,212,655,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738878.11/warc/CC-MAIN-20200812053726-20200812083726-00237.warc.gz | 239,069,613 | 11,887 | Explore BrainMass
# Relevant Information for Decision Making, Pecos Printers, Inc.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Pecos Printers, Inc. is a small manufacturing firm in Houston, Texas that manufactures color ink jet printers for the small business market. It has just launched the PP 7500.
A 50% markup is standard in this industry so that Pecos must sell to distributors below \$400 per printer to keep the retail price below the industry top of \$600 (\$400 * 150% = \$600). Paul Pecos, the founder and CEO of Pecos Printers, wants to keep the price to distributors as low as possible so he has carefully engineered his manufacturing process to be as efficient as possible.
The model PP 7500 is an exceptionally desirable model with the following features:
A monthly capacity of 10,000 copies
A print speed of 10 copies per minute for black and white and 5 copies per minute for color.
A lifetime capacity of 120,000 copies.
The ability to accept readily available HP ink cartridges.
Lester Ledger, the Pecos Controller has developed the following cost sheet for the model 7500:
Cost Category Cost per Unit
Direct Materials (Variable) \$145
Direct Labor (Variable) 60
Total Unit Costs \$290
*This is determined on a per unit basis as followed. Lester assumes that the annual fixed overhead costs for this product will be \$450,000 and that approximately 10,000 Model 7500's will be produced during the current year. Pecos has the capacity to produce 20,000 units per year without increasing fixed costs.
Paul has determined that approximately 20% of the total manufacturing costs are necessary for a decent profit.
Based on these data, Paul has developed the following pricing rule for his sales staff: Accept any offer from distributors of \$300 or more and reject any offer below \$300.
The sales staff is on salary with no commission paid for any sale. The salesmen negotiate with distributors who make firm offers which the Pecos salesmen then either accept or reject. Last month the three salesmen reported the following offers and results:
Offer (per unit) Number of Units Accepted?
Sam Smoothtalk
Offer No. 1 \$310 200 Yes
Offer No. 2 \$305 150 Yes
Offer No. 3 \$295 300 No
Harry Hustler
Offer No. 1 \$305 50 Yes
Offer No. 2 \$200 250 No
Offer No. 3 \$300 100 Yes
Offer No. 4 \$330 75 Yes
Offer No. 1 \$305 250 Yes
Offer No. 2 \$245 400 No
Offer No. 3 \$325 100 Yes
In addition, Ms. Glenda Goodperson, the office assistant manager received an offer from a new distributor for 700 units at \$290. She felt this would be advantageous for Pecos and accepted the offer. When Paul Pecos found out about this transaction, he was furious that Ms. Goodperson had violated his decision rule and fired her on the spot. He then cancelled the order with the new distributor.
Overall, Paul was satisfied with the month's sales results. His sales staff had sold 925 units which translated to an annual rate of over 11,000 units. This was 10% above his estimate of 10,000 annual sales.
Evaluate Paul Pecos' decision rule. Evaluate Paul Pecos' reaction to Ms. Goodperson's sale.
Prepare a contribution margin income statement for the month with two columns: in the first column, show the results following Paul's decision rule. In the second column, show what the results would have been if you chose to revise the decision rule and your revised decision rule had been followed. For simplicity sake, ignore non-manufacturing costs and taxes.
Do you have any other recommendations for Paul to improve his operations?
© BrainMass Inc. brainmass.com June 4, 2020, 1:33 am ad1c9bdddf | 860 | 3,671 | {"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-2020-34 | latest | en | 0.941388 |
https://nethercraft.net/a-parallel-plate-air-capacitor-has-a-capacitance-of-920-pf-the-charge-on-each-plate-is-2-55-uc/ | 1,695,446,309,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00880.warc.gz | 476,180,928 | 12,419 | # A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.55 uC?
0
A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.55 uC
What is the potential difference between the plates?
If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled?
How much work is required to double the separation?
• 1st answer is correct with respect to the potential differences before and after doubling the separation. However, work is equal to the change in potential energy stored in the capacitor.
PE = V^2C/2.
The separation doubles PE because it quadruples V^2 while halving C.
PE1 = V1^2*C1/2 = 3.5339674E-03 J
C2 = C1/2
V2 = 2V1
PE2 = V2^2*C2/2 = 4V1^2*C1/4 = V1^2*C = 2PE1 = 7.0679348E-03 J
Work = PE2-PE1 = PE1 = 3.5339674E-03 J
• V = Q/C = 2.55 * 10^-6/920 * 10^-12 = 0.00277 * 10^6 = 2.77 * 10^3 V
V = Q/C = Qd/AEo
if seperation is doubled, potential also doubles!
W = Q(V2 – V1) = QV = 2.55 * 10^-6 * 2.77 * 10^3 = 7.0635 * 10^-3 J
• Q is constant:
C1 = eo A/d1
C2 = eo A/(2*d1) = (C1 / 2)
U1 = Q^2/C1/2
U2 = Q^2/C1
delta_work = U2 – U1 = Q^2/C1/2
delta_work = 0.00353396739 J
• (a) By C = εo x A/d =>d = (εo x A)/C =>d = (8.85 x 10^-12 x 1.2 x 10^-4)/(5.7 x 10^-12) =>d = 1.86 x 10^-4 m =>d = 0.186 mm (b) By E = 1/2CV^2 =>E = 1/2 x 5.7 x 10^-12 x (220)^2 =>E = 1.38 x 10^-7 J (c) By C’ = C x k =>C’ = 5.7 x 10^-12 x 3.5 =>C’ = 19.95 (pF) By C = Q/V =>Q = V x C’ =>Q = 220 x 19.95 x 10^-12 =>Q = 4.39 x 10^-9 C
Also Check This do you need a box spring if you are going to use a memory foam mattress? | 686 | 1,649 | {"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} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.855474 |
https://www.shaalaa.com/question-bank-solutions/laws-exponents-simplify-express-result-power-notation-positive-exponent-1-2-3-square2_15484 | 1,575,605,320,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484477.5/warc/CC-MAIN-20191206023204-20191206051204-00381.warc.gz | 872,752,096 | 10,276 | Share
# Simplify and Express the Result in Power Notation with Positive Exponent. (1/2^3)Square2 - CBSE Class 8 - Mathematics
ConceptLaws of Exponents
#### Question
Simplify and express the result in power notation with positive exponent.
(1/2^3)^2
#### Solution
(1/2^3)^2 = 1/(2^3)^2 = 1/2^6 " "((a^m)^n = a^(mn))
Is there an error in this question or solution?
#### APPEARS IN
NCERT Solution for Mathematics Textbook for Class 8 (2018 to Current)
Chapter 12: Exponents and Powers
Ex. 12.10 | Q: 2.2 | Page no. 197
Solution Simplify and Express the Result in Power Notation with Positive Exponent. (1/2^3)Square2 Concept: Laws of Exponents.
S | 208 | 662 | {"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} | 3.359375 | 3 | CC-MAIN-2019-51 | latest | en | 0.740619 |
https://forum.knime.com/t/java-snippet/72671 | 1,695,650,537,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508977.50/warc/CC-MAIN-20230925115505-20230925145505-00582.warc.gz | 281,799,832 | 10,867 | Java Snippet
Hey guys!
I need help on a topic, have an order table already handled, need to manage stock.
This table will arrive in the Java Snippet and need to treat it:
ORDER CLIENT QT (ORDER) POSSIBLE STOCK QT (STOCK)
1 A 200 A1234B 50
1 A 200 A1335C 150
2 B 150 A123A 100
2 B 150 A124B 50
2 B 150 A125B 50
3 C 1250 A0972 250
3 C 1250 A0973 500
3 C 1250 A0974 250
3 C 1250 A0975 1500
3 C 1250 A0976 2000
4 D 100 B123 90
4 D 100 B124 5
In this table, grouped the orders and replicated them to possible stocks, so that within the Java Snippet I could calculate the QUANTITY (ORDER) - QUANTITY (STOCK).
In ORDER 1, we need stock A1234B and A1335C.
In ORDER 2, we need stocks A123A and A124B and we will not need stock A125B
…
I would like this result in output Java Snippet:
ORDER CLIENT QT (ORDER) POSSIBLE STOCK QT (STOCK) RESULT
1 A 200 A1234B 50 50
1 A 200 A1335C 150 150
2 B 150 A123A 100 100
2 B 150 A124B 50 50
2 B 150 A125B 50 0
3 C 1250 A0972 250 250
3 C 1250 A0973 500 500
3 C 1250 A0974 250 500
3 C 1250 A0975 1500 0
3 C 1250 A0976 2000 0
4 D 100 B123 90 90
4 D 100 B124 5 5
4 D 100 - INSUFFICIENT STOCK
For order 4, there was not enough stock and you must add a line and include this message “INSUFFICIENT STOCK”.
Remembering that there may be many stocks, but only what is necessary should be consumed.
And another question, I need to import the libraries in code? How do this?
Thanks!
Hi @Victor_Marguti , is there a particular reason why you want to use Java snippet for this?
Are you looking for some kind of cumulative stock calculation or just working on a row by row basis in terms of available stock?
Whilst the calculating can be done with Java snippet, you won’t be able to use it to add the additional rows. Along with many nodes, Java snippet cannot add new rows or delete them so at least part of this processing will need to be using other nodes. So back to my question about why the specific need to use Java snippet… There may be other better solutions, or at least some hybrid.
1 Like
Hello!
Thanks for the answer, there is no specific reason to use the Java Snippet, but I have this code in Python and I wouldn’t like to use it because it needs the extension, I wanted to use something native to knime.
If there is any other solution other than Java that would also work, but it cannot be Python either.
I would recommend to try the integrated python extension:
1 Like
Going back to your original question, when you say “we will not need stock A125”, what is it in the data that informs us of this? How would the solution know?
Likewise what does “only what is necessary should be consumed” mean? It feels like there are some additional pieces of information that we would need to determine “necessary”.
And now that I’m actually looking at your example numbers, I don’t actually understand your example “result” table.
You said it was `QUANTITY (ORDER) - QUANTITY (STOCK)`
but that would result in the follows calcs (using Math Formula node), so why wouldn’t all the negative lines here also need to say “insufficient stock”
The more I look at your sample results table, the less I understand it. Sorry if I’m being slow today, but could you explain the requirement in more detail please.
Finally, which version of KNIME are you using (i.e. 4.7 or 5.1) .
btw, welcome to the KNIME community!
1 Like
No problem!
Maybe I’m not very clear with the information, sorry
Order 2 = Customer A - 150 Qt
To meet the quantity needed to supply Order 2, I have stock A123A (100 qt) A124B (50 qt) and A125B (50 qt). If the order is for 150 Qt, stocks A123A and A124B already resolve the issue and it would not be necessary to use A125B, because the required order quantity (150 Qt) has already been supplied with both stocks (A123A and A124B)
To define what would be only what is necessary, the order and customer column must be the basis, the order is 150 and I have two first stocks that meet the requirements, I will not need to take any more stock, if I have several stocks but the first one already meets the required quantity (order quantity) I no longer need to look at any other stock, basically I need the new column to inform exactly the quantity that I will use from that stock to fulfill the order and bring the difference in the bottom line, until the order is completely fulfilled and so on. Once the order is fulfilled, it should show 0 in the lines of other stocks available for that order, meaning that we will not use them.
The order and customer column are repeated for each available stock, if I have an order of 1000 qt and I have 10 stocks, the order and customer line will be repeated 10 times, this information is handled in previous stages of the flow, where I already search for a key and bring what the possible stock would be.
The functionality should be like this:
Step 1 - If stock is greater than or equal to the order, you must use the order quantity (completed order)
Step 2 - If the stock is less than the order, you must use the stock quantity and generate a new order quantity to be used in the line below the same order. (Next step)
Step 3 - If the stock is less than the new order quantity, you must repeat step 2. (Next step)
Step 4 - If stock is equal to or greater than the order quantity, you must use the new order quantity (Finished order)
Step 5 - Look at the next order and repeat the steps.
Remembering that it should be based on the order and the customer
Math formula seemed like the ideal functionality at first, but it didn’t work because it had these rules.
Yes, but in this situation have no way to install the extension, so it needs to be something native.
Why not just build this as an iterative loop using just the native nodes? How many rows of data per run?
If you want some help building it out, then we would need some data. Do you have 1 table with the inventory and another table with the sale / stock adjustments? Can you upload files with sample data but correct column names and structure?
3 Likes
It can also work, I thought Java would be the most coherent alternative but other alternatives can work too.
The number of lines is variable, but can reach thousands.
I just built an much more complex iterative joining / processing loop that handles hundreds to thousands of iterations from the join table against a wide transaction table with 100k-500k of rows. It completes in 5 to 10 minutes. (I am running a pretty powerful PC though) I would expect your project to come in under that performance wise. If that processing time is a reasonable trade off for easy process auditing and editing then we could help you quickly build something out to test.
3 Likes
Okay!
What do you have in mind?
Upload an excel document to get us started.
Also, the input table appears to come in pre-grouped with the available stock counts at the time of order. If so, then the solution doesn’t have to be iterative (or even loop based at all). Does the calculated reduced stock info from above orders need to become the basis for future orders, or is that updated stock info already in the table for each order?
1 Like
There is no need for there to be new stock, as this table will update every day, meaning there is no need to “Update” the stock.
It doesn’t look like the file uploaded to the forum.
I’m sorry.
Use this similar archive.
STOCK_java.xlsx (17.7 KB)
2 Likes
Edit:
Sample Workflow.knwf (2.7 MB)
Here it is cleaned up with some annotations to explain how I calculated it. There is no group capability in the Moving Aggregation node at this point, so a group loop was required for this one. The loop should calculate pretty quickly with this minimal loop processing, so run time shouldn’t be an issue.
2 Likes
Have you given it a test run? I am curious as to the run time on your data using this native looping approach Vs your prior Python code.
Hello, sorry for the delay!
I couldn’t use it because when opening the flow it asks to install extensions.
On this computer I have some restrictions to install external sources, I can only use native nodes.
Ok. I can swap those out for base nodes next time I have KNIME open.
1 Like
Sample Workflow (1).knwf (2.8 MB)
Try this one. | 2,099 | 8,285 | {"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} | 2.71875 | 3 | CC-MAIN-2023-40 | latest | en | 0.794152 |
https://webmaths.wordpress.com/2009/04/08/pythagoras/ | 1,532,144,553,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592309.94/warc/CC-MAIN-20180721032019-20180721052019-00605.warc.gz | 798,943,496 | 20,722 | ## Pythagoras
Here I am on term holiday marking my year 9’s Pythagoras’ test. It wasn’t a particularly hard or long test and I am pleased to see that most students have done well. Because the test was short I decided to also test student’s problem solving ability with a short investigation into pythagorean triads:
Below are two rules to generate pythagorean triads. Use these rules to generate as many different triads as you can.
1. Start with an odd number greater than 1. Square the number then halve the answer. The two whole numbers either side of this result, together with the original odd number, form a Pythagorean triad.
2. Start with an even number greater than 2. Square this number then divide the answer by 4. The two whole numbers either side of this result, together with the original even number, form a Pythagorean triad.
This caused even my better students trouble. “I can’t do it sir”, “I don’t know what to do”, “How many do I have to do?”, were some comments. I checked that students knew what odd and even numbers were. They could also square numbers and do the simple arithmetic required. Some had trouble writing down two whole numbers either side of say 12.5. Eventually a list of triads started to appear. One student found four triads and then wrote on their paper: “Sorry, I don’t feel like doing more of these.”
I can see that next term I will have to give these students more practice at solving word problems where they have to read, read again, interpret, try a simpler problem, etc, etc.
This statue was erected in the harbour of Pythagorio, on the island of Samos, where Pythagoras and Aristarchus were born. | 377 | 1,652 | {"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} | 3.765625 | 4 | CC-MAIN-2018-30 | latest | en | 0.961738 |
https://programmer.ink/think/from-getting-started-to-mastering-python-100-days-is-enough-cognitive-function.html | 1,720,965,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00570.warc.gz | 421,455,408 | 5,766 | # From getting started to mastering Python, 100 days is enough—— Cognitive function
Posted by Yesideez on Wed, 29 Dec 2021 12:10:35 +0100
In the previous study, if we need to calculate the factorial of three numbers, we need to write repeated code three times to calculate, which is very troublesome. World class programming master Mr. Martin Fowler once said:“ Code has many bad smells, and repetition is the worst! ". To write high-quality code, the first thing to solve is the problem of repeated code. For the code mentioned at the beginning, we can package the function of calculating factorial into a code block called "function". Where factorial needs to be calculated, we only need to "call function".
#### Define function
Mathematical functions usually take the form of y = f(x) or z = g(x, y). In y = f(x), f is the name of the function, X is the independent variable of the function, and Y is the dependent variable of the function; In z = g(x, y), G is the function name, X and y are the independent variables of the function, and z is the dependent variable of the function. Functions in Python are consistent with this structure. Each function has its own name, argument and dependent variable. We usually call the argument of a function in Python as the parameter of the function, and the dependent variable as the return value of the function.
In Python, we use the def keyword to define a function. After the keyword space, write the function name. After the function name, write the parameters passed to the function in parentheses, that is, the argument just mentioned, followed by a colon. The code block to be executed (what to do) of a function is also represented by indentation, which is the same as the code block of the previous branch and loop structure. After the function is executed, we will return the execution result of the function through the return keyword, which is the dependent variable of the function we just mentioned. For example:
```
c = a + b
return c
# When using, you can directly enter parameters
```
**Note: * * the whole function needs to be left blank before and after.
#### Parameters of function
##### Default value of function
If there is no return statement in the function, the function returns None representing a null value by default. In addition, when defining a function, the function can also have no arguments, but there must be parentheses after the function name. Python also allows function parameters to have default values. If no new parameters are passed in when using a function, the default values are used by default. For example:
```
# Defines the function of chromophore shaking. n represents the number of chromophores. The default value is 2
import random
def roll_dice(n=2):
"""The chromophore returns the total number of points"""
total = 0
for _ in range(n): # n = 2
total += random.randrange(1, 6)
# If no parameter is specified, then n uses the default value of 2, which means shaking two dice
print(roll_dice()) # 8
```
##### Variable parameters
We can also implement an add function that sums any number of numbers, because functions in Python language can support variable parameters through asterisk expression syntax. The so-called variable parameter means that 0 or any number of parameters can be passed into the function when calling the function. In the future, when we develop commercial projects in a team way, it is likely to design functions for others to use, but sometimes we do not know how many parameters the caller of the function will pass to the function. At this time, variable parameters can be used. The following code demonstrates the add function of summing any number with variable parameters.
```# Use an asterisk expression to indicate that args can receive 0 or any number of parameters
total = 0
# Variable parameters can be placed in a for loop to get the value of each parameter
for val in args:
total += val
# You can pass in 0 or any number of parameters when calling the add function
print(add(1, 3, 5, 7, 9)) # 25
```
#### Global and local variables
Global variables (variables not written in any function)
Local variables (variables defined inside a function)
The local variable and global variable in the function are two different variables, which have no relationship with each other and do not affect each other.
Searching for a variable in a Python program is performed in LEGB order:
Local scope - > nested scope - > global scope
- > build in (built-in scope)
If none of the four are found, an error will be reported: NameError: name... not defined
Therefore, we can use global keyword declaration to use global variables or define a local variable to put it into the global scope; nonlocal keyworddeclare variables that use nested scopes (no local variables). Example:
```x = 100
def foo():
x = 200
print(x) # Output the value 200 of the local variable x
foo()
print(x) # Output the value 100 of the global variable x
```
If I don't want to define the local variable x in the function foo and want to use the global variable x directly, what should I do???
Use global declaration!
```x = 100
def foo():
global x
print(x) # Output the value 100 of the global variable x
foo()
print(x) # Output the value 100 of the global variable x
```
If I don't want to define the local variable x in the function bar and want to directly use X in the nested scope, what should I do???
```x = 100
def foo():
x = 200
x = x + 1
def bar():
nonlocal x
x = 300
print(f'1.{x}') # 1.300
bar()
print(f'2.{x}') #2.300
foo()
print(x) # 100
```
#### Manage functions with modules
If there are two identical functions in the same file, problems will easily occur when we call them. However, if the project is developed by team collaboration and multiple people, multiple programmers in the team may have defined a function called foo. In this case, how to solve the naming conflict?
In fact, the answer is very simple. Each file in Python represents a module. We can have functions with the same name in different modules. When using the function, we import the specified module through the import keyword, and then use the call method of fully qualified name to distinguish which module foo function to use. The code is as follows.
module1.py
```def foo():
print('hello, world!')
```
module2.py
```def foo():
print('goodbye, world!')
```
test.py
```import module1
import module2
# Call the function in the mode of "module name. Function name" (fully qualified name),
module1.foo() # hello, world!
module2.foo() # goodbye, world!
```
When importing a module, you can also use the as keyword to alias the module, so that we can use a shorter fully qualified name.
test.py
```import module1 as m1
import module2 as m2
m1.foo() # hello, world!
m2.foo() # goodbye, world!
```
In the above code, we import the module that defines the function. We can also use from import... Syntax directly imports the functions to be used from the module. This method can also use the as keyword to alias the module.
test.py
```from module1 import foo as f1
from module2 import foo as f2
f1() # hello, world!
f2() # goodbye, world!
```
Functions are packages of code that are relatively independent and reusable. Learn to use definitions and functions, you can write better code. Of course, the standard library of Python language has provided us with a large number of modules and commonly used functions. If we make good use of these modules and functions, we can do more with less code.
Topics: Python Functional Programming | 1,727 | 7,621 | {"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} | 3.4375 | 3 | CC-MAIN-2024-30 | latest | en | 0.888656 |
https://forum.allaboutcircuits.com/threads/multiplexer-verilog-code.100064/ | 1,561,029,637,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999210.22/warc/CC-MAIN-20190620105329-20190620131329-00471.warc.gz | 442,530,252 | 17,681 | multiplexer verilog code
Discussion in 'General Electronics Chat' started by vead, Aug 2, 2014.
Nov 24, 2011
712
11
multiplexer
module full (sel, i1, i2, i3, i4, o1); input [1:0] sel; input [1:0] i1, i2, i3, i4; output [1:0] o1; reg [1:0] o1; always @(sel or i1 or i2 or i3 or i4) begin case (sel) 2'b00: o1 = i1; 2'b01: o1 = i2; 2'b10: o1 = i3; 2'b11: o1 = i4; endcase end endmodule
case (sel) 2'b00: o1 = i1; 2'b01: o1 = i2; 2'b10: o1 = i3; 2'b11: o1 = i4; endcase end [/FONT] [/FONT]
what is meaning of this table
I think if we put value of i1 then we will get output but what is 2'b00
Jul 27, 2014
189
10
Looks like a crock of s*it to me mate.
3. gagwd New Member
Aug 2, 2014
6
1
Where did this code come from?
It needs some help.
However in the case statement the 2'b00 is or is supposed to be the case match of the first condition. sel is being used to match with one of the 2'xx values and on a match, and there must be one since all possibilities are covered, the statement after the colon is executed.
Nov 24, 2011
712
11
look another example
multiplexer 2 to 1
s d1 d0 q 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 1
verilog code
Code ( (Unknown Language)):
1. module ( s,d1, d0,q)
2. input s, d1, d0;
3. output q; reg q;
4. always @( s or d0, d1);
5. begin
6. case (s)
7. 1'b0 : q= d0;
8. 1'b1 : q= d1;
9. end case
10. end
11. endmodule
I tried to understand in this way
1'b0 : q= d0
q d0 s 0 0 0 1 1 0 0 0 0 1 1 0
I don't know its correct or wrong
Last edited: Aug 3, 2014 | 659 | 1,526 | {"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-2019-26 | latest | en | 0.723034 |
http://mathforum.org/library/drmath/view/58079.html | 1,498,189,350,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319992.22/warc/CC-MAIN-20170623031127-20170623051127-00608.warc.gz | 256,219,755 | 3,792 | Associated Topics || Dr. Math Home || Search Dr. Math
### Multiplying Fractions
```
Date: 02/06/2001 at 23:22:56
From: Alysha Houston
Subject: Multiplying fractions
I need help on how to solve this:
8 * 3
- -
9 5
My teacher said something about cross-cancelling, but I didn't
understand. Can you please explain? Thanks.
```
```
Date: 02/07/2001 at 08:47:02
From: Doctor Rick
Subject: Re: Multiplying fractions
Hi, Alysha.
You know the basic method for multiplying fractions, right? You
multiply the numerators, and that's the numerator of the product.
Multiply the denominators, and that's the denominator of the product.
Let's do that, but DON'T do the multiplications yet - just write them
where they go.
8 3 8*3
- * - = ---
9 5 9*5
Now, to reduce the fraction to lowest terms, we want to look for a
common factor in the numerator and denominator. By not multiplying
first, we make this job easier: we have already done some factoring of
the numerator and denominator. I'll finish the factoring, but keep the
factors together to show where they came from:
8*3 (2*2*2)*3
--- = ---------
9*5 (3*3)*5
Now you can see a common factor: 3. We can "pull them out" of the
fraction:
8*3 (2*2*2) 3
--- = ------- * -
9*5 (3)*5 3
But 3/3 = 1, so all that's left is
8*3 (2*2*2) 8
--- = ------- = --
9*5 (3)*5 15
That's the answer. What we call "cancelling" is really "pulling out"
the same number in the numerator and denominator, making a factor of
1, as I did above when I pulled out a factor of 3/3.
Now we can talk about "cross-cancelling." Notice that the 3 in the
numerator came from the 3 of 8*3, while the 3 in the denominator came
from the 9 of 9*5. In other words, the 3 in the numerator was from the
numerator of the second fraction, 3/5, while the 3 in the denominator
was from the denominator of the first fraction, 8/9.
You don't need to write nearly as much as I did. You can just look at
the problem
8 3
- * - = ?
9 5
and visualize it like this:
8*3
--- = ?
9*5
Then look for common factors in one number of the numerator and the
other number of the denominator. We see that 3 is a factor of both 3
(in the numerator) and 9 (in the denominator). Divide each of these
numbers by 3 (on paper, you'd cross out the 3 and write a 1 above it,
and cross out the 9 and write a 3 below it):
8*1
--- = ?
3*5
You can look again for other common factors. Do the 8 and 5 have a
common factor? No. Therefore we're done, and the product is
8*1 8
--- = --
3*5 15
Do you see why it's called cross-cancellation? You look for common
factors in these pairs:
8 3
\ /
__ \/ __
/\
/ \
9 5
It's possible you might find common factors vertically, too - between
the 8 and 9, or between the 3 and 5. But if you did, it would mean
that the fractions you started with weren't in lowest terms. If the
fractions are in lowest terms, then you only need to look for common
factors in the "cross-terms."
You can find more help in our Dr. Math Archives. Go to the Search Dr.
Math page at http://mathforum.org/mathgrepform.html , and try
searching for the words
multiply fractions cross cancel
- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Elementary Fractions
Middle School Factoring Numbers
Middle School Fractions
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search | 1,067 | 3,690 | {"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} | 4.625 | 5 | CC-MAIN-2017-26 | longest | en | 0.930882 |
https://shopsolarkits.com/blogs/learning-center/convert-kwh-to-mwh | 1,696,410,779,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00752.warc.gz | 580,017,112 | 73,551 | Convert kWh to MWh - Understanding the Conversion Process - ShopSolar.com
# Convert kWh to MWh
The conversion process between kWh and MWh is a simple one. The complexities show up when we don’t understand what the units of power are used for.
Once you know what each unit is used for and how you derive it, converting one to another is as easy as pie.
## Understanding Kilowatt-Hours
We use kWh usage calculators to help scale our solar system. If your monthly power bill is continuously rising due to all the appliances you’re powering, then it’s best to take the solar route.
We must distinguish between kilowatts and kilowatt hours.
Kilowatts (kW) are a measure of the amount of energy a piece of equipment either uses or produces at a specific point. Kilowatt-hours (kWh) indicate how much of that energy is built up over time.
For example, a toaster will likely have a higher kilowatt rating than a fridge, but since your fridge is running pretty much the entire day, it’s going to use more kWh to stay powered.
So when we use a kW to kWh calculator, we’re simply looking at the amount of energy used over time.
### kW and W
You often see a kilowatt or watt rating on an appliance’s packaging, describing how much power is needed to use the appliance.
If we solely want to use kilowatts as our unit of measurement in calculations, but the appliance is measured in watts, we can do some simple kW to watts conversions.
### kWh
Kilowatt-hours factor in time as a variable. We use the energy formula to obtain the kilowatt hours of an appliance or household.
This formula states that energy is equivalent to multiplying power by time. So, let’s say we have a heater that’s rated at 2 kW, and we keep it running for 4 hours.
If we follow our formula, we’re left with 6 kWh of energy being consumed.
Now that we’re clued up with kWh, using a kWh cost calculator or sizing system for our solar needs is less complicated.
## Understanding Megawatt-Hours
After diving into kilowatts and watts, we can say that power is the rate of energy that’s transferred and is measured in watts.
If 1,000 W is equivalent to 1 kW, then 1,000 kW is a single megawatt. This is a large amount of power usually used when dealing with industrial amounts of electricity.
Chances are you will hear megawatts come up in a conversation when people speak about the rated capacity of a generation unit.
That value mentioned in megawatts refers to the maximum instantaneous output. But what about megawatt hours?
### MWh
One megawatt hour is one megawatt of power being sustained for one hour.
In solar power generation units, chances are you will need a battery to store the energy.
This means that both megawatts and megawatts hours are used to describe the capacity of batteries for larger solar power system kits.
But now that we’ve understood the variables we’re working with, how do we go about converting one to the other?
## How to convert kWh to MWh
In the same way that a single megawatt is equal to 1,000 kW, 1 MWh is equivalent to 1,000 kWh.
If you want to build a solar system made up of many arrays of solar panels, you use a kWh per square foot calculator to size the system adequately.
However, the bigger the land, the smaller you want the values you’re working with to be.
This means you can convert your kWh to MWh and use an MWh per square foot calculator instead.
## Conclusion
When it comes to converting one unit of power to another, in most cases you would rely on laws or calculations.
However, when it comes to kW to MW, or kWh to MWh, it’s simply a matter of multiplying or dividing by 1,000 to get the unit you’re looking for.
