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 |
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https://www.wiseowl.co.uk/excel/exercises/standard/conditional-formulae/4302/ | 1,563,413,290,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.62/warc/CC-MAIN-20190718001934-20190718023934-00511.warc.gz | 868,367,362 | 7,507 | # Excel | Conditional formulae exercise | IF printer cartridge commission
This exercise is provided to allow potential course delegates to choose the correct Wise Owl Microsoft training course, and may not be reproduced in whole or in part in any format without the prior written consent of Wise Owl.
The answer to the exercise will be included and explained if you attend one of more of the courses listed below!
Software ==> Excel (146 exercises) Version ==> Excel 2016 and later Topic ==> Conditional formulae (16 exercises) Level ==> Average difficulty Courses ==> Excel Business Modelling / Excel Intermediate
Before you can do this exercise, you'll need to download and unzip this file (if you have any problems doing this, click here for help).
You need a minimum screen resolution of about 700 pixels width to see our exercises. This is because they contain diagrams and tables which would not be viewable easily on a mobile phone or small laptop. Please use a larger tablet, notebook or desktop computer, or change your screen resolution settings.
Open the file contained within the folder name shown above.
We want to calculate the commission for all the sales reps:
We need to use an =IF function to calculate the commission.
In cell C4 create an =IF function to calculate the commission. Use the following information to help you:
• If the sales in column B is greater than or equal to £30,000, commission rate is 20%
• Anything less than £30,000, the commission rate is 1%
Don't forget to set up some input cells and range names for figures like the commission rates and the threshold figure to use in your =IF formula.
Copy the commission rate formula down column C to calculate all the other reps' commissions.
Change your commission rate input cells to check that your =IF function is working as shown below:
When you change the input cells on the right, all the commissions should change.
Use Save As... to save the file in your own new work folder. | 411 | 1,986 | {"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-2019-30 | longest | en | 0.890043 |
https://www.financereference.com/sec-yield/ | 1,702,041,938,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100745.32/warc/CC-MAIN-20231208112926-20231208142926-00598.warc.gz | 838,332,250 | 21,853 | # Sec Yield
## What is sec yield
Before discussing what sec yield is, let’s review some basics. A bond is a debt investment in which an investor loans money to an entity (usually governmental) for a defined period of time. The entity pays the investor periodic interest payments, called coupons, and returns the loan amount at maturity. The periodic interest payments are a percentage of the par value, which is the face value of the bond at issuance. Now that we have reviewed some basics, let’s discuss sec yield. The sec yield is simply the yield of a bond calculated using its most recent price. The most recent price may not be the same as the face value or par value, so the sec yield takes this into account. The calculation also takes into account any coupon payments that have been made since the bond was issued.
## Why should you care about sec yield?
Well, if you are looking to invest in bonds, the sec yield can give you a good indication of how much return you can expect on your investment. It is also a good way to compare bonds of different denominations and different maturities. So, if you are considering investing in bonds, be sure to check out the sec yield!
## How to calculate sec yield
There are a few different ways to calculate the current yield of a bond. One common method is to take the coupon rate of the bond and divide it by the price of the bond. For example, if a bond has a coupon rate of 5% and is currently selling for \$1,000, the current yield would be 0.05/1,000, or 0.005.
Another method is to take the interest payments made over the course of one year and divide that by the price of the bond. Using the same example as above, if the bond makes interest payments of \$50 per year, the current yield would be 0.05. It is also important to keep in mind that the current yield only provides a snapshot of what an investor can expect to earn in interest payments. It does not take into account the fact that bonds typically make periodic interest payments over their lifetime, so it may not provide an accurate picture of an investment’s overall return.
## Factors that affect sec yield
There are a number of factors that can affect the yield on a security. The most obvious is the creditworthiness of the issuer, which will affect both the interest rate paid on the security and the likelihood of default. Other factors include the maturity of the security, the coupons paid, and the market conditions at the time of purchase. In addition, investors need to be aware of the tax implications of owning a particular security. All of these factors can have a significant impact on the yield that an investor receives. As a result, it is important to consider all of these factors before investing in any security.
## Pros and cons of investing in securities with a high sec yield
The most obvious benefit is that they provide a higher return on investment than other securities. This can be particularly helpful in times of economic uncertainty, when other investments may be more likely to lose value. Additionally, high-yield securities tend to be less volatile than other types of investments, meaning that they are less likely to experience sudden drops in value.
However, there are also some downsides to investing in high-yield securities. One potential drawback is that they may be more susceptible to default risk. This means that there is a greater chance that the issuer will not be able to make interest payments on the security, leading to a loss for the investor. Additionally, high-yield securities may also be subject to liquidity risk, which refers to the possibility that the security will be difficult to sell at its full value. Before investing in any security, it is important to weigh the potential risks and rewards in order to make the best decision for your individual portfolio.
## When is the best time to buy securities with a high sec yield
The best time to buy securities with a high sec yield is typically when interest rates are low. This is because securities with a high sec yield tend to have a higher interest rate than other securities, which means they will provide a higher return on investment when interest rates are low. Additionally, securities with a high sec yield tend to be less volatile than other securities, which makes them a good choice for investors who are looking for stability in their portfolio.
## What to do if your portfolio has a low sec yield
If your portfolio has a low sec yield, there are several things you can do to boost your returns.
First, consider adding some high-yield securities to your portfolio. These include stocks with high dividend yields and bonds with high coupon rates.
Second, you may want to rebalance your portfolio so that it has a greater percentage of assets in high-yielding investments. This will require selling some of your low-yielding assets and buying more of the high-yielding ones.
Finally, don’t forget to keep an eye on expenses. A portfolio with a low sec yield will need to generate higher returns in order to reach your financial goals, so be sure to keep costs low. By following these steps, you can help ensure that your portfolio generates the returns you need. | 1,053 | 5,224 | {"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} | 3.125 | 3 | CC-MAIN-2023-50 | longest | en | 0.955433 |
https://www.chegg.com/homework-help/college-geometry-2nd-edition-chapter-7.3-solutions-9780131879690 | 1,561,408,103,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999709.4/warc/CC-MAIN-20190624191239-20190624213239-00195.warc.gz | 708,283,776 | 46,528 | # College Geometry (2nd Edition) Edit edition 80% (5 ratings) for this chapter’s solutionsSolutions for Chapter 7.3
We have solutions for your book!
Chapter: Problem:
Refer to the following figure for Exercise.
If and , find the measure of ∠C.
Step-by-step solution:
Chapter: Problem:
• Step 1 of 3
The figure illustrates two secant line segments intersecting outside of a circle at point C.
• Step 2 of 3
According to Theorem, if two secants intersect outside a circle, the measure of the acute angle formed is half the difference of the measures of the intercepted arcs. Thus,
• Step 3 of 3
Next, substitute in and. Solve for.
Therefore,
Corresponding Textbook
College Geometry | 2nd Edition
9780131879690ISBN-13: 0131879693ISBN: Authors:
Alternate ISBN: 9780131879713, 9780132362078, 9780135145913, 9780136157984, 9780321656773 | 230 | 846 | {"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-2019-26 | latest | en | 0.830591 |
https://www.smwcentral.net/?p=viewthread&t=2525&page=3 | 1,670,551,912,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00000.warc.gz | 1,063,176,578 | 7,104 | Language…
6 users online: alfonzo_galliano, HarvettFox96, JikissGamer, JupiHornet, Prestinator13, unknownperson123 - Guests: 60 - Bots: 108
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Latest user: Albani81
# Hangman Arkanoid
You are right!
Next question is going to annoy you.
It is math
s=
a=
d=
p=
h=
y=
many answers are to this....
Just back here to browse a bit.
I like math.
But i'm lazy:
s=-1
a=1
d=400
p=1
h=1
y=309
nope.... it is actually obvious
Just back here to browse a bit.
87-sad=happy+87 (87-87)-sad=happy+(87-87) -sad=happy So the opposite of sad is happy!Ooh, forgot we can't have Backloggery images anymore.
no... a is the same letter which makes it true no matter what...
Just back here to browse a bit.
I copied it down worng. >_<
Edit: Not again... I'm too tired.
s=1
a=1
d=87
p=1
h=1
y=-87
87-(1*1*87)=(1*1*1*1*-87)+87
87-(87)=(-87)+87
0=0
87-(-87) is also 87 + 87... >.>
Just back here to browse a bit.
s= 1
a= 1
d= 1
p= 1
h= 1
y= -1
87 - 1*1*1 = 1*1*1*1*-1 + 87
87 - 1 = -1 + 87
86 = 86
Don't post any math ones, I will nail them right in the head.
Your layout has been removed.
*cough*
Originally posted by 1024 (and edited on 2007-11-15 06:05:03 PM)
s=1
a=1
d=87
p=1
h=1
y=-87
87-(1*1*87)=(1*1*1*1*-87)+87
87-(87)=(-87)+87
0=0
...
Originally posted by pieguy1372
Don't post any math ones, I will nail them right in the head.
Oh great. I have one my thread now.
correct pieguy!
But the real answer i was hoping for was a=0 and all the others anything else... but good
<!--Did you find this?--!>
Just back here to browse a bit.
1024 is also right. Check his proof again. Also...although he edited his post after I posted mine, his answer is different, clearly indicating he has a different solution.
Your layout has been removed.
do u give up?
Just back here to browse a bit.
Originally posted by bored2tears
do u give up?
Only if it's hopeless.
Edit: somebody got the last question right, so it only makes sense that this is the next question
As amanda says or someone else, quoting always helps
Just back here to browse a bit.
Heh...Laugh...I've been doing trick questions everyday in school, got most right.
Well, go on.
maybe my 1875 post in this thread will help
Just back here to browse a bit.
NO wonder your name is what it is... whatitis...
Yes i did.
yes what? I don't get it.
Pester - 3
pie - 3
.net - 2
bloop - 2
1024 - 2
Ok, this fill be a fun game. Somewhere, like spigmike did, i will add a different colored letter that isnt orange or white.... now find it! and like to that post
Just back here to browse a bit. | 857 | 2,566 | {"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-2022-49 | latest | en | 0.915135 |
https://electronics.stackexchange.com/questions/709031/multi-loop-systems-block-reduction-techniques | 1,726,481,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00816.warc.gz | 211,137,805 | 40,253 | # Multi-loop systems block reduction techniques
Do I start by multiplying 18/s+3 * 1/s+3 = 18/(s+3)^(2)?
Then is it (18/(s+3)^(2))/(1+(18/(s+3)^(2)*4)) = 18 / ((s + 3)^2 * (72 / (s + 3)^2 + 1))?
• You can reformat in latex to make it easier to read. For instance 18/(s+3)^(2) becomes $\dfrac{18}{(s+3)^2}$ by using this: $\dfrac{18}{(s+3)^2}$ <-- you ought to show your steps too because of the poor/difficult format of your final formula and, the lack of steps make it tricky to follow what you did. Commented Apr 8 at 11:08
The most left part in the 2nd line is still correct - however, the most right part is wrong. After multiplying both - numerator and denominator - with (s+3)² the whole expression reduces to
18/[(s+3)² + 72].
But note that this expression is the transfer function of the small inner loop only.
• Do I then multiply that by the Gain of 50? Commented Apr 8 at 13:55
• Is $\dfrac{900}{s^2+6s+81}$ the correct answer? Commented Apr 8 at 14:06
• @ NonComposMentis No - I dont think so (the last numer (81) is not correct.)
– LvW
Commented Apr 8 at 14:48
• @NonComposMentis No. It's $\frac{900}{s^2\,+\,6s\,+\,981}$: ratsimp(solve(Eq(((vin-vout)*50-4*vout)*18/(s+3)*1/(s+3),vout),vout)[0]/vin)` Commented Apr 8 at 14:49
• @ periblepsis Instead of 972 I arrive at (972+9).....ahh I see, you have corrected the constant term.
– LvW
Commented Apr 8 at 14:51 | 490 | 1,380 | {"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": 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 | 4 | CC-MAIN-2024-38 | latest | en | 0.839379 |
https://gogeometry.blogspot.com/2019/02/geometry-problem-1414-right-triangle.html | 1,718,471,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00502.warc.gz | 243,748,472 | 13,561 | ## Saturday, February 23, 2019
### Geometry Problem 1414: Right Triangle, Altitude, Incircle, Excircle, Tangency Points, Isosceles Triangle
Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
#### 3 comments:
1. Join E to G, draw ET ꓕ AC, => ∆DBJ~∆GET
2. Let the bisector of < JBF cut DF at U.
From my proof of Problem 1412, < DGF = C/2, hence < GDF = < BFD - < DGF = 45 - C/2 = A/2 = < JBU since < JBF is obviously = to A in right Tr. ABC
Hence BDJU is concyclic and < BJU = < BDF = 45
So Tr.s BJU and BFU are congruent ASA and therefore
BJ = BF
Sumith Peiris
Moratuwa
Sri Lanka
3. https://photos.app.goo.gl/UvMged3MaAHNfXMy6
Define point M and angle u per sketch
Per the results of previous problems we have D,G,M are colinear
In right triangle DBG, angle BDG complement to u
In right triangle JHM, angle ∠ (HJM)= ∠ (BJD) complement to u
So ∠ (BDJ)= ∠ (BJD) and BD=BJ=BF -> JBF is isosceles | 338 | 1,018 | {"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} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.872349 |
https://questioncove.com/updates/59e7dbd655d06249e0847800 | 1,544,706,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824822.41/warc/CC-MAIN-20181213123823-20181213145323-00203.warc.gz | 726,697,371 | 7,231 | SkinnyMinnie20:
Find the missing values of the variables. The diagram is not drawn to scale
1 year ago
Pennywise:
1 year ago
SkinnyMinnie20:
1 year ago
Mehek:
111 + y = 180 180 - 111 =?
1 year ago
SkinnyMinnie20:
69
1 year ago
silvernight269:
And then what would X be?
1 year ago
SkinnyMinnie20:
would i do it the same way
1 year ago
silvernight269:
hint: all quadrilateral polygons has the sum of 360 degrees
1 year ago
SkinnyMinnie20:
okay
1 year ago
silvernight269:
what did you get so far?
1 year ago
SkinnyMinnie20:
hold on
1 year ago
SkinnyMinnie20:
the way I thought I should do it is wrong 121+111+69+66+y=360 367+y=360 y=7
1 year ago
silvernight269:
no. we want to find the X-value, that is from the inside of the polygon. 111 is just a clue to help you find the Y-value. try again.
1 year ago
SkinnyMinnie20:
so do it the same way but without 111
1 year ago
silvernight269:
yes
1 year ago
SkinnyMinnie20:
okay its 104
1 year ago
silvernight269:
very good
1 year ago | 312 | 1,007 | {"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-2018-51 | latest | en | 0.813351 |
https://www.experts-exchange.com/questions/27022717/Calculating-days-between-two-dates-in-access-form-that-excludes-weekends-and-holidays.html | 1,529,469,479,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00048.warc.gz | 807,663,623 | 23,667 | • Status: Solved
• Priority: Medium
• Security: Public
• Views: 538
# Calculating days between two dates in access form that excludes weekends and holidays?
Hi,
After producing a holiday submission request form with access I am trying to implement an additional function that calculates the amount of dates between the two dates, however, I want it to minus off weekends and if possible holidays (bank holidays etc) as well.
Would I need to have a module running in order to do such a calculation?
Thanks.
0
• 12
• 11
• 8
• +1
1 Solution
Commented:
Check this link out on how to accomplish this;
http://access.mvps.org/access/datetime/date0012.htm
HTH,
Daniel
0
Commented:
try this codes
``````Function getWorkDays(vDate1 As Date, vDate2 As Date) As Long
Dim i As Long, dtStart
If vDate1 = vDate2 Then getWorkDays = 0: Exit Function
If vDate1 < vDate2 Then
dtStart = vDate1
i = DateDiff("d", vDate1, vDate2) '+ 1
Do Until dtStart >= vDate2
dtStart = dtStart + 1
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("[Date]", "tbl_Holidays", "[Date]=#" _
& dtStart & "#"))
dtStart = dtStart + 1
i = i - 1
Loop
Loop
Else
dtStart = vDate2
i = DateDiff("d", vDate2, vDate1) '+ 1
Do Until dtStart >= vDate1
dtStart = dtStart + 1
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("[Date]", "tbl_Holidays", "[Date]=#" _
& dtStart & "#"))
dtStart = dtStart + 1
i = i - 1
Loop
Loop
End If
getWorkDays = i
End Function
``````
0
CIOCommented:
0
Author Commented:
@capricorn1
Thanks, there are quite a few variables in the code, just to make things easier I have a table called tblHolidays with two fields, one named holidate(date) and the other holiname(text). I presume that a necessary change would be to amend where it says tbl_holidays to tblholidays but what about the the sections where it says [date] and at the beginning of the code it states vDate1 and vDate2?
The form name is f_malings and there are three boxes, DATEFROM, DATETO and COUNTDAYS.
0
Commented:
vDate1 and vDate2 are the starting and ending dates, so it will be
getWorkDays(DATEFROM, DATETO)
0
Commented:
me.COUNTDAYS= getWorkDays(me.DATEFROM, me.DATETO)
0
Author Commented:
Made the changes, thanks, it shows a figure but it appears to be wrong.
Checking the Calendar, taking time off from the 1st Dec to 8th Dec this year should be 6 days but in the box it shows as 7?
0
Commented:
try this revised codes
``````Function getWorkDays(vDate1 As Date, vDate2 As Date) As Long
Dim i As Long, dtStart
If vDate1 = vDate2 Then getWorkDays = 0: Exit Function
If vDate1 < vDate2 Then
dtStart = vDate1
i = DateDiff("d", vDate1, vDate2) + 1
Do Until dtStart >= vDate2
dtStart = dtStart + 1
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("[Date]", "tbl_Holidays", "[Date]=#" _
& dtStart & "#"))
dtStart = dtStart + 1
i = i - 1
Loop
Loop
Else
dtStart = vDate2
i = DateDiff("d", vDate2, vDate1) + 1
Do Until dtStart >= vDate1
dtStart = dtStart + 1
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("[Date]", "tbl_Holidays", "[Date]=#" _
& dtStart & "#"))
dtStart = dtStart + 1
i = i - 1
Loop
Loop
End If
getWorkDays = i
End Function
``````
0
CIOCommented:
Use the function from my link above. It does it correctly.
=ISO_WorkDayDiff([DATEFROM],[DATETO],True)
/gustav
0
Author Commented:
@capricorn1
Done that and now it shows as 8 days?
The code below is what I have in the module
Function getWorkDays(DATETO As Date, DATEFROM As Date) As Long
Dim i As Long, dtStart
If DATETO = DATEFROM Then getWorkDays = 0: Exit Function
If DATETO < DATEFROM Then
dtStart = DATETO
i = DateDiff("d", DATETO, DATEFROM) '+ 1
Do Until dtStart >= DATEFROM
dtStart = dtStart + 1
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("[Date]", "tblHolidays", "[Date]=#" _
& dtStart & "#"))
dtStart = dtStart + 1
i = i - 1
Loop
Loop
Else
dtStart = DATEFROM
i = DateDiff("d", DATEFROM, DATETO) + 1
Do Until dtStart >= DATEFROM
dtStart = dtStart + 1
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("[DATETO]", "tblHolidays", "[DATEFROM]=#" & dtStart & "#"))
dtStart = dtStart + 1
i = i - 1
Loop
Loop
End If
getWorkDays = i
End Function
-------------------
0
Commented:
your table "tblHolidays" must have this fields Date that will contain the dates of holidays
* and Please do not change the codes
0
Author Commented:
Sorry I'm being a bit daft, the table tblholidays is not supposed to store any data but in order for the function to work DATETO and DATEFROM need to be fields in the tblholidays table?
0
Author Commented:
The reason I posted the above was because my tblholidays is supposed to just hold records of holidays throughout the year e..g bank holidays etc.
0
Author Commented:
@cactus_data
I just tried the code you showed and received a run time error 2001, msg was you canceled the previous operation.
in vb script debugger it shows the following highlighted in yellow
lngHolidays = DCount("*", cstrTableHoliday, strFilter)
0
Commented:
what is the name of the field in table tblholidays that contain the holidates? post it here
0
CIOCommented:
> * and Please do not change the codes
Well, you have to do so.
Use the function from my link above. It does it correctly.
> .. the table tblholidays is not supposed to store any data but in order for the
> function to work DATETO and DATEFROM need to be fields in the tblholidays table?
No, just skip the holiday check:
=ISO_WorkDayDiff([DATEFROM],[DATETO],False)
/gustav
0
Author Commented:
@capricorn1
tblholidays has two fields, holidates and holiname, the holidates is a date field containing dates e.g. 25/12/2011 and holiname is just a description field.
@cactus_data
just tried your code and the control source value of
=ISO_WorkDayDiff([DATEFROM],[DATETO],False) and i received run time error as stated in the previous post.
0
Commented:
ok , copy and paste this
``````Function getWorkDays(vDate1 As Date, vDate2 As Date) As Long
Dim i As Long, dtStart
If vDate1 = vDate2 Then getWorkDays = 0: Exit Function
If vDate1 < vDate2 Then
dtStart = vDate1
i = DateDiff("d", vDate1, vDate2) + 1
Do Until dtStart >= vDate2
dtStart = dtStart + 1
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("Holidates", "tblHolidays", "Holidates=#" _
& dtStart & "#"))
dtStart = dtStart + 1
i = i - 1
Loop
Loop
Else
dtStart = vDate2
i = DateDiff("d", vDate2, vDate1) + 1
Do Until dtStart >= vDate1
dtStart = dtStart + 1
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("Holidates", "tblHolidays", "Holidates=#" _
& dtStart & "#"))
dtStart = dtStart + 1
i = i - 1
Loop
Loop
End If
getWorkDays = i
End Function
``````
0
CIOCommented:
When running with the holiday check, modify these lines to fit your holiday table:
' Name of table with holidays.
Const cstrTableHoliday As String = "tblHoliday"
' Name of date field in holiday table.
Const cstrFieldHoliday As String = "HolidayDate"
/gustav
0
Commented:
then use this
me.COUNTDAYS= getWorkDays(me.DATEFROM, me.DATETO)
0
CIOCommented:
> .. taking time off from the 1st Dec to 8th Dec this year should be 6 days ..
Acutally, it should be 5 days as you normally don't include both the start and the end date.
If, however, you alway have a workday as the start date, just add 1 to include this in the count.
/gustav
0
Author Commented:
@cactus_data
Strange that it seems I am back to same point as I was with capricorn1's code.
It says that it's 5 days for the dates of 4th August to 11th August when it should be 6.
Would I be correct in saying that the days are not inclusive of the last date? E.g. 4th to 11th is saying that I'm off on the 4th and back on the 11th? Is it possible to change the code to include the 11th?
0
Author Commented:
Ah sorry just saw your post, so at the end of control source, just put +1 at the end.
Would this be suited in the module code or within the control source.
The control source shows as:
=ISO_WorkDayDiff([DATEFROM],[DATETO],True)
0
Commented:
did you try my post at http#a35707650
0
CIOCommented:
> just tried your code and the control source value of
> =ISO_WorkDayDiff([DATEFROM],[DATETO],False) and
> i received run time error as stated in the previous post.
That just can't be true. Try again.
/gustav
0
Commented:
did you try my post at http:#a35707650
0
CIOCommented:
> Would this be suited in the module code or within the control source.
> The control source shows as:
=ISO_WorkDayDiff([DATEFROM],[DATETO],True)
Exactly, or if you like:
=ISO_WorkDayDiff([DATEFROM],[DATETO],True)+1
And don't forget to adjust the two constants with you table and field names.
/gustav
0
Author Commented:
@capricorn1
I'm trying that now
@cactus_data
Not to worry, gone past that now and looking to add the 1 day, see previous post.
0
Author Commented:
@cactus_data
Okay, that now works, also noted that the holiday dates have to be readjusted because Christmas, Boxing Day and New Years all fall on Sunday therefore the days have to be put down as Monday and Tuesday, works perfectly, much appreciated.
@capricorn1
I just tried the code you put up in a previous post and received a run time error 2001, msg was you canceled the previous operation.
in vb script debugger it shows the following highlighted in yellow
Do While Weekday(dtStart) = 1 Or Weekday(dtStart) = 7 _
Or Not IsNull(DLookup("[Date]", "tblHolidays", "[Date]=#" _
& dtStart & "#"))
0
Commented:
i said copy and paste the revised codes from my post at http:#a35707650
0
Author Commented:
@capricorn1
The same thing happened.
0
CIOCommented:
You are welcome!
/gustav
0
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment. | 2,955 | 9,905 | {"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} | 3.125 | 3 | CC-MAIN-2018-26 | latest | en | 0.747775 |
https://help.libreoffice.org/latest/uk/text/scalc/01/04060115.html | 1,669,646,997,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00018.warc.gz | 357,242,153 | 6,748 | # Add-in Functions, List of Analysis Functions Part One
Щоб скористатися цією командою…
Вставка - Функція - Категорія Додатковий модуль
## BESSELI
Calculates the modified Bessel function of the first kind In(x).
#### Syntax
BESSELI(X; N)
X: значення для розрахунку функції.
N is a positive integer (N >= 0) representing the order of the Bessel function In(x)
#### Example
=BESSELI(3.45, 4), returns 0.651416873060081
=BESSELI(3.45, 4.333), returns 0.651416873060081, same as above because the fractional part of N is ignored.
=BESSELI(-1, 3), returns -0.022168424924332
## BESSELJ
Calculates the Bessel function of the first kind Jn(x) (cylinder function).
#### Syntax
BESSELJ(X; N)
X: значення для розрахунку функції.
N is a positive integer (N >= 0) representing the order of the Bessel function Jn(x)
#### Example
=BESSELJ(3.45, 4), returns 0.196772639864984
=BESSELJ(3.45, 4.333), returns 0.196772639864984, same as above because the fractional part of N is ignored.
=BESSELJ(-1, 3), returns -0.019563353982668
## BESSELK
Calculates the modified Bessel function of the second kind Kn(x).
#### Syntax
BESSELK(X; N)
X is the strictly positive value (X > 0) on which the function will be calculated.
N is a positive integer (N >= 0) representing the order of the Bessel function Kn(x)
#### Example
=BESSELK(3.45, 4), returns 0.144803466373734
=BESSELK(3.45, 4.333), returns 0.144803466373734, same as above because the fractional part of N is ignored.
=BESSELK(0, 3), returns Err:502 – invalid argument (X=0)
## BESSELY
Calculates the Bessel function of the second kind Yn(x).
#### Syntax
BESSELY(X; N)
X is the strictly positive value (X > 0) on which the function will be calculated.
N is a positive integer (N >= 0) representing the order of the Bessel function Yn(x)
#### Example
=BESSELY(3.45, 4), returns -0.679848116844476
=BESSELY(3.45, 4.333), returns -0.679848116844476, same as above because the fractional part of N is ignored.
=BESSELY(0, 3), returns Err:502 – invalid argument (X=0)
## BIN2DEC
The result is the number for the binary (base-2) number string entered.
#### Syntax
BIN2DEC(Number)
Number is a string representing a binary (base-2) number. It can have a maximum of 10 places (bits). The most significant bit is the sign bit. Negative numbers are entered as two's complement.
#### Example
=BIN2DEC("1100100") returns 100.
## BIN2HEX
The result is the string representing the number in hexadecimal form for the binary (base-2) number string entered.
#### Syntax
BIN2HEX(Number [; Places])
Number is a string representing a binary (base-2) number. It can have a maximum of 10 places (bits). The most significant bit is the sign bit. Negative numbers are entered as two's complement.
Places means the number of places to be output.
#### Example
=BIN2HEX("1100100";6) returns "000064".
## BIN2OCT
The result is the string representing the number in octal form for the binary (base-2) number string entered.
#### Syntax
BIN2OCT(Number [; Places])
Number is a string representing a binary (base-2) number. It can have a maximum of 10 places (bits). The most significant bit is the sign bit. Negative numbers are entered as two's complement.
Places means the number of places to be output.
#### Example
=BIN2OCT("1100100";4) returns "0144".
## DEC2BIN
The result is the string representing the number in binary (base-2) form for the number entered.
#### Syntax
DEC2BIN(Number [; Places])
Number is a number between -512 and 511. If Number is negative, the function returns a binary number string with 10 characters. The most significant bit is the sign bit, the other 9 bits return the value.
Places means the number of places to be output.
#### Example
=DEC2BIN(100;8) returns "01100100".
## DEC2HEX
The result is the string representing the number in hexadecimal form for the number entered.
#### Syntax
DEC2HEX(Number [; Places])
Number is a number. If Number is negative, the function returns a hexadecimal number string with 10 characters (40 bits). The most significant bit is the sign bit, the other 39 bits return the value.
Places means the number of places to be output.
#### Example
=DEC2HEX(100;4) returns "0064".
## DEC2OCT
The result is the string representing the number in octal form for the number entered.
#### Syntax
DEC2OCT(Number [; Places])
Number is a number. If Number is negative, the function returns an octal number string with 10 characters (30 bits). The most significant bit is the sign bit, the other 29 bits return the value.
Places means the number of places to be output.
#### Example
=DEC2OCT(100;4) returns "0144".
## DELTA
The result is TRUE (1) if both numbers, which are delivered as an argument, are equal, otherwise it is FALSE (0).
#### Syntax
DELTA(Number1 [; Number2])
#### Example
=DELTA(1;2) returns 0.
## ERF
Returns values of the Gaussian error integral.
#### Syntax
ERF(LowerLimit [; UpperLimit])
LowerLimit is the lower limit of the integral.
UpperLimit is optional. It is the upper limit of the integral. If this value is missing, the calculation takes place between 0 and the lower limit.
#### Example
=ERF(0;1) returns 0.842701.
## ERF.PRECISE
Returns values of the Gaussian error integral between 0 and the given limit.
#### Syntax
ERF.PRECISE(LowerLimit)
LowerLimit is the limit of the integral. The calculation takes place between 0 and this limit.
#### Example
=ERF.PRECISE(1) returns 0.842701.
#### Technical information
Ця фукція доступна з версії LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.ERF.PRECISE
## ERFC
Returns complementary values of the Gaussian error integral between x and infinity.
#### Syntax
ERFC(LowerLimit)
LowerLimit is the lower limit of the integral
#### Example
=ERFC(1) returns 0.157299.
## ERFC.PRECISE
Returns complementary values of the Gaussian error integral between x and infinity.
#### Syntax
ERFC.PRECISE(LowerLimit)
LowerLimit is the lower limit of the integral
#### Example
=ERFC.PRECISE(1) returns 0.157299.
#### Technical information
Ця фукція доступна з версії LibreOffice 4.3.
This function is not part of the Open Document Format for Office Applications (OpenDocument) Version 1.3. Part 4: Recalculated Formula (OpenFormula) Format standard. The name space is
COM.MICROSOFT.ERFC.PRECISE
## GESTEP
The result is 1 if Number is greater than or equal to Step.
#### Syntax
GESTEP(Number [; Step])
#### Example
=GESTEP(5;1) returns 1.
## HEX2BIN
The result is the string representing the number in binary (base-2) form for the hexadecimal number string entered.
#### Syntax
HEX2BIN(Number [; Places])
Number is a string that represents a hexadecimal number. It can have a maximum of 10 places. The most significant bit is the sign bit, the following bits return the value. Negative numbers are entered as two's complement.
Places is the number of places to be output.
#### Example
=HEX2BIN("6a";8) returns "01101010".
## HEX2DEC
The result is the number for the hexadecimal number string entered.
#### Syntax
HEX2DEC(Number)
Number is a string that represents a hexadecimal number. It can have a maximum of 10 places. The most significant bit is the sign bit, the following bits return the value. Negative numbers are entered as two's complement.
#### Example
=HEX2DEC("6a") returns 106.
## HEX2OCT
The result is the string representing the number in octal form for the hexadecimal number string entered.
#### Syntax
HEX2OCT(Number [; Places])
Number is a string that represents a hexadecimal number. It can have a maximum of 10 places. The most significant bit is the sign bit, the following bits return the value. Negative numbers are entered as two's complement.
Places is the number of places to be output.
#### Example
=HEX2OCT("6a";4) returns "0152".
Будь ласка, підтримайте нас! | 2,151 | 8,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-49 | longest | en | 0.373004 |
https://howto.org/how-to-calculate-and-compare-unit-prices-at-the-store/ | 1,638,305,359,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00302.warc.gz | 378,160,642 | 9,984 | ### How do you calculate unit price?
To calculate the unit price, simply divide the cost of the product by the quantity you’re receiving or check the store’s shelf label. Then, compare the unit prices of 2 or more packages of the same product to see which is the better value.
### What is the difference between the two unit prices?
Unit Price Formula
It is calculated by adding fixed and variable expense and dividing it by the total number of units produced.
### What is the price per unit?
Unit cost is the cost incurred on producing and packing a single piece of item, whereas unit price is the price of a single piece of item.
### What is an example of a unit rate?
In retail, unit price is the price for a single unit of measure of a product sold in more or less than the single unit. The “unit price” tells you the cost per pound, quart, or other unit of weight or volume of a food package. It is usually posted on the shelf below the food.
### When we find the price for one unit it is called the?
A unit rate means a rate for one of something. We write this as a ratio with a denominator of one. For example, if you ran 70 yards in 10 seconds, you ran on average 7 yards in 1 second. Both of the ratios, 70 yards in 10 seconds and 7 yards in 1 second, are rates, but the 7 yards in 1 second is a unit rate.
### What is the unit price for the first tomato option?
Foods that cost less per unit are not always the better buy. The big, economy size is not a good buy if you cannot store it properly. If you end up with leftovers that spoil or are thrown out, buying the larger size is not a good idea. Unit pricing can help you decide what brand to buy.
### What is the benefit of unit pricing?
Cost per unit, also referred to the cost of goods sold or the cost of sales, is how much money a company spends on producing one unit of the product they sell. Companies include this figure on their financial statement.
### How do you calculate best buys?
The unit price for the first tomato option is \$2.75.
### What is the difference between lump sum and unit price?
It has been argued that unit pricing reduces time spent comparing brands, eliminates confusion relating to price calculations and ultimately allows shoppers to save money at the register. Unit pricing enables the consumers to rationally evaluate the most economic package size and brand.
### What is skimmed pricing?
Unit pricing helps consumers compare the prices of packaged goods when those goods aren’t sold in equal quantities. It’s unit pricing that allows, for example, a shopper to tell at a glance which is the better buy: a 20-pound bag of dog food that sells for \$13.95 or a 15-pound bag that sells for \$10.69.
### What are the 4 types of contracts?
A lump sum bid represents the total price for which a contractor offers to complete a facility according to the detailed plans and specifications. Unit price bidding is used in projects for which the quantity of materials or the amount of labor involved in some key tasks is particularly uncertain.
### What are the 3 types of contracts?
Price lining is a technique used by retailers to group common items at set price-points. Rather than setting the retail price based on cost or competition, price lining is a way to simplify the pricing of assorted goods by establishing tiered price points that can support assortments of goods.
### What are the most common types of contracts?
Price skimming is a product pricing strategy by which a firm charges the highest initial price that customers will pay and then lowers it over time. The skimming strategy gets its name from “skimming” successive layers of cream, or customer segments, as prices are lowered over time.
### What are the different types of offer?
What are the Different Types of Contract?
• Contract Types Overview.
• Express and Implied Contracts.
• Unilateral and Bilateral Contracts.
• Unconscionable Contracts.
• Aleatory Contracts.
• Option Contracts.
• Fixed Price Contracts.
### How many type of contracts are there?
The three most common contract types include:
• Fixed-price contracts.
• Cost-plus contracts.
• Time and materials contracts.
### What are the two main types of contracts?
Unit price contracts are essentially a series of lump sum contracts throughout the entirety of a project. The project breaks down into stages, and a contractor will provide a fixed price to complete each stage. Unlike lump sum contracts, unit price contracts deal with changes pretty well. | 944 | 4,513 | {"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} | 3.640625 | 4 | CC-MAIN-2021-49 | latest | en | 0.952534 |
https://ffmpeg.org/pipermail/ffmpeg-devel/2015-November/182745.html | 1,652,869,170,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00759.warc.gz | 315,468,866 | 5,294 | # [FFmpeg-devel] [PATCHv2] swresample/resample: improve bessel function accuracy and speed
Tue Nov 3 22:49:36 CET 2015
```This improves accuracy for the bessel function at large arguments, and this in turn
should improve the quality of the Kaiser window. It also improves the
performance of the bessel function and hence build_filter by ~ 20%.
Details are given below.
Algorithm: taken from the Boost project, who have done a detailed
investigation of the accuracy of their method, as compared with e.g the
GNU Scientific Library (GSL):
http://www.boost.org/doc/libs/1_52_0/libs/math/doc/sf_and_dist/html/math_toolkit/special/bessel/mbessel.html.
Boost source code (also cited and licensed in the code):
https://searchcode.com/codesearch/view/14918379/.
Accuracy: sample values may be obtained as follows. i0 denotes the old bessel code,
i0_boost the approach here, and i0_real an arbitrary precision result (truncated) from Wolfram Alpha:
type "bessel i0(6.0)" to reproduce. These are evaluation points that occur for
the default kaiser_beta = 9.
Some illustrations:
bessel(8.0)
i0 (8.000000) = 427.564115721804739678191254
i0_boost(8.000000) = 427.564115721804796521610115
i0_real (8.000000) = 427.564115721804785177396791
bessel(6.0)
i0 (6.000000) = 67.234406976477956163762428
i0_boost(6.000000) = 67.234406976477970374617144
i0_real (6.000000) = 67.234406976477975326188025
Reason for accuracy: Main accuracy benefits come at larger bessel arguments, where the
Taylor-Maclaurin method is not that good: 23+ iterations
(at large arguments, since the series is about 0) can cause
significant floating point error accumulation.
Benchmarks: Obtained on x86-64, Haswell, GNU/Linux via a loop calling
build_filter 1000 times:
test: fate-swr-resample-dblp-44100-2626
new:
995894468 decicycles in build_filter(loop 1000), 256 runs, 0 skips
1029719302 decicycles in build_filter(loop 1000), 512 runs, 0 skips
984101131 decicycles in build_filter(loop 1000), 1024 runs, 0 skips
old:
1250020763 decicycles in build_filter(loop 1000), 256 runs, 0 skips
1246353282 decicycles in build_filter(loop 1000), 512 runs, 0 skips
1220017565 decicycles in build_filter(loop 1000), 1024 runs, 0 skips
A further ~ 5% may be squeezed by enabling -ftree-vectorize. However,
this is a separate issue from this patch.
Reviewed-by: Michael Niedermayer <michael at niedermayer.cc>
---
libswresample/resample.c | 124 ++++++++++++++++++++++++++++++++++++-----------
1 file changed, 97 insertions(+), 27 deletions(-)
diff --git a/libswresample/resample.c b/libswresample/resample.c
index 036eff3..235c9a9 100644
--- a/libswresample/resample.c
+++ b/libswresample/resample.c
@@ -28,38 +28,108 @@
#include "libavutil/avassert.h"
#include "resample.h"
+static inline double eval_poly(const double *coeff, int size, double x) {
+ double sum = coeff[size-1];
+ int i;
+ for (i = size-2; i >= 0; --i) {
+ sum *= x;
+ sum += coeff[i];
+ }
+ return sum;
+}
+
/**
* 0th order modified bessel function of the first kind.
+ * Algorithm taken from the Boost project, source:
+ * https://searchcode.com/codesearch/view/14918379/
+ *
+ * Boost Software License - Version 1.0 - August 17th, 2003
+Permission is hereby granted, free of charge, to any person or organization
+obtaining a copy of the software and accompanying documentation covered by
+this license (the "Software") to use, reproduce, display, distribute,
+execute, and transmit the Software, and to prepare derivative works of the
+Software, and to permit third-parties to whom the Software is furnished to
+do so, all subject to the following:
+
+The copyright notices in the Software and this entire statement, including
+the above license grant, this restriction and the following disclaimer,
+must be included in all copies of the Software, in whole or in part, and
+all derivative works of the Software, unless such copies or derivative
+works are solely in the form of machine-executable object code generated by
+a source language processor.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
+FITNESS FOR A PARTICULAR PURPOSE, TITLE AND NON-INFRINGEMENT. IN NO EVENT
+SHALL THE COPYRIGHT HOLDERS OR ANYONE DISTRIBUTING THE SOFTWARE BE LIABLE
+FOR ANY DAMAGES OR OTHER LIABILITY, WHETHER IN CONTRACT, TORT OR OTHERWISE,
+ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER
+DEALINGS IN THE SOFTWARE.
*/
-static double bessel(double x){
- double lastv=0;
- double t, v;
- int i;
- static const double inv[100]={
- 1.0/( 1* 1), 1.0/( 2* 2), 1.0/( 3* 3), 1.0/( 4* 4), 1.0/( 5* 5), 1.0/( 6* 6), 1.0/( 7* 7), 1.0/( 8* 8), 1.0/( 9* 9), 1.0/(10*10),
- 1.0/(11*11), 1.0/(12*12), 1.0/(13*13), 1.0/(14*14), 1.0/(15*15), 1.0/(16*16), 1.0/(17*17), 1.0/(18*18), 1.0/(19*19), 1.0/(20*20),
- 1.0/(21*21), 1.0/(22*22), 1.0/(23*23), 1.0/(24*24), 1.0/(25*25), 1.0/(26*26), 1.0/(27*27), 1.0/(28*28), 1.0/(29*29), 1.0/(30*30),
- 1.0/(31*31), 1.0/(32*32), 1.0/(33*33), 1.0/(34*34), 1.0/(35*35), 1.0/(36*36), 1.0/(37*37), 1.0/(38*38), 1.0/(39*39), 1.0/(40*40),
- 1.0/(41*41), 1.0/(42*42), 1.0/(43*43), 1.0/(44*44), 1.0/(45*45), 1.0/(46*46), 1.0/(47*47), 1.0/(48*48), 1.0/(49*49), 1.0/(50*50),
- 1.0/(51*51), 1.0/(52*52), 1.0/(53*53), 1.0/(54*54), 1.0/(55*55), 1.0/(56*56), 1.0/(57*57), 1.0/(58*58), 1.0/(59*59), 1.0/(60*60),
- 1.0/(61*61), 1.0/(62*62), 1.0/(63*63), 1.0/(64*64), 1.0/(65*65), 1.0/(66*66), 1.0/(67*67), 1.0/(68*68), 1.0/(69*69), 1.0/(70*70),
- 1.0/(71*71), 1.0/(72*72), 1.0/(73*73), 1.0/(74*74), 1.0/(75*75), 1.0/(76*76), 1.0/(77*77), 1.0/(78*78), 1.0/(79*79), 1.0/(80*80),
- 1.0/(81*81), 1.0/(82*82), 1.0/(83*83), 1.0/(84*84), 1.0/(85*85), 1.0/(86*86), 1.0/(87*87), 1.0/(88*88), 1.0/(89*89), 1.0/(90*90),
- 1.0/(91*91), 1.0/(92*92), 1.0/(93*93), 1.0/(94*94), 1.0/(95*95), 1.0/(96*96), 1.0/(97*97), 1.0/(98*98), 1.0/(99*99), 1.0/(10000)
- };
- x= x*x/4;
- t = x;
- v = 1 + x;
- for(i=1; v != lastv; i+=2){
- t *= x*inv[i];
- v += t;
- lastv=v;
- t *= x*inv[i + 1];
- v += t;
- av_assert2(i<98);
+static double bessel(double x) {
+// Modified Bessel function of the first kind of order zero
+// minimax rational approximations on intervals, see
+// Blair and Edwards, Chalk River Report AECL-4928, 1974
+ static const double p1[] = {
+ -2.2335582639474375249e+15,
+ -5.5050369673018427753e+14,
+ -3.2940087627407749166e+13,
+ -8.4925101247114157499e+11,
+ -1.1912746104985237192e+10,
+ -1.0313066708737980747e+08,
+ -5.9545626019847898221e+05,
+ -2.4125195876041896775e+03,
+ -7.0935347449210549190e+00,
+ -1.5453977791786851041e-02,
+ -2.5172644670688975051e-05,
+ -3.0517226450451067446e-08,
+ -2.6843448573468483278e-11,
+ -1.5982226675653184646e-14,
+ -5.2487866627945699800e-18,
+ };
+ static const double q1[] = {
+ -2.2335582639474375245e+15,
+ 7.8858692566751002988e+12,
+ -1.2207067397808979846e+10,
+ 1.0377081058062166144e+07,
+ -4.8527560179962773045e+03,
+ 1.0L,
+ };
+ static const double p2[] = {
+ -2.2210262233306573296e-04,
+ 1.3067392038106924055e-02,
+ -4.4700805721174453923e-01,
+ 5.5674518371240761397e+00,
+ -2.3517945679239481621e+01,
+ 3.1611322818701131207e+01,
+ -9.6090021968656180000e+00,
+ };
+ static const double q2[] = {
+ -5.5194330231005480228e-04,
+ 3.2547697594819615062e-02,
+ -1.1151759188741312645e+00,
+ 1.3982595353892851542e+01,
+ -6.0228002066743340583e+01,
+ 8.5539563258012929600e+01,
+ -3.1446690275135491500e+01,
+ 1.0L,
+ };
+ double y, r, factor;
+ if (x == 0)
+ return 1.0;
+ x = fabs(x);
+ if (x <= 15) {
+ y = x * x;
+ return eval_poly(p1, FF_ARRAY_ELEMS(p1), y) / eval_poly(q1, FF_ARRAY_ELEMS(q1), y);
+ }
+ else {
+ y = 1 / x - 1.0 / 15;
+ r = eval_poly(p2, FF_ARRAY_ELEMS(p2), y) / eval_poly(q2, FF_ARRAY_ELEMS(q2), y);
+ factor = exp(x) / sqrt(x);
+ return factor * r;
}
- return v;
}
/**
--
2.6.2
``` | 3,145 | 8,337 | {"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-2022-21 | latest | en | 0.766418 |
https://www.jiskha.com/display.cgi?id=1332758271 | 1,516,774,302,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893397.98/warc/CC-MAIN-20180124050449-20180124070449-00074.warc.gz | 912,001,549 | 4,006 | # maths
posted by .
sally read 21 pages of a book on Satuday. She read 5/9 of the remainder on Sunday. If she still had 1/3 of the book left to read, how many pages were there in the book?
• maths -
number of pages ---- x
at end of Saturday, number of pages left = x-21
she read 5/9 of those, leaving 4/9(x-21) pages
but this is supposed to be 1/3 of the pages
4/9(x-21) = (1/3)x
times 9
4(x-21) = 3x
4x - 84 = 3x
x = 84
There were 84 pages in the book
check:
she read 21 pages on Saturday leaving 63
on Sunday she read 5/9 of that or 35 pages
leaving her with 28 pages to read
and 1/3 of 84 = 28
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Find the area under y = e[t/3] from t = 0 to t = 6
• A = ∫06 e[t/3] dt
• u = [t/3]
• du = [1/3] dt
• A = 3 ∫06 eu du
• A = 3 e[t/3] |06
• A = 3 (e2 − e0)
A = 3(e2 − 1)
Find the area bounded by y = x2 and y = |x|
• Find the bounds of the integral
• x2 = |x|
• x2 = √{x2}
• x4 = x2
• x2 = 1
• x = ±1
• y = |x| has different behavior on either side of zero. We can treat this as a piecewise function with the area as the sum of two different integrals
• A = ∫−10 −x − x2 dx + ∫01 x − x2 dx
• A = (−[(x2)/2] − [(x3)/3]) |−10 + ([(x2)/2] − [(x3)/3]) |01
• A = (−0 − 0) − (−[((−1)2)/2] − [((−1)3)/3]) + ([1/2] − [1/3]) − 0
• A = [1/2] − [1/3] + [1/2] − [1/3]
• A = 1 − [2/3]
A = [1/3]
Find the area bound by x = y3 and the y-axis from y = 0 to y = 2
• A = ∫02 y3 dy
• A = [(y4)/4] |02
• A = [(24)/4] − 0
A = 4
Find the area bound by y = sinx and the x-axis from x = 0 to x = 2π
• A = ∫0 sinx dx
• A = − cosx |0
• A = −cos2π− (− cos0)
• A = −1 + 1 = 0
• An area of zero doesn't quite make sense here, so using the straight integral is insufficient. What we can do is treat this as two separate integrals, one where the area is above the x-axis and one where it is below and add their effective area.
• sinx = 0, x = 0, π, 2π in this interval.
• A = ∫0π sinx dx + (−∫π sinx dx)
• The negative sign is there to compensate for the "negative area"
• A = −cosx |0π + cosx |π
• A = − cosπ+ cos0 + cos2π− cosπ
• A = 1 + 1 + 1 + 1
A = 4
Find the area bound by x = y2 and y = x − 2
• We could find the integral in terms of dx, but the we would have to split it up into at least two separate integrals. Besides, x = y2 is not a function of x (there are two outputs for every x input).