Did You Find Our Blog Helpful? Then Consider Checking: | 844 | 3,716 | {"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} | 3.265625 | 3 | CC-MAIN-2023-40 | longest | en | 0.924349 |
http://mathhelpforum.com/calculus/207930-using-logarithm-expression-sinh-1-a-print.html | 1,527,392,678,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867995.55/warc/CC-MAIN-20180527024953-20180527044953-00205.warc.gz | 197,218,062 | 2,780 | # using logarithm expression of sinh^-1
• Nov 18th 2012, 11:52 AM
carla1985
using logarithm expression of sinh^-1
Hi all,
I'm really stuck on a question and hoping someone can point me in the right direction.
By using the logarithm expression of sinh-1(y), show that sinh-1(-y)=-sinh-1y
So far im at
sinh-1(-y)=ln((-y)+sqrt((-y)2+1))
Thanks :)
• Nov 18th 2012, 03:36 PM
MarkFL
Re: using logarithm expression of sinh^-1
So, you have:
$\displaystyle \sinh^{-1}(-y)=\ln(\sqrt{y^2+1}-y)$
My next step would be:
$\displaystyle \sinh^{-1}(-y)=-\ln\left(\frac{1}{\sqrt{y^2+1}-y} \right)$
Now, rationalize the denominator of the argument of the log function on the right.
• Nov 19th 2012, 06:04 AM
carla1985
Re: using logarithm expression of sinh^-1
Finished it, thanks. That was a big help :)
• Nov 26th 2012, 10:05 PM
happysoso
Re: using logarithm expression of sinh^-1
I'm really stuck on a question and hoping someone can point me in the right direction. cheap nfl hats hello kitty hats snapback for cheap | 335 | 1,009 | {"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} | 3.515625 | 4 | CC-MAIN-2018-22 | latest | en | 0.782876 |
http://blog.thewhitegoddess.us/2020/04/13/science-definition-is-definitely-an-field/ | 1,723,236,848,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00271.warc.gz | 4,313,928 | 6,250 | # Temperature’s science explains the connections amongst those measurements and believes several unique sorts of temperature measurements.
Temperature measurement is one of one of the fundamental concepts.
Temperature science significance describes that the relationship between heat and vitality , equally inside their phase-changes and in their own nature for being a pressure. In thermodynamics, this romance describes. It clarifies that the shift in the temperature of an individual from 1 portion of an individual system to some other component of an individual machine. This isn’t limited by a machine , but applies to systems which are near one another.
Temperature science definition defines the phase change, an physical or chemical phenomenon that may take place in any procedure in case some chemical is warmed or chilled. https://expert-writers.net/ It refers to the procedure in. You will find numerous distinct kinds of compound responses.
Chemical reactions happen at several unique degrees. These ranges include the response that takes place such as the compound compound’s presence results in a response which transforms it or increases the levels of both compounds in solution. In addition, there are reactions which happen out the chemical compound, such as the creation of the compound. There are also reactions which occur between the outside of a compound and also the inside a compound, like when a chemical is shifted in to an chemical or the following chemical.
Science definition defines how the size of the stress is about the https://math.hws.edu/web/department/math_courses.html magnitude of its own temperatures. In order to complete this, it describes the essentials of thermodynamics. It claims that the larger the amount of your system, the lower its own temperature. It also states the more expensive the warmth, the bigger the stress that is exerted on an individual machine.
Science definition utilizes two theories that are distinct to be defined by these principles. It states that the pressure exerted onto a machine is directly proportional to the system’s volume. It claims the temperature of a system is directly proportional to the kinetic power of an individual system. The temperatures changes, as do all levels, when the total amount of energy is changed.
Science significance also states that inorder for a bodily or chemical system to experience equilibrium, its temperature has to be equal to the vitality. It clarifies the ways that temperature may be affected by temperature, including evaporation. It also describes the changes in temperatures which occur place.
Temperature science definition can be used to spell out the surroundings from that physical system or any given chemical exists. In a system without the system being forced to exist temperature may be changed or modified. Temperature may be utilised to induce a strategy to exist to ruin a chemical system. | 526 | 2,929 | {"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-2024-33 | latest | en | 0.943796 |
http://bipolarii.org/forex-in-usa/overpriced-put-option-arbitrage.php | 1,527,295,104,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00391.warc.gz | 37,129,741 | 11,408 | Sep 5, - Question: The current price of stock ABC is USD 42 and the call option with a strike at USD 44 is trading at USD 3. Expiration is in one year. The put option with the same exercise price and same expiration date is priced at USD 2. Assume that the annual risk-free rate is 10% and that there is a risk-free bond.
### bipolarii.org › Money › Options. Put-Call Parity. A portfolio consisting of stock and a protective put on the stock establishes a minimum amount of value for the portfolio that also has an unlimited upside potential. If the stock declines below the strike of the put, the put increases in value by a dollar for every dollar decline of. If you have an overpriced Call, the arbitrage strategy would be to Sell the Call and Buy the underlying, right? Yes. Quite intuitive. Similarly, if you have an overpriced Put, the arbitrage strategy would be to Sell the Put and Buy the underlying, right? Wrong, though intuitively it looks correct. The correct strategy Arbitrage opportunity. In the options market, arbitrage trades are often performed by firm or floor traders to earn small profits with little or no risk. To setup an arbitrage, the options trader would go long on an underpriced position and sell the equivalent overpriced position. If puts are overpriced relative to calls, the arbitrager would sell a naked put Dividend Arbitrage · Long Box · Conversion · Reversal.
Related terms:
May 25, - Table 1 – Arbitrage opportunity when put option is overpriced. \left[\begin{array}{llll} \text{Year 1. \text{ }. The above table shows that buying a synthetic put (shorting shares and lending $) and selling a put will have no loss at the end of 1 year. Yet, the time 0 cash flow is$ (= – ). Jul 27, - If you trade options, price is paramount. An overpriced option, like an overpriced stock, can adjust downward without warning, reducing or eliminating possibilities for resale. Even if it doesn't adjust, paying too much for an option contract will reduce any profit you receive, whether you sell the option itself or. In the video example it would coincidentally work out that way. However, let's assume that the stock has a.
Options Grandeur As derivative securities, pairs bolt from futures in a very giant respect. They exploit rights rather than times — beginners gives overpriced put option arbitrage the expert to buy and hours gives you the weekdays to sell. On, a key use of options is that the afternoons on an area position are countless to what you up for the time, if you are a trading.
After there is towards an vacant overpriced put option arbitrage that is selected, you can, as with futures, friday positions that essentially are riskfree by thinking options with the paramount force.
Exercise Arbitrage The last arbitrage opportunities in the time market exist when options violate simple speaking bounds. No it, for thursday, should impression for less than its indigence value. With a call week: The kids then become: Long what happens a footstep from now: In other pairs, you invest nothing closer and are countless a costs and training forex means in the paramount.
You could calendar a similar direction with folks. The business bounds popular loosen for non-dividend plan fits and for outs that can be drawn only at purpose In options. Most bounces in the paramount record can be put only at all Probability options and are on evenings that pay outs. Even with these beginners, though, you should not see to search options trading violating these times by dim margins, about because are is so off even with put American options and means tend to be selected.
As options become low area and times become larger and more traffic, you may very well find pairs that loosen these generation bounds, but you may not be required to facilitate off them. Taking Sunday One of the key cash that Fischer Encourage and Myron Scholes had about forex world clock software in the s that designed option pricing was that a consequence vacant of the paramount conception and the riskless being could be paced to have about the same repeat factors as a call or put aim.
This possible is called the timing taking. In fact, Exploit and Scholes similar the grandeur argument to encourage their option just model by speaking that since the sleeping portfolio and the mentioned consequence had the same lunch flows, they would have to situation at the same conception.
To record how replication works, let us point a very facility opposite for stop prices where means can giant to one of two pairs in each everyday period.
One model, which is prohibited a capacious bunk, allows us to exchange the replicating portfolio analisa teknikal forex marketiva collect.
Average that the time is to facility a call with a few depart of 50, which is trying to facilitate in two selected hours: Since we point the cashflows on the superlative with cash at you, it is trading to time with the last speaking and work back through the paramount tree.
Draw with the end weekdays and work no. The week of the call therefore has to be the same as the paced of opening this position. Out the cashflows on the two outs are countless, you would be selected to no show and make a unhurried repeat.
Again, you would not have been in to any can. You could desire a unhurried generation using puts. The speaking portfolio in that person overpriced put option arbitrage be created by situation similar on the paramount reference and fix the money at the riskless situation.
Again, if opportunities are countless at a trading over from the timing watch, you could capture the time and be required to no risk. Each are the reversals that underlie this money. The first is that both the scheduled quantity and the whole are traded and that you can off simultaneously in both evenings, thus outcome in your requires. The valuable is that there are no or at least commodities and futures forex low pairs costs.
If weekdays costs are large, cash will have to move double the band created by these fits for arbitrage to be selected. The third is that you can parch at the riskless extent and sell short, if state. If sydney forex review cannot, grandeur may no slower be feasible.
Scrutiny across forms Friday you have multiple no listed on the same love, you may be selected to take bunk of relative mispricing — how one generation is trying how to another - and bolt in riskless kids. We will bed first at the direction of calls thinking to relies and then use how options with popular exercise prices and beginners should be selected, alternative to each other. Put-Call Consequence When you have a put and a call are with the same trade lunch and the same cash, you can create a riskless with by draw the call, opening the put and happening the paramount custom at the same situation.
To see why, strive thinking a call and trying a put with friday price K and popular date t, and maybe timing the underlying asset at the paramount price S. The proceeding from this tourism australia indonesia market profile forex is riskless and always hours K at expiration t. The last on each of the comes in the portfolio can be selected as evenings:
.
## Mophie frinstaforexship
Our job here is to calculate. Again, if puts are priced at a value different from the replicating portfolio, you could capture the difference and be exposed to no risk. However, an important lesson to learn from here is that the actions by floor traders doing reversals and conversions quickly restore the market to equilibrium, keeping the price of calls and puts in line, establishing what is known as the put-call parity. The key to finding the value of the option is to compare the payoff of the put to that of a portfolio consisting of the following investments: The following diagram shows the calculation. The put option owner sells the stock only when he makes money. What are the assumptions that underlie this arbitrage?
Here are the top 10 option concepts you should understand before making your first real trade: | 1,684 | 8,018 | {"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} | 2.53125 | 3 | CC-MAIN-2018-22 | latest | en | 0.911156 |
https://de.mathworks.com/matlabcentral/cody/problems/44379-one-track-five-lanes/solutions/1353823 | 1,600,938,036,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00693.warc.gz | 314,432,070 | 25,445 | Cody
# Problem 44379. One track five lanes
Solution 1353823
Submitted on 22 Nov 2017 by Nicolas Schoonbroodt
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
2 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0]),0))
3 Pass
assert(isequal(fivelanes([0 0 0 0 1 1 0 0 0 1;0 1 0 0 0 1 0 1 0 0;0 1 0 0 0 0 0 0 0 0;0 0 1 1 0 0 0 0 0 1;0 0 0 0 0 1 0 1 0 0]),1))
4 Pass
assert(isequal(fivelanes([1 0 1 1 1 1 1 1 0 1 1 0;1 1 1 0 1 0 1 0 0 0 0 1;0 0 0 1 0 1 1 0 0 0 0 0;0 0 0 0 0 1 1 0 1 0 0 0;0 0 1 0 0 0 0 0 0 0 1 0]),2))
5 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 1 1 0 0 1 0 0 0 1 1 0;0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0;0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0;0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 1 0]),3))
6 Pass
assert(isequal(fivelanes([0 0 1 1 0 0 0 0 1 0 0 0 0;0 0 0 1 0 0 1 0 1 1 1 1 0;0 1 0 0 0 1 1 0 0 1 0 1 1;0 0 1 0 0 1 0 0 0 1 0 0 0;1 0 1 1 1 0 1 1 0 0 0 0 0]),4))
7 Pass
assert(isequal(fivelanes([1 1 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0;1 0 1 1 0 1 0 1 0 1 1 1 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0;0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0;0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0]),5))
8 Pass
assert(isequal(fivelanes([0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 1 0 0 0 1 0 0 1;0 0 0 0 0 0 0 1 1 1 0 0 1 0 1 1 0 1 1 0 1 0 0 0 0 0;0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0;0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0;0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1]),6))
9 Pass
assert(isequal(fivelanes([0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0;1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0;0 1 1 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1;0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]),7))
10 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0;0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0;0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1;0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0;0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0]),8))
11 Pass
assert(isequal(fivelanes([0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 1 1 0 0 0 0 0;0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0;0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0;1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 0 0]),9))
12 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 1 0 0 1 0 0;0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 0 0 1;0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 1 1 0;0 0 1 0 1 1 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0;0 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 1 1 0 0 1 0 1]),10))
13 Pass
assert(isequal(fivelanes([0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 0 1 0;0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 0 0 1 0;0 0 1 1 0 0 1 1 1 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0;0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0;0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0]),4))
14 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0;0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0;0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0;0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0]),5))
15 Pass
assert(isequal(fivelanes([0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0;0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 1 1 0 1 0 0 1 0 0 0 1 0 1 0 0 0 1 0 0;0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1;0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]),6))
16 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0;0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0;0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1]),12))
17 Pass
assert(isequal(fivelanes([0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0;0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0;1 1 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 0 0;0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 1 0]),13))
18 Pass
assert(isequal(fivelanes([1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 0 0 0 1 1 0 0 1 1 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0;1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 0 0 0 1;1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0]),14))
19 Pass
assert(isequal(fivelanes([0 1 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 1;1 1 0 0 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1;0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1;0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1 0 0 0 0 0 1 0 1 1;0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0]),15))
20 Pass
assert(isequal(fivelanes([0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 1 0 1 1 0 0 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0;0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 1 0 1 1 1 0 1 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1;0 0 0 1 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0;0 1 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0;0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0]),16))
21 Pass
assert(isequal(fivelanes([1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 1;0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1 1 0 0 0 0 1;0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0;0 1 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0]),3))
22 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]),0))
23 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0]),1))
24 Pass
assert(isequal(fivelanes([1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 0 1 0;1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0]),2))
25 Pass
assert(isequal(fivelanes([0 0 0 0 0 0 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0;0 0 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 1;0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 1 0 0 0 1 1 0 0 0 1 0 0 1 0 0 1 0 0 1 1 1 0 0 0 0;0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0]),8))
26 Pass
assert(isequal(fivelanes([0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0;0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0;1 1 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0;1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1]),9))
27 Pass
assert(isequal(fivelanes([0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0;1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0;0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 0 0 1]),10))
28 Pass
assert(isequal(fivelanes([0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0;0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 1 0 0;0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0;0 0 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0;0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0]),11))
29 Pass
assert(isequal(fivelanes([1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0;0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 0 0 0 1 1 1 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1;0 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0 0 0 1 0 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 1 0;0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0]),17))
30 Pass
assert(isequal(fivelanes([1 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0;1 0 0 1 0 1 0 1 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 0 1 0 0 0 1 1 1 1 1 0 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 1;0 1 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0;0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0;0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 1 0 0 0 1 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1]),18))
31 Pass
assert(isequal(fivelanes([0 0 0 1 1 0 1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1;1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 1 1 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 1;0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0;0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 0 0 1 1 0 0 0 0;0 0 0 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0]),19))
32 Pass
assert(isequal(fivelanes([0 0 1 0 0 1 0 0 1 1 1 1 1 0 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 0 0;1 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1 0 0 1 0 1 1 0 0 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 1 1 0;0 1 1 1 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 1 1 0 1 0 0 0 0 1 0 0 1 1 1;1 0 0 0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 1 0 1 0 1 1 1 0 0 1 0 0 0 0 0 0 1 0;0 1 0 0 0 0 1 1 0 1 0 1 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0]),20))
33 Pass
assert(isequal(fivelanes(reshape('1010010000000000000010000100000010000000100000000000000100000000110100000001000000000010000100010000000000000000000000000000000001001100000000000000000001110001000000001100000010000000010000001001001000000000000011000000100100000000100000000000001100000010001000001000000000001000010000000000000010100001101100000000001000010000000000000000000000000100000100000010001000000010000000000000000000000000000000000010100110101000001000000010010000000010000000001001010001010000001000000010001000000010000100100000100000110000000000000100000010000000100001000000000000001000001000000000000000000011001010000000000001001000000000000001101000000001100000011000001100000000000001000001010100000010101000000010000000000000000100000000000000000011000000000000000000000001000000000100000000000010000010000000001000001000001000000011100001000001001000000000000011000000000000000000100000100000000000000000000100000100000000110000100010010010001000000000000011000110000010000001011100000010000000001000000000001000000000000111000000010000001000010000000000010000000000000000001101000000000000000000000101000000000010001000100000000000000000000000000000000000000000101000000000100000010011000001010000001000011001000100000101000000000000001000011100000000000010100010001000001000001100000000010001000000100000100000000010000000000001100000101010000011001010100000000001000000001001000000001000000000001000010000000000000000110000010100001000000100000000001000000000110000000000000000010010000000010101001000000101000001000100000000000000000000000000000001000000001100000000000000000110000000000000001101000000010000000000110000000000100000001000101000010000000000000000000000101000000010000000000000000000000100010100000010000000101000000001000011000001001000000001000100000000010001000000000010000110010000010000000010001100000000000000010000000000000001000000000010010000010001000101100010000000100100100000000000000000011000000110000010010000000001010000110100101000000000000000010000000010001000000001000000000001000000001000010000000000000000000000010100000010000000000100100001000000010010000010001000001000000100001000000000001010100000000000000000010000000001000000000100000100001000000001000000000000000000000001000000010110001000000000000000010000101000000000001101000001010000000000000000000010010000000000011000100000000000000000000000010000010001001000101100100000000100000010001000010000000000000000000000000000011000000100000100000001000001010000000001000011000001000000000111000000000000010000101000000000000000000011000000001001010001010000100000000001010000000000000010000100010100010100001000001000000010000000000000000100000100100000000111010010001000000000000000010000000010000000000000000010100000010001000001001000100000000000100000001000010100110000010001010010000000001000100100001000001000000000010000000010000000000000000001000000001000000000000100000000000000001100000000110000000100010000001100010001101010010000000000000101000000001000000000000000110000000010000000010000100100000000001000000000000000100000000000100010000100000000000000000000000000001000000000000000001000000000000001010000100000000010001000100000011000000000000000001001000010010000010000001000000100100000010101000010000010001010000000000000010001001000000000000000000000010000000010000000000001000000001010000000000000001000001000000001000000000100000100000000000000110101101010100001000100000010001001100100000000000000010101000000001100000001000010000001000010000010100100000100000110010000000000000000010000000100100010000000000000100000000000000100110000100000000100000011000100000000000000000000000000000000000101110010001000001000000000001000101100001000110000001000000100001000000110001001000010001000000000000000100000000010000000100001001000000000100000000100000001100000001000000000000001000000000100000000000000000000100000100000000000000000000000010000000110010000001010000000001000000010001000000000001000000000101000000010000010000000000000001000001000000000000001000011010000101000000010000001000000000000000100001000000000000000001010000000010000000010010000001010010010000000100000001000101100010000100000000000100000011000000001000000000000000000'-'0',5,[])),76))
34 Pass
assert(isequal(fivelanes(reshape('11000000100000000000000000010000000000000010001000001010000001000000000000000000100010101001000010000001000001100000100010100000001000100100010000010000000000010010101000001000000000000000000000000010010000000000001001010000110000000000000000000101000000001000000010100000000000000100010000000001000000000010000000000000000000001100000000000001001000000010000000000000000000000000000010100000101000110000010000000000100000000000000000000001100100000000010000000000100000000000000100000010010010000001000101001000001000100101001000010000001001000100100000010000010100000110000000000001100001010000000000010001000000000000000001110000100000000000000011000001000001100110010100100000000101000000010000000000001000010110010010000000001001000001001000000000000000000000010000100000011000000000000100110000000000000000000010000010000000001000001000000000000000100001000000000100000000000000000001001010000000000000000101000100000000010010000000100000000000001100101100000010000000000000000000100000000000001010000100000000010010000010000000000010000000000000000000000000100000000101000000000000000010000000000100100000000010001000000001000011000000100011010000100000000100001000100000001010000100000000000000011000100000000000000010000000000000000000001000100000010000000000000101000000010001000000001000001101000010001000000100000000010000000000000000000000000001000110000000000100000000000000010000000000000000000000100000000000000000000100000100100001001000100000000000001000100010001001100000000111100100000000000010000100000001001101000001000000000001'-'0',5,[])),33))
35 Pass
assert(isequal(fivelanes(reshape('01000100000100000000000100000000001011100000000000000110000010000100000100100100000100000010000000111001000000000000000000011001010000000000000000000000010000000000000101100000000110000000100000000000000000100010000000000000000000001101001100000000000001011000101000000000101000001000000010000000000101000000010010000000000000101000000000000000000010000000010000000000000100000000001000100010000000000000000000000001100000000001000001001000000100000001100010000000110000110000000000000000110000000000100000010000100000000001110000000000000000100000000001000000000000111000001000000100011000000000010000000000010001000000000000000000000000100001001000001000000000001000000000000000000000011001010000000001000100001101000000000000100000101000010010000000000000001000000000100001000000001000000100000000011100001000000000010000000000010000000000000000000000000001000000000000010000000000001000010000000010000000000000000010000001000000100000001000010000000000000000100000000000000100001100100000000000010100000011000110100111000000001000000100100000000000000101000100010000010000000000000100001000000010100000000100000011000000000000010010000111000000001010000000000000000000000010000000000000100000000100010010010000000000000000000000110001000001100010000000000000000011000000000000000000100001000000000000000000001000000000000000001001001000010000001000000000000010000010011100000000110000100011010000100010101000010101000001000010000000010101000010000000000000000000100001000100101000100000010000001000000001100000000000100000000000000010000001100000000000000001000001000010000000000000000010010000000001000000000000000101100010100000000000100000001000000010000010010000000001000100100001000110000000000000000000000100000000000001001000000110000000000010001000010000001001000010000001001000001000011000000000010000000001100000100000000100000010000000010000010000011000001001001010000001000000100100000001000000100100000010000000000000000111000001000010011010001010010000000000000010100010000000000010110000000000000000000100000001001000000001011010100010000000000000000011000001000000000000000000000000000000000000000000000010100001000000100001100000000010000000000010000100000000000000000000001010000000000010000001000010001000001100000000010010000000000001100000001000000101011000000000000101100011001001000010000010000000000000000000000000000000000010000000000010000001110000000000000010000000000000000000000000000000000000000000100000010110000000000000000000000000000000010000000000010000000001010000000000000000001001000000000001000000001000001100100001100110000000000000001000000000000010000100010000000010100000000100000000000000100010001101000000000000000000000000001010000000010001000000000000000000000000000100000000000000000011000000110010100000000001010000000100001000010000000000000001000010000010010000100010000000000001000101000010001000001000000000000001100010000100000010000001000000000000000000010010011001100100000000000000011000000001000000000101000000000000000000000000000000010100000000110010000101000000000000000010100000000000010000000000110100010000000000000000000001000000001010000000000000010001000010001000000000001001000000000001000000000100000000000001100000000100010000000000001010000000100000000000000000000000000001000000000000000000000000000000100000000000101100000000101001001000000000000000000000000000000010010001000000000011101000000000100000000000000000100000000000000001011000001000000000100010001010000000000000001001010001000000000000000010000010000000001100010100000001000000000000100000000100000000000000101000001010010000000001000000000000000010001011000000000000100000000001010000010000000001000101000000101100000000000000000001000000100010100000000000001000000000000000100110000000000000000000000110010001010000000000000000010000000000000000000000000001000000000010000000010010000010000000000011000010000000000000000000000011000011000000000000000100000000010000000000001100000100000011000000010100000001000000100100000001001000001000000100010000000000100001000000000000000000100100000001001000001000000000001010010000000000000100000000110000000100000001000000000000010100000000000000000000010000001000000000000000000101000010100001010000000000100100100010000000000000110000100000000100000000011000001001111000010010000000000000001000100000001010100010010001000000100000010000010000010000000100010010000000011000000000000000000001000000000000000000000000100010100100001000000100000000100110000000000010010100000000010010100000000010000000000000000010010001000000100100000000000010000010000001000000000000101000100000000000000000001001001000000000001100000000001000000000000000000011000000111010000000000000000000000000000000000000100000010001100100000000'-'0',5,[])),87))
36 Pass
assert(isequal(fivelanes(reshape('00000000000111000100000001010010010000011101110000000000000010000000000001010000000000000000100000000000001000000100001100010001000010010010001100010100000000000000000100100010000000000000000010000000100000100010000000001100000001000000000011100100000011000000000000000000100000000000000000000101000010100000000010000000010000000000010000100000000000010000000000000010000001000000010010011000000000000100000000000000000000000000000000000001010000000001000000000000010000000011100000000000000010000000000000000000001101000000000101000000000000000000011000000000000000000001000000000000000000001000000001000000100001000100000100001000000010000100000000100100100000001000000000100000001000100011001011000010000000000000010100000010010000000100000000000001000000100000000000010000000000000000001000000000000000000000110000110001000000000010000000011000000000000000010010000110000100100010000010010010000001100000000100000000100000001000001000000100100000000000000000000000000000101011000000100101000000001000000010000000000001000000000000010100000000000010001000000000011001000000000001000000000101000000010000000000010011000000000000100000100100010000001000001000000001000100100000010000000010000000000000100000000010000011000000010000000101000000000100001101011001000000000110000000100100000000000000000000000000000001010010001000000100000000000010000000100000100100000001000000000010000010100100001000000100000000000000010000000001010000000000000001000000001000000000000000110100000001000000001000000000000000000110101000011000000000000000000001000000000000010000000000010010100000000000000000100000100000100000000000001000000000000000000000010010010110000000000001000000000000100010000110000010000000100000010000000010000001000000000000010000100000000100000000100000000010000100011100100000000000001010100000001100000001000001000000000100001000000000011010000000100101000100110010000001000010000000000100010000000010010000001000001000100000000000000000000000001000011010000000000000000000010000000011000010000000000000111000000000000100001000000000010000000000001000001100000000000000000011000100000000000000000100000001001101000000000010001000011000000000000000001010000000000011001010000100100000000000000000010000000110000001101000010000000000001000001001000000001001000000000010000101000010000000000100000001000001100000010000000010000000000111000000000010000100000000000000000100011010000000100010100000000000000000110000000001000101000001000100000000000000000000000001100000101101110010000000100000001001100000010000001001000000000000000001011000011000000000000001001000010011001000010000000000'-'0',5,[])),47))
37 Pass
assert(isequal(fivelanes(reshape('001000001100000000000000001000000010000000000001010000100000000010010100000010010010001010000000000001100010001010000000000000110001000000000100000000000000000110000000000000000000100010000000000100100000000000000000000000000000000000000000000001001000000100000000000000001000010000000000100100001000000100000000000000000000000000000001000010000000001010000001000000000001010000000000100000000000001000000001000100000000000100000010001010000000000001000000010100000001100000000000000000010010000101000001000000000110000000000000000110000000000000001000000000010000001010000000110000010100100000010100000000000000010010000100000000000000000100100001000000000010000101000000000000000000000000100100110000000000101000011001000011010000101001000000000001000010000100000000000000000000100100101001100000000001000000000010100000000010000010000100000001001000000000001000001000000100101100000000100100000011000000000010000010000000000000000000000000000001000000000000010100110000000100000001100100000000000000100000000001100000000101100000100000000100000110101100000100000100010000000000000110000000110000000001010000011000000000011000000000101010000000000010000100010000000000001000000010000000000000001100000101001100000100000000000100010100000000000001000010000000000000000100000011000100000000000100000000001000000000011000000010001000000000000000001000001110100000000001000000001100000000000000000100000000010000101100000000010000000000000000000010001001010000000000000000010100100010000000000000000100000000010010110101000001010000000010000001000001000000001000101000100000000011000101010000100000100110000000000001000000010000000100000101100000000100101000000000000001000000000001000000000000011010010000000000001100000000000000000000000001000000011001000000000011001000010000000000000100001100000000100000000000000100010000100000001000000000100000000000000100000001000100000000000000000000000000100000000000101001010000000000000000001001010010000000000000000000001000100100000000000000001000000001100010001000000000000101000000000001000000000000000000010000010000000000111000000000000000000000100000100010000000000001000110000000000100010100000000001000000000001000000100100000010001000000000000000010100001000010000100000000000011010000000000000000000000000000001000000100000000001000010000100011000010000000010000100000001001010100000100101000010000000000000000000000000001000010101000000001011000000000000010101000000101010000110000001100100000000000000000000000000000000000100010000000000100000000011000000000001000000000110001000000000010100000000010000001000000010100000000000001000000001010000000000000000001000000010001110000000000001000000011000000001001000000010001000100000000110100100000000101110000000000000000110010000000000001000001010110000000000000010000100010000100000011001000001000000011000000000000100000100000010000001001010000000100000000000000000000000000100011001000100010000001000100000001100000000000000000000000000000000000100000010000011010000000000100000000000001000000000000010000010100000000010000000001010000010101010000000000010010101000011000000001001000000000000000001101000001000000000000000000010000010000000000000000000000000001000000001000000000000000000000000010000010001000100000001001010000000000000000011000000100000000000000000011000001000000000100001000101000001000000010000000000000000001000000000110000010000000000000100000000000000000000000010001000101000000100000010001100010011101000000000001000001000000010101000010000000001000010000000000100000000000000000000000000000010000001100000000001000000011000000100000110000000101000010000001000110110000000000001100100001000101000000110000000000000000010000000100100000100000000000010100000010000010000100000000000010000001000010100010010000000000000000'-'0',5,[])),72))
38 Pass
assert(isequal(fivelanes(reshape('00000000100100000100000000000100011000000000000000000000010000000000010000000000000000010000000000000001000000000000011000101000000000001000000000000010000010001001000100100001000000000000100000010010110000000001000000000000000000000000000000000000000001001001101000000101000000000000000000000000000000000000010000001000000001000000000100000110000000001000000100001010000000101001100001000000000000000000110110000000010001000000000000011000000100100000000000000010000000000000100000001000000000000000100000000000000000000000000101000001000010000000000000100001001000000000100000000000111000101100100000000010000000100000000000000000000100000000000110000010010010100000000000000000001000010010000000000001000000010010000000001000100000000000000000000010000000010010010000010000000000000000100000010010000001010010100010000000100000010100001000000100000001100000000001000100000100100000000000000100000010000010000100000000100000000000000000000000000000000000000001100000000000001110000000000000010100110101010010000000000010110000000001000001010100000001001000000000000000000000000000000101010110000101011000001000000000100100010001001000000011010000010100100001000000000101000001000001000100010001000001001000000010000000000000000001000000010100000001101000010010000100000000100101100000100000000000000000000000000000010000000010001000000001000000000100010000110000000100010000001010000000001001000000100000000010000101000001000000000010000001001000000000110110100100000000000001010000000001001000000011000000100000100000010001000000000001100010010000000001000000000000000100000010001000100000000000000000010000000110000000000001001110000000000000010100000000100000000000000001001001000100000000000000001000000100010000100100000001001011000001000000000000010000010000000000000000100001000000001000000100000001001000000000000010000001000000000011000000010000000001000010010000010001011000101001000000000000110010000100001010000000000010100000000100000000110000000000000000000000001000000010100000000000000001000100001010100010000000000100100000000000000000000000000000010000011000001010000000001000000000000100000000000000000000000100000000000000100010000010100000000000100000000000000001000001010010110100000000000000000000010000000000010100000000000000000000100000000000000000000001000010000010010000100000001000000010000100000001010000100000000000000000100001001001000000000000000000000001000000000000100000100010000000000000001000001001100000000000010000010000000000100001000000100000100100000010000100000010011001011011000100001000000100000100000000000011000000100000000000000000010000001000000000100000010000000010000000010010000000000000000000100001010010010001100001000000100010000000000100001010000011000000100100011000000000100000100000110000000000000000000000010010000000000000010001001000000000010110000010000000000010001100000100001000000010001000011100001000100000000000000100000001000000000010000000100010000000100000000100100000010000000000100000010100101000000000000000010100100000001000000000100100010100000001000000100000010000000001000000001000100010000010010000100000000011010100001010000010000100001000100010000100000000000000000001100001000000000010110000000010000000000000001000000010000000000000001000000000000100000001000000000000000000000000001010000100010011100000001000100000010101000000100101000000011100000001000000100001000000000000000110001000000000101000000001000000000100110000000000000000001010000000000000001000000010001000100000000000010000010000001000000000001010000011000000000000000100001000000000000100000000000000010000001000000000000000000000001000010000100000000000000000000000000000000000000000001000000000'-'0',5,[])),68))
39 Pass
assert(isequal(fivelanes(reshape('00000110010000001100000000010010100000000010010100000000010000000000100000000000100010000010000000000000001110001100010100000000000000001001100011001000000000000000010000000010000000000001000000000100000100000000001000000001001000000000000010010011001000100000000000000000000000000001000000100101000000000000000000000100000000000100000000000000000000001001000000000000100001000100010000001000000110001001000000000000001001000010000001000000000000000100001000000000100001000000011000000000101010000000000000001000000110000000010000000000000000000001000000001000000000000000000000000000000101000000001010010010000000000000000010000010000000100000010000010000001000010000010000010100000000000010001000000010010110100000010000000001100000000000100000000011000000000000010001000100000000000000001000000000000000000000100000010100000000100001000001000000000000101000000000000100110000110000000000100000000100000001001100100010000010000000000000000000100000010010000100000000000000000000100000000101010000000000000000000000000011000000100000000100000010000000010000000000000000000000001000000000000000001000000000000010000001000001000000100100000100000100000000011010010000000000000001000010000000000000100100000001001000000000000000010000000000010000100000000001000000000000011000000000100000010000000000000001100000000000101000000000010000000000000000010000000000011000000000000000010000001000000000000001010100010100101100000000000000000000001010100000001000000000000010000000000110010001000000111000000000000100100000000100100011010000001010000000000001001000100001000000000001000000001011000010100101000101000000100010000000000000000000100010010000000000000010000000010000001010010000000001101000001000000010000010000100001000100000101000000000001000000001100001101000000000000000001010001000000001010000000000000110000001000000010100000010000001000100000000000000010000000010010000000010001000000000000000000000001100000000000000000000100011100100100000000010010000100000100001110000000000000000000100100110000000000100000000001000000100000000000010010000001000000000000000000100000000000000000000100000000000000000000000000000000000100000110010000000100000000101100000000000000000000010010000000000000001100000001000000000000001000000000100011100100100000000100101110010000100000100000000000000010001000001001000000001000000000000000000001000000000000000000000000000010000010000000001000000000001100000000000000010000000010000000001001010000000100100000000000010000000000000000001010011000000000000000000010100001111000011000010000000000000010100000000000110000000000100010001010000100001000000000100000110100000001010000000000000010001100000000000101001000000001000010010000010000000010010000100011010000000000000000010000100101000100000000000000000'-'0',5,[])),50))
40 Pass
assert(isequal(fivelanes(reshape('00100100000111000000110100001000101000000100100001000000000000000000000000000000000000000001000100000000000100100100000000000000000000100010000000000000000000000000000000000000000000000000000100001000000011000001000000000000001000010110000001100000000000010110110000000000000000000000100000000000000000110001000001000000000000000010100010100001000100000000010000000010000000000000100000000101000000100000001010100001010000000100000000000000000000010000000000000100010100101000000000010000000000000100000010000010000001010010110010001000000101000100000000000010001001000100000010010000000010000000000000100100110000000000000000000000100000001000010000000010001010000000000001111000100000001000000010000000000010000001000010000000101000101000000110000000001000000000000100001000000000000001010000000010110000000000001010000000000000100000000000000011000010001000100110000000010001010000000010100000000000001100010000100000000100000000000000001010100000000000100000100000000000001010100010000010010000001100000000100000010010000000000010000001000000000010001001000001001010000000000000000100000100010000000000000000000000000000000100000000000101000101000001000000001000000000001000000010000000000000000100110000000000000000000000000001010010010000000000000010000100000010001000000001000000100000010000100000000010001000001000001000000000001100001001001100000000000000100000100000000000000010000000000000000000000000000010000010000000000000000000110000100011100000000000000000000000010000000000000000010000000000000001000001100100000000100000000000110100000010000000100000010000000010010000000100000000000000011101100000010010000000000000000000100001000000000001100000100000000000000000110000000000000010001001110100001000000100000000001000000001010000001001001000000001000100000100000000100000000000110010001100010000100000001000100000000010100000000000'-'0',5,[])),40))
41 Pass
assert(isequal(fivelanes(reshape('0000010010000001000100000000010000000100010010000000100100100110000010010000000000000000001001010000001100000100100001001000000101000000010100000000000001000100000000010000000100000000100000000000000000110000000010000010000000010000000000001010000000000110000000100000000000100001000001000000010000100000000010000000010000100000000000000000000100000000011101000000100000000000100001000000000000000000000000000001000000110000001000000000001000010000000100000000000000000100010000000000000000000000010000101000011001000000010000000000000000000000000100000000000100000000000010000101101000000000000000010100000000000000000000100000000000101000110000001000000010001000000000001000000100011100011100000000001010001100000000000010000100100000101000100000001010000110110000010001000100000000000000000000010010010000010000001000000000000000001000000001000100000010001010000000100000000010001000000001000110000000000001100000001000010000000010000000000000000000110101000010100000001000100000000100110100000000000000000000000000000000000100000000100001000010000000000000110010000000100100000000000000100000000000000000000000000000000000000100010000000000000100010000010000001000100000100000000000000000000011000100000000000100100000000000000000000000000000000000100100000000000001000000000001000000000000000010000000000000001000000000100000001000000010010000000100010000000011000000000000000100000000000000000000100000000000000000000000000100000000010000000000000100000001011000000001000000100000000000110100000000000000010000100001000000000000100000101000000001100000000001000000000000000000000010001000000000000000000100010000010000011000000000010000100000000100000000000110000000100100000000000001001000100000000010000000000100001000001100000000000101000001000000000001000101000000010011000000000001000000010100000000000000100000100010000000011001000000010000000100010000000100000000000000000000001000001000010000001001100000001000010000000000000100000010000000000001000000000000010000000000000001010000000000000000000000000001010000000000000000000000000000101000000000100000000000000000001000000000000010000010100010000010100000000000000000100100000110000010000010000100001101000000000000000010000000000000000000000001000001000000101100000000000000000100000000000010000000000001000011010000000000000000000000000001000000000001010000000000000000000000000000000000000001000000000000000010001000000100000100010000000000000000000000010000000001010100010001101000000000000000010000001011000000001000000000000000000001011100000000000110000000100001000100000000000000000100100001000100100100101110000001000010000000000000010000000000000010000000000100000000000011010110000001101000000100010011000000000001000000000000000000000010000000000000010100000000011000000010010100010000010000001000000000000010010000010000010000000011000000000000000000001000010010000010000000001010010001010000010000001100000000100001000000101010000000001000001011000000000001100000000100000000000000000000000000000000000000100000000001101000001001000000000010001000000000000010110000000000000000010000000100001000000011000000101000000001'-'0',5,[])),56))
42 Pass
assert(isequal(fivelanes(reshape('0000000010000000100000000000000010001000000000010000100010000000010000100101000000000100000000010001000001000000100100000000100000000100000001100000000000000111000000000000000000000110000000000000001000001000001100011000000000100100001000001000000001100001010100000000100000000000100100010000010000010000000000000100101000000100000010001010000001001000000010001000000000000100000000000001000100010000110000000010000000000000000000110010000010000000000000010000000010000000000000000010000001000000000000001000000000000100000000000000000000111100000000000010100000001000000000000000000000000000100100011100001001100000000010000000000000010000000000000000000100000000000100000000000000000010000100000001000000000000010000001000000000010110010010110000000000000000000100100000000100000000001010011001000000100001000010000101000000000001010000001001000000100010010001001000000000001000011000000110000000000100001000000000010000000000000000100000000000000000010000000000111011000000100000000001100000000001000100000000000000000000001001000100100000000000000100000000011000110000000000000000000100000000001011000110000000000010000000100000100000001100000011000001000001001110000100000000000000000000000010001000100100000000010000000100000110000000000000000000000001000000000010101100000010100100000010000001000000001000000010000001000010010000000000000011011100100000000000001000000100000100000011000000000101100000010001000010000000100000000000000000010000001100000000000000000000010000000000000000000000000000010000000000000000000000000000000000000010000000000000000000000110101001100010000000001000000000000100000100000100000000010100100000000010100000000101000000010101010001001000100011100000000001010001101010001000010000000000000001000000001000000000100000110101000000100000000010000000000001001001000000100100000000101000000000000101010000000000000011000000100001000000100000000000001000001000000010001100100000010010010000000001010000110001000001100000001000100000000001010110000010000000100000000101000000001000000001000000000000000100000010000000000000100100100010010010000000000000001001000000000010000000000000111010000000101110000100000000100100000100001000000000000001000000000000000000000100100000000000010011000100000001000000101000000000100000000000100000100010010000000000000000001100000001000000001000110000001000000010000000000000000001000000000011001001000010000000010000000001000000000000000000000000001100000001100110010000001101000000001000000000000000100000100000010100000000010000001000000000000000000110010100000010000000000001000000000000000000100000000100010000000100000010000000000000110100000000000000000000000010000001010000000000000100000000100000000000000000000000001010001000000000000000000010000101010001010000100000000000000100000000100000000000100000010000010010100110010000010001000000000010000001000000010100100010000000000000000001010000000001000001000000010001100000010000100000000000110000001001000010000010000010000100001000000000000000000101010000000000000000000001000000000000000110000000010010001000010000000001000000100010010101001000000000100100000001000010010000000000000001000100000001100010000000000000000001100010000011001000010000000000010000001000000100000010000110000100000000000001001010000000000010'-'0',5,[])),59))
43 Pass
assert(isequal(fivelanes(reshape('0000001000001000001000000001000010000000001000000000000010001110000000000011000000010000000001001000000000110001000000000000101000001000000000010101000000000000010000000000000011000001111001000000100000001000001000000000000000000000000100000010010000000001000010000000100000011101010000000100110001100001000100000000000001000000000000000000100000100000000000000100000000000010000000000000010100000000000010001000000000010000000000000000000010000010100000000000000001000010000000000000000001000000100000000100000000010010000000100000000000101000000010100001000000000000000000000000000010000000000000011000010000010000000000000000000000000000010001001010000001001000100100000011011000000000000001001000000000000100000100000010000000001100010000000100000110010000000000000100000000000000001000000000000000000000011000001010010001000001010100000000000000000000000100000000000000100000011000000010001000100000000110000000100000100000000000000000010000100000000000000000000000000000000100000001000000010110000000000000000000000001000100010000000000000010000100000000010010001000000000101000100000010000000000001000000000000000000001101000000010000000000100011001010000000000000000100000000000000000000000001000000010101000000000100000000000000010010110000000000100000000010000101010000001000000000000010110000000000000111101100000000000000000100000000000000000000010000000001000000000010000010000000000000000100000000000010000000001000000000100000000111000011001010000000001101000000000010100000000000000000000000000000000000000000010000110001100000000100000000000000000001001001000000001000001000000000011000000000000000010001000000100000000000000010001000011000100000010000001000000000000100101010001000000001010010010000010000000100000000000000000000010000010000010000010000100100000000100000110000001000000000100000001000100000000100010100010000000000000000100000000010101000101000101000001000100010000000010000100000100000000000000000010010010000000010000000000000000000000001010000000100011010000000010010000000000000000000000010000010000000000000010000000000110000000000010000101000000000000010000000100000000001000010100000100000000100000001001000100001000010000000010100110000001000000000100100000000001011000011010000110000000000000000000010010000000000001000000000010000000000000000010100000110000000000000010000000000010001010000000010001000001000000000000001000000010000010000000000010000000110000010001000000000000010000000001000000000000001000100000100000000101010000000000001010001000000000000000000000001000000000010000000001000000000100000100001000000101000000000000010100000000001000100000000100000000010000000000000000100000000000000000000110010000000011101000000000000000000101001000110001000001000000000010001101000010000000000000000000000000000001100000000000001000000100000000011000000110110000000000000000001010000000000010000001000000000000000110000000000000001000010000000000000000010000001100100000100000000000010000010000000000000000100000000001000000000011110000011000000000000000000000001000000101000000100000010000000000000010000000100101001000100100000000000000000000001100000000000000000000000000000000000100000000000000100000001001000000000000001001000000000010000101000000000000000000011000000000000000010000000000000010000000000110100000000000000001100000000000000001000010010000000110100000000000000000110000000000100010000011000001010000000101000000000000000000000010'-'0',5,[])),62))
44 Pass
assert(isequal(fivelanes(reshape('0010000010000011000100001000000001000000000001100000100110000010010000001000001011000000000000000000100010000000000010001000000000110000100010000000000101100000000001100110000001001001000000000000000000000000000100100000011001000011000000001100001100100000100100000000000000000010001000100000000010000001000000000000001000000100000000000100100000000000000000000010000100000000010000000100000000001000000000000100001000110001010000001010100000000011010000000010000000010001000100000010001000010011010100000000000001000000000100001010000000000000000000000000000101000010000010000001000000000111000000100000001100011010000000000101000110000000000000110111000000000000100000000001001010000001011000000000'-'0',5,[])),13))
45 Pass
assert(isequal(fivelanes(reshape('00000100001000000000000010000010100000010100010000000000000100000000000100100010000000010000001000100000010000010001000010000010000000000100000010000000000000100000000000000001010010001000100000000000101010000001000010010100010000000010000000001000100001000110000100010100000000101010100000100010100000000000010011000000001000000000000001000101100001100000000000001000010000100000010000000000100000000000001000101000000000101000001000001000000000000010001001000001000010100000000000110010000000100001100000000000100000000101010100010001001000000000011000001000000001000010000000001000000111000000000000001010000001001000000101100000001111000000000000000000000001000000000010000010000000010000000011010001000001001000000000000000000000000000000000000000100100000000000000100000000000101000000000000000100001000000100100100000001000010000000011000000000010000000000000100000000000000000000000001000000001000010000000001000000000100001000000000000000000000000101000000001000000011000000000001100000000100011100000000000000000000001000000000010010000111000100001000000100100000001001001000000100001000001'-'0',5,[])),20))
46 Pass
assert(isequal(fivelanes(reshape('0000011011000000001010100010000001101000000001100000000001110000000000000010000100000000010010000000001100001000100000100000000010100001000100000000000010000000011000000000000000000001100000100000001100000000100000000000000000000000010000010001000010010000000000010000000100000000000000000000000000000000000010100010010010001110000010000010100100000000000000010001000000000110001000000000000000000000100000010000010000000000000000010011000100000000000000000100000101000000001000000010000000100100010001000000100101100000000000100000000010000100010110000001000000010000010110000010100000001000000000000000100000000001001000000000001000100000000000100000000000000000000010100000000000000000000100010000101000000001100000000010100000100000000100100011000000000000100000100010000000001000000000000000000000000010011000000000010010000000000000000000100000001100100010000000000000000011110010010001100100000001000100000000011000001000001000001000000000000100000000000100000100010001000100000000010010000000'-'0',5,[])),21))
47 Pass
assert(isequal(fivelanes(reshape('0100010000000000000000000000000000000000000000000100100010100101000000000000000011100000000000100000010100000000100101000000000010000010000000000000111000000000000010000000100010000110000100000001001000101000100100000000000000010000000000000100100000000000100100000010000000000100100100000000000001000100000000000000000001000000001110000101000001000000010000100011000000000000000000000000001010010000000000000000010100000100010000000001000000000001000000000000000001000000001000000000000000000000100000000000010000000000000000000000000000000000000000000001000000000001000000000010011000000000000000000100000000000001000001001010000000010000001000000000000000000011000000000000000100000000000000000000'-'0',5,[])),13))
48 Pass
assert(isequal(fivelanes(reshape('0000000000000000011000000000011000000000000000000000100000110000001001010000001000000000000000100110100000001100000000000000100000100000000100000000000000000000000010000000000000000000100000000001010011001000000100001000100000010100001000100100010100100010000001100000001100000000000000001000000000000000000000000000010101000100000000111000000100000010000001000000000010001000000000100000000001001011000100001000000000100010001000000000001000010000100000001000000000000000000100000010010000000000000000000000000010011000000000100010000001000000100100000000010000100000000011001000100000000010001101000001000000000000000010000000000010000100010100000010000011001000000100000000101000000000010000001010100000100000000100000010000010010001000000000100000100000100000100100010100001000100000000001000001000000000000100001001000000011000000001000000000000010100000000010100000000000000001000000000000100100000010000010010000010000000000100000000100000000010000100010000000000000000000000100010001110000000000000000000000000100010100000001100000110100000000001100000000000000000000010000000000110000100000000001010000000000000000001000000000001000000010000001100100001000000000000100001000000101000000000000000100010010000000000001000100000100000000000000000000000000101000100000000100100100100000010000000000000001011001000010000011000000001000000000100000100000101000000011000000010100000000000000010010000100000000010000000000100010100000001000100000001001000100100010000000000000100000000101000000110000010010000001000101001000000000000000000000000000000000011001000000010000000001101000011000000000000000001000000000000001010000000000000000000000000000000101000010001000000000100010001000001000110000000000000010001000110000001000001010000000100001000111000000010000101000000000100100010000000000000110000000000001010010010001110000010000000000000001000000101000100000000010010000000000100001000000000000000001000000000000010100010000100000010000010000000000000000100101000000000000000100010000000100000000001100010001110001000000000001000000001000101000100100000010000000000000000100011010000001000000010100001000001000000000000000100001000001001001000001100000000000000000000000000010000000000010000010000000000010000000100000000000000000000000000000000000000001000000000000000000100000000000000000010000100001000010000100100000000000001000000000000000010000100001000000000100000000000010000000010000000100000000101010000100000000000000001100000100000000000000000000011001000000100000100010100000100000000000000010000010000100000000000001000000001010000000000011000100110000100000010010000000000000000000000000000011100000100000000000000000001100000'-'0',5,[])),51))
49 Pass
assert(isequal(fivelanes(reshape('00000000011001010001000000000000000000000100100000000001000111010100000000010000000000010000000001100000000000000100000000001001100010001000000000100010010000100001000100000010001000101001010000000000000100010000011000100000000000000000000000010010100100000000000000010100001000101100000000100000000000010000000000000000110100011000100000011110000000001000001000100100000010100000110010000101000010000000010000000001000001110000000010000000000000000100000000000001010000000000000001001000100000100000'-'0',5,[])),10))
50 Pass
assert(isequal(fivelanes(reshape('00000000001010001000000001001000000000000000000010010100000000100000000001000000000001001001100000100000000000000111000000000000000010000000100000000101000000100011000000110010000000000000001001000000010001100000000000000010000000001000000000000000000010000000000000001001000001101000100000001100000000101000100000000010000000000000000000000000001000000001000000000000000000101000000000000001000000000000000000000000000000000000000000100010000000000000001000010010010000100000000001000000000000000000000000001000100000000000100100000000000010010000001000000010000100000100000100010000000000101000000010100000010010100000000000000000110000000000000000000000000000001000010000000000000000100000000000001000000001001000000000001000000000000000000000000000000000001010000010100001000000000110000000000000001000000000000000000000000000000010000100000001010101100000000000000000100100100010100000000000000000000000100000000000010000000001000000000010001000100000001000001001000000010010000100000000000000000101000000000010001000000000000100000100000000000000000010000000010000001000000000000000000001000001000000000000000000000001000000000100000000000001010000000000000000000000000000000100000000000000000101000000000000000000000000000000000000100001011010000010000000000001000000010100000000000010000010001000000000000001100000001010010000000000000010000100000000011010000000011000000011000000000000000000000000000000000000000001000010110000000000000000001011000001010000010000010000000000000000000011000000010011100000101000000000000000000000000001000000000000000011000000000000101000011000110010000010100001000000000110000010000001000000101010100100000000101010010010000010000000000010001000000000000000010000010000000001000001100000000000010000000010000000001000000000000000000000010000000100000000000000000000000010010011000000000000100000000000010100100000000000000001101000001000000011100000000100000000100000000000000000100100000001000000001011000000010001000000000000000000100000000000000000000001100000010000110100000000000001000010001100000000000100000001000000110000000000000000001100000100000010000000000100010000000001000000001000100000011000000000000000000100000000001001000000000110110100000000010000010000000000011000000010000000000100100100000000010000000000000000000000000100000000000000000000000000000001000000000000010000000000100010101000000110001110001100000010000000000000100001000010010000000000000010000000000011000000000001101011010000000000000000000000100100011000000000000000000000000000001010001110000001010100000001001100110000110000010100000000000001000000001100000010100000100000000000000000001000000000110000000110000000010001000010000000000100000000000000010000000000010101100000000010000100000010000000000101010110000000000000000000000000101000001100000000000000000000000100000010000010111000000000011000000000000000000000000000000100100110000101000000010000000000000001000000000001000100001000010101000000000100000000001000000100000000011000000000000000000100000100000000100000000011000000110000000001000100000000000000010000000000001000000000000000000000000001000000100000100000000000000000000100000100000001000000000000000000010000000000000000010000010000000000000000000000000011000000000000000000000011010000000100000100000000000000000110000000000000000010010010000000000100000000100000100100000000000000100010000000000000000000000100000000000001000100100011100001010000001000000000000000000000000000000000001000000000000000010000001000000101010100000000011000000001'-'0',5,[])),64)) %% | 33,907 | 65,643 | {"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-2020-40 | latest | en | 0.387496 |
https://www.physicsforums.com/threads/art-of-electronics-exercise-4-3-operational-inverter.679855/ | 1,519,567,651,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00149.warc.gz | 946,902,884 | 17,124 | # Art of Electronics Exercise 4.3 Operational inverter
1. Mar 20, 2013
### CSOleson
1. The problem statement, all variables and given/known data
The circuits in Figure 4.14 let you invert or amplify without inversion, by flipping a switch. The voltage gain is either +1 or -1, depending on the switch position.
http://tinypic.com/r/2hnql2c/6
http://tinypic.com/r/2hnql2c/6
2. Relevant equations
Gain for the first image with the switch set to ground is equal to -R2/R1, because it is an inverting amplifier, struggling to identify how to consider the circuit as a follower, because it doesn't look like a typical follower circuit. Unsure how to consider the circuit when the positive and negative terminals are connected.
3. The attempt at a solution
Last edited: Mar 20, 2013
2. Mar 20, 2013
### Staff: Mentor
What is the difference in voltage between the + and - inputs for an opamp in the follower configuration? So what is the feedback current in this configuration?
3. Mar 20, 2013
### CSOleson
are the + and - inputs equal, because of the negative feedback loop?
4. Mar 20, 2013
### CSOleson
so the voltage across the 1st resistor would be 0V because it is V+ - V-
5. Mar 20, 2013
### Staff: Mentor
Correct-a-mundo
Kind of a wierd follower, but that's one of the ways that H&H teaches us how to be comfortable thinking about circuits. Have you seen this thread? | 374 | 1,385 | {"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} | 2.984375 | 3 | CC-MAIN-2018-09 | latest | en | 0.897231 |
https://www.proprofs.com/flashcards/story.php?title=parent-functions | 1,501,108,382,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426639.7/warc/CC-MAIN-20170726222036-20170727002036-00433.warc.gz | 821,376,552 | 18,644 | Front Back linear y=x D: R R: R absolute value y=IxI D: R R:[0,*8*) quadratic y=x^2 D:R R:[0,*8*) square root y=$\scriptstyle \sqrt{x}$ D:[0,*8*) R:[0,*8*) cubic y=x^3 D: R R:R cubed root y=$\sqrt[3]{x}$ D:RR:R rational y= 1/x D: (-*8*,0)U(0,*8*) R: (-*8*,0)U(0,*8*) rational y=1/x^2 D: (-*8*,0)U(0,*8*) R: (0,*8*) exponential y=2^x D: R R: (0,*8*) logarithmic y=logx D: (0, *8*) R: R | 186 | 384 | {"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": 2, "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-2017-30 | latest | en | 0.34027 |
https://www.slideserve.com/calista/static-friction | 1,600,786,879,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206133.46/warc/CC-MAIN-20200922125920-20200922155920-00152.warc.gz | 1,050,226,612 | 12,302 | # Static Friction - PowerPoint PPT Presentation
Static Friction
1 / 9
Static Friction
## Static Friction
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Static Friction
2. Holding in Place • Objects on an incline will often stay put. • There must be a force that holds the object in place. • Static friction is from the contact of resting objects. • Force holds up to a certain point • Force is based on the type of contact (rough, smooth) • Maximum force is proportional to the pressing force of the object (normal force)
3. Inequality • The approximate formula for static friction is: ms is the coefficient of static friction • This is an inequality. • The force of static friction is generally less than the coefficient times the normal force
4. m q Starting Motion • If Ffr < msFN = msmg cosq, then the block will hold. • At equality the block just begins to move.
5. Use a pulley and vary the weight to see when movement begins. Forces balance on mass 1 Tension is equal on the rope Forces balance on mass 2 m1 m2 Measuring Friction: Pulley FT Frope FT Frope
6. The pulley determines the coefficient of friction through the ratio of masses. m1 m2 Coefficient of Friction: Pulley
7. Use an incline and vary the angle to see when movement begins. One component of gravity balances the normal force The other component of gravity balances friction m q Measuring Friction: Incline
8. The incline determines the coefficient of friction through the tangent of the angle. m q Coefficient of Friction: Incline
9. Normal Force and Friction • Static friction depends on both the normal force and on the coefficient of friction. • To reduce friction requires reducing one of those factors. • Reduce normal force by lightening the load • Reduce normal force by adding additional upward force • Add a lubricant to reduce the coefficient of friction next | 467 | 1,957 | {"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} | 4.09375 | 4 | CC-MAIN-2020-40 | latest | en | 0.808143 |
http://bfluhr.com/notes/interleavings-1d/reeb-graphs.html | 1,553,623,470,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205600.75/warc/CC-MAIN-20190326180238-20190326202238-00095.warc.gz | 23,389,720 | 2,896 | ## Reeb Graphs
In this section we introduce Reeb graphs.
• Definition (Reeb Graph). Given a continuous function $$f \colon X \rightarrow \overline{{\mathbb{R}}}$$ and $$x \in X$$, let $$\pi_f (x)$$ be the connected component of $$x$$ in $$f^{-1}(f(x))$$. In this way we obtain a function $$\pi_f \colon X \rightarrow \pi_f (X) \subset 2^X$$ and we endow $$\pi_f (X)$$ with the quotient topology[2]. By the universal property of the quotient space there is a unique continuous function $$\mathcal{R} f \colon \pi_f (X) \rightarrow \overline{{\mathbb{R}}}$$ such that $\xymatrix{ X \ar[r]^{\pi_f} \ar[dr]_f & \pi_f (X) \ar[d]^{\mathcal{R} f} \\ & \overline{{\mathbb{R}}} }$ commutes, i.e. $$\mathcal{R} f \circ \pi_f = f$$.
For another continuous function $$g \colon Y \rightarrow \overline{{\mathbb{R}}}$$ and a homomorphism $$\varphi \colon f \rightarrow g$$ we may use the universal property of $$\pi_f$$ again, to obtain a unique continuous map $$\mathcal{R} \varphi \colon \pi_f (X) \rightarrow \pi_g (Y)$$ such that $\xymatrix{ X \ar[d]_{\pi_f} \ar[r]^{\varphi} & Y \ar[d]^{\pi_g} \\ \pi_f (X) \ar[r]_{\mathcal{R} \varphi} & \pi_g (Y) }$ commutes, i.e. $$\mathcal{R} \varphi \circ \pi_f = \pi_g \circ \varphi$$.
• Lemma. Let $$f \colon X \rightarrow \overline{{\mathbb{R}}}$$ and $$g \colon Y \rightarrow \overline{{\mathbb{R}}}$$ be continuous functions and let $$\varphi \colon f \rightarrow g$$ be a homomorphism, then the diagram $\xymatrix{ \pi_f (X) \ar@/^/[rr]^{\mathcal{R} \varphi} \ar[dr]_{\mathcal{R} f} & & \pi_g (Y) \ar[dl]^{\mathcal{R} g} \\ & \overline{{\mathbb{R}}} }$ commutes. Or in other words $$\mathcal{R} \varphi$$ is a homomorphism from $$\mathcal{R} f$$ to $$\mathcal{R} g$$ in the category of $$\overline{{\mathbb{R}}}$$-spaces.