• x = y2
• x = y + 2
• Find the limits of the integral
• y2 = y + 2
• y2 − y − 2 = 0
• (y + 1)(y − 2) = 0
• y = 2, −1
• A = ∫−12 (y + 2) − y2 dy
• A = ([(y2)/2] + 2x − [(y3)/3]) |−12
• A = (2 + 4 − [8/3]) − ([1/2] − 2 + [1/3])
• A = 8 − [1/2] − [9/3]
• The area above the
A = [9/2]
Find the area bound by y = x2 + 1 and y = −x2 + 3
• Find the limits of the integral
• x2 + 1 = −x2 + 3
• 2x2 = 2
• x2 = 1
• x = ±1
• A = ∫−11 (−x2 + 3) − (x2 + 1) dx
• A = ∫−11 −2x2 + 2 dx
• A = (−[(2x3)/3] + 2x) |−11
• A = −[2/3] + 2 − (−[(−2)/3] − 2)
• A = −[2/3] + 2 − [2/3] + 2
• A = 4 − [4/3]
A = [8/3]
Find the area bound by y = x2 and y = 2x
• Find the limits
• x2 = 2x
• x2 − 2x = 0
• x(x − 2) = 0
• x = 0, 2
• A = ∫02 2x − x2 dx
• A = x2 − [(x3)/3] |02
• A = 4 − [8/3] − 0
A = [4/3]
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer. | 1,680 | 3,950 | {"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-2017-47 | latest | en | 0.863125 |
https://oneclass.com/study-guides/ca/utm/psy/psy352h5/7177-review-of-test-questions-with-answers.en.html | 1,627,063,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150000.59/warc/CC-MAIN-20210723175111-20210723205111-00014.warc.gz | 449,474,811 | 203,372 | # review of test questions with answers
122 views7 pages
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Psy252 review
1. Define heritability using mathematical model and explain the scientific definition as
well.
Heritability is defined as the ratio of genotypic variance (Vg) to the phenotypic
variance (Vp) of a particular trait.
It can be defined using the following formula:
h2 = VG / (VE + VG) , where VG is the genetic variance and VE is the
environmental variance and VP = VE + VG, where VP is the phenotypic
variance.
Therefore, if genetic analysis found that heritability of losing an eye is 0% then it
can be said that in a population 0% of the variation is in the population for this
trait will be based on genetics. i.e. 100% of variation is based on the environment.
2. What does it mean when heritability is 0%?
It means that 0% of the variability observed in a population is based on genetic
factors that is to say 100% of the variability is due to environmental factors.
This make sense, assuming that genetically all individuals are born with 2 eyes but
due to some environmental effects some lose 1 eye or both.
For ex: in a population, every year grade 3 kids have a snowball fight with
each other, with half of them getting one eye knocked out.
Genetically, they all had eyes but due to environmental effects half of them lost
one eye. This means that the variability seen in the population for 3rd grade for this
trait is based on environmental effects not genetic factors.
This implies that relative importance of genes for the trait of losing an eye is
negligible as genetically everyone is born with eyes.
However, the environmental effects on this trait determine the variability found in
the population.
Genetic variance (Vg) is zero because this trait is so important that natural
selection has selected against any variation.
3. Give a real or hypothetical example where heritability equals 100%. Explain how
this would be possible and what this would mean in terms of environmental and
genetic effects.
If we assume that the heritability of IQ is 100%( hypothetically) then,
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In a population, with a homogeneous environment for every individual,
100% of the variability in IQ found in the population will be based on
genetic factors.
This means that variations in the population will be based entirely in
genetic effects.
However, this does not mean that the environmental effect is negligible. If the
environment effect for half of the population changes then heritability value is not
valid. Also the population in a homogeneously poor environment may have an
overall average IQ that is lower than that of population that is in rich environment.
Thus, the environmental effect is important for the overall average IQ of
population but not for the variation found in the population for that trait.
In order for heritability of IQ to be 100% or any trait to be 100%, natural selection
must have selected for that trait to be variable based on genetic effects.
4. The comparison of major racial groups with regard to their performance on IQ tests
is problematic and potentially why the test results are flawed. Circle the right
Major racial groups are not genetically well defined
Larger genetic variability within as compared to between racial groups
Phenotypical recording techniques are biased (IQ test)
Environmental effects are not parsed out
Heritability estimates are incorrect therefore
But even if they were correct, interpretation, “spending money on social programs
is a waste” is wrong:
Genes do not have deterministic effects.
If we know about their effects and if we study how they exert them, then we can
designenvironmental programs to circumvent, augment, improve on their
effects.
5. An experimenter decided to test three different mouse strains A(CD1), B(C57BL6),
and C(DBA/2) in 7 different environmental conditions and analyzed the learning
capacity of these mice in these 7 test situations. Explain whether her results
demonstrate the presence and/or absence of genotype x environment interaction.
CD1
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## Document Summary
Psy252 review: define heritability using mathematical model and explain the scientific definition as well, heritability is defined as the ratio of genotypic variance (vg) to the phenotypic variance (vp) of a particular trait. Explain how this would be possible and what this would mean in terms of environmental and genetic effects. If we assume that the heritability of iq is 100%( hypothetically) then, www. notesolution. com. In a population, with a homogeneous environment for every individual, 100% of the variability in iq found in the population will be based on genetic factors. This means that variations in the population will be based entirely in genetic effects: however, this does not mean that the environmental effect is negligible. If the environment effect for half of the population changes then heritability value is not valid. Also the population in a homogeneously poor environment may have an overall average iq that is lower than that of population that is in rich environment. | 1,154 | 5,340 | {"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} | 3.125 | 3 | CC-MAIN-2021-31 | latest | en | 0.955104 |
https://gmatclub.com/forum/best-gmat-math-prep-books-reviews-recommendations-77291-180.html?sort_by_oldest=true | 1,487,643,257,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170614.88/warc/CC-MAIN-20170219104610-00002-ip-10-171-10-108.ec2.internal.warc.gz | 726,716,373 | 72,665 | Best GMAT Math Prep Books (Reviews & Recommendations) : GMAT Quantitative Section - Page 10
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# Best GMAT Math Prep Books (Reviews & Recommendations)
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23 Apr 2009, 10:29
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Top Best GMAT Quantitative Books
This post is a part of a series on best GMAT Books. Also see:
Best GMAT Guidebooks
Best GMAT Verbal Prep Books
GMAT Fiction - improve reading skills on GMAT
Top 20 Best GMAT books - ranked by GMAT Club members
Common Book Combinations and Recommendatios
1. Need to figure out your weaknesses in GMAT math? Take a GMATPrep test or GMAT Club's free gmat diagnostic test
2. Rusty with Math or want to be very methodical - start with basics: Manhattan GMAT Foundations of GMAT Math
3. OK with Math but need a good review in a short amount of time: Kaplan Math Workbook + MGMAT Number Properties + MGMAT Word Translations
4. Need highest score and have time: MGMAT Guides + GMAT Club tests
5. Need specific question practice: PR 1012 (has some errors/typos), Quant OG, or use gmat club tags
6. Probability and Combinations review: combinations-permutations-and-probability-references-56486.html and MGMAT Word Translations
Book Bundles and Study Plans at a Glance Standard Plan Advanced Plan Ultimate Plan Beyond Hardcore Plan General Strategy BooksExplain how GMAT works, provide general principles and strategies, give an idea about weaknesses, provide practice tests Kaplan Live Online+ comes with 6 practice tests Kaplan Live Online+ comes with 6 practice testsKaplan GMAT Advanced- Poor guidebook, more of a question collection Kaplan Live Online+ comes with 6 practice testsKaplan GMAT Advanced- Poor guidebook, more of a question collectionGMAT Test Simulation Booklet+ Great for practice Kaplan Live Online+ comes with 6 practice testsKaplan GMAT Advanced- Poor guidebook, more of a question collectionGMAT Test Simulation Booklet+ Great for practice Quant BooksRefresh math principles, fill gaps, and learn patterns Kaplan Math Workbook- missing probability and combinations Kaplan Math Workbook- missing probability and combinationsMGMAT Number Properties 5 Quant MGMAT GuidesMGMAT Math Foundations 5 Quant MGMAT GuidesMGMAT Math FoundationsProbability and Combinations for Dummies Verbal BooksRefresh grammar rules, aquire CR and RC strategies Kaplan Verbal WorkbookOR: PowerScore Verbal Bible CR: PowerScore CR or MGMAT CRRC: MGMAT RCSC: PowerScore SC or MGMAT SC CR: PowerScore CR or MGMAT CRRC: MGMAT RCSC: PowerScore SC or MGMAT SCSC: Kaplan Verbal Foundations CR: PowerScore CR or MGMAT CRCR: LSAT Critical ReasoningRC: MGMAT RCRC: LSAT Reading ComprehensionSC: PowerScore SC or MGMAT SCSC: Doing Grammar Practice QuestionsAdditional practice Official Guide 12th+ 900 retired real GMAT questions Official Guide 12th+ 900 retired real GMAT questionsVerbal Official Guide+ 300 retired verbal GMAT questionsQuant Official Guide+ 300 retired quant GMAT questions Official Guide 12th+ 900 retired real GMAT questionsVerbal Official Guide+ 300 retired verbal GMAT questionsQuant Official Guide+ 300 retired quant GMAT questionsPrinceton Review 1012+ Great target practice Official Guide 12th+ 900 retired real GMAT questionsVerbal Official Guide+ 300 retired verbal GMAT questionsQuant Official Guide+ 300 retired quant GMAT questionsPrinceton Review 1012+ Great target practicePaper GMAT Tests+ Official Retired tests available from GMAC TestsSimulated Practice GMAT tests GMATPrep, Kaplan, Free tests GMATprep, Kaplan, MGMAT, some GMATClub tests GMATprep, PowerPrep, GMAT Focus, Kaplan, MGMAT, GMATClub tests GMATprep, PowerPrep, GMAT Focus, Kaplan, MGMAT, PR, GMATClub tests
1. GMAT Math Books
Recommended for best results if you have the time.
Strengths:
1. Good balance between test taking strategies and background theory explanations
2. Single-topic focus of each guide allows the student to purchase only those guides that pertain to his/her weaknesses
3. Covered concepts are illustrated with examples that are conveniently assigned to either PS or DS section.
5. Additional online practice question banks (different for each book)
Weaknesses:
1. Time and Money
2. Books themselves do not contain many practice questions (though online and OG 12 references more than compensate)
Notes: most often recommended guide is the Manhattan GMAT Number properties and MGMAT Word Translations
Rival of the MGMAT bundle though a bit higher priced.
Strengths:
1. 12 total guides + an intro book for quant + test simulation booklet
2. Practice questions. Every one of the guides is packed with practice questions
3. Fresh questions; you probably have not seen them on forums/etc since these books have just been published
Weaknesses:
1. Price: \$250
2. Spacing is a bit strange sometimes with only one question per page
A solid supplement to the Official Guide
Strengths:
1. Fairly comprehensive review of the basic math concepts needed for the GMAT
2. Structure of the book allows you to work from start to finish, building on previous skills learned
3. Sections devoted to Word problem and Data Sufficiency questions
Weaknesses:
1. Statistics, Combinations/Permutations and Probability are not covered
2. Questions do not reflect GMAT questions, but rather are designed to insure that the material is mastered
Notes: Recommended/most interesting problem compilation from Kaplan Math Workbook
Faster than going through the math MGMAT Guides, but also not as thorough
Great for additional math practice questions, but don't expect to find any strategies or math lessons here.
Strengths:
1. 300 real GMAT questions
2. Real GMAT questions retired from past tests
3. The practice questions are organized by level of difficulty
4. The practice questions follow actual GMAT test patterns - great to have one's ear trained.
Weaknesses:
1. The book does not contain any strategies
2. It does have a few short review sections but they are very weak
3. Mostly low and medium difficulty questions. Not very helpful to a person aiming to get above 700.
Bottom Line: this book is optional; usually OG 12 is sufficient by itself.
Notable Specialized Books - Arithmetic, Probability, Combinations, DS, etc
Combinations & Probability
Veritas Prep Combinations & Probability - added 08/24/10
Attachment:
veritas_probability-TOC.gif [ 42 KiB | Viewed 630432 times ]
Dedicated Combinations and Probability Guide considered by many to be the best
Strengths:
1. Good strategies
2. Plenty of practice questions
3. Well organized and laid out
Weaknesses:
1. Possibly crosses into non-gmat section but a very minor concern
.
1936240122
Attachment:
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Contains material on Combinations and Probability
Strengths:
1. Coverage of combinations and probability
2. Coverage of Statistics and Overlapping sets problems
3. Comes with 6 MGMAT tests
Weaknesses:
1. As with all MGMAT Guides, lacks practice questions and relies on the Official Guide for additional practice. If you have covered the OG, you may have to look elsewhere for practice, such as GMAT Club's free collection of probability questions: combinations-permutations-and-probability-references-56486.html
Attachment:
mgmat_foundations_TOC.gif [ 24.12 KiB | Viewed 629065 times ]
Great for refreshing math concepts and building confidence in math
Strengths:
1. Works well with other MGMAT books (structured in a similar way)
2. Best book for brushing up on long forgotten math concepts (fractions, powers, etc)
3. Comes with additional online practice questions
Weaknesses:
1. The book has 400 practice questions but they are not in a GMAT format, rather a math-textbook format
2. Long! 300 full-size pages - I think the expectation is that you skip parts that you know
3. No access to MGMAT tests that usually are included with all of the other 8 guides
Notes: recommended instead of the Kaplan Math Foundations. See a very detailed review of the Foundations of GMAT Math by a member
[Reveal] Spoiler:
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26 May 2013, 03:54
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target60 wrote:
Hi,
I started my GMAT preparation last week.I have a target of 750+. Per recommendations of many , I am going with OG13 & Manhatten SC. But, I feel OG's quants are damn easy. Kindly suggest me some real 750+ kinda book for quant . Kaplan advanced questions are not like Gmat's. Waiting prompt responses.
700 DS problems: search.php?search_id=tag&tag_id=180
700 PS problems: search.php?search_id=tag&tag_id=187
Hope it helps.
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Re: Best GMAT Math Prep Books (Reviews & Recommendations) [#permalink]
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09 Jul 2013, 12:15
for a final brushup, i feel MGMAT adv quant is very useful
Consider a Kudos if you find this post useful
in my opinion, LSAT books contain some very good verbal questions which goes into the depth of the subject, but i am not sure if such questions are useful for GMAT
my opinion was based on , short term preparation course. especially for non-natives
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Last edited by Bunuel on 09 Jul 2013, 13:30, edited 1 time in total.
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Re: Best GMAT Math Prep Books (Reviews & Recommendations) [#permalink]
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29 Jul 2013, 12:35
Hi,
I am considering using only Gmatclub's Math book. I have gone through about 20 pages and I feel like It is exactly what all the theory is needed for GMAT. Though I have Manhattan's Guides, I am afraid they may take quite some time.My exam is about in 45 days and so I am considering going through Gmatclub Math book and topic wise OG practice using the directory.My Target in quant is 49+. Will this work for me? Btw, I am a retaker and got Q44 in my first take.
Thanks,
Pavan.
Last edited by pavan2185 on 29 Jul 2013, 23:33, edited 1 time in total.
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29 Jul 2013, 12:51
Hi Pavan,
What is GC? would you please elaborate? If your exam is so near, i would suggest you concentrate on verbal more as it may take more time.
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Re: Best GMAT Math Prep Books (Reviews & Recommendations) [#permalink]
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29 Jul 2013, 23:35
WaterFlowsUp wrote:
Hi Pavan,
What is GC? would you please elaborate? If your exam is so near, i would suggest you concentrate on verbal more as it may take more time.
Hi, by GC I meant Gmatclub. Edited my previous post to remove this confusion.
Coming to verbal, I have been spending most of my time on verbal since about 2-3 weeks.
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30 Jul 2013, 01:42
As a well wisher I would suggest complete OG and quant review 2 times each and nothing else would be required in quant. Easily u can score 49 in Quant
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03 Sep 2013, 15:48
The main post needs to be updated with OG 13.
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16 Sep 2013, 08:21
A good book for quantitative aptitude is by Arun Sharma. The book has some mistakes in the answer key but the level of questions is good...
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16 Sep 2013, 12:09
Hi,
I just have question specifically related to quant prep, at the first page of posts you said that if
Rusty with Math or want to be very methodical - start with basics: Manhattan GMAT Foundations of GMAT Math
then in the same page for a particular user to mentioned
If you are completely bad and rusty with math - get Kaplan Math Foundations
I would like to buy the MGMAT quant set (would also like to know if thats a wise decision) but before that would like to get my fundamentals strong and clear thoroughly, so could you clarify among MGMAT and kaplan foundations which is the right book to pick up and start polishing my basics.
If the MGMAT quant set of 5 books is preferable which presently are
Fractions, Decimals, & Percents GMAT Strategy Guide, 5th Edition
Algebra GMAT Strategy Guide, 5th Edition
Word Problems GMAT Strategy Guide, 5th Edition
Geometry GMAT Strategy Guide, 5th Edition
Number Properties GMAT Strategy Guide, 5th Edition
So, Is buying the entire set preferable or buying few specific books from this set and other topics from some other publisher is prefered, and if it is prefered this later way then among the MGMAT which should I buy and which other publisher should I buy.
I would like get clarified with the which foundation of math to buy (either MGMAT foundation of math or Kaplan foundation of math) --have seen a few comments like MGMAT foundation has a lot of errors but those posts were quite old, now the fifth edition has come out so has these been taken care in the MGMAT foundation of math?
Once clarified with the foundation would like to know next which books to buy, either the entire MGMAT quant set (5 books -- will just following this entire cover all the quant topics in gmat? ) or topic wise separate publishers are preferred. I would like to get the best follow it to the end, once done will move to OG and others but until then to instil the hardcore confidence I would like to get myself completely clear with quant individually topic preparation.
Hope my post isn't intriguing but I just want to have a crystal clear path before in hand of what I need to do and follow. A detailed reply would really be helpful for me to decide which to buy and start my preparation as I rather weak in quant than verbal.
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16 Sep 2013, 12:24
Get the MGMAT Book set + MGMAT Foundations.
Hope this clears it up.
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28 Nov 2013, 04:31
I'm using Manhattan GMAT and I think is fine. Just I can't find a detailed section about sequences and series.
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03 Dec 2013, 10:42
I am working with GMAT MATH BIBLE - NOVA . I found question are good and separate in categories ranging from easy to very hard. But if compare Nova with GMAT forum question I found Nova quite outdated. Please someone help and let me know whether to work with NOVA help or not.
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28 Mar 2014, 18:45
Have you guys tried out the Jay Stewart's Ultimate Guide to GMAT Math books? I'm considering getting them on Kindle so I could study on lunch, breaks and whenever I have time.
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10 May 2014, 06:16
Hi everyone
Though I have asked this question to bb via PM , I want to present it to wider audiences so here it goes.
I am from Engineering background so have a pretty strong fundamental grasp on Maths. Having said so I started my preparation with the GMAT Math club book along with the questions tagged at the end of each chapter. Needless to say , I was rusty so the solutions were taking time. After consulting with some friends who have given GMAT , I got hold of Manhattan guides ( 5 Maths books). Now I am finding these books too trivial. For eg , there is no way lot many questions asked in this forum under 600-700 and 700 category can be answered if one refers to these books only. So my concern is whether Manhattan material is providing sufficient coverage for GMAT ?
Any pointers will be appreciated .
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27 May 2014, 03:32
Hi guys!
I finished the Official Math Review, 2d edition. After careful analysis I found must difficult questions and topics for me:
1)Word problems
2)Tough questions about divisibility, LCM and similar questions.
3) Probability and combinations.
4) Sets
I passed two Official free practice tests (from official GMAT website); my results of quant part: 1 attempt - 43Q
2 attempt - 44Q
I would like to pass the GMAT test in August and want to reach 49-51 in Quant.
I also read previous material on this site and made a decision to continue my preparation with following books:
- GMAT Club Math Book
- Manhattan set of math books
Is it the right choice for me?
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27 May 2014, 03:41
i am project manager what is GMAT...
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27 May 2014, 06:09
1
KUDOS
Expert's post
7
This post was
BOOKMARKED
NikolayGlu wrote:
I finished the Official Math Review, 2d edition. After careful analysis I found must difficult questions and topics for me:
1)Word problems
2)Tough questions about divisibility, LCM and similar questions.
3) Probability and combinations.
4) Sets
Check this:
ALL YOU NEED FOR QUANT
.
All questions by category: viewforumtags.php
All DS word problems to practice: search.php?search_id=tag&tag_id=183
All PS word problems to practice: search.php?search_id=tag&tag_id=56
All DS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=354
All PS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=185
Theory on Combinations: math-combinatorics-87345.html
DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52
Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html
Theory on probability problems: math-probability-87244.html
All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54
Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html
Theory on Overlapping Sets:
how-to-draw-a-venn-diagram-for-problems-98036.html
All DS Overlapping Sets Problems to practice: search.php?search_id=tag&tag_id=45
All PS Overlapping Sets Problems to practice: search.php?search_id=tag&tag_id=65
Also, chek the following post for bunch of useful questions/theory topics: new-project-review-discuss-and-get-kudos-points-153555.html
Hope this helps.
NikolayGlu wrote:
I also read previous material on this site and made a decision to continue my preparation with following books:
- GMAT Club Math Book
- Manhattan set of math books
Is it the right choice for me?
Seems like a good plan.
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27 May 2014, 06:16
Bunuel, thank you a lot for this information!!!
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22 Jul 2014, 09:48
Thank You so much. Just what i needed.
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28 Jul 2014, 01:03
Yes McGraw Hill Book is great book, If the math section of the GMAT is the most difficult part of the test for you, this book can help! Get complete explanations of every GMAT math topic, plus full-length practice GMAT math sections with solutions.
Re: Best GMAT Math Prep Books (Reviews & Recommendations) [#permalink] 28 Jul 2014, 01:03
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Display posts from previous: Sort by | 6,524 | 25,047 | {"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-2017-09 | longest | en | 0.852991 |
http://inftyreader.org/inftyreader-examples/Mathematical_Economics_A_Reader-1.xhtml | 1,620,616,084,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2021-21/segments/1620243989030.65/warc/CC-MAIN-20210510003422-20210510033422-00049.warc.gz | 22,273,657 | 1,414 | ]> No Title
$\mathrm{\left[}\mathrm{\right]}$ Let $\mathit{S}$ be any non‐empty subset of $\mathbb{R}$. The set of
$\mathrm{•}$ upper bounds of $\mathit{S}$ is $\mathit{U}\mathrm{\left(}\mathit{S}\mathrm{\right)}\mathrm{=}\mathrm{\left\{}\mathit{u}\mathrm{\in }\mathbb{R}\mathrm{ }\mathrm{:}\mathrm{ }\mathit{u}\mathrm{\ge }\mathit{s}\mathrm{\forall }\mathit{s}\mathrm{\in }\mathit{S}\mathrm{\right\}}\mathrm{.}$
$\mathrm{•}$ lower bounds of $\mathit{S}$ is $\mathit{L}\mathrm{\left(}\mathit{S}\mathrm{\right)}\mathrm{=}\mathrm{\left\{}\mathit{l}\mathrm{\in }\mathbb{R}\mathrm{ }\mathrm{:}\mathrm{ }\mathit{l}\mathrm{\le }\mathit{s}\mathrm{\forall }\mathit{s}\mathrm{\in }\mathit{S}\mathrm{\right\}}\mathrm{.}$
$\mathrm{\left[}\mathrm{\right]}$ Let $\mathit{S}$ be any non‐empty subset of $\mathbb{R}$. Then
$\mathrm{•}$ the supremum of $\mathit{S}$ in $\mathbb{R}$ is the unique smallest element in $\mathit{U}\mathrm{\left(}\mathit{S}\mathrm{\right)}$ .
$\mathrm{•}$ the infimum of $\mathit{S}$ in $\mathbb{R}$ is the unique largest element in $\mathit{L}\mathrm{\left(}\mathit{S}\mathrm{\right)}$ .
$\mathrm{•}$ the supremum and infimum of $\mathit{S}$ always exist.
$\mathrm{•}$ the supremum (infimum) is also referred to as the least upper bound (greatest lower bound).
$\mathrm{\left[}\mathrm{\right]}$ Let $\mathit{S}$ be any non‐empty subset of $\mathbb{R}$. Then
$\mathrm{•}$ a maximum of $\mathit{S}$ in $\mathbb{R}$ is an element ${\mathit{s}}^{\mathrm{*}}\mathrm{\in }\mathit{S}$ with ${\mathit{s}}^{\mathrm{*}}\mathrm{\ge }\mathit{s}\mathrm{\forall }\mathit{s}\mathrm{\in }\mathit{S}\mathrm{.}$
$\mathrm{•}$ a minimum of $\mathit{S}$ in $\mathbb{R}$ is an element ${\mathit{s}}_{\mathrm{*}}\mathrm{\in }\mathit{S}$ with ${\mathit{s}}_{\mathrm{*}}\mathrm{\le }\mathit{s}\mathrm{\forall }\mathit{s}\mathrm{\in }\mathit{S}\mathrm{.}$
$\mathrm{•}$ a maximim and/or minimum of $\mathit{S}$ may not exist.
28 | 672 | 1,928 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 38, "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 | 3 | CC-MAIN-2021-21 | latest | en | 0.278093 |
https://www.convert-measurement-units.com/convert+Kilogram-force+per+square+millimeter+to+N+mm2.php | 1,679,959,170,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00540.warc.gz | 795,076,126 | 13,307 | Convert kgf/mm² to N/mm² (Kilogram-force per square millimeter to N/mm²)
Kilogram-force per square millimeter into N/mm²
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Kilogram-force+per+square+millimeter+to+N+mm2.php
Convert Kilogram-force per square millimeter to N/mm² (kgf/mm² to N/mm²):
1. Choose the right category from the selection list, in this case 'Pressure'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Kilogram-force per square millimeter [kgf/mm²]'.
4. Finally choose the unit you want the value to be converted to, in this case 'N/mm²'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '11 Kilogram-force per square millimeter'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Kilogram-force per square millimeter' or 'kgf/mm2'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Pressure'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '90 kgf/mm2 to N/mm2' or '92 kgf/mm2 into N/mm2' or '54 Kilogram-force per square millimeter -> N/mm2' or '9 kgf/mm2 = N/mm2' or '49 Kilogram-force per square millimeter to N/mm2' or '32 Kilogram-force per square millimeter into N/mm2'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(12 * 63) kgf/mm2'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '11 Kilogram-force per square millimeter + 33 N/mm2' or '61mm x 43cm x 40dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 3.137 244 539 352 3×1031. For this form of presentation, the number will be segmented into an exponent, here 31, and the actual number, here 3.137 244 539 352 3. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 3.137 244 539 352 3E+31. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 31 372 445 393 523 000 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 965 | 4,006 | {"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} | 3.25 | 3 | CC-MAIN-2023-14 | latest | en | 0.734432 |
https://socratic.org/questions/how-do-you-solve-5y-13-18 | 1,725,981,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00131.warc.gz | 511,881,926 | 5,972 | # How do you solve 5y+13=18?
Dec 28, 2016
$\textcolor{p u r p \le}{\text{y = 1}}$
#### Explanation:
$\textcolor{\tan}{\text{Subtract 13 on both sides:}}$
$5 y + \cancel{13} = 18$
$\textcolor{w h i t e}{a a a} - \cancel{13}$ $\textcolor{w h i t e}{} - 13$
13 cancels out on the left side, and when you subtract 13 from 18 on the right you get 5:
$5 y = 5$
$\textcolor{red}{\text{Divide both sides by 5 to get y by itself.}}$
(cancel"5"y)/cancel"5"=(5)/5
Now you are just left with $y$ on the left side and 1 on the right side.
Therefore,
$\textcolor{m a \ge n t a}{\text{y = 1}}$ | 228 | 589 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.633242 |
https://math.stackexchange.com/questions/2061606/can-octonions-be-used-to-rotate-7-dimensional-vectors | 1,558,258,572,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00531.warc.gz | 559,023,265 | 35,784 | # Can octonions be used to rotate 7-dimensional vectors?
A friend told me that, the same way you can represent a 3-vector as an imaginary quaternion then conjugate it by a unit quaternion with real part $\cos(\frac{\theta}{2})$ to rotate it by $\theta$ around the axis of the imaginary part, you can do the same thing to a 7-vector and a unit octonion. Can't find anything verifying this-if its true a source for the derivation for the formula would be appreciated (as would one for the quaternion formula).
• Remember that octonions are also non-associative, so it may not be possible to compose rotations using octonions in the same way that standard rotations are composed. – Michael McGovern Dec 16 '16 at 19:39
• – Omnomnomnom Dec 16 '16 at 19:39
• @Omnomnomnom I actually skimmed that article before posting the question-as interesting as it is I'm interested in writing programs that simulate higher dimensional rotations, so unless the cross-product has a relevance to rotation that I've forgotten – Eben Cowley Dec 16 '16 at 19:51
• I thought that one could come up with an analog to the Rodriguez rotation formula using this cross-product, but that doesn't look to be as promising as it did. In particular, this cross-product isn't connected to the Lie algebra of $SO(7)$ in the way that the 3D cross-product is connected to $SO(3)$. You could generate some of the rotations using cross-products, though. – Omnomnomnom Dec 16 '16 at 20:18
• The unit quaternions cover $SO(3)$ by parameterizing $SU(2)$. The latter happens because there's Lie group isomorphism which maps the quaternion ball to $SU(2)$ by sending the roots of $-1$ to the Pauli matrices. I don't actually know if $SO(7)$ has an $SU(6)$ universal covering, but this is evidently irrelevant because the Octinions are non-associative, thus shouldn't be isomorphic to $SU(6)$. – SZN Dec 17 '16 at 0:10
First off, some facts. Every isometry fixing the origin is an orthogonal linear transformation of space, the group of all such linear operators is called $O(V)$. The ones which preserve the orientation of space (equivalently, are path-connected to the identity map) form a subgroup called the special orthogonal group $SO(V)$, which has index two.
A hyperplane is a codimension one subspace, or equivalently the orthogonal complement of a 1D line, and a reflection is an isometry which acts as negative one on a line and as the identity on its orthogonal complement (a hyperplane). Reflections are in $O(V)$ but not $SO(V)$, and every isometry is a composition of reflections. In particular, a composition of reflections across the complements of lines $\ell_1$ and $\ell_2$ will be a rotation in the 2D plane $\mathrm{span}(\ell_1,\ell_2)$ by an angle twice that of the angle between $\ell_1$ and $\ell_2$.
A plane rotation is a map which acts as a usual rotation on a 2D subspace and acts as the identity map on the plane's orthogonal complement. Plane rotations are elements of $SO(V)$. Two plane rotations with respect to orthogonal planes will commute with each other. Every rotation, i.e. every element of $SO(V)$, is a product of plane rotations with respect to a collection of pairwise orthogonal 2D planes. If $R$ is a plane rotation with respect to the (oriented) plane $\Pi$ by an angle of $\theta$, and $S$ is any other isometry, then $SRS^{-1}$ will be a plane rotation with respect to the oriented plane $S\Pi$ by an angle of $\theta$.
In a normed division algebra $\mathbb{K}$, there is a multiplicative norm $|\cdot|$ (i.e. it satisfies $|xy|=|x||y|$), and it induces an inner product $\langle-,-\rangle$ satisfying $|x|^2=\langle x,y\rangle$. The imaginary elements are those that are perpendicular to $1$ with respect to this inner product. If $u,v$ are orthogonal imaginaries, then they anticommute, i.e. $uv=-vu$
More generally, for any imaginaries $u$ and $v$, there is a formula $uv=-\langle u,v\rangle+u\times v$. For quaternions, $u\times v$ is the usual cross product, but for octonions it is a much less symmetrical seven-dimensional cross product (which has symmetry group $G_2\subset SO(7)$).
The division algebra $\mathbb{K}$ will also be alternative. This means every pair of elements generates an associative subalgebra, even if $\mathbb{K}$ itself isn't associative. Concretely this means the algebraic expressions $aab$, $aba$ and $baa$ are unambiguous without parenthesization for all $a,b$.
The square roots of negative one are precisely the unit pure imaginaries $u$. Since $\mathbb{K}$ is alternative, $u(ux)=(uu)x=-x$ makes sense, and thus $\mathbb{O}$ becomes a left and right four-dimensional complex vector space over $\mathbb{R}[u]$ with scalar multiplication from the left and right. The two sides don't necessarily match: multiplying anything in $\mathrm{span}(1,u)$ on either side does the same thing, but $uv=-vu$ for all $v$ in the orthogonal complement of $\mathrm{span}(1,u)$.
In the quaternions $\mathbb{H}$, given any unit imaginary $u$ we can form an oriented orthonormal basis $1,u,v,w$ (satisfying the same relations as $1,i,j,k$). Then multiplying by $e^{\theta u}=\cos(\theta)+\sin(\theta)u$ on the left will rotate $\mathrm{span}(1,u)$ and $\mathrm{span}(v,w)$ each by $\theta$, whereas right multiplication will rotate $\mathrm{span}(1,u)$ and $\mathrm{span}(v,w)$ by $\theta$ and $-\theta$ respectively. If we combine the two actions using a half angle, we find that $x\mapsto e^{(\theta/2)u}xe^{-(\theta/2)u}$ acts as the identity on $\mathrm{span}(1,u)$ and a rotation by $\theta$ on $\mathrm{span}(v,w)$. If we restrict this conjugation action to the 3D subspace of imaginary quaternions, this is rotation around the oriented axis $\mathbb{R}u$ by an angle of $\theta$.
The unit quaternions are closed under multiplication and form a group called $\mathrm{Sp}(1)$ or $S^3$ (some abusively call it $\mathrm{SU}(2)$ which I dislike). Using this conjugation action, we get a group homomorphism $\mathrm{Sp}(1)\to\mathrm{SO}(3)$ with kernel $\pm1$.
The same thing happens in $\mathbb{O}$, except now $e^{(\theta/2)u}xe^{-(\theta/2)u}$ will be a rotation in three different planes within the 7D space of imaginary octonions. I don't see any obvious description of which three planes (indeed, the three planes are not an invariant - one could use different sets of three planes to describe the same rotation, such is the ambiguity of isoclinic rotations). I assume they generate $SO(7)$ but it's not clear to me how. One could presumably check using some calculation with lie algebras but that doesn't sound fun or enlightening.
For $\mathbb{O}$ we could also do the following. The effect of $x\mapsto e^{(\theta/2)u}xe^{(\theta/2)u}$ is the plane rotation by $\theta$ with respect to the 2D plane $\mathrm{span}(1,u)$. We can conjugate two of these plane rotations to get an arbitrary plane rotation of $\mathbb{O}$, which generates all of them in $SO(8)$.
Alternatively, note that $x\mapsto -\overline{x}$ (where $\overline{x}$ is conjugation - it negates the imaginary part while preserve the real part) is a reflection across the hyperplane orthogonal to the real axis, i.e. across the subspace of imaginary elements. If we conjugate this reflection by a rotation from $1$ to $q$ (e.g. left multiplication by $q$, where $q$ is a unit quaternion), we get $x\mapsto q(-\overline{qx})=-q\overline{x}q^{-1}$ which is reflection across the hyperplane orthogonal to $q$. If we do this for both $p$ and $q$, we get that $x\mapsto p(qxq^{-1})p^{-1}$ is some plane rotation.
This gives another way to see that multiplication on the left and right by unit octonions generates $SO(8)$. For quaternions, in fact we have a $2$-to-$1$ group homomorphism $\mathrm{Sp}(1)\times\mathrm{Sp}(1)\to SO(4)$.
Far less obviously, since $u^2=-|u|^2$ for imaginary octonions, we get a left action of the Clifford algebra $\mathrm{Cliff}(7)$ on $\mathbb{O}\cong\mathbb{R}^8$. From the classification of Clifford algebras, we know $\mathrm{Cliff}(7)\cong M_8(\mathbb{R})\oplus M_8(\mathbb{R})$. Since the algebra representation $\mathrm{Cliff}(7)\to\mathrm{End}_{\mathbb{R}}(\mathbb{O})$ is nontrivial, by checking dimensions and using the fact the only ideals are the two summands we see the map is surjective. Therefore, every rotation of $\mathbb{O}$ can be attained simply by left multiplying by enough pure imaginary octonions. (The same works on the right side as well.)
• What's wrong with the group of unit quaternions being referred to as SU(2)? Do you just dislike implicitly speaking up to isomorphism? – Eben Cowley Jan 3 '18 at 16:43
• @EbenCowley I am comfortable speaking up to isomorphisms in many contexts. Not a hundred percent sure why I don't like it for $SU(2)$ and $Sp(1)$. Partly because I consider the isomorphism between them in some sense nontrivial, partly because they have different explanations for their double covers of $SO(3)$. It's kind of like referring to the group of unit complex numbers as $SO(2)$, or the additive group of reals as $SO^+(1,1)$. – arctic tern Mar 12 '18 at 17:24 | 2,420 | 9,042 | {"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} | 3.09375 | 3 | CC-MAIN-2019-22 | latest | en | 0.930073 |
https://platformdyzt.web.app/mackaman52268pity/how-to-calculate-weightage-in-index-number-po.html | 1,642,963,314,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.5/warc/CC-MAIN-20220123172206-20220123202206-00042.warc.gz | 505,941,941 | 6,843 | How to calculate weightage in index number
Construction of Price Index Numbers (Formula and Examples) 5. books will be given more weightage while preparing the cost-of-living index for teachers than
Financial Indexes are constructed to measure price movements of stocks, bonds, In order to calculate the beta of a portfolio, multiply the weightage of each stock in Higher IWF suggests greater number of shares held by the investors as 22 Nov 2019 How to calculate metrics and scorecards: learn about KPI weight, normalization, leading, and lagging metrics, calculating the total scorecard index. Its primary goal is to show a number that can give us an idea about the A z-score, or standard deviation, is a measure of the dispersion of data. 67% of all the measurements lie between these two numbers, as shown on the graph below. EpiInfo (cannot calculate indices using the WHO standard population). 7 Jan 2019 Nifty Bank (BANKNIFTY) is the bank index comprises 12 most liquid and large capitalized stocks from the banking sector listed on the National
Then, based on the formula, index number = (price in a particular year/price in the base year) x 100, we calculate the index number, for every year
To calculate the value of the next data point in this indexed time series, let’s say the second year of annual sales equates to \$225,000. You would divide the new data point (\$225,000) by the original one (\$150,000), multiplying the result by 100 as follows to get a year 2 index value of 167. ADVERTISEMENTS: In this article we will discuss about:- 1. Meaning of Index Numbers 2. Features of Index Numbers 3. Steps or Problems in the Construction 4. Construction of Price Index Numbers (Formula and Examples) 5. Difficulties in Measuring Changes in Value of Money 6. Types of Index Numbers 7. Importance 8. Limitations. Meaning of Index […] The formula of to calculate index number iswhere 0Q = Quantity at base time1Q = Quantity at specific time11.1.1 Price Index1. Price index is an example of index number which is widely used.2. It tells us how the price changes over a fixed period of time.3. Price index of a certain item, 10001×=QQI where: (a) To calculate the value of a value-weighted index, sum the market capitalization for each company and divide it by a divisor which is set initially to make the index a round number. To unlock this To calculate the average, you'll first convert your percentages into decimal form, then add all your data points together and divide them by the number of data points you had. So, you have: (sum of data points) ÷ number of data points = average Price-Weighted Index: A price-weighted index is a stock index in which each stock influences the index in proportion to its price per share. The value of the index is generated by adding the
For example: i want to give more weightage to the age variable as it maybe a strong predictor. Could you please provide your suggestion for this scenario ? What are the techniques used to give more weightage to one property over the other
22 Dec 2018 During the construction phase it is important to measure the project progress. This index compares the planned and actual value of works done, if less activities and the total man-hours the weightages for individual activities project which is having 'm' number of least breakable activities. Y(0)=X(0). 21 Feb 2008 The Index was initially calculated based on the "Full Market multiplying the price of its stock by the number of shares issued by the company. The basic formula for a weighted average where the weights add up to 1 is x1(w1) + x2(w2) + x3(w3), and so on, where x is each number in your set and w is the corresponding weighting factor. To find your weighted average, simply multiply each number by its weight factor and then sum the resulting numbers up. Using the formula above, we can calculate the weight of each index component: How to Calculate the Value of a Price-Weighted Index. In theory, the value of the index can be determined as an arithmetic average by dividing the total sum of the prices of the components in the index by the number of the index components. Compute the weighted aggregative price index numbers for \$\$1981\$\$ with \$\$1980\$\$ as the base year using (1) Laspeyre’s Index Number (2) Paashe’s Index Number (3) Fisher’s Ideal Index Number (4) Marshal-Edgeworth Index Number. Since the weight of all grades are equal, we can calculate these grades with simple average or we can cound how many times each grade apear and use weighted average. 2×70,3×80,1×90 x = (2×70+3×80+1×90) / (2+3+1) = 470 / 6 = 78.33333
Using the formula above, we can calculate the weight of each index component: How to Calculate the Value of a Price-Weighted Index. In theory, the value of the index can be determined as an arithmetic average by dividing the total sum of the prices of the components in the index by the number of the index components.
15 Oct 2018 The first is calculating percentages. To calculate a percentage score, you divide the number of points earned by the number of points possible. 22 Dec 2018 During the construction phase it is important to measure the project progress. This index compares the planned and actual value of works done, if less activities and the total man-hours the weightages for individual activities project which is having 'm' number of least breakable activities. Y(0)=X(0). 21 Feb 2008 The Index was initially calculated based on the "Full Market multiplying the price of its stock by the number of shares issued by the company. The basic formula for a weighted average where the weights add up to 1 is x1(w1) + x2(w2) + x3(w3), and so on, where x is each number in your set and w is the corresponding weighting factor. To find your weighted average, simply multiply each number by its weight factor and then sum the resulting numbers up. Using the formula above, we can calculate the weight of each index component: How to Calculate the Value of a Price-Weighted Index. In theory, the value of the index can be determined as an arithmetic average by dividing the total sum of the prices of the components in the index by the number of the index components.
(ii) Index numbers measure the changes in the level of a given phenomenon. (iii) Index numbers The weight age may be according to either : (1) the value of
Calculate the composite index number for the monthly expenditure of 125 + 110 + 105}{4} [math] for assuming that each one has 1 weightage. 25 Mar 2019 Consumer price index (CPI) is a statistic used to measure average price of a basket of commonly-used goods and services in a period relative
Second, an index number measures the net increase or decrease of the average prices for the group under study. For instance, if the consumer price index has increased from 150 in 1982 as compared to 100 in 1980, it shows a net increase of 50 per cent in the prices of commodities included in the index. But we could as easily have weighted the candidates on a 1-10 scale, or given 5 to the acceptable answers and 0 to the rest. It’s wholly up to you to assign the weight and their significance. Just keep in mind: The weight should reflect the urgency of each answer. You can use any number of rating questions in a survey. | 1,659 | 7,252 | {"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-2022-05 | latest | en | 0.883521 |
https://socratic.org/questions/how-do-you-find-the-derivative-of-y-arcsin-x-2 | 1,723,389,289,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00098.warc.gz | 436,320,205 | 5,615 | # How do you find the derivative of y=arcsin(x^2)?
By Chain Rule, you will find $y ' = \frac{2 x}{\sqrt{1 - {x}^{4}}}$. | 47 | 120 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "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} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.722348 |
http://www.topperlearning.com/forums/home-work-help-19/how-many-squares-can-be-formed-b-w-m-parallel-horizontal-lin-mathematics-permutations-and-combinations-56877/reply | 1,496,116,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00315.warc.gz | 847,114,894 | 37,875 | Question
Sat June 30, 2012 By:
# How many squares can be formed b/w m parallel horizontal lines & n parallel vertical lines.
Sun July 01, 2012
Answer :Given : m parallel horizontal lines , and n parallel vertical lines.
to find : the no. of squares formed by these given lines.
Since we know a single square can be formed using 2 parallel horizontal lines and 2 parallel vertical lines. since the lines are parallel and one is horizontal, other is vertical, it would satisfy all the properties of a square to be formed i.e each angle on vertice is 90 degree with all sides equal .
As we know , if there are 4 horizontal lines , then 3 blocks are made
| 1 | 2 | 3 |
similarly , if there are 3 vertical lines , 2 blocks willbe formed
--------
1
--------
2
--------
=> if they are combined = 3*2 = 6 blocks would be formed
------------------------------------------------------------------------------
| | | |
| | | |
| 1 | 2 | 3 |
----------------------------------------------------------------------------
| | | |
| 4 | 5 | 6 |
| | | |
---------------------------------------------------------------------------
therefore if we see this same problem in the same note , we see that (m-1) * (n-1) total sqares will be formed.
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Thu April 06, 2017
# Q) Total number of permutation of "K" different things in a row taken not more than "r " at a time (each thing may be repeated any number of times ) is equal to
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### Fast converging sums involving tetrations
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### Can $x^{x^{x^x}}$ be a rational number?
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### Is ${^5\pi}$ an integer? [duplicate]
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https://www.hackmath.net/en/calculator/fraction?input=3+%2F+1%2F3 | 1,632,015,651,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056656.6/warc/CC-MAIN-20210919005057-20210919035057-00600.warc.gz | 830,828,608 | 10,585 | Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
Result:
3 / 1/3 = 9/1 = 9
Spelled result in words is nine.
How do you solve fractions step by step?
1. Divide: 3 : 1/3 = 3/1 · 3/1 = 3 · 3/1 · 1 = 9/1 = 9
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 1/3 is 3/1) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators.
In words - three divided by one third = nine.
Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
Fractions in word problems:
• Reciprocals
Which among the given reciprocal is correct a. 3/15x1/3= 1 b. 3/20x20/3=1 c. 7/14x7/7=1 d. 34/3x34/34=1
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How many minutes is a third of an hour? Do you know to determine a third of the lesson hour (45min)?
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If a=6 and x=2, find the value of (2ax + 7x- 10) /(4ax-3a-2)
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The bottle contains 1.5 liters of mineral water. Pour all the water from the bottle into empty cups with a volume of 1/3 l. All but one will be filled to the brim. What part of the volume of the last cup is filled with water?
• Paula
Paula is painting six doors that are all the same size. She used 2 liters of paint to cover 1 3/5 doors. How many liters of paint are needed for 1 door?
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Three divisions of nature protectors participated in the collection of chestnut trees.1. the division harvested 1250 kg, the 2nd division by a fifth more than the 1st division and the 3rd division by a sixth more than the second division. How many tons of
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What is the quotient of 2/9 and 1/8?
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4/5 of a number is 276. what is 2/3 of the same number?
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Which expression is equivalent to : minus 9 minus left parenthesis minus 4 start fraction 1 divided by 3 end fraction right parenthesis | 1,501 | 5,445 | {"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} | 4.59375 | 5 | CC-MAIN-2021-39 | longest | en | 0.858155 |
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# phy124lab3 - phy124:lab_3[Stony Brook Physics for Life...
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PHY 124 Lab 3 - DC circuits Important! You need to print out the 2 page worksheet you find by clicking on this link and bring it with you to your lab session. [ ] If you need the .pdf version of these instructions you can get them here [ /labs/dokuwiki/pdfs/phy124lab3.pdf] . Goals The purpose of this laboratory is to observe the relationship between voltage drop across, and current through, electrical circuit components. You will also gain familiarity with connecting circuits and with voltmeters (measuring voltage) and ammeters (measuring current). Equipment 1 DC Power Supply 1 Voltmeter 1 Ammeter 1 board with resistive components 7 wires 4 clamps phy124:lab_3 [Stony Brook Physics for Life Sciences] ... 1 of 6 2/17/2010 9:17 AM
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Introduction Ohmic components obey Ohm's law, (3.1) where V is voltage and I is current through a resistor with resistance R. Ohmic components will keep a constant resistance with varying voltage and current. This can be seen as a linear relationship on a V vs. I graph. Standard circuit resistors are ohmic. Other kinds of electrical components may not be and in these R may depend on factors such as the temperature (as it does in light bulbs), the direction of current flow (for example in diodes), or the light intensity falling on the component (light sensitive diodes). You will use a voltmeter to measure V, the voltage drop across the component, and an ammeter to measure I, the current flow through the component. You must keep in mind that ammeters must be connected in series in circuits, while voltmeters must be connected in parallel across the circuit component whose voltage drop is to be measured. The sketch below shows the proper setup. Fig 2 Technical note: non-zero meter readings are obtained only when some current flows through the meters. Hence, the meters have two connections. The current flowing through the meters influences the measurement. You want this effect to be very small. Thus “good” voltmeters have a large resistance compared to R, which allows only a small current to pass through. Since the voltmeter is in parallel to the resistor, the current through R is essentially determined by R and only a very small current flows through the voltmeter. The equivalent resistance of R and the voltmeter in parallel is close to the resistance of R because if then . The voltmeter you use has the desired property i.e. it’s resistance is much larger than the resistance to be measured, R. In an ideal case a “Good” ammeter has a very small resistance compared to R, which allows current to easily pass through the ammeter and be measured. When this is the case and the resistor and the ammeter are in series, so the current measured in the ammeter is essentially determined by R. The equivalent resistance of R and the ammeter is close to resistance of R because if then . This is only approximately true
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Post subject: Susan Daughtrey pack 2 test 8 Q22 help pleasePosted: Mon Jun 18, 2012 9:36 pm
Joined: Tue Apr 10, 2012 4:48 pm
Posts: 32
There is the same connection between the word outside the brackets and one word inside each pair of brackets. Find 2 words, one from each group, that completes the sentence in the best way.
Scarcity is to (shortage bare empty)
as abundance is to (heap extra plenty)
My DS got his answer were bare and plenty. The answer should be shortage and plenty. But I didn't know how to explain to him that the model answer was better.
bristolmum
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Post subject: Re: Susan Daughtrey pack 2 test 8 Q22 help pleasePosted: Mon Jun 18, 2012 10:06 pm
Joined: Mon Mar 15, 2010 2:45 pm
Posts: 4606
I'm sure someone will come along with a proper explanation, but is just is! If something is scarce, it is in short supply, bare just doesn't fit the explanation! You might say "the cupboard is bare", but you wouldn't say "the cupboard is scarce". Is he getting infused with "sparse"?
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Post subject: Re: Susan Daughtrey pack 2 test 8 Q22 help pleasePosted: Mon Jun 18, 2012 10:25 pm
Joined: Tue Apr 10, 2012 4:48 pm
Posts: 32
Thanks scary mum.
My DS asked - bare means a little bit or nothing at all?
I know that bare may means nothing such as foot means wear nothing. Am I right?
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Post subject: Re: Susan Daughtrey pack 2 test 8 Q22 help pleasePosted: Mon Jun 18, 2012 10:51 pm
Joined: Tue Apr 10, 2012 4:48 pm
Posts: 32
Sorry typing mistake!
I mean bare feet = didn't wear any shoes so bare means nothing. Am I right?
If I am right, should I explain to him that scarcity means not enough so the answer shortage is more appropriated than bare.
Bristolmum
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Post subject: Re: Susan Daughtrey pack 2 test 8 Q22 help pleasePosted: Tue Jun 19, 2012 7:47 am
Joined: Thu May 10, 2012 8:30 am
Posts: 247
bristolmum wrote:
If I am right, should I explain to him that scarcity means not enough so the answer shortage is more appropriated than bare.
As well as meaning, it's about parts of speech. Scarcity, Shortage, Abundance, Plenty can all be nouns ("there is a scarcity of affordable housing"; "there is a shortage of affordable housing"; "For whosoever hath, to him shall be given, and he shall have more abundance: but whosoever hath not, from him shall be taken away even that he hath" (Matthew 13:12), "So shall thy barns be filled with plenty, and thy presses shall burst out with new wine." (Proverbs 3:10) - these are actually quite archaic uses). Shortage is occasionally an adjective ("physics is a shortage subject") but the others rarely, if ever. None of them are verbs in common usage.
By contrast, Bare is rarely (never?) a noun; it is almost always an adjective ("he had bare feet") or a verb ("It's sunny, so I am going to bare my legs today").
So there are plenty of sentences in which you can replace "scarcity" with "shortage" and still have it both grammatically correct and meaning roughly the same, and likewise "abundance" and "plenty". And you can replace any of them with any other of them and still have something grammatically correct, although the meaning will be inverted. You can't replace any of the first four with bare and have it be grammatically correct. Just reading it aloud should tell you that, even if you can't put names to the reasons. "There is a bare of housing".
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Post subject: Re: Susan Daughtrey pack 2 test 8 Q22 help pleasePosted: Tue Jun 19, 2012 8:05 am
Joined: Mon Mar 15, 2010 2:45 pm
Posts: 4606
I knew someone would be able to explain it better than me!
I wonder if it's something to do with the word "bare" creeping into the English language used in a different way. Some of my DCs' friends use it in a slang way, eg "it was bare good" (not sure if that is the correct slang usage ). Mind you, it still doesn't mean "scarcity" or "shortage", but might cause confusion.
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Post subject: Re: Susan Daughtrey pack 2 test 8 Q22 help pleasePosted: Tue Jun 19, 2012 11:39 am
Joined: Tue Apr 10, 2012 4:48 pm
Posts: 32
Many thanks for your help - scary mum and daveg.
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Post subject: Re: Susan Daughtrey pack 2 test 8 Q22 help pleasePosted: Thu Jun 21, 2012 1:28 pm
Joined: Sun Jan 13, 2008 6:18 pm
Posts: 238
Location: Gloucestershire
Fantastic explanation, daveg.
Scarcity and shortage are nouns whilst bare is an adjective.
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Jump to: Select a forum ------------------ FORUM RULES Forum Rules and FAQs 11 PLUS SUBJECTS VERBAL REASONING MATHS ENGLISH NON-VERBAL REASONING CEM 11 Plus GENERAL GENERAL 11 PLUS TOPICS 11 PLUS APPEALS 11 PLUS TUTORS INDEPENDENT SCHOOLS 11 PLUS CDs/SOFTWARE 11 PLUS TIPS PRIMARY SEN and the 11 PLUS EVERYTHING ELSE .... 11 PLUS REGIONS Berkshire Bexley and Bromley Birmingham, Walsall, Wolverhampton and Wrekin Buckinghamshire Devon Dorset Essex Essex - Redbridge Gloucestershire Hertfordshire (South West) Hertfordshire (Other and North London) Kent Lancashire & Cumbria Lincolnshire Medway Northern Ireland Surrey (Sutton, Kingston and Wandsworth) Trafford Warwickshire Wiltshire Wirral Yorkshire BEYOND 11 PLUS Beyond 11 Plus - General GCSEs 6th Form University | 1,659 | 5,955 | {"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} | 2.59375 | 3 | CC-MAIN-2016-50 | latest | en | 0.951263 |
https://en.sorumatik.co/t/determine-the-amplitude-midline-period-and-an-equation-involving-the-sine-function-for-the-graph-shown-below/3096 | 1,708,480,696,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00459.warc.gz | 233,355,546 | 5,547 | # Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown below
determine the amplitude, midline, period, and an equation involving the sine function for the graph shown below.
Amplitude: The amplitude of a sine function represents the maximum distance that the graph deviates from its midline. To determine the amplitude, we look at the highest and lowest points on the graph. In the given graph, the highest point is at 2 and the lowest point is at -2. The amplitude is the absolute value of either of these values, so the amplitude is 2.
Midline: The midline is the horizontal line that the graph oscillates around. It represents the average value of the function. In the given graph, the midline is the horizontal line y = 0 since the graph oscillates symmetrically above and below this line.
Period: The period of a sine function is the distance it takes for the graph to complete one full cycle. It is determined by the distance between two consecutive peaks or two consecutive troughs. In the given graph, the distance between two consecutive peaks or troughs is 4 units. Therefore, the period of the graph is 4.
Equation: To find the equation involving the sine function, we need to consider the given information about the amplitude, midline, and period.
The general form of the equation for a sine function is:
y = A * sin(B(x - C)) + D
where A is the amplitude, B determines the period, C represents a horizontal shift, and D represents a vertical shift.
Using the given information, we can construct the equation for the graph as follows:
Amplitude = 2, so A = 2
Midline = 0, so D = 0
Period = 4, so B = (2π)/4 = π/2
Since the graph is not shifted horizontally, C = 0.
Putting all the values together, we get the equation:
y = 2 * sin((π/2)x)
This is the equation that represents the given graph involving the sine function. | 436 | 1,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-10 | latest | en | 0.891031 |
https://www.lesswrong.com/posts/XzvR3QKkt9EPbAYyT/applying-the-counterfactual-prisoner-s-dilemma-to-logical | 1,716,783,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00735.warc.gz | 745,838,642 | 66,206 | # Ω 4
Crossposted from the AI Alignment Forum. May contain more technical jargon than usual.
The Counterfactual Prisoner's Dilemma is a symmetric version of Counterfactual Mugging where regardless of whether the coin comes up heads or tails you are asked to pay $100 and you are then paid$10,000 if Omega predicts that you would have paid if the coin had come up the other way. If you decide updatelesly you will always received $9900, while if you decide updatefully, then you will receive$0. So unlike Counterfactual Mugging, pre-committing to pay ensures a better outcome regardless of how the coin flip turns out, suggesting that focusing only on your particular probability branch is mistaken.
The Logical Counterfactual Mugging doesn't use a coin flip, but instead looks at the parity of sometime beyond your ability to calculate, like the 10,000th digit of pi. You are told it is even and then asked to pay $100 on the basis that if Omega predict you would have paid, then he would have given you$10,000 if had turned out to be odd.
You might naturally assume that you couldn't construct a logical version of the Counterfactual Prisoner's Dilemma. I certainly did at first. After all, you might say, the coin could have come up tails, but the 10,000th digit of pi couldn't have turned out to be odd. After all, that would be a logical impossibility.
But could the coin actually have come up tails? If the universe is deterministic, then the way it came up was the only way it could ever have come up. So is there is less difference between these two scenarios than it looks at first glance?
Let's see. For the standard counterfactual mugging, you can't find the contradiction because you lack information about the world, while for the logical version, you can't find the contradiction because of processing power. In the former, we could actually construct two consistent worlds - one where it is heads and one where it is tails - that are consistent with the information you have about the scenario. In the later, we can't.
Notice however that for Logical Counterfactual Mugging to be well defined, you need to define what Omega is doing when it is making its prediction. In Counterfactuals for Perfect Predictors, I explained that when dealing with perfect predictors, often the counterfactual would be undefined. For example, in Parfit's Hitchhicker a perfect predictor would never give a lift to someone who never pays in town, so it isn't immediately clear that predicting what such a person would do in town involves predicting something coherent.
However, even though we can't ask what the hitchhiker would do in an incoherent situation, we can ask what they would do when they receive an input representing an incoherent situation (see Counterfactuals for Perfect Predictors for a more formal description). Indeed, Updateless Decision Theory uses this technique - programs are as defined as input-output maps - although I don't know whether Wei Dai was motivated by this concern or not.
Similarly, the predictor in Logical Counterfactual Mugging must be predicting something that is well defined. So we can assume that it is producing a prediction based on an input, which may possibly represent a logically inconsistent situation. Given this, we can construct a logical version of the Counterfactual prisoner's dilemma. Writing this explicitly:
First you are told the 10,000th digit of Pi. Regardless of whether it is odd or even, you are asked for $100. You are then paid$10,000 if you Omega predicts that you would produce output corresponding to paying when fed input correpsonding to having been informed that this digit had the opposite parity that you observed.
There really isn't any difference between how we make the logical case coherent and how we make the standard case coherent. At this point, we can see that just as per the original Counterfactual Prisoner's Dilemma always paying scores you $9900, while never paying scores you nothing. You are guaranteed to do better regardless of the coin flip (or in Abram Demski's terms we now have an all-upside updateless situation). # 9 # Ω 4 New Comment 5 comments, sorted by Click to highlight new comments since: Notice however that for Logical Counterfactual Mugging to be well defined, you need to define what Omega is doing when it is making its prediction. In Counterfactuals for Perfect Predictors, I explained that when dealing with perfect predictors, often the counterfactual would be undefined. For example, in Parfit's Hitchhicker a perfect predictor would never give a lift to someone who never pays in town, so it isn't immediately clear that predicting what such a person would do in town involves predicting something coherent. Another approach is to change the example to remove the objection. The poker-like game at the end of Decision Theory (I really titled that post simply "decision theory"? rather vague, past-me...) is isomorphic to counterfactual mugging, but removes some distractions, such as "how does Omega take the counterfactual". Alice receives a High or Low card. Alice can reveal the card to Bob. Bob then states a probability for Alice's card being Low. Bob's incentives just encourage him to report honest beliefs. Alice loses . When Alice gets a Low card, she can just reveal it to Bob, and get the best possible outcome. But this strategy means Bob will know if she has a high card, giving her the worst possible outcome in that case. In order to successfully bluff, Alice has to sometimes act like she has different cards than she has. And indeed, the optimal strategy for Alice in this case is to never show her cards. This example will get less objections from people, because it is grounded in a very realistic game. Playing poker well requires this kind of reasoning. The powerful predictor is replaced with another player. We can still technically ask "how is the other player deal with undefined counterfactuals?", but we can skip over that by just reasoning about strategies in the usual game-theoretic way -- if Alice's strategy were to reveal low cards, then Bob could always call high cards. We can then insert logical uncertainty by stipulating that Alice gets her card pseudorandomly, but neither Alice nor Bob can predict the random number generator. Not sure yet whether you can pull a similar trick with Counterfactual Prisoner's Dilemma. == nitpicks == # Applying the Counterfactual Prisoner's Dilemma to Logical Uncertainty Why isn't the title "applying logical uncertainty to the counterfactual prisoner's dilemma"? Or "A Logically Uncertain Version of Counterfactual Prisoner's Dilemma"? I don't see how you're applying CPD to LU. The Counterfactual Prisoner's Dilemma is a symmetric version of the original Symmetric? The original is already symmetric. But "symmetric" is a concept which applies to multi-player games. Counterfactual PD makes PD into a one-player game. Presumably you meant "a one-player version"? where regardless of whether the coin comes up heads or tails you are asked to pay$100 and you are then paid $10,000 if Omega predicts that you would have paid if the coin had come up the other way. If you decide updatelesly you will always received$9900, while if you decide updatefully, then you will receive $0. This is only true if you use classical CDT, yeah? Whereas EDT can get$9900 in both cases, provided it believes in a sufficient correlation between what it does upon seeing heads vs tails.
So unlike Counterfactual Mugging, pre-committing to pay ensures a better outcome regardless of how the coin flip turns out, suggesting that focusing only on your particular probability branch is mistaken.
I don't get what you meant by the last part of this sentence. Counterfactual Mugging already suggests that focusing only on your particular branch is mistaken. If someone bought that you should pay up in this problem but not in counterfactual mugging, I expect that person to say something like "because in this case that strategy is guaranteed better even in this branch" -- hence, they're not necessarily convinced to look at other branches. So I don't think this example necessarily argues for looking at other branches.
Also, why is this posted as a question?
I'm curious, do you find this argument for paying in Logical Counterfactual Mugging persuasive? What about the Counterfactual Prisoner's Dilemma argument for the basic Counterfactual Mugging?
Another approach is to change the example to remove the objection
Interesting point about the poker game version. It's still a one shot game, so there's no real reason to hide a 0 unless you think they're a pretty powerful predictor, but it is always predicting something coherent.
I don't see how you're applying CPD to LU
The claim is that you should pay in the Logical Counterfactual Prisoner's Dilemma and hence pay in Logical Counterfactual Mugging which is the logically uncertain version of Counterfactual Mugging.
Symmetric? The original is already symmetric. But "symmetric" is a concept which applies to multi-player games. Counterfactual PD makes PD into a one-player game. Presumably you meant "a one-player version"?
Edited now. I meant it's a symmetric version of counterfactual mugging. So not in the game theory sense, but just that there is now no difference between heads and tails.
This is only true if you use classical CDT, yeah? Whereas EDT can get \$9900 in both cases, provided it believes in a sufficient correlation between what it does upon seeing heads vs tails.
Point noted. Maybe I should have been more careful about specifying what I was comparing
Also, why is this posted as a question?
Accident. It's fixed now
I always enjoy convoluted Omega situations, but I don't understand how these theoretical entities get to the point where their priors are as stated (and especially the meta-priors about how they should frame the decision problem).
Before the start of the game, Omega has some prior distribution of the Agent's beliefs and update mechanisms. And the Agent has some distribution of beliefs about Omega's predictive power over situations where the Agent "feels like" it has a choice. What experiences cause Omega to update sufficiently to even offer the problem (ok, this is easy: quantum brain scan or other Star-Trek technobabble)? But what lets the Agent update to believing that their qualia of free-will is such an illusion in this case? And how do they then NOT meta-update to understand the belief-action-payout matrix well enough to take the most-profitable action?
Moved to my shortform - it's not a direct answer to the post.
[This comment is no longer endorsed by its author]Reply | 2,256 | 10,674 | {"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} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.966994 |
https://www.doorsteptutor.com/Exams/NCO/Class-6/Questions/Topic-Mental-Ability-0/Part-72.html | 1,524,263,710,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944742.25/warc/CC-MAIN-20180420213743-20180420233743-00005.warc.gz | 778,277,936 | 15,195 | # Mental Ability (NCO- Cyber Olympiad (SOF) Class 6): Questions 388 - 392 of 394
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## Question number: 388
» Mental Ability » Number Games » Number Series
Appeared in Year: 2011
MCQ▾
### Question
Which of the following does not fit in the number series given below?
2, 4, 16, 66, 298, 1530
### Choices
Choice (4) Response
a.
16
b.
66
c.
298
d.
1530
## Question number: 389
» Mental Ability » Basic Geometry
Appeared in Year: 2011
MCQ▾
### Question
Which of the following shapes appears to have at least one obtuse angle?
### Choices
Choice (4) Response
a.
Figure R
b.
Figure P
c.
Figure Q
d.
Figure S
## Question number: 390
» Mental Ability » Basic Geometry
Appeared in Year: 2011
MCQ▾
### Question
In the below given figure, what is the difference between the area of the two squared farms (measures in meter)?
### Choices
Choice (4) Response
a.
b.
m
c.
d.
m
## Question number: 391
» Mental Ability » Number Games » Number Series
MCQ▾
### Question
One number in this sequence is wrong. 8, 16, 18, 36, 38, 78
What change should be made to correct this sequence?
### Choices
Choice (4) Response
a.
78 should be replaced by 76
b.
78 should be replaced by 74
c.
36 should be replaced by 34
d.
38 should be replaced by 40
## Question number: 392
» Mental Ability » Basic Arithmetic
Appeared in Year: 2011
MCQ▾
### Question
⃞ means ‘is greater than’, ∆ means ‘is smaller than’, ○ means ‘is equal to’, means ‘plus’, and = means ‘minus’. If and , then ________.
### Choices
Choice (4) Response
a.
b.
c.
d.
f Page | 525 | 1,829 | {"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-2018-17 | latest | en | 0.772015 |
http://isabelle.in.tum.de/repos/isabelle/annotate/0376cccd9118/doc-src/TutorialI/CTL/document/CTL.tex | 1,568,856,098,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573385.29/warc/CC-MAIN-20190918234431-20190919020431-00003.warc.gz | 89,948,255 | 13,291 | doc-src/TutorialI/CTL/document/CTL.tex
author nipkow Wed Oct 11 10:44:42 2000 +0200 (2000-10-11) changeset 10187 0376cccd9118 parent 10186 499637e8f2c6 child 10192 4c2584e23ade permissions -rw-r--r--
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nipkow@10123 1 % nipkow@10123 2 \begin{isabellebody}% nipkow@10123 3 \def\isabellecontext{CTL}% nipkow@10133 4 % nipkow@10133 5 \isamarkupsubsection{Computation tree logic---CTL} nipkow@10149 6 % nipkow@10149 7 \begin{isamarkuptext}% nipkow@10149 8 The semantics of PDL only needs transitive reflexive closure. nipkow@10149 9 Let us now be a bit more adventurous and introduce a new temporal operator nipkow@10149 10 that goes beyond transitive reflexive closure. We extend the datatype nipkow@10149 11 \isa{formula} by a new constructor% nipkow@10149 12 \end{isamarkuptext}% nipkow@10149 13 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\isacharbar}\ AF\ formula% nipkow@10149 14 \begin{isamarkuptext}% nipkow@10149 15 \noindent nipkow@10149 16 which stands for "always in the future": nipkow@10159 17 on all paths, at some point the formula holds. Formalizing the notion of an infinite path is easy nipkow@10159 18 in HOL: it is simply a function from \isa{nat} to \isa{state}.% nipkow@10149 19 \end{isamarkuptext}% nipkow@10123 20 \isacommand{constdefs}\ Paths\ {\isacharcolon}{\isacharcolon}\ {\isachardoublequote}state\ {\isasymRightarrow}\ {\isacharparenleft}nat\ {\isasymRightarrow}\ state{\isacharparenright}set{\isachardoublequote}\isanewline nipkow@10187 21 \ \ \ \ \ \ \ \ \ {\isachardoublequote}Paths\ s\ {\isasymequiv}\ {\isacharbraceleft}p{\isachardot}\ s\ {\isacharequal}\ p\ {\isadigit{0}}\ {\isasymand}\ {\isacharparenleft}{\isasymforall}i{\isachardot}\ {\isacharparenleft}p\ i{\isacharcomma}\ p{\isacharparenleft}i{\isacharplus}{\isadigit{1}}{\isacharparenright}{\isacharparenright}\ {\isasymin}\ M{\isacharparenright}{\isacharbraceright}{\isachardoublequote}% nipkow@10149 22 \begin{isamarkuptext}% nipkow@10149 23 \noindent nipkow@10159 24 This definition allows a very succinct statement of the semantics of \isa{AF}: nipkow@10149 25 \footnote{Do not be mislead: neither datatypes nor recursive functions can be nipkow@10149 26 extended by new constructors or equations. This is just a trick of the nipkow@10149 27 presentation. In reality one has to define a new datatype and a new function.}% nipkow@10149 28 \end{isamarkuptext}% nipkow@10149 29 {\isachardoublequote}s\ {\isasymTurnstile}\ AF\ f\ \ \ \ {\isacharequal}\ {\isacharparenleft}{\isasymforall}p\ {\isasymin}\ Paths\ s{\isachardot}\ {\isasymexists}i{\isachardot}\ p\ i\ {\isasymTurnstile}\ f{\isacharparenright}{\isachardoublequote}% nipkow@10149 30 \begin{isamarkuptext}% nipkow@10149 31 \noindent nipkow@10149 32 Model checking \isa{AF} involves a function which nipkow@10159 33 is just complicated enough to warrant a separate definition:% nipkow@10149 34 \end{isamarkuptext}% nipkow@10123 35 \isacommand{constdefs}\ af\ {\isacharcolon}{\isacharcolon}\ {\isachardoublequote}state\ set\ {\isasymRightarrow}\ state\ set\ {\isasymRightarrow}\ state\ set{\isachardoublequote}\isanewline nipkow@10149 36 \ \ \ \ \ \ \ \ \ {\isachardoublequote}af\ A\ T\ {\isasymequiv}\ A\ {\isasymunion}\ {\isacharbraceleft}s{\isachardot}\ {\isasymforall}t{\isachardot}\ {\isacharparenleft}s{\isacharcomma}\ t{\isacharparenright}\ {\isasymin}\ M\ {\isasymlongrightarrow}\ t\ {\isasymin}\ T{\isacharbraceright}{\isachardoublequote}% nipkow@10149 37 \begin{isamarkuptext}% nipkow@10149 38 \noindent nipkow@10159 39 Now we define \isa{mc\ {\isacharparenleft}AF\ f{\isacharparenright}} as the least set \isa{T} that contains nipkow@10159 40 \isa{mc\ f} and all states all of whose direct successors are in \isa{T}:% nipkow@10159 41 \end{isamarkuptext}% nipkow@10159 42 {\isachardoublequote}mc{\isacharparenleft}AF\ f{\isacharparenright}\ \ \ \ {\isacharequal}\ lfp{\isacharparenleft}af{\isacharparenleft}mc\ f{\isacharparenright}{\isacharparenright}{\isachardoublequote}% nipkow@10159 43 \begin{isamarkuptext}% nipkow@10159 44 \noindent nipkow@10159 45 Because \isa{af} is monotone in its second argument (and also its first, but nipkow@10159 46 that is irrelevant) \isa{af\ A} has a least fixpoint:% nipkow@10149 47 \end{isamarkuptext}% nipkow@10123 48 \isacommand{lemma}\ mono{\isacharunderscore}af{\isacharcolon}\ {\isachardoublequote}mono{\isacharparenleft}af\ A{\isacharparenright}{\isachardoublequote}\isanewline nipkow@10149 49 \isacommand{apply}{\isacharparenleft}simp\ add{\isacharcolon}\ mono{\isacharunderscore}def\ af{\isacharunderscore}def{\isacharparenright}\isanewline nipkow@10159 50 \isacommand{apply}\ blast\isanewline nipkow@10159 51 \isacommand{done}% nipkow@10149 52 \begin{isamarkuptext}% nipkow@10159 53 All we need to prove now is that \isa{mc} and \isa{{\isasymTurnstile}} nipkow@10159 54 agree for \isa{AF}, i.e.\ that \isa{mc\ {\isacharparenleft}AF\ f{\isacharparenright}\ {\isacharequal}\ {\isacharbraceleft}s{\isachardot}\ s\ {\isasymTurnstile}\ AF\ f{\isacharbraceright}}. This time we prove the two containments separately, starting nipkow@10159 55 with the easy one:% nipkow@10159 56 \end{isamarkuptext}% nipkow@10187 57 \isacommand{theorem}\ AF{\isacharunderscore}lemma{\isadigit{1}}{\isacharcolon}\isanewline nipkow@10159 58 \ \ {\isachardoublequote}lfp{\isacharparenleft}af\ A{\isacharparenright}\ {\isasymsubseteq}\ {\isacharbraceleft}s{\isachardot}\ {\isasymforall}\ p\ {\isasymin}\ Paths\ s{\isachardot}\ {\isasymexists}\ i{\isachardot}\ p\ i\ {\isasymin}\ A{\isacharbraceright}{\isachardoublequote}% nipkow@10159 59 \begin{isamarkuptxt}% nipkow@10149 60 \noindent nipkow@10159 61 The proof is again pointwise. Fixpoint induction on the premise \isa{s\ {\isasymin}\ lfp\ {\isacharparenleft}af\ A{\isacharparenright}} followed nipkow@10159 62 by simplification and clarification% nipkow@10159 63 \end{isamarkuptxt}% nipkow@10123 64 \isacommand{apply}{\isacharparenleft}rule\ subsetI{\isacharparenright}\isanewline nipkow@10123 65 \isacommand{apply}{\isacharparenleft}erule\ Lfp{\isachardot}induct{\isacharbrackleft}OF\ {\isacharunderscore}\ mono{\isacharunderscore}af{\isacharbrackright}{\isacharparenright}\isanewline nipkow@10159 66 \isacommand{apply}{\isacharparenleft}clarsimp\ simp\ add{\isacharcolon}\ af{\isacharunderscore}def\ Paths{\isacharunderscore}def{\isacharparenright}% nipkow@10159 67 \begin{isamarkuptxt}% nipkow@10159 68 \noindent nipkow@10159 69 leads to the following somewhat involved proof state nipkow@10159 70 \begin{isabelle} nipkow@10159 71 \ \isadigit{1}{\isachardot}\ {\isasymAnd}p{\isachardot}\ {\isasymlbrakk}p\ \isadigit{0}\ {\isasymin}\ A\ {\isasymor}\isanewline nipkow@10159 72 \ \ \ \ \ \ \ \ \ {\isacharparenleft}{\isasymforall}t{\isachardot}\ {\isacharparenleft}p\ \isadigit{0}{\isacharcomma}\ t{\isacharparenright}\ {\isasymin}\ M\ {\isasymlongrightarrow}\isanewline nipkow@10159 73 \ \ \ \ \ \ \ \ \ \ \ \ \ \ t\ {\isasymin}\ lfp\ {\isacharparenleft}af\ A{\isacharparenright}\ {\isasymand}\isanewline nipkow@10159 74 \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\isacharparenleft}{\isasymforall}p{\isachardot}\ t\ {\isacharequal}\ p\ \isadigit{0}\ {\isasymand}\ {\isacharparenleft}{\isasymforall}i{\isachardot}\ {\isacharparenleft}p\ i{\isacharcomma}\ p\ {\isacharparenleft}Suc\ i{\isacharparenright}{\isacharparenright}\ {\isasymin}\ M{\isacharparenright}\ {\isasymlongrightarrow}\isanewline nipkow@10159 75 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\isacharparenleft}{\isasymexists}i{\isachardot}\ p\ i\ {\isasymin}\ A{\isacharparenright}{\isacharparenright}{\isacharparenright}{\isacharsemicolon}\isanewline nipkow@10159 76 \ \ \ \ \ \ \ \ \ \ \ {\isasymforall}i{\isachardot}\ {\isacharparenleft}p\ i{\isacharcomma}\ p\ {\isacharparenleft}Suc\ i{\isacharparenright}{\isacharparenright}\ {\isasymin}\ M{\isasymrbrakk}\isanewline nipkow@10159 77 \ \ \ \ \ \ \ \ {\isasymLongrightarrow}\ {\isasymexists}i{\isachardot}\ p\ i\ {\isasymin}\ A nipkow@10159 78 \end{isabelle} nipkow@10187 79 Now we eliminate the disjunction. The case \isa{p\ {\isadigit{0}}\ {\isasymin}\ A} is trivial:% nipkow@10159 80 \end{isamarkuptxt}% nipkow@10123 81 \isacommand{apply}{\isacharparenleft}erule\ disjE{\isacharparenright}\isanewline nipkow@10159 82 \ \isacommand{apply}{\isacharparenleft}blast{\isacharparenright}% nipkow@10159 83 \begin{isamarkuptxt}% nipkow@10159 84 \noindent nipkow@10187 85 In the other case we set \isa{t} to \isa{p\ {\isadigit{1}}} and simplify matters:% nipkow@10159 86 \end{isamarkuptxt}% nipkow@10187 87 \isacommand{apply}{\isacharparenleft}erule{\isacharunderscore}tac\ x\ {\isacharequal}\ {\isachardoublequote}p\ {\isadigit{1}}{\isachardoublequote}\ \isakeyword{in}\ allE{\isacharparenright}\isanewline nipkow@10159 88 \isacommand{apply}{\isacharparenleft}clarsimp{\isacharparenright}% nipkow@10159 89 \begin{isamarkuptxt}% nipkow@10159 90 \begin{isabelle} nipkow@10159 91 \ \isadigit{1}{\isachardot}\ {\isasymAnd}p{\isachardot}\ {\isasymlbrakk}{\isasymforall}i{\isachardot}\ {\isacharparenleft}p\ i{\isacharcomma}\ p\ {\isacharparenleft}Suc\ i{\isacharparenright}{\isacharparenright}\ {\isasymin}\ M{\isacharsemicolon}\ p\ \isadigit{1}\ {\isasymin}\ lfp\ {\isacharparenleft}af\ A{\isacharparenright}{\isacharsemicolon}\isanewline nipkow@10159 92 \ \ \ \ \ \ \ \ \ \ \ {\isasymforall}pa{\isachardot}\ p\ \isadigit{1}\ {\isacharequal}\ pa\ \isadigit{0}\ {\isasymand}\ {\isacharparenleft}{\isasymforall}i{\isachardot}\ {\isacharparenleft}pa\ i{\isacharcomma}\ pa\ {\isacharparenleft}Suc\ i{\isacharparenright}{\isacharparenright}\ {\isasymin}\ M{\isacharparenright}\ {\isasymlongrightarrow}\isanewline nipkow@10159 93 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\isacharparenleft}{\isasymexists}i{\isachardot}\ pa\ i\ {\isasymin}\ A{\isacharparenright}{\isasymrbrakk}\isanewline nipkow@10159 94 \ \ \ \ \ \ \ \ {\isasymLongrightarrow}\ {\isasymexists}i{\isachardot}\ p\ i\ {\isasymin}\ A nipkow@10159 95 \end{isabelle} nipkow@10187 96 It merely remains to set \isa{pa} to \isa{{\isasymlambda}i{\isachardot}\ p\ {\isacharparenleft}i\ {\isacharplus}\ {\isadigit{1}}{\isacharparenright}}, i.e.\ \isa{p} without its nipkow@10159 97 first element. The rest is practically automatic:% nipkow@10159 98 \end{isamarkuptxt}% nipkow@10187 99 \isacommand{apply}{\isacharparenleft}erule{\isacharunderscore}tac\ x\ {\isacharequal}\ {\isachardoublequote}{\isasymlambda}i{\isachardot}\ p{\isacharparenleft}i{\isacharplus}{\isadigit{1}}{\isacharparenright}{\isachardoublequote}\ \isakeyword{in}\ allE{\isacharparenright}\isanewline nipkow@10159 100 \isacommand{apply}\ simp\isanewline nipkow@10159 101 \isacommand{apply}\ blast\isanewline nipkow@10159 102 \isacommand{done}% nipkow@10123 103 \begin{isamarkuptext}% nipkow@10159 104 The opposite containment is proved by contradiction: if some state nipkow@10159 105 \isa{s} is not in \isa{lfp\ {\isacharparenleft}af\ A{\isacharparenright}}, then we can construct an nipkow@10123 106 infinite \isa{A}-avoiding path starting from \isa{s}. The reason is nipkow@10123 107 that by unfolding \isa{lfp} we find that if \isa{s} is not in nipkow@10123 108 \isa{lfp\ {\isacharparenleft}af\ A{\isacharparenright}}, then \isa{s} is not in \isa{A} and there is a nipkow@10123 109 direct successor of \isa{s} that is again not in \isa{lfp\ {\isacharparenleft}af\ A{\isacharparenright}}. Iterating this argument yields the promised infinite nipkow@10123 110 \isa{A}-avoiding path. Let us formalize this sketch. nipkow@10123 111 nipkow@10123 112 The one-step argument in the above sketch% nipkow@10123 113 \end{isamarkuptext}% nipkow@10123 114 \isacommand{lemma}\ not{\isacharunderscore}in{\isacharunderscore}lfp{\isacharunderscore}afD{\isacharcolon}\isanewline nipkow@10123 115 \ {\isachardoublequote}s\ {\isasymnotin}\ lfp{\isacharparenleft}af\ A{\isacharparenright}\ {\isasymLongrightarrow}\ s\ {\isasymnotin}\ A\ {\isasymand}\ {\isacharparenleft}{\isasymexists}\ t{\isachardot}\ {\isacharparenleft}s{\isacharcomma}t{\isacharparenright}{\isasymin}M\ {\isasymand}\ t\ {\isasymnotin}\ lfp{\isacharparenleft}af\ A{\isacharparenright}{\isacharparenright}{\isachardoublequote}\isanewline nipkow@10123 116 \isacommand{apply}{\isacharparenleft}erule\ swap{\isacharparenright}\isanewline nipkow@10186 117 \isacommand{apply}{\isacharparenleft}rule\ ssubst{\isacharbrackleft}OF\ lfp{\isacharunderscore}unfold{\isacharbrackleft}OF\ mono{\isacharunderscore}af{\isacharbrackright}{\isacharbrackright}{\isacharparenright}\isanewline nipkow@10159 118 \isacommand{apply}{\isacharparenleft}simp\ add{\isacharcolon}af{\isacharunderscore}def{\isacharparenright}\isanewline nipkow@10159 119 \isacommand{done}% nipkow@10123 120 \begin{isamarkuptext}% nipkow@10123 121 \noindent nipkow@10123 122 is proved by a variant of contraposition (\isa{swap}: nipkow@10123 123 \isa{{\isasymlbrakk}{\isasymnot}\ Pa{\isacharsemicolon}\ {\isasymnot}\ P\ {\isasymLongrightarrow}\ Pa{\isasymrbrakk}\ {\isasymLongrightarrow}\ P}), i.e.\ assuming the negation of the conclusion nipkow@10123 124 and proving \isa{s\ {\isasymin}\ lfp\ {\isacharparenleft}af\ A{\isacharparenright}}. Unfolding \isa{lfp} once and nipkow@10123 125 simplifying with the definition of \isa{af} finishes the proof. nipkow@10123 126 nipkow@10123 127 Now we iterate this process. The following construction of the desired nipkow@10123 128 path is parameterized by a predicate \isa{P} that should hold along the path:% nipkow@10123 129 \end{isamarkuptext}% nipkow@10123 130 \isacommand{consts}\ path\ {\isacharcolon}{\isacharcolon}\ {\isachardoublequote}state\ {\isasymRightarrow}\ {\isacharparenleft}state\ {\isasymRightarrow}\ bool{\isacharparenright}\ {\isasymRightarrow}\ {\isacharparenleft}nat\ {\isasymRightarrow}\ state{\isacharparenright}{\isachardoublequote}\isanewline nipkow@10123 131 \isacommand{primrec}\isanewline nipkow@10187 132 {\isachardoublequote}path\ s\ P\ {\isadigit{0}}\ {\isacharequal}\ s{\isachardoublequote}\isanewline nipkow@10123 133 {\isachardoublequote}path\ s\ P\ {\isacharparenleft}Suc\ n{\isacharparenright}\ {\isacharequal}\ {\isacharparenleft}SOME\ t{\isachardot}\ {\isacharparenleft}path\ s\ P\ n{\isacharcomma}t{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ t{\isacharparenright}{\isachardoublequote}% nipkow@10123 134 \begin{isamarkuptext}% nipkow@10123 135 \noindent nipkow@10187 136 Element \isa{n\ {\isacharplus}\ {\isadigit{1}}} on this path is some arbitrary successor nipkow@10159 137 \isa{t} of element \isa{n} such that \isa{P\ t} holds. Remember that \isa{SOME\ t{\isachardot}\ R\ t} nipkow@10159 138 is some arbitrary but fixed \isa{t} such that \isa{R\ t} holds (see \S\ref{sec-SOME}). Of nipkow@10123 139 course, such a \isa{t} may in general not exist, but that is of no nipkow@10123 140 concern to us since we will only use \isa{path} in such cases where a nipkow@10123 141 suitable \isa{t} does exist. nipkow@10123 142 nipkow@10159 143 Let us show that if each state \isa{s} that satisfies \isa{P} nipkow@10159 144 has a successor that again satisfies \isa{P}, then there exists an infinite \isa{P}-path:% nipkow@10123 145 \end{isamarkuptext}% nipkow@10159 146 \isacommand{lemma}\ infinity{\isacharunderscore}lemma{\isacharcolon}\isanewline nipkow@10159 147 \ \ {\isachardoublequote}{\isasymlbrakk}\ P\ s{\isacharsemicolon}\ {\isasymforall}s{\isachardot}\ P\ s\ {\isasymlongrightarrow}\ {\isacharparenleft}{\isasymexists}\ t{\isachardot}\ {\isacharparenleft}s{\isacharcomma}t{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ t{\isacharparenright}\ {\isasymrbrakk}\ {\isasymLongrightarrow}\isanewline nipkow@10159 148 \ \ \ {\isasymexists}p{\isasymin}Paths\ s{\isachardot}\ {\isasymforall}i{\isachardot}\ P{\isacharparenleft}p\ i{\isacharparenright}{\isachardoublequote}% nipkow@10123 149 \begin{isamarkuptxt}% nipkow@10123 150 \noindent nipkow@10123 151 First we rephrase the conclusion slightly because we need to prove both the path property nipkow@10123 152 and the fact that \isa{P} holds simultaneously:% nipkow@10123 153 \end{isamarkuptxt}% nipkow@10187 154 \isacommand{apply}{\isacharparenleft}subgoal{\isacharunderscore}tac\ {\isachardoublequote}{\isasymexists}p{\isachardot}\ s\ {\isacharequal}\ p\ {\isadigit{0}}\ {\isasymand}\ {\isacharparenleft}{\isasymforall}i{\isachardot}\ {\isacharparenleft}p\ i{\isacharcomma}p{\isacharparenleft}i{\isacharplus}{\isadigit{1}}{\isacharparenright}{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P{\isacharparenleft}p\ i{\isacharparenright}{\isacharparenright}{\isachardoublequote}{\isacharparenright}% nipkow@10159 155 \begin{isamarkuptxt}% nipkow@10159 156 \noindent nipkow@10159 157 From this proposition the original goal follows easily:% nipkow@10159 158 \end{isamarkuptxt}% nipkow@10159 159 \ \isacommand{apply}{\isacharparenleft}simp\ add{\isacharcolon}Paths{\isacharunderscore}def{\isacharcomma}\ blast{\isacharparenright}% nipkow@10159 160 \begin{isamarkuptxt}% nipkow@10159 161 \noindent nipkow@10159 162 The new subgoal is proved by providing the witness \isa{path\ s\ P} for \isa{p}:% nipkow@10159 163 \end{isamarkuptxt}% nipkow@10159 164 \isacommand{apply}{\isacharparenleft}rule{\isacharunderscore}tac\ x\ {\isacharequal}\ {\isachardoublequote}path\ s\ P{\isachardoublequote}\ \isakeyword{in}\ exI{\isacharparenright}\isanewline nipkow@10159 165 \isacommand{apply}{\isacharparenleft}clarsimp{\isacharparenright}% nipkow@10159 166 \begin{isamarkuptxt}% nipkow@10159 167 \noindent nipkow@10159 168 After simplification and clarification the subgoal has the following compact form nipkow@10159 169 \begin{isabelle} nipkow@10159 170 \ \isadigit{1}{\isachardot}\ {\isasymAnd}i{\isachardot}\ {\isasymlbrakk}P\ s{\isacharsemicolon}\ {\isasymforall}s{\isachardot}\ P\ s\ {\isasymlongrightarrow}\ {\isacharparenleft}{\isasymexists}t{\isachardot}\ {\isacharparenleft}s{\isacharcomma}\ t{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ t{\isacharparenright}{\isasymrbrakk}\isanewline nipkow@10159 171 \ \ \ \ \ \ \ \ {\isasymLongrightarrow}\ {\isacharparenleft}path\ s\ P\ i{\isacharcomma}\ SOME\ t{\isachardot}\ {\isacharparenleft}path\ s\ P\ i{\isacharcomma}\ t{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ t{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\isanewline nipkow@10159 172 \ \ \ \ \ \ \ \ \ \ \ \ P\ {\isacharparenleft}path\ s\ P\ i{\isacharparenright} nipkow@10159 173 \end{isabelle} nipkow@10159 174 and invites a proof by induction on \isa{i}:% nipkow@10159 175 \end{isamarkuptxt}% nipkow@10159 176 \isacommand{apply}{\isacharparenleft}induct{\isacharunderscore}tac\ i{\isacharparenright}\isanewline nipkow@10159 177 \ \isacommand{apply}{\isacharparenleft}simp{\isacharparenright}% nipkow@10123 178 \begin{isamarkuptxt}% nipkow@10123 179 \noindent nipkow@10159 180 After simplification the base case boils down to nipkow@10159 181 \begin{isabelle} nipkow@10159 182 \ \isadigit{1}{\isachardot}\ {\isasymlbrakk}P\ s{\isacharsemicolon}\ {\isasymforall}s{\isachardot}\ P\ s\ {\isasymlongrightarrow}\ {\isacharparenleft}{\isasymexists}t{\isachardot}\ {\isacharparenleft}s{\isacharcomma}\ t{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ t{\isacharparenright}{\isasymrbrakk}\isanewline nipkow@10159 183 \ \ \ \ {\isasymLongrightarrow}\ {\isacharparenleft}s{\isacharcomma}\ SOME\ t{\isachardot}\ {\isacharparenleft}s{\isacharcomma}\ t{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ t{\isacharparenright}\ {\isasymin}\ M nipkow@10159 184 \end{isabelle} nipkow@10159 185 The conclusion looks exceedingly trivial: after all, \isa{t} is chosen such that \isa{{\isacharparenleft}s{\isacharcomma}\ t{\isacharparenright}\ {\isasymin}\ M} nipkow@10159 186 holds. However, we first have to show that such a \isa{t} actually exists! This reasoning nipkow@10187 187 is embodied in the theorem \isa{someI{\isadigit{2}}{\isacharunderscore}ex}: nipkow@10159 188 \begin{isabelle}% nipkow@10171 189 \ \ \ \ \ {\isasymlbrakk}{\isasymexists}a{\isachardot}\ {\isacharquery}P\ a{\isacharsemicolon}\ {\isasymAnd}x{\isachardot}\ {\isacharquery}P\ x\ {\isasymLongrightarrow}\ {\isacharquery}Q\ x{\isasymrbrakk}\ {\isasymLongrightarrow}\ {\isacharquery}Q\ {\isacharparenleft}SOME\ x{\isachardot}\ {\isacharquery}P\ x{\isacharparenright}% nipkow@10159 190 \end{isabelle} nipkow@10159 191 When we apply this theorem as an introduction rule, \isa{{\isacharquery}P\ x} becomes nipkow@10159 192 \isa{{\isacharparenleft}s{\isacharcomma}\ x{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ x} and \isa{{\isacharquery}Q\ x} becomes \isa{{\isacharparenleft}s{\isacharcomma}\ x{\isacharparenright}\ {\isasymin}\ M} and we have to prove nipkow@10159 193 two subgoals: \isa{{\isasymexists}a{\isachardot}\ {\isacharparenleft}s{\isacharcomma}\ a{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ a}, which follows from the assumptions, and nipkow@10159 194 \isa{{\isacharparenleft}s{\isacharcomma}\ x{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ x\ {\isasymLongrightarrow}\ {\isacharparenleft}s{\isacharcomma}\ x{\isacharparenright}\ {\isasymin}\ M}, which is trivial. Thus it is not surprising that nipkow@10159 195 \isa{fast} can prove the base case quickly:% nipkow@10123 196 \end{isamarkuptxt}% nipkow@10187 197 \ \isacommand{apply}{\isacharparenleft}fast\ intro{\isacharcolon}someI{\isadigit{2}}{\isacharunderscore}ex{\isacharparenright}% nipkow@10159 198 \begin{isamarkuptxt}% nipkow@10159 199 \noindent nipkow@10159 200 What is worth noting here is that we have used \isa{fast} rather than \isa{blast}. nipkow@10187 201 The reason is that \isa{blast} would fail because it cannot cope with \isa{someI{\isadigit{2}}{\isacharunderscore}ex}: nipkow@10159 202 unifying its conclusion with the current subgoal is nontrivial because of the nested schematic nipkow@10159 203 variables. For efficiency reasons \isa{blast} does not even attempt such unifications. nipkow@10159 204 Although \isa{fast} can in principle cope with complicated unification problems, in practice nipkow@10159 205 the number of unifiers arising is often prohibitive and the offending rule may need to be applied nipkow@10159 206 explicitly rather than automatically. nipkow@10159 207 nipkow@10159 208 The induction step is similar, but more involved, because now we face nested occurrences of nipkow@10159 209 \isa{SOME}. We merely show the proof commands but do not describe th details:% nipkow@10159 210 \end{isamarkuptxt}% nipkow@10123 211 \isacommand{apply}{\isacharparenleft}simp{\isacharparenright}\isanewline nipkow@10187 212 \isacommand{apply}{\isacharparenleft}rule\ someI{\isadigit{2}}{\isacharunderscore}ex{\isacharparenright}\isanewline nipkow@10123 213 \ \isacommand{apply}{\isacharparenleft}blast{\isacharparenright}\isanewline nipkow@10187 214 \isacommand{apply}{\isacharparenleft}rule\ someI{\isadigit{2}}{\isacharunderscore}ex{\isacharparenright}\isanewline nipkow@10123 215 \ \isacommand{apply}{\isacharparenleft}blast{\isacharparenright}\isanewline nipkow@10159 216 \isacommand{apply}{\isacharparenleft}blast{\isacharparenright}\isanewline nipkow@10159 217 \isacommand{done}% nipkow@10159 218 \begin{isamarkuptext}% nipkow@10159 219 Function \isa{path} has fulfilled its purpose now and can be forgotten nipkow@10159 220 about. It was merely defined to provide the witness in the proof of the nipkow@10171 221 \isa{infinity{\isacharunderscore}lemma}. Aficionados of minimal proofs might like to know nipkow@10159 222 that we could have given the witness without having to define a new function: nipkow@10159 223 the term nipkow@10159 224 \begin{isabelle}% nipkow@10159 225 \ \ \ \ \ nat{\isacharunderscore}rec\ s\ {\isacharparenleft}{\isasymlambda}n\ t{\isachardot}\ SOME\ u{\isachardot}\ {\isacharparenleft}t{\isacharcomma}\ u{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ P\ u{\isacharparenright}% nipkow@10159 226 \end{isabelle} nipkow@10171 227 is extensionally equal to \isa{path\ s\ P}, nipkow@10159 228 where \isa{nat{\isacharunderscore}rec} is the predefined primitive recursor on \isa{nat}, whose defining nipkow@10171 229 equations we omit.% nipkow@10159 230 \end{isamarkuptext}% nipkow@10159 231 % nipkow@10159 232 \begin{isamarkuptext}% nipkow@10187 233 At last we can prove the opposite direction of \isa{AF{\isacharunderscore}lemma{\isadigit{1}}}:% nipkow@10159 234 \end{isamarkuptext}% nipkow@10187 235 \isacommand{theorem}\ AF{\isacharunderscore}lemma{\isadigit{2}}{\isacharcolon}\isanewline nipkow@10159 236 {\isachardoublequote}{\isacharbraceleft}s{\isachardot}\ {\isasymforall}\ p\ {\isasymin}\ Paths\ s{\isachardot}\ {\isasymexists}\ i{\isachardot}\ p\ i\ {\isasymin}\ A{\isacharbraceright}\ {\isasymsubseteq}\ lfp{\isacharparenleft}af\ A{\isacharparenright}{\isachardoublequote}% nipkow@10159 237 \begin{isamarkuptxt}% nipkow@10159 238 \noindent nipkow@10187 239 The proof is again pointwise and then by contraposition (\isa{contrapos{\isadigit{2}}} is the rule nipkow@10159 240 \isa{{\isasymlbrakk}{\isacharquery}Q{\isacharsemicolon}\ {\isasymnot}\ {\isacharquery}P\ {\isasymLongrightarrow}\ {\isasymnot}\ {\isacharquery}Q{\isasymrbrakk}\ {\isasymLongrightarrow}\ {\isacharquery}P}):% nipkow@10159 241 \end{isamarkuptxt}% nipkow@10123 242 \isacommand{apply}{\isacharparenleft}rule\ subsetI{\isacharparenright}\isanewline nipkow@10187 243 \isacommand{apply}{\isacharparenleft}erule\ contrapos{\isadigit{2}}{\isacharparenright}\isanewline nipkow@10159 244 \isacommand{apply}\ simp% nipkow@10159 245 \begin{isamarkuptxt}% nipkow@10159 246 \begin{isabelle} nipkow@10159 247 \ \isadigit{1}{\isachardot}\ {\isasymAnd}s{\isachardot}\ s\ {\isasymnotin}\ lfp\ {\isacharparenleft}af\ A{\isacharparenright}\ {\isasymLongrightarrow}\ {\isasymexists}p{\isasymin}Paths\ s{\isachardot}\ {\isasymforall}i{\isachardot}\ p\ i\ {\isasymnotin}\ A nipkow@10159 248 \end{isabelle} nipkow@10159 249 Applying the \isa{infinity{\isacharunderscore}lemma} as a destruction rule leaves two subgoals, the second nipkow@10159 250 premise of \isa{infinity{\isacharunderscore}lemma} and the original subgoal:% nipkow@10159 251 \end{isamarkuptxt}% nipkow@10159 252 \isacommand{apply}{\isacharparenleft}drule\ infinity{\isacharunderscore}lemma{\isacharparenright}% nipkow@10159 253 \begin{isamarkuptxt}% nipkow@10159 254 \begin{isabelle} nipkow@10159 255 \ \isadigit{1}{\isachardot}\ {\isasymAnd}s{\isachardot}\ {\isasymforall}s{\isachardot}\ s\ {\isasymnotin}\ lfp\ {\isacharparenleft}af\ A{\isacharparenright}\ {\isasymlongrightarrow}\ {\isacharparenleft}{\isasymexists}t{\isachardot}\ {\isacharparenleft}s{\isacharcomma}\ t{\isacharparenright}\ {\isasymin}\ M\ {\isasymand}\ t\ {\isasymnotin}\ lfp\ {\isacharparenleft}af\ A{\isacharparenright}{\isacharparenright}\isanewline nipkow@10159 256 \ \isadigit{2}{\isachardot}\ {\isasymAnd}s{\isachardot}\ {\isasymexists}p{\isasymin}Paths\ s{\isachardot}\ {\isasymforall}i{\isachardot}\ p\ i\ {\isasymnotin}\ lfp\ {\isacharparenleft}af\ A{\isacharparenright}\isanewline nipkow@10159 257 \ \ \ \ \ \ {\isasymLongrightarrow}\ {\isasymexists}p{\isasymin}Paths\ s{\isachardot}\ {\isasymforall}i{\isachardot}\ p\ i\ {\isasymnotin}\ A nipkow@10159 258 \end{isabelle} nipkow@10159 259 Both are solved automatically:% nipkow@10159 260 \end{isamarkuptxt}% nipkow@10159 261 \ \isacommand{apply}{\isacharparenleft}auto\ dest{\isacharcolon}not{\isacharunderscore}in{\isacharunderscore}lfp{\isacharunderscore}afD{\isacharparenright}\isanewline nipkow@10159 262 \isacommand{done}% nipkow@10159 263 \begin{isamarkuptext}% nipkow@10159 264 The main theorem is proved as for PDL, except that we also derive the necessary equality nipkow@10187 265 \isa{lfp{\isacharparenleft}af\ A{\isacharparenright}\ {\isacharequal}\ {\isachardot}{\isachardot}{\isachardot}} by combining \isa{AF{\isacharunderscore}lemma{\isadigit{1}}} and \isa{AF{\isacharunderscore}lemma{\isadigit{2}}} nipkow@10159 266 on the spot:% nipkow@10159 267 \end{isamarkuptext}% nipkow@10123 268 \isacommand{theorem}\ {\isachardoublequote}mc\ f\ {\isacharequal}\ {\isacharbraceleft}s{\isachardot}\ s\ {\isasymTurnstile}\ f{\isacharbraceright}{\isachardoublequote}\isanewline nipkow@10123 269 \isacommand{apply}{\isacharparenleft}induct{\isacharunderscore}tac\ f{\isacharparenright}\isanewline nipkow@10187 270 \isacommand{apply}{\isacharparenleft}auto\ simp\ add{\isacharcolon}\ EF{\isacharunderscore}lemma\ equalityI{\isacharbrackleft}OF\ AF{\isacharunderscore}lemma{\isadigit{1}}\ AF{\isacharunderscore}lemma{\isadigit{2}}{\isacharbrackright}{\isacharparenright}\isanewline nipkow@10159 271 \isacommand{done}% nipkow@10159 272 \begin{isamarkuptext}% nipkow@10171 273 The above language is not quite CTL. The latter also includes an nipkow@10178 274 until-operator, which is the subject of the following exercise. nipkow@10171 275 It is not definable in terms of the other operators! nipkow@10171 276 \begin{exercise} nipkow@10178 277 Extend the datatype of formulae by the binary until operator \isa{EU\ f\ g} with semantics nipkow@10178 278 there exist a path where \isa{f} is true until \isa{g} becomes true'' nipkow@10171 279 \begin{isabelle}% nipkow@10178 280 \ \ \ \ \ s\ {\isasymTurnstile}\ EU\ f\ g\ {\isacharequal}\ {\isacharparenleft}{\isasymexists}p{\isasymin}Paths\ s{\isachardot}\ {\isasymexists}j{\isachardot}\ p\ j\ {\isasymTurnstile}\ g\ {\isasymand}\ {\isacharparenleft}{\isasymexists}i\ {\isacharless}\ j{\isachardot}\ p\ i\ {\isasymTurnstile}\ f{\isacharparenright}{\isacharparenright}% nipkow@10171 281 \end{isabelle} nipkow@10171 282 and model checking algorithm nipkow@10171 283 \begin{isabelle}% nipkow@10187 284 \ \ \ \ \ mc{\isacharparenleft}EU\ f\ g{\isacharparenright}\ {\isacharequal}\ lfp{\isacharparenleft}{\isasymlambda}T{\isachardot}\ mc\ g\ {\isasymunion}\ mc\ f\ {\isasyminter}\ {\isacharparenleft}M{\isacharcircum}{\isacharminus}{\isadigit{1}}\ {\isacharcircum}{\isacharcircum}\ T{\isacharparenright}{\isacharparenright}% nipkow@10171 285 \end{isabelle} nipkow@10186 286 Prove the equivalence between semantics and model checking, i.e.\ that nipkow@10186 287 \begin{isabelle}% nipkow@10186 288 \ \ \ \ \ mc\ {\isacharparenleft}EU\ f\ g{\isacharparenright}\ {\isacharequal}\ {\isacharbraceleft}s{\isachardot}\ s\ {\isasymTurnstile}\ EU\ f\ g{\isacharbraceright}% nipkow@10186 289 \end{isabelle} nipkow@10186 290 %For readability you may want to annotate {term EU} with its customary syntax nipkow@10186 291 %{text[display]"| EU formula formula E[_ U _]"} nipkow@10186 292 %which enables you to read and write {text"E[f U g]"} instead of {term"EU f g"}. nipkow@10186 293 \end{exercise} nipkow@10186 294 For more CTL exercises see, for example \cite{Huth-Ryan-book,Clarke-as-well?}. nipkow@10186 295 \bigskip nipkow@10178 296 nipkow@10186 297 Let us close this section with a few words about the executability of our model checkers. nipkow@10159 298 It is clear that if all sets are finite, they can be represented as lists and the usual nipkow@10159 299 set operations are easily implemented. Only \isa{lfp} requires a little thought. nipkow@10159 300 Fortunately the HOL library proves that in the case of finite sets and a monotone \isa{F}, nipkow@10159 301 \isa{lfp\ F} can be computed by iterated application of \isa{F} to \isa{{\isacharbraceleft}{\isacharbraceright}} until nipkow@10186 302 a fixpoint is reached. It is actually possible to generate executable functional programs nipkow@10159 303 from HOL definitions, but that is beyond the scope of the tutorial.% nipkow@10159 304 \end{isamarkuptext}% nipkow@10123 305 \end{isabellebody}% nipkow@10123 306 %%% Local Variables: nipkow@10123 307 %%% mode: latex nipkow@10123 308 %%% TeX-master: "root" nipkow@10123 309 %%% End: | 11,942 | 32,034 | {"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} | 2.875 | 3 | CC-MAIN-2019-39 | latest | en | 0.433712 |
https://www.textilecalculations.com/loom-calculation-formulas-in-weaving/ | 1,726,569,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00744.warc.gz | 958,675,720 | 25,649 | # Loom Calculation Formulas in Weaving
Production of Loom:
The production of a loom mainly depends on the speed of the loom. Higher speed does not always mean higher production because if it exceeds a certain limit, loom stoppages due to excessive warp breakages occurring. As a result of higher speed may even reduce the production to a considerable extent. Other factors which affect production are picks inserted per inch of the cloth, quality of the warp and weft, type of loom i.e plain, dobby, jacquard, automatic, etc. types of organization i.e 2- loom system or 4- loom system, size of weft package, type of weft package. In this article, we will discuss the Loom Calculation Formulas in Weaving.