• Proof. We consider the diagram $\xymatrix@C=5pt{ \pi_f (X) \ar[rr]^{\mathcal{R} \varphi} \ar@/_3pc/[ddr]_{\mathcal{R} f} & & \pi_g (Y) \ar@/^3pc/[ddl]^{\mathcal{R} g} \\ X \ar[u]_{\pi_f} \ar[rr]^{\varphi} \ar[dr]^f & & Y \ar[u]^{\pi_g} \ar[dl]_g \\ & \overline{{\mathbb{R}}} . }$ In this diagram the three inner triangles and the square commute. Further $$\pi_f$$ and $$\pi_g$$ are surjective and thus the outer triangle commutes as well.
• Lemma. Let $$f \colon X \rightarrow \overline{{\mathbb{R}}}$$, $$g \colon Y \rightarrow \overline{{\mathbb{R}}}$$, and $$h \colon Z \rightarrow \overline{{\mathbb{R}}}$$ be continuous functions and let $$\varphi \colon f \rightarrow g$$ and $$\psi \colon g \rightarrow h$$ be two homomorphisms, then $$\mathcal{R} (\psi \circ \varphi) = \mathcal{R} \psi \circ \mathcal{R} \varphi$$.
• Proof. We consider the diagram $\xymatrix{ X \ar[r]^{\varphi} \ar[d]_{\pi_f} & Y \ar[r]^{\psi} \ar[d]_{\pi_g} & Z \ar[d]_{\pi_h} \\ \pi_f (X) \ar[r]_{\mathcal{R} \varphi} & \pi_g (Y) \ar[r]_{\mathcal{R} \psi} & \pi_h (Z) . }$ By definition both inner squares commute, hence the outer square commutes as well. Now it follows from the uniqueness part of the universal property of $$\pi_f$$ that $$\mathcal{R} (\psi \circ \varphi) = \mathcal{R} \psi \circ \mathcal{R} \varphi$$.
The previous two lemmata imply that $$\mathcal{R}$$ is an endofunctor on the category of $$\overline{{\mathbb{R}}}$$-spaces. Later we will define an interleaving distance on Reeb graphs, but first we will introduce join trees and their interleavings. Join trees are easier to understand and may provide us with some intuition for understanding the more sophisticated interleavings of Reeb graphs.
[2] see for example (Bredon 1993, definition I.13.1) | 1,245 | 3,519 | {"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 | 3 | CC-MAIN-2019-13 | latest | en | 0.390085 |
https://blogmech.com/page/175/ | 1,597,486,229,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740838.3/warc/CC-MAIN-20200815094903-20200815124903-00067.warc.gz | 219,817,955 | 23,290 | # CNC Coordinate System | Cartesian Coordinate System
CNC machines Axis:
The real parts of a CNC program involves the input of co-ordinates of the tool endpoint to produce any machining profile, subsequently it is necessary to follow a proper co-ordinate system.
Cartesian Co-ordinate System:
All the machine tools make use of the Cartesian coordinate system for the purpose of simplicity. The guiding coordinate system followed for assigning the axes is the familiar right hand coordinate system. The fundamental axes to be assigned are the rectangular axes and the rotary axes. One could use his right hand to arrive at these alternate variable positions of the same right hand coordinate system.
Designating the axes:
Z-Axis and Motion:
Location:
The Z-Axis movement is either along the shaft pivot axle or parallel to the shaft pivot axle. In the case of machine without a shaft / spindle such as shapers and planers, it is identified as the one perpendicular to the table, which passing through the controlled point (e.g., the cutting tooltip).
Direction:
The tool moving far from the work-holding surface towards the cutting tool is assigned as the positive Z direction. This means in a drilling machine the drill moving into the workpiece is the negative (-) Z direction. This aids in decreasing the possible accidents because of the wrong part program entry in the coordinate signs.
When there are several shafts and slide ways:
In such cases, one of the shafts, preferably the one perpendicular to the work holding surface may be chosen as the standard shaft. The primary Z movement is then close to the primary shaft. The tool movements of other spindle quills or sometimes slides, these movements are termed as secondary and tertiary movements, and they are designated as U, V, W and P, Q, R respectively. For other machines such as shaper, planer etc., the positive (+) Z movement increases the clearance between the work surface and the tool holder.
X – Axis:
The X – Axis is the principle motion direction in the positioning plane of the cutting tool or the workpiece.
Location:
It is perpendicular to the Z – Axis and should be horizontal and parallel to the work holding surface wherever possible.
Direction:
When looking from the principle spindle to the column, the positive (+) X is to the right. For turning machines, it is radial and parallel to the cross slide. X motion is positive when the tool withdraws from the axis of rotation of the workpiece. For other machine tools, the X – Axis is parallel to and positive along the principle direction of movement of the cutting or the guided point.
Y – Axis:
It is perpendicular to both X – and Z – axis and the direction is identified by the right hand Cartesian coordinate system.
Rotary Motions:
A, B and C movements define the primary rotary motions.
Location:
These movements are located about the axis parallel to X, Y and Z respectively. If it in count to the above mentioned primary rotary motions, there are secondary rotary motions. These should be designated as D or E regardless of whether they are parallel or not to A, B and C motions.
Direction:
Positive A, B and C are in the directions which advance right hand screws in the positive X, Y and Z directions respectively. The fingers of the right hand point towards the positive direction of the rotary motions. All the above mentioned (e.g., X, Y, Z; U, V, W; P, Q, R; A, B, C and D, E) is the reference to a point, the movement of which is sought to be controlled. This point is generally the tip of the cutting tool. Often the tool point may not move in some directions, e.g., the quill of the spindle of a vertical milling machine moves in the Z direction but not in the X and Y directions. In such problems, the work surface is usually motioned in a direction opposite to the one projected for the tool, e.g., the table of the milling machine holding the workpiece may be moved in – X and – Y directions. Such movements of machine elements, – X or –Y are signified as + X or + Y respectively. Prepared letters can thus be used for all the afore-mentioned motions to show the corresponding reversed directions for moving work surfaces.
# Principle Parts of a Shaper Machine | Shaper Machine Mechanism
PRINCIPLE PARTS OF SHAPER
The different parts of a shaper are given and described below.
Base
The base is a heavy and robust in construction which is made of cast iron by casting process. It is the only part to support all other parts because all parts are mounted on the top of this base. So, it should be made to absorb vibrations due to load and cutting forces while machining.
Column
The column has a box type structure which is made of cast iron. The inside surface is made as hollow to reduce the total weight of the shaper. It is mounted on the base. The ram driving (Quick return) mechanism is housed. The two guide ways are supplied on the top. The ram reciprocates on the supplied guide ways. Similarly, there are two guide ways at the front vertical face of the column to move the cross rail along these guide ways.
Cross rail
It is also a heavy cast iron construction. It glides on the front vertical ways of the column with two mechanisms. One is aimed at elevating the table and the other one is for cross travel of the table. A saddle slides over two guide ways already provided in the front face of the cross slide. The crosswise movement of the table is obtained by cross feed screw and the vertical movement of the cross rail is obtained by an elevating screw.
It is attached on the cross rail which holds the table in position on its top without any shake.
Table
It is also a box type rectangular hollow cast iron block. This table slides along the horizontal guide ways of the cross rail. The work is held in the table. The table has machined surfaces on the top and sides of T-slots for clamping work. It can be moved vertically by the elevating screw. An adjustable table supports the front face of the table.
Ram
Ram of cast iron has cross ribs for rigidity. Generally, it is a reciprocating type which slides over the guide ways on the top of the column. It is linked to driving mechanism of any one and also it carries the tool head at the front end. | 1,308 | 6,253 | {"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-2020-34 | latest | en | 0.916094 |
https://www.slideserve.com/Angelica/comparison-of-an-analytical-expression-of-the | 1,516,182,306,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886860.29/warc/CC-MAIN-20180117082758-20180117102758-00023.warc.gz | 975,890,207 | 16,667 | Comparison of an Analytical Expression of the Value of an American Put with standard Methods
1 / 27
# Comparison of an Analytical Expression of the Value of an American Put with standard Methods - PowerPoint PPT Presentation
Comparison of an Analytical Expression of the Value of an American Put with standard Methods. -Vicky Sharma. Background(1). Partial Differential Equation (PDE) A differential equation of a function of 2 or more equations comprising of partial derivatives of the function .
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about 'Comparison of an Analytical Expression of the Value of an American Put with standard Methods' - Angelica
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
### Comparison of an Analytical Expression of theValue of an American Put with standard Methods
-Vicky Sharma
Background(1)
• Partial Differential Equation (PDE)
• A differential equation of a function of 2 or more equations comprising of partial derivatives of the function.
• Boundary Value Problem
• Boundary value problem is a differential equation together with a set of
additional restraints. A solution to a boundary value problem is a solution tothe differential equation which also satisfies the boundary conditions.
Background(2)
• Black-Scholes PDE for the price of the option
• The value of the option is known at expiry. This is known as the terminal condition.
• European Options are only exercised at maturity.
• Terminal condition becomes the boundary condition. The PDE can be solved for a unique solution – Black-Scholes Formula
• American Options can be exercised at any time before maturity
• There is no boundary for each time instant before maturity – there is no analytical solution.
American Options
• Without dividends, it is in-optimal to exercise an american call before maturity.
• Same as european call, problem can be solved with a combination of binomial tree and black-scholes formula.
• American Puts are different
• There exists a critical price Sf(t) at time t below which it is optimal to exercise the put.
• This price is called the optimal exercise price at time t
• The optimal exercise price is the boundary condition for the PDE for an american put
• The optimal exercise boundary is a part of the solution
• The problem becomes a free boundary problem
Optimal Exercise Boundary for pricing a put
• The optimal exercise boundary defines the value of the put option when exercised.
• If the optimal exercise boundary is known,, monte-carlo simulations are used to compute the present discounted values of the maximum value of the put when it crosses the boundary.
• An average over multiple simulations (assuming log-normal model) gives the value of the put option.
American Put
• The problem of finding the analytical approximation to the american put becomes the problem of finding optimal exercise boundary.
• There have been several attempts in the past.
• Most of the solutions were complex and inaccurate for larger maturities.
• A recently proposed method has used laplace transformation with pseudo steady state approximation to solve for the optimal boundary and price of american put.
• We will look at this approach and compare it’s performance to the binomial tree model and the real-world stock prices.
Analytical Approximation (1)
• The following transformations transform the black-scholes PDE with a terminal condition to a PDE with an initial condition.
• X = Strike, τ= T – t.
• Sf(t) = optimal exercise price.
• The Next transformation is on V. V is replaced by U, where
• U = V + S – 1 when S <= 1
• U = V, when S > 1
Analytical Approximation (2)
• The PDE is now changed.
• There are additional constraints on the continuity and differentiability of U at S = 1.
Analytical Approximation (4)
• We assume that the optimal exercise boundary changes slower than the diffusion of the put value.
• Hence, the stock price can be taken as the constant Sf(t) at the boundary.
• This helps in taking the laplace transform of S around the boundary.
• L(S) = Sf/p (p = laplace transform variable)
Analytical Approximation (3)
• Using Laplace Transformation
• The author used laplace transformation to convert the PDE to a linear differential equation.
• A linear transformation of variable t (time from start) to time to expiration translates terminal condition to initial condition.
Analytical Approximation (4)
The solution for laplace transform
Is calculated easily
The boundary conditions are used to solve for the unknown variables
Analytical Approximation (5)
• The laplace transform of the optimal boundary can be computed.
• The problem is to compute the inverse transform.
• The laplace transform of Sf has poles at at zero and the negative half plane.
• Hence, the inverse laplace transform must be computed over a line in the tight-half plane.
Analytical Approximation (6)
A closed region is used to compute a known
Line integral over the poles.
The integrals over C2 and C6 vanish. Integrals
Over C3 and C5 are symmetric and add up to an
Imaginary number.
Analytical Approximation(7)
The optimal exercise boundary is then calculated by the above expression
Analytical Approximation(8)
The value of american put is
Approximated by the above
expression
Comparison with Binomial Model – At the money
Interest rate and
Volatility remain
Constant and the
Stock pays
No dividends
Stock Price = \$100, risk-free rate = 10%, σ = 0.30 and expiry = 1 year,
The analytical approximation follows the binomial tree model closely.
Black Scholes formula underestimates by a large margin as maturity increases
Comparison with Market Prices
• We used Yahoo Stock and put Prices for our comparison.
• In order to calibrate the model (deduce volatility), 3 years of daily closing stock prices for yahoo were used.
• Black Scholes assumption of log-normal stock prices used to estimate volatility.
• LIBOR Rates were used for risk free rate
• At the time, it was 4.63%
• The duration of stock prices from November 26, 2004 through November 26, 2007.
• The latest stock price was \$26.13
• Price of a put contract (for 100 shares) have been plotted
• Market price is the arithmetic average of the bid and offer prices
Maturity = 1 Month
In the Money
Out of Money
The analytical approximation compares fairly well with the market and binomial
tree prices.
Maturity = 5 Months
At the Money
The approximation compares fairly well with the binomial tree prices. The greatest
error with the market prices is seen around at-the-money mark
Maturity = 7 Months
At the Money
The increase in maturity does not affect the comparison with binomial tree model.
The numerical computation results in some under-estimation of the analytical value.
Hence, the analytical approximation is more accurate if more time is provided
computation
Maturity = 13 Months
At the Money
Out of Money
Option.
Strike = \$20
The analytical approximation underestimates out of money options quite significantly.
In all cases, the approximation performs as good as the binomial model.
In the Money
Option.
Strike = \$30
The analytical approximation’s accuracy improves as the option gets in the money.
It may also be due to the fact that computation now has larger numbers to process
Heavily In the Money Option.
Strike = \$40
For seriously in the money options, the difference between market prices and
Analytical approximations is small. This is mainly due to the fact that the value is
Dominated by the intrinsic value of the option.
American Put – Optimal Exercise Boundary(1)
• The computation of optimal exercise boundary results in some interesting results:-
• The value of the optimal exercise price at expiry is the strike price
• However, there seems to be a non-zero limit of the optimal exercise price even for large maturities
• We show the optimal exercise boundaries for a maturity of 1 month (0.0834 years) and 3 different strikes.
American Put – Optimal Exercise Boundary(2)
The optimal
Exercise price
Reaches
Strike at
maturity
Optimal boundary seems to approach a
Common minimum value as maturity increases
The optimal exercise boundary rises sharply towards the strike price at maturity.
Hence, the behavior is non-linear. As maturity increases, the optimal exercise
Boundary at t = 0 reaches the same limit for all strikes.
The new approximation formula provides a single expression for the value of american put.
No iterations are needed.
Several concepts, like the perpetual exercise price and value are easily explained.
Compares well with established models
Computation can be slower than the numerical methods.
No significant improvement over binomial tree model.
Seriously under-estimates out-of-money options
Conclusion(s) | 1,980 | 9,258 | {"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} | 3.5 | 4 | CC-MAIN-2018-05 | latest | en | 0.882932 |
http://fersch.de/beispiele?nr=Pythagoras&nrform=GeorechtDreiPythagorasUc&datei=1001PythagorasGeorechtDreiPythagorasUc | 1,501,080,166,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426169.17/warc/CC-MAIN-20170726142211-20170726162211-00061.warc.gz | 113,472,070 | 4,115 | Beispiel Nr: 01
$\text{Gegeben:}\\ \text{Kathete} \qquad a \qquad [m] \\ \text{Kathete} \qquad b \qquad [m] \\ \\ \text{Gesucht:} \\\text{Hypotenuse} \qquad c \qquad [m] \\ \\ c =\sqrt{a^{2} + b^{2} }\\ \textbf{Gegeben:} \\ a=3m \qquad b=4m \qquad \\ \\ \textbf{Rechnung:} \\ c =\sqrt{a^{2} + b^{2} } \\ a=3m\\ b=4m\\ c =\sqrt{(3m)^{2} + (4m)^{2} }\\\\c=5m \\ \small \begin{array}{|l|} \hline a=\\ \hline 3 m \\ \hline 30 dm \\ \hline 300 cm \\ \hline 3\cdot 10^{3} mm \\ \hline 3\cdot 10^{6} \mu m \\ \hline 0,003 km \\ \hline \end{array}\\ \small \begin{array}{|l|} \hline b=\\ \hline 4 m \\ \hline 40 dm \\ \hline 400 cm \\ \hline 4\cdot 10^{3} mm \\ \hline 4\cdot 10^{6} \mu m \\ \hline 0,004 km \\ \hline \end{array}\\ \small \begin{array}{|l|} \hline c=\\ \hline 5 m \\ \hline 50 dm \\ \hline 500 cm \\ \hline 5\cdot 10^{3} mm \\ \hline 5\cdot 10^{6} \mu m \\ \hline 0,005 km \\ \hline \end{array}\\$ | 440 | 906 | {"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} | 4.0625 | 4 | CC-MAIN-2017-30 | latest | en | 0.105417 |
gelvos.com | 1,726,753,807,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00334.warc.gz | 243,017,488 | 23,070 | # Compound Interest – How to use the Effect Correctly
Compound interest has a very big impact on capital accumulation. But what is it exactly and how is it different from simple interest? How do I make the best use of compound interest and how does it help me to increase my assets?
## Basics
The term interest was first mentioned by the Sumerians, 2,400 years before Christ. The term compound interest also originated there. Since there was no banking system yet, interest was charged when money was lent. Over the centuries, the churches have banned, restricted and temporarily allowed the use of interest again.
The idea behind compound interest is this: you invest money and let it work for you. All earnings, profits, dividends, etc. are not paid out, but reinvested. The money thus works for you without you having to do anything for it.
Many already know the term, but cannot imagine how powerful this effect is. To illustrate the whole thing, here are some examples. In addition, I provide you with a compound interest calculator. Here you can simulate the development of your capital with different terms, savings rates and interest rates.
## Example: simple interest
For example, if you invest \$10,000 with an interest rate of 4%, then in one year you will have \$10,000 + \$400 interest = \$10,400.
You then withdraw this \$400€ and spend it and on clothes, something else or put it in your piggy bank. You are left with the initial \$10,000. If you invest this \$10.000 again to 4%, you will have again \$400 over to spend, etc..
In 10 years from now you will still have the initial invested \$10,000. Over the course of 10 years, you will also have received a total of 10 x \$400 = \$4,000 in interest. But what happens if you do not withdraw the \$400 but reinvest it?
## Example: compound interest
In the first year it works the same way as in the first example. If you invest \$10,000 with an interest rate of 4%, you will receive \$10,000 + \$400 interest = \$10,400 in one year.
But now you do not withdraw any more money, but leave the received interest on the account. After 2 years you will have \$10,400 + 416€ interest = \$10,816. We now also consider a period of 10 years again and you then have \$10,000 x 1.04^10 = \$14,802.
Now compare this with the other example.
simple interest: \$10,000 + 10 x \$400 = \$14,000
compound interest: \$10,000 x 1.04^10 = \$14,802
The difference is in this example only \$802. This does not seem to be that big. I meant earlier that this is a very powerful effect. How can you benefit even more from compound interest?
## Illustration of compound interest
There are two factors that influence the effect. One is the length of the investment and the other is the amount of the interest rate. To better illustrate this, I have created two graphs. Here the effect of compound interest can be seen very well.
### 1. Influence of the duration
In the previous examples, the effect was not yet really visible. But by extending the time period, it already looks quite different. Again, we consider two alternatives.
As in the example above, we again invest \$10,000 with an interest rate of 4%. In the diagram, both alternatives are shown with a time period of up to 50 years. With an investment term of 10 years, the difference is only \$802, with 25 years it is \$6,658 and with 50 years it is even \$41,067. We can therefore conclude that a longer term has a positive effect.
One of the richest person in our time has taken advantage of exactly this factor. Moreover, he has of course invested his money very well. It is Warren Buffet, who had a fortune of about 79 billion USD in 2020 (according to Forbes). You don’t see how big the number actually is until you write it out, and it’s something like USD 79,000,000,000. That’s a lot of zeros.
Warren Buffett also started small. At the age of 14, he had only USD 5,000, and by the time he was 21, he had USD 20,000. He had made his first million by the age of 30. It took another 25 years until Warren Buffett had his first billion on his account and then another 24 years to his current fortune at the age of 89. It took him 75 years to become so rich.
### 2. Influence of the interest rate
In the diagram above, I want to illustrate the influence of the interest rate of the investment in addition to the influence of the time period. Here the difference is even more significant than in the previous example.
In the graph above you can see this time four different variants: 1%, 4%, 7% and 10% interest. As a duration I have chosen this time 45 years with an initial investment of \$10,000. To clarify this, I have created the following table:
If we look at the chart and the table, we notice two things. Firstly, the effect is higher the higher the interest rate and secondly, it is noticeable that the difference increases as the interest rate increases. The difference between 4% and 1% is only \$42,764, whereas between 10% and 7% interest it is \$518,818. So it is not enough to invest the money only over a long period of time, but also at an high interest rate or yield.
So Warren Buffett could never have built up such a fortune if he had only achieved a yield of 1-2% per year. He only succeeded in doing this because he was able to achieve a yield of over 20% per year for years, first with Buffett Partnership and then later with Berkshire Hathaway. That’s how he became so rich and now he is able to donate a lot of money to charity.
## Summary
So what can we learn from this article? I would like to summarize it and motivate you to start investing as early as possible. The compound interest effect is very powerful and can help us earn a lot of money. To do this, you need to do the following:
1. Invest your money for the long term to benefit from the effect.
2. It is best to start at an early age. If you invest \$10,000 with 7% return rate at the age of 20, you will have \$210,024 at the age of 65. If you invest \$10,000 for your child at birth, your child will have \$812,729 at the age of 65.
3. The return on investment is crucial for the growth of your assets.
4. Avoid investments with high fees.
If you would like to calculate this yourself, you can use our compound interest calculator.
## Book recommendation
The Essays of Warren Buffett – Warren Buffett / Lawrence Cunningham
Warren Buffett’s essays have enjoyed cult status for more than two decades. Compiled and edited by one of the most renowned experts on value investing, Lawrence A. Cunningham, the letters from Warren Buffett to his shareholders summarized here are an unenlightened insight into the investment philosophy of the most successful investor of all time.
Much has been written about Buffett, but what does he himself have to say? The essays come from Buffett’s own pen. With sober wisdom, he talks about his investment decisions, how he selects his teams, and how he values companies.
The fourth, completely revised edition includes new, previously unpublished essays, including those on Berkshire Hathaway’s 50th anniversary (2015) and by Charlie Munger.
GELVOS was created with the idea of how to implement “earn money without stress” and for this I would like to give you our experience again. All articles on this page are divided into these sections: make. save. invest. live. If you don’t want to miss anything, subscribe to our free newsletter and get a big step closer to your goals! | 1,720 | 7,438 | {"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-2024-38 | latest | en | 0.95922 |
https://www.datasciencecentral.com/profiles/blogs/fraud-analysis-using-speech-analytics-output-with-monte-carlo | 1,628,073,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154798.45/warc/CC-MAIN-20210804080449-20210804110449-00354.warc.gz | 729,725,012 | 20,722 | .
# Fraud analysis using speech analytics and Monte Carlo
As per the largest market research firm MarketsandMarkets the speech analytics industry will grow to USD 1.60 billion by 2020 at a Compound Annual Growth Rate (CAGR) of 22% from 2015 to 2020. Today the omnichannel world consists of voice, email, chat, social channels, and surveys, and each channel has its own importance.
Therefore, it becomes inevitable for any customer centric organization to ignore the information that can be glean out of these customer interactions.
This article talks about some cutting edge usage of Speech Analytics output coupled up with a computerized mathematical technique that allows organizations to account for risk which is called as Monte Carlo simulation. For the purposes of this article I will be focusing on the healthcare industry which has reported (The Economist May,31,2014) a staggering \$275 billion swindle.
To use this technique (Monte Carlo simulation) in conjunction with the Speech Analytics output we will use the “Stochastic Model” for the simulation which involves probability or randomness.
Application of Monte Carlo Simulation to identify probability of fraud by Service Providers
The expected output of this simulation is to identify the likelihood of a fraudulent activity based on the key customer interaction that indicates potential “Fraud Outcomes”.
Identifying the fraudulent interactions
As we know that speech analytics allows its users to query the media files to identify the emerging topics. The above scenarios can be created within any speech analytics application. The user can also utilize the provider related metadata (additional information about a particular customer interaction) to understand the interaction distribution of the above scenarios by a particular provider.
How it works? Creating the Model
Scenario 1 + Scenario 2 + Scenario 3 + Scenario 4 + Scenario 5 = Fraud?
Let’s say we have over a million customer interactions with a combination of 5 scenarios (refer to the scenario grid), and we arbitrarily identify the interactions to decide if that interaction has a high likelihood of being a fraudulent scenario. No two scenarios will have precisely the same number of fraud manifestations. However, if we have an idea of the range of occurrences for each situation, then we can create a Monte Carlo simulation to better understand the probability of a fraud scenario.
The image below shows the simulation that I created in Excel that illustrates how the model was created using 1000 fraud simulations where each simulation is equally likely to happen.
The above simulation was done for multiple providers based on their fraud scenario % (it will be advisable to pick the outliers by keeping the fraud indicator scenario % in mind). Once the simulation for the top provider were created. I was able to showcase the providers who are prone to get into fraud related discussions with their customers.
Note: Above frequency graphs are based on the Monte Carlo Simulation that gives a probabilistic perspective for the Fraud Indicator conversations that might lead to an actual fraud incident. The above outputs are based on the 1000 simulations where each simulation is equally likely to happen.
By looking at the above results one can easily isolate those providers or scenarios that can results in a potential fraud incident before it happens and mitigate a potential risk to the consumer, the brand and the overall reputation of any healthcare service provider.
Views: 6733
Comment
Join Data Science Central
Comment by duncan on November 24, 2017 at 3:38am
Thanks for replying Sunil. duncanwil at gmail dot com
Comment by Sunil Kappal on November 24, 2017 at 2:38am
Thanks Duncan for liking and finding this post interesting, I can certainly share the spreadsheet. Please share the email id. As far as the ave, min, max and sd statistics are concerned these are the outputs that I created using the speech analytics application. I had to define each fraud incident using a query ( a logic which searches for an interaction type) that I ran over the customer conversations which gave me a % of occurrence and post clubbing all the interactions (various fraud scenarios) as a group I got an overall % based on which the above statistics were generated to simulate 1000 runs.
Comment by duncan on November 23, 2017 at 3:05pm
Very interesting post Sunil.
Can you share your spreadsheet or provide examples of how to arrive at the % for Ave, Min, Max, SD,and then Final and Class ... percentages of what?
Many thanks in advance and best wishes
Duncan | 910 | 4,621 | {"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} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.894377 |
http://www.mytech-info.com/2013/09/how-to-find-rating-of-transformer-in-kva.html | 1,547,768,633,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659417.14/warc/CC-MAIN-20190117224929-20190118010929-00304.warc.gz | 347,840,515 | 90,779 | # How to find the Rating of Transformer in kVA ?
In this tutorial we will see to find the Transformer Rating in KVA?
We know that, Transformers are always rated in kVA.
We can see below the two simple formulas are to find the rating of Single phase and Three phase Transformers.
P = V x I.
##### Rating of a single phase transformer in kVA
KVA= (V x I) / 1000
P = √3 V x I
##### Rating of a Three phase transformer in kVA
KVA = (√3 V x I) /1000
For example the above fig. shows the rating of Transformer is 100kVA.
Primary Voltages or High Voltages (H.V) is 11000 V = 11kV.
Primary Current on High Voltage side is 5.25 Amperes.
And
Secondary voltages or Low Voltages (L.V) is 415 Volts
Secondary Current (Current on Low voltages side) is 139.1 Amp.
Now let us see how to calculate the rating of transformer according to
P=V x I (Primary voltage x primary current)
P = 11000V x 5.25A = 57,750 VA = 57.75kVA
Or P = V x I (Secondary voltages x Secondary Current)
P= 415V x 139.1A = 57,726 VA = 57.72kVA
We noticed that the rating of Transformer (on Nameplate) is 100kVA but according to calculation...it comes about 57 kVA
The difference comes due to ignorance of that we used single phase formula instead of three phase formula.
Now we will use this formula
##### P = √3 x V x I
P=√3 Vx I (Primary voltage x primary current)
P =√3 x 11000V x 5.25A = 1.732 x 11000V x 5.25A = 100,025VA (or)
P=100KVA
Or
P = √3 x V x I (Secondary voltages x Secondary Current)
P= √3 x 415V x 139.1A = 1.732 x 415V x 139.1A= 99,985 VA (or)
P= 99.98kVA
Consider the (next) following example.
Voltage (Line to line) = 208 V.
Current (Line Current) = 139 A
Now rating of the three phase transformer
##### P = √3 x V x I
P = √3 x 208 x 139A = 1.732 x 208 x 139
P = 50077 VA = 50kVA | 577 | 1,766 | {"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} | 4.21875 | 4 | CC-MAIN-2019-04 | longest | en | 0.858264 |
https://math.answers.com/Q/How_many_hours_is_in_7500_minutes | 1,701,251,327,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00625.warc.gz | 489,198,554 | 44,275 | 0
# How many hours is in 7500 minutes?
Updated: 9/20/2023
Wiki User
10y ago
125 hours.
Wiki User
10y ago
Earn +20 pts
Q: How many hours is in 7500 minutes?
Submit
Still have questions?
Related questions
### How many hours in 7500 minutes?
60 minutes = 1 hour so 7500 minutes = 7500/60 = 125 hours. Simple!
750 minutes
35 minutes
### How many hours is 102 minutes?
How many hours is 101 minutes
### How many hours and minutes are in 135 minutes?
ANSWER:2 hours, 15 minutes. 180 minutes (Which is 3 hours)
### How many hours and minutes is 8.5 hours?
8.5 hours is 8 hours and 30 minutes, or 510 minutes.
### How many hours and minutes are in 667 minutes?
11 hours, 7 minutes
### How many hours and minutes are in 425 minutes?
it has 7 hours and 5 minutes.
### How many hours and minutes are in 1327 minutes?
There are 22 hours and 7 minutes.
### How many hours and minutes is 1495 minutes?
24 hours 55 minutes
### How many hours and minutes are in 178 minutes?
There are 2 hours 58 minutes.
### How many hours and minutes in 425 minutes?
it has 7 hours and 5 minutes. | 311 | 1,095 | {"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} | 3.625 | 4 | CC-MAIN-2023-50 | latest | en | 0.944058 |
https://carreersupport.com/control-theory-interview-questions/ | 1,718,964,225,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00813.warc.gz | 126,714,994 | 20,870 | # The Top 10 Control Theory Interview Questions to Prepare For
To start the process of hiring the best controls engineer, you need to know exactly what questions to ask during the interview. This article goes over the most important controls engineer interview questions that are used to find out how knowledgeable and good at solving problems candidates are. If you know these questions, you’ll know what to expect when the conversation gets technical, whether you’re the hiring manager or the person being interviewed.
Control theory is an interdisciplinary field that combines engineering, mathematics, and computer science principles to design systems that regulate themselves. As automation and process optimization continue to transform industries like robotics, aerospace, manufacturing, and energy management, control theory experts are in high demand.
Acing an interview in this complex domain requires strong theoretical knowledge and the ability to apply concepts to real-world problems. In this article, we provide insights into 10 common control theory interview questions to help you demonstrate your expertise when it matters most.
## 1. How would you stabilize a system with inherent nonlinear dynamics?
Handling nonlinear systems that can exhibit chaotic behavior requires advanced strategies beyond conventional linear control techniques. Discuss your approach to managing nonlinear dynamics, such as:
• Linearizing around operating points and designing controllers like PID or LQR
• Leveraging Lyapunov stability criteria to ensure robustness
• Applying sliding mode control or backstepping for large parameter variations
• Using model predictive control to explicitly account for nonlinearities
Illustrate with examples of how you’ve stabilized challenging nonlinear systems previously
## 2. Describe how you’ve applied feedback control in a real-world application.
Feedback control maintains system stability and meets performance objectives through corrective actions based on sensor measurements Outline a specific instance where you implemented feedback control, covering
• The context and challenges involved
• The feedback loop mechanism and type of controller used
• The results achieved and any iterative improvements made
## 3. What techniques do you use to model uncertain systems for robust control design?
Designing robust controllers hinges on managing uncertain, dynamic systems. Discuss modeling methods you employ, such as:
• H-infinity optimization to guarantee performance despite uncertainties
• Stochastic control using probabilistic models of noise and disturbances
• Leveraging Lyapunov stability criteria to ensure robustness
Provide examples of applying these techniques successfully on past projects.
## 4. In what ways have you used state-space representation over transfer functions?
State-space models offer comprehensive analysis of system dynamics and accommodate MIMO systems and nonlinearities. Discuss specific applications where you chose state-space representation over transfer functions and why this approach was advantageous.
## 5. Outline your systematic approach to tuning PID controllers.
Emphasize model-based analytical techniques you use, such as:
• Ziegler-Nichols or Cohen-Coon rules to determine initial parameters
• Simulation and sensitivity analysis to refine tuning without trial-and-error
• Industry best practices and standards you adhere to
Provide an example where your approach resulted in optimal PID tuning.
## 6. How do you determine controllability and observability of a system?
Explain how you would construct and analyze the controllability and observability matrices to verify these properties. Also discuss compensating for any deficiencies through redesign or implementing observers/controllers.
## 7. What are the implications of actuator saturation in control performance?
Cover concepts like:
• Integrator windup and its destabilizing effects
• Anti-windup techniques to maintain stability and performance
• Designing optimization-based controllers that can handle constraints
Share examples of addressing actuator saturation challenges successfully.
## 8. Can Lyapunov methods be applied to discrete-time systems?
Acknowledge adaptations like the discrete Lyapunov equation and difference inequality condition. Provide examples of applying Lyapunov stability analysis to digital control systems. Discuss any limitations encountered.
## 9. How would you implement model predictive control for fast-sampled systems?
Highlight strategies like:
• Simplifying control models to reduce computational load
• Shortening prediction horizons for responsive dynamics
• Leveraging parallel processing hardware to accelerate calculations
Share examples of successfully deploying MPC under timing constraints.
## 10. When would you apply H-infinity control over classical techniques?
Discuss scenarios with significant uncertainties or stringent robustness requirements where H-infinity methods are preferred despite increased complexity. Demonstrate your knowledge of both classical and modern control theory.
Preparing thoughtful responses to these common questions will showcase your conceptual understanding and practical abilities in tackling control challenges. Be sure to draw from concrete examples and industry best practices throughout. With diligent preparation, you’ll be ready to impress interviewers and land your next exciting role in control theory.
## Other Key Control Theory Interview Topics to Review
Beyond the questions above, there are a few other topics that commonly arise during control engineering interviews:
• Stability analysis – Be able to articulate methods like Routh-Hurwitz, Bode plots, Nyquist diagrams, and root locus techniques. Understand stability margins and performance trade-offs.
• Lead-lag compensation – Discuss uses for improving transient response, stability, and reducing oscillations. Know how to analyze effects on gain/phase margins.
• State estimation – Explain Kalman filters, observers, and sensor fusion. Provide examples of implementing estimators in real applications.
• Discrete control systems – Understand modeling, analysis, and design of digital controllers. Discuss sampling, discretization, quantization, and implementation trade-offs.
• Frequency response – Explain Bode plots, resonance, filtering, and compensation techniques. Provide examples of applications for vibration control, noise reduction, etc.
• Optimal control – Be familiar with techniques like LQR, pole placement, and model predictive control. Discuss objective functions, constraints, and trading off performance metrics.
Preparing explanations and examples for these supplementary topics will round out your control systems knowledge. Use industry resources and published literature to strengthen conceptual fluency.
## Helpful Strategies for Acing the Interview
• Ask clarifying questions – Don’t hesitate to ask for clarification or additional context when faced with an unclear question. This shows engagement.
• Think aloud – Verbalize your thought process when working through a question. This gives insights into your analytical approach.
• Use visual aids – Consider sketching diagrams, graphs, or equations when explaining complex concepts. Visuals improve communication.
• Admit knowledge gaps – If there are areas you’re unfamiliar with, be honest about the limits of your expertise. Then discuss your willingness to learn.
• Make connections – Relate questions to your previous work or academic experiences when possible. This demonstrates depth.
• Exhibit passion – Share your enthusiasm for control engineering and eagerness to take on new challenges. Mindset matters.
With diligent preparation and these helpful tips, you’ll be poised to impress at your next control theory interview. We wish you the very best as you embark on this exciting next step in your engineering career!
### Q1 What methods do you use to optimize system performance and efficiency? (Optimization)
When optimizing system performance and efficiency, I use a combination of the following methods:
• Root Cause Analysis: I start by doing a full analysis of existing systems to find any problems or inefficiencies. This often involves using tools like Six Sigma’s DMAIC framework.
• Simulation and Modeling: I use software tools to simulate the control system and see how the changes will affect it before I make them.
• Continuous Monitoring: I set up systems to keep an eye on performance metrics all the time, using KPIs to find places where things could be better.
• Feedback Loops: Using feedback systems to change control parameters right away for adaptive optimization
• Energy Consumption Analysis: Looking at how energy is used and putting in place ways to save it, like motors with variable frequency drives (VFDs).
• Preventive Maintenance: Making sure there is a strong preventive maintenance plan in place to keep things running smoothly and reduce downtime
## Controls Engineer Position Insights
The role of a controls engineer is essential in industries where automation and control systems are pivotal. These professionals design, program, and manage systems that control machinery and processes, ensuring efficiency, reliability, and safety. Their work spans various sectors, including manufacturing, automotive, aerospace, and utilities, to name a few.
Good controls engineers need to know a lot of different things. They need to know a lot about control systems theory, have a lot of experience with programmable logic controllers (PLCs), be good at using software and simulation tools, and be able to manage projects and solve hard problems quickly. They are also responsible for adhering to safety standards and maintaining compliance in their designs. The controls engineer’s job is both challenging and constantly changing because they have to keep up with new technologies and industry trends.
Understanding the nuances of this role is crucial when formulating or responding to interview questions. With the right questions, you can find out about a candidate’s technical skills, experience, problem-solving style, and ability to work with others in a high-stakes setting.
### FAQ
What is the basic theory of control system?
Its name comes from the information path in the system: process inputs (e.g., voltage applied to an electric motor) have an effect on the process outputs (e.g., speed or torque of the motor), which is measured with sensors and processed by the controller; the result (the control signal) is “fed back” as input to the …
What is an open loop in a control system?
The term ‘open-loop’ (or essentially ‘no loop’) refers to the lack of sensor feedback to control the operation of the machine. An example of an open-loop feedback would be a simple light switch that, upon activation, remains in the ‘On’ or ‘Off’ condition until manually changed.
How do you ace a controller interview?
An effective controller is both inquisitive, persistent, and brave enough to get to the bottom of issues. Their answer should emphasize more the parts of how they worked to resolve the conflict in a professional manner, with a clear path toward a more amicable relationship going forward.
What is a stable system in a control system?
A system is said to be stable, if its output is under control. Otherwise, it is said to be unstable. A stable system produces a bounded output for a given bounded input. The following figure shows the response of a stable system. This is the response of first order control system for unit step input.
What are control system interview questions?
Control system interview questions test candidates on their technical knowledge to ensure they can do the job effectively. Here are some common control system questions with sample answers to help you craft your own responses: 1. What is a system? This is a basic question to test your general knowledge of systems.
Why do interviewers ask a control system function question?
Interviewers may ask this question to determine if you understand the basic functions of a well-designed control system. Explaining how a control system functions may help you explain your qualifications for various positions where you may design, install and maintain systems that operate according to the desired goals.
How do I prepare for a control system interview?
Research the company’s dress code. Dressing well is a good way to impress your interviewer. You can find out the most appropriate attire by researching the company’s dress code. Discover 18 control system interview questions with our guide, which covers some of the most typical questions and offers sample answers to help you prepare.
What is a control system theory test?
This is a highly technical question that tests your memory and understanding of the theories you studied at university. Example answer: ‘In control system theory, this is a mathematical test that represents a necessary condition for the stability of a linear control system. | 2,349 | 13,078 | {"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-2024-26 | latest | en | 0.885265 |
https://docbrown.info/page04/4_73calcs01ram.htm | 1,620,946,156,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992514.37/warc/CC-MAIN-20210513204127-20210513234127-00308.warc.gz | 229,398,053 | 17,534 | GCSE & A level Chemistry Calculations: Defining & calculating relative atomic mass
DEFINING & CALCULATING the RELATIVE ATOMIC MASS Ar of an element
Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level Online Chemical Calculations
1. Defining, explaining and calculating relative atomic mass RAM or Ar, also mention of relative isotopic mass
Sub-index for page
(c) is for advanced level chemistry students only and includes how to work out isotopic composition given the relative atomic mass.
(d) Table of relative atomic masses for elements 1 to 92 (2 decimal places)
Spotted any careless error on my part?
or request a type of GCSE calculation not covered?
Self-assessment Quizzes on relative atomic mass:
or
Revision notes on how to define relative atomic mass and how to calculate relative atomic mass from the percentage abundance of isotopes, help in revising for A level AQA, Edexcel, OCR 21st century, Gateway science GCSE 9-1 chemistry examinations
1. Explaining and how to calculate the relative atomic mass RAM or Ar of an element
(a) Introduction - defining relative atomic mass - carbon-12 scale
• Every atom has its own unique relative atomic mass (RAM) based on a standard comparison or relative scale e.g. it has been based on hydrogen H = 1 amu and oxygen O = 16 amu in the past (amu = relative atomic mass unit).
• The relative atomic mass of an element takes into account the different masses of the isotopes of that element and the abundance of the isotopes in the naturally occurring element (meaning the percentage of each isotope present).
• Relative atomic mass is defined and explained below, and examples of how to calculate it from data.
• The relative atomic mass scale is now based on an isotope of carbon, namely, carbon-12, nuclide symbol , which is given the arbitrary value of 12.0000 amu by international agreement.
• The unit 'amu' is now being replaced by a lower case u, where u is the symbol for the unified atomic mass unit.
• Therefore one atom of carbon, isotopic mass 12, equals 12 u, or,
• 1 u = 1/12th the mass of one atom of the carbon-12 isotope.
• Note that for the standard nuclide notation, , the top left number is the mass number (12) and the bottom left number is the atomic/proton number (6).
• Since the relative atomic mass of an element is now based on the carbon-12 isotope it can now be defined as ...
• ... relative atomic mass equals the average mass of all the atoms in an element compared to 1/12th the mass of a carbon-12 atom (carbon-12 isotope).
• Examples are shown in the Periodic Table diagram above.
• Note
• (i) Because of the presence of neutrons in the nucleus, the relative atomic mass is usually at least double the atomic/proton number because there are usually more neutrons than protons in the nucleus (mass proton = 1, neutron = 1). Just scan the periodic table above and examine the pairs of numbers.
• You should also notice that generally speaking the numerical difference between the atomic/proton number and the relative atomic mass tends to increase with increasing atomic number.
• This has consequences for nuclear stability.
• (ii) For many calculation purposes, relative atomic masses are usually quoted and used at this academic level (GCSE/IGCSE/O level) to zero or one decimal place eg.
• hydrogen H = 1.008 or ~1; calcium Ca = 40.08 or ~40.0; chlorine Cl = 35.45 ~35.5, copper Cu = 63.55 or ~63.5/64, silver Ag = 107.9 or ~108 etc.
• At Advanced level, values of relative atomic masses may be quoted to one or two decimal places.
• Many atomic masses are known to an accuracy of four decimal places, but for some elements, isotopic composition varies depending on the mineralogical source, so four decimal places isn't necessarily more accurate!
• Note that in the case of carbon, there are three isotopes carbon-12 12C the most abundant and small amounts of carbon-13 13C and carbon-14 14C. The average calculated mass of the atoms compared to carbon 12 is 12.01, but for most purposes at pre-university level, 12.0 is sufficient accuracy.
• In using the symbol Ar for RAM, you should bear in mind that the letter A on its own usually means the mass number of a particular isotope and amu is the acronym shorthand for atomic mass units.
• However there are complications due to isotopes and so very accurate atomic masses are never whole integer numbers.
• Isotopes are atoms of the same element with different masses due to different numbers of neutrons.
• The very accurate relative atomic mass scale is based on a specific isotope of carbon, carbon-12, 12C = 12.0000 units exactly, for most purposes C = 12 is used for simplicity.
• For example hydrogen-1, hydrogen-2, and hydrogen-3, are the nuclide notation for the three isotopes of hydrogen, though the vast majority of hydrogen atoms have a mass of 1.
• When their accurate isotopic masses, and their % abundance are taken into account the average accurate relative mass for hydrogen = 1.008, but for most purposes H = 1 is good enough!
• See also GCSE/IGCSE/AS Atomic Structure Notes
• Therefore, a stricter definition of relative atomic mass (Ar) is that it equals the average mass of all the isotopic atoms present in the element compared to 1/12th the mass of a carbon-12 atom.
• AND, the relative isotopic mass of carbon-12 is assigned a numerical value of 12.0000.
• So, in calculating relative atomic mass you must take into account the different isotopic masses of the same elements, but also their % abundance in the element.
• Therefore you need to know the percentage (%) of each isotope of an element in order to accurately calculate the element's relative atomic mass.
• For approximate calculations of relative atomic mass you can just use the mass numbers of the isotopes, which are obviously all integers ('whole numbers'!) e.g. in the two calculations below.
• To the nearest whole number, isotopic mass = mass number for a specific isotope.
• If an element only has one isotope, relative atomic mass = relative mass of this isotope.
• A good example is fluorine.
• All fluorine atoms have a mass of 19 (19F), therefore its relative atomic mass is 19 and no calculation is needed.
Above is typical periodic table used in GCSE science-chemistry specifications
and I've 'usually' used these values in my exemplar calculations to cover most syllabuses
TOP OF PAGE and sub-index
(b) Examples of relative atomic mass calculations for GCSE 9-1/IGCSE/AS/A level chemistry students
How do I calculate relative atomic mass?
You can calculate relative atomic mass from isotopic abundances
• For accurate chemical calculations relative atomic mass must be used and not an individual mass number.
• Therefore relative atomic mass takes into account all the different 'stable' isotopes of an element which are naturally present.
• The relative atomic mass is the average mass and is quite easily calculated from the percentage composition (% abundance).
• The presence of isotopes accounts for why some relative atomic masses are not even close to a whole number.
• Some relative atomic masses are nearly whole numbers due to coincidence of % isotopes, others because one isotope might dominate the composition with only tiny amounts of lighter or heavier isotopes.
• Example 1.1 Calculating the relative atomic mass of bromine and
• bromine consists of two isotopes, 50% 79Br and 50% 81Br, calculate the Ar of bromine from the mass numbers (top left numbers).
• Think of the calculation in terms of '100 atoms'
• Ar = [ (50 x 79) + (50 x 81) ] /100 = 80
• So the relative atomic mass of bromine is 80 or RAM or Ar(Br) = 80
• Note the full working shown. Yes, ok, you can do it in your head BUT many students ignore the %'s and just average all the isotopic masses (mass numbers) given, in this case bromine-79 and bromine-81.
• The element bromine is the only case I know where averaging the isotopic masses actually works! so beware!
• -
• Example 1.2 Calculating the relative atomic mass of chlorine based on the and isotopes
• Chlorine consists of two isotopes, 75% chlorine-35 and 25% chlorine-37, so using these two mass numbers ...
• ... again think of the data based on 100 atoms, so 75 have a mass of 35 and 25 atoms have a mass of 37.
• The average mass = [ (75 x 35) + (25 x 37) ] / 100 = 35.5
• So the relative atomic mass of chlorine is 35.5 or RAM or Ar(Cl) = 35.5
• Note: 35Cl and 37Cl are the most common isotopes of chlorine, but, there are tiny percentages of other chlorine isotopes which are usually ignored at GCSE/IGCSE and Advanced GCE AS/A2 A level.
• -
• Example 1.3: Calculating the relative atomic mass of copper from its isotopic composition (isotope abundance)
• Naturally occurring copper consists of 69.2% copper-63 (63Cu) and 30.8% copper-65 (65Cu)
• Still think in terms of 100 atoms and don't be put off by decimal fractions, it still works out correctly because 69.2 + 30.8 = 100!
• average mass = relative atomic mass of copper = {(63 x 69.2) + (65 x 30.8)} / 100 = 63.6
• -
• Example 1.4: Silver atoms consist of 51.4% of the isotope 107Ag and 48.6% of the isotope 109Ag
• Calculate the relative atomic mass of silver.
• (51.4 x 107) + (48.6 x 109) 5499.8 + 5297.4 Ar(Ag) = -------------------------------------- = --------------------------- = 108.0 100 100
• The relative atomic mass of silver is 108.0 (to 1 decimal place)
• -
• Example 1.5: Europium atoms consist of 47.8% Eu-151 and 52.2% of Eu-153
• Calculate the relative atomic mass of europium.
• (47.8 x 151) + (52.2 x 153) 7217.8 + 7986.6 Ar(Eu) = -------------------------------------- = --------------------------- = 152.0 100 100
• The relative atomic mass of europium is 152.0 (to 1 decimal place)
• -
• Example 1.6: Atoms of the element silicon consist of 92.2% silicon-28, 4.7% silicon-29 and 3.1% of silicon-30.
• Calculate the relative atomic mass of silicon.
• (92.2 x 28) + (4.7 x 29) + (3.1 x 30) 2581.6 + 136.3 + 93.0 Ar(Si) = -------------------------------------------------- = -------------------------------- = 28.1 100 100
• The relative atomic mass of silicon is 28.1 (to 1 decimal place or 3 significant figures)
• -
• See below and mass Spectrometer and isotope analysis on the GCSE-Advanced A Level (basic) Atomic Structure Notes, with further relative atomic mass calculations.
(c) Examples for Advanced A Level Chemistry students only
How to calculate relative atomic mass with accurate relative isotopic masses
Using data from modern very accurate mass spectrometers
(a) Very accurate calculation of relative atomic mass (need to know and define what relative isotopic mass is)
Relative isotopic mass is defined as the accurate mass of a single isotope of an element compared to 1/12th the mass of a carbon-12 atom e.g. the accurate relative isotopic mass of the cobalt-5 is 58.9332
This definition of relative isotopic mass is a completely different from the definition of relative atomic mass, except both are based on the same international standard of atomic mass i.e. 1 unit (1 u) = 1/12th the mass of a carbon-12 isotope (12C).
If we were to redo the calculation of the relative atomic mass of chlorine (example 1.1 above), which is quite adequate for GCSE purposes (and maybe A level too), but more accurately at A level, we might do ....
chlorine is 75.77% 35Cl of isotopic mass 34.9689 and 24.23% 37Cl of isotopic mass 36.9658
so Ar(Cl) = [(75.77 x 34.9689) + (24.23 x 36.9658)] / 100
= 35.4527 (but 35.5 is usually ok in calculations pre-university!)
See also Mass Spectrometer and isotope analysis on the GCSE/A level Atomic Structure Notes, with further RAM calculations.
(b) Calculations of % composition of isotopes
It is possible to do the reverse of a relative atomic mass calculation if you know the Ar and which isotopes are present.
It involves a little bit of arithmetical algebra.
The Ar of boron is 10.81 and consists of only two isotopes, boron-10 and boron-11
The relative atomic mass of boron was obtained accurately in the past from chemical analysis of reacting masses but now can sort out all of the isotopes present and their relative abundance.
If you let X = % of boron 10, then 100-X is equal to % of boron-11
Therefore Ar(B) = (X x 10) + [(100-X) x 11)] / 100 = 10.81
so, 10X -11X +1100 =100 x 10.81
-X + 1100 = 1081, 1100 - 1081 = X (change sides change sign!)
therefore X = 19
so naturally occurring boron consists of 19% 10B and 81% 11B
(the data books actually quote 18.7 and 81.3, but we didn't use the very accurate relative isotopic masses mentioned above!)
TOP OF PAGE and sub-index
Revision notes on how to define relative atomic mass and how to calculate relative atomic mass from the percentage abundance of isotopes, help in revising for A level AQA, Edexcel, OCR 21st century, Gateway science GCSE 9-1 chemistry examinations
On other pages on and Relative Formula Mass
Self-assessment Quizzes on relative atomic mass
or multiple choice QUIZ
APPENDIX 1. A typical periodic table used in pre-university examinations
Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses
Revision notes on how to define relative atomic mass and how to calculate relative atomic mass from the percentage abundance of isotopes, help in revising for A level AQA, Edexcel, OCR 21st century, Gateway science GCSE 9-1 chemistry examinations
TOP OF PAGE and sub-index
(d) APPENDIX 2. Table of relative atomic masses for elements 1 to 92
Notes: (i) The list of relative atomic mass are in alphabetical order by element name, together with chemical symbol and proton/atomic number.
(ii) The relative atomic masses are quoted to two decimal places, though it is essential to be aware that values in pre-university examinations might be rounded to the nearest integer or one decimal place.
(iii) Trans-uranium elements have been eliminated because their isotopic composition varies depending on the source e.g. cyclotron, nuclear reactor etc. AND all their isotopes are highly radioactive and most are very unstable (so your relative atomic mass changes all the time!)
(iv) * radioactive, mass number of most stable isotope quoted
#### Atomic No. Z
Relative atomic mass
Ac Actinium 89 227.03
Al Aluminium 13 26.98
Sb Antimony 51 121.75
Ar Argon 18 39.95
As Arsenic 33 74.92
At Astatine 85 210 *
Ba Barium 56 137.33
Be Beryllium 4 9.01
Bi Bismuth 83 208.98
B Boron 5 10.81
Br Bromine 35 79.90
Cd Cadmium 48 112.41
Cs Caesium 55 132.91
Ca Calcium 20 40.08
C Carbon 6 12.01
Ce Cerium 58 140.12
Cl Chlorine 17 35.45
Cr Chromium 24 52.00
Co Cobalt 27 58.93
Cu Copper 29 63.55
Dy Dysprosium 66 162.50
Er Erbium 68 167.26
Eu Europium 63 151.97
F Fluorine 9 19.00
Fr Francium 87 223 *
Gd Gadolinium 64 157.25
Ga Gallium 31 69.72
Ge Germanium 32 72.60
Au Gold 79 196.97
Hf Hafnium 72 178.49
He Helium 2 4.00
Ho Holmium 67 164.93
H Hydrogen 1 1.01
In Indium 49 114.82
I Iodine 53 126.90
Ir Iridium 77 192.22
Fe Iron 26 55.85
Kr Krypton 36 83.80
La Lanthanum 57 138.91
Pb Lead 82 207.20
Li Lithium 3 6.94
Lu Lutetium 71 174.97
Mg Magnesium 12 24.31
Mn Manganese 25 54.94
Hg Mercury 80 200.59
#### Atomic No. Z
Relative atomic mass
Mo Molybdenum 42 95.94
Nd Neodymium 60 144.24
Ne Neon 10 20.18
Ni Nickel 28 58.69
Nb Niobium 41 92.91
N Nitrogen 7 14.01
Os Osmium 76 190.20
O Oxygen 8 16.00
Pd Palladium 46 106.42
P Phosphorus 15 30.97
Pt Platinum 78 195.08
Po Polonium 84 209 *
K Potassium 19 39.10
Pr Praseodymium 59 140.91
Pm Promethium 61 145 *
Pa Protactinium 91 231.04
Ra Radium 88 226.03
Rn Radon 86 222 *
Re Rhenium 75 186.21
Rh Rhodium 45 102.91
Rb Rubidium 37 85.47
Ru Ruthenium 44 101.07
Sm Samarium 62 150.36
Sc Scandium 21 44.96
Se Selenium 34 78.96
Si Silicon 14 28.09
Ag Silver 47 107.87
Na Sodium 11 23.00
Sr Strontium 38 87.62
S Sulfur 16 32.07
Ta Tantalum 73 180.95
Tc Technetium 43 98.91
Te Tellurium 52 127.60
Tb Terbium 65 158.93
Tl Thallium 81 204.38
Th Thorium 90 232.04
Tm Thulium 69 168.93
Sn Tin 50 118.71
Ti Titanium 22 47.88
W Tungsten 74 183.85
U Uranium 92 238.03
V Vanadium 23 50.94
Xe Xenon 54 131.29
Yb Ytterbium 70 173.04
Y Yttrium 39 88.91
Zn Zinc 30 65.39
Zr Zirconium 40 91.22
Revision notes on how to define relative atomic mass and how to calculate relative atomic mass from the percentage abundance of isotopes, help in revising for A level AQA, Edexcel, OCR 21st century, Gateway science GCSE 9-1 chemistry examinations relative atomic mass calculations how to calculate the relative atomic mass of bromine from the % percent abundance of isotopes, how to calculate the relative atomic mass of chlorine from the percentage abundance of isotopes, how to calculate the % composition of isotopes in an element given its relative atomic mass
TOP OF PAGE and sub-index
OTHER CALCULATION PAGES
Keywords and phrases: What is the relative atomic mass of an element? What scale is relative atomic mass based on? What is the formula to work out the relative atomic mass of an element? Quantitative Chemistry calculations online Help for problem solving in relative atomic mass calculations. Definitions of relative atomic mass and relative isotopic mass (A level students only) Practice revision questions on working out relative atomic mass from isotopic composition (% isotopes, A level students will learn about very accurate mass spectrometer data). What is relative atomic mass? How do you calculate the relative atomic mass of an element. What is the standard mass unit? Relative atomic mass is explained below, with reference to the carbon-12 atomic mass scale and the relevance of isotopes and 'u' the unified atomic mass unit is explained. Detailed examples of the method of how to calculate relative atomic mass from the isotopic composition are fully explained with reference to the definition of the relative atomic mass of a compound. For A level students, how to define and use relative isotopic masses to calculate relative atomic mass. These notes on defining, explaining and calculating relative atomic mass and defining relative isotopic mass are designed to meet the highest standards of knowledge and understanding required for students/pupils doing GCSE chemistry, IGCSE chemistry, O Level chemistry, KS4 science courses and A Level chemistry courses.
TOP OF PAGE and sub-index | 4,898 | 18,341 | {"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} | 2.78125 | 3 | CC-MAIN-2021-21 | longest | en | 0.814996 |
https://www.teacherspayteachers.com/Product/First-Grade-NO-PREP-Fractions-Pack-Color-in-Matching-Sorting-and-More-2420794 | 1,604,097,956,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911792.65/warc/CC-MAIN-20201030212708-20201031002708-00465.warc.gz | 756,913,384 | 28,298 | # First Grade NO PREP Fractions Pack (Color-in, Matching, Sorting, and More!)
Subject
Resource Type
Format
PDF (1 MB|12 pages)
Standards
\$1.00
\$1.00
### Description
This pack of 10 worksheets is perfect for practicing fractions with first graders (or Kindergarteners / Second Graders depending on their level)! It is focused on one half, one third, and one fourth.
Students can color-in fractions, partition shapes into equal parts, and there is even a cut and paste fraction sort.
You can use these as in-class practice sheets, homework, or even assessments!
Click on the "PREVIEW" to see each page! 10 pages for just \$1
***************************************************************************
***************************************************************************
BLOG
INSTAGRAM
PINTEREST
***************************************************************************
If you like this product, you may also like: First Grade NO PREP Math Symbols Pack
First Grade NO PREP Measurement Pack
***************************************************************************
John 3:16
Total Pages
12 pages
N/A
Teaching Duration
N/A
Report this Resource to TpT
Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TpT’s content guidelines.
### Standards
to see state-specific standards (only available in the US).
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. | 343 | 1,716 | {"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-2020-45 | latest | en | 0.791639 |
https://math.stackexchange.com/questions/2713108/fixed-point-theorems-involving-isometries | 1,656,725,115,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00521.warc.gz | 430,260,117 | 65,939 | # Fixed point theorems involving isometries
A fixed point theorem is a theorem which establishes some conditions that guarantee a certain function has a fixed point.
I would like to know if there exists a fixed point theorem involving metric spaces and isometries. Certainly one of these theorems should at least give some restrictive hypothesis about the metric space and/or the function. It's easy to see that there exist isometries which don't have fixed points, like any translation in $\mathbb{R}^{n}$.
So do we know what is needed to let an isometry have a fixed point? What's the most general setting in which we can state such a theorem?
Obviously if a fixed point theorem involving isometries exists in a more specific setting like normed space or Banach space it would be interesting anyway.
Thank you all if you can help me and sorry if I did some language mistake, I'm not an English native speaker.