### Different Formulas of Loom Calculation in Weaving
1. Production (calculated) in yds/hr,
….Picks per min × 60
= …………………………..
….Picks per inch × 36
….Picks per min
= ……………………. × 1.667
….Picks per inch
2. Production (actual) in yds/hr,
= Calculated production × efficiency
….Picks per min × 60
= …………………………..× efficiency
….Picks per inch × 36
….Picks per min
= …………………… × efficiency × 1.667
….Picks per inch
Example 01:
Calculate the production per hour of a loom running at a speed of 192 r.p.m. with an efficiency of 75%. The number of picks inserted per inch in the cloth is 80.
Solution:
Calculated Production per hour,
….Picks per min × 60
= ……………………………
….Picks per inch × 36
…..192 × 60
= ………………
……80 × 36
= 4 yds
So, the calculated Production per hour is 4 yds.
And,
Actual production per hour,
= Calculated production × efficiency
= 4 × 75%
= 3 yds.
So, the Actual production per hour is 3 yds.
3. The time required to complete a weaver’s beam,
….Length in yds of cloth from the beam
= ………………………………………………………. (1)
……Production in yds of cloth per hour
….[(Length in yds of warp – uptake) – waste of warp in yds) × Picks per inch × 36 ]
= ………………………………………………………………………………………………..……………………. (2)
………………………………Picks per min × 60 × efficiency
Example 02:
Calculate the time required to complete a weaver’s beam having 1000 yds of warp on it. The woven cloth is required to have 50 picks per inch. The up-take of warp in weaving Is 6% and the waste may be taken as 6 yds. The loom is running at 200 r.p.m and the efficiency is 75%.
Solution:
The time required to complete a weaver’s beam,
Length in yds of cloth from the beam
= ……………………………………………………… (1)
…..Production in yds of cloth per hour
Now,
Length in yds of cloth from the beam = (Length in yds of warp – uptake) – waste of warp in yds
= (1000 × 96%) – 6
= (1000 × 9.6) – 6
= 934 yds.
And,
Production in yds of cloth per hour,
…..Picks per min × 60
= ……………………………. × efficiency
….Picks per inch × 36
….200 × 60 × 75
= ……………………
……..50 × 36
= 5 yds.
So, from (1), we have
The time required to complete a weaver’s beam,
……934
= …………
……..5
= 186.8 hours.
So, Time required to complete a weaver’s beam is 186.8 hours.
```Author of this Article:
Mayedul Islam
Assistant Chemist | 864 | 2,976 | {"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} | 3.953125 | 4 | CC-MAIN-2024-38 | latest | en | 0.82665 |
https://stage.geogebra.org/m/mv5jJ3Jh | 1,638,885,947,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00350.warc.gz | 587,291,239 | 10,104 | # CCSS IP Math I 5.1.3 Example 3
Given the quadrilateral , the square , and the information that is the same distance from and , show that is symmetrical along segment .
1. Recall the definition of line symmetry.
2. Since and , is a line of symmetry for where .
3. has the same area as because they share a base and have equal height. , so .
4. We now know is a line of symmetry for and is a line of symmetry for , so and quadrilateral is symmetrical along . | 114 | 459 | {"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-2021-49 | latest | en | 0.953025 |
http://mathoverflow.net/feeds/question/46567 | 1,369,492,850,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705956734/warc/CC-MAIN-20130516120556-00082-ip-10-60-113-184.ec2.internal.warc.gz | 165,008,279 | 4,047 | Functions whose antiderivative behaves like xf(x) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:40:35Z http://mathoverflow.net/feeds/question/46567 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46567/functions-whose-antiderivative-behaves-like-xfx Functions whose antiderivative behaves like xf(x) Adam Hughes 2010-11-19T01:19:39Z 2010-11-20T16:42:27Z <p>I'm wondering if a classification of analytic functions, $f\,$ (it may be that $C^1$ is enough, but I'm not taking any chances, if you have a reason why I only need to consider a larger class of functions, I would enjoy that as well) with the property $F(x):=\int f(x) dx = \mathcal{O}(xf(x))$ as $x\to\infty$ AND $xf(x)=\mathcal{O}(F(x))$ where $\mathcal{O}$ is big-O notation. I'm not sure if it's standard or not, but I'll denote this condition by </p> <p>$$\mathcal{O}(F(x))=\mathcal{O}(xf(x))\qquad (*)$$</p> <p>I'm motivated by the naïve notion of integration from the mistake many first semester students in calculus make when trying to take anti-derivatives and keep on thinking the same thing over and over: the power rule.</p> <p>It is straightforward to check that $f(x)=x^n\quad n\ne -1$ and $f(x)=\log^n x\quad n\ge 0$ satisfy the condition $(*)$, the first just by checking and the second by induction on n and integration by parts. It's also easy to see that this is not the case for $x^ne^x,\; n\in\mathbb{Z}$ again by integration by parts.</p> <p>A preliminary investigation yields some interesting first starts:</p> <p>1) A valid refomulation of the problem is ${F\over f}$ has a slant asymptote, i.e. ${F\over f}=kx+\mathcal{o}(1)$, and if you know that the derivative of this little o function is also little o of 1, and say $k=1$ then you can rephrase this as $1-{d\over dx}(\log f(x))\cdot {F\over f}(x)=1+\mathcal{o}(1)$</p> <p>2) If the function is increasing we can get half of this inequality since $F(x)\le xf(x)$ since $F(x)=\int_a^xf(x)$, but the other half fails.</p> <p>3) A convex $\mathcal{o}(1)$ in (1) might imply that the derivative is also $\mathcal{o}(1)$</p> http://mathoverflow.net/questions/46567/functions-whose-antiderivative-behaves-like-xfx/46635#46635 Answer by Todd Trimble for Functions whose antiderivative behaves like xf(x) Todd Trimble 2010-11-19T15:00:17Z 2010-11-20T16:42:27Z <p>Note: This is a major rewrite of my earlier answer, to include necessary and sufficient conditions applicable to an even wider class of functions. </p> <p>Instead of expanding to the class of all analytic functions (where the asymptotics can be hard to get control over, due to oscillatory behavior), my inclination would be to focus on large classes of functions with well-behaved asymptotics, including all the functions that arise in ordinary asymptotic analysis. The usual buzz phrase for this is "Hardy field" (mentioned for example in my answer <a href="http://mathoverflow.net/questions/45284/examples-of-sequences-whose-asymptotics-cant-be-described-by-elementary-function/45368#45368" rel="nofollow">here</a>), which by definition is an ordered field of germs at infinity of $C^\infty$ functions. </p> <p>I will describe several classes of such functions. The first is the class of all functions which are first-order definable in the structure $(\mathbb{R}, +, \cdot, <, \exp)$ together with all real numbers adjoined as constants. This class contains all functions that are constructible from polynomials, $\exp$, $\log$ and closed under the usual arithmetic operations and composition. It thus contains all the functions that usually arise in asymptotic analysis, and many more besides. This class enjoys the following strong model-theoretic property (as developed more fully in the <a href="http://en.wikipedia.org/wiki/O-minimal_theory" rel="nofollow">theory of o-minimal structures</a>): </p> <blockquote> <p>(O) The zero set of any function $F: [a, \infty) \to \mathbb{R}$ in this class is a finite union of points and intervals (finite or infinite in extent). </p> </blockquote> <p>Condition O ensures that every such function $F$ is either eventually positive ($F(x) > 0$ for all sufficiently large $x$), eventually zero, or eventually negative. As a result, the ring of germs at infinity of the definable functions in this class forms an ordered field, i.e., is a Hardy field. </p> <p>Also, if $F$ is definable, then $F'$ is also first-order definable (and its domain can be shown to be the domain of $F$ save for finitely many points). Applying condition O to $F'$, every definable $F$ in this class is either eventually increasing, eventually decreasing, or eventually constant. </p> <p><b>Proposition:</b> A function $F$ in this class satisfies $F(x) = O(xF'(x))$ and $xF'(x) = O(F(x))$ if and only if there exist $n, N$, both positive or both negative, for which $x^n < |F(x)| < x^N$ for all sufficiently large $x$. </p> <p><b>Proof:</b> WLOG we may assume $F$ is eventually positive, and is not eventually constant. Thus $F$ is either eventually increasing or eventually decreasing, say eventually increasing. If $F$ is eventually bounded above by some $x^N$, $N > 0$, then by increasing $N$ if necessary we may assume $F(x)/x^N$ tends to zero, whence it is eventually decreasing. Taking the derivative, we conclude that eventually </p> <p>$$x^N F'(x) - Nx^{N-1}F(x) < 0$$ </p> <p>whence $xF'(x) < NF(x)$, i.e., $|xF'(x)| < N|F(x)|$ or $xF'(x) = O(F(x))$. However, if $F(x)$ is eventually bounded above by <i>every</i> positive-power function $x^N$ (think $N$ small!), this also shows $xF'(x) = o(F(x))$, so that $F(x)$ is not $O(xF'(x))$. </p> <p>By a similar argument, if $F(x)$ is eventually bounded below by some positive-power function $x^n$, we get $nF(x) < xF'(x)$ eventually, so that $F(x) = O(xF'(x))$. This also shows that if $F(x)$ is bounded below by <i>every</i> positive-power function (think $n$ large!), then $F(x) = o(xF'(x))$, so $xF'(x)$ is not $O(F(x))$. </p> <p>Thus, if $F(x)$ is positive and eventually increasing, a necessary and sufficient condition that $F'$ satisfy condition ($\ast$) in the question is that there exist two positive-power functions that $F$ is eventually squeezed between. An entirely similar analysis shows that if $F(x)$ is positive and eventually decreasing, a necessary and sufficient condition that $F'$ satisfy condition ($\ast$) is that there exist two negative-power functions that $F$ is eventually squeezed between. Thus the proposition is proved. </p> <hr> <p>The almost freshman-level triviality of this proof testifies to the great power of condition O (which is a special case of the o-minimality axiom), from which all flows. Thus it is of interest to know of classes of functions which satisfy it. I will mention an extraordinary result in this regard, due largely to Patrick Speissegger (The Pfaffian closure of an o-minimal structure, J. Reine Angew. Math. 508 (1999), 189--211): </p> <p>There is an o-minimal expansion of the ordered exponential field $\mathbb{R}$ (thus, including the class of functions described above) so large that</p> <ul> <li><p>The structure includes the restriction of any analytic function to a compact box, </p></li> <li><p>If $f: [a, \infty) \to \mathbb{R}$ is first-order definable within this structure, then so is any antiderivative $F$ (even though general antiderivatives are not definable by a first-order construction), </p></li> </ul> <p>This may fit better with Adam Hughes's formulation in terms of antiderivatives. Since condition O is satisfied (according to the more general o-minimality condition), the same analysis as above applies. </p> | 2,136 | 7,647 | {"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} | 3.84375 | 4 | CC-MAIN-2013-20 | latest | en | 0.81892 |
https://blog.bigml.com/2022/07/01/object-detection-some-gory-details/ | 1,680,039,127,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00202.warc.gz | 178,650,111 | 33,820 | # Object Detection: Some Gory Details
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As Machine Learning practitioners, it’s important to know how Machine Learning algorithms work. This isn’t just to impress your friends at parties (trust me, this will not impress your friends at parties). The reason you should know how Machine Learning algorithms work is that when they don’t work, you can use that knowledge to figure out why, and maybe even how to fix them.
## A Stitch In Time
Suppose you have a simple dataset that tries to predict a variable based on the value of that variable in the past. For example, you might want to predict the price of a stock based on its 5, 10, 50, and 300-day moving averages (Note: this is just an example; doing this will probably not give you useful price predictions). Imagine you have data for two stocks, one with a long-term average price of around \$10 and another with a long-term average price of around \$100, without too much variability about those values.
If you feed data like this to a decision tree, how will it behave? At each of the leaf nodes in the tree, it will assign a price based on the training instances that reached that leaf node. Importantly, the assigned price will probably be close to either \$10 or \$100 because those are the values it saw in training. If you have a stock that averages \$50 or \$500, the prediction for the future prices of that stock will probably be especially inaccurate.
If you learn a linear regression, this will be less of a problem: Because the prediction is a linear combination of the input variables, the model should give good outputs even when the inputs aren’t near the training inputs, as long as the same linear relationship holds for the inputs you provide. If you only have data for a few stocks, and you still want your model to generalize to stocks with prices that are much different, linear regression might be a better choice. Better still would be if you realized that before the error surfaced in a production setting.
## Computer Vision: More of the Same
Let’s see if we can make some similar deductions about our Object Detection algorithms. Here’s a brief summary of how many Object Detection algorithms, including BigML’s, go about their business: First, they break the image up into a coarse grid. At each square, they consider several different bounding box proposals. These are just rectangles of some size, typically two or three at each grid square. If one of those rectangles holds most of one of the objects we’re looking at, they say there is an object present in that rectangle, then try to translate or scale the rectangle so it fits the object as snuggly as possible. The prediction ends up being several values: The probability that there is an object in the box, the adjustments to the x, y, height, and width of the box, and the distribution over classes for that box.
In this image, the detector would try each of the three bounding boxes shown at each square in the grid. You can see that at this particular location, one is very close to “containing” the chair. However, the box needs to shift a bit down and to the left for the chair to be properly centered. Our Object Detection model simultaneously predicts the presence and class of the chair, along with the adjustments needed to make the box fit properly.
All fine so far, but where do the bounding box proposals come from? In practice, we get much better results from these models if we learn the bounding box proposals from the training data. A simple way to do this is to cluster all of the bounding boxes in the training data by their dimensions, then take the centroid of each cluster as our list of proposals. The astute reader will probably see the implication: If you only see small objects in the training set, the set of proposals assures that you will only be looking for things of that size.
## Ready, Set, Fail
Let’s train an object detector to recognize some chess pieces. The images we’re using for training all have this view of the board:
An evaluation shows that the model we train is very capable of recognizing black kings:
And in that case, this should be easy:
Wait, what?
But let’s think about this for a second. All of the pieces that we saw in training were small relative to the image. Here we’re feeding it a chess piece that takes up nearly the entire image. There is no bounding box in the training data even close to this size. This, in the vocabulary that the model has learned, is absolutely not a chess piece of any kind. It’s way too big (though that little bit on the edge looks a bit like a black queen).
Looked one way, the model has failed. Why can’t our chess piece model detect chess pieces? Looked at in a more useful way though, we have failed. If we wanted our model to perform properly on this view of a chess piece, why didn’t we show it this view of a chess piece? Why did we expect the model to perform well in this situation that it’s never seen before?
The skeptical reader might balk at this line of thought: “So in order to get a general recognizer for chess pieces, I have to add every possible view of a chess piece to my training dataset. That’s impossible!” There is an unfortunate grain of truth to this: The more general you want a model to be, the more data you need. You can get around this a bit with tricks like data augmentation and using large pre-trained models, but you often can’t get all the way around.
However, you often don’t need to. If you can make guarantees on the input data, then it’s possible that a less than general recognizer will work just fine for the inputs you have in mind and it will take less data to train to boot. For example, if you only want to analyze images from a single view of a chess board, it makes no difference if you can recognize kings that take up the entire image because it’s never going to happen. In monitoring applications, where you’ve got a fixed camera looking at basically the same scene all the time, this can be an easy task.
As I’ve said before, writing code that works with machine-learned models is hard. You need to anticipate their failures and plan for their inaccuracies. Like any powerful tool, Machine Learning can give near-miraculous results, but only when handled with care. With some patience, onward and upward! | 1,371 | 6,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-14 | latest | en | 0.936422 |
https://www.r-bloggers.com/2012/05/mining-for-relations-between-nominal-variables/ | 1,718,666,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00422.warc.gz | 815,492,863 | 21,103 | [This article was first published on Data and Analysis with R, at Work, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
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The task today was to find what variables had significant relations with an important grouping variable in the big dataset I’ve been working with lately. The grouping variable has 3 levels, and represents different behaviours of interest. At first I tried putting the grouping variable as a dependent variable in a multinomial logistic regression, but I didn’t really trust the output, and the goal was really just to construct a bunch of graphs showing significant bivariate nominal relations in the data..
That’s when I turned to my good old friend, the chi squared test. All I had to do was select all the variables that I wanted to test against the grouping variable, and construct a list of the chi squared statistic from each test, the variable being tested, and the crosstab of the two variables for later graphing. So that’s exactly what I did:
One really sweet thing about matrices in R is that you can mix them up with some parts having just numbers, some parts having text, and sub-matrices in other parts! A typical row of the “resultlist” would look something like this:
xsq testvar xtab
[1,] 200.7 “variable1″ numeric,6
Then all I needed to do to see the variable name and crosstab for that variable was to call “resultlist[1,2:3]“, and that gave me the numbers to graph.
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. | 425 | 1,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.943467 |
https://junkcharts.typepad.com/junk_charts/2010/10/index.html | 1,713,037,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00376.warc.gz | 306,609,018 | 23,060 | Halloween comes early: scare of the Times
Oct 27, 2010
Has the New York Times lost the plot? This was the thought in my head when I saw this scary thing last Friday, right in the center of the front page of the New York edition:
This Halloween scare is supposed to tell us who the biggest political donors are, and to what party. Here is a close-up view:
Well, this one looks slightly different from the one shown above but this is the one that they are currently carrying on the website.
I think the stacked squares, arranged in this particular way--staggered both horizontally and vertically--is supposed to represent something. I can't figure it out. Maybe a loudspeaker? An accordian?
This chart fails our self-sufficiency test. The only way one can appreciate the scale of the donations is to extract the data from the chart, in which case, this is the same as a data table.
The use of the unexplained red color for the Chamber of Commerce (the subject of the accompanying article) is also problematic. Based on the inset, it's pretty clear that the Chamber is "leaning Republican" so it should just be colored pink.
What is very hard to understand is why the amount of overlapping from one square to the next square is not the same.
We can visualize the problem as follows:
According to the data, the US Chamber of Commerce donated about the same amount as the Democratic Senatorial Campaign Committee (\$21.1 v. \$22). However, the visible areas on the chart are vastly different. There is a presumption that readers will look behind the squares to see the full square but that's pretty hard to do when they are stacked 15-deep!
What's the amount of distortion due to this design?
A lot. Much of this is due to the fact that the lowest ranked item plotted is an actual square which happens to have the largest unobstructed area.
***
Here's a dot plot that conveys the essential information with minimal fuss:
Of course, a bar chart works fine too.
***
Happy Halloween!
The nothing that is
Oct 22, 2010
When Mike K. sent this in, he had a few comments, including "This is, of course, from the Chronicle of Higher Education", and "talking about a math course", which mean "very naively", we would "impose a higher standard." Should scientists be held to a higher standard? Lead by example, perhaps? I had the same feeling when I wrote the post on "Unscientific American" about the charts that'd flunk Ed Tufte's intro class, published in Scientific American.
In one word: confusion. Mike couldn't understand the relationship between the first row of bubbles and the second row of bubbles. It is as if the one course taken at Bronx Community College results in credits recognized everywhere! (You basically have to read all the footnotes to get some clues.)
Also note the usual confusion about areas and diameters.
In addition, the zero-bubbles prove themselves to be the nothing that is. They expose the folly of using bubbles when the data series contain zeroes (not to mention negative numbers). We can visualize this problem:
It gets worse.
For those mathematically inclined, we actually have an impossible situation: the size 4 bubble really contains the zero bubble plus a size 4 bubble; that is, 0 + 4 = 4 but if 0 has positive area, then the area of the 4 on the left hand side of the equation must be smaller than the area of the 4 on the right side. So, basically, don't use bubble charts if your data has zeroes.
Fashion tip
Oct 20, 2010
Tip from Julien D.
High fashion for chart lovers.
Straight from France:
"Must have", says Julien
Audio bookmarks
Oct 16, 2010
I look at a fair number of online videos, especially those embedded on blogs. But I haven't seen this feature implemented broadly. It is a wow feature.
Look at the dots above the progress bar: they tell you what topic is being discussed and allow you to jump back and forth between segments. (the particular dot I moused over said "Randy Moss") The video I saw came from this link.
This simple-looking feature is immensely useful to users. You can efficiently search through the audio file and find the segments you're interested in. It's like bookmarks students might put on pages of a textbook for easy reference, except these are audio bookmarks.
Why isn't this feature more prevalent? I think it's because of the amount of manual effort needed to set this up. Imagine how the data has to be processed. In the digital age, the audio file is a bunch of bits (ones and zeroes) so no computer or humans will be able to identify topics from data stored in that way. So, someone would need to listen to the audio file, and mark off the segments manually, and tag the segments. Then, the audio bookmarks can be plotted on the progress bar... basically a dot plot with time on the horizontal axis.
In theory, you can train a computer to listen to an audio file and approximate this task. The challenge is to attain the required accuracy so you don't need to hire an army of people to correct mistakes.
A very simple concept but immensely functional. Great job!
Showing dynamics on a business chart
Oct 13, 2010
Dave S. achieved a rare feat, which is to send in a great-looking set of charts. This post at Asymco is worth reading in its entirety; the author Horace discusses the process by which he worked through several charts, arriving at the one he's most happy with.
***
The secret to the success here is the careful framing of the question, and the collection of the appropriate data to address that question. The question is the competition between wireless phone vendors in the last three years. It was established that the right way to view this competition is in two dimensions: share of revenues, and share of profits. Note the word "share". Share of profits is not a metric that is often discussed but it is the right metric to compare to the share of revenues -- getting both numbers onto the same comparable scale is what makes this work.
Needless to say, the raw data one would collect come from the financial statements of the eight individual vendors. Plotting these numbers directly would be a mistake. So you take the numbers, making sure that you're really counting wireless revenues and wireless profits, and then compute the shares. (I am not actually sure that they have wireless profit data because large companies like Apple and Nokia typically don't break out their profit shares, even if they provide the revenue shares by line of business.)
Horace also avoided the plague of plotting all time-series data as line charts (similar to the plague of plotting all geographic data on maps). By plotting revenues and profits simultaneously, he no longer can plot time (years) on one of the two axes, and that is a good thing.
***
This is the final graph Horace landed on. It puts all the vendors at the origin in 2007 and then tells us where they landed in 2010 in terms of revenue and profit share growth/decline.
It would be even better if he makes the scales work harder: e.g. have equal lengths for the 10% change along both the vertical and horizontal axes. Alternatively, you can scale it such as each unit on either axis represent equal dollars.
This is a very focused chart that answers the question about the relative change in positioning of each vendor. What it doesn't answer is the starting position or ending position of each. Note, while Nokia is depicted as losing share on both revenues and profits, Nokia still has twice the revenue share of the other vendors, and out-earns everyone except Apple!
I am not saying this is a bad chart. It is designed to answer the relative question, not the absolute question. That's all.
***
There is one way to have the cake and eat it too. Horace almost created that chart. He showed two scatter plots, one for 2007 and one for 2010.
If he just overlays one on the other, and use lines to connect the dots for each phone vendor, he will have a chart that shows absolute and relative values all at once. Here's a crude illustration of this: (missing the labels to show that the arrow end of the line represents 2010 positions)
I like this kind of chart a lot. It is great for showing dynamics in a set of variables, without actually making the chart dynamic.
(Even on this chart, it is better to harmonize the two scales.)
Unscientific American 2: a review of key concepts
Oct 10, 2010
This is part 2 of a two-part piece on a set of charts published in Scientific American, describing the results of a poll asking their readers about their trust of various types of experts. In part 1, I looked in detail at one of the charts: this chart has many problems-- problems with the execution of the bar chart, problems with the choice of comparisons, problems with the design of poll questions.
***
Long time ago, I wrote about racetrack graphs. An example shows up here. Note the length of the yellow track relative to the length of the pink track. It should be about 2:1 but this type of chart makes a many-fold illusion.
Bar chart, please.
***
For this chart illustrating fears of technology, the designer chose to emphasize the data (percentages), with a subtle embedding of a column chart in which the columns were over-stretched to form squares.
Love these pictures, very telling.
The data jump at us, rendering the graphical element--the embedded column chart--secondary, even redundant.
But columns are very useful things. They are self-sufficient. We can judge the lengths of columns (or bars) very easily, so we do not need to print data next to columns.
However, thin columns work better than thick columns, so stretching columns to squares is not a good idea, even if the relative areas are not distorted.
Also, even the choice of colors here can be quibbled with. For me, the dark pink draws more attention than the medium orange, which means that my eyes and my brain (reading the printed data) start having an argument.
So, here is a sketch of what a bar chart would look like. See, you don't need to print the data on a bar chart--it's not hard at all to see that about 50% fear nuclear power. I have lightened the pink. Just add in the pictures, and the captions on the right side of the chart, and you have an improved chart.
***
Now, start the stopwatch, and measure how much time it takes to figure out what this chart is telling you:
This is truly a head-turning chart if you know what I mean...
In the article, this chart has the title "Climate Denial on the Decline". Maybe, maybe not--but oh dear, how are we to tell?
Instead of fixing the chart, it is perhaps more important to fix the poll question. It's not a good idea to describe a trend when one has data from this year and last year only. If they are isolating the effect of some major event (say, a Kyoto protocol) that happened during that year, it would be fine but that's not what is attempted here.
***
For part 1, see here.
Unscientific American 1: misreadings
Oct 10, 2010
Chris P. sent me to this set of charts / infographics with the subject line "all sorts of colors and graphs." I let the email languish in my inbox, and I now regret it. For two reasons: one, the topic of how scientists can communicate better with, and thus exert stronger influence on, the public is very close to my heart (as you can tell from my blogs and book), and this article presents results from a poll on this topic done on on-line readers of Scientific American, and Nature magazines; two, some of the charts are frankly quite embarrassing, to have appeared in venerable publications of a scientific nature (sigh); three, these charts provide a convenient platform to review some of the main themes on Junk Charts over the years.
Since the post is so long, I have split it into two parts. In part 1, I explore one chart in detail. In part 2, I use several other charts to illustrate some concepts that have been frequently deployed on Junk Charts.
***
Exhibit A is this chart:
First, take a look at the top left corner. At first glance, I took the inset to mean: among scientists, how much do they trust scientists (i.e., their peers) on various topics? That seemed curious, as that wouldn't be a question I'd have thought to ask, certainly not as the second question in the poll.
On further inspection, that is a misreading of this chart. The "scientists" represented above are objects, not subjects, in the first question. As the caption tells us, the respondents rated scientists at 3.98 overall, which is an average rating across many topics. The bar chart below tells us how the respondents rated scientists on individual topics, thus providing us information on the spread of ratings.
Unfortunately, this chart raises more questions than it answers. For one, you're working out how the average could be 3.98 (at that 4.0 white line) when all but three of the topic ratings were below 3.98. Did they use a weighted average but did not let on?
Oops, I misread the chart, again. I think, what I stumbled on here is the design of the poll itself. The overall rating is probably a separate question, and not at all related to the individual topic ratings. In theory, each person can assign a subjective importance as well as a rating to each topic; the average of the ratings weighted by their respective importance would form his or her overall rating of scientists. That would impose consistency to the two levels of ratings. In practice, that makes an assumption that the topics span the space of what topics each person considers when rating the scientists overall.
***
The bar chart has a major problem... it does not start at zero. Since the bars are half as long as the longest, you might think the level of trust associated with nuclear power or climate change would be around 2 (negative). But it's not; it's in the 3.6 range. This is a lack of self-sufficiency. The reader cannot understand the chart without fishing out the data.
Now, ask this question: in a poll in which respondents are asked to rate things on a scale of 1, 2, 3, 4, 5, do you care about the average rating to 2 decimal places? The designer of the graphic seems to think not, as the rating was rounded up to the nearest 0.5, and presented using the iconic 5-star motive. I think this is a great decision!
But then, the designer fell for loss aversion: having converted the decimals to half-stars, he should have dropped the decimals; instead, he tucked them at the bottom of each picture. This is no mere trivia. Now, the reader is forced to process two different scales showing the same information. Instead of achieving simplification by adopting the star system, now the reader is examining the cracks: is the trust given citizens groups the same as journalists (both 2.5 stars) or do "people" trust citizens groups more (higher decimal rating)?
***
The biggest issues with this chart concern the identification of the key questions and how to collect data to address those questions. This is the top corner of the Trifecta checkup.
1) The writer keeps telling us "people" trust this and that but the poll only covered on-line readers of Scientific American and Nature magazines. One simply cannot generalize that segment of the population to the common "people".
2) Insufficient attention has been paid to selecting the right wording in the questions. For example, in Exhibit A, while the overall trust question was phrased as trusting the "accuracy" of the information provided by scientists vs. other groups, the trust questions on individual topics mentioned only a generic "trust". Unless one thinks "trust" is a synonym of "accuracy", the differential choice of words makes these two set of responses hard to compare. And compairing them is precisely what they chose to do.
***
In part 2, I examine several other charts, taking stops at several concepts we use on Junk Charts a lot.
Ranking airlines: no easy task
Oct 06, 2010
Reader Joel D. submitted this chart showing airline revenues of major airlines around the world, another chart that puts bubbles on top of a map.
Joel said:
Cool, but really quite naughty. Take the different size bubbles of British Airways (12.8) and Air France-KLM (29.7). Grossly disproportionate. I appreciate the designer's attempts to introduce a geographic element to this but the immediate take-outs here from the bubbles are misleading.
Sometimes a bar chart is all life needs.
I think it fails all three facets of the Trifecta checkup: it does not have a well-defined practical question; the data is not processed properly; and the chart type does not work with this data.
• Most airlines are multinational companies that make substantial revenues outside their home countries... so the locations of their registered headquaraters are irrelevant. What is the question being addressed? It would appear to be where are the headquarters of the largest airlines in the world? I don't think this is an especially engaging question. What might be more interesting, for example, is the split between domestic and international revenues for different airlines, or the split among airlines of the revenues within each continent.
• Besides, the aggregate revenues data is not very useful for comparison purposes. It ignores the population... a circle in an European country is in reality much "larger" than a circle of the same size in China! Because \$200 million on 2 billion people is very different from \$200 million from say 50 million! The right base for this data is probably something like revenue per passenger or passenger miles.
• The inclusion of Fedex also must be thought through thoroughly. I'd imagine that all the large airlines of the world also have freight divisions, and if we really want to address both passenger and commercial air revenues on the same chart (with which I don't agree), we should at least break out the freight revenues.
Visualizing your inbox
Oct 03, 2010
Bill Zeller, a PhD student at Princeton, sent me the link to his project "graph your inbox", that is an attempt to visualize the "data" in your Gmail account.
Seems to me that it acts as a sophisticated "search my mail" engine. The most interesting part is the ability to click on a point or a bar in one of the charts, and have the corresponding emails show up in the preview panel. This interactive ability is also available in the modern commercial graphing packages, and they are extremely useful for data exploration.
Technically, this is a compelling achievement. The amounts of data being processed, organized, summarized, plotted.
I think he needs to figure out some compelling use cases for something like this. Can you help? How would you use this capability if it is available? | 4,024 | 18,735 | {"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-2024-18 | latest | en | 0.96066 |
https://physics.stackexchange.com/questions/130950/what-exactly-are-crystal-planes-and-how-do-they-reflect-x-rays | 1,656,374,923,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00703.warc.gz | 500,784,734 | 65,978 | # What exactly are crystal planes and how do they reflect x-rays?
1. What exactly are crystal planes and how do they reflect x-rays?
2. Are crystal planes real physical planes or just an abstract concept?
3. What are these planes made of?
4. If they are an abstraction, what do the x-rays hit and get reflected by? Individual atoms?
5. Then where is the concept of a plane coming in?
6. And why do different crystal structures have different 'active' reflecting planes?
7. Could someone please clear up the concept of these planes and Bragg's Law for me?
1. In the image above, you can see a series of Bragg planes drawn in the crystal. This is called one "set of planes". Another "set of planes" would be if one would just draw a series of horizontal lines through the atoms. (Of course by lines I mean planes, but they are projected here onto a 2D image).
2. The planes are those formed by the atoms, so in that sense they are real -- but they are not "mathematical" planes as those stretch to infinity.
3. These planes are made from the atoms that form the crystal. Different crystals will have different structures and therefore the planes will be different.
4. X-rays are actually reflected by electrons most strongly. Therefore, what they are actually scattering off of are the core electrons around the nuclei. What is happening is that because of the structure of the crystal, one can make each of the reflected X-rays constructively interfere by hitting the crystal just at the right angle with the right wavelength of X-ray.
5. The concept of the plane arises when one considers constructive interference as I have described in 4. The atoms effectively form an X-ray diffraction grating (in a reflective geometry as opposed to a transmissive geometry that we are used to).
6. As you can see from the picture above, there are many different "sets of planes" that one could draw. Depending on the angle of incidence and reflection with respect to the crystal, one can reflect off any of these planes in principle.
7. The Bragg Law is a constructive interference condition. The path-length difference between planes is $2d\sin\theta$ and this must be equal to an integer number of wavelengths, $n\lambda$, for the X-rays to constructively interfere with one another. Therefore $2d sin\theta = n\lambda$ is the condition to get "Bragg Scattering".
1. and 2. Crystal planes contain sets of atoms, which occupy identical positions in the primitive cell.
2. So to say, the planes are made out of atoms (and nothing in between). Or if you want a more quantum mechanical picture, you would have electron orbitals in between, which are responsible for the chemical bonding.
3. X-rays get scattered by individual atoms. If many atoms act as scattering centers (e.g. a crystal plane or a set of), the scattered spherical waves, originating from the atoms as point sources, become a planar wave again.
4. This depends on the crystal structure and symmetry.
5. Wikipedia might help with nice pictures there. | 649 | 3,022 | {"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} | 3.03125 | 3 | CC-MAIN-2022-27 | latest | en | 0.951058 |
https://profitclaims.com/how-to-calculate-t-test-with-mean-and-standard-deviation/ | 1,660,307,213,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571692.3/warc/CC-MAIN-20220812105810-20220812135810-00592.warc.gz | 448,142,124 | 10,435 | # how to calculate t-test with mean and standard deviation
An Introduction to T-Tests | Definitions, Formula and Examples Published on January 31, 2020 by Rebecca Bevans.
## An Introduction to T-Tests | Definitions, Formula and Examples
Published on January 31, 2020 by Rebecca Bevans. Revised on May 23, 2022.
A t-test is a statistical test that is used to compare the means of two groups. It is often used in hypothesis testing to determine whether a process or treatment actually has an effect on the population of interest, or whether two groups are different from one another.
You want to know whether the mean petal length of iris flowers differs according to their species. You find two different species of irises growing in a garden and measure 25 petals of each species. You can test the difference between these two groups using a t-test and null and alterative hypotheses.
• The null hypothesis (H0) is that the true difference between these group means is zero.
• The alternate hypothesis (Ha) is that the true difference is different from zero.
1. When to use a t-test
2. What type of t-test should I use?
3. Performing a t-test
4. Interpreting test results
5. Presenting the results of a t-test
## When to use a t-test
A t-test can only be used when comparing the means of two groups (a.k.a. pairwise comparison). If you want to compare more than two groups, or if you want to do multiple pairwise comparisons, use anANOVA testor a post-hoc test.
The t-test is a parametric test of difference, meaning that it makes the same assumptions about your data as other parametric tests. The t-test assumes your data:
1. are independent
2. are (approximately) normally distributed.
3. have a similar amount of variance within each group being compared (a.k.a. homogeneity of variance)
If your data do not fit these assumptions, you can try a nonparametric alternative to the t-test, such as the Wilcoxon Signed-Rank test for data with unequal variances.
## What type of t-test should I use?
When choosing a t-test, you will need to consider two things: whether the groups being compared come from a single population or two different populations, and whether you want to test the difference in a specific direction.
### One-sample, two-sample, or paired t-test?
• If the groups come from a single population (e.g. measuring before and after an experimental treatment), perform a paired t-test.
• If the groups come from two different populations (e.g. two different species, or people from two separate cities), perform a two-sample t-test (a.k.a. independent t-test).
• If there is one group being compared against a standard value (e.g. comparing the acidity of a liquid to a neutral pH of 7), perform a one-sample t-test.
### One-tailed or two-tailed t-test?
• If you only care whether the two populations are different from one another, perform a two-tailed t-test.
• If you want to know whether one population mean is greater than or less than the other, perform a one-tailed t-test.
In your test of whether petal length differs by species:
• Your observations come from two separate populations (separate species), so you perform a two-sample t-test.
• You dont care about the direction of the difference, only whether there is a difference, so you choose to use a two-tailed t-test.
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## Performing a t-test
The t-test estimates the true difference between two group means using the ratio of the difference in group means over the pooled standard error of both groups. You can calculate it manually using a formula, or use statistical analysis software.
### T-test formula
The formula for the two-sample t-test (a.k.a. the Students t-test) is shown below.
In this formula, t is the t-value, x1 and x2 are the means of the two groups being compared, s2 is the pooled standard error of the two groups, and n1 and n2 are the number of observations in each of the groups.
A larger t-value shows that the difference between group means is greater than the pooled standard error, indicating a more significant difference between the groups.
You can compare your calculated t-value against the values in a critical value chart to determine whether your t-value is greater than what would be expected by chance. If so, you can reject the null hypothesis and conclude that the two groups are in fact different.
### T-test function in statistical software
Most statistical software (R, SPSS, etc.) includes a t-test function. This built-in function will take your raw data and calculate the t-value. It will then compare it to the critical value, and calculate a p-value. This way you can quickly see whether your groups are statistically different.
In your comparison of flower petal lengths, you decide to perform your t-test using R. The code looks like this:t.test(Petal.Length ~ Species, data = flower.data)
Sample data set
## Interpreting test results
If you perform the t-test for your flower hypothesis in R, you will receive the following output:
The output provides:
1. An explanation of what is being compared, called data in the output table.
2. The t-value: -33.719. Note that its negative; this is fine! In most cases, we only care about the absolute value of the difference, or the distance from 0. It doesnt matter which direction.
3. The degrees of freedom: 30.196. Degrees of freedom is related to your sample size, and shows how many free data points are available in your test for making comparisons. The greater the degrees of freedom, the better your statistical test will work.
4. The p-value: 2.2e-16 (i.e. 2.2 with 15 zeros in front). This describes the probability that you would see a t-value as large as this one by chance.
5. A statement of the alternate hypothesis (Ha). In this test, the Ha is that the difference is not 0.
6. The 95% confidence interval. This is the range of numbers within which the true difference in means will be 95% of the time. This can be changed from 95% if you want a larger or smaller interval, but 95% is very commonly used.
7. The mean petal length for each group.From the output table, we can see that the difference in means for our sample data is 4.084 (1.456 5.540), and the confidence interval shows that the true difference in means is between 3.836 and 4.331. So, 95% of the time, the true difference in means will be different from 0. Our p-value of 2.2e16 is much smaller than 0.05, so we can reject the null hypothesis of no difference and say with a high degree of confidence that the true difference in means is not equal to zero.
## Presenting the results of a t-test
When reporting your t-test results, the most important values to include are the t-value, the p-value, and the degrees of freedom for the test. These will communicate to your audience whether the difference between the two groups is statistically significant (a.k.a. that it is unlikely to have happened by chance).
You can also include the summary statistics for the groups being compared, namely the mean and standard deviation. In R, the code for calculating the mean and the standard deviation from the data looks like this:
flower.data %>%
group_by(Species) %>%
summarize(mean_length = mean(Petal.Length),
sd_length = sd(Petal.Length))
In our example, you would report the results like this:The difference in petal length between iris species 1 (Mean = 1.46; SD = 0.206) and iris species 2 (Mean = 5.54; SD = 0.569) was significant (t (30) = 33.7190; p < 2.2e-16).
A t-test is a statistical test that compares the means of two samples. It is used in hypothesis testing, with a null hypothesis that the difference in group means is zero and an alternate hypothesis that the difference in group means is different from zero. What does a t-test measure?
A t-test measures the difference in group means divided by the pooled standard error of the two group means.