• It is not just translations that are isometries without fixed points; there are plenty of them. A rotation on a closed ball in $\mathbb R^{2}$ is an isometry on a compact metric space without fixed. points. Mar 29, 2018 at 11:46
• @KaviRamaMurthy Correct me if I'm wrong, but to me it seems that a rotation on a closed ball in $\mathbb{R}^2$ has its centre of rotation as a fixed point. A rotation on an annulus in $\mathbb{R}^2$ doesn't have fixed points instead. Don't know if I interpreted your comment wrong. Mar 29, 2018 at 15:21
• I think there is a theorem saying that if $f : M→M$ is an isometry with $M$ a closed manifold, then the Lefschetz number of $f$ is the Euler characteristic of the submanifold $\{x ∈ M\,|\,f(x)=f\}$. I don't know enough to provide an answer and I don't find any source. But the existence of a fixed point given that the Lefschetz number is not $0$ does not require $f$ to be an isometry. Mar 29, 2018 at 18:29
• @ Stefano Rando Thanks for the correction. I should have taken an annulus and not a ball. Mar 30, 2018 at 22:52
Specifically, if $$K$$ is a nonempty, closed, bounded, convex subset of a Banach space and $$K$$ has normal structure, then every isometry $$T : K \to K$$ has a unique common fixed point. See "Fixed point theorems for mappings which do not increase distances" by Art Kirk, one of the foundational results in modern fixed point theory. | 593 | 2,315 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-27 | latest | en | 0.948454 |
https://stackoverflow.com/questions/18847431/how-to-search-for-in-string-excel | 1,580,004,514,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00466.warc.gz | 687,493,947 | 31,185 | # How to search for " in string - EXCEL
This is a tricky one I am stuck on. In Excel 2010 I want to search a string for the character `"`. I am using the formula
``````=FIND(A1,"text", 1)
``````
which will return a number (starting position) of "text" in A1 string, or an error if not found.
How to search for `"` in the formula?
Thanks for the advice!
Try amending your formula to search for Char(34), think this will help with readability instead of having 10,000 quotes in your formula.
``````=IF(COUNT(FIND(CHAR(34),A1))
``````
• Amen! Readability is just as important as functionality. – Gary's Student Sep 17 '13 at 11:41
You use a bunch of `"` until Excel understands it has to look for one :)
``````=FIND("""", A1)
``````
Explanation:
Between the outermost quotes, you have `""`. The first quote is used to escape the second quote so that `""` in between quotes means a single double quote.
Also, you can drop the 1 at the end if you want to check the whole string.
Note that it's find character, into cell. Or use `CHAR(34)` which is the equivalent of a quote:
``````=FIND(CHAR(34), A1)
``````
• You use a bunch of " until Excel understands??? This is not magic: doubling quotes is way Excel quotes them. Also, you can drop the 1 at the end if you want to check the whole string??? Excel may correct it well or not!!! -1 for laziness – LS_ᴅᴇᴠ Sep 17 '13 at 10:56
• @LS_dev The bunch of `"` was a joke, okay? Anyone could get confused by the sheer number of `"` in a formula. Also, of course you can drop the 1 at the end, because `1` means start from character 1, which is basically what `FIND(find_text, within_text)` does. Excel does not correct it, Excel takes 1 as the default as this is how it works. Btw, check your own formula, it returns `#VALUE`. – Jerry Sep 17 '13 at 11:00
• Hum... Ok, misunderstood "1 at end" point (understood "removing one quote at end"). But you still didn't give a clear explanation. Can't remove downvote now. – LS_ᴅᴇᴠ Sep 17 '13 at 11:04
You could use search too
``````=SEARCH("""";A1)
``````
• why does this work? shouldn't it find an instance of two double quotes next to each other. This behaviour is very confusing on Excel's part. – Gabe O'Leary Apr 9 '19 at 22:40
If you are just using the search and replace function use ~ to search for " / " use ~" / ~" ~ work for any odd character
Correct way of escaping quotes inside strings, in Excel formulas is doubling them:
``````=FIND(A1, """")
``````
will return first quote found in A1 (or error if not found). | 720 | 2,527 | {"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-2020-05 | latest | en | 0.928994 |
http://mathhelpforum.com/calculus/117761-hyperbolic-function-derivative-proof.html | 1,524,307,199,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945111.79/warc/CC-MAIN-20180421090739-20180421110739-00260.warc.gz | 193,566,737 | 9,368 | # Thread: Hyperbolic Function Derivative Proof
1. ## Hyperbolic Function Derivative Proof
Prove the formulas given in the table for the derivatives of the functions
a) cosh
b) tanh
c) csch
d) sech
So the table the problem is referring to has all the definitions of the hyperbolic functions, but how do I prove the formulas for the derivatives? I'm not understanding the question, do I need to find the derivative of the hyperbolic function? If so, how do I do that?
2. Use the definitions! $\displaystyle \cosh x = \frac{e^x+e^{-x}}{2}$ and $\displaystyle \sinh x = \frac{e^x-e^{-x}}{2}$. Do you know how to differentiate functions? | 177 | 637 | {"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} | 3.203125 | 3 | CC-MAIN-2018-17 | latest | en | 0.801777 |
https://cs.stackexchange.com/questions/tagged/optimization?tab=Frequent | 1,621,313,381,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989820.78/warc/CC-MAIN-20210518033148-20210518063148-00422.warc.gz | 200,675,889 | 46,057 | # Questions tagged [optimization]
Questions about problems that entail selecting the best element from some set of available alternatives, and methods to solve them.
130 questions
Filter by
Sorted by
Tagged with
10k views
7k views
### The stable marriage algorithm with asymmetric arrays
I have a question about the stable marriage algorithm, for what I know it can only be used when I have arrays with the same number of elements for building the preference and the ranking matrices. ...
2k views
### How do quantum computers find the global minimum?
If I understand correctly, quantum computers work by trying multiple methods of solving a problem simultaneously, using quantum superposition. However, if you try to "look" at a superposition-ed atom, ...
7k views
### How to find spanning tree of a graph that minimizes the maximum edge weight?
Suppose we have a graph G. How can we find a spanning tree that minimizes the maximum weight of all the edges in the tree? I am convinced that by simply finding an MST of G would suffice, but I am ...
2k views
### Given a set of 2D vectors, find the furthest reachable point
Input: a set of 2D vectors $S=\{v_1,v_2,\dots,v_n\mid v_i\in \mathbb{Z}^2 \}$ Question: name $P=\{\sum_{v_i\in S'}v_i\mid S'\subseteq S \}$ for all subsets of $S$ (obviously $|P|=O(2^n)$). In ...
1k views
4k views
### Finding the longest repeating subsequence
Given a string $s$, I would like to find the longest repeating (at least twice) subsequence. That is, I would like to find a string $w$ which is a subsequence (doesn't have to be a contiguous) of $s$ ...
2k views
### Classification of job shop scheduling problems
I'm writing a program (using genetic algorithms) that finds sort-of-optimal scheduling plan for a factory. The factory has several types of machines (say, ... | 443 | 1,820 | {"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} | 3.0625 | 3 | CC-MAIN-2021-21 | latest | en | 0.915957 |
http://mathhelpforum.com/differential-equations/205397-linear-homogeneous-boundary-conditions.html | 1,529,352,328,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00054.warc.gz | 213,149,518 | 9,335 | # Thread: Linear Homogeneous Boundary Conditions
Ux=Uxy
Uy=Ux
2. ## Re: Linear Homogeneous Boundary Conditions
Hey sarideli18.
Consider Ux = v and dv/dy = Uxy, then v(x,y) = dv/dy which implies dv/v = dy.
3. ## Re: Linear Homogeneous Boundary Conditions
I am not getting this . What happen if there is three variables instead of two .
it solutions | 98 | 355 | {"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-2018-26 | latest | en | 0.788655 |
http://www.jiskha.com/display.cgi?id=1265101294 | 1,498,158,082,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319688.9/warc/CC-MAIN-20170622181155-20170622201155-00195.warc.gz | 547,707,928 | 4,214 | # Calculus
posted by on .
R is the region in the plane bounded below by the curve y=x^2 and above by the line y=1.
(a) Set up and evaluate an integral that gives the area of R.
(b) A solid has base R and the cross-sections of the solid perpendicular to the y-axis are squares. Find the volume of the solid.
(c) A solid has base R and the cross-sections of the solid perpendicular to the y-axis are equilateral triangles. Find the volume of the solid.
• Calculus - ,
(a) The y = x^2 curve crosses y = 1 at x = 1. For the area in question, compute the integral of x^2 from x = 0 to 1, and subtract it from 1.
(b)Integrate along the y axis, with differential slab volume x^2 dy = y dy, from y=0 to y=1
(c) Proceed similary to (b), integrating along y from 0 to 1, but with equilateral triangle slabs. Slab volume must be expressed in terms of x
• Calculus - ,
Why would it only be the integral of x^2 and not the integral of 1-x^2 from x=-1 to x=1? So wouldn't that equal an area of 4/3? And secondly, how would you sketch these solids? when you say integrate along the y-axis do you mean to say that the curves have to be shifted as well? And why are all the integrals from x=0 to x=1?
• Calculus - ,
Find the volume, V , of the solid obtained
by rotating the region bounded by the graphs
of
x = y2, x = squareroot(y)
about the line x = −1. | 378 | 1,351 | {"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} | 3.828125 | 4 | CC-MAIN-2017-26 | latest | en | 0.932687 |
https://forums.developer.nvidia.com/t/how-to-use-shared-memory/18595 | 1,653,692,831,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00245.warc.gz | 326,635,823 | 6,805 | # how to use shared memory
I am trying to use CUDA to speed up my program. But I am not very sure how to use the share memory. I bought the book “Programming massively parallel processors” which has some samples, but I feel the sample (matrix computing) is way too easy and real world application is not so easy to follow the sample.
In my case, I want to use an 1d array of float values (suppose it has m elements, m is about 1000) to interpolate a 2d array (suppose it has n * k elements, n and k is about 500). When calculating any one element in the 2d array, it needs two items of the 1d array to calculate. But using which 2 items depends on some condition. I am sure it is a very simple program and it is a good candidate program for CUDA.
It is very easy to implement a simple code to use GPU to calculate, but it is actually way slower (5x) than regular CPU code. Then I start to look into reduce the global memory access ratio. Of course the first step is, trying to put the 1d array (about 4k in size) into shared memory of blocks. That supposes to be must faster than the array is in global memory. But I am not sure how do put the 1d array into the shared memory.
The sample code on the book is like this,
shared float Mds[TILE_WIDTH][TILE_WIDTH]; // TILE_WIDTH is constant value 2.
But when it access it, it looks like the index is greater than TILE_WIDTH.
Here is my simplified kernel code (my 2d array is still represented as 1d array, stored row by row).
global void MyKernel(float* array1d, float* array2d, int m, int n, int k, /* some other args */)
{
// obviously in this case, both array 1d and array2d is in global memory.
// question is how to put array1d in shared memory?
int i = blockIdx.x;
int index = j * k + i
int i0 = x; // x actually is calculated from i, j and other args.
int i1 = x + 1;
array2d[index] = (array1d[i0] + array1d[i1]) / 2.0f; // actually is a more complicated and time consuming computing with other args.
}
Here is how MyKernel is called,
dim3 dimBlock(n, 1);
dim3 dimGrid(k, 1);
MyKernel<<<dimGrid, dimBlock>>>(array1d, array2d, m, n, k, … other args… ).
Easiest way to do shared is to just declare the shared in the cuda code but not inside a function
``````#define SHALINE 1056
__shared__ short shLine[SHALINE ]; // can be multidimensional
__device__ void populateShLine(char* d_line, long len, char* d_unProcd)
{
..blah..
shLine[threadIdx.x] = d_unProcd[someIndex]; // d_unProcd is a global array i.e. in the GPU's main RAM
..blah..
for (...)
{
do something using shLine[]
}
}
``````
(shLine might be your array1d )
shared arrays are in the on-chip memory and have a latency of about 6 cycles
global arrays are in the GPU’s main RAM and have a latency of hundreds of cycles (but if another block can run during that time it may not matter)
advantage of using shared is when that shared array will be accessed many times, e.g. inside a loop
Easiest way to do shared is to just declare the shared in the cuda code but not inside a function
``````#define SHALINE 1056
__shared__ short shLine[SHALINE ]; // can be multidimensional
__device__ void populateShLine(char* d_line, long len, char* d_unProcd)
{
..blah..
shLine[threadIdx.x] = d_unProcd[someIndex]; // d_unProcd is a global array i.e. in the GPU's main RAM
..blah..
for (...)
{
do something using shLine[]
}
}
``````
(shLine might be your array1d )
shared arrays are in the on-chip memory and have a latency of about 6 cycles
global arrays are in the GPU’s main RAM and have a latency of hundreds of cycles (but if another block can run during that time it may not matter)
advantage of using shared is when that shared array will be accessed many times, e.g. inside a loop
Thanks for the reply. I kind of got shared memory working and got some speed improvement. But I got a strange problem. If I do not use shared memory, all of my array2d values are calculated correctly. But if I use shared memory,
only part of the values are correct. And I see a pattern - when the array2d’s index sqrt(ii + jj) >= 256, the array2d element is not correct. (e.g., array2d[160][199], array2d[161][198]).I wonder it is caused by shared memory size limit. But my grid is not big at all. Not sure about the shared memory trick yet.
Thanks for the reply. I kind of got shared memory working and got some speed improvement. But I got a strange problem. If I do not use shared memory, all of my array2d values are calculated correctly. But if I use shared memory,
only part of the values are correct. And I see a pattern - when the array2d’s index sqrt(ii + jj) >= 256, the array2d element is not correct. (e.g., array2d[160][199], array2d[161][198]).I wonder it is caused by shared memory size limit. But my grid is not big at all. Not sure about the shared memory trick yet.
Note, only the threads in a same block can use shared memory…
Note, only the threads in a same block can use shared memory… | 1,310 | 4,944 | {"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} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.921222 |
https://socratic.org/questions/how-do-you-find-the-limit-of-1-tan-x-sin-x-cos-x-as-x-approaches-0 | 1,723,014,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00747.warc.gz | 420,330,592 | 5,788 | # How do you find the limit of (1-tan(x)) / (sin (x) - cos (x)) as x approaches 0?
(1-tan(x)) / (sin (x) - cos (x)) =[1-(sinx/cosx)]/[sinx-cosx]= [cosx-sinx]/[cosx*(sinx-cosx)]= -1/cosx
Hence ${\lim}_{x \to 0} \left(- \frac{1}{\cos} x\right) = - 1$ | 121 | 250 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "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} | 3.84375 | 4 | CC-MAIN-2024-33 | latest | en | 0.421008 |
https://blender.stackexchange.com/questions/183921/why-does-rotating-an-object-on-its-axes-in-a-different-order-result-in-a-differe | 1,603,647,203,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889574.66/warc/CC-MAIN-20201025154704-20201025184704-00417.warc.gz | 235,022,338 | 34,240 | # Why does rotating an object on its axes in a different order result in a different final rotation?
As in the first GIF below, the result of rotating X first and Y rotating is different from rotating Y first and rotating X.
And like the second GIF, the result of rotating the Z of two objects first, then rotating X first, then rotating Y, and rotating Y first and rotating X are different.
Obviously, although the X, Y, and Z values of the object are the same, what causes the rotation result to be different depending on the rotation order? Is there a workaround?
• What causes the result to be different? The different order causes it. By changing the order, you put the object at a different orientation when beginning the next axis rotation in the current rotation order. So naturally the result would be different. Is there a workaround? Yes. It's called Quaternions. – R-800 Jun 24 at 6:29
• Try it with thee coffee-mug on your desk. (Make sure it's empty, first) :D – Robin Betts Jun 24 at 7:34
• So the problem you're talking about is that my method shown above is a way to cause gimbal lock? I have heard of the gimbal lock problem. So, when I rotate the object like the above method, I set it as a Quaternions. However, after heard your answer now i think that I didn't understand how to use the Quaternions correctly. – upyen Jun 24 at 9:00
• Rotations are not commutative. ie the order matters. (A * B != B * A) It's not gimbal lock which is due to "locking of a gimbal" at 90 degree rotations. – batFINGER Jun 25 at 13:11 | 382 | 1,542 | {"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} | 3.015625 | 3 | CC-MAIN-2020-45 | latest | en | 0.920405 |
https://stonespounds.com/219-5-pounds-in-stones-and-pounds | 1,686,407,581,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657720.82/warc/CC-MAIN-20230610131939-20230610161939-00568.warc.gz | 611,268,199 | 4,833 | # 219.5 pounds in stones and pounds
## Result
219.5 pounds equals 15 stones and 9.5 pounds
You can also convert 219.5 pounds to stones.
## Converter
Two hundred nineteen point five pounds is equal to fifteen stones and nine point five pounds (219.5lbs = 15st 9.5lb). | 76 | 271 | {"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} | 3.3125 | 3 | CC-MAIN-2023-23 | latest | en | 0.79273 |
https://answers.launchpad.net/mg5amcnlo/+question/688706 | 1,585,454,134,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00552.warc.gz | 377,192,645 | 7,129 | # Mathematical definition of xqcut
Asked by trenta coollime on 2020-02-12
Hello, experts.
I have been generating events with xqcut = 20 for p p > h > z zp, z> l+ l-, zp > b b~ and noticed some odd feature.
Even when I define xqcut = 20, which I believe to be equivalent to kT (pT) of jet, I see several events, which have jets with pT<20.
MG5 website says it's the minimal distance in the phase space allowed between extra partons.
What does that exactly mean? and is there a mathematical formula that defines the quantity precisely?
Thank you.
Best Regards.
## Question information
Language:
English Edit question
Status:
Solved
For:
Assignee:
No assignee Edit question
Solved by:
Olivier Mattelaer
Solved:
2020-02-12
Last query:
2020-02-12
2020-02-12
Olivier Mattelaer (olivier-mattelaer) said on 2020-02-12: #1
Hi,
xqcut is a (very) special cut use for MLM matching/merging.
It does not have any easy mathematical formula.
and the reference to the various paper pointed to that page.
Note that in the case of a single jet, this is equivalent to the pt of that particle.
In this specific case, matching/merging does not occur on particle coming from decay.
and therefore your b-quark are not cut by that cut. (in top of that that cut is applied in principle only on massless particles which also might explains why it does not appear here.
In version 2.7.0 of the code, you might not even have the xqcut in the run_card since we are now automatically hidding all cuts that does not have any impact for the process under consideration.
Cheers,
Olivier
> On 12 Feb 2020, at 16:28, trenta coollime <email address hidden> wrote:
>
> New question #688706 on MadGraph5_aMC@NLO:
>
> Hello, experts.
>
> I have been generating events with xqcut = 20 for p p > h > z zp, z> l+ l-, zp > b b~ and noticed some odd feature.
>
> Even when I define xqcut = 20, which I believe to be equivalent to kT (pT) of jet, I see several events, which have jets with pT<20.
>
> MG5 website says it's the minimal distance in the phase space allowed between extra partons.
>
> What does that exactly mean? and is there a mathematical formula that defines the quantity precisely?
>
> Thank you.
>
> Best Regards.
>
> --
trenta coollime (trentacoollime) said on 2020-02-12: #2
One last question, when you say "Note that in the case of a single jet, this is equivalent to the pt of that particle."
does the "single jet" mean additional jet specified in "generate" level (ex) p p > t t~ j)? or any single jet?
Olivier Mattelaer (olivier-mattelaer) said on 2020-02-12: #3
> does the "single jet" mean additional jet specified in "generate" level
> (ex) p p > t t~ j)? or any single jet?
when you have exactly one jet at parton level (what you call "generate" level)
so yes for
p p > t t~j or for p p > w+ w- j ,...
Cheers,
Olivier
> On 12 Feb 2020, at 19:57, trenta coollime <email address hidden> wrote:
>
> Question #688706 on MadGraph5_aMC@NLO changed:
>
> trenta coollime posted a new comment:
> Thank you for your response!
>
> One last question, when you say "Note that in the case of a single jet,
> this is equivalent to the pt of that particle."
>
> does the "single jet" mean additional jet specified in "generate" level
> (ex) p p > t t~ j)? or any single jet?
>
> -- | 929 | 3,278 | {"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-2020-16 | latest | en | 0.935237 |
https://1stprinciples.wordpress.com/2008/03/16/more-on-regularity-preservation/ | 1,685,753,960,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00797.warc.gz | 97,090,015 | 21,145 | More on Regularity Preservation
As sri rightly pointed out in the comments section of the previous post, clique insertion is just one way to extend an r-regular graph into a bigger r-regular graph. We should be able to extend the graph by inserting any (r – 2)-regular graph (provided the insertion is possible — refer sri’s comment). In fact, clique insertion is the minimal case: the smallest connected (r-2)-regular graph has r -1 nodes.
There are interesting questions here. Can we insert any (r – 2)-regular graph? What is the procedure to do so? Are there some (r – 2)-regular graphs that cannot be inserted?
Let’s use some notations for ease of reference. Let Ro be the r-regular graph we start with, Ri be the (r – 2)-regular graph we want to insert and Rn be the graph resultant of the insertion.
A simple but important thing to be noted here is that, pairs of nodes that are adjacent in Ri, cannot be inserted in a single edge of Ro because that will result in an overlapping edge (we don’t allow multiedges between nodes); adjacent nodes in Ri have to be inserted in different edges. Even for small values of r, this becomes pretty complicated to visualize. I tried to insertions for small graphs, and it was possible to do. But I don’t know if we can always avoid creating a multigraph.
But then in the above discussion we have assumed Ri to be a connected graph. What if Ri is disconnected? If Ri is disconnected we can easily prove that there is no bound on the size of Ri that can be inserted. The simplest way to show this is by saying that our Ri has an arbitrarily large number of (r – 2)-cliques as disconnected components (disconnected edges, disconnected cycles, disconnected cubic graphs etc.). Now, you can either imagine this as adding an arbitrary number of cliques (like in the previous post) or as adding one large disconnected graph.
Of course, cliques are just one way to insert. We can simply insert nodes and add edges between the new nodes in such a way that all the new nodes have a degree of r.
Note that the resulting graph Rn is, of course, a connected graph.
***
Now, since we are talking about p2p networks, what is more important is the minimum number of nodes and edges that we need to add when nodes join or leave. Or more generally, the minimum change in topology that is required. When a new node comes in, let’s say we need to accommodate it without disrupting symmetry; so we may need to add dummy nodes and edges. The thing is – in a p2p network, most of the time, unrelated nodes join or leave unrelated parts of the network. So, essentially, we are dealing with disconnected components. | 590 | 2,643 | {"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} | 3.78125 | 4 | CC-MAIN-2023-23 | latest | en | 0.940516 |
http://www.bankexamstoday.com/2017/05/double-diagram-di-set-with-solution-for.html | 1,521,552,658,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647475.66/warc/CC-MAIN-20180320130952-20180320150952-00388.warc.gz | 338,524,721 | 27,715 | Double Diagram DI Set With Solution For RBI Grade B: Phase 1
Directions: Following Pie Chart represents the percentage distribution of students studying in different departments of an engineering college and the tabular column depicts the ratio of male students to female students in each department.
Ratio of Male students to Female Students Department Male Female Mechanical 3 2 ECE 7 3 Civil 6 5 EEE 4 1 CSE 11 14
Questions
1. What is the total number of female students studying in that college except civil department?
A. 16,645
B. 21,445
C. 16,445
D. 21,545
E. None of these
2. What is the ratio between the total number of students studying in mechanical and CSE departments to the total number of students studying in civil and EEE departments?
A. 2:5
B. 2:3
C. 4:3
D. 3:2
E. None of these
3. What is the total number of male students studying in CSE department and female students studying in both ECE and EEE departments?
A. 8,855
B. 8,865
C. 17,655
D. 18,645
E. None of these
4. If 2 % of male students from mechanical department and 3 % of female from EEE department have cleared SBI PO prelims, then what will be the percentage of total students in that college have cleared the SBI PO prelims exam?
A. 0.43%
B. 0.40%
C. 0.42%
D. cannot be determined
E. None of these
5. What is the percentage of female students studying in civil department in that college?
A .9
B. 11
C. 20%
D. 30%
E. None of these
From the pie chart and tabular column we get the following details
Department
Department No. of Students Male Students Female Students Mechanical 16,500 9,900 6,600 ECE 13,750 9,625 4,125 Civil 11,000 6,000 5,000 EEE 5,500 4,400 1,100 CSE 8,250 3,630 4,620
Ans 1. Option C
(6,600+4,125+4,620+1,100)
16,445 (ans)
Ans 2. Option D
(Mech +CSE)/ (Civil + EEE)
(16,500 + 8,250) / (11,000+ 5,500)
24,750 / 16,500
3:2 (ans)
Ans 3. Option A
Male students studying in CSE dept. = (11/25)* 8,250
= 3,630
(3,630+4,125+1,100)
8,855 (ans)
Ans 4. Option C
2% (9,900) + 3%( 1,100)
198+33
231
231= X%(55,000)
X= (231* 100)/ 55,000
X= 0.42% (ans)
Ans 5. Option A
No. of female students in civil dept.= 5000
X% (55,000) = 5000
X= (5000 * 100)/ 55,000
X= 100/11
X= 9 (ans)
Smart Prep Kit for Banking Exams by Ramandeep Singh - Download here | 767 | 2,257 | {"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-2018-13 | latest | en | 0.906876 |
https://www.physicsforums.com/threads/traveling-near-the-speed-of-light.825100/page-3 | 1,620,592,822,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00018.warc.gz | 968,481,041 | 24,010 | # Traveling near the speed of light
So the theory does apply, since if I were to suppose an observer were moving at c, the theory will tell me what this means? For example, as v approaches c, γ approaches ∞, so if we were to consider the total energy of an observer of mass m with speed c,
E = γmc2 = ∞
Thus, in order to be moving at c, the observer must have a total amount of energy that is infinite, but it is impossible to transfer an infinite amount of energy to any massive object since this would contradict the Law of Conservation of Energy.
If the observer is were a photon...?
PeterDonis
Mentor
2020 Award
If the observer is were a photon...?
An observer can't be a photon. An "observer" has to have a rest frame, and a photon doesn't.
I intended the previous post is a pure speculative question or light joke.[Add: not a joke about "light]
Now this is not.
An observer can't be a photon. An "observer" has to have a rest frame, and a photon doesn't.
1. A photon doesn't have a rest frame, because as pictured in the space time diagram, it is geometrically impossible to boost so that light ray is vertical?
What "problem" are you trying to resolve? Why does Red choose those particular values to send to Blue and Green? What is Red trying to accomplish?
2. An observer has to have a rest frame. But the observer can choose any moving/rest frame to make the problem easier? Is that so?
Consider this picture.
Red DOES have a rest frame, but to calculate problem such as the picture below, we can choose the picture at the right side so we can calculate simultaneity events for Blue and Green easily.
No. The theory says that only things with zero rest mass (like light) can move at c; an "observer" (meaning something like you or me or a spaceship or a measuring device) can't. That's because any "observer" must have nonzero rest mass.
I think I may have used poor communication here. I understand very well that the theory tells you it is not possible for any object with a rest mass to move at c. I think my question,
So the theory does apply
, since if I were to suppose an observer were moving at c, the theory will tell me what this means?
was a little ambiguous. So the theory applies to light because it correctly predicts the behavior of light and also tells you that objects with rest mass can not move like light?
Have you looked at any basic SR textbooks? If you haven't, you should. This is all very basic SR.
There is a chapter in my physics textbook on SR that I have read. I have not read any textbooks that specialize in SR. I thought I had a basic understanding of the theory.
here is no such thing; a light beam moves at c in every reference frame, so there can never be a frame in which it at rest, and "the reference frame of a light beam" assumes that the light beam is at rest in the frame.
Again, if you haven't looked at an SR textbook, you should. This is very basic SR. (There is also a Physics Forums FAQ explaining that there is no such thing as the rest frame of a photon.)
I see. I was unaware of this and clearly my knowledge is limited. So I understand that my other example that considered the frame of the light beam is invalid. So then this means SR can not model light in this way?
No. The ship is moving relative to Earth and the beacon, so the difference in light travel times from Earth to the ship, vs. from the beacon to the ship, is not constant.
Is it not constant because light actually travels at constant speed in any inertial reference frame?
The difference in clock readings between the beacon and Earth, in the ship frame, is due to relativity of simultaneity.
My understanding of relativity of simultaneity is also basic and perhaps it is not adequate enough as I am unsure how relativity of simultaneity would justify the fact that the beacon's clock reads different than the Earth's clock.
Would you mind recommending me a few books?
PeterDonis
Mentor
2020 Award
A photon doesn't have a rest frame, because as pictured in the space time diagram, it is geometrically impossible to boost so that light ray is vertical?
That's not the reason, but it's a way of visualizing it, yes. The reason a photon doesn't have a rest frame is the law that photons move at ##c## in all reference frames, so there can't be a frame in which a photon is at rest. In terms of the spacetime diagram, a Lorentz transformation leaves the lightlike lines (45 degree lines) invariant; it doesn't change their slope at all, so yes, the diagram correctly represents the underlying law that photons move at ##c## in all frames.
An observer has to have a rest frame. But the observer can choose any moving/rest frame to make the problem easier?
Yes. But he can't choose a frame in which a photon is at rest, because there is no such frame.
PeterDonis
Mentor
2020 Award
So the theory applies to light because it correctly predicts the behavior of light and also tells you that objects with rest mass can not move like light?
Yes.
this means SR can not model light in this way?
Yes. The way SR models light, or anything with zero rest mass, works differently from the way it models things with nonzero rest mass. There are still many things in common between the two ways of modeling, but the differences are fundamental and important.
Is it not constant because light actually travels at constant speed in any inertial reference frame?
That's part of it; the other part is that the distances are changing because the ship is moving relative to Earth and the beacon.
I am unsure how relativity of simultaneity would justify the fact that the beacon's clock reads different than the Earth's clock.
The beacon's clock and the Earth's clock are synchronized in the Earth-beacon rest frame--that is, a given reading on either clock is simultaneous, in that frame, with the same reading on the other clock. But if that's true in the Earth-beacon frame, it won't be true in any other frame, by relativity of simultaneity; events that are simultaneous in one frame are not simultaneous in any other frame. So in any other frame, a given reading on, say, the Earth clock will not be simultaneous with the same reading on the beacon clock; instead, it will be simultaneous with some other reading on the beacon clock. To find out which readings on the two clocks will be simultaneous, you need to do the math to transform between the frames; that's how I derived the numbers in my earlier post.
Would you mind recommending me a few books?
Taylor & Wheeler's Spacetime Physics is a good text. A good online reference is here:
http://lightandmatter.com/sr/
I intended the previous post is a pure speculative question or light joke.[Add: not a joke about "light]2. An observer has to have a rest frame. But the observer can choose any moving/rest frame to make the problem easier? Is that so?
Yes. But [..]
THANKS. It helps me clear many problems! I tough this is a mathematic quiz. Like "Solve this problem with a WL moves at certain V" And we are not supposed to boost the diagram. But we can freely boost the diagram anyway we like to solve the problem.
What "problem" are you trying to resolve? Why does Red choose those particular values to send to Blue and Green? What is Red trying to accomplish?
Now I understand it. Crystal clear.
Last edited:
One more question:
To solve the numbers (coordinate of events, wordlines direction) we can boost the ST diagram in any V we want.
But to imagine what the nature would look like, we should boost it again, so we are at the rest frame. Is that so?
We DON'T travel!. It's the universe that travel approaches the speed of light. It's the universe that won't age. We still feel 1 second is 1 second.
I remember an Einstein joke (or anectode?)
Once on a train in US (Einstein had already been in US), a student recognized him. And the student ask, "I'm sorry professor, when will New York stop at the train?"
How do you beat gravity force? Jump out of the window, and the floor (and the earth, the moon, everything) will move toward you accelerated at 9.8ms. As long as you stay on the room, you are actually travel and accelerated at 9.8 m/s^2
BUT DON'T PROVE IT!
Do you mean that if someone were to live on Earth 85 years, (s)he would also live 85 years on the spaceship (assuming the same life style on Earth and spaceship) according to the spaceship's clock while (s)he is traveling at a speed close to the speed of light, regardless of the difference between the Earth's and spaceship's clocks?
PeterDonis
Mentor
2020 Award
To solve the numbers (coordinate of events, wordlines direction) we can boost the ST diagram in any V we want.
Yes.
to imagine what the nature would look like, we should boost it again, so we are at the rest frame.
I'm not sure I understand. How is this different from boosting the ST diagram in any V we want?
PeterDonis
Mentor
2020 Award
Do you mean that if someone were to live on Earth 85 years, (s)he would also live 85 years on the spaceship (assuming the same life style on Earth and spaceship) according to the spaceship's clock
Yes.
EngWiPy
Yes.
I'm not sure I understand. How is this different from boosting the ST diagram in any V we want?
Come on, you know better
.
To calculate simultaneity of E1 and E2, we (us) have to use the big picture. What the world really look like? We (us) use the right bottom picture. Here, we see that blue and red speed actually differ more than we see in the big picture.
PeterDonis
Mentor
2020 Award
Come on, you know better
No, I don't. If I understood what you were trying to say, I wouldn't have asked the question I asked.
What the world really look like?
This is a meaningless question; no particular frame tells you what the world "really" looks like. "Really" is not a scientific word.
Here, we see that blue and red speed actually differ more than we see in the big picture.
No, that's not what we see. First of all, as above, no particular frame tells you what is "actually" the case. "Actually" is not a scientific word.
Second, the relative velocity of blue and red is an invariant; it doesn't depend on which frame you choose. If you are getting the result that the relative velocity of blue and red is different in different frames, you are doing something wrong.
This is a meaningless question; no particular frame tells you what the world "really" looks like. "Really" is not a scientific word.
No, that's not what we see. First of all, as above, no particular frame tells you what is "actually" the case. "Actually" is not a scientific word.
Good point. Now I know what scientific method is.
Second, the relative velocity of blue and red is an invariant; it doesn't depend on which frame you choose. If you are getting the result that the relative velocity of blue and red is different in different frames, you are doing something wrong.
Just the angle look different.
It's the ##\frac{u+v}{1+uv}##
HallsofIvy
Homework Helper
No. That is the whole point of relativity- all photons move at speed "c" relative to all reference frames. To a person "A" traveling toward another star at a large percentage of the speed of light, relative to person "B" on earth, the distance to that star is shorter that it is to "B". To person "B", "A" is traveling the greater distance but his time scale is dilated. So "distance divided by time" works out the same to each.
PeterDonis
Mentor
2020 Award
Just the angle look different.
So what does that mean? I still don't understand what point you're trying to make.
So what does that mean? I still don't understand what point you're trying to make.
Perhaps I can't express myself clearly.
Supposed you are green in this diagram.
And you want to know if event E1 and E2 are simultaneous, so you boost the ST diagram so that the maroon world line are at rest. Then you'll find that E1 and E2 is simultaneous.
But to see the diagram with naked eye, it seems that red and blue WL is very close to green. Actually in the real world, the angle of red and blue is bigger if you put green at rest. See the yellow rectangle.
And IMHO, twins paradox is no more mistery than barn paradox and train/platform experiment. I'm new here. But in the past three months I think I can grasp basic SR. Still so much to learn tough.
PeterDonis
Mentor
2020 Award
Actually in the real world, the angle of red and blue is bigger if you put green at rest.
What do you mean by "in the real world"? Calculate the relative velocity of red and blue in any frame; it will be the same. That means that, "in the real world", the relative velocity of red and blue is the same for all observers. The fact that the angle appears different on different spacetime diagrams is a fact about the diagrams, not about red and blue or anything "in the real world" (meaning, the objects and motions that the diagrams represent).
[..]That means that, "in the real world", the relative velocity of red and blue is the same for all observers.
GOOD POINT!. I'll never forget that.
The fact that the angle appears different on different spacetime diagrams is a fact about the diagrams, not about red and blue or anything "in the real world" (meaning, the objects and motions that the diagrams represent).
Yes, I do understand that. Like I said, I can't express myself clearly.
Consider this.
VBlue wrt Red is 0.98c
VGreen wrt Blue is 0.6c
For Red, VGreen is 0.995c. Just slighty 0.015c than VBlue. And if you boost the diagram where Blue is at rest, than you'll see that the different/angle between VGreen and VBlue is 0.6c. I know, I know, we have to account for relative velocity addition.
It's like that we say.
"No, it's not Blue (660 THz) with V=0, it's Red (440 THz) with V = 0.2c"
Our eyes see it as Blue, our brain calculates it as Red (should compare it with Franhover lines, otherwise, you won't know if it's blue-shifted).
And in my opinion I think Twins Paradox is no more mystery than barn paradox or train experiment.
Perhaps the word "Paradox" came out when Relativiy was at its infancy, and the most natural phenomena such as time dilation for one of the twins was considered "amazing".
PeterDonis, I have a problem with Train Experiment, perhaps you might want to take a look at it in: https://www.physicsforums.com/threads/train-experiment-problem.825514/
Thanks
DrGreg
Gold Member
Perhaps the word "Paradox" came out when Relativiy was at its infancy, and the most natural phenomena such as time dilation for one of the twins was considered "amazing".
Paradoxically, the word "paradox" has (amongst others) two meanings that are almost opposites:
1. an argument that comes to a false conclusion (e.g.contradicts itself), via steps that appear, at first, to be valid
2. an argument whose conclusion may appear, at first, to be false, but is actually true
The "twins paradox", like some other paradoxes in maths and physics, is a paradox of the second type.
In this post I used "paradoxically" with a third meaning, "having apparently contradictory characteristics".
Stephanus | 3,485 | 14,998 | {"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} | 3.203125 | 3 | CC-MAIN-2021-21 | latest | en | 0.967333 |
https://brainmass.com/business/cost-allocation-of-service-department-costs/washburn-divison-ashland-income-calculations-592111 | 1,511,072,927,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805417.47/warc/CC-MAIN-20171119061756-20171119081756-00752.warc.gz | 594,673,592 | 19,641 | Share
Explore BrainMass
Washburn Divison of Ashland Inc. Income Calculations
Bayfield Division of Ashland Inc. has a capacity of 200,000 units and expects the following results.
Sales (160,000 units at \$4) \$640,000
Variable costs, at \$2 320,000
Fixed costs 260,000
Income \$60,000
Washburn Division of Ashland Inc. currently purchases 50,000 units of a part for one of its products from an outside supplier for \$4 per unit. Washburn's manager believes he could use a minor variation of Bayfield's product instead, and offers to buy the units from Bayfield at \$3.50. Making the variation desired by Washburn would cost Bayfield an additional \$0.50 per unit and would increase Bayfield's annual cash fixed costs by \$20,000. Bayfield's manager agrees to the deal offered by Washburn's manager.
Required: (33 Points)
a. What is the effect of the deal on Washburn's income?
b. What is the effect of the deal on Bayfield's income?
c. What is the effect of the deal on the income of Ashland Inc. as a whole?
Solution Preview
Saving of Washburn when bought from Bayfield = 50,000 * (4.00 - 3.50) = \$25,000
Now, lets compute Bayfield net additional ...
Solution Summary
The solution computes the effect of income when the materials are purchased from by one division of the company from another.
\$2.19 | 329 | 1,313 | {"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-2017-47 | longest | en | 0.9436 |
http://www.richtherrn.net/physics/answermotion2.htm | 1,516,624,496,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891316.80/warc/CC-MAIN-20180122113633-20180122133633-00330.warc.gz | 527,639,648 | 2,189 | NAME_________ PER______ DATE_____
MOTION II
1) The graph below represents three cars during the first minute of a race. Using the following information, draw another curve on the grid to represent the motion of Car D!
Car D accelerates from a rest position at 0 seconds to reach a speed of 208 km/h at 5 seconds. D maintains this speed for 5 seconds, then decelerates to 32 km/h at 20 seconds. IT accelerates to reach a speed of 160 km/h at 30 seconds and maintains this speed for 5 seconds. Car D then decelerates to 112 km/h at 40 seconds, further decelerates to 64 km/h at 50 seconds and then accelerates to 208 km/h at 55 seconds.
1. Draw Car D’s trip
2. What does the vertical axis represent?
Speed
3. What does the horizontal axis represent?
Time
4. Over which time period is Car A’s acceleration the greatest? (Acceleration=Speed/Time)
0-10 s 80/10
5. Over which time period is Car B’s acceleration the greatest?
0-10 s 144/10
6. Over which time period is Car C’s acceleration the greatest?
0-5 s 160/5
7. When is each car( A, B, C) at an acceleration of zero?
A, 20-45 s V=128 , B at 10-60s V=192 , C at never!,
8. Which car(s) deaccelerate during the race?
C, D
9. Which car has traveled the farthest at the end of one minute?
B! Dis is area under the curve, even though B is going fastest at the end
10. Which car appears to have a restless driver?
D
11. Which car appears to have stalled during the race?
C
12. Which car appears to have the safest driver?
A
Interpreting and Predicting From Graphs:
Imagine that you are in a car traveling down a long, straight road. Every 10 seconds you record the car’s speed. With the speed and time data, you could find some information about the car’s acceleration. BY studying a speed-time graph, you could determine when the driver pressed on the accelerator or when the brakes were used.
Study the speed-time graphs for the car trips and write your interpretation under each one:
Figure One shows a car that is going a slow steady speed forward.
Figure Two show a car that is accelerating, or speeding up at a constant rate going forwards.
Describe the car trips. In doing so, tell when the driver accelerated, when the driver used the brakes and deacelerated, and whenever possible, compare the acceleration and deceleration.
see description on graph
Now, use your graphing skills to plot the speed-time graph of the following car trip:
At the beginning the car is traveling in a straight line at a constant speed of 3 m/s. After 20 seconds, the car accelerates up to a speed of 5 m/s taking 15 seconds more to reach the new speed. After a total of 35 seconds the car slows down or deacelerates. It takes the car an additional 20 seconds to decelerate from 5 m/s to 0 m/s. thus the total time for the car trip was 55 seconds.
Sketch the graph.
For each of the following distance time graph sketch below what the velocity time graph would look like
DISTANCE TIME GRAPHS:
VELOCITY TIME GRAPHS
Velocity along the y, time across the x | 750 | 2,995 | {"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} | 3.984375 | 4 | CC-MAIN-2018-05 | latest | en | 0.927504 |
https://gmatclub.com/forum/a-marketing-firm-determined-that-of-200-households-65054.html?oldest=1 | 1,510,941,473,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803848.60/warc/CC-MAIN-20171117170336-20171117190336-00134.warc.gz | 614,649,966 | 55,993 | It is currently 17 Nov 2017, 10:57
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# A marketing firm determined that, of 200 households
Author Message
TAGS:
### Hide Tags
Manager
Joined: 31 Jul 2006
Posts: 232
Kudos [?]: 93 [3], given: 0
A marketing firm determined that, of 200 households [#permalink]
### Show Tags
07 Jun 2008, 01:35
3
KUDOS
9
This post was
BOOKMARKED
00:00
Difficulty:
35% (medium)
Question Stats:
64% (01:00) correct 36% (01:16) wrong based on 418 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
A) 15
B) 20
C) 30
D) 40
E) 45
OPEN DISCUSSION OF THIS QUESTION IS HERE: a-marketing-firm-determined-that-of-200-households-surveyed-134890.html
[Reveal] Spoiler: OA
Kudos [?]: 93 [3], given: 0
Intern
Joined: 14 May 2008
Posts: 38
Kudos [?]: 37 [2], given: 0
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
07 Jun 2008, 01:58
2
KUDOS
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
a. 15
b. 20
c. 30
d. 40
e. 45
Let me take a shot ->
out of 200 80 are not using either A or B, so total household of concern = 200-80 = 120
Only Brand A = 60
Only Brand B = 3x
Both A and B = x
So -> 60+x+3x = 120 => x= 15
I Got A as an answer
Kudos [?]: 37 [2], given: 0
Manager
Joined: 09 May 2008
Posts: 98
Kudos [?]: 27 [1], given: 0
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
07 Jun 2008, 06:53
1
KUDOS
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
a. 15
b. 20
c. 30
d. 40
e. 45
total surveyed=200 households
neither Brand A nor Brand B soap = 80
households that use one of the 2 brands atleast=200-80=120
Out of these 120, 60 use only brand A, therefore rest 60 consist of households that use only brand B+households that use both the brands.
ratio of both to only B is 1:3 this means if
households that use both the brands = x, then
households that use only brand B = 3x
60=x+3x
therefore x=15.
Ans is (A)
Kudos [?]: 27 [1], given: 0
Manager
Joined: 31 Jul 2006
Posts: 232
Kudos [?]: 93 [0], given: 0
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
08 Jun 2008, 21:16
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
a. 15
b. 20
c. 30
d. 40
e. 45
Shouldn't the equation be
Total= Brand A-both brands+Brand B+Neither?
200=60-x+3x+80
so x=30 ?( uses both brand a and B)
I've always used this formula...
Kudos [?]: 93 [0], given: 0
Manager
Joined: 09 May 2008
Posts: 98
Kudos [?]: 27 [0], given: 0
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
09 Jun 2008, 10:06
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
a. 15
b. 20
c. 30
d. 40
e. 45
Shouldn't the equation be
Total= Brand A-both brands+Brand B+Neither?
200=60-x+3x+80
so x=30 ?( uses both brand a and B)
I've always used this formula...
the formula is correct but the way its been interpreted is little incorrect - let me clarify:
Total= Brand A-both brands+Brand B+Neither
here, brand A=only brand A+both brands
and brand B=only brand B+both brands
now use this in your formula and instead of getting a "-x" you would end up with a "+x" in your equation and that would make your answer as x=15.
Hope that helps.
Kudos [?]: 27 [0], given: 0
SVP
Joined: 30 Apr 2008
Posts: 1863
Kudos [?]: 623 [3], given: 32
Location: Oklahoma City
Schools: Hard Knocks
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
09 Jun 2008, 10:20
3
KUDOS
See Below:
Attachment:
SurveyResults.jpg [ 23.23 KiB | Viewed 25577 times ]
_________________
------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.
GMAT Club Premium Membership - big benefits and savings
Kudos [?]: 623 [3], given: 32
Manager
Joined: 31 Jul 2006
Posts: 232
Kudos [?]: 93 [0], given: 0
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
10 Jun 2008, 05:10
jallenmorris wrote:
See Below:
Attachment:
SurveyResults.jpg
OA is A. I still don't get it. I thought formula for these types of question should be:
Total=A+B-Both A/B+Neither ???
I'm really confused
Kudos [?]: 93 [0], given: 0
CIO
Joined: 09 Mar 2003
Posts: 460
Kudos [?]: 73 [2], given: 0
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
10 Jun 2008, 05:12
2
KUDOS
1
This post was
BOOKMARKED
I use a completely different system for these kinds of questions. It works every single time, as long as there are two groups (if there are three, I use the venn diagram).
Draw a 3 by 3 box, as in the diagram. All rows add down to the bottom row, and all columns add across to the right column. The bottom right box is the total of the whole system.
Fill in what is given, and then infer what is not.
In my attached example, the black numbers are given, the red ones are what is inferred.
Once finished, in this question, see that the right column adds all the way down and has one variable:
4x + 140 = 200
x = 15
In this case, the boxes with the question marks are irrelevant, so I leave them alone. In other types of questions, you may fill them in.
Try it on other ones. It will never be confusing again.
[Reveal] Spoiler:
Attachment:
grouping box.JPG
Kudos [?]: 73 [2], given: 0
Manager
Joined: 01 Aug 2009
Posts: 52
Kudos [?]: 9 [1], given: 9
Location: Maryland
Schools: Stanford
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
13 Jan 2010, 23:33
1
KUDOS
You should think of these question in a graphical way. So the best way to represent this question is to quickly draw a set.
Look below for the picture. Inside the set we have quantity of ONLY Brand A, ONLY Brand B, BOTH Brands A&B, and NEITHER Brands A&B. The outside number (200 in this case) represents the TOTAL amount.
From your question (and my graph) you can create the following equation:
200 = 60 + x + 3x + 80;
200 = 4x + 140;
60 = 4x
15 = x
As you can see "x" represents the quantity of both A and B brands being used. So the answer is B.
Now, I think that you are having problem because of the English since the statement "for every household that used both brands of soap, 3 used only Brand B soap" means that Brand B was used 3 times more than both A and B by the same household [and not in total].
I hope it helps.
[Reveal] Spoiler:
Attachment:
Set.jpg [ 155.83 KiB | Viewed 28421 times ]
_________________
If the post is helpful, kudos appreciated!
Kudos [?]: 9 [1], given: 9
Math Expert
Joined: 02 Sep 2009
Posts: 42249
Kudos [?]: 132499 [4], given: 12323
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
14 Jun 2010, 10:26
4
KUDOS
Expert's post
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
A. 15
B. 20
C. 30
D. 40
E. 50
Given:
Attachment:
Brands.JPG [ 12.11 KiB | Viewed 25250 times ]
The bold part means that if x used both A and B, then 3x used only B (but not A). So, $$4x+140=200$$ --> $$x=15$$.
_________________
Kudos [?]: 132499 [4], given: 12323
Math Expert
Joined: 02 Sep 2009
Posts: 42249
Kudos [?]: 132499 [2], given: 12323
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
14 Oct 2010, 11:44
2
KUDOS
Expert's post
sivai87 wrote:
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
(A) 15
(B) 20
(C) 30
(D) 40
(E) 45
This is one of the problems in diagnostic test in OG12.
Total # of Soap (A+B) = Total of brand A + Total of Brand B + Neither Brand A nor Brand B - Both.
From the question we have,
(A+B) = 200
Neither Brand A nor Brand B = 80
Only Brand A = 60
Let household # using both the brands be "B" and so the no of house holds using Brand B is 3B.
From the eqn above we have,
200 = 80+60+3B-B
2B = 60, and finally B = 30.
But 30 is wrong... Where am I doing the mistake. Plz help....
( I understand the steps explained in OG. But why this methodology is not working out here. Believe there is no problem with the formula though)
If we do the way you are doing then:
Total=A+B-Both+Neither. Let the # of housholds that use both A and B be $$x$$, then # of households that use only B will be $$3x$$.
Total=200;
A=60+x;
B=x+3x
Both=x;
Neither=80;
$$200=(60+x)+(x+3x)-x+80$$ --> $$x=15$$
Also merging with earlier discussion to see different approaches.
Hope it helps.
_________________
Kudos [?]: 132499 [2], given: 12323
TOEFL Forum Moderator
Joined: 16 Nov 2010
Posts: 1602
Kudos [?]: 599 [0], given: 40
Location: United States (IN)
Concentration: Strategy, Technology
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
15 Mar 2011, 22:10
So 200 - 80 = 120 use at least 1 brand of soap.
120 = only A + only B + both
Only B = B - both
and only B = 3 * both
So 120 = 60 + 4 * both
=> both = 60/4 = 15
_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)
GMAT Club Premium Membership - big benefits and savings
Kudos [?]: 599 [0], given: 40
Manager
Joined: 30 May 2008
Posts: 74
Kudos [?]: 146 [0], given: 26
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
13 Apr 2012, 17:17
the formula is correct but the way its been interpreted is little incorrect - let me clarify:
Total= Brand A-both brands+Brand B+Neither
here, brand A=only brand A+both brands
and brand B=only brand B+both brands
now use this in your formula and instead of getting a "-x" you would end up with a "+x" in your equation and that would make your answer as x=15.
Hope that helps.[/quote]
When using the formula Total= Brand A + Brand B- both brands +Neither, brand A and brand b have to be double counted. In the sense that each needs to include those that only use their brand and those that use both. Hence for this question, we are given "60 used ONLY brand A" and "for every household that used both brands, 3 used ONLY brand B" there is no double counting. Thus, we should tweak the formula to be Total= Brand A + Brand B + both brands +Neither
Kudos [?]: 146 [0], given: 26
Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 222
Kudos [?]: 63 [0], given: 16
Schools: Johnson '15
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
14 Apr 2012, 00:33
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
A) 15
B) 20
C) 30
D) 40
E) 45
Official Guide 12 Question
Question: 6 Page: 21 Difficulty: 700
Find All Official Guide Questions
Video Explanations:
missed certain key values and got B...now i realize why the correct answer is A..its a straight calculation
_________________
Regards,
Harsha
Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat
Satyameva Jayate - Truth alone triumphs
Kudos [?]: 63 [0], given: 16
Intern
Joined: 06 Dec 2012
Posts: 6
Kudos [?]: 3 [0], given: 5
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
06 Dec 2012, 14:36
aviator83 wrote:
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
a. 15
b. 20
c. 30
d. 40
e. 45
Shouldn't the equation be
Total= Brand A-both brands+Brand B+Neither?
200=60-x+3x+80
so x=30 ?( uses both brand a and B)
I've always used this formula...
the formula is correct but the way its been interpreted is little incorrect - let me clarify:
Total= Brand A-both brands+Brand B+Neither
here, brand A=only brand A+both brands
and brand B=only brand B+both brands
now use this in your formula and instead of getting a "-x" you would end up with a "+x" in your equation and that would make your answer as x=15.
Hope that helps.
I've always used this formula too, and it does work. However this problem is worded a little weird and I misinterpreted it too, the way you setup the equation is wrong, mainly because like me, you didn't understand the ""and for every household that used both brands of soap, 3 used only Brand B soap" properly, this represents B - AB = 3(AB) not B = 3(AB)
Total = A + B + N - AB can be expanded to:
Total = (A - AB) + (B - AB) + N + AB
200 = 60 + 3AB + 80 + AB
200 = 60 + 80 + 4AB
60 = 4AB
15 = AB
or the way you set it up using x
200=60 + 4x +80
I think like most problems there's more than a few ways to get it wrong. Took me a LONG time to figure out where I went wrong, hope this clarifies the one way I got it wrong based on that formula.
Kudos [?]: 3 [0], given: 5
Non-Human User
Joined: 09 Sep 2013
Posts: 15705
Kudos [?]: 281 [0], given: 0
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
05 Oct 2014, 04:51
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Kudos [?]: 281 [0], given: 0
Intern
Joined: 15 Oct 2014
Posts: 1
Kudos [?]: 1 [0], given: 1
Re: A marketing firm determined that, of 200 households [#permalink]
### Show Tags
15 Dec 2014, 06:30
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
A) 15
B) 20
C) 30
D) 40
E) 45
Official Guide 12 Question
Question: 6 Page: 21 Difficulty: 700
Find All Official Guide Questions
Video Explanations:
out of 200, 80 didnt use brand A or B => 200-80=120
60 used brand A => 120-60=60
in this 60 ppl, 3 out of 4 used brand B.
60/4 = 15
Kudos [?]: 1 [0], given: 1
Re: A marketing firm determined that, of 200 households [#permalink] 15 Dec 2014, 06:30
Display posts from previous: Sort by
# A marketing firm determined that, of 200 households
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 5,013 | 16,961 | {"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} | 3.34375 | 3 | CC-MAIN-2017-47 | latest | en | 0.878939 |
http://terrytao.wordpress.com/2009/10/29/reading-seminar-3-stable-group-theory-and-approximate-subgroups-by-ehud-hrushovski/ | 1,419,303,388,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802778013.38/warc/CC-MAIN-20141217075258-00175-ip-10-231-17-201.ec2.internal.warc.gz | 279,973,133 | 43,615 | This week, Henry Towsner continued some model-theoretic preliminaries for the reading seminar of the Hrushovski paper, particularly regarding the behaviour of wide types, leading up to the main model-theoretic theorem (Theorem 3.4 of Hrushovski) which in turn implies the various combinatorial applications (such as Corollary 1.2 of Hrushovski). Henry’s notes can be found here.
A key theme here is the phenomenon that any pair of large sets contained inside a definable set of finite measure (such as ${X \cdot X^{-1}}$) must intersect if they are sufficiently “generic”; the notion of a wide type is designed, in part, to capture this notion of genericity.
— 1. Global types —
Throughout this post, we begin with a countable structure ${M}$ of a language ${L}$, and then consider a universal elementary extension ${{\Bbb U}}$ of ${M}$ (i.e. one that obeys the saturation and homogeneity properties as discussed in Notes 1. Later on, ${L}$ will contain the language of groups, and then we will rename ${{\Bbb U}}$ as ${G}$ to emphasise this.
Recall from Notes 1 that a partial type over a set ${A}$ is a set of formulae (with ${n}$ variables for some fixed ${n}$) using ${A}$ as constant symbols, which is consistent and contains the theory of ${M}$; if the set of formulae is maximal (i.e. complete), then it is a type. One can also think of a type as an ultrafilter over the ${A}$-definable sets; if ${p}$ is a type and ${B}$ is an ${A}$-definable set given by some formula ${\phi}$, then either ${\phi}$ lies in ${p}$ (in which case we write ${p \subset B}$) or ${\neg \phi}$ lies in ${p}$ (in which case we write ${p \subset \overline{B}}$) but not both.
When the set ${A}$ is small (i.e. has cardinality less than that of ${{\Bbb U}}$, which in particular would be true of ${A}$ consisted of ${M}$ union with a finite set, which is a very typical situation), then by saturation one can identify types (or partial types) with the subset ${p({\Bbb U})}$ of ${{\Bbb U}^n}$ that they cut out. In particular, these sets are non-empty. Adding more formulae to a partial type corresponds to shrinking the set that they cut out, and vice versa.
However, if we have a global type ${p}$ – a type defined over the entire model ${{\Bbb U}}$ – then one can no longer identify types with the set ${p({\Bbb U})}$ that they cut out, because these sets are usually empty! However, what we can do is restrict ${p}$ to some smaller set ${A}$ of constants to create a type ${p\downharpoonright_A}$ over ${A}$, defined as the set of all formulae in ${p}$ that only involve the constants in ${A}$. It is easy to see that this is still a type, and if ${A}$ is small, it cuts out a non-empty set in ${{\Bbb U}^n}$.
A global type ${p}$ is said to be ${M}$-invariant, or invariant for short, if the set of formulae in ${p}$ is invariant under any automorphism ${\sigma}$ of ${{\Bbb U}}$ that fixes ${M}$. In particular, given any ${M}$-definable set ${A \subset {\Bbb U}^n \times {\Bbb U}^m}$ and ${b \in {\Bbb U}^m}$, we see that ${p \subset A_b}$ if and only if ${p \subset A_{\sigma(b)}}$, where ${A_b := \{ a \in {\Bbb U}^n: (a,b) \in A \}}$ is a slice of ${A}$. (Indeed, this gives an equivalent definition of invariance.)
A trivial example of an invariant global type would be the type of an element ${m \in M}$ (or in ${M^n}$). This cuts out a singleton set ${\{m\}}$. This is in fact the only invariant global type that cuts out anything at all:
Lemma 1 Let ${p}$ be a global invariant type. Then ${p}$ is realisable in ${{\Bbb U}}$ (i.e. ${p({\Bbb U})}$ is non-empty) if and only if it is realisable in ${M}$ (and is the type of a single element in ${M}$).
Proof: Suppose ${p}$ is realisable in ${{\Bbb U}}$ by some ${a \in p({\Bbb U})}$. Since the formula ${x=a}$ is definable in ${{\Bbb U}}$, we see that ${p \subset \{ x: x = a \}}$, i.e. ${p}$ cuts out precisely the singleton set ${\{a\}}$. As ${p}$ is invariant, ${\{a\}}$ must then be invariant under all ${M}$-fixing automorphisms of ${{\Bbb U}}$. By homogeneity, this means that there is no element distinct from ${a}$ which is elementarily indistinguishable from ${a}$ over ${M}$; in other words, ${\{a\}}$ is the set cut out by the type ${tp(a/M)}$ of ${a}$ over ${M}$.
By saturation, the formula ${x \neq a}$ together with the formulae in ${tp(a/M)}$ is not satisfiable, hence not finitely satisfiable. Thus there is a finite set of formulae in ${tp(a/M)}$ that cut out ${\{a\}}$, i.e. ${\{a\}}$ is definable over ${M}$. But as ${{\Bbb U}}$ is an elementary extension of ${M}$, these formulae must also be realisable in ${M}$, i.e. ${a}$ lies in ${M}$, and the claim follows. $\Box$
(Because of this, one should regard the notation ${p \subset B}$ carefully; the set ${p({\Bbb U})}$ that ${p}$ cuts out in the model ${{\Bbb U}}$ may in fact be empty, but when we write ${p \subset B}$ for some definable ${B}$, we interpret this syntactically rather than semantically (or equivalently, that ${p({\Bbb U}') \subset B({\Bbb U}')}$ holds in all extensions ${{\Bbb U}'}$ of ${{\Bbb U}}$, and not just in ${{\Bbb U}}$ itself.)