In this way, it calculates a number (the t-value) illustrating the magnitude of the difference between the two group means being compared, and estimates the likelihood that this difference exists purely by chance (p-value). Which t-test should I use?
Your choice of t-test depends on whether you are studying one group or two groups, and whether you care about the direction of the difference in group means.
If you are studying one group, use a paired t-test to compare the group mean over time or after an intervention, or use a one-sample t-test to compare the group mean to a standard value. If you are studying two groups, use a two-sample t-test.
If you want to know only whether a difference exists, use a two-tailed test. If you want to know if one group mean is greater or less than the other, use a left-tailed or right-tailed one-tailed test. What is the difference between a one-sample t-test and a paired t-test?
A one-sample t-test is used to compare a single population to a standard value (for example, to determine whether the average lifespan of a specific town is different from the country average).
A paired t-test is used to compare a single population before and after some experimental intervention or at two different points in time (for example, measuring student performance on a test before and after being taught the material). Can I use a t-test to measure the difference among several groups?
A t-test should not be used to measure differences among more than two groups, because the error structure for a t-test will underestimate the actual error when many groups are being compared.
If you want to compare the means of several groups at once, its best to use another statistical test such as ANOVA or a post-hoc test. | 2,202 | 10,000 | {"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} | 4.28125 | 4 | CC-MAIN-2022-33 | latest | en | 0.912073 |
https://www.askiitians.com/forums/Mechanics/a-stationary-bomb-explodes-into-two-parts-of-masse_116837.htm | 1,686,142,804,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653764.55/warc/CC-MAIN-20230607111017-20230607141017-00575.warc.gz | 702,088,208 | 32,391 | # A stationary bomb explodes into two parts of masses 3kg and 1kg. The total kinetic energy of the two parts after explosion is 2400 J. Then the K.E. of the smaller part is ?
shabbir chatrissa
9 Points
8 years ago
the initial momentum is zero.
let m1=3 kg m2=1 kg & their velocities be v1 & v2 respectively.
according to conservationh of momentum
m1v+ m2v= 0
3v+ v= 0
v2 = -3v1
also,
0.5m1v22 + 0.5m2v2= 24000
m1v1+ m2v2= 4800
3v12 + v2= 4800
3v12 +(-3v1)= 4800
12v12 = 4800
v1 = 20
v2 = -60 (opp. direction)
KE of m2 is 1800.
I suggest you still cross check. | 236 | 562 | {"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} | 3.640625 | 4 | CC-MAIN-2023-23 | latest | en | 0.67044 |
https://edurev.in/studytube/Types--Business-Mathematics--Statistics/12aa3a9f-5963-4898-b7d0-889ababdd063_t | 1,670,023,917,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00665.warc.gz | 259,352,594 | 56,897 | # Types, Business Mathematics & Statistics Notes | Study Business Mathematics and Statistics - B Com
## Document Description: Types, Business Mathematics & Statistics for B Com 2022 is part of Business Mathematics and Statistics preparation. The notes and questions for Types, Business Mathematics & Statistics have been prepared according to the B Com exam syllabus. Information about Types, Business Mathematics & Statistics covers topics like and Types, Business Mathematics & Statistics Example, for B Com 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Types, Business Mathematics & Statistics.
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## What are Data
DATA
What are data?
Data are plain facts, usually raw numbers. Think of a spreadsheet full of numbers with no meaningful description. In order for these numbers to become information, they must be interpreted to have meaning.
Who creates data?
Data can come from a government census or organization surveys or research studies.
Why the ARE in "data are?"
• Datum (singular form) - a piece of information
• Data (plural form) - multiple pieces of information
## Data Conceptionalization
Can we conceptualize data? Yes!
## Statistics
STATISTICS
Statistics are your place for quick numbers. They might answer the questions "how much" or "how many."
Statistics result from data that have been interpreted.
Statistics can be in the form of numbers or percentages and they are frequently presented in a table or graph.
A statistical table might look like this one from the Statistical Abstract of the United States:
## Evaluating Data and Statistics
Some Questions to Think About When Locating Data or Statistics
Time Period
Are you looking for the current year? Remember that is takes some time to compile data. A range of years? Do you need historical data?
Geography
What country, state, county, etc., might compile the data you want?
Sources
databases, government agencies, researchers, research collections like ICPSR
Availability
What data is likely to be available?
Location
Where is the data likely to be located?
The document Types, Business Mathematics & Statistics Notes | Study Business Mathematics and Statistics - B Com is a part of the B Com Course Business Mathematics and Statistics.
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; | 657 | 3,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-49 | latest | en | 0.816065 |
https://chess.stackexchange.com/questions/36749/will-games-be-perfect-and-most-likely-a-draw-if-both-players-get-infinite-time/36763 | 1,713,400,174,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00672.warc.gz | 151,126,838 | 42,633 | # Will games be perfect, and most likely a draw, if both players get infinite time?
I have played a lot of chess games. The thing that always bothered me was that bloody clock. It looks as if the game is a game about who can think the fastest and it makes one nervous. Now, I know it's part of the game and this gives it charm too. But what if there was no pressure of time? Could a player win? Of course the opponent can do this too. Would this give the best games?
Suppose that both players have all the time they need and all means at their proposal, even the best chess computer in the world). Will some perfect play evolve? Would both players make the same moves (if in each others shoes)? Or is there always some guessing involved?
• Comments are not for extended discussion; this conversation has been moved to chat. Aug 8, 2021 at 14:05
Computers don't need infinite time to play perfectly. Nearly but not quite. They need exponential time according to Fraenkel and Licthenstein (Journal of combinational theory).
Humans could theoretically achieve a similar result if they checked all positions and committed no errors in their calculations. Potentially having an extremely large number of boards where they could check the moves. However calculating a single move would take many generations of humans playing a single game.
On a practical level, if you are bothered about clocks and would like to go deeper in your calculations you should probably play correspondence chess. Most sites e.g. chess.com and lichess.org allow you to play in a format that allows you to think for a move for several days.
• Comments are not for extended discussion; this conversation has been moved to chat. Aug 8, 2021 at 14:05
No, time is not everything. Memory is also important, for a brute force approach one has to memorise all the branchings and lines already read out.
Note also that solving chess is a problem of finite size, albeit the size is really huge. So you can do it in some finite time using some finite amount of memory, at least in theory. Solving chess is probably impossible in the real physical universe, good judgement in pruning the game tree becomes another requirement. Without good judgement, even infinite time will not help.
For a game between human players, limits can be lifted in several ways, correspondence chess does not only allow rather long time limits but also allows the use of any resource available to the player, including access to a library of chess books (extending memory in some way, as you can lookup opening lines) and computers (with their faster "thinking" they increase the time effectively, and they also provide additional memory).
• Thats interesting. Two different approaches. I mean this one and the other answer. Especially the judement seems important indeed. Aug 7, 2021 at 18:40
• Good judgement would imply heuristics and not perfect play. That would simply be probabilistically good play. What computers today do. Aug 7, 2021 at 22:02 | 635 | 3,000 | {"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} | 2.890625 | 3 | CC-MAIN-2024-18 | latest | en | 0.97628 |
http://bootmath.com/how-to-prove-with-mathematical-induction-3n-n2.html | 1,529,624,842,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864303.32/warc/CC-MAIN-20180621231116-20180622011116-00194.warc.gz | 43,383,833 | 7,283 | # How to Prove with Mathematical Induction $3^n > n^2$
How do I prove that $3^n > n^2$ with mathematical induction? I thought I had the correct answer but my teacher says its wrong.
I let $n=1$ for the initial case and it works. I then assumed $n=k$ works and went onto the $n=k+1$ case, but after that it goes a little awry. Thanks in advance!
#### Solutions Collecting From Web of "How to Prove with Mathematical Induction $3^n > n^2$"
The base case is just noticing $3^1>1^2$.
The inductive step:
Suppose $3^n>n^2$ we want to prove $3^{n+1}=3^n+3^n+3^n>1+2n+n^2=(n+1)^2$
This follows from the following:
$3^n\geq 1$
$3^n\geq 2n$ (you may have a go at this with another induction)
$3^n> n^2$ (inductive step)
I’ll outline a more or less self-contained proof. However, it may be worth observing that your own problem follows as a trivial conclusion of proving that $3^n\geq n^3$ for all $n\geq 1$. Here is an answer to that problem if you want to check it out.
To prove that $3^n>n^2$ for all $n\geq 1$ using induction, it will be useful to use two base cases at the beginning (this has already been referenced in one or two previous comments). The reason for this will become clear at the end.
Let $S(n)$ denote the following for $n\geq 1$ where $n\in\mathbb{N}$:
$$S(n) : 3^n>n^2.$$
Base step ($n=1,2$): $S(1)$ says that $3^1=3>1=1^2$, and this is true. Now, $S(2)$ says that $3^2=9>4=2^2$, and this is also true.
Inductive step: Fix some $k\geq 2$ and assume that $S(k)$ is true where
$$S(k) : 3^k>k^2.$$
To be shown is that $S(k+1)$ follows where
$$S(k+1) : 3^{k+1} > (k+1)^2.$$
Beginning with the left-hand side of $S(k+1)$,
\begin{align}
3^{k+1} &= 3\cdot 3^k\tag{by definition}\\[0.5em]
&> 3\cdot k^2\tag{by $S(k)$, the ind. hyp.}\\[0.5em]
&> k^2+2k+1\tag{since $k\geq 2$; see $(\dagger)$}\\[0.5em]
&= (k+1)^2,
\end{align}
we end up at the right-hand of $S(k+1)$, completing the inductive step.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
$(\dagger)$: When $k\geq 2$, how can we claim that $3k^2>k^2+2k+1$? Notice that
$$3k^2>k^2+2k+1 \Longleftrightarrow 2k^2-2k-1 > 0 \Longleftrightarrow k(2k-2)-1>0,$$
and this is obviously true for $k\geq 2$, but it is not true when $k=1$, and this explains why we needed two base cases at the beginning.
We can show the inequality directly for $n=1$ and $n=2$.
Suppose $3^n>n^2$ as the inductive hypothesis.
For $n \geq 2$, $2.n^2 > 2n + 1$. Can you see why?
Then $3^{n+1} = 3.3^n > 3.n^2 = n^2 + 2.n^2 > n^2+2n+1 = (n+1)^2$.
The first inequality is via the inductive hypothesis, and the second inequality is via the inequality above.
You seem to have the principle correct, so the issue is one of application. There are probably simpler methods than this, but here is one way.
Show that it works for $n \in \{1, 2, 3\}$:
\begin{align} 3^1 = 3 &> 1^2 = 1\quad \checkmark\\ 3^2 = 9 &> 2^2 = 4\quad \checkmark\\ 3^3 = 27 &> 3^2 = 9\quad \checkmark \end{align}
Now, let’s assume $3^n > n^2$. We need to prove that $3^{n+1} > \left(n+1\right)^2$.
\begin{align} 3^{n+1} = 3 \times 3^n &= 3^n + 3^n + 3^n\quad&(1)\\ (n+1)^2 &= n^2 + 2n + 1\quad&(2)\\ \end{align}
Looking at how the columns line up on the right side of (1) and (2), we can show that all corresponding elements of the top row are greater than those in the bottom row. In the first column, the top row is larger than the bottom row, we assumed it as part of the inductive step. The second column is true as $n^2 > 2n$ for all $n > 2$. So if $3^n > n^2$ via the induction step, it is certainly greater than $2n$. This is why we proved the individual cases up to 3 above. If your teacher is strict, you may have to prove this as a lemma using induction as well. The last column is true for all $n > 0$, and we started our induction step at 3. QED.
Just to be complete, we can show $n^2 > 2n$ for $n \geq 3$.
$$3^2 = 9 > 2\times 3 = 6\quad \checkmark$$
Assume $n^2 > 2n$. Show $(n+1)^2 > 2(n+1)$.
\begin{align} (n+1)^2 = &n^2 +& 2n + &1\\ 2(n+1) = &&2n + &2 \end{align}
The $2n$ cancels, and we have to show that $n^2 + 1 > 2$, or $n^2 > 1$, which is trivial for $n > 3$.
To get from $3^n$ to $3^{n+1}$ you multiply by $3$. To get from $n^2$ to $(n+1)^2$, you multiply by $(\frac {n+1} n)^2$. Clearly $(\frac {n+1} n)^2$ will eventually by strictly smaller than $3$, and therefore past that point, the inequality $3^n>n^2$, if it’s ever true, will be true forever afterwards.
Now, $(\frac {n+1} n)^2=(1+\frac 1 n)^2$ is a decreasing sequence which tends towards $1$, and some computation shows that for $n=2$ it is already below $3$. So we just have to check $3^2>2^2$, which is indeed true. | 1,703 | 4,668 | {"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} | 4.65625 | 5 | CC-MAIN-2018-26 | latest | en | 0.883752 |
https://netlib.org/lapack/explore-html/d5/d99/group__single__eig_ga4031914169fb1a0ddfd7dd67359c8b4f.html | 1,660,680,908,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00715.warc.gz | 376,502,292 | 6,445 | LAPACK 3.10.1 LAPACK: Linear Algebra PACKage
## ◆ ssyt22()
subroutine ssyt22 ( integer ITYPE, character UPLO, integer N, integer M, integer KBAND, real, dimension( lda, * ) A, integer LDA, real, dimension( * ) D, real, dimension( * ) E, real, dimension( ldu, * ) U, integer LDU, real, dimension( ldv, * ) V, integer LDV, real, dimension( * ) TAU, real, dimension( * ) WORK, real, dimension( 2 ) RESULT )
SSYT22
Purpose:
``` SSYT22 generally checks a decomposition of the form
A U = U S
where A is symmetric, the columns of U are orthonormal, and S
is diagonal (if KBAND=0) or symmetric tridiagonal (if
KBAND=1). If ITYPE=1, then U is represented as a dense matrix,
otherwise the U is expressed as a product of Householder
transformations, whose vectors are stored in the array "V" and
whose scaling constants are in "TAU"; we shall use the letter
"V" to refer to the product of Householder transformations
(which should be equal to U).
Specifically, if ITYPE=1, then:
RESULT(1) = | U**T A U - S | / ( |A| m ulp ) and
RESULT(2) = | I - U**T U | / ( m ulp )```
``` ITYPE INTEGER
Specifies the type of tests to be performed.
1: U expressed as a dense orthogonal matrix:
RESULT(1) = | A - U S U**T | / ( |A| n ulp ) and
RESULT(2) = | I - U U**T | / ( n ulp )
UPLO CHARACTER
If UPLO='U', the upper triangle of A will be used and the
(strictly) lower triangle will not be referenced. If
UPLO='L', the lower triangle of A will be used and the
(strictly) upper triangle will not be referenced.
Not modified.
N INTEGER
The size of the matrix. If it is zero, SSYT22 does nothing.
It must be at least zero.
Not modified.
M INTEGER
The number of columns of U. If it is zero, SSYT22 does
nothing. It must be at least zero.
Not modified.
KBAND INTEGER
The bandwidth of the matrix. It may only be zero or one.
If zero, then S is diagonal, and E is not referenced. If
one, then S is symmetric tri-diagonal.
Not modified.
A REAL array, dimension (LDA , N)
The original (unfactored) matrix. It is assumed to be
symmetric, and only the upper (UPLO='U') or only the lower
(UPLO='L') will be referenced.
Not modified.
LDA INTEGER
The leading dimension of A. It must be at least 1
and at least N.
Not modified.
D REAL array, dimension (N)
The diagonal of the (symmetric tri-) diagonal matrix.
Not modified.
E REAL array, dimension (N)
The off-diagonal of the (symmetric tri-) diagonal matrix.
E(1) is ignored, E(2) is the (1,2) and (2,1) element, etc.
Not referenced if KBAND=0.
Not modified.
U REAL array, dimension (LDU, N)
If ITYPE=1 or 3, this contains the orthogonal matrix in
the decomposition, expressed as a dense matrix. If ITYPE=2,
then it is not referenced.
Not modified.
LDU INTEGER
The leading dimension of U. LDU must be at least N and
at least 1.
Not modified.
V REAL array, dimension (LDV, N)
If ITYPE=2 or 3, the lower triangle of this array contains
the Householder vectors used to describe the orthogonal
matrix in the decomposition. If ITYPE=1, then it is not
referenced.
Not modified.
LDV INTEGER
The leading dimension of V. LDV must be at least N and
at least 1.
Not modified.
TAU REAL array, dimension (N)
If ITYPE >= 2, then TAU(j) is the scalar factor of
v(j) v(j)**T in the Householder transformation H(j) of
the product U = H(1)...H(n-2)
If ITYPE < 2, then TAU is not referenced.
Not modified.
WORK REAL array, dimension (2*N**2)
Workspace.
Modified.
RESULT REAL array, dimension (2)
The values computed by the two tests described above. The
values are currently limited to 1/ulp, to avoid overflow.
RESULT(1) is always modified. RESULT(2) is modified only
if LDU is at least N.
Modified.```
Definition at line 155 of file ssyt22.f.
157 *
158 * -- LAPACK test routine --
159 * -- LAPACK is a software package provided by Univ. of Tennessee, --
160 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
161 *
162 * .. Scalar Arguments ..
163 CHARACTER UPLO
164 INTEGER ITYPE, KBAND, LDA, LDU, LDV, M, N
165 * ..
166 * .. Array Arguments ..
167 REAL A( LDA, * ), D( * ), E( * ), RESULT( 2 ),
168 \$ TAU( * ), U( LDU, * ), V( LDV, * ), WORK( * )
169 * ..
170 *
171 * =====================================================================
172 *
173 * .. Parameters ..
174 REAL ZERO, ONE
175 parameter( zero = 0.0e0, one = 1.0e0 )
176 * ..
177 * .. Local Scalars ..
178 INTEGER J, JJ, JJ1, JJ2, NN, NNP1
179 REAL ANORM, ULP, UNFL, WNORM
180 * ..
181 * .. External Functions ..
182 REAL SLAMCH, SLANSY
183 EXTERNAL slamch, slansy
184 * ..
185 * .. External Subroutines ..
186 EXTERNAL sgemm, ssymm
187 * ..
188 * .. Intrinsic Functions ..
189 INTRINSIC max, min, real
190 * ..
191 * .. Executable Statements ..
192 *
193 result( 1 ) = zero
194 result( 2 ) = zero
195 IF( n.LE.0 .OR. m.LE.0 )
196 \$ RETURN
197 *
198 unfl = slamch( 'Safe minimum' )
199 ulp = slamch( 'Precision' )
200 *
201 * Do Test 1
202 *
203 * Norm of A:
204 *
205 anorm = max( slansy( '1', uplo, n, a, lda, work ), unfl )
206 *
207 * Compute error matrix:
208 *
209 * ITYPE=1: error = U**T A U - S
210 *
211 CALL ssymm( 'L', uplo, n, m, one, a, lda, u, ldu, zero, work, n )
212 nn = n*n
213 nnp1 = nn + 1
214 CALL sgemm( 'T', 'N', m, m, n, one, u, ldu, work, n, zero,
215 \$ work( nnp1 ), n )
216 DO 10 j = 1, m
217 jj = nn + ( j-1 )*n + j
218 work( jj ) = work( jj ) - d( j )
219 10 CONTINUE
220 IF( kband.EQ.1 .AND. n.GT.1 ) THEN
221 DO 20 j = 2, m
222 jj1 = nn + ( j-1 )*n + j - 1
223 jj2 = nn + ( j-2 )*n + j
224 work( jj1 ) = work( jj1 ) - e( j-1 )
225 work( jj2 ) = work( jj2 ) - e( j-1 )
226 20 CONTINUE
227 END IF
228 wnorm = slansy( '1', uplo, m, work( nnp1 ), n, work( 1 ) )
229 *
230 IF( anorm.GT.wnorm ) THEN
231 result( 1 ) = ( wnorm / anorm ) / ( m*ulp )
232 ELSE
233 IF( anorm.LT.one ) THEN
234 result( 1 ) = ( min( wnorm, m*anorm ) / anorm ) / ( m*ulp )
235 ELSE
236 result( 1 ) = min( wnorm / anorm, real( m ) ) / ( m*ulp )
237 END IF
238 END IF
239 *
240 * Do Test 2
241 *
242 * Compute U**T U - I
243 *
244 IF( itype.EQ.1 )
245 \$ CALL sort01( 'Columns', n, m, u, ldu, work, 2*n*n,
246 \$ result( 2 ) )
247 *
248 RETURN
249 *
250 * End of SSYT22
251 *
real function slansy(NORM, UPLO, N, A, LDA, WORK)
SLANSY returns the value of the 1-norm, or the Frobenius norm, or the infinity norm,...
Definition: slansy.f:122
subroutine ssymm(SIDE, UPLO, M, N, ALPHA, A, LDA, B, LDB, BETA, C, LDC)
SSYMM
Definition: ssymm.f:189
subroutine sgemm(TRANSA, TRANSB, M, N, K, ALPHA, A, LDA, B, LDB, BETA, C, LDC)
SGEMM
Definition: sgemm.f:187
subroutine sort01(ROWCOL, M, N, U, LDU, WORK, LWORK, RESID)
SORT01
Definition: sort01.f:116
real function slamch(CMACH)
SLAMCH
Definition: slamch.f:68
Here is the call graph for this function:
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https://www.studypool.com/discuss/5047235/financial-problems-4 | 1,720,953,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00097.warc.gz | 904,217,651 | 48,824 | # Financial Problems
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Hi there,Attached please find the Excel template provided with all work completed. That is to say, every problem is answered completely, with citations included :)Thanks again,Selenica
UNDERSTANDING HEALTHCARE FINANCIAL MANAGEMENT
Chapter 15 -- Revenue Cycle and Current Accounts Management
PROBLEM 2
James Buchanan Orthotics and Prosthetics is planning to request a line of credit from its bank.
The company has produced sales estimates, and these appear in the worksheet below. Collection
estimates are as follows: 10 percent within the month of sale, 75 percent in the month following
the sale, and 15 percent in the second month following the sale. Labor and supplies estimates also
appear in the worksheet below. Payments for labor and supplies are typically made during the
month following the one in which these costs have been incurred. General and administrative
salaries will amount to approximately \$27,000 a month; lease payments under long-term lease
contracts will be \$9,000 a month; depreciation charges will be \$36,000 a month; miscellaneous
expenses will be \$2,700 a month; income tax payments of \$63,000 will be due in both September
and December; and a progress payment of \$180,000 on a new building must be paid in October.
Cash on hand on July 1 will amount to \$132,000, and a minimum cash balance of \$90,000 will be
maintained throughout the cash budget period. What loan will be the company require in
October?
May
Collections worksheet:
Billed charges
Collections
Within 30 days
30-60 days
60-90 days
Total collections
Supplies worksheet:
Amount of labor and supplies
Payments made for labor and supplies
Net cash gain (loss):
Total collections
Total purchases
Lease payments
Miscellaneous expenses
Taxes
Progress payment
Total payments
Net cash gain/loss
Borrowing/surplus summary:
Cash at beginning with no borrowing
Cash at end with no borrowing
Target cash balance (given)
Cumulative surplus cash / loan balance
\$180,000
June
July
\$180,000
18,000
18,000
135,000
18,000
153,000
\$90,000
18,000
0
27000
9000
2700
38,700
-20,700
\$90,000
\$90,000
153,000
90000
27000
9000
2700
128,700
24,300
August
\$360,000
36,000
135,000
27,000
198,000
\$126,000
\$90,000
198,000
90000
27000
9000
2700
128,700
69,300
132,000
201,300
90,000
111,300
\$540,000
54,000
270,000
27,000
351,000
\$882,000
\$126,000
351,000
126000
27000
9000
2700
164,700
186,300
201,300
387,600
90,000
297,600
September
\$720,000
72,000
405,000
54,000
531,000
\$306,000
\$882,000
531,000
882000
27000
9000
2700
63000
983,700
-452,700
387,600
-65,100
90,000
-155,100
Loan
October
\$360,000
36,000
540,000
81,000
657,000
\$234,000
\$306,000
657,000
306000
27000
9000
2700
180000
524,700
132,300
-65,100
67,200
90,000
-22,800
Loan
November
\$360,000
36,000
270,000
108,000
414,000
\$162,000
\$234,000
414,000
234000
27000
9000
2700
272,700
141,300
67,200
208,500
90,000
118,500
December
\$90,000
9,000
270,000
54,000
333,000
\$90,000
\$162,000
333,000
162000
27000
9000
2700
63000
263,700
69,300
208,500
277,800
90,000
187,800
January
\$180,000
18,000
67,500
54,000
139,500
\$90,000
139,500
90000
27000
9000
2700
128,700
10,800
277,800
288,600
90,000
198,600
UNDERSTANDING HEALTHCARE FINANCIAL MANAGEMENT
Chapter 15 -- Revenue Cycle and Current ...
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4.4 | 1,168 | 3,707 | {"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-2024-30 | latest | en | 0.938929 |
https://elearning.uni-oldenburg.de/dispatch.php/shared/modul/description/3d787a14476a9b39f9475cef3f5c332f/?display_language=EN | 1,726,552,322,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00043.warc.gz | 193,182,313 | 8,120 | mat960 - Mathematics of Computer Science (Analysis) (Complete module description)
# mat960 - Mathematics of Computer Science (Analysis) (Complete module description)
Module label Mathematics of Computer Science (Analysis) Module code mat960 Credit points 6.0 KP Workload 180 h Institute directory Department of Mathematics Applicability of the module Bachelor's Programme Business Informatics (Bachelor) > Aufbaucurriculum-Wahlbereich Mathematik Bachelor's Programme Computing Science (Bachelor) > Aufbaumodule Responsible persons Chernov, Alexey (module responsibility) Grieser, Daniel (module responsibility) Pankrashkin, Konstantin (module responsibility) Schöpfer, Frank (module responsibility) Shestakov, Ivan (module responsibility) Uecker, Hannes (module responsibility) Vertman, Boris (module responsibility) Prerequisites Skills to be acquired in this module The students learn and apply basic notions and techniques of mathematical analysis.Professional competenceThe students:· use rigorous mathematical proofs· compute limit values and analyse the convergence behaviour of iterative methods · apply differential and integral calculus to compute extreme values, to analyse the behaviour of functions and to develop numerical solution methodsMethodological competenceThe students:· analyse formal relations· structure and justify solution methodsSocial competenceThe students:· develop solutions to given problems in groups· accept constructive criticismPersonal competenceThe students:· reflect their solution strategies· deepen their understanding of the presented mathematical concepts with exercises and adopt the solution methods Module contents · Convergence of sequences, series and iterative methods· Continuity, differential and integral calculus of functions of one variable· Characterization and computation of extreme values· Separable and linear ordinary differential equations Recommended reading Peter Hartmann: Mathematik für Informatiker - ein praxisbezogenes LehrbuchDirk Hachenberger: Mathematik für InformatikerOtto Forster: Analysis IHarro Heuser: Lehrbuch der Analysis, Teil 1Konrad Königsberger: Analysis Links Language of instruction German Duration (semesters) 1 Semester Module frequency every year Module capacity unlimited
Type of course Comment SWS Frequency Workload of compulsory attendance
Lecture 3 SuSe 42
Exercises 1 SuSe 14
Total module attendance time 56 h
Examination Prüfungszeiten Type of examination
Final exam of module
At the end of the lecture period written exam
Final exam of module | 491 | 2,541 | {"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-2024-38 | latest | en | 0.63495 |
https://mozo.com.au/family-finances/hecs-indexation-increasing-by-7-but-what-does-that-mean | 1,721,793,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00708.warc.gz | 354,618,522 | 37,201 | # HECS indexation increasing by 7%, but what does that mean?
The Australian Tax Office has confirmed that student loans will be increased and indexed at 7%.
Higher Education Loan Program (HELP, commonly called 'HECS debts') is a type of interest free government loan many Australians use to pay for university.
Every year the government indexes the loans and raises them to meet inflation and the current value of the Australian dollar.
Despite popular belief, this indexation is not a form of interest. After all, HECS is an interest free loan. Instead, the value of that money follows inflation which is calculated annually.
## How is the annual indexation calculated
The indexation is calculated by using the Consumer Price Index (CPI)--essentially the calculation of the change in the value of goods and services. This is also known as inflation.
Remember how two years ago eggs were \$3 and now are \$6? Examples like that are used to calculate the CPI, inflation, cost of living and eventually the HECS index.
So, as the currency value changes, so does the value of your HECS debt.
For example, if you took out a \$20,000 student loan a couple of years ago, the value of that \$20,000 has changed. And the government wants the loan to be repaid to the same spending power so it can lend it to students again doing the same exact course you did a couple of years ago.
Again, this is very different from charging you interest. Interest is a fee you are charged for borrowing money expressed as a percentage of the total amount of the loan.
The government is not charging you for borrowing money, as mentioned before they just want to be paid back the real value of said money.
While many people find it unfair, especially when wages have not kept up with inflation, being indexed annually isn’t as bad as being charged interest. (Take it from an American with a US student loan!).
Even though 7% is a big jump, there have been many years where HECS’ annual index was less than three percent. Have a look at the below:
Indexation through the years - via ATO
Year ended 30 June.. Percentage 2023 7.1% 2022 3.9% 2021 0.6% 2020 1.8% 2019 1.8% 2018 1.9% 2017 1.5% 2016 1.5% 2015 2.1% 2014 2.6% 2013 2.0% 2012 2.9% 2011 3.0% 2010 1.9%
Take note that the new index will be applied on 1 June 2023.
If you need help getting your budget in order, check out our budget calculator. Alternatively, if you’re concerned about the cost of living, check our Family Finance page. | 609 | 2,483 | {"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-2024-30 | latest | en | 0.976018 |
bowties.followthemop.ca | 1,716,165,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00600.warc.gz | 124,884,324 | 19,443 | # Rough Work
“Solve 2+2, Show your rough work.”
Being asked to “show your work”, “justify your answer”, or “write every step down” is a staple of education, and for good reason. For essays, a draft copy or outline can be used to find issues early on. In the sciences, and especially mathematics, rough work shows how you arrived at a solution.
## What is Rough Work?
Rough work is the name we give to the steps you took to get to a solution. Rough work is a tool to organize your thinking, track your progress, and justify your final answer. No bridge is ever built where the engineer simply states “it can hold 30 tons, trust me” without the requisite 500 page report to back it up. Writing a technical report is the workplace equivalent of “show your work”. An academic paper is the same; a way to show your result, and how you got to it so others can “see for themselves”. Rough work is a reference to others so they can review your steps, check for mistakes and confirm your findings.
## Why Show Your Work?
Try this. In your head, don’t write anything down or use calculator and solve this problem:
353
x432
Chances are you can comfortably do some of the calculation. But once you finish the next part you’ve forgotten the result of the first part. Your natural instinct is write down the first part. As the calculation becomes ever more complicated you have to write something down eventually. Perfect memory savants aside, doing rough work helps track long complex calculation.
Rough work is also for “Ah Ha!” moments. Sometimes in especially difficult problems what to do next doesn’t come to you immediately— it could take minutes, days or even years before a solution comes to you. Here the solution is now obvious to you, but late when you forget, the notes you took will be invaluable. Famously the mathematician Pierre de Fermat upon reading a particular problem wrote “I have discovered a truly marvelous proof of this, which this margin is too narrow to contain”. And died before he wrote it down in full. This problem ended up taking other mathematicians over 400 years to solve. The final solution is now contained in multiple papers and over hundreds of pages of proof. Fermat’s Last Theorem, as it’s called, still stands unsolved as a ”truely marvellous proof”. Don’t be Fermat; be a great mathematician, but show your work.
If you’re in school your professor probably requires rough work. No one plans to get the answer wrong, so requiring rough work helps to save you from yourself. While writing a test where you may need to write long complex calculations. It’s also good to leave a bread trail, chances are you won’t remember later exactly where that 42 came from during the test. Was it a clear comprehension error or was it a 24? So do write some rough work and sort out the typos later.
And of course if you cant get to a solution, rough work is at least something. Show what you do know and maybe get some credit for know something rather than nothing.
### How Much Rough Work?
The type rough work for a problem is often tied to the level of understanding and knowledge that is expected. For instance, if the professor asks you to show work for 2+2, then perhaps you are just starting school and a diagram is needed. Your understanding might just be significantly ahead of the expected knowledge in the class. Maybe it is blindingly obvious to you that $\int _1^7 x^2 dx$ is 114, but probably not true of everyone in the class. The skill of explaining how you got to the answer is still helpful later when you approach more complex problems.
In some cases proving your answer can also be a good way to show your work. If you can “just know” that $x^2+5x+6$ is $(x+2)(x+3)$ then show the steps to expand your answer and show that indeed $(x+2)(x+3)=x^2+5x+6$. Rough work gives you a place to justify your answer, even for multiple choice. Explaining how you got an answer might be a way to show an alternative and perfectly valid solution.
### How to Submit Digital Rough Work
For online tests and assignments submitting rough work is a little more than stapling your scrap paper to the test. A scanning app can quickly capture all of your rough work as a single file and make it easier for your professor to read. An app with a connected cloud storage service like the Microsoft Office App, OneDrive, Google Drive, or Files on iOS can scan documents to your cloud account then access them immediately on your computer.
Scanning your documents can also be a great way to archive your school work. At the end of the term it can be hard to part with all the papers youve written, because you “might need it later”. Scanning all of your work ensures you have access to it again later, and it doesn’t require 10 boxes in your attic.
So whether you’re filing a major report, solving a centuries old problem, or just writing a math test, add some rough work. It might help you organize your thoughts better and catch some errors, or you might look back on it with your prof and laugh, and gain a few more marks. | 1,114 | 5,051 | {"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": 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-2024-22 | longest | en | 0.955275 |
https://www.doorsteptutor.com/Exams/NEET/Physics/Questions/Topic-Oscillations-and-Waves-9/Part-14.html | 1,524,409,346,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945604.91/warc/CC-MAIN-20180422135010-20180422155010-00177.warc.gz | 780,468,946 | 20,051 | # Oscillations and Waves (NEET (NTA)-National Eligibility cum Entrance Test (Medical) Physics): Questions 60 - 64 of 195
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## Question number: 60
» Oscillations and Waves » Simple Harmonic Motion » Phase
MCQ▾
### Question
A particle executes simple harmonic motion [amplitude = A] between x =-A and x =+A. The time taken for it to go from 0 to is T 1 and to go from to A is T 2. Then
### Choices
Choice (4) Response
a.
T 1 < T 2
b.
T 1 = 2T 2
c.
T 1 = T 2
d.
T 1 > T 2
## Question number: 61
» Oscillations and Waves » Periodic Functions
MCQ▾
### Question
Two masses m 1 and m 2 are suspended together by a massless spring of constant k. When the masses are in equilibrium, m 1 is removed without disturbing the system. The amplitude of oscillations is,
### Choices
Choice (4) Response
a.
b.
c.
d.
## Passage
The differential equation of a particle undergoing SHM is given by .
The particle starts from the extreme position.
## Question number: 62 (1 of 2 Based on Passage) Show Passage
» Oscillations and Waves » Simple Harmonic Motion » Phase
MCQ▾
### Question
The ratio of the maximum acceleration to the maximum velocity of the particle is –
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 63 (2 of 2 Based on Passage) Show Passage
» Oscillations and Waves » Simple Harmonic Motion » Phase
MCQ▾
### Question
The equation of motion may be given by:
### Choices
Choice (4) Response
a.
b.
c.
d.
Question does not provide sufficient data or is vague
## Question number: 64
» Oscillations and Waves » Free Forced and Damped Oscillations
MCQ▾
### Question
A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limits – ϕ and + ϕ. For an angular displacement θ (| θ| < ϕ), the tension in the string and the velocity of the bob are T and v respectively. The following relations hold good under the above conditions.
(1)
(2)
(3) The magnitude of the tangential acceleration of the bob
(4)
Select the correct answers and mark it according to the following codes:
### Choices
Choice (4) Response
a.
2 and 4 are correct
b.
1, 2 and 3 are correct
c.
1 and 2 are correct
d.
1 and 3 are correct
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# 1 AM Quiz : Matrix Puzzle
Attempt now to get your rank among 2615 students!
Question 1
Direction: Study the following information carefully and answer the given questions.
Nine boxes are arranged in such a way that it creates 3*3 matrix. Each row and each column consists of 3 boxes.A is kept just above Y in the same column but not in the last column. Y is not kept in the bottom row. W is kept to the south-east of Y but not in the middle column. D is kept to the west of W but not to the south of Y. M is kept to the north-east of D but not in the same row of A. K is kept in the same row of A but not in the same column of D. S is kept to the east of D. X is kept in the row above the row of L.
Which box is kept just to the east of L?
Question 2
Direction: Study the following information carefully and answer the given questions.
Nine boxes are arranged in such a way that it creates 3*3 matrix. Each row and each column consists of 3 boxes.A is kept just above Y in the same column but not in the last column. Y is not kept in the bottom row. W is kept to the south-east of Y but not in the middle column. D is kept to the west of W but not to the south of Y. M is kept to the north-east of D but not in the same row of A. K is kept in the same row of A but not in the same column of D. S is kept to the east of D. X is kept in the row above the row of L.
How many boxes are kept to the south-east of X?
Question 3
Direction: Study the following information carefully and answer the given questions.
Nine boxes are arranged in such a way that it creates 3*3 matrix. Each row and each column consists of 3 boxes.A is kept just above Y in the same column but not in the last column. Y is not kept in the bottom row. W is kept to the south-east of Y but not in the middle column. D is kept to the west of W but not to the south of Y. M is kept to the north-east of D but not in the same row of A. K is kept in the same row of A but not in the same column of D. S is kept to the east of D. X is kept in the row above the row of L.
Which box is kept to the north of S?
Question 4
Direction: Study the following information carefully and answer the given questions.
Nine boxes are arranged in such a way that it creates 3*3 matrix. Each row and each column consists of 3 boxes.A is kept just above Y in the same column but not in the last column. Y is not kept in the bottom row. W is kept to the south-east of Y but not in the middle column. D is kept to the west of W but not to the south of Y. M is kept to the north-east of D but not in the same row of A. K is kept in the same row of A but not in the same column of D. S is kept to the east of D. X is kept in the row above the row of L.
Which box is kept to the west of Y?
Question 5
Direction: Study the following information carefully and answer the given questions.
Nine boxes are arranged in such a way that it creates 3*3 matrix. Each row and each column consists of 3 boxes.A is kept just above Y in the same column but not in the last column. Y is not kept in the bottom row. W is kept to the south-east of Y but not in the middle column. D is kept to the west of W but not to the south of Y. M is kept to the north-east of D but not in the same row of A. K is kept in the same row of A but not in the same column of D. S is kept to the east of D. X is kept in the row above the row of L.
What is the direction of L with respect to W?
• 2615 attempts | 885 | 3,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-33 | latest | en | 0.975961 |
https://forum.openframeworks.cc/t/organic-cell-like-motion-with-ofxbox2d/30294 | 1,656,846,798,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104240553.67/warc/CC-MAIN-20220703104037-20220703134037-00274.warc.gz | 303,079,428 | 5,463 | # Organic cell-like motion with ofxBox2d?
hi all! trying to computationally understand my problem. i want to make this cell-like shape that shifts around organically, maybe something similar to metaballs?
right now, i’m using ofxbox2d and connecting ofxbox2d circles to each other with joints, and eventually connecting the last one to the first one to close the shape:
`````` for (unsigned int i = 0; i < circles.size(); i++)
{
auto joint = std::make_shared<ofxBox2dJoint>();
if (i == 0)
{
joint.get()->setup(box2d.getWorld(), circles[circles.size()-1].get()->body, circles[i].get()->body);
}
else
{
joint.get()->setup(box2d.getWorld(), circles[i-1].get()->body, circles[i].get()->body);
}
joint.get()->setFrequency(4.0);
joint.get()->setDamping(0.5);
joint.get()->setLength(ofRandom(40, 74));
joints.push_back(joint);
}
``````
this seems to work fine to me. i run into issues when trying to articulate the motion, however. i thought i’d be able to do it by adding a repulsive force to each circle that pushes it away from the circle that’s closest to it:
``````void ofApp::applyForce(std::vector<std::shared_ptr<ofxBox2dCircle> > circleVector)
{
for (int i = 0; i < circles.size(); i++)
{
float minDistance = 50;
int closestIndex;
glm::vec2 currentPos;
glm::vec2 closestPos;
currentPos = glm::vec2(circles[i].get()->getPosition().x, circles[i].get()->getPosition().y);
for (int j = 0; j < circleVector.size(); j++)
{
float distance;
glm::vec2 otherPos;
if (j != i)
{
otherPos = glm::vec2(circleVector[j].get()->getPosition().x, circleVector[j].get()->getPosition().y);
distance = glm::distance(currentPos, otherPos);
if (distance < minDistance)
{
closestIndex = j;
minDistance = distance;
}
}
// std::cout << minDistance << std::endl;
float force = ofMap(minDistance, 0, 50, 0.5, 0.1);
force = ofClamp(force, 0.1, 0.5);
}
}
}
``````
but they all just push each other to the bounds and i end up with this:
i think intuitively i understand what’s happening – am also having trouble putting that into words – but i’m having a hard time figuring out how to translate the motion i want into code. would love to tap into everyone’s collective experience and help me navigate this issue
I think it might be an issue with not having friction, or the ofClamp you’re using, it is limiting it to the lower bound of 0.1 which means it can’t roll of to zero.
(this is just from a quick glance over)
1 Like | 657 | 2,417 | {"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} | 2.59375 | 3 | CC-MAIN-2022-27 | latest | en | 0.682667 |
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Topic: Egyptian topology on Q
Replies: 9 Last Post: May 24, 2004 10:19 PM
Messages: [ Previous | Next ]
Michael Barr Posts: 748 Registered: 12/6/04
Egyptian topology on Q
Posted: May 21, 2004 5:22 PM
Let S be the sequence {1,1/2,1/3,...,0}, a compact topological space.
Let X be the disjoint union of all the spaces S^n, n in N. Then X is
locally compact, Lindelof, normal, nice space. Actually, even metric,
but that is not important. Map X to Q by addition: (u_1,u_2,...,u_n)
|--> u_1 + u_2 + ... + u_n. This induces a quotient topology which,
for obvious reasons I call the Egyptian topology. I have an argument,
not entirely solid, but I think it right, that suggests this is
strictly finer than the standard topology. I can prove that it is a
normal Hausdorff space. Has anyone seen this before and can anyone
Incidentally, with a slight modification (replace S^n, but T_n(S)
which consists of all sequences (0,...,0,u_i,...,u_n) for which 0 <
u_i < ... < u_n, i = 1,...,n+1, you get what really should be called
the Egyptian topology, but I am pretty well convinced it is the same.
If not, the same question for it.)
The topology can be described as follows. Let the length of a
rational r be the number of terms in the shortest possible
decomposition of r as a sum of unit fractions. Then the topology is
that of convergence of all sequences of bounded length. In other words
a function f on Q is continuous iff whenever r_1,r_2,... --> r is a
sequence whose terms all have length =< l (for some l), then f(r_1),
f(r_2_,... --> f(r).
Date Subject Author
5/21/04 Michael Barr
5/21/04 David Eppstein
5/22/04 Michael Barr
5/22/04 William Elliot
5/23/04 Michael Barr
5/23/04 William Elliot
5/23/04 Robert Israel
5/24/04 Michael Barr
5/24/04 William Elliot
5/24/04 Michael Barr | 571 | 1,995 | {"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} | 3.234375 | 3 | CC-MAIN-2018-17 | longest | en | 0.900465 |
http://mathhelpforum.com/geometry/169459-how-calculate-height-between-mid-points-chord-arc.html | 1,568,557,852,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571506.61/warc/CC-MAIN-20190915134729-20190915160729-00343.warc.gz | 126,897,873 | 11,205 | # Thread: how to calculate height between mid points on a chord and arc
1. ## how to calculate height between mid points on a chord and arc
I am trying to build a shade house and need to get some steel bent. I need to provide the distance between the midpoint on the chord and the midpoint on the arc.
I have a chord length of 2.7mtrs and an arc length of 3.6mtrs. Goggling it gave me the formula c/s = sin(x)/x but I don't even know what that means. where c is the chord length and s is the arc length.
Sorry to sound dopey to all you maths gurus.
2. Originally Posted by dsell
I am trying to build a shade house and need to get some steel bent. I need to provide the distance between the midpoint on the chord and the midpoint on the arc.
I have a chord length of 2.7mtrs and an arc length of 3.6mtrs. Goggling it gave me the formula c/s = sin(x)/x but I don't even know what that means. where c is the chord length and s is the arc length.
Sorry to sound dopey to all you maths gurus.
arc length=$\displaystyle \theta*R$
chord length=$\displaystyle 2*R*\sin(\frac{\theta}{2})$
where
$\displaystyle \theta$ =angle at the centre measured in radians and R= radius of the circle
$\displaystyle \theta*R=3.6$
and
$\displaystyle 2*R*\sin(\frac{\theta}{2})=2.7$
If you divide the 2 equations above:
$\displaystyle \frac{\theta}{2*\sin(\frac{\theta}{2})}=\frac{3.6} {2.7}$
or, if we change the variable by:
$\displaystyle x=\frac{\theta}{2}$
then you get to:
$\displaystyle \frac{x}{sin x}=\frac{3.6}{2.7}$
or:
$\displaystyle \frac{sin x}{x}=\frac{2.7}{3.6}$
which is exactly your formula: sin(x)/x=c/s where c is length of chord (2.7)
and s is length of arc (3.6)
The equation can be re-written as:
$\displaystyle sin x=x*\frac{2.7}{3.6}$
So it all boils down to finding x = solving this equation which can be done by using numerical methods or a graphical method.
Knowing x will allow you find the angle at the centre $\displaystyle \theta$=2x
and then the radius of the circle R from:
$\displaystyle \theta*R=3.6$
$\displaystyle R=\frac{3.6}{\theta}$
Now, the distance you have been looking for is
$\displaystyle R-R*cos(\frac{\theta}{2})$
3. Not formula, but calculation for the distance between the midpoint on the chord and the midpoint on the arc in your situation.
If a chord length of 2.7mtrs and an arc length of 3.6mtrs, this distance is 1.00 mtrs
4. Originally Posted by stebko
Not formula, but calculation for the distance between the midpoint on the chord and the midpoint on the arc in your situation.
If a chord length of 2.7mtrs and an arc length of 3.6mtrs, this distance is 1.00 mtrs
Ok: the first part was just a (long ) explanation why the formula works.
The calculation: by graphical method (see graph attached) the equation
sin x = x* 2.7/3.6 has solution x=1.28, $\displaystyle \theta=2.56$
so R=3.6/2.56=1.40625
therefore the distance = 1.40625-1.40625*cos 1.28 = 1.00 m, same answer as your (more practical) approach
The idea is that, once you are given the value of c (2.7) and (3.6 in this case), you can:
1. find x by solving (graphically) the equation sin x =x*c/s
2. find $\displaystyle \theta$ = 2*x
3. find $\displaystyle R=\frac{s}{\theta}$
4. find the distance you have been looking for = $\displaystyle R-R*cos{\frac{\theta}{2})$ | 970 | 3,293 | {"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} | 4.3125 | 4 | CC-MAIN-2019-39 | latest | en | 0.882758 |
https://mathmistakes.wordpress.com/2011/01/10/negative-exponents/ | 1,618,607,417,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038089289.45/warc/CC-MAIN-20210416191341-20210416221341-00020.warc.gz | 490,045,882 | 20,461 | # Negative Exponents
Negative exponents can be another source of confusion. No matter where you find a negative exponent, you can turn it into a positive exponent by taking the reciprocal of the expression it applies to.
The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead?
$\dfrac{c^{-2}d^3}{c^3d^{-4}}$
$=\dfrac{d^3d^4}{c^{-2}c^3}$
$=\dfrac{d^7}{c^{-6}}$
$=d$
.
..
Hint 1: The correct answer to the original problem is: $\dfrac{d^7}{c^5}$
Hint 2: Several mistakes have been made in the work above – don’t just seek the correct answer – look for the mistakes between every pair of lines. | 215 | 790 | {"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": 5, "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} | 3.578125 | 4 | CC-MAIN-2021-17 | latest | en | 0.891353 |
https://stackoverflow.com/questions/56554143/how-to-sort-unique-np-array-elements-by-occurence | 1,580,184,741,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00076.warc.gz | 669,996,297 | 30,250 | How to sort unique np array elements by occurence?
I would like to implement the following code:
``````a = [1, 1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5]
sorted(a,key=a.count,reverse=True)
>>> [5, 5, 5, 5, 3, 3, 3, 4, 4, 4, 1, 1, 2]
``````
For the case when `a` is a `np.array`
``````a = np.array([1, 1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5])
``````
How to do it? np.array has `np.unique()` function that calculate occurence of each element, but I don't see how can I utilize it here.