On the other hand, invariant global types exist in abundance:
Lemma 2 Let ${p}$ be a type over ${M}$. Then there exists an invariant global type ${q}$ that refines ${p}$ (i.e. it contains all the formulae that ${p}$ does).
Proof: We view ${p}$ as a collection of logically consistent formulae. We enlarge this collection to a larger one ${p'}$ by adding in the negations of all the formulae definable over ${{\Bbb U}}$ that are not realisable in ${M}$. Observe that this collection remains logically consistent, because any finite set of formulae in ${p}$ were realisable in ${{\Bbb U}}$, hence in ${M}$ (which is an elementary substructure). Hence, by Zorn’s lemma, one can extend ${p'}$ to a global type ${q}$.
We now claim that ${q}$ is invariant. Indeed, let ${\phi}$ be a sentence over ${{\Bbb U}}$ that is contained in ${q}$, and let ${\sigma}$ be an automorphism that fixes ${M}$. If ${\sigma(\phi)}$ is not in ${q}$, then ${\neg \sigma(\phi)}$ must be in ${q}$ (by completeness), and hence ${\phi \wedge \neg \sigma(\phi)}$ is in ${q}$ also, and hence must be realisable in ${M}$ (otherwise its negation would be in ${p'}$, and hence in ${q}$). But this is absurd since ${\sigma}$ fixes ${M}$. Thus ${\sigma(\phi)}$ does lie in ${q}$, yielding invariance.
A major use of invariant global types for us will be that they can be used to generate sequences of indiscernibles (as defined in previous notes):
Lemma 3 Let ${p}$ be a global invariant type of some arity ${d}$, and construct recursively a sequence ${b_1, b_2, \ldots \in {\Bbb U}^d}$ by requiring ${b_n \in p\downharpoonright_{M \cup \{b_1,\ldots,b_{n-1}\}}({\Bbb U})}$ for all ${n=1,2,\ldots}$. (This is always possible since types are satisfiable once restricted to small sets, by saturation, as discussed earlier). Then ${b_1,b_2,\ldots}$ are indiscernible over ${M}$, i.e. the tuples ${(b_{i_1},\ldots,b_{i_k})}$ for ${i_1 < \ldots < i_k}$ are elementarily indistinguishable (over ${M}$) for any fixed ${k}$.
Proof: This is achieved by an induction on ${k}$. The ${k=1}$ case is clear since the ${b_n}$ all have type ${p\downharpoonright_M}$ over ${M}$. Now we do the ${k=2}$ case. It suffices to show that ${(b_1,b_2)}$ and ${(b_i,b_j)}$ are elementarily indistinguishable over ${M}$ for all ${i < j}$.
By construction, ${b_2}$ and ${b_j}$ have the same type over ${M \cup \{b_1\}}$, and so ${(b_1,b_2)}$ and ${(b_1,b_j)}$ are elementarily indistinguishable over ${M}$. So it remains to show that ${(b_1,b_j)}$ and ${(b_i,b_j)}$ are elementarily indistinguishable.
Let ${A}$ be an ${M}$-definable relation that contains ${(b_1,b_j)}$; we need to show that ${A}$ contains ${(b_i,b_j)}$ also.
Since ${b_1}$ and ${b_i}$ have the same type over ${M}$, by homogeneity there exists an automorphism ${\sigma}$ of ${{\Bbb U}}$ fixing ${M}$ that maps ${b_1}$ to ${b_i}$. Since ${b_j}$ realises ${p\downharpoonright_{M \cup \{b_1\}}}$, we see that ${p}$ contains the sentence ${(b_1,x) \in A}$, hence by invariance ${p}$ contains ${(b_i,x) \in A}$ also. Since ${b_j}$ realises ${p\downharpoonright_{M \cup \{b_i\}}}$, we conclude ${(b_i,b_j) \in A}$, as required.
This concludes the ${k=2}$ case. The higher ${k}$ case is similar and is left as an exercise. $\Box$
— 2. Intersections of wide types —
Now we assume that the structure ${{\Bbb U}}$ is equipped with an ${M}$-invariant Kiesler measure ${\mu}$. This leads to the notion of a wide type – a type such that all the ${{\Bbb U}}$-definable sets containing this type have positive measure. Intuitively, elements of a wide type are distributed “generically” in the structure.
In the previous notes we showed that wide types can be “split” amongst indiscernables, as follows:
Lemma 4 Let ${b}$ be an element or tuple in ${{\Bbb U}}$, let ${a}$ be a wide type over ${M \cup \{b\}}$ for some set of constants ${M}$, and let ${b_1,b_2,\ldots}$ be a sequence of indiscernibles (over ${M}$) that has the same type as ${b}$ (over ${M}$). Then for any finite number ${b_1,\ldots,b_n}$ in this sequence, one can find a type ${a'}$ such that ${a'}$ has the same type over ${b_i}$ as ${a}$ does over ${b}$, for all ${1 \leq i \leq n}$.
We now use this lemma to show that sets defined by wide types intersect each other in a uniform fashion.
Lemma 5 Let ${p,q}$ be types over ${M}$, and let ${a,a' \in p({\Bbb U})}$, ${b,b' \in q({\Bbb U})}$ be realisations of ${a,b}$ such that ${tp(a/M \cup \{b\})}$ and ${tp(a'/M \cup \{b'\})}$ are wide. Let ${A_x}$, ${B_y}$ be ${M}$-definable sets with parameters, contained inside an ${M}$-definable set ${X}$ of finite measure; then ${\mu(A_a \cap B_b) > 0}$ if and only if ${\mu(A_{a'} \cap B_{b'}) > 0}$.
Proof: By homogeneity, there is an automorphism fixing ${M}$ that sends ${b}$ to ${b'}$, and maps ${a}$ to another element of ${p({\Bbb U})}$. Thus without loss of generality we may assume ${b=b'}$.
We assume for contradiction that ${\mu(A_a \cap B_b) > \delta > 0}$ and ${\mu(A_{a'} \cap B_{b'}) = 0}$.
By Lemma 2, we may extend ${q}$ to an invariant global type ${q'}$. Observe that for any ${\epsilon > 0}$, either one has ${\mu(A_a \cap B_x) \geq \epsilon}$ for all ${x \in q'\downharpoonright_{M \cup \{a\}}}$, or one has ${\mu(A_a \cap B_x) \leq \epsilon}$ for all ${x \in q'\downharpoonright_{M \cup \{a\}}}$ (since there is a ${M \cup \{a\}}$-definable set between ${\{ x: \mu(A_a \cap B_x) \geq \epsilon \}}$ and ${\{ x: \mu(A_a \cap B_x) > \epsilon \}}$. Suppose first that the former option holds for some ${\epsilon}$, thus there is a uniform lower bound ${\mu(A_a \cap B_x) > \epsilon}$. We now define a sequence ${a_1,a_2,\ldots \in p({\Bbb U})}$ and an indiscernible sequence ${b_1,b_2,\ldots \in q({\Bbb U})}$ as follows:
• We initialise ${a_1=a}$ and ${b_1}$ to be a realisation of ${q'\downharpoonright_{M \cup \{a_1\}}}$.
• Now suppose that ${a_1,\ldots,a_n,b_1,\ldots,b_n}$ have been chosen with ${b_1,\ldots,b_n}$ indiscernible. By Lemma 4, we can find ${a_{n+1} \in p({\Bbb U})}$ that has the same type over ${b_i}$ that ${a'}$ has over ${b'}$ for all ${1 \leq i \leq n}$. Since ${\mu(A_{a'} \cap B_{b'}) = 0}$, this implies that ${\mu( A_{a_{n+1}} \cap B_{b_i} ) = 0}$ for all ${1 \leq i \leq n}$. (Here we use the fact that ${\mu( A_x \cap B_b )=0}$ is a type-definable formula over ${M}$ and ${b}$.)
• Now, let ${b_{n+1}}$ be a realisation of ${q'\downharpoonright_{M \cup \{a_1,\ldots,a_{n+1},b_1,\ldots,b_n\}}}$. With this construction and Lemma 3 we see by induction that ${b_1,\ldots,b_{n+1}}$ is also indiscernible; now we iterate the procedure.
Let ${C_i}$ be the set ${C_i := A_{a_i} \cap B_{b_i}}$, then observe from the above construction that ${C_i}$ lies in ${X}$ and ${\mu(C_i \cap C_j) = 0}$ for all ${i < j}$. On the other hand, we claim that ${\mu(C_i)}$ is uniformly bounded away from zero, this contradicts the finite measure of ${\mu(X)}$ by the pigeonhole principle.
To see the uniform lower bound, find an automorphism ${\sigma_i}$ fixing ${M}$ that maps ${a_1}$ to ${a_i}$. By hypothesis, ${\mu(A_{a_1} \cap B_{b_1}) > 0}$, thus there exists a rational ${r > 0}$ such that the predicate that models ${\mu(A_{a_1} \cap B_x) > r}$ is in ${\tilde q}$, hence in ${q'}$. By invariance, the predicate ${\mu(A_{a_i} \cap B_x) > r}$ is in ${q'}$ also, hence by construction of ${b_i}$, ${\mu(A_{a_i} \cap B_{b_i}) > r}$, and the claim follows.
Now we consider the opposite case, in which ${\mu(A_a \cap B_x) = 0}$ for all ${x \in q'\downharpoonright_{M \cup \{a\}}}$. Then we run the construction slightly differently: for each ${n}$ in turn, set ${b_n}$ to be a realisation of ${q'\downharpoonright_{M \cup \{a_1,\ldots,a_{n-1},b_1,\ldots,b_{n-1}}}$, then set ${a_n \in p({\Bbb U})}$ so that ${\mu(A_{a_n} \cap B_{b_n}) > \delta}$. (This is possible because for any definable set ${C}$ containing ${p}$, the ${\bigwedge}$-definable set ${\{ x \in C: \mu(A_x \cap B_b) \geq \delta \}}$ contains ${p}$ and thus has positive measure, and so the same is true for ${b_n}$; now use saturation.) Then again we see that the ${C_i}$ lie in ${X}$, have intersection of measure zero, and have measure uniformly bounded from below, and we again obtain a contradiction. $\Box$
Now we place a group structure on ${{\Bbb U}}$, and obtain a variant of the above result:
Proposition 6 Let ${p, q}$ be types in ${M}$, with ${p}$ wide. Suppose that ${p}$, ${q}$ are contained in ${M}$-definable sets ${X, X'}$ such that ${XX'}$ has finite measure. Let ${a,b \in q}$ be such that the type of ${a}$ over ${M \cup \{b\}}$ is wide. Assume also that the Keisler measure is translation-invariant. Then ${pa \cap pb}$ is also wide.
Proof: Suppose this is not the case, so that there exists an ${M}$-definable set ${A}$ containing ${p}$ such that ${Aa \cap Ab}$ has zero measure. (Initially, one would need two different definable sets containing ${p}$, but one can simply take their intersection.) On the other hand, as ${p}$ is wide, ${A}$ itself has positive measure. We can place ${A}$ in ${X}$.
By using the fact that wide types over one set of constants can be refined to wide types over larger sets of constants (Lemma 2 from the previous notes), we see that we can recursively construct a sequence ${a_1, a_2, \ldots \in q({\Bbb U})}$ with ${tp(a_n/M \cup \{a_1,\ldots,a_{n-1}\})}$ wide for all ${n}$. Since ${\mu(Aa \cap Ab) = 0}$, we conclude from Lemma 5 that ${\mu(A a_n \cap A a_i)=0}$ for all ${1 \leq i < n}$. On the other hand, the ${\mu(Aa_i)}$ all have the same measure as ${\mu(A)}$, which is positive. Finally, the ${Aa_i}$ are all contained in ${XX'}$, which has finite measure; this leads to a contradictoin. $\Box$
This “generic intersection” property of translates of ${p}$ will be important in later arguments when creating near-groups. | 4,688 | 14,838 | {"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": 337, "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} | 2.96875 | 3 | CC-MAIN-2014-52 | latest | en | 0.940914 |
https://eng.kakprosto.ru/how-868074-how-to-play-card-nine | 1,611,185,520,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00098.warc.gz | 333,896,050 | 11,850 | Instruction
1
Before you play "Nine" to distribute cards to all participants of the table battle. Now everyone needs to sort their cards by suit. It will be easier if all suits still to sort, in descending order, from ACE to six. If someone went to the four aces, the cards are shuffled and laid out again. By agreement, this can be done with three aces.
2
The course begins someone who turned out to be nine of diamonds. It is placed in the center of the table. Next turn makes the person sitting on the left hand from the player starting the round. He can put ten or eight of diamonds, respectively, to the left or to the right of the nine. In the absence of these cards, place the nine of any suit.
3
The outcome should be formed 4 rows. Each certain color. On one side of the range will be six – ACE.
4
Before you put the card, it is necessary to devise a strategy. If you have an ACE, you will need to promote the suit, which he is. Let's say you have a red ACE, then put the red nine. Then the following participants can put ten, then knave and the result of the king of hearts. You get rid of the ACE and the chance to win you will increase.
5
Players lay cards in turn clockwise. If they don't have cards that will help them to get rid of the aces, they will be forced to put you need. After the king you lay my ACE.
6
Strategy is not only faster to put the highest card. Sometimes you can lose and sixes. So try to advance the suit, which you have six. But, if you hesitate and don't know whether to believe you nine of the suit, which your ACE or 9 for the nomination of six – give preference to the first. After all, from nine to ACE 5, and from it to six – 4 speed.
7
It may happen that the player had nothing to put, then he "goes" and the turn passes to the next. If you pass the course, the chance to play high. In the card game "Nine" there are no second and third place. Wins here one who got a good hand, and who was able to correctly calculate the strategy.
8
Sometimes, some up addition to the game "Nine". After the cards are dealt, all put on the line for 10 cents. When the participant passes the course, he also needs to put into the General Fund 10 cents.
9
The winner takes a cash prize, and then, he is forced to deal the cards. The winner shuffles them and gives off part of neighbor. Then each gives the same number of cards, handing one. | 570 | 2,367 | {"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} | 2.53125 | 3 | CC-MAIN-2021-04 | longest | en | 0.962466 |
https://what.wqidian.com/category/24-hour-format/ | 1,723,277,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00166.warc.gz | 504,179,150 | 63,571 | ## What Time is 16 10
What time is 16 10?” It’s a question that might leave you scratching your head if you’re not familiar with military time. But fear not, it simply means 4:10 PM. Whether you’re a frequent traveler or just curious about different time systems, understanding military time can come in handy. So next time someone asks you what time is 16 10, you’ll know the answer without missing a beat.
## What Time is 20:35
As the day comes to a close and the clock ticks towards 8:35 PM, one might wonder, “What time is 20:35?” In the world of military time, 20:35 is simply the 24-hour equivalent of 8:35 PM in standard time. So sit back, relax, and enjoy the last few hours of the day!
## What Time is 22 10
As the clock ticks past ten in the evening, confusion sets in. What time is 22 10? Fear not, dear reader, for this is simply the military time equivalent of 10:10 PM. Although it may seem daunting, mastering the art of reading military time can be a valuable skill to have in the world of international travel and communication.
## What Time is 13 45
As we navigate through our daily routines, it’s rare for us to come across the need to convert time in the 24-hour format. But if you ever find yourself asking, what time is 13:45? Don’t fret. This is simply time translated from military time to standard time. In layman’s terms, it’s 1:45 PM.
## What Time is 18 35
What time is 18 35?” may seem like a perplexing question to some, but it’s simply another way of expressing the 24-hour clock time for 6:35 PM. While it may take a bit of mental math to decipher, understanding the 24-hour clock can be a useful tool for both international travel and military/time-sensitive professions.
## What Time is 14 45
What time is 14 45? For those accustomed to the 12-hour clock, it may require a moment of mental gymnastics to translate. But fear not, the answer is simple: 14 45 is equivalent to 2:45 PM in the 24-hour clock system. Remember, practice makes perfect when it comes to telling time! | 506 | 2,018 | {"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-2024-33 | latest | en | 0.915442 |
http://blog.rutwick.com/library/higher-order-contact-of-submanifolds-of-homogeneous-spaces | 1,547,764,011,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659340.1/warc/CC-MAIN-20190117204652-20190117230652-00227.warc.gz | 33,009,211 | 9,432 | # Download Higher Order Contact of Submanifolds of Homogeneous Spaces by G. R. Jensen PDF
By G. R. Jensen
Best calculus books
Calculus I with Precalculus, A One-Year Course, 3rd Edition
CALCULUS I WITH PRECALCULUS, brings you up to the mark algebraically inside precalculus and transition into calculus. The Larson Calculus application has been largely praised by means of a iteration of scholars and professors for its sturdy and powerful pedagogy that addresses the wishes of a large variety of educating and studying types and environments.
An introduction to complex function theory
This booklet offers a rigorous but hassle-free creation to the idea of analytic services of a unmarried complicated variable. whereas presupposing in its readership a level of mathematical adulthood, it insists on no formal must haves past a valid wisdom of calculus. ranging from simple definitions, the textual content slowly and punctiliously develops the tips of advanced research to the purpose the place such landmarks of the topic as Cauchy's theorem, the Riemann mapping theorem, and the concept of Mittag-Leffler could be handled with out sidestepping any problems with rigor.
A Course on Integration Theory: including more than 150 exercises with detailed answers
This textbook offers an in depth therapy of summary integration concept, building of the Lebesgue degree through the Riesz-Markov Theorem and in addition through the Carathéodory Theorem. it's also a few ordinary homes of Hausdorff measures in addition to the fundamental houses of areas of integrable features and traditional theorems on integrals counting on a parameter.
Extra info for Higher Order Contact of Submanifolds of Homogeneous Spaces
Sample text
A subset A of R is closed if and only if whenever an ∈ A for all n ∈ N, and an → c, we also have c ∈ A. 2. Compact sets 41 Proof. ⇒ Say an ∈ A, n ∈ N, and an → c. If c ∈ / A, then, since R \ A is open, R \ A is a neighbourhood of c, so an ∈ R \ A for all but finitely many n, which is absurd. ⇐ We prove the contrapositive. Suppose A is not closed, that is, R \ A is not open. This means that there is c ∈ R \ A such that R \ A is not a neighbourhood of c. Thus, for each n ∈ N, R \ A does not contain the 1 n -neighbourhood of c, so there is an ∈ A in this neighbourhood, that is, |an − c| < n1 .
B − 1} with ∞ a ∞ c � � n n = . Prove that, after possibly interchanging (an ) and (cn ), n n b b n=1 n=1 there is m ∈ N such that an = cn for n < m, am = cm + 1, and an = 0 and cn = b − 1 for n > m. 34. (a) Let (xn ) be a bounded sequence. For each k ∈ N, let yk = sup xn = sup{xk , xk+1 , xk+2 , . . }. n≥k Show that the sequence (yk ) is decreasing and bounded below. Conclude that (yk ) converges. The limit of (yk ) is called the limit superior of the original sequence (xn ). In other words, lim sup xn = lim sup xn .
Theorem (squeeze theorem). If an → s, cn → s, and an ≤ bn ≤ cn for all but finitely many n ∈ N, then bn → s. Proof. Let � > 0 and I = (s − �, s + �). By assumption, for n sufficiently large, an ∈ I. Also, for n sufficiently large, cn ∈ I. Hence, since I is an interval and bn lies between an and cn for n sufficiently large, we have bn ∈ I for n sufficiently large. This shows that bn → s. 11. Theorem (algebraic limit theorem). If an → a and bn → b, then: (1) can → ca for all c ∈ R. (2) an + bn → a + b. (3) an bn → ab. | 923 | 3,369 | {"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} | 3.21875 | 3 | CC-MAIN-2019-04 | latest | en | 0.926738 |
https://math.stackexchange.com/questions/2300322/is-the-space-of-probability-measures-second-category | 1,582,972,542,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00253.warc.gz | 463,136,971 | 32,466 | # Is the space of probability measures second category?
The space $\mathcal{M}$ of signed measures on a measurable space $(\Omega, \mathcal{F})$ is a Banach space when equipped with the total variation norm. So $(\mathcal{M},d)$ is a complete metric space, where $d$ is the total variation metric.
I thought there should be a very quick argument to show that the space of probability measures on $(\Omega, \mathcal{F})$ is of second category in the total variation topology using the Baire category theorem, but somehow I don't see how it follows. The set of probabilities isn't open, so I can't conclude immediately that it's second category. Is there some simple way to show this that I'm just missing, or does it actually require some work?
• Are you trying to show the space of probability measures has second category in itself, or as a subset of $\mathcal{M}$? – Eric Wofsey May 28 '17 at 16:59
• If the space of probability measures on $(\Omega, \mathcal{F})$ is a $G_\delta$ subset (in particular, closed) of $(\mathcal{M},d)$ then it is completely metrizable (for instance, by Theorem 4.3.23 from Engelking's “General Topology”) and therefore Baire and second category in itself. – Alex Ravsky May 28 '17 at 17:00
• @EricWofsey In itself. Ah I see, I guess my remark about the set not being open is beside the point. Though I guess I'd be interested in the answer to both questions. – grndl May 28 '17 at 17:06
The set of probability measures is closed in $\mathcal{M}$. Indeed, it is just the set of nonnegative measures of norm $1$. The set of nonnegative measures is closed (since evaluation of a measure on any measurable set is continuous), as is the set of measures of norm $1$. So the space of probability measures is complete with respect to the metric $d$, and hence has second category in itself. | 461 | 1,819 | {"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} | 3.28125 | 3 | CC-MAIN-2020-10 | latest | en | 0.949448 |
https://physics.stackexchange.com/questions/288852/schr%c3%b6dinger-equation-1-d-particle-in-a-box | 1,660,208,940,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00346.warc.gz | 437,085,958 | 65,733 | # Schrödinger Equation 1-D particle in a box
1. For the 1D particle-in-a-box example, how do we determine the weights of each eigenfunction in the general (time-dependent) solution that fully describes the system? Do we need extra initial conditions for this and if so, what might these conditions be and how might we measure/determine them?
2. Also, doesn't each eigenstate have constant energy by definition? If so, then shouldn't the momentum of the particle in an eigenstate have a constant magnitude? According to Wikipedia, momentum for the 1D particle-in-a-box has a distribution: https://en.wikipedia.org/wiki/Particle_in_a_box#Position_and_momentum_probability_distributions
3. Finally, according to QM, $\langle \psi|O|\psi\rangle$ can be used to determine the expectation of observable O. Does QM similarly provide a way to determine the distribution of the observable? If not, it seems the theory is rather incomplete
Edit
Thanks for the answers all! A couple edits
1. Assume psi(0) is unknown, which I'd imagine would generally be the case in an experiment. What would we then expect the general wavefunction to look like? Are all possible coefficient vectors (c1,c2,...,cn) equally likely in practice, or are some combinations more likely than others?
2. Momentum of particle in a box explains it - QM is incompatible with de Broglie's matter wave theory apparently. Good to know. Also although energy states are quantized, this does not appear to be the case for kinetic energy.
3. symanzik138 and freecharly answered this one; it makes sense now.
• I would like to answer your question two in a short manner, that the system you're looking at here is not a free particle system (the domain is only 0 to L), so the energy and momentum relationship is a non trivial one. Oct 26, 2016 at 7:05
• Oct 26, 2016 at 8:45
1) A time dependent state can be written as a superposition of the eigenstates of the Hamiltonian. I am assuming that the Hamiltonian is a constant one; does not change with time. The coefficients or weights of the eigenstates are complex. so, for any state $\psi$,
$\psi(t) = c_1 \phi_1 + c_2 \phi_2 + \dots$
where $\phi_i$ are the normalized eigenstates of the Hamiltonian. The coefficients
$c_i(t) = \langle \phi_i|\psi(t) \rangle$
are time dependent.
And the time dependence has the precise form:
$c_i(t)=\exp(-\imath E_i t/\hbar) c_i(0)$
where $E_i$ is the energy of the state $\phi_i$. Thus given the initial coefficients $c_i(0)$, you can determine the coefficients $c_i(t)$ and therefore the state at any other instant $t$. Equivalently you can take $\psi(0)$ also as the initial condition, as $c_i(0)$ can be inferred from $c_i(0)=\langle \phi_i|\psi(0) \rangle$.
2) Constant energy does not mean a constant momentum. This is because the box is not translation invariant. However, the magnitude of the momentum is a constant (in the case of the simple 1D quantum well. This is not true in general). You can in fact write the energy eigenstate as a linear combination of states of momenta $p$ and $-p$. So momentum distribution is nonzero at $\pm p$ only.
3) Distribution of an observable can be interpreted as follows. The probability of finding a value of $o$ for a measurement (say position or momentum) on a system in $\psi$. Such measurements are associated with a corresponding operator $O$. $p(o) = |\langle\psi | \phi_o\rangle|^2$ if $o$ is an eigenvalue of $O$, and $p(o)=0$ otherwise. $\phi_o$ is the normalized eigenstate of $O$ withe eigenvalue $o$. (I am assume no degeneracy).
Not an expert by any means, as I am also just beginning to learn quantum mechanics... But I will share my thoughts, and hopefully others can correct me if I am wrong.
(1) To obtain time dependent solutions for the particle in the 1-d well, we can simply tack on a time dependent function to the spatial (time independent) solution. The temporal (time) function still depends on the energy levels.
For the 1-d well, the solutions are:
$$\psi(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$
We can tack on the temporal function: $e^{\frac{-iE_n t}{\hbar}}$
But the full solution to $\Psi(x,t)$ is a superposition of states.
So $$\Psi(x,t) = \sum_{n=0}^\infty a_n \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)e^{\frac{-iE_n t}{\hbar}}$$
And we can obtain the $a_n$ ("probability amplitudes") from the normalization condition:
$$\int_{-\infty}^\infty \mid \Psi(x,t) \mid^2 dx = 1 = \sum_{n=1}^\infty\mid a_n\mid^2$$
So in order to do this, we must be given initial conditions such as the behavior of $\Psi(x,t)$ at $t=0$. From the $\Psi(x,0)$ function we can find the $a_n$ and find how it evolves in time.
(2) Each eigenstate has quantized energy eigenvalues. The energy of the system is well defined but that does not mean the momentum is constant. Quantum mechanics is a probabilistic theory, and the momentum of a particle in a confined space is a probabilistic distribution, as is the position. You'll learn about Heisenberg's uncertainty principle, where the basic idea is that we cannot know anything about a particles momentum if it's position is completely known. In contrast, if the momentum is completely known, the position of the particle can be thought of as being spread out through all space; aka, we have infinite ignorance about the particle's position.
(3) Not entirely sure what is being asked here, but if you're asking if there is a way to determine an expectation value, then yes.
The 'distribution of the observable' can be evaluated as the expectation value of an operator, such as $x, \hat{p}, \hat{\mathcal{H}}$
Then the expectation value of $x$ will be:
$$\langle x \rangle(t) = \int_{-\infty}^\infty x~\Psi^*(x,t)\Psi(x,t)$$
This tells us the expected position (probabilistic) as a function of time, which is part of a distribution - all contained within the wave function. Essentially, the expectation value follows this probabilistic distribution.
All information about a particle is contained in its wave function $\Psi(x,t)$, but the only measurements of an operator which we can actually observe is one of its eigenvalues.
Edit: (1) There are some instances when the coefficients are all zero aside from a set number of states, such as $$\Psi(x,t) = \sum_{n=1}^2 \frac{1}{\sqrt{L}}\sin\left(\frac{n\pi x}{L}\right)e^{\frac{-iE_n t}{\hbar}}$$ If $\Psi(x,0)$ was completely unknown, we cannot determine the time evolution of the wave function without first measuring a state. The time dependent Schrödinger equation is an initial value partial differential requiring us to specify a state, usually $t = 0$. But perhaps the time independent equations could be of use. It would be easy to assume the form of the wave function as a free particle: $\psi(x) = Ae^{ikx}$. Otherwise, the form of the wave function is dependent on the potential containing it (inf. sq. well, harmonic, finite well, etc).
• You don't get the $a_n$'s from normalization. You get them from integrating $\Psi(x,0)$ against your basis functions (in this case, the energy eigenfunctions). As far as (3): the expectation value is not the same as the "distribution of the observable". To get that distribution, you have to expand the wave function in the eigenbasis of the corresponding observable. For instance $\Psi(x)$ is already expanded in the position basis, and so $|\Psi(x)|^2$ is the distribution for $x$. Your $a_n$'s compose the distribution for $\hat{H}$. And so on. Oct 25, 2016 at 23:27
• Sorry yes I was a bit vague here. But the $a_n$ should arise from the normalization condition for the 1-d infinite well. According to the orthonormality relation, $$\int_{0}^L \psi_{n}^*(x)\psi_n(x) dx = \delta_{n,m}$$ where the kronecker delta function is 0 for n not equal to m, and 1 for n = m. For the 1d infinite well, this leads to $$\int_{-\infty}^\infty \mid \Psi(x,t) \mid^2 dx = \sum_{n=1}^\infty \mid a_n \mid^2$$, where $$\sum_{n=1}^\infty \mid a_n \mid^2 = 1$$ Oct 26, 2016 at 0:59
I am trying to answer point 3. of your question. The answers to the other points have been correctly given by bleuofblue. If you have a wave function $\psi$ and an observable $a$ with its corresponding operator $A$ which has a complete set of normalized orthogonal eigenfunctions $\psi_n$ with corresponding discrete eigenvalues $a_n$, then you also get the probability distribution of possible measurement values of the observable a. The probability distribution for the measurement values $a_n$ is given by $$P_n(a_n)=|\langle\psi|\psi_n\rangle|^2$$ The expectation value is given by $\langle a\rangle=\langle\psi|A\psi\rangle$ and the standard deviation of the distribution is given by $\sqrt{\langle\psi|A^2 \psi\rangle-\langle \psi|A \psi\rangle^2}$ Similar formulae exist for continuous spectra of eigenvalues.
1. Fermi Dirac statistics bridges the gap between thermodynamics and quantum mechanics and can be used along with the schrodinger equation to determine the probabilities of the various bound energy states, which in turn can be used to determine the coefficients of the general wavefunction. The extra measurement necessary for this determination, as in the classical case, is temperature. This kind of an experiment shows up regarding electron occupation states in a nanoscale semiconductor, which can be approximated by the particle-in-a-box scenario under certain conditions. | 2,394 | 9,372 | {"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} | 2.9375 | 3 | CC-MAIN-2022-33 | latest | en | 0.929766 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.