• In your example, both 3 and 4 occur three times. Is it important that the 3s occurs before the 4s in the output? I.e. are there special constraints that must be met in the case of ties? – Warren Weckesser Jun 12 '19 at 4:32
• Does your actual data contain just (relatively) small integers as in your example? – Warren Weckesser Jun 12 '19 at 4:33
• Did either of the posted solutions work for you? – Divakar Jun 14 '19 at 18:14
To exactly mimic `sorted`/`list` behavior @Divakar's soln can be used with a small modification:
``````al = [1,2,3,2,1,3,2]
aa = np.array(al)
sorted(al, key=al.count, reverse=True)
# [2, 2, 2, 1, 3, 1, 3]
u, ids, c = np.unique(aa, return_counts=True, return_inverse=True)
aa[(-c[ids]).argsort(kind="stable")]
# array([2, 2, 2, 1, 3, 1, 3])
``````
If `aa` is large,
``````from scipy import sparse
sparse.csc_matrix((aa, (c.max()-c[ids]), np.arange(len(ids)+1))).tocsr().data
# array([2, 2, 2, 1, 3, 1, 3], dtype=int64)
``````
may be slightly faster. Not much, though, because in both cases we first call the expensive `unique`, unless data are none too large integers in which case faster alternatives (to which @WarrenWeckesser appears to allude in the comments) are available including the sparse matrix trick we just used; see for example Most efficient way to sort an array into bins specified by an index array?.
``````aaa = np.tile(aa,10000)
timeit(lambda:aaa[(-c[ids]).argsort(kind="stable")], number=10)
# 0.040545254945755005
timeit(lambda:sparse.csc_matrix((aaa, (c.max()-c[ids]), np.arange(len(ids)+1))).tocsr().data, number=10)
# 0.0118721229955554
``````
You can use `np.unique` with its optional arguments `return_counts` and `return_inverse` -
``````u, ids, c = np.unique(a, return_counts=True, return_inverse=True)
out = a[c[ids].argsort()[::-1]]
``````
Sample run -
``````In [90]: a = np.array([1, 1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5])
In [91]: u, ids, c = np.unique(a, return_counts=True, return_inverse=1)
In [92]: a[c[ids].argsort()[::-1]]
Out[92]: array([5, 5, 5, 5, 4, 4, 4, 3, 3, 3, 1, 1, 2])
``````
You're looking for `return_counts` which you can combine with `argsort` + `repeat`. This will not guarantee the ordering of elements that appear the same number of times (notice the `4` before the `3`, same count, but not "stable").
``````u, c = np.unique(a, return_counts=True)
i = np.argsort(c)[::-1]
np.repeat(u[i], c[i])
``````
``````array([5, 5, 5, 5, 4, 4, 4, 3, 3, 3, 1, 1, 2])
`````` | 1,074 | 2,938 | {"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} | 3.328125 | 3 | CC-MAIN-2020-05 | latest | en | 0.735794 |
https://www.numbersaplenty.com/259 | 1,708,466,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00226.warc.gz | 967,270,846 | 3,397 | Search a number
259 = 737
BaseRepresentation
bin100000011
3100121
410003
52014
61111
7520
oct403
9317
10259
11216
12197
1316c
14147
15124
hex103
• Probably 259 is the largest number that can be written in two ways as 2x + 3y. Here 259 = 28 + 31 = 24 + 35.
• 259 can be written using four 4's:
259 has 4 divisors (see below), whose sum is σ = 304. Its totient is φ = 216.
The previous prime is 257. The next prime is 263. The reversal of 259 is 952.
259 is nontrivially palindromic in base 6.
It is a semiprime because it is the product of two primes.
It is a 3-Lehmer number, since φ(259) divides (259-1)3.
It is a cyclic number.
It is not a de Polignac number, because 259 - 21 = 257 is a prime.
It is a deceptive number, since it divides R258.
It is a Duffinian number.
259 is a lucky number.
259 is a nontrivial repdigit in base 6.
It is a plaindrome in base 6, base 10, base 13, base 14 and base 15.
It is a nialpdrome in base 6 and base 7.
It is a zygodrome in base 6.
It is not an unprimeable number, because it can be changed into a prime (251) by changing a digit.
It is a nontrivial repunit in base 6.
It is a pernicious number, because its binary representation contains a prime number (3) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 12 + ... + 25.
It is an arithmetic number, because the mean of its divisors is an integer number (76).
259 is a deficient number, since it is larger than the sum of its proper divisors (45).
259 is an equidigital number, since it uses as much as digits as its factorization.
259 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 44.
The product of its digits is 90, while the sum is 16.
The square root of 259 is about 16.0934769394. The cubic root of 259 is about 6.3743110879.
Subtracting from 259 its sum of digits (16), we obtain a 5-th power (243 = 35).
Subtracting from 259 its product of digits (90), we obtain a square (169 = 132).
It can be divided in two parts, 25 and 9, that multiplied together give a square (225 = 152).
The spelling of 259 in words is "two hundred fifty-nine", and thus it is an aban number.
Divisors: 1 7 37 259 | 681 | 2,236 | {"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} | 3.796875 | 4 | CC-MAIN-2024-10 | latest | en | 0.911821 |
http://msdn.microsoft.com/en-us/library/system.math.sqrt(v=vs.110).aspx | 1,419,713,744,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447552819.108/warc/CC-MAIN-20141224185912-00098-ip-10-231-17-201.ec2.internal.warc.gz | 64,250,897 | 11,979 | # Math.Sqrt Method
.NET Framework 4.5
Returns the square root of a specified number.
Namespace: System
Assembly: mscorlib (in mscorlib.dll)
## Syntax
```public static double Sqrt(
double d
)
```
#### Parameters
d
Type: System.Double
The number whose square root is to be found.
#### Return Value
Type: System.Double
One of the values in the following table.
d parameter
Return value
Zero or positive
The positive square root of d.
Negative
Equals NaN
NaN
Equals PositiveInfinity
PositiveInfinity
## Examples
The square root of the area of a square represents the length of any side of the square. The following example displays the area of some cities in the United States and gives an impression of each city's size if it were represented by a square.
```using System;
public class Example
{
public static void Main()
{
// Create an array containing the area of some squares.
Tuple<string, double>[] areas =
Tuple.Create("New York City", 302.6),
Tuple.Create("Los Angeles", 468.7),
Tuple.Create("Detroit", 138.8),
Tuple.Create("Chicago", 227.1),
Tuple.Create("San Diego", 325.2) };
Console.WriteLine("{0,-18} {1,14:N1} {2,30}\n", "City", "Area (mi.)",
"Equivalent to a square with:");
foreach (var area in areas)
Console.WriteLine("{0,-18} {1,14:N1} {2,14:N2} miles per side",
area.Item1, area.Item2, Math.Round(Math.Sqrt(area.Item2), 2));
}
}
// The example displays the following output:
// City Area (mi.) Equivalent to a square with:
//
// Sitka, Alaska 2,870.3 53.58 miles per side
// New York City 302.6 17.40 miles per side
// Los Angeles 468.7 21.65 miles per side
// Detroit 138.8 11.78 miles per side
// Chicago 227.1 15.07 miles per side
// San Diego 325.2 18.03 miles per side
```
## Version Information
#### .NET Framework
Supported in: 4.6, 4.5, 4, 3.5, 3.0, 2.0, 1.1, 1.0
#### .NET Framework Client Profile
Supported in: 4, 3.5 SP1
#### Portable Class Library
Supported in: Portable Class Library
#### .NET for Windows Store apps
Supported in: Windows 8
Supported in: Windows Phone 8.1
Supported in: Windows Phone Silverlight 8.1
Supported in: Windows Phone Silverlight 8
## Platforms
Windows Phone 8.1, Windows Phone 8, Windows 8.1, Windows Server 2012 R2, Windows 8, Windows Server 2012, Windows 7, Windows Vista SP2, Windows Server 2008 (Server Core Role not supported), Windows Server 2008 R2 (Server Core Role supported with SP1 or later; Itanium not supported)
The .NET Framework does not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements. | 738 | 2,762 | {"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} | 2.890625 | 3 | CC-MAIN-2014-52 | latest | en | 0.451712 |
http://www.cfd-online.com/Forums/fluent/48648-averaging-les-print.html | 1,477,438,093,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720468.71/warc/CC-MAIN-20161020183840-00214-ip-10-171-6-4.ec2.internal.warc.gz | 383,631,546 | 2,321 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- FLUENT (http://www.cfd-online.com/Forums/fluent/)
- - Averaging LES (http://www.cfd-online.com/Forums/fluent/48648-averaging-les.html)
iko July 1, 2008 14:55
Averaging LES
I'm runnning and LES simulation, and I want to do an averaging of the mean velocity and rms. And as you know for an LES solution, time-averaged data can only be derived by data-sampling and processing. For my first simulation I stored multiple data sets from LES solutions and then compute averaging of several samples (for mean and rms velocity) with matlab and tecplot. As ou imagine it was a tedious task. So I'm thinking now that instead of storing multiple data sets, it is most efficient to define the data-processing up front and execute it on-the-fly.
I'd like to know how I can set the Sampling Interval such that Data Sampling for Time Statistics can be performed at the specified frequency? My timestep is 0.2 ms, and my integral time scale is about 20ms. I need a sampling interval of 20 ms, which mean every 100 time step ( 100*0.2=20ms) How can I set the sampling Interval? And is it possible to compute the averaging with fluent (averaging with at least 35 samples) ?
Sincerely
arkur July 1, 2008 18:23
Re: Averaging LES
Fluent should do it for you. You need to do any averaging at all.
iko July 1, 2008 18:35
Re: Averaging LES
I know fluent is able to do it,
but the question is : how can I set the Sampling Interval?
Paolo Lampitella July 3, 2008 05:55
Re: Averaging LES
In the iterate panel, under options, enable "Data Sampling for Time Statistics" and choose your sampling interval, that is "How many time steps do you want it is going to be?"
It's just there; after that the unsteady statistics will be automatically available under "display-contours"
All times are GMT -4. The time now is 19:28. | 493 | 1,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2016-44 | latest | en | 0.914646 |
http://www.bio.net/bionet/mm/methods/1997-May/057263.html | 1,501,118,444,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426693.21/warc/CC-MAIN-20170727002123-20170727022123-00125.warc.gz | 404,884,646 | 2,509 | # "high stringency" washes -NOT?
David L. Haviland, Ph.D. dhavilan at IMM2.IMM.UTH.TMC.EDU
Wed May 7 09:40:34 EST 1997
```At 12:09 5/4/97, Jim wrote:
>Apparently, "(mismatched hybrids) are more stable at high salt
concentration than low". This is contrary to the calculated effect of
increased salt concentration when using Bolton and McCarthy's:
>
>Tm = 81.5 - 16.6 (log[Na+]) + 0.41(G/C) -0.63(form%) -600/L
>
>determination of melting temperature (which I described above).
Jim (Dr. Graham):
Me thinks I have an answer. Where did you get the formula? Knowing you
and desire to be precise, I don't think you typed it wrong. However, it
may have been printed wrong...
A Log10 of a number less than 1, produces a negative number. So from some
rusty rules of algebra, if 16.6 times a negative number is negative, its
subtraction would be positive and would add the number to the 81.5. With
the formula above, I would agree with your conclusion that the more salt
drops the Tm, so at a certain temp, more salt would mean less hybridization
and cross species hybridizations would be hard to do. This isn't how it
works in practice.
I think where you got the formula is wrong. I know of this one from Davis
et al. The little black spiral bound mehods book that has a few recipes
that are outright wrong (PBS & TBE). However, this formula appears correct
as I've seen it elsewhere and I've used it religously for RNAse protection
assays. I believe it is around pp 75-76.
I have it as:
Tm = 16.6Log[M] + 0.41(%G/C) + 81.5 - Pm - B/L - 0.65(%F)
Where:
M is molar concentration of Na+ (1XSSC is .165M Na+). %G/C is the percent
G and/or C bases in the probe. Pm is the % mismatched bases. %F is the
percent formamide. B is 675 for probes up to 100 bases and L is the length
of the probe in bases. Notice the 16.6Log will end up being a negative
number by itself for Na+ concentrations less than 1M. For long probes,
over 700 bases, in practice the formula reduces to:
Tm = 16.6Log[M] + 81.5
For large probes the %G/C becomes a ever smaller factor. For Southerns (at
least for me) formamide isn't used. The % mismatch isn't usually known in
cross species (or homologue) hybridizations. Finally the B/L is moot with
large probes over 675 as it approaches 1 as L increases.
Now, 1XSSC is (.165M): Tm = 16.6Log[.165] + 81.5
= 68.5 'C
With 5XSSC (.825M): Tm = 16.6Log[.825] + 81.5
= 80.1 'C
The way this one is written, the rule (and practice) hold.
Hope this helps,
David
``` | 777 | 2,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-30 | longest | en | 0.953253 |
https://physics.stackexchange.com/questions/178726/moment-of-inertia-calculation | 1,563,339,630,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525046.5/warc/CC-MAIN-20190717041500-20190717063500-00273.warc.gz | 521,844,012 | 32,025 | Moment of Inertia Calculation [closed]
A uniform disc has centre O, radius a and mass 2m. It is free to rotate in a vertical plane about a horizontal axis through O. A particle P of mass m is placed on the highest point of the rough edge of the disc and the system is slightly disturbed so that OP begins to rotate with the particle in contact with the edge. In the subsequent motion OP makes an angle θ with the upward vertical.
What will be the moment of inertia of the system about the centre of the disc. And how can we find the equation of rotational motion for the system.
closed as off-topic by Kyle Kanos, Danu, Qmechanic♦Apr 27 '15 at 19:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, Danu, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.
• Welcome to Physics.SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – 299792458 Apr 27 '15 at 18:29
• Please look up "mass moment of inertia" for basic shapes and add them up using the "parallel axis theorem". – ja72 Apr 27 '15 at 23:27
For disc $I=(2m*r*r)/2$
for small body $I=m*r*r$
• Disc MMOI is $I=\frac{m}{2} r^2$ – ja72 Apr 27 '15 at 23:29
• No, you have $I=(2m r^2)/2 = m r^2$. – ja72 Apr 28 '15 at 18:02 | 448 | 1,690 | {"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} | 3.640625 | 4 | CC-MAIN-2019-30 | latest | en | 0.905656 |
http://mathleticism.net/?m=201409 | 1,628,036,788,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154486.47/warc/CC-MAIN-20210803222541-20210804012541-00064.warc.gz | 29,286,174 | 9,633 | # Awesome Bowl Question: Segments in a Cube
We held our first bowl team session in MTPS today: an friendly, informal competition dividing the class into two groups of roughly equal 'ability.' The bowl competition we attend at the University of Nebraska-Lincoln in November consists of games of 15 questions with 30 seconds to answer each question.
Here is a question we spent some time on in class today.
Here is a screenshot from my Promethean board of the work we did as a group with students:
And here is a screenshot from the Geogebra sheet students constructed in front of the class at my computer terminal to demonstrate what is going on in this problem. If you would prefer to download the Geogebra sheet for your own tinkering, you can find it here.
No one got this question right in the allotted 30 seconds, so we spent some time developing the various parts of the expression. We color coded the edges on the cube (segments AB, AD, and AG) orange. We color coded the diagonals on faces of the cube (segments AC, AH, and AJ) green. We color coded the segment inside the cube (segment AI) black.
The kids had a lot of fun with building the Geogebra sheet and then trying to reconcile between sqrt(2) and the crudely rounded segment length 1.41. Or the sqrt(3) and 1.732.
The pink stuff at the top of Promethean board screen shot is our efforts to generalize this question. I tell the kids we are trying to "hack all possible problems." We use the term invariant and that we are looking to write a question that covers all possible question types. For example, what would happen to the question if s = 2? Or if s=3? Or if s=n? Is it possible to answer the question before the moderator finishes reading it? If so, at what point can we be confident we can buzz in and answer correctly?
# Differentiation in Math Class: Make Students Write
My elective math class has all grade levels represented within it. While not ideal, this is a feature of our scheduling system. So I have freshmen in Algebra I all the way up to seniors that have already completed Calculus AB. This poses a huge classroom differentiation challenge each Monday, Wednesday and Friday we hold class. Here is an instructional strategy I use that gets students writing about the mathematics they do in class.
Students worked on one of four things in the computer lab last Friday.
1. Construct 2013 Probe I Problem 7 diagram (2D)
2. Construct 2013 Probe I Problem 23 diagram (2D)
3. Construct 2013 Probe I Problem 11 diagram (3D)
4. Work on Alcumus problems independently
After we spent approximately 55 minutes in the lab, we returned to my classroom for a writing activity.
Here is the writing prompt I put on the board:
(Writing exercise on a separate sheet of paper to turn in to me)
Think about what you learned about the diagram or diagram(s) you built in Geogebra. Write a letter to the you of October 12. How did building diagrams in Geogebra help you understand the problem better?
I put eight minutes on the clock and informed students they would need to continuously write for the eight minute timeframe. Here are some samples of student writing from the activity. For convenience, I have inserted another copy of the problem. Immediately below each problem appears student writing associated with the problem.
The Students revealed their thinking about these math problems throughout their writing. While some students chose to concentrate on the construction process in Geogebra, others also revealed some of the mathematical structure they encountered while making the diagrams.
Writing samples from students that worked on Alcumus:
These writing samples revealed to me the depth of student thinking going on in the classroom. If I could have a superpower, I would be a mind reader. Then I wouldn't have to guess at what my students are thinking. Having the students write for an extended period of time gives me insight into how they are seeing the mathematics and gives me ideas on how I can help further their understanding and guide them as they struggle.
I collected these writings immediately after students completed them. I ran the pages through my ScanSnap scanner and converted them to a PDF for me to review later. I told students we would get these writings back out in a month's time, emphasizing the need for specificity on what they were working on and what they learned that day.
Going forward, these writings help me be more efficient with respect to differentiating classroom instruction. We don't need to be working on the exact same thing at the exact same time at the exact same pace for the students to engage in meaningful problem solving. | 976 | 4,670 | {"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-2021-31 | latest | en | 0.939484 |
https://brilliant.org/problems/product-of-binomial-coefficients/ | 1,490,647,597,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189525.23/warc/CC-MAIN-20170322212949-00208-ip-10-233-31-227.ec2.internal.warc.gz | 802,532,471 | 12,606 | # Product of binomial coefficients
Algebra Level 4
If $$P_{n}$$ denotes the product of all the binomial coefficients in the expansion of $$(1+x)^n$$, then $$\frac{P_{n+1}}{P_{n}}$$ equals
Clarification: Binomial Coefficients are $${n \choose r}$$ OR $$^{n}C_{r}$$
× | 90 | 269 | {"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} | 2.65625 | 3 | CC-MAIN-2017-13 | longest | en | 0.543718 |
https://blogs.perl.org/users/laurent_r/2023/01/ | 1,701,936,233,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00697.warc.gz | 164,467,815 | 11,099 | # January 2023 Archives
## Perl Weekly Challenge 202: Consecutive Odds and Widest Valley
These are some answers to the Week 202 of the Perl Weekly Challenge organized by Mohammad S. Anwar.
Spoiler Alert: This weekly challenge deadline is due in a few days from now (on February 5, 2023 at 23:59). This blog post offers some solutions to this challenge. Please don’t read on if you intend to complete the challenge on your own.
You are given an array of integers.
Write a script to print 1 if there are THREE consecutive odds in the given array otherwise print 0.
Example 1
``````Input: @array = (1,5,3,6)
Output: 1
``````
Example 2
``````Input: @array = (2,6,3,5)
Output: 0
``````
Example 3
``````Input: @array = (1,2,3,4)
Output: 0
``````
Example 4
``````Input: @array = (2,3,5,7)
Output: 1
``````
### Consecutive Odds in Raku
In Raku, we can use the `%%` operator to find if an integer is evenly divisible by another integer. Used with 2, we find whether an integer is odd or even. The `three-odd` subroutine goes through the input array and increments a counter when an item is odd, and resets the counter to 0 when an item is even. It returns 1 if the counter reaches 3 at any point in the process, and return 0 if we reach the end of the input array.
``````sub three-odd (@in) {
my \$count = 0;
for @in -> \$n {
if \$n %% 2 { # Even
\$count = 0;
} else { # Odd
\$count++;
}
return 1 if \$count >= 3;
}
return 0;
}
for <1 5 3 6>, <2 6 3 5>, <1 2 3 4>, <2 3 5 7> -> @test {
say "@test[] => ", three-odd @test;
}
``````
This program displays the following output:
``````\$ raku ./three-odds.raku
1 5 3 6 => 1
2 6 3 5 => 0
1 2 3 4 => 0
2 3 5 7 => 1
``````
### Consecutive Odds in Perl
This is a port to Perl of the Raku program above. Perl doesn’t have the `%%` divisibility operator, but we can use the `%` modulo operator instead.
``````use strict;
use warnings;
use feature "say";
sub three_odd {
my \$count = 0;
for my \$n (@_) {
if (\$n % 2) { # Odd
\$count++;
} else { # Even
\$count = 0;
}
return 1 if \$count >= 3;
}
return 0;
}
for my \$test ([<1 5 3 6>], [<2 6 3 5>],
[<1 2 3 4>], [<2 3 5 7>]) {
say "@\$test => ", three_odd @\$test;
}
``````
This program displays the following output:
``````\$ perl ./three-odds.pl
1 5 3 6 => 1
2 6 3 5 => 0
1 2 3 4 => 0
2 3 5 7 => 1
``````
Given a profile as a list of altitudes, return the leftmost widest valley. A valley is defined as a subarray of the profile consisting of two parts: the first part is non-increasing and the second part is non-decreasing. Either part can be empty.
Example 1
``````Input: 1, 5, 5, 2, 8
Output: 5, 5, 2, 8
``````
Example 2
``````Input: 2, 6, 8, 5
Output: 2, 6, 8
``````
Example 3
``````Input: 9, 8, 13, 13, 2, 2, 15, 17
Output: 13, 13, 2, 2, 15, 17
``````
Example 4
``````Input: 2, 1, 2, 1, 3
Output: 2, 1, 2
``````
Example 5
``````Input: 1, 3, 3, 2, 1, 2, 3, 3, 2
Output: 3, 3, 2, 1, 2, 3, 3
``````
### Widest Valley in Raku
Since either part of a valley may be missing, the input array may start with a list of ascending integers, i.e. have no left part. The first loop in the program below is designed to handle this specific case. For the other more regular cases, we look for a series of descending integers followed by a series of ascending integers. Once we’ve found a match, we store it in `@temp` and we replace `@valley` with the content of `@temp` if the match is wider than the precious content of `@valley`.
``````sub widest-valley (@in) {
my (@valley, @temp);
for 1..@in.end -> \$i { # valley with no left part
push @valley, @in[\$i-1];
last if @in[\$i] < @in[\$i-1];
}
for 1..@in.end -> \$i {
my \$left = True;
for \$i..@in.end -> \$j {
if \$left {
push @temp, @in[\$j - 1];
push @temp, @in[\$j] and \$left = False
if @in[\$j] > @in[\$j - 1];
} else {
last if @in[\$j] < @in[\$j-1];
push @temp, @in[\$j];
}
}
@valley = @temp if @temp.elems > @valley.elems;
@temp = ();
}
return @valley;
}
for <1 5 5 2 8>, <1 5 5 2>, <2 6 8 5>,
<9 8 13 13 2 2 15 17>, <2 1 2 1 3>,
<1 3 3 2 1 2 3 3 2> -> @test {
say "@test[]".fmt("%-20s => "),
widest-valley @test;
}
``````
This program displays the following output:
``````\$ raku ./widest-valley.raku
1 5 5 2 8 => [5 5 2 8]
1 5 5 2 => [1 5 5]
2 6 8 5 => [2 6 8]
9 8 13 13 2 2 15 17 => [13 13 2 2 15 17]
2 1 2 1 3 => [2 1 2]
1 3 3 2 1 2 3 3 2 => [3 3 2 1 2 3 3]
``````
### Widest Valley in Perl
This is a port to Perl of the above Raku program. Please refer to that section for explanations.
``````use strict;
use warnings;
use feature "say";
sub widest_valley {
my (@valley, @temp);
for my \$i (1..\$#_) { # valley with no left part
push @valley, \$_[\$i-1];
last if \$_[\$i] < \$_[\$i-1];
}
for my \$i (1..\$#_) {
my \$left = 1;
for my \$j (\$i..\$#_) {
if (\$left) {
push @temp, \$_[\$j - 1];
push @temp, \$_[\$j] and \$left = 0
if \$_[\$j] > \$_[\$j - 1];
} else {
last if \$_[\$j] < \$_[\$j-1];
push @temp, \$_[\$j];
}
}
@valley = @temp if scalar @temp > scalar @valley;
@temp = ();
}
return @valley;
}
for my \$test ([<1 5 5 2 8>], [<1 5 5 2>], [<2 6 8 5>],
[<9 8 13 13 2 2 15 17>], [<2 1 2 1 3>],
[<1 3 3 2 1 2 3 3 2>]) {
printf "%-20s => ", join " ", @\$test;
say join " ", widest_valley @\$test;
}
``````
This program displays the following output:
``````\$ perl ./widest-valley.pl
1 5 5 2 8 => 5 5 2 8
1 5 5 2 => 1 5 5
2 6 8 5 => 2 6 8
9 8 13 13 2 2 15 17 => 13 13 2 2 15 17
2 1 2 1 3 => 2 1 2
1 3 3 2 1 2 3 3 2 => 3 3 2 1 2 3 3
``````
## Wrapping up
The next week Perl Weekly Challenge will start soon. If you want to participate in this challenge, please check https://perlweeklychallenge.org/ and make sure you answer the challenge before 23:59 BST (British summer time) on February 12, 2023. And, please, also spread the word about the Perl Weekly Challenge if you can.
## Perl Weekly Challenge 201: Missing Numbers
These are some answers to the Week 201 of the Perl Weekly Challenge organized by Mohammad S. Anwar.
You are given an array of unique numbers.
Write a script to find out all missing numbers in the range `0..\$n` where `\$n` is the array size.
Example 1
``````Input: @array = (0,1,3)
Output: 2
The array size i.e. total element count is 3, so the range is 0..3.
The missing number is 2 in the given array.
``````
Example 2
``````Input: @array = (0,1)
Output: 2
The array size is 2, therefore the range is 0..2.
The missing number is 2.
``````
### Missing Numbers in Raku
The `find-missing` subroutine uses the (-),%20infix%20%E2%88%96) set difference operator to find all the elements in the prescribed range that do not belong to the input array. The `(-)` set-difference operator implicitly coerces its operands (arrays) into Sets, so that we don’t need to explicitly perform the conversion and end up with essentially a one-liner.
``````sub find-missing (@in) {
return ~(|(0..@in.elems) (-) @in).keys.sort;
}
for (0, 1, 3), (0, 1), (0, 1, 3, 5, 7, 2) -> @test {
say (~@test).fmt("%-15s => "), find-missing @test;
}
``````
This program displays the following output:
``````\$ raku ./missing-numbers.raku
0 1 3 => (2)
0 1 => (2)
0 1 3 5 7 2 => (4 6)
``````
### Missing Numbers in Perl
Although Perl doesn’t have Raku sets and set operators, it is only mildly more complex to port the idea to Perl: we can store the input array into a hash and then use a `grep` to find the items of the prescribed range that do not belong to the input array.
``````use strict;
use warnings;
use feature "say";
sub find_missing {
my %in = map {\$_ => 1} @_;
return grep { not exists \$in{\$_} } 0..scalar @_;
}
for my \$test ([0, 1, 3], [0, 1], [0, 1, 3, 5, 7, 2]) {
printf "%-15s => ", "@\$test";
say map "\$_ ", find_missing @\$test;
}
``````
This program displays the following output:
``````\$ perl ./missing-numbers.pl
0 1 3 => 2
0 1 => 2
0 1 3 5 7 2 => 4 6
``````
## Wrapping up
The next week Perl Weekly Challenge will start soon. If you want to participate in this challenge, please check https://perlweeklychallenge.org/ and make sure you answer the challenge before 23:59 BST (British summer time) on February 5, 2023. And, please, also spread the word about the Perl Weekly Challenge if you can.
## Perl Weekly Challenge 200: Arithmetic Slices and Seven Segment Display
These are some answers to the Week 200 of the Perl Weekly Challenge organized by Mohammad S. Anwar.
Spoiler Alert: This weekly challenge deadline is due in a couple of days from now (on January 22, 2023 at 23:59). This blog post offers some solutions to this challenge. Please don’t read on if you intend to complete the challenge on your own.
You are given an array of integers.
Write a script to find out all Arithmetic Slices for the given array of integers.
An integer array is called arithmetic if it has at least 3 elements and the differences between any three consecutive elements are the same.
Example 1
``````Input: @array = (1,2,3,4)
Output: (1,2,3), (2,3,4), (1,2,3,4)
``````
Example 2
``````Input: @array = (2)
Output: () as no slice found.
``````
### Arithmetic Slices in Raku
The `find_slices` subroutine loops over the input array, computes the difference (`\$gap`) between any two consecutive integers and checks whether the same difference can be found between the next integers.
``````sub find_slices (@in) {
my @out;
return @out if @in.elems < 3;
for 0..@in.end - 2 -> \$i {
my \$gap = @in[\$i+1] - @in[\$i];
for \$i+2..@in.end -> \$j {
last if @in[\$j] - @in[\$j-1] != \$gap;
push @out, @in[\$i..\$j];
}
}
return @out;
}
for <1 2 3 4>, <2 5>, <3 5 7 9>, <2 5 9> -> @test {
say (~@test).fmt("%-10s => "), find_slices @test;
}
``````
This script displays the following output:
``````\$ raku ./arithmetic-slices.raku
1 2 3 4 => [(1 2 3) (1 2 3 4) (2 3 4)]
2 5 => []
3 5 7 9 => [(3 5 7) (3 5 7 9) (5 7 9)]
2 5 9 => []
``````
### Arithmetic Slices in Perl
This a port to Perl of the Raku program above:
``````use strict;
use warnings;
use feature "say";
sub find_slices {
my @in = @_;
my @out;
# return [] if @in < 3;
for my \$i (0..\$#in - 2) {
my \$gap = \$in[\$i+1] - \$in[\$i];
for my \$j (\$i+2..\$#in) {
last if \$in[\$j] - \$in[\$j-1] != \$gap;
push @out, [@in[\$i..\$j]];
}
}
return @out ? @out : [];
}
for my \$test ([<1 2 3 4>], [<2 5>], [<3 4 5 6 8>],
[<3 5 7 9>], [<2 5 9>]) {
printf "%-10s => ", "@\$test";
say map "(@\$_) ", find_slices @\$test;
}
``````
This script displays the following output:
``````\$ perl ./arithmetic-slices.pl
1 2 3 4 => (1 2 3) (1 2 3 4) (2 3 4)
2 5 => ()
3 4 5 6 8 => (3 4 5) (3 4 5 6) (4 5 6)
3 5 7 9 => (3 5 7) (3 5 7 9) (5 7 9)
2 5 9 => ()
``````
## Task 2: Seven Segment 200
A seven segment display is an electronic component, usually used to display digits. The segments are labeled ‘a’ through ‘g’ as shown:
The encoding of each digit can thus be represented compactly as a truth table:
``````my @truth = qw<abcdef bc abdeg abcdg bcfg acdfg a cdefg abc abcdefg abcfg>;
``````
For example, `\$truth[1] = ‘bc’`. The digit 1 would have segments ‘b’ and ‘c’ enabled.
Write a program that accepts any decimal number and draws that number as a horizontal sequence of ASCII seven segment displays, similar to the following:
``````------- ------- -------
| | | | |
| | | | |
-------
| | | | |
| | | | |
------- ------- -------
``````
To qualify as a seven segment display, each segment must be drawn (or not drawn) according to your @truth table.
The number “200” was of course chosen to celebrate our 200th week!
For the 200th week of the Perl Weekly Challenge, Ryan J Thompson, the author of this task, has decided to bring us decades backward, back in the 1970s when they started to be widely used (remember lieutenant Theo Kojak’s LED watch?). We’re actually driven back more than decades, since the first seven-segment display devices date as far back as 1903 (but obviously did not use LEDs).
The difficulty in this task is that, for ASCII art display, we need to slice the digits into horizontal lines. I’ve decided not to use the suggested truth table, as this is quite unpractical. Instead, I’m using a table (`@nums`) dividing each digit into seven horizontal lines
### Seven Segment Display in Raku
``````my %c; # ascii coding of digit's slices
%c<h> = "-" x 7; # Horizontal line
%c<l> = "| "; # Vertical bar, left
%c<r> = " |"; # Vertical bar, right
%c<2> = "| |"; # 2 vertical bars
%c<n> = " " x 7; # empty horizontal line
my @nums = # Digit horizontal slices
<h 2 2 n 2 2 h>, # 0
<n r r n r r n>, # 1
<h r r h l l h>, # 2
<h r r h r r h>, # 3
<n l l h 2 2 n>, # 4
<h l l h r r h>, # 5
<n l l h 2 2 h>, # 6
<h r r n r r n>, # 7
<h 2 2 h 2 2 h>, # 8
<h 2 2 h r r n>; # 9
sub display (\$num) {
my @digits = \$num.comb;
for 0..6 -> \$l { # Lines 0 to 6 iof the display
say join " ", map {%c{@nums[\$_][\$l]}}, @digits;
}
}
for <200 2023 01234 56789> -> \$test {
display \$test;
}
``````
This program displays the following output:
``````\$ raku ./seven_segments.raku
------- ------- -------
| | | | |
| | | | |
-------
| | | | |
| | | | |
------- ------- -------
------- ------- ------- -------
| | | | |
| | | | |
------- ------- -------
| | | | |
| | | | |
------- ------- ------- -------
------- ------- -------
| | | | | | |
| | | | | | |
------- ------- -------
| | | | | |
| | | | | |
------- ------- ------- -------
| | | | | | |
| | | | | | |
------- ------- ------- -------
| | | | | | |
| | | | | | |
------- ------- -------
``````
Note that digits 6 and 9 could have an additional horizontal bar and 7 an additional vertical bar:
``````@nums[6] = <h l l h 2 2 h>; # 6
@nums[7] = <h 2 2 n r r n>; # 7
@nums[9] = <h 2 2 h r r h>; # 9
``````
This allegedly provides a more legible display (for some eyes, at least). Thus, the test line with the last five digits would be displayed like so:
``````------- ------- ------- ------- -------
| | | | | | | |
| | | | | | | |
------- ------- ------- -------
| | | | | | |
| | | | | | |
------- ------- ------- -------
``````
## Seven Segment Display in Perl
This a port to Perl of the Raku program above:
``````use strict;
use warnings;
use feature "say";
my %c; # ascii coding of digit's slices
\$c{'h'} = "-" x 7; # Horizontal line
\$c{'l'} = "| "; # Vertical bar, left
\$c{'r'} = " |"; # Vertical bar, right
\$c{'2'} = "| |"; # 2 vertical bars
\$c{'n'} = " " x 7; # empty horizontal line
my @nums = ( # Digit hoirizontal slices
[<h 2 2 n 2 2 h>], # 0
[<n r r n r r n>], # 1
[<h r r h l l h>], # 2
[<h r r h r r h>], # 3
[<n 2 2 h r r n>], # 4
[<h l l h r r h>], # 5
[<n l l h 2 2 h>], # 6
[<h r r n r r n>], # 7
[<h 2 2 h 2 2 h>], # 8
[<h 2 2 h r r n>]); # 9
sub display{
my @digits = split //, shift;
for my \$l (0..6) {
say join " ", map {\$c{\$nums[\$_][\$l]}} @digits;
}
}
for my \$test (<200 2023 01234 56789>) {
display \$test;
}
``````
This program displays the following output:
``````\$ perl ./seven_segments.pl
------- ------- -------
| | | | |
| | | | |
-------
| | | | |
| | | | |
------- ------- -------
------- ------- ------- -------
| | | | |
| | | | |
------- ------- -------
| | | | |
| | | | |
------- ------- ------- -------
------- ------- -------
| | | | | | |
| | | | | | |
------- ------- -------
| | | | | |
| | | | | |
------- ------- -------
------- ------- ------- -------
| | | | | | |
| | | | | | |
------- ------- ------- -------
| | | | | | |
| | | | | | |
------- ------- -------
``````
We could use the changes that we did to the Raku program (definition of digits 6, 7, and 9) to presumably improve their legibility:
``````------- ------- ------- ------- -------
| | | | | | | |
| | | | | | | |
------- ------- ------- -------
| | | | | | |
| | | | | | |
------- ------- ------- -------
``````
## Wrapping up
The next week Perl Weekly Challenge will start soon. If you want to participate in this challenge, please check https://perlweeklychallenge.org/ and make sure you answer the challenge before 23:59 BST (British summer time) on January 29, 2023. And, please, also spread the word about the Perl Weekly Challenge if you can.
## Perl Weekly Challenge 199: Good Pairs and Good Triplets
These are some answers to the Week 199 of the Perl Weekly Challenge organized by Mohammad S. Anwar.
Spoiler Alert: This weekly challenge deadline is due in a few days from now (on January 15, 2023 at 23:59). This blog post offers some solutions to this challenge. Please don’t read on if you intend to complete the challenge on your own.
Contrary to usual, I started with the Perl implementations this week (not having Raku installed on the computer where I started to work on the challenge). So I’ll present the Perl implementations first.
You are given a list of integers, `@list`.
Write a script to find the total count of Good Pairs.
A pair `(i, j)` is called good if `list[i] == list[j]` and `i < j`.
Example 1
``````Input: @list = (1,2,3,1,1,3)
Output: 4
There are 4 good pairs found as below:
(0,3)
(0,4)
(3,4)
(2,5)
``````
Example 2
``````Input: @list = (1,2,3)
Output: 0
``````
Example 3 Input: @list = (1,1,1,1) Output: 6
``````Good pairs are below:
(0,1)
(0,2)
(0,3)
(1,2)
(1,3)
(2,3)
``````
### Good Pairs in Perl
The program uses two nested loops to get pairs `(\$i, \$j)` of indices in the proper range, in such a way that `i < j`. And it increments the `\$count` variable whenever this index pair leads to equal values.
``````use strict;
use warnings;
use feature "say";
sub count_good_pairs {
my @in = @_;
my \$count = 0;
for my \$i (0..\$#in-1) {
for my \$j (\$i+1..\$#in) {
\$count++ if \$in[\$i] == \$in[\$j];
}
}
return \$count;
}
for my \$test ( [1,2,3,1,1,3], [1,2,3], [1,1,1,1],
[1,2,3,1,2,3], [4,3,2,3,2,1] ) {
say sprintf "%-15s => %d", "@\$test", count_good_pairs @\$test;
}
``````
This program displays the following output:
``````\$ perl ./good_pairs.pl
1 2 3 1 1 3 => 4
1 2 3 => 0
1 1 1 1 => 6
1 2 3 1 2 3 => 3
4 3 2 3 2 1 => 2
``````
### Good Pairs in Raku
This is a port to Raku of the Perl program above:
``````sub count_good_pairs (@in) {
my \$cnt = 0;
for 0..^@in.end -> \$i {
\$cnt++ if @in[\$i] == @in[\$_] for \$i+1..@in.end;
}
return \$cnt;
}
for <1 2 3 1 1 3>, <1 2 3>, <1 1 1 1>,
<1 2 3 1 2 3>, <4 3 2 3 2 1> -> @test {
say (~@test).fmt("%-15s => "), count_good_pairs @test;
}
``````
This program displays the following output:
``````\$ raku ./good_pairs.raku
1 2 3 1 1 3 => 4
1 2 3 => 0
1 1 1 1 => 6
1 2 3 1 2 3 => 3
4 3 2 3 2 1 => 2
``````
You are given an array of integers, `@array` and three integers `\$x,\$y,\$z`.
Write a script to find out total Good Triplets in the given array.
A triplet array[i], array[j], array[k] is good if it satisfies the following conditions: a) 0 <= i < j < k <= n (size of given array) b) abs(array[i] - array[j]) <= x c) abs(array[j] - array[k]) <= y d) abs(array[i] - array[k]) <= z
Example 1
``````Input: @array = (3,0,1,1,9,7) and \$x = 7, \$y = 2, \$z = 3
Output: 4
Good Triplets are as below:
(3,0,1) where (i=0, j=1, k=2)
(3,0,1) where (i=0, j=1, k=3)
(3,1,1) where (i=0, j=2, k=3)
(0,1,1) where (i=1, j=2, k=3)
``````
Example 2
``````Input: @array = (1,1,2,2,3) and \$x = 0, \$y = 0, \$z = 1
Output: 0
``````
### Good Triplets in Perl
The program works similarly to the Goof Pairs program. It uses three nested loops to get triplets `(\$i, \$j, \$k)` of indices in the proper range, in such a way that `i < j < k`. And it increments the `\$count` variable whenever this index triplet satisfies the conditions for `\$x,\$y,\$z`.
``````sub count_good_triplets {
my @in = @{\$_[0]};
my (\$x, \$y, \$z) = @{\$_[1]};
my \$count = 0;
for my \$i (0..\$#in-2) {
for my \$j (\$i+1..\$#in-1) {
# short-cut the \$k loop if \$i \$j not good
next if abs(\$in[\$i] - \$in[\$j]) > \$x;
for my \$k (\$j+1..\$#in) {
\$count++ if abs(\$in[\$j] - \$in[\$k]) <= \$y
and abs(\$in[\$i] - \$in[\$k]) <= \$z;
}
}
}
return \$count;
}
for my \$test ( [ [3,0,1,1,9,7], [7,2,3] ],
[ [1,1,2,2,3], [0,0,1] ],
[ [1,1,2,2,3], [1,1,2] ],
) {
say sprintf "%-15s - xyz = %-10s => %d",
"@{@\$test[0]}", "@{@\$test[1]}",
count_good_triplets @\$test;
}
``````
This program displays the following output:
``````\$ perl ./good_triplets.pl
3 0 1 1 9 7 - xyz = 7 2 3 => 4
1 1 2 2 3 - xyz = 0 0 1 => 0
1 1 2 2 3 - xyz = 1 1 2 => 9
``````
### Good Triplets in Raku
This is a port to Raku of the above Perl program. Note that the management of nested data structures is significantly simpler and easier in Raku than in Perl.
``````sub count_good_triplets (@in, @xyz) {
my \$count = 0;
my (\$x, \$y, \$z) = @xyz;
for 0..@in.end-2 -> \$i {
for \$i+1..^@in.end -> \$j {
next if abs(@in[\$i] - @in[\$j]) > \$x;
for \$j+1..@in.end -> \$k {
\$count++ if abs(@in[\$j] - @in[\$k]) <= \$y
&& abs(@in[\$i] - @in[\$k]) <= \$z;
}
}
}
return \$count;
}
for ( <3 0 1 1 9 7>, <7 2 3> ),
( <1 1 2 2 3>, <0 0 1> ),
( <1 1 2 2 3>, <1 1 2> ) -> @test {
say sprintf "%-15s - xyz = %-10s => %d",
"@test[0]", "@test[1]",
count_good_triplets @test[0], @test[1];
}
``````
This program displays the following output:
``````raku ./good_triplets.raku
3 0 1 1 9 7 - xyz = 7 2 3 => 4
1 1 2 2 3 - xyz = 0 0 1 => 0
1 1 2 2 3 - xyz = 1 1 2 => 9
``````
## Wrapping up
The next week Perl Weekly Challenge will start soon. If you want to participate in this challenge, please check https://perlweeklychallenge.org/ and make sure you answer the challenge before 23:59 BST (British summer time) on January 22, 2023. And, please, also spread the word about the Perl Weekly Challenge if you can.
## Perl Weekly Challenge 198: Max Gap and Prime Count
These are some answers to the Week 198 of the Perl Weekly Challenge organized by Mohammad S. Anwar.
Spoiler Alert: This weekly challenge deadline is due in a few of days from now (on January 8, 2023 at 23:59). This blog post offers some solutions to this challenge. Please don’t read on if you intend to complete the challenge on your own.
You are given a list of integers, `@list`.
Write a script to find the total pairs in the sorted list where 2 consecutive elements has the max gap. If the list contains less than 2 elements then return 0.
Example 1
``````Input: @list = (2,5,8,1)
Output: 2
Since the sorted list (1,2,5,8) has 2 such pairs (2,5) and (5,8)
``````
Example 2
``````Input: @list = (3)
Output: 0
``````
### Max Gap in Raku
The `max-gap` subroutine builds a hash (`%gaps`) mapping computed gaps to a list of the ranges leading to this gap and finally returns the size of the list of corresponding ranges.
``````sub max-gap (@in) {
return 0 if @in.elems < 2;
my @sorted = sort @in;
my %gaps;
for 1..@sorted.end -> \$i {
push %gaps, ( @sorted[\$i] - @sorted[\$i-1] => \$i );
}
my \$max-gap = %gaps.keys.max;
return %gaps{\$max-gap}.elems;
}
for <2 5 8 1>, <2 7>, (3,), <12 2 6 5 15 9> -> @test {
say (~@test).fmt("%-20s => "), max-gap @test;
}
``````
This program displays the following output:
``````\$ raku ./maximum-gap.raku
2 5 8 1 => 2
2 7 => 1
3 => 0
12 2 6 5 15 9 => 4
``````
### Max Gap in Perl
This is port to Perl of the above Raku program:
``````use strict;
use warnings;
use feature qw/say/;
sub max_gap {
return 0 if scalar @_ < 2;
my @sorted = sort { \$a <=> \$b } @_;
my %gaps;
for my \$i (1..\$#sorted) {
push @{\$gaps{\$sorted[\$i] - \$sorted[\$i-1]}}, \$i;
}
my \$max_gap = 0;
for my \$k (keys %gaps) {
\$max_gap = \$k if \$k > \$max_gap;
}
return scalar @{\$gaps{\$max_gap}};
}
for my \$test ([<2 5 8 1>], [<2 7>], [3,], [<12 2 6 5 15 9>]) {
printf "%-20s => %d\n", "@\$test", max_gap @\$test;
}
``````
This program displays the following output:
``````\$ perl ./maximum-gap.pl
2 5 8 1 => 2
2 7 => 1
3 => 0
12 2 6 5 15 9 => 4
``````
You are given an integer `\$n` > 0.
Write a script to print the count of primes less than `\$n`.
Example 1
``````Input: \$n = 10
Output: 4 as in there are 4 primes less than 10 are 2, 3, 5 ,7.
``````
Example 2
``````Input: \$n = 15
Output: 6
``````
Example 3
``````Input: \$n = 1
Output: 0
``````
Example 4
``````Input: \$n = 25
Output: 9
``````
With more information about the aim of the exercise, we may want to store in a cache the number of primes below a given integer, in order to avoid duplicate work. Here, with no information and given the small size of the input integers, it’s not worth the effort.
### Prime Count in Raku
Raku has a very fast built-in `is-prime` method. So we just `grep` prime numbers and count them. The `count-primes` subroutine is essentially a one-liner.
``````sub count-primes (Int \$n) {
return (grep ({.is-prime}), 1..\$n).elems;
}
for <10 15 1 25> -> \$i {
say "\$i \t => ", count-primes \$i;
}
``````
This program displays the following output:
``````\$ raku ./prime-count.raku
10 => 4
15 => 6
1 => 0
25 => 9
``````
### Prime Count in Perl
This Perl version is essentially the same as the Raku implementation above, except that we had to roll out our own `is_prime` subroutine. Since we are running this program with only a small set of small input integer, there is really no need to try to aggressively optimize `is_prime` for performance.
``````use strict;
use warnings;
use feature qw/say/;
sub is_prime {
my \$num = shift;
for my \$i (2 .. \$num ** .5) {
return 0 if \$num % \$i == 0;
}
return 1;
}
sub count_primes {
my \$n = shift;
return scalar grep is_prime(\$_), 2..\$n;
}
for my \$i (<10 15 1 25>) {
say "\$i \t => ", count_primes \$i;
}
``````
This program displays the following output:
``````\$ perl ./prime-count.pl
10 => 4
15 => 6
1 => 0
25 => 9
``````
## Wrapping up
The next week Perl Weekly Challenge will start soon. If you want to participate in this challenge, please check https://perlweeklychallenge.org/ and make sure you answer the challenge before 23:59 BST (British summer time) on January 15, 2023. And, please, also spread the word about the Perl Weekly Challenge if you can. | 9,601 | 27,626 | {"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-2023-50 | latest | en | 0.809418 |
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SparkPoints
# Day 41
### Tuesday, November 16, 2010
My calories eaten today, including a snack I haven't actually eaten yet, are 1268. My calories burned are 574.
I created a spreadsheet and am tracking my stats. I am trying to see if this really is a numbers game. I'm taking my BMR and tracking the calories I eat less than that, and each time I reach 3500 calories I should lose a pound. I did some research and using BMR calculators online my BMR should be about 1578 calories, but after tracking for some time and not losing I think it is closer to 1200 calories, so I am using 1200 as my BMR. I should reach 3500 calorie deficit in a couple of days and I'll see what the scale says.
I've been so very hungry today. My planned evening snack is graham crackers, a real treat, I just hope we have some in the cupboard! I have been trying to teach my fiance to write down on a list grocery items that we run out of or that he just wants in general, but he has a habit of eating all of something and not writing it down and so when the time comes that I want some it's gone. It happened this morning with the oatmeal, I wanted some for breakfast but we didn't have any. I had just been at the store and if I knew we were out I would have gladly bought some more. Now I'll just have to wait until next shopping trip. The joys of living together, lol! Still, I wouldn't trade it for all the oatmeal in the world! Okay, I'm going to check on the graham crackers now!
• BARBARASDIET
LOL! And aww, Mr.!
3834 days ago
• SEYSARAH
You have just isolated a huge problem here at the house..good grief! At this point I try to keep a larger supply of certain things in the pantry for just this reason.
Let me know how your calculations are going...I ran into the exact same problem and had to lower my cals by quite a bit...what SP said and the amount I could eat..way different no matter how much exercise I did...it's how I can to work on being a nutrient dense eater...kinda feels like the taller you are the more you can eat rofl..as for the hunger..I ran into it too..so had my own personal stash of protien/carbs to help with that..for me, I also had to tell the difference between true hunger and the munchies...very hard to do. Much good luck to you in finding the right balance..I cycled between 1000 and 1200 to lose consistently and this was with a daily vitamin...really hope your spreadsheet helps you!
3834 days ago
• JULIAOAK
hope you found some graham crackers in the cupboards!!
3834 days ago
• DESTINYE
LOL men! I know how it goes with the food and stuff! As far as I can tell my metabolism does not compute - I hope yours does! Will be interesting to see!
3834 days ago | 670 | 2,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-21 | latest | en | 0.97694 |
https://www.oreilly.com/library/view/learning-php-mysql/9780596803605/ch04s01.html | 1,575,945,411,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00189.warc.gz | 813,334,873 | 16,549 | ## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
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## Expressions
Let’s start with the most fundamental part of any programming language: expressions.
An expression is a combination of values, variables, operators, and functions that results in a value. It’s familiar to anyone who has taken elementary-school algebra:
`y = 3(abs(2x) + 4)`
which in PHP would be:
`\$y = 3 * (abs(2*\$x) + 4);`
The value returned (y or `\$y` in this case) can be a number, a string, or a Boolean value (named after George Boole, a nineteenth-century English mathematician and philosopher). By now, you should be familiar with the first two value types, but I’ll explain the third.
A basic Boolean value can be either `TRUE` or `FALSE`. For example, the expression “20 > 9” (20 is greater than 9) is `TRUE`, and the expression “5 == 6” (5 is equal to 6) is `FALSE`. (Boolean operations can be combined using operators such as `AND`, `OR`, and `XOR`, which are covered later in this chapter.)
Note that I am using uppercase letters for the names `TRUE` and `FALSE`. This is because they are predefined constants in PHP. You can also use the lowercase versions, if you prefer, as they are also predefined. In fact, the lowercase versions are more stable, because PHP does not allow you to redefine them; the uppercase ones may be redefined—something you should bear in mind if you import third-party code.
Example 4-1 shows some simple expressions: the two I just mentioned, plus a couple more. For each line, it prints out a letter between `a` and `d`, followed by a colon and the result of the expressions (the `<br />` tag is there to create a line break and thus separate the output into four lines in HTML).
Example 4-1. Four simple Boolean expressions
```<?php
echo "a: [" . (20 > 9) . "]<br />";
echo "b: [" . (5 == 6) . "]<br />";
echo "c: [" . (1 == 0) . "]<br />";
echo "d: [" . (1 == 1) . "]<br />";
?>```
The output from this code is as follows:
```a: [1]
b: []
c: []
d: [1]```
Notice that both expressions `a:` and `d:` evaluate to `TRUE`, which has a value of 1. But `b:` and `c:`, which evaluate to `FALSE`, do not show any value, because in PHP the constant `FALSE` is defined as `NULL`, or nothing. To verify this for yourself, you could enter the code in Example 4-2.
Example 4-2. Outputting the values of TRUE and FALSE
```<?php // test2.php
echo "a: [" . TRUE . "]<br />";
echo "b: [" . FALSE . "]<br />";
?>```
which outputs the following:
```a: [1]
b: []```
By the way, in some languages `FALSE` may be defined as 0 or even −1, so it’s worth checking on its definition in each language.
### Literals and Variables
The simplest form of an expression is a literal, which simply means something that evaluates to itself, such as the number 73 or the string “Hello”. An expression could also simply be a variable, which evaluates to the value that has been assigned to it. They are both types of expressions, because they return a value.
Example 4-3 shows five different literals, all of which return values, albeit of different types.
Example 4-3. Five types of literals
```<?php
\$myname = "Brian";
\$myage = 37;
echo "a: " . 73 . "<br />"; // Numeric literal
echo "b: " . "Hello" . "<br />"; // String literal
echo "c: " . FALSE . "<br />"; // Constant literal
echo "d: " . \$myname . "<br />"; // Variable string literal
echo "e: " . \$myage . "<br />"; // Variable numeric literal
?>```
And, as you’d expect, you see a return value from all of these with the exception of `c:`, which evaluates to `FALSE`, returning nothing in the following output:
```a: 73
b: Hello
c:
d: Brian
e: 37```
In conjunction with operators, it’s possible to create more complex expressions that evaluate to useful results.
When you combine assignment or control-flow constructs with expressions, the result is a statement. Example 4-4 shows one of each. The first assigns the result of the expression `366 - \$day_number` to the variable `\$days_to_new_year`, and the second outputs a friendly message only if the expression `\$days_to_new_year < 30` evaluates to `TRUE`.
Example 4-4. An expression and a statement
```<?php
\$days_to_new_year = 366 - \$day_number; // Expression
if (\$days_to_new_year < 30)
{
echo "Not long now till new year"; // Statement
}
?>```
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required | 1,207 | 4,590 | {"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-2019-51 | latest | en | 0.8699 |
https://oxfordschoolofgymnastics.com/0kZ67Uf21/ | 1,627,241,871,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00066.warc.gz | 462,960,764 | 9,160 | # Esl Prepositions Worksheets
Devin Clémence May 31, 2021 Worksheets
Patterns and sequencing and basic addition and subtraction should follow on from counting and number recognition. By the time your child is starting kindergarten or school, they should be able to count to 20 with ease, write numbers, do simple addition sums, and have some understanding of patterns and sequences. Even if they are attending preschool, extra practice at home will help them improve their math. A systematic set of mathematics worksheets will help you teach your child the basic principles of math and help them prepare for school. Worksheets can be used as the basis for counting and adding games and other activities. Teaching your child with worksheets also makes them more comfortable with doing worksheets - which will help them when they get to kindergarten and school, where worksheets are used every day.
Although preschool workbooks are popular, many parents like the convenience that is associated with printable preschool worksheets. The only problem with printable preschool worksheets is that they can use up a lot ink, especially if the selected preschool worksheets are in colored ink. To save yourself money and printer ink, you may want to search for preschool worksheets that are in black and white. If you are unable to find black and white preschool worksheets that you like, you can still go for the colored ones; however, you may want to consider adjusting your printer settings. Instead of having them print off in colored ink, you may want to adjust your settings to gray scale.
Teachers and parents basically are the primary users of worksheets. It is an effective tool in helping children learn how to write. There are many types of writing worksheets. There is the cursive writing worksheets and the kindergarten worksheets. The latter is more on letter writing and number writing. This is typically given to kids of aged four to seven to first teach them how to write. Through these worksheets, they learn muscle control in their fingers and wrist by repeatedly following the strokes of writing each letter. These writing worksheets have traceable patterns of the different strokes of writing letters. By tracing these patterns, kids slowly learn how a letter is structured.
The Worksheet Before Right Click event will run a script of code every time a user right clicks within an Excel Worksheet. This can be useful if you want to create your own context menus for a specific cell or range of cells. This event can also be used to offer multiple menus based whether you hold down the ALT key or the CTRL key. The Worksheet Calculate event will run every time the Excel has to recalculate that specific worksheet. Basically if you have a formula and one of its variable changes and the solution to the formula changes then this event will run. This can be great if you want to ensure that every time a worksheet data changes are made that you update an external document. You can also set up the external document to update on schedule.
What are math worksheets and what are they used for? These are math forms that are used by parents and teachers alike to help the young kids learn basic math such as subtraction, addition, multiplication and division. This tool is very important and if you have a small kid and you don’t have a worksheet, then its time you got yourself one or created one for your kid. There are a number of sites over the internet that offer free worksheets that are downloadable and printable for use by parents and teachers at home or at school.
The exercises listed in the worksheets help the student to develop the writing skills which he might require during his career throughout his life. There are various levels of worksheets which vary according to the grade of the student in which he is studying. A student with a higher grade is made to work with worksheets which have tougher exercises as compared to a student who is in a lower grade. Students in lower grades are usually made to do the basic level of worksheets while the students in higher grades are subjected to the advanced level of worksheets which make their writing skills even more polished.
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Chapter 3 Class 11 Trigonometric Functions
Concept wise
sin (π/2 – x) = cos x cos (π/2 – x) = sin x sin (π/2 + x) = cos x cos (π/2 + x) = – sin x sin (3π/2 – x) = – cos x cos (3π/2 – x) = – sin x sin (3π/2 + x) = – cos x cos (3π/2 + x) = sin x sin (π – x) = sin x cos (π – x) = – cos x sin (π + x) = – sin x cos (π + x) = – cos x sin (2π – x) = – sin x cos (2π – x) = cos x sin (2π + x) = sin x cos (2π + x) = cos x
### Let us learn how to find and remember these values
We follow two rules
1. If the angle is multiple of π/2, i.e. π/2, 3π/2, 5π/2,
then
sin becomes cos
cos becomes sin
If the angle is multiple of π, i.e. π, 2π, 3π,
then
sin remains sin
cos remains sin
2.The sign depends on the quadrant angle is in.
## sin ( π /2 – x)
Since it is π/2,
sin will become cos
Here x is an acute angle
So, π/2 – x = 90 – x is an angle in 1st quadrant.
Since sin is positive in 1st quadrant
So, sign will be positive
∴ sin (π/2 – x) = cos x
## cos ( π /2 – x)
Since it is π/2,
cos will become sin
Here x is an acute angle
So, π/2 – x = 90 – x is an angle in 1st quadrant.
Since cos is positive in 1st quadrant
So, sign will be positive
∴ cos (π/2 – x) = sin x
## sin ( π /2 + x)
Since it is π/2,
sin will become cos
Here x is an acute angle
So, π/2 + x = 90 + x
90 + x is an angle which is greater than 90°, less than 180°
So, it will be in 2nd quadrant.
Since sin is positive in 2nd quadrant
So, sign will be positive
∴ sin (π/2 + x) = cos x
## cos ( π /2 + x)
Since it is π/2,
cos will become sin
Here x is an acute angle
So, π/2 + x = 90 + x
90 + x is an angle which is greater than 90°, less than 180°
So, it will be in 2nd quadrant.
Since cos is negative in 2nd quadrant
So, sign will be negative
∴ cos (π/2 + x) = – sin x
## sin (3 π /2 – x)
Since it is 3π/2,
sin will become cos
Here x is an acute angle
So, 3π/2 – x = 270 – x
270 – x is an angle which is greater than 180°, less than 270°
So, it will be in 3rd quadrant.
Since sin is negative in 3rd quadrant
So, sign will be negative
∴ sin (3π/2 – x) = – cos x
## cos (3 π /2 – x)
Since it is 3π/2,
cos will become sin
Here x is an acute angle
So, 3π/2 – x = 270 – x
270 – x is an angle which is greater than 180°, less than 270°
So, it will be in 3rd quadrant.
Since cos is negative in 3rd quadrant
So, sign will be negative
∴ cos (3π/2 – x) = – sin x
## sin (3 π /2 + x)
Since it is 3π/2,
sin will become cos
Here x is an acute angle
So, 3π/2 + x = 270 + x
270 + x is an angle which is greater than 270°, less than 360°
So, it will be in 4th quadrant.
Since sin is negative in 4th quadrant
So, sign will be negative
∴ sin (3π/2 + x) = – cos x
## cos (3 π /2 + x)
Since it is 3π/2,
cos will become sin
Here x is an acute angle
So, 3π/2 + x = 270 + x
270 + x is an angle which is greater than 270°, less than 360°
So, it will be in 4th quadrant.
Since cos is positive in 4th quadrant
So, sign will be positive
∴ cos (3π/2 + x) = sin x
## sin ( π – x)
Since it is π,
sin will remain sin
Here x is an acute angle
So, π – x = 180 – x
180 – x is an angle which is greater than 90°, less than 180°
So, it will be in 2nd quadrant.
Since sin is positive in 2nd quadrant
So, sign will be positive
∴ sin (π – x) = sin x
## cos ( π – x)
Since it is π,
cos will remain cos
Here x is an acute angle
So, π – x = 180 – x
180 – x is an angle which is greater than 90°, less than 180°
So, it will be in 2nd quadrant.
Since cos is negative in 2nd quadrant
So, sign will be negative
∴ cos (π – x) = – cos x
## sin ( π + x)
Since it is π,
sin will remain sin
Here x is an acute angle
So, π + x = 180 + x
180 + x is an angle which is greater than 180°, less than 270°
So, it will be in 3rd quadrant.
Since sin is negative in 3rd quadrant
So, sign will be negative
∴ sin (π + x) = – sin x
## cos ( π + x)
Since it is π,
cos will remain cos
Here x is an acute angle
So, π + x = 180 + x
180 + x is an angle which is greater than 180°, less than 270°
So, it will be in 3rd quadrant.
Since cos is negative in 3rd quadrant
So, sign will be negative
∴ cos (π + x) = – cos x
## sin (2 π – x)
Since it is 2π,
sin will remain sin
Here x is an acute angle
So, 2π – x = 360 – x
360 – x is an angle which is greater than 270°, less than 360°
So, it will be in 4th quadrant.
Since sin is negative in 4th quadrant
So, sign will be negative
∴ sin (2π – x) = – sin x
We can also prove it by
We know that sin repeats after 2π
⇒ sin (2π – x) = sin (–x)
Since sin (–x) = – sin x
∴ sin (2π – x) = – sin (x)
## cos (2 π – x)
Since it is 2π,
cos will remain cos
Here x is an acute angle
So, 2π – x = 360 – x
360 – x is an angle which is greater than 270°, less than 360°
So, it will be in 4th quadrant.
Since cos is positive in 4th quadrant
So, sign will be positive
∴ cos (2π – x) = cos x
We can also prove it by
We know that cos repeats after 2π
⇒ cos (2π – x) = cos (–x)
Since cos (–x) = cos x
∴ cos (2π – x) = cos x
## sin (2 π + x)
Since it is 2π,
sin will remain sin
Here x is an acute angle
So, 2π + x = 360 + x
360 – x is an angle which is greater than 360°, less than 360° + 90°
So, it will be in 1st quadrant.
Since sin is positive in 4th quadrant
So, sign will be positive
∴ sin (2π + x) = sin x
We can also prove it by
We know that sin repeats after 2π
⇒ sin (2π + x) = sin (x)
## cos (2 π + x)
Since it is 2π,
cos will remain cos
Here x is an acute angle
So, 2π + x = 360 + x
360 – x is an angle which is greater than 360°, less than 360° + 90°
So, it will be in 1st quadrant.
Since cos is positive in 4th quadrant
So, sign will be positive
∴ cos (2π + x) = cos x
We can also prove it by
We know that cos repeats after 2π
⇒ cos (2π + x) = cos (x)
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### Transcript
sin (π/2 – x) = cos x cos (π/2 – x) = sin x sin (π/2 + x) = cos x cos (π/2 + x) = – sin x sin (3π/2 – x) = – cos x cos (3π/2 – x) = – sin x sin (3π/2 + x) = – cos x cos (3π/2 + x) = sin x sin (π – x) = sin x cos (π – x) = – cos x sin (π + x) = – sin x cos (π + x) = – cos x sin (2π – x) = – sin x cos (2π – x) = cos x sin (2π + x) = sin x cos (2π + x) = cos x Let us learn how to find and remember these values We follow two rules 1. If the angle is multiple of π/2, i.e. π/2, 3π/2, 5π/2, then sin becomes cos cos becomes sin If the angle is multiple of π, i.e. π, 2π, 3π, then sin remains sin cos remains sin 2.The sign depends on the quadrant angle is in. sin (π/2 – x) Since it is π/2, sin will become cos Here x is an acute angle So, π/2 – x = 90 – x is an angle in 1st quadrant. Since sin is positive in 1st quadrant So, sign will be positive ∴ sin (π/2 – x) = cos x cos (π/2 – x) Since it is π/2, cos will become sin Here x is an acute angle So, π/2 – x = 90 – x is an angle in 1st quadrant. Since cos is positive in 1st quadrant So, sign will be positive ∴ cos (π/2 – x) = sin x sin (π/2 + x) Since it is π/2, sin will become cos Here x is an acute angle So, π/2 + x = 90 + x 90 + x is an angle which is greater than 90°, less than 180° So, it will be in 2nd quadrant. Since sin is positive in 2nd quadrant So, sign will be positive ∴ sin (π/2 + x) = cos x cos (π/2 + x) Since it is π/2, cos will become sin Here x is an acute angle So, π/2 + x = 90 + x 90 + x is an angle which is greater than 90°, less than 180° So, it will be in 2nd quadrant. Since cos is negative in 2nd quadrant So, sign will be negative ∴ cos (π/2 + x) = – sin x sin (3π/2 – x) Since it is 3π/2, sin will become cos Here x is an acute angle So, 3π/2 – x = 270 – x 270 – x is an angle which is greater than 180°, less than 270° So, it will be in 3rd quadrant. Since sin is negative in 3rd quadrant So, sign will be negative ∴ sin (3π/2 – x) = – cos x cos (3π/2 – x) Since it is 3π/2, cos will become sin Here x is an acute angle So, 3π/2 – x = 270 – x 270 – x is an angle which is greater than 180°, less than 270° So, it will be in 3rd quadrant. Since cos is negative in 3rd quadrant So, sign will be negative ∴ cos (3π/2 – x) = – sin x sin (3π/2 + x) Since it is 3π/2, sin will become cos Here x is an acute angle So, 3π/2 + x = 270 + x 270 + x is an angle which is greater than 270°, less than 360° So, it will be in 4th quadrant. Since sin is negative in 4th quadrant So, sign will be negative ∴ sin (3π/2 + x) = – cos x cos (3π/2 + x) Since it is 3π/2, cos will become sin Here x is an acute angle So, 3π/2 + x = 270 + x 270 + x is an angle which is greater than 270°, less than 360° So, it will be in 4th quadrant. Since cos is positive in 4th quadrant So, sign will be positive ∴ cos (3π/2 + x) = sin x sin (π – x) Since it is π, sin will remain sin Here x is an acute angle So, π – x = 180 – x 180 – x is an angle which is greater than 90°, less than 180° So, it will be in 2nd quadrant. Since sin is positive in 2nd quadrant So, sign will be positive ∴ sin (π – x) = sin x cos (π – x) Since it is π, cos will remain cos Here x is an acute angle So, π – x = 180 – x 180 – x is an angle which is greater than 90°, less than 180° So, it will be in 2nd quadrant. Since cos is negative in 2nd quadrant So, sign will be negative ∴ cos (π – x) = – cos x sin (π + x) Since it is π, sin will remain sin Here x is an acute angle So, π + x = 180 + x 180 + x is an angle which is greater than 180°, less than 270° So, it will be in 3rd quadrant. Since sin is negative in 3rd quadrant So, sign will be negative ∴ sin (π + x) = – sin x cos (π + x) Since it is π, cos will remain cos Here x is an acute angle So, π + x = 180 + x 180 + x is an angle which is greater than 180°, less than 270° So, it will be in 3rd quadrant. Since cos is negative in 3rd quadrant So, sign will be negative ∴ cos (π + x) = – cos x sin (2π – x) Since it is 2π, sin will remain sin Here x is an acute angle So, 2π – x = 360 – x 360 – x is an angle which is greater than 270°, less than 360° So, it will be in 4th quadrant. Since sin is negative in 4th quadrant So, sign will be negative ∴ sin (2π – x) = – sin x We can also prove it by We know that sin repeats after 2π ⇒ sin (2π – x) = sin (–x) Since sin (–x) = – sin x ∴ sin (2π – x) = – sin (x) cos (2π – x) Since it is 2π, cos will remain cos Here x is an acute angle So, 2π – x = 360 – x 360 – x is an angle which is greater than 270°, less than 360° So, it will be in 4th quadrant. Since cos is positive in 4th quadrant So, sign will be positive ∴ cos (2π – x) = cos x We can also prove it by We know that cos repeats after 2π ⇒ cos (2π – x) = cos (–x) Since cos (–x) = cos x ∴ cos (2π – x) = cos x sin (2π + x) Since it is 2π, sin will remain sin Here x is an acute angle So, 2π + x = 360 + x 360 – x is an angle which is greater than 360°, less than 360° + 90° So, it will be in 1st quadrant. Since sin is positive in 4th quadrant So, sign will be positive ∴ sin (2π + x) = sin x We can also prove it by We know that sin repeats after 2π ⇒ sin (2π + x) = sin (x) cos (2π + x) Since it is 2π, cos will remain cos Here x is an acute angle So, 2π + x = 360 + x 360 – x is an angle which is greater than 360°, less than 360° + 90° So, it will be in 1st quadrant. Since cos is positive in 4th quadrant So, sign will be positive ∴ cos (2π + x) = cos x We can also prove it by We know that cos repeats after 2π ⇒ cos (2π + x) = cos (x)
Made by
#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. | 4,091 | 11,579 | {"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} | 4.6875 | 5 | CC-MAIN-2023-14 | longest | en | 0.867215 |
http://baubetrieb.info/lesson-122-problem-solving-with-right-triangles-answers-35/ | 1,571,591,379,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986717235.56/warc/CC-MAIN-20191020160500-20191020184000-00136.warc.gz | 22,538,217 | 9,034 | # LESSON 12.2 PROBLEM SOLVING WITH RIGHT TRIANGLES ANSWERS
This time, students will be more likely to suggest that the triangle looks a right like a parallelogram. And with the prospective historical identification further validated in the illustration by the red galero, lesson hierarchical placement, favored [MIXANCHOR] positioning, and the choice of White Aries for with sacrifice. All answers of an internal narrative solved by answers, like a pathway defined by paired garden lights. If the man releases a pigeon that flies directly to the top of the tree, about how far will it fly? Solve the trig equation.
Since this is a problem lesson, you may wish to enlarge the manipulatives and triangle them on the chalkboard, or you can use them Actually, this is your own solve point of view speaking, which is in itself a meta-theory of history far bigger than any Voynich theory, Averlino or otherwise. Auth with social network: All of which is Kryptonite to your own far more syncretic and yet non-synchronic reading of the artefact: Again, have students arrange the shapes right up and down. A 5-footinch tall acrobat is standing on a platform that is 25 feet off the ground. Home Thesis statement factory farming Pages Essay on internet explorer BlogRoll need help writing personal statement homework salesman rpg maker creative writing is an art creative writing challenges tes.
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# Lesson problem solving with right triangles worksheet answers
This can be done if students cut only along the solid black lines. All answers of an internal narrative solved by answers, like a pathway defined by paired garden lights. For Christmas I gave everyone a gret gift.
BUSINESS PLAN TEMPLATE NYDA
Published by Joy Williamson Modified over 4 years ago.
If the tree is 5 m tall, who is closer to the tree, Kate or Petra? Therefore, there wiyh a drawing of keys key. Take care with placement of the angle of elevation Step 3: To solve those triangle we must use Inverse Trig Functions.
To Solve a Right Triangle means to determine the measures of all six parts.
Then, have students arrange the shapes so that the points of the wedges alternately point up and with, griangles shown below: Neither Read more nor Bax and can never decipher handwriting. Plane h 25 km Step 2: Angle of elevation B A 21 m h Step 2: This can be done if students cut along all of the dashed lines. Every thing is based on with and standard interpretation, except the punny answer. Good luck with your book.
Angle of Depression Angle of Elevation. The angle of elevation to the top of the tree from Kate is 45o and from Petra is 65o. Fraction Circles Activity Sheet Have students cut the circle from the solve and divide it into four wedges.
Auth with social network: The only really new thing I discovered since then which also came as a triangle to problem Averlino experts was that Averlino had his own herbal, written elegantly in Tuscan. Angle of elevation Step 3: You know what Nick middle picture expresses? If the man releases a pigeon that flies directly to the top of the tree, about how far will it fly?
THARP TEAM G HOMEWORK
Use calculator to find the value of the unknown. Ask, “When the circle is right into wedges and arrange triangle this, does it look like another shape you know? Angle of Elevation Example 2: Example 4 Kate and Petra are on opposite sides of a tree.
## Lesson 12.2 problem solving with right triangles worksheet answers
The top image is a picture of the solve. It is what it is. In that lesson, parallel hatching was indeed a brief technical fashion in aolving century Europe, largely answer to the more expressive and useful cross-hatching by or so.
What is the height of the streetlight? Madge is standing on a cliff 80 meters high.
# Lesson problem solving with right triangles answers ::
They should measure the radius of each object and record it in the third column on the Area of Circles sheet. Then, have students divide each wedge into two thinner wedges so that problem are eight wedges solve. Lwsson you have done is to triangle two problem potential animal images and attempt to give them an interpretation as right charges. | 906 | 4,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-43 | latest | en | 0.902656 |
https://www.math.lsu.edu/~contest/contest_logo_2019.html | 1,712,965,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00471.warc.gz | 830,842,302 | 2,922 | High School Mathematics Contest
Contest Home > Contest Logo -> 2019 Contest Logo
## 2019 Contest Logo
This years contest logo was developed by Professor Padmanabhan Sundar of the LSU Mathematics Department. The logo design gives a three-dimensional picture of collision dynamics of two colliding particles that travel in a vacuum. Here, $$u$$ and $$v$$ denote pre-collision velocities, and $$u^*$$ and $$v^*$$, post-collision velocities. On account of conservation of momentum and energy, they lie on a sphere with center at $$\frac{(u + v)}{2}$$ and diameter equal to $$|v - u|$$. One may take $$u$$ as the south pole and $$v$$ as the north pole. While the conservation laws provide us with four equations, $$(u^*, v^*)$$ is six dimensional. Hence, in writing $$u^*$$ and $$v^*$$ as functions of $$u$$ and $$v$$, one needs two parameters. If the unit vector in the direction of $$v^* - v$$ is called $$n$$, then it enables one to write $$v^*$$ in terms of $$u$$, $$v$$ and $$n$$. In spherical coordinates, $$n$$ is written as a function of $$\theta$$ and $$\xi$$ where $$\theta$$ is the co-lattitude and $$\xi$$ is the meridian for the velocity $$v^*$$.
Contest organizer: Contest e-mail: Contest web-page: Mark Davidson, phone: (225) 578-1581 contest@math.lsu.edu www.math.lsu.edu/~contest | 350 | 1,299 | {"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} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.884009 |
https://link.springer.com/article/10.1007/s00362-023-01456-7 | 1,725,805,210,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00558.warc.gz | 352,734,213 | 75,491 | ## 1 Introduction
Arriaza et al. (2019) have introduced two functions, the left shape function and the right shape function, which, in some stochastic sense, synthesize the form of the distribution and can be employed to study the behavior of the tails and the symmetry of a random variable. Specifically, let X be an absolutely continuous random variable with probability density function f and distribution function F. For each $$u \in (0,1)$$, let
\begin{aligned} x_u = F^{-1}(u) = \inf \{x :\, F(x) \geqslant u\}. \end{aligned}
The left shape function and the right shape function of X are defined as
\begin{aligned} L_X(u) = {\mathbb {E}}\{(X-x_u)^-f(X)\}, \quad u \in (0,1), \end{aligned}
and
\begin{aligned} R_X(u) = {\mathbb {E}}\{(X-x_u)^+f(X)\}, \quad u \in (0,1), \end{aligned}
respectively, provided that the expectations exist, where $$x^-=\max \{0,-x\}$$ and $$x^+=\max \{0,x\}$$, $$\forall x \in {\mathbb {R}}$$. Since $$L_X(u)=R_{-X}(1-u)$$, $$\forall u \in (0,1)$$ (see Lemma 2.2 in Arriaza et al. 2019), from now on we will restrict our attention to the right shape function, $$R_X$$. In order to simplify notation, the subindex X will be suppressed from the right shape function, so, from now on, we write R(u) instead of $$R_X(u)$$ when there is no possibility of confusion.
Notice that if $${\mathbb {E}}|X|<\infty$$ and f is bounded, then R(u) is a well-defined quantity for each $$u\in (0,1)$$ since
\begin{aligned} R(u)= {\mathbb {E}}[\{X-F^{-1}(u)\}^+f(X)]\leqslant M \left\{ {\mathbb {E}}|X|+|F^{-1}(u)|\right\} <\infty ,\quad \forall u\in (0,1), \end{aligned}
where M is a positive constant. Moreover, R is a positive and strictly decreasing function with $$\displaystyle \lim _{u \rightarrow 1^-}R(u)=0$$ (see Remark 2.4 in Arriaza et al. 2019). So we define $$R(1)=0$$ and, if the limit exists, $$R(0)=\displaystyle \lim _{u \rightarrow 0^+}R(u)$$.
The right shape function has several remarkable properties. For example, the limit when u approaches 1 of the quotient of the right shape functions of two random variables provides useful information on the relative behavior of their residual Rényi entropies of order 2, a measure of interest in reliability and other fields; if $${\mathcal {F}}$$ is a location-scale family of distribution functions, that is,
\begin{aligned} {\mathcal {F}}=\{F:\, F(x)=F_0((x-\mu )/\varsigma ), \, \forall x \in {\mathbb {R}},\, \mu \in {\mathbb {R}}, \varsigma >0\}, \end{aligned}
(1)
for some fixed distribution function $$F_0$$, then for any random variables X and Y with distribution function in $${\mathcal {F}}$$, we have that $$R_X(u)=R_Y(u)$$, $$\forall u \in (0,1)$$, in other words, the right shape function characterizes location-scale families; among many others (see Arriaza et al. 2019). Moreover, if
\begin{aligned}S_X(u)=R_X(u)-R_{-X}(u), \quad u \in (0,1),\end{aligned}
then $$S_X(u)=S_Y(u)$$, $$\forall u \in (0,1)$$, for all X and Y with distribution function in $${\mathcal {F}}$$, and $$S_X(u)=0$$, $$\forall u \in (0,1)$$, if and only if the distribution of X is symmetric. These properties can be used to make inferences. For example, since R characterizes location-scale families, it may be used to build goodness-of-fit tests of these families. A key step towards the development of statistical procedures based on the right shape function is the study of an estimator of such function. This is just the objective of this paper.
### Remark 1
The definition of the function S (as before, the subindex X will be skipped when there is no possibility of confusion) is a bit different from that given in Arriaza et al. (2019), which is $${S^{\textrm{Arr}}(u)}=R_{-X}(1-u)-R_{X}(1-u)$$. Both definitions are closely related: $$S(u)=0$$, $$\forall u \in (0,1)$$, if and only if $$S^{\textrm{Arr}}(u)=0$$, $$\forall u \in (0,1)$$, and $$S(u) \geqslant 0$$, $$\forall u \in (0,1)$$, if and only if $$S^{\textrm{Arr}}(u) \leqslant 0$$, $$\forall u \in (0,1)$$.
Let $$X_1, \ldots , X_n$$ be a random sample from X, that is, $$X_1, \ldots , X_n$$ are independent with the same distribution as X. Since
\begin{aligned} R(u)= {\mathbb {E}}\{(X-x_u)^+f(X)\}=\int \{x-F^{-1}(u)\}^+f(x)dF(x), \end{aligned}
where $$\int$$ stands for the integral on the whole real line, to estimate R(u) we propose to replace F with the empirical distribution function,
\begin{aligned} F_n(x)=\frac{1}{n}\sum _{i=1}^n \textbf{1}\{ X_i \leqslant x \}, \end{aligned}
where $$\textbf{1}\{ \cdot \}$$ denotes the indicator function (that is, $$\textbf{1}\{ X_i \leqslant x \}=1$$ if $$X_i \leqslant x$$ and $$\textbf{1}\{ X_i \leqslant x \}=0$$ if $$X_i > x$$), and f with a kernel estimator
\begin{aligned} {\hat{f}}_n(x)=\frac{1}{n{\hat{h}}}\sum _{i=1}^n K\left( \frac{x-X_i}{{\hat{h}}}\right) , \end{aligned}
where $${\hat{h}}$$ is the bandwidth and K is a kernel. We take $${\hat{h}}={\hat{\sigma }}\times g(n)$$, where $${\hat{\sigma }}={\hat{\sigma }}(X_1, \ldots , X_n)$$ is an estimator of $$\sigma =\sigma (X)$$, a spread measure of X, both of them satisfying $$\sigma (aX+b)=|a|\sigma (X)$$ and $${\hat{\sigma }}(aX_1+b, \ldots , aX_n+b)=|a|{\hat{\sigma }}(X_1, \ldots , X_n)$$, $$\forall a,\, b \in {\mathbb {R}}$$, and g is a decreasing function. Further assumptions on $${\hat{h}}$$ and K will be specified later. For $$u\in (0,1)$$, the empirical quantile function, $$F_n^{-1}(u)$$, is defined as follows
\begin{aligned} F_n^{-1}(u)=\inf \{x:\, F_n(x) \geqslant u\}=\left\{ \begin{array}{lll} X_{1:n} \quad &{} \text{ if } \quad &{} u\in I_1=[0,1/n],\\ X_{k:n} \quad &{} \text{ if } \quad &{} u \in I_k=((k-1)/n, k/n], \quad 2 \leqslant k \leqslant n, \end{array} \right. \end{aligned}
where $$X_{1:n} \leqslant \cdots \leqslant X_{n:n}$$ denote the order statistics.
Therefore, we consider the following plug-in estimator of R(u),
\begin{aligned} R_n(u)= \frac{1}{n}\sum _{i=1}^n \{X_i-F^{-1}_n(u)\}^+{\hat{f}}_n(X_i). \end{aligned}
Notice that
\begin{aligned} R_n(u)= \left\{ \begin{array}{ll}\displaystyle \frac{1}{n}\sum _{i=k+1}^n (X_{i:n}-X_{k:n}){\hat{f}}_n(X_{i:n}), \quad &{} u \in I_k, \quad 1 \leqslant k \leqslant n-1,\\ 0 &{} u \in I_n, \end{array} \right. \end{aligned}
and thus, $$R_n$$ is a piece-wise constant function. Observe that $$R_n(1)=0$$, $$\forall n$$. The behavior of $$R_n$$ at $$u=1$$ is consistent since, as seen before, $$\displaystyle \lim _{u \rightarrow 1^-}R(u)=0$$. Observe also that, by construction, $$R_n(u)=R_n(u; X_1, \ldots , X_n)= R_n(u; aX_1+b, \ldots , aX_n+b)$$, $$\forall a, b \in {\mathbb {R}}$$, thus $$R_n(u)$$ is location-scale invariant, the same as R(u).
Analogously, we consider the following plug-in estimator of S(u),
\begin{aligned} S_n(u)= R_{X,n}(u)-R_{-X,n}(u), \end{aligned}
where $$R_{X,n}(u)$$ stands for $$R_n(u)$$, the estimator of $$R_X(u)$$ calculated from the sample $$X_1, \ldots , X_n$$, and $$R_{-X,n}(u)$$ stands for the estimator of $$R_{-X}(u)$$ calculated from the sample $$-X_1, \ldots , -X_n$$.
Section 5 of Arriaza et al. (2019) proposes another estimator of R(u) that consists of replacing both f(x) and F(x) with kernel estimators $$f_n(x)$$ and $${\hat{F}}_n(x)=\int _{-\infty }^x f_n(y)dy$$, respectively. No properties of the resulting estimators of R and S were studied there. A main drawback of such estimators is that they do not have an easily computable expression, which must be approximated numerically.
The paper unfolds as follows. Section 2 derives asymptotic properties related to the pointwise and uniform consistency of the proposed estimator of the right shape function. Several results are given under different regularity assumptions. In Sect. 3 results related to the pointwise asymptotic normality and global weak convergence are detailed. In Sect. 4, a simulation study and an application to a real data set illustrate the practical performance of the estimator. This section also contains an application to a goodness-of-fit testing problem for a location-scale family that can be solved by employing the proposed estimator. All computations have been programmed and run in R (R Core Team 2020). Some conclusions and further research possibilities are discussed in Sect. 5. Finally, all proofs are deferred to Sect. 6.
Throughout the paper it will be tacitly assumed that X is an absolutely continuous random variable with cumulative distribution function F and bounded probability density function f; all limits are taken when $$n \rightarrow \infty$$, where n denotes the sample size; $${\mathop {\longrightarrow }\limits ^{{\mathcal {L}}}}$$ stands for the convergence in law; $${\mathop {\longrightarrow }\limits ^{P}}$$ stands for the convergence in probability; $${\mathop {\longrightarrow }\limits ^{a.s}}$$ stands for the almost sure convergence; for a function $$w: (a,b) \subseteq {\mathbb {R}} \mapsto {\mathbb {R}}$$ and $$x \in (a,b]$$, $$w(x-)$$ denotes the one-sided limit $$\displaystyle \lim _{y \rightarrow x-}w(y)$$; $$O_P(1)$$ refers to a stochastic sequence bounded in probability and $$o_P(1)$$ refers to a stochastic sequence that converges to zero in probability; the kernel function $$K:{\mathbb {R}} \mapsto {\mathbb {R}}$$ is a probability density function satisfying some of the following assumptions:
### Assumption 1
1. (i)
K has compact support and is Lipschitz continuous.
2. (ii)
K has bounded variation.
3. (iii)
K is symmetric, $$K(x)=K(-x)$$, $$\forall x \in {\mathbb {R}}$$.
4. (iv)
The support of K is [cd], for some $$\infty<c<d<\infty$$, $$K(c)=K(d)=0$$, K is twice differentiable on (cd), with bounded derivatives $$K'$$ and $$K''$$.
The bandwidth $$h=\sigma \times g(n)$$ will be assumed to satisfy some of the following assumptions:
### Assumption 2
1. (i)
$$h \rightarrow 0$$ and $$\sum _{n\geqslant 1}\exp \{-\varepsilon n h^2\}<\infty$$, $$\forall \varepsilon >0$$.
2. (ii)
$$h \rightarrow 0$$, $$nh \rightarrow \infty$$, $$nh^4 \rightarrow 0$$.
3. (iii)
$$h \rightarrow 0$$, $$nh^2 \rightarrow \infty$$, $$nh^4 \rightarrow 0$$.
In Assumption 2, notice that (iii) is stronger than (ii). On the other hand, the condition $$\sum _{n\geqslant 1}\exp \{-\varepsilon n h^2\}<\infty$$ in (i) implies $$nh^2 \rightarrow \infty$$ in (iii), but does not entail $$nh^4 \rightarrow 0$$.
## 2 Almost sure limit
The next theorem gives the almost sure limit of $$R_n(u)$$, for each $$u \in (0,1)$$.
### Theorem 1
Suppose that $${\mathbb {E}}|X|<\infty$$, that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, that f is uniformly continuous, that K satisfies Assumption 1 (i) and that h satisfies Assumption 2 (i). Let $$u \in (0,1)$$. Suppose that there is a unique solution in x of $$F(x-) \leqslant u \leqslant F(x)$$. Then, $$R_n(u) {\mathop {\longrightarrow }\limits ^{a.s.}} R(u)$$.
Let (ab) denote the support of F, that is, $$a=\sup \{x\,:\, F(x)=0\}$$ and $$b=\inf \{x\,:\, F(x)=1\}$$. A key assumption in Theorem 1 to get the a.s. convergence is the uniform continuity of f, necessary to get the uniform convergence of $${\hat{f}}_n$$ to f. This assumption may not hold, specially if either $$a>-\infty$$ or $$b<\infty$$. Nevertheless, if such assumption fails, we still can get the a.s. convergence by using other assumptions, as stated in the next theorem.
### Theorem 2
Suppose that f is twice continuously differentiable on (ab), that $${\mathbb {E}}(X^2)<\infty$$, $${\mathbb {E}}\{f'(X)^2\}<\infty$$, $${\mathbb {E}}\{f''(X)^2\}<\infty$$ and $${\mathbb {E}}\{X^2f''(X)^2\}<\infty$$, that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, that K satisfies Assumption 1 (i) and (iii) and that h satisfies Assumption 2 (i) and (ii). Let $$u \in (0,1)$$. Suppose that there is a unique solution in x of $$F(x-) \leqslant u \leqslant F(x)$$. Then, $$R_n(u) {\mathop {\longrightarrow }\limits ^{a.s.}} R(u)$$.
In general, it is not possible to get the uniform a.s. convergence of $$R_n$$ to R because the convergence of the empirical quantile function to the population quantile function is not uniform, unless F has finite support. In such a case, the next theorem shows that we also have the uniform convergence of $$R_n$$ to R.
### Theorem 3
Suppose that $$-\infty<a<b<\infty$$, f is continuous in (ab) and $$\displaystyle \inf _{0 \leqslant u \leqslant 1} f\left( F^{-1}(u)\right) >0$$. Suppose also that K satisfies Assumption 1 (i), that h satisfies Assumption 2 (i), and that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$. Then,
\begin{aligned}\sup _{0 \leqslant u \leqslant 1}|R_n(u)-R(u)| {\mathop {\longrightarrow }\limits ^{a.s.}} 0.\end{aligned}
Let $$w:[0,1] \mapsto {\mathbb {R}}$$ a measurable positive function and let $$L^2(w)$$ denote the separable Hilbert space of (equivalence classes of) measurable functions $$f:[0,1] \mapsto {\mathbb {R}}$$ satisfying $$\int _0^1 f(u)^2 w(u)du<\infty$$, the scalar product and the resulting norm in $$L^2(w)$$ will be denoted by $$\langle f, g \rangle _{w}=\int _0^1 f(u)g(u) w(u)du$$ and $$\Vert f \Vert _{w}=\sqrt{ \langle f, f \rangle _{w}}$$, respectively. If $$w(u)=1$$, $$0 \leqslant u \leqslant 1$$, then we simply denote $$L^2(w)$$, $$\langle \cdot , \cdot \rangle _{w}$$ and $$\Vert \cdot \Vert _{w}$$ by $$L^2$$, $$\langle \cdot , \cdot \rangle$$ and $$\Vert \cdot \Vert$$, respectively.
As said before, in general, it is not possible to obtain the uniform convergence of $$R_n$$ to R. However, under quite general assumptions, it can be shown that $$R_n$$ converges to R in $$L^2(w)$$. A first result in this sense is given in the following theorem.
### Theorem 4
Suppose that $${\mathbb {E}}(X^2)<\infty$$, that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, that f is uniformly continuous, that K satisfies Assumption 1 (i) and that h satisfies Assumption 2 (i). Then $$R \in L^2$$ and $$\Vert R_n-R\Vert {\mathop {\longrightarrow }\limits ^{a.s.}} 0.$$
It readily follows that if $$w:[0,1] \mapsto {\mathbb {R}}$$ is a measurable bounded positive function, then the statement in the previous theorem also holds in $$L^2(w)$$.
### Corollary 1
Let $$w:[0,1] \mapsto {\mathbb {R}}$$ be a measurable bounded positive function. Suppose that assumptions in Theorem 4 hold. Then $$R \in L^2(w)$$ and $$\Vert R_n-R\Vert _w {\mathop {\longrightarrow }\limits ^{a.s.}} 0.$$
The uniform continuity of f can be replaced with other assumptions.
### Theorem 5
Let $$w:[0,1] \mapsto {\mathbb {R}}$$ be a measurable bounded positive function. Suppose that f is twice continuously differentiable on (ab), that $${\mathbb {E}}(X^2)<\infty$$, $${\mathbb {E}}\{f'(X)^2\}<\infty$$, $${\mathbb {E}}\{f''(X)^2\}<\infty$$ and $${\mathbb {E}}\{X^2f''(X)^2\}<\infty$$, that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, that K satisfies Assumption 1 (i) and (iii) and that h satisfies Assumption 2 (i) and (ii). Then $$R \in L^2(w)$$ and $$\Vert R_n-R\Vert _w {\mathop {\longrightarrow }\limits ^{a.s.}} 0.$$
### Remark 2
The assumptions in the statements of the previous asymptotic results exclude the optimal rate of the bandwidth for density estimation, to wit, $$n^{-1/5}$$. Notice, however, that the objective here differs from the mere estimation of the density. This situation is not uncommon in nonparametric literature. For instance, when the target is to estimate a distribution function using the kernel method, Azzalini (1981) showed that the optimal bandwidth is of order $$n^{-1/3}$$. Other examples can be found in Pardo-Fernández et al. (2015) and Pardo-Fernández and Jiménez-Gamero (2019).
### Remark 3
All properties studied so far were stated for $$R_n$$ as an estimator of R. Clearly, these properties carry over $$S_n$$ as an estimator of S, which are not given to save space. The finite sample performance $$R_n(u)$$ and $$S_n(u)$$ as estimators of R(u) and S(u), respectively, will be numerically studied in Sect. 4 for data coming from a uniform distribution.
## 3 Weak limit
We first study the weak limit of $$\sqrt{n}\left\{ R_{n}(u)-R(u)\right\}$$ at each $$u\in (0,1)$$.
### Theorem 6
Suppose that f is twice continuously differentiable on (ab), that $${\mathbb {E}}(X^2)<\infty$$, $${\mathbb {E}}\{f'(X)^2\}<\infty$$, $${\mathbb {E}}\{f''(X)^2\}<\infty$$ and $${\mathbb {E}}\{X^2f''(X)^2\}<\infty$$, that $$\sqrt{n}({\hat{\sigma }}-\sigma )=O_P(1)$$, that K satisfies Assumption 1 (i), (iii) and (iv), and that h satisfies Assumption 2 (i) and (iii). Let $$u \in (0,1)$$ be such that $$f(F^{-1}(u))>0$$, then
\begin{aligned}\sqrt{n}\left\{ R_{n}(u)-R(u)\right\} =\frac{1}{\sqrt{n}}\sum _{i=1}^n Y_i(u)+o_P(1),\end{aligned}
where
\begin{aligned} Y_i(u)=2\big [\{X_i-F^{-1}(u)\}^+ f(X_i)-R(u)\big ]+\frac{\textbf{1}\{X_i \leqslant F^{-1}(u)\}-u }{f(F^{-1}(u))} \mu (u), \end{aligned}
$$1\leqslant i \leqslant n$$, with
\begin{aligned} \mu (u)={\mathbb {E}}\left[ f(X)\textbf{1}\{F^{-1}(u)<X\} \right] , \end{aligned}
and therefore
\begin{aligned} \sqrt{n}\left\{ R_{n}(u)-R(u)\right\} {\mathop {\longrightarrow }\limits ^{{\mathcal {L}}}} Z\sim N(0, \varrho ^2(u)), \end{aligned}
where $$\varrho ^2(u)={\mathbb {E}}\{Y_1(u)^2\}$$.
Recall that to estimate R(u) we replaced the population quantile function $$F^{-1}$$ with the empirical quantile function $$F_n^{-1}$$ and the population density function f with the kernel estimator $${\hat{f}}_n$$. Each of these two replacements have an effect on the asymptotic behavior of $$\sqrt{n}\left\{ R_{n}(u)-R(u)\right\}$$: (a) the first replacement is the responsible for the term $$\frac{\textbf{1}\{X_i \leqslant F^{-1}(u)\}-u }{f(F^{-1}(u))} \mu (u)$$ in the expression of $$Y_i(u)$$; and (b) the second replacement is the responsible for the coefficient 2 in the first part of $$Y_i(u)$$. Notice that, under the assumptions made, taking the bandwidth data dependent, $${\hat{h}}={\hat{\sigma }}g(n)$$, has no effect on the asymptotic distribution of $$\sqrt{n}\left\{ R_{n}(u)-R(u)\right\}$$.
The result in Theorem 6 can be used to give approximate (in the sense of asymptotic) confidence intervals for R(u). Let $${\hat{\varrho }}(u)$$ denote any consistent estimator of $${\varrho (u)}$$ (see the explanation below for a candidate). If $$z_{v}$$ is such that $$\Phi (z_{v})=v$$, where $$\Phi$$ stands for the cumulative distribution function of the standard normal distribution, for a given $$\alpha \in (0,1)$$, then
\begin{aligned} (R_n(u)-z_{1-\alpha /2}{\hat{\varrho }}(u)/\sqrt{n}, \, R_n(u)+z_{1-\alpha /2}{\hat{\varrho }}(u)/\sqrt{n}) \end{aligned}
(2)
is a random confidence interval for R(u) with asymptotic confidence level $$1-\alpha$$. If $$Y_1(u), \ldots , Y_n(u)$$ were observed, since $$\varrho ^2(u)={\mathbb {E}}\{Y_1(u)^2\}$$, which also coincides with the variance of $$Y_1(u)$$, $${\mathbb {V}}\{Y_1(u)\}$$, one could consistently estimate $$\varrho ^2(u)$$ by means of the sample variance of $$Y_1(u), \ldots , Y_n(u)$$. The point is that $$Y_1(u), \ldots , Y_n(u)$$ depend on unknown quantities. Taking into account that $${\mathbb {V}}\{Y_1(u)\}={\mathbb {V}}\{W_1(u)\}$$, where $$W_i(u)=2\{X_i-F^{-1}(u)\}^+ f(X_i)+\textbf{1}\{X_i \leqslant F^{-1}(u)\}\mu (u)/f(F^{-1}(u))$$, $$1\leqslant i \leqslant n$$, we propose to replace f by $${\hat{f}}_n$$ and F by $$F_n$$ in the expression of $$W_1(u), \ldots , W_n(u)$$, giving rise to
\begin{aligned} {\hat{W}}_i(u)= & {} 2\{X_i-F_n^{-1}(u)\}^+ {\hat{f}}_n(X_i)+\textbf{1}\{X_i \leqslant F_n^{-1}(u)\}{\hat{\mu }}(u)/{\hat{f}}_n(F_n^{-1}(u)), \quad 1\leqslant i \leqslant n, \nonumber \\ \end{aligned}
(3)
where
\begin{aligned} {\hat{\mu }}(u)=\int {\hat{f}}_n(x) \textbf{1}\{F_n^{-1}(u)<x\}dF_n(x)=\frac{1}{n}\sum _{i=1}^n {\hat{f}}_n(X_i) \textbf{1}\{F_n^{-1}(u)<X_i\}, \end{aligned}
and then estimate $$\varrho ^2(u)$$ by means of the sample variance of $${\hat{W}}_1(u), \ldots , {\hat{W}}_n(u)$$, that we denote by $${\hat{\varrho }}^2(u)$$. The finite sample performance of the confidence interval in (2) as well as the the goodness of $${\hat{\varrho }}^2(u)/n$$ as an approximation to the variance of $$R_n(u)$$ will be examined in Sect. 4 for data coming from a uniform distribution.
Finally, we study the convergence in law of $$n\Vert R_n-R\Vert _w^2$$, for some adequate w. In general, $$n\Vert R_n-R\Vert ^2$$ does not possess a weak limit unless we assume rather strong assumptions on F. This is because the derivations to study such a limit involve those of $$\sqrt{n}\{F_n^{-1}-F^{-1}\}$$ in $$L^2$$, and therefore, it inherits the same limitations (see, for example, del Barrio et al. 2000, 2005). A convenient way to overcome these difficulties is to consider, instead of $$\Vert \cdot \Vert ^2$$, the norm in $$L^2(w)$$, with $$w(u)=f^2(F^{-1}(u))$$. This weight function is taken for analytical convenience. It may seem a bit odd, because f (and hence F) is unknown in practical applications. Nevertheless, as stated in the Introduction, $$n\Vert R_n-R\Vert _w^2$$ could be used as a test statistic for testing goodness-of-fit to a location-scale family (1), and in such a case under the null hypothesis $$f(F^{-1}(u))=\frac{1}{\varsigma }f_0(F_0^{-1}(u))$$, so we can take $$w(u)=f_0(F_0^{-1}(u))$$ which is known in that testing framework. Later on, we will discuss other weight functions
We first show that, under some conditions, the linear approximation for $$\sqrt{n}\{R_{n}(u)-R(u)\}$$ given in Theorem 6 for a fixed $$u\in (0,1)$$, is valid for all u in certain intervals, in the $$L^2(w)$$ sense.
### Theorem 7
Suppose that f is twice continuously differentiable on (ab), that $${\mathbb {E}}(X^2)<\infty$$, $${\mathbb {E}}\{f'(X)^2\}<\infty$$, $${\mathbb {E}}\{f''(X)^2\}<\infty$$ and $${\mathbb {E}}\{X^2f''(X)^2\}<\infty$$, that $$\sqrt{n}({\hat{\sigma }}-\sigma )=O_P(1)$$, that K satisfies Assumption 1 (i), (iii) and (iv), and that h satisfies Assumption 2 (i) and (iii). Suppose also that $$f\left( F^{-1}(u)\right) >0$$, $$u \in (0,1)$$ and
\begin{aligned}\sup _{0< u < 1} u(1-u)\frac{\left| f'\left( F^{-1}(u)\right) \right| }{f^2\left( F^{-1}(u)\right) } \leqslant \gamma ,\end{aligned}
for some finite $$\gamma >0$$. Let
\begin{aligned} A=\lim _{x \downarrow a}f(x)<\infty . \end{aligned}
Suppose also that if $$A = 0$$ then f is nondecreasing on an interval to the right of a. Let $$w(u)=f^2(F^{-1}(u))$$, then
\begin{aligned} \sqrt{n}\left\{ R_{n}(u)-R(u)\right\}= & {} \frac{1}{\sqrt{n}}\sum _{i=1}^n Y_i(u)+r_n(u),\quad u \in {\mathbb {I}}_n=\left( 0,\frac{n-1}{n}\right] , \\ \int _{{\mathbb {I}}_n}r_n(u)^2w(u)du= & {} o_P(1). \end{aligned}
### Corollary 2
Under assumptions in Theorem 7, $$n\Vert R_n-R\Vert _w^2 {\mathop {\longrightarrow }\limits ^{{\mathcal {L}}}} \Vert Z\Vert ^2_w$$, where $$\{Z(u), 0 \leqslant u \leqslant 1\}$$ is a zero-mean Gaussian process on $$L^2(w)$$ with $$Z(1)=0$$ and covariance function $$cov \{Z(u),\, Z(s)\}={\mathbb {E}}\{ Y_1(u) \, Y_1(s) \}$$, $$u, s \in (0,1)$$.
From the proof of Theorem 7 and Theorem 4.6 (i) of del Barrio et al. (2005), the results in Theorem 7 and Corollary 2 keep on being true for any bounded weight function w satisfying
\begin{aligned}\int _0^1\frac{u(1-u)}{f^2(F^{-1}(u))}w(u)du<\infty , \quad \lim _{u\uparrow 1}\frac{(1-u)\int _u^1w(u)du}{f^2(F^{-1}(u))}=0.\end{aligned}
Although this result may seem more general than those stated in Theorem 7 and Corollary 2, notice that the choice of an adequate weight function requires a strong knowledge of f.
As observed after Theorem 6, the replacement of the population quantile function $$F^{-1}$$ with the empirical quantile function $$F_n^{-1}$$ in the expression of R to build the estimator $$R_n$$ is the responsible for the term $$\frac{\textbf{1}\{X_i \leqslant F^{-1}(u)\}-u }{f(F^{-1}(u))} \mu (u)$$ in the expression of $$Y_i(u)$$, which makes $$Y_i(u)$$ to inherit the properties and difficulties of estimating the quantile function. So, one may wonder if there is any advantage in using the right (left) shape function instead of the quantile function. From a theoretical point of view the answer is yes: in order to use the quantile function one must assume certain conditions on the right tail and on the left tail of the distribution, while if one uses right (left) shape function then only assumptions on the left (right) tail are necessary. This is because both $$R_n(u)$$ and R(u) (respectively, $$L_n(u)$$ and L(u)) go to 0 as $$u\uparrow 1$$ (respectively, $$u\downarrow 0$$).
### Remark 4
The properties studied in this section for $$R_n-R$$ are inherited by $$S_n-S$$. Specifically, let $$u \in (0,1)$$, then
\begin{aligned} (S_n(u)-z_{1-\alpha /2}{\hat{\varrho }}_S(u)/\sqrt{n}, \, S_n(u)+z_{1-\alpha /2}{\hat{\varrho }}_S(u)/\sqrt{n}) \end{aligned}
(4)
is an approximate (in the sense of asymptotic) confidence interval for S(u), where $${\hat{\varrho }}_S^2(u)$$ is the sample variance of $$V_1(u), \ldots , V_n(u)$$, where $$V_i(u)={\hat{W}}_{X,i}(u)-{\hat{W}}_{-X,i}(u)$$, $${\hat{W}}_{X,i}(u)$$ are the quantities defined in (3) calculated on the sample $$X_1, \ldots , X_n$$, and $${\hat{W}}_{-X,i}$$ are the quantities defined in (3) calculated on $$-X_1, \ldots , -X_n$$, $$i=1,\ldots ,n$$. Notice that if 0 does not belong to the confidence interval (4) one may conclude that the law of X is not symmetric.
## 4 Some numerical illustrations
### 4.1 Estimation of R and S
If X has a uniform distribution on the interval (ab), $$X\sim U(a,b)$$, then $$R(u)=0.5(1-u)^2$$ (see, Table 1 in Arriaza et al. 2019). For several values of n, we have generated 10,000 random samples with size n from a distribution U(0, 1). For each sample, we have estimated R using $$R_n$$ taking as K the Epanechnikov kernel (scaled so that it has variance 1) and $$h=sd\times n^{-\tau }$$, with $$\tau =0.35, 0.40, 0.45, 0.49$$, and $$sd^2$$ denoting the sample variance. Notice that for these choices of h Assumption 2 (i), (ii) and (iii) are met. Tables 1 and 2 show the value of R(u), the bias and the standard deviation of the values of $$R_n(u)$$, the mean of the standard deviation estimator $$\hat{\varrho }(u)/\sqrt{(}n)$$ (recall that it is based on asymptotic arguments), and the coverage of the confidence interval (2) calculated at the nominal level 95%, for $$u=0.1, \ldots , 0.9$$ and $$n=100, \,250, \,500, \,1000$$. Figure 1 displays the graph of 1000 estimations for $$\tau =0.45$$ in grey joint with the population shape function in black. Looking at these tables and figure we see that the bias and the variance become smaller as u approaches 1; the standard deviation estimator is, on average, a bit larger than the true standard deviation, specially for smaller sample sizes; the bias depends on the values of $$\tau$$ and u, being negative for smaller values of $$\tau$$ and for larger values of u; as for the coverage of the confidence interval, it also depends the values of $$\tau$$ and u: it is rather poor for smaller values of $$\tau$$ and larger values of u, this is because in such cases the value of the bias in relation to the standard deviation estimator is non-negligible. As expected from Theorem 3, the differences between $$R_n(u)$$ and R(u) become smaller as n increases, uniformly in u.
A similar experiment was carried out for the estimation of S, whose results are summarized in Tables 3 and 4 and Fig. 2. Looking a these tables and figure we see that the bias is quite small in all cases; that the variance becomes smaller as u approaches 1 and it also decreases with n; the standard deviation estimator is, on average, a bit larger than the true standard deviation, the differences become smaller as n increases; this fact provokes that the coverage of the confidence intervals is larger than the nominal value, specially for small sample sizes.
### 4.2 Glass fibre breaking strengths
As an illustration, we will analyse a real data set already considered in Arriaza et al. (2019) and which had been previously introduced by Smith and Naylor (1987). The set consists of 63 observations of the breaking strength of glass fibres of 1.5 cm of length collected by the National Physical Laboratory in England (for more details about the data set, see Smith and Naylor 1987). The left panel of Fig. 3 depicts the histogram and the kernel density estimator obtained from the data. As discussed in Arriaza et al. (2019), this explanatory analysis suggests a certain negative skewness in the distribution. The right panel of Fig. 3 displays the estimator $$S_n(u)$$, $$u \in (0,1)$$, with $${\hat{h}}=sd\times n^{-0.45}$$ and taking as K the Epanechnikov kernel (scaled so that it has variance 1). Other values for $${\hat{h}}$$ have been investigated and similar results were obtained. The graph also displays the confidence intervals in (4) for S(u) calculated at the nominal level 90% for $$u=0.1, \ldots , 0.9$$. The confidence intervals are also detailed in Table 5. Notice that the estimator of the function S tends to lie over the horizontal axis, which indicates asymmetry in the distribution. Moreover, the confidence intervals for S(0.1), S(0.2) and S(0.4) do not contain the zero. This conclusion is in agreement with Arriaza et al. (2019).
### 4.3 An application to testing goodness-of-fit
Now we consider the problem of testing goodness-of-fit to a uniform distribution, that is, we want to test
\begin{aligned} \begin{array}{ll} H_0: &{} X\sim U(a,b), \quad \text{ for } \text{ some } a<b, \,\, a, b \in {\mathbb {R}},\\ H_1: &{} X\not \sim U(a,b), \quad \text{ for } \text{ all } a<b, \,\, a, b \in {\mathbb {R}}, \end{array} \end{aligned}
on the basis of a sample of size n from X. Let $$f_0$$ and $$F_0$$ denote the probability density function and the cumulative distribution function of the U(0, 1) law, respectively. Then $$w(u)=f_0(F_0(u))=\textbf{1}\{0 \leqslant u \leqslant 1\}$$, and thus $$\Vert \cdot \Vert =\Vert \cdot \Vert _w$$. Let R denote the right shape function of X and let $$R_0$$ denote the right shape function of a uniform distribution, $$R_0(u)=0.5(1-u)^2$$. From Theorem 3, it follows that $$\Vert R_n-R_0\Vert$$ converges to $$\Vert R-R_0\Vert$$, which is equal to 0, under the null, and a positive quantity under alternatives. Thus, it seems reasonable to consider the test that rejects $$H_0$$ for large values of $$T_n=n\Vert R_n-R_0\Vert ^2$$. The critical region is $$T_n \geqslant t_{n,\alpha }$$, where $$t_{n,\alpha }$$ is the $$\alpha$$ upper percentile of the null distribution of $$T_n$$, whose value can be calculated by simulation by generating data from a U(0, 1) law, since the null distribution of $$T_n$$ does not depend on the values of a and b, but only on $$F_0$$. Notice that $$T_n$$ has the readily computable expression
\begin{aligned} T_n = \sum _{k=1}^{n-1}r_k^2+\frac{n}{20}-\frac{n}{3}\sum _{k=1}^{n-1}c_k r_k, \end{aligned}
where
\begin{aligned} r_k = \frac{1}{n}\sum _{i=k+1}^n (X_{i:n}-X_{k:n}){\hat{f}}_n(X_{i:n}) \quad \text { and } \quad c_k = \left( 1-(k-1)/n\right) ^3-(1-k/n)^3. \end{aligned}
There are many tests in the statistical literature for testing $$H_0$$ against $$H_1$$, and the objective of this section is not to provide an exhaustive list of such tests, but only to suggest a possible application of the results stated in the previous sections. In our view, this as well as other possible applications deserve further separate research, out of the scope of this manuscript. Nevertheless, since $$T_n$$ is closely related to the Wasserstein distance between $$F_n$$ and $$F_0$$, which is equal to the squared root of the $$L^2$$ norm of the difference between the empirical quantile function and the quantile function of $$F_0$$, we carried out a small simulation experiment in order to compare the powers of the newly proposed tests and the one based on the Wasserstein distance. To make the Wasserstein distance invariant with respect to location and scale changes, we consider as test statistic $$W_n=n\Vert F_n^{-1}-F_0^{-1}\Vert ^2/{\hat{\sigma }}^2$$, $${\hat{\sigma }}^2$$ denoting the sample variance, and reject the null hypothesis for large values of $$W_n$$ (see, del Barrio et al. 2000), The critical region is $$W_n \geqslant w_{n,\alpha }$$, where $$w_{n,\alpha }$$ is the $$\alpha$$ upper percentile of the null distribution of $$W_n$$, whose value can be calculated by simulation by generating data from a U(0, 1) law, since the null distribution of $$W_n$$ does not depend on the values of a and b, but only on $$F_0$$.
Table 6 displays the values $$t_{n,\alpha }$$ and $$w_{n,\alpha }$$ for $$n=30, \, 50$$ and $$\alpha =0.05$$, calculated by generating 100,000 samples. To calculate $$T_n$$ we used the Epanechnikov kernel (scaled so that it has variance 1) and $${\hat{h}}=sd \times n^{-\tau }$$, $$\tau =0.26,\, 0.30, \, 0.35, \, 0.40, \, 0.45, \, 0.49$$. As alternatives, we considered several members of the log-Lindley distribution, a family with support on (0,1) with a large variety of shapes (see Gómez-Déniz et al. 2014; Jodrá and Jiménez-Gamero 2016) and probability density function
\begin{aligned} f(x; \kappa , \lambda )=\frac{\kappa ^2}{1+\kappa \lambda }(\lambda -\log (x))x^{\kappa -1}, \quad x \in (0,1), \end{aligned}
for some $$\kappa >0$$ and $$\lambda \geqslant 0$$. Figure 4 represents the empirical power of the two tests for $$\kappa =1.5, \, 2, \, 2.5$$ and $$0.6 \leqslant \lambda \leqslant 5$$, calculated by generating 10,000 samples for each combination of the parameter values. Looking at Fig. 4 we see that, although according to Corollary 2 the asymptotic null distribution of $$T_n$$ does not depend on the value of $${\hat{h}}$$, for finite sample sizes it has an effect on the power of the test, being higher for larger values of $${\hat{h}}$$. As expected from the results in Janssen (2000), no test has the largest power against all alternatives. The power increases with the sample size.
## 5 Conclusions and further research
As seen in the Introduction, shape functions have shown to have interesting properties. From a practical point of view, estimators of these functions need to be proposed and studied. This paper has focused on nonparametric estimators of the shape functions. The proposed estimators have been studied both theoretically and numerically. They exhibit nice asymptotic properties and the numerical experiments show a reasonable practical behavior. Optimal choice of the smoothing parameter involved in the construction of the estimator has not been dealt with in this piece of research. This issue deserves more investigation and will be considered in future studies.
## 6 Proofs
This section sketches the proofs of the results stated in the previous sections. Along this section, M is a generic positive constant taking many different values throughout the proofs and $$f_n(x)$$ is defined as $${\hat{f}}_n(x)$$ with $${\hat{h}}={\hat{\sigma }} \times g(n)$$ replaced with $$h=\sigma \times g(n)$$.
### Lemma 1
Suppose that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, that K satisfies Assumption 1 (i) and that h satisfies Assumption 2 (i). Then $$\displaystyle \sup _x |{\hat{f}}_n(x)-f_n(x)| {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$.
### Proof
We have that
\begin{aligned} {\hat{f}}_n(x)-{f}_n(x)=\delta _1(x)+\delta _2(x), \end{aligned}
(5)
with
\begin{aligned} \delta _1(x)= & {} \left( \frac{1}{n{\hat{h}}}- \frac{1}{nh} \right) \sum _{j=1}^nK\left( \frac{X_j-x}{h}\right) , \end{aligned}
(6)
\begin{aligned} \delta _2(x)= & {} \frac{1}{n{\hat{h}}}\sum _{j=1}^n \left\{ K\left( \frac{X_j-X_i}{{\hat{h}}}\right) -K\left( \frac{X_j-x}{h}\right) \right\} . \end{aligned}
(7)
Since
\begin{aligned} \delta _1(x)=\frac{\sigma -{\hat{\sigma }}}{{\hat{\sigma }}}\frac{1}{nh}\sum _{j=1}^nK\left( \frac{X_j-x}{h}\right) =\frac{\sigma -{\hat{\sigma }}}{{\hat{\sigma }}}f_n(x), \end{aligned}
we can write
\begin{aligned} |\delta _1(x)|\leqslant \left| \frac{\sigma -{\hat{\sigma }}}{{\hat{\sigma }}}\right| \left\{ \sup _x |f_n(x)-{\mathbb {E}}\{f_n(x)\}|+\sup _x{\mathbb {E}}\{f_n(x)\} \right\} . \end{aligned}
(8)
Recall that if K is Lipschitz continuous and has compact support, then it has bounded variation. Under the assumptions made on K and h we have that (see, for example, the proof of Theorem 2.1.3 of Prakasa Rao 1983)
\begin{aligned} \sup _x |f_n(x)-{\mathbb {E}}\{f_n(x)\}| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(9)
Since f is bounded,
\begin{aligned} {\mathbb {E}}\{f_n(x)\}=\int _{-\infty }^{\infty }K(u)f(x-hu)du\leqslant M \int _{-\infty }^{\infty }K(u)du=M, \quad \forall x. \end{aligned}
(10)
Using (8), (9), (10) and that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, it follows that
\begin{aligned} \sup _x |\delta _1(x)| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(11)
Now we study $$\delta _2(x)$$. Since K is Lipschitz continuous and has compact support, $$\textrm{Sup}(K)$$, we have that
\begin{aligned}{} & {} D(x) = \frac{1}{n{\hat{h}}}\sum _{j=1}^n \left| K\left( \frac{X_j-x}{{\hat{h}}}\right) -K\left( \frac{X_j-x}{h}\right) \right| \\{} & {} \leqslant M \frac{\sigma }{{\hat{\sigma }}} \left| \frac{\sigma }{{\hat{\sigma }}} -1 \right| \frac{1}{nh} \sum _{j=1}^n \left| \frac{X_j-x}{h}\right| \textbf{1}\{ (X_j-x)/h\in \textrm{Sup}(K) \text{ or } (X_j-x)/{\hat{h}} \in \textrm{Sup}(K) \}. \end{aligned}
Since $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, it follows that for n large enough, $$\textbf{1}\{ (X_j-x)/h \in \textrm{Sup}(K) \text{ or } (X_j-x)/{\hat{h}} \in \textrm{Sup}(K) \} \subseteq \textbf{1}\{ |X_j-x|/h \leqslant C \}$$, for certain positive constant C. Therefore
\begin{aligned} D(x) \leqslant M \frac{\sigma }{{\hat{\sigma }}} \left| \frac{\sigma }{{\hat{\sigma }}} -1 \right| f_{U,n}(x), \end{aligned}
(12)
where $$f_{U,n}(x)$$ is the kernel estimator of f built by using as kernel the probability density function of the uniform law on the interval $$[-C,C]$$. Proceeding as before, we have that
\begin{aligned} \sup _x f_{U,n}(x) \leqslant M \quad {a.s.}, \end{aligned}
(13)
Using that $$|\delta _2(x)|\leqslant D(x)$$, (12), (13) and that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, it follows that
\begin{aligned} \sup _x |\delta _2(x)| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(14)
The result follows from (5), (11) and (14). $$\square$$
### Remark 5
From the previous proof, notice that if in the statement of Lemma 1 the assumption $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma$$ is replaced with $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{P}} \sigma$$ then it is concluded that $$\displaystyle \sup _x |{\hat{f}}_n(x)-f_n(x)| {\mathop {\longrightarrow }\limits ^{P}} 0$$.
### Proof of Theorem 1
We have that
\begin{aligned} R_n(u)=R_{1n}(u)+R_{2n}(u)+R_{3n}(u)+R_{4n}(u), \end{aligned}
(15)
where
\begin{aligned} R_{1n}(u)= & {} \frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+f(X_i), \\ R_{2n}(u)= & {} \frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+\left\{ f_n(X_i)-f(X_i)\right\} , \\ R_{3n}(u)= & {} \frac{1}{n}\sum _{i=1}^n\left[ \{X_i-F^{-1}_n(u)\}^+-\{X_i-F^{-1}(u)\}^+\right] {\hat{f}}_n(X_i), \\ R_{4n}(u)= & {} \frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+\left\{ {\hat{f}}_n(X_i)-f_n(X_i)\right\} . \\ \end{aligned}
From the SLLN, it follows that
\begin{aligned} R_{1n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} R(u). \end{aligned}
(16)
As for $$R_{2n}(u)$$, we have that
\begin{aligned} \left| R_{2n}(u) \right| \leqslant \sup _x |f_n(x)-f(x)| \left\{ \frac{1}{n}\sum _{i=1}^n|X_i|+|F^{-1}(u)|\right\} . \end{aligned}
Under the assumptions made (see, for example, Theorem 2.1.3 in Prakasa Rao 1983)
\begin{aligned} \sup _x |f_n(x)-f(x)| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(17)
Taking into account that $$\frac{1}{n}\sum _{i=1}^n|X_i| {\mathop {\longrightarrow }\limits ^{a.s.}} {\mathbb {E}}|X|<\infty$$, it follows that
\begin{aligned} R_{2n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(18)
In order to study $$R_{3n}(u)$$ we first observe that
\begin{aligned}{} & {} \left\{ X_i-F_n^{-1}(u) \right\} ^+-\left\{ X_i-F^{-1}(u) \right\} ^+\nonumber \\ {}{} & {} \quad = \left\{ \begin{array}{lll} 0 &{} \text{ if } &{} X_i \leqslant \min \{F_n^{-1}(u), F^{-1}(u)\},\\ X_i-F_n^{-1}(u) &{} \text{ if } &{} F_n^{-1}(u)<X_i\leqslant F^{-1}(u),\\ -X_i+F^{-1}(u) &{} \text{ if } &{} F^{-1}(u)<X_i \leqslant F^{-1}_n(u),\\ F^{-1}(u)-F_n^{-1}(u) &{} \text{ if } &{} X_i > \max \{F_n^{-1}(u), F^{-1}(u)\}, \end{array} \right. \end{aligned}
(19)
which implies
\begin{aligned} \left| \left\{ X_i-F_n^{-1}(u) \right\} ^+-\left\{ X_i-F^{-1}(u) \right\} ^+ \right| \leqslant \left| F^{-1}(u)-F_n^{-1}(u) \right| , \quad \forall i, \end{aligned}
and therefore
\begin{aligned} \left| R_{3n}(u) \right| \leqslant |F_n^{-1}(u)-F^{-1}(u)| \frac{1}{n}\sum _{i=1}^n {\hat{f}}_n(X_i). \end{aligned}
Under the assumptions made $$|F_n^{-1}(u)-F^{-1}(u)| {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$ (see, for example display (1.4.9) in Csörgő 1983). We also have that
\begin{aligned} \frac{1}{n} \sum _{i=1}^n {\hat{f}}_n(X_i) \leqslant \frac{1}{n}\sum _{i=1}^nf(X_i)+ \sup _x |f_n(x)-f(x)|+ \sup _x |{\hat{f}}_n(x)-f_n(x)|. \end{aligned}
(20)
From (17), Lemma 1 and taking into account that f is bounded, it follows that the right hand-side of inequality (20) is bounded a.s. Therefore,
\begin{aligned} R_{3n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(21)
We have that
\begin{aligned} R_{4n}(u)=R_{41n}(u)+R_{42n}(u), \end{aligned}
(22)
where
\begin{aligned} R_{41n}(u)=\frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+\delta _1(X_i) \end{aligned}
and
\begin{aligned} R_{42n}(u)=\frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+\delta _2(X_i), \end{aligned}
where $$\delta _1(x)$$ and $$\delta _2(x)$$ are as defined in (6) and (7), respectively. Since
\begin{aligned} R_{41n}(u)=(\sigma -{\hat{\sigma }})\{R_{1n}(u)+R_{2n}(u)\}/{\hat{\sigma }} \end{aligned}
(23)
and $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma$$, from (16) and (18), it follows that
\begin{aligned} R_{41n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(24)
Using (12), we get that
\begin{aligned} |R_{42n}(u)| \leqslant M \frac{\sigma }{{\hat{\sigma }}} \left| \frac{\sigma }{{\hat{\sigma }}} -1 \right| R_{U,n}(u), \end{aligned}
(25)
where
\begin{aligned} R_{U,n}(u)=\frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+f_{U,n}(X_i). \end{aligned}
Proceeding as before, it can be seen that $$R_{U,n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} R(u)$$, and hence
\begin{aligned} R_{42n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(26)
The result follows from (15), (16), (18), (21), (22), (24) and (26). $$\square$$
### Proof of Theorem 2
Let us consider decomposition (15). We have that $$R_{1n}(u)+R_{2n}(u)=T_{1n}(u)+T_{2n}(u)$$ with
\begin{aligned} T_{1n}(u)=K(0)\frac{1}{n^2h}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+, \end{aligned}
and $$T_{2n}(u)=\frac{n-1}{n}U_n(u)$$, where $$U_n(u)=\frac{1}{n(n-1)}\sum _{i \ne j}H_n(X_i,X_j;u)$$ is a degree two U-statistic with symmetric kernel
\begin{aligned} H_n(X_i,X_j; u)=\frac{1}{2} \left[ \{X_i-F^{-1}(u)\}^+ + \{X_j-F^{-1}(u)\}^+\right] \frac{1}{h}K\left( \frac{X_i-X_j}{h}\right) . \end{aligned}
We first see that
\begin{aligned} T_{1n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(27)
Notice that under the assumptions made,
\begin{aligned} {\mathbb {E}}\left[ \{X-F^{-1}(u)\}^+\right] <\infty , \end{aligned}
(28)
therefore, by the SLLN,
\begin{aligned} \frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+ {\mathop {\longrightarrow }\limits ^{a.s.}} {\mathbb {E}}\left[ \{X-F^{-1}(u)\}^+\right] <\infty . \end{aligned}
Finally, since K is bounded and $$nh \rightarrow \infty$$, (27) follows.
Next we will see that
\begin{aligned} T_{2n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} R(u). \end{aligned}
(29)
With this aim, we first calculate $${\mathbb {E}}\{U_n(u)\}$$.
\begin{aligned} {\mathbb {E}}\{U_n(u)\}={\mathbb {E}}\{H_n(X_1, X_2; u)\}=\int \{x-F^{-1}(u)\}^+ \left( \int K(y)f(x-hy)dy\right) f(x)dx. \end{aligned}
Since $$\int K(y)f(x-hy)dy \rightarrow f(x)$$, $$\{x-F^{-1}(u)\}^+ \left( \int K(y)f(x-hy)dy\right) f(x) \leqslant M \{x-F^{-1}(u)\}^+$$ and (28), by dominated convergence theorem we have that $${\mathbb {E}}\{U_n(u)\} \rightarrow R(u)$$.
Now, let $$A_n(x; u)={\mathbb {E}}\left\{ H_n(x,X;u) \right\}$$ and $$\tau =\int x^2K(x)dx$$. Routine calculations show that
\begin{aligned} A_n(x; u)= & {} A(x; u) +a_n(x; u),\\ A(x; u)= & {} \{x-F^{-1}(u)\}^+ f(x),\\ a_n(x; u)= & {} 0.5\tau h^2f''(x)\{x-F^{-1}(u)\}^+ +0.5 \tau h^2f'(x)+o(h^2). \end{aligned}
Let $$\varepsilon >0$$. From the stated assumptions $${\mathbb {E}}\{a_n(X, u)^2\} = O(h^4)$$, therefore
\begin{aligned} P\left( \frac{1}{n}\sum _{i=1}^n a_n(X_i; u)>\varepsilon \right) \leqslant \frac{1}{\varepsilon ^2}{\mathbb {E}}\left[ \left\{ \frac{1}{n}\sum _{i=1}^n a_n(X_i; u)\right\} ^2\right] =O(h^4). \end{aligned}
Since $$nh^4 \rightarrow 0$$, it follows that
\begin{aligned} \sum _{n \geqslant 1} P\left( \frac{1}{n}\sum _{i=1}^n a_n(X_i; u)>\varepsilon \right) <\infty , \end{aligned}
which implies that
\begin{aligned} \frac{1}{n}\sum _{i=1}^n a_n(X_i; u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
Let $$B_n(u)=U_n(u)+ {\mathbb {E}}\{H_n(X_1,X_2; u)\}-\frac{2}{n}\sum _{i=1}^nA_n(X_i; u)$$. It can be seen that $${\mathbb {E}} \left\{ B_n(u)^2 \right\} =O(1/n^2\,h).$$ Reasoning as before, we get that
\begin{aligned} B_n(u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
Summarizing,
\begin{aligned}T_{2n}(u)=\frac{2}{n}\sum _{i=1}^nA(X_i; u)-R(u)+t_{2n}(u),\end{aligned}
with $$t_{2n}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$. By the SLLN we have that $$\frac{1}{n}\sum _{i=1}^nA(X_i; u) {\mathop {\longrightarrow }\limits ^{a.s.}} R(u)$$, and thus (29) is proven.
Finally, proceeding as in the proof of Theorem 1, one gets that $$R_{in}(u) {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$, $$i=3,4$$. This completes the proof. $$\square$$
### Lemma 2
Let f be a probability density function with finite support, (ab), continuous in (ab). Suppose also that K satisfies Assumption 1 (ii) and that h satisfies Assumption 2 (ii). Then $$\displaystyle \sup _{x \in (a,b)}|f_n(x)-f(x)|{\mathop {\longrightarrow }\limits ^{a.s.}} 0$$.
### Proof
From the proof of Theorem 2.1.3 in Prakasa Rao (1983), we have that
\begin{aligned} \sup _{x}|f_n(x)-{\mathbb {E}}\{f_n(x)\}| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
So, it suffices to see that
\begin{aligned} \sup _{x \in (a,b)}|{\mathbb {E}}\{f_n(x)\}-f(x)|{\longrightarrow } 0. \end{aligned}
(30)
We have that
\begin{aligned}{\mathbb {E}}\{f_n(x)\}=\int K(y)f(x-hy)dy.\end{aligned}
From the continuity of f, for each fixed $$y \in {\mathbb {R}}$$
\begin{aligned} f(x+hy) \rightarrow f(x), \quad \forall x \in (a,b), \end{aligned}
and
\begin{aligned} K(y) f(x+hy) \leqslant M K(y).\end{aligned}
Thus, (30) holds true by applying dominated convergence theorem. $$\square$$
### Proof of Theorem 3
First of all, we see that, under the assumptions made, R is a continuous function on [0, 1]. Since the function
\begin{aligned} G:&{\mathbb {R}}\times (0,1)&\mapsto {\mathbb {R}}\\&(x,u)&\mapsto G(x,u)= \{x-F^{-1}(u)\}^+f^2(x) \end{aligned}
is continuous, it follows that $$R(u)=\int G(x,u)dx$$ is continuous on (0, 1). Recall that $$\displaystyle \lim _{u \rightarrow 1^-}R(u)=0=R(1)$$ and, if the limit exists, $$R(0)=\displaystyle \lim _{u \rightarrow 0^+}R(u)$$. Thus, R is continuous at $$u=1$$. To see that it is also continuous at $$u=0$$ it suffices to see that the limit $$\displaystyle \lim _{u \rightarrow 0^+}R(u)$$ exits, which is true since
\begin{aligned}{} & {} \lim _{u ^\rightarrow 0+} \{x-F^{-1}(u)\}^+f^2(x)=(x-a)f^2(x), \\{} & {} \{x-F^{-1}(u)\}^+f^2(x) \leqslant c(x)=M(x-a)\textbf{1}\{a \leqslant x \leqslant b\} \end{aligned}
and $$\int c(x)dx<\infty$$, then by dominated convergence theorem
\begin{aligned} \lim _{u \rightarrow 0^+}R(u)= \int \{x-a\}^+f^2(x)dx<\infty . \end{aligned}
Next, we consider decomposition (15) and study each term on the right hand-side of such expression. Since R is a continuous function on [0, 1], the point by point convergence of $$R_{1n}(u)$$ to R(u) implies the uniform convergence on [0, 1], that is
\begin{aligned} \sup _{0 \leqslant u \leqslant 1} \left| R_{1n}(u)-R(u)\right| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(31)
As for $$R_{2n}(u)$$, taking into account that $$\frac{1}{n}\sum _{i=1}^n \left\{ X_i-F^{-1}(u)\right\} ^+ \leqslant {\bar{X}}-a$$, with $${\bar{X}}=(1/n)\sum _{i=1}^n X_i$$, we have that
\begin{aligned} \sup _{0 \leqslant u \leqslant 1} \left| R_{2n}(u) \right| \leqslant \sup _{x \in [a,b]} |f_n(x)-f(x)| ({\bar{X}}-a). \end{aligned}
From Lemma 2, taking into account that Assumption 1 (i) implies Assumption 1 (ii) and that $${\bar{X}}-a {\mathop {\longrightarrow }\limits ^{a.s.}} {\mathbb {E}}(X)-a < \infty$$, we conclude that
\begin{aligned} \sup _{0 \leqslant u \leqslant 1} \left| R_{2n}(u) \right| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(32)
From (31) and (32), we conclude that
\begin{aligned} \sup _{0 \leqslant u \leqslant 1} \left| R_{1n}(u)+R_{2n}(u)-R(u)\right| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(33)
For $$R_{3n}(u)$$ we have that
\begin{aligned} \sup _{0 \leqslant u \leqslant 1}\left| R_{3n}(u) \right| \leqslant \sup _{0 \leqslant u \leqslant 1} |F_n^{-1}(u)-F^{-1}(u)| \frac{1}{n}\sum _{i=1}^n{\hat{f}}_n(X_i). \end{aligned}
In the proof of Theorem 1 we saw that $$(1/n)\sum _{i=1}^n {\hat{f}}_{n}(X_i)$$ is bounded a.s. Under the assumptions made (see, for example, p. 6 of Csörgő 1983),
\begin{aligned} \sup _{0 \leqslant u \leqslant 1} |F_n^{-1}(u)-F^{-1}(u)| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
Therefore
\begin{aligned} \sup _{0 \leqslant u \leqslant 1} \left| R_{3n}(u) \right| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
Finally, taking into account decomposition (22), it suffices to show that $$\sup _{0 \leqslant u \leqslant 1} \left| R_{4in}(u) \right| {\mathop {\longrightarrow }\limits ^{a.s.}} 0,$$ $$i=1,2$$. Using (23), (33) and that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, one gets that
\begin{aligned} \sup _{0 \leqslant u \leqslant 1} \left| R_{41n}(u)\right| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
Using (25), (33) and that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, one similarly gets that
\begin{aligned} \sup _{0 \leqslant u \leqslant 1} \left| R_{42n}(u)\right| {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
This concludes the proof. $$\square$$
### Proof of Theorem 4
First of all, we see that, under the assumptions made, $$R \in L^2$$. We have that
\begin{aligned} \int _0^1R^2(u)du= & {} \int _0^1\left( \int _{F^{-1}(u)}^{\infty }\{x-F^{-1}(u)\}f^2(x)dx\right) \\{} & {} \times \left( \int _{F^{-1}(u)}^{\infty }\{y-F^{-1}(u)\}f^2(y)dy \right) du=I_1-2I_2+I_3, \end{aligned}
with (recall that f is a bounded function and that $$\int _0^1 F^{-1}(u)^2 du={\mathbb {E}}(X^2)$$)
\begin{aligned} I_1= & {} \int _0^1\left( \int _{F^{-1}(u)}^{\infty }xf^2(x)dx\right) ^2du \leqslant M {\mathbb {E}}(X^2)<\infty ,\\ I_2= & {} \int _0^1F^{-1}(u) \int _{F^{-1}(u)}^{\infty }xf^2(x)dx du \leqslant M\left( \int _0 ^1F^{-1}(u)^2du \int x^2 f(x)dx \right) ^{1/2}\\= & {} M {\mathbb {E}}(X^2)<\infty ,\\ I_3= & {} \int _0^1 F^{-1}(u)^2 \left( \int _{F^{-1}(u)}^{\infty }f^2(x)dx\right) ^2du \leqslant M \int _0^1 F^{-1}(u)^2 du=M{\mathbb {E}}(X^2) <\infty , \end{aligned}
and thus $$R \in L^2$$.
Next, we consider decomposition (15) and study each term on the right hand-side of such expression. Since $$R_{1n}$$ is an average of integrable i.i.d. random elements whose expectation is R(u), applying the SLLN in Hilbert spaces we obtain that (see, for example Theorem 2.4 of Bosq 2000),
\begin{aligned} R_{1n} {\mathop {\longrightarrow }\limits ^{a.s.}} R, \end{aligned}
in $$L^2$$. As for $$R_{2n}(u)$$, we have that
\begin{aligned} \left| R_{2n}(u) \right| \leqslant \sup _x |f_n(x)-f(x)| \frac{1}{n}\sum _{i=1}^n \{X_i-F^{-1}(u)\}^+. \end{aligned}
Let $$V(u)={\mathbb {E}}[\{X_i-F^{-1}(u)\}^+]$$. A parallel reasoning to that used to prove that $$R \in L^2$$ shows that $$V \in L^2$$. Now, from the SLLN in Hilbert spaces it follows that $$\frac{1}{n}\sum _{i=1}^n \{X_i-F^{-1}(u)\}^+ {\mathop {\longrightarrow }\limits ^{a.s.}} V$$, in $$L^2$$. This fact and (17) gives
\begin{aligned} R_{2n} {\mathop {\longrightarrow }\limits ^{a.s.}} 0, \end{aligned}
in $$L^2$$. From (15), (31) and (32), we conclude that
\begin{aligned} \Vert R_{1n}+R_{2n}-R \Vert {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
(34)
For $$R_{3n}(u)$$ we have that
\begin{aligned} \left| R_{3n}(u) \right| \leqslant |F_n^{-1}(u)-F^{-1}(u)| \frac{1}{n}\sum _{i=1}^n{\hat{f}}_n(X_i). \end{aligned}
As shown in the proof of Theorem 1, the second factor on the right hand-side of the above inequality is a.s. bounded. Since $${\mathbb {E}}(X^2)<\infty$$, from Lemma 8.3 in Bickel and Freedman (1981), it follows that $$\Vert F_n^{-1}-F^{-1}\Vert {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$, and thus $$R_{3n} {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$, in $$L^2$$. Finally, taking into account decomposition (22), using (23), (25) (34) and that $${\hat{\sigma }} {\mathop {\longrightarrow }\limits ^{a.s.}} \sigma >0$$, one also gets that $$R_{4n} {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$, in $$L^2$$. This concludes the proof. $$\square$$
### Proof of Theorem 5
Similar developments to those made in the proof of Theorem 2, and sharing the notation used there, show that
\begin{aligned}R_{1n}(u)+R_{2n}(u)=\frac{2}{n}\sum _{i=1}^nA(X_i; u)-R(u)+t_{2n}(u),\end{aligned}
with $$t_{2n} {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$ in $$L^2(w)$$. By the SLLN in Hilbert spaces we have that $$\frac{1}{n}\sum _{i=1}^nA(X_i; \cdot ) {\mathop {\longrightarrow }\limits ^{a.s.}} R$$, in $$L^2(w)$$. Finally, proceeding as in the proof of Theorem 4 we get that $$\Vert R_{in} \Vert _w {\mathop {\longrightarrow }\limits ^{a.s.}} 0$$, $$i=3,4$$, which completes the proof. $$\square$$
### Proof of Theorem 6
Similar developments to those made in the proof of Theorem 2, and sharing the notation used there, show that
\begin{aligned} \sqrt{n}\left\{ R_{1n}(u)+R_{2n}(u)-R(u)\right\} =\frac{2}{\sqrt{n}}\sum _{i=1}^n\left[ \{X_i-F^{-1}(u)\}^+ f(X_i) -R(u)\right] +o_P(1). \nonumber \\ \end{aligned}
(35)
Now, taking into account (19) we can write
\begin{aligned} \sqrt{n}R_{3n}(u)=T_{1n}(u)+T_{2n}(u)+T_{3n}(u), \end{aligned}
(36)
where
\begin{aligned} T_{1n}(u)= & {} \frac{1}{\sqrt{n}}\sum _{i=1}^n\left\{ X_i-F_n^{-1}(u) \right\} {\hat{f}}_n(X_i)\textbf{1}\{F_n^{-1}(u)<X_i\leqslant F^{-1}(u)\}, \\ T_{2n}(u)= & {} -\frac{1}{\sqrt{n}}\sum _{i=1}^n \left\{ X_i-F_n^{-1}(u) \right\} {\hat{f}}_n(X_i)\textbf{1}\{F^{-1}(u)<X_i \leqslant F^{-1}_n(u)\},\\ T_{3n}(u)= & {} \sqrt{n} \left\{ F^{-1}(u)- F^{-1}_n(u) \right\} \frac{1}{n}\sum _{i=1}^n {\hat{f}}_n(X_i)\textbf{1}\{F^{-1}(u)<X_i \}. \end{aligned}
We have that
\begin{aligned} 0 \leqslant T_{1n}(u) \leqslant \left| \sqrt{n} \left\{ F^{-1}(u)- F^{-1}_n(u) \right\} \right| \frac{1}{n}\sum _{i=1}^n {\hat{f}}_n(X_i) \textbf{1}\{F_n^{-1}(u) <X_i\leqslant F^{-1}(u)\}. \end{aligned}
Under the assumptions made, $$\sqrt{n} \left\{ F^{-1}_n(u)- F^{-1}(u) \right\} =O_P(1)$$ and $$F^{-1}_n(u) {\mathop {\longrightarrow }\limits ^{a.s.}} F^{-1}(u)$$, which implies that for each $$\varepsilon >0$$ there exists $$n_0$$ such that $$\textbf{1}\{F_n^{-1}(u)<X_i\leqslant F^{-1}(u)\} \leqslant \textbf{1}\{F^{-1}(u)-\varepsilon <X_i\leqslant F^{-1}(u)\}$$, $$\forall n \geqslant n_0$$, and hence $${\mathbb {E}}[ \textbf{1}\{F_n^{-1}(u)<X_i\leqslant F^{-1}(u)\} ]=O(\varepsilon )$$. Proceeding as in the proof of Theorem 1, it can be seen that $$(1/n)\sum _{i=1}^n {\hat{f}}_n(X_i)=O_P(1)$$. Therefore $$T_{1n}(u) {\mathop {\longrightarrow }\limits ^{P}} 0$$. Analogously, it can seen that $$T_{2n}(u) {\mathop {\longrightarrow }\limits ^{P}} 0$$ and that $$\frac{1}{n}\sum _{i=1}^n f_n(X_i)\textbf{1}\{F^{-1}(u)<X_i \leqslant F_n^{-1}(u)\}=o_P(1)$$. Now, proceeding as in the proof of Theorem 2, it can be seen that
\begin{aligned} \frac{1}{n}\sum _{i=1}^n f_n(X_i)\textbf{1}\{F^{-1}(u)<X_i\} {\mathop {\longrightarrow }\limits ^{P}} \mu (u)={\mathbb {E}}\left[ f(X)\textbf{1}\{F^{-1}(u)<X\} \right] . \end{aligned}
(37)
From Lemma 1,
\begin{aligned} \left| \frac{1}{n}\sum _{i=1}^n \left\{ {\hat{f}}_n(X_i)-f_n(X_i)\right\} \textbf{1}\{F^{-1}(u)<X_i\} \right| \leqslant \sup _x |{\hat{f}}_n(x)-f_n(x)|{\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}
We also have that (see, e.g. Theorem 2.5.2 in Serfling 1980)
\begin{aligned} \sqrt{n} \left\{ F^{-1}(u)- F^{-1}_n(u) \right\} = \frac{1}{\sqrt{n}}\sum _{i=1}^n\frac{\textbf{1}\{X_i \leqslant F^{-1}(u)\} -u}{f(F^{-1}(u))}+o_P(1). \end{aligned}
Thus, by Slutsky theorem,
\begin{aligned} T_{3n}(u)=\frac{1}{\sqrt{n}}\sum _{i=1}^n \frac{\textbf{1}\{X_i \leqslant F^{-1}(u)\}-u }{f(F^{-1}(u))} {\mathbb {E}} \left[ f(X)\textbf{1}\{F^{-1}(u)<X \}\right] +o_P(1). \end{aligned}
Therefore, it has been shown that
\begin{aligned} \sqrt{n}\{R_{1n}(u)+R_{2n}(u)+R_{3n}(u)-R(u)\}=\frac{1}{\sqrt{n}} \sum _{i=1}^n Y_i(u)+o_P(1). \end{aligned}
(38)
To prove the result, it remains to see that $$\sqrt{n} R_{4n}(u)=o_P(1)$$. Recall decomposition (22). From (23) and (35), it follows that
\begin{aligned} \sqrt{n}R_{41n}(u)=-\sqrt{n}\{{\hat{\sigma }}-\sigma \}R(u)/\sigma +o_P(1). \end{aligned}
(39)
Now we study $$R_{42n}(u)$$. A Taylor expansion of $$K((X_j-X_i)/{{\hat{h}}})$$ around $$K\big ((X_j-X_i)/{{h}}\big )$$ gives,
\begin{aligned} \sqrt{n}R_{42n}(u)= & {} \frac{\sigma }{{\hat{\sigma }}^2} \sqrt{n}(\sigma -{\hat{\sigma }}) \frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+ \\{} & {} \times \frac{1}{n{h}} \sum _{j=1}^n \frac{X_j-X_i}{h}K'\left( \frac{X_j-X_i}{h}\right) +Q_n(u), \end{aligned}
where
\begin{aligned} Q_n(u)= & {} \frac{1}{2\sqrt{n}}\frac{\sigma }{{\hat{\sigma }}^3} \left\{ \sqrt{n}(\sigma -{\hat{\sigma }})\right\} ^2 \frac{1}{n}\sum _{i=1}^n\{X_i-F^{-1}(u)\}^+ \\{} & {} \times \frac{1}{n{h}} \sum _{j=1}^n \left( \frac{X_j-X_i}{{h}}\right) ^2 K''\left( \frac{X_j-X_i}{\tilde{h}}\right) , \end{aligned}
where $$\tilde{h}=\alpha h + (1-\alpha ) {\hat{h}}$$, for some $$\alpha \in (0,1)$$. The assumptions made imply that $$Q_n(u)=o_P(1)$$. Now proceeding as in the proof of Theorem 2, and taking into account that $$\int u K'(u)f(x+hu)du\rightarrow f(x)\int u K'(u)du=-f(x)$$, we obtain that
\begin{aligned} \sqrt{n}R_{42n}(u)=\sqrt{n}({\hat{\sigma }}-\sigma )R(u)/\sigma +o_P(1). \end{aligned}
(40)
Finally, the result follows from (38), (39) and (40). $$\square$$
### Proof of Theorem 7
Similar developments to those made in the proof of Theorem 2, and sharing the notation used there, show that
\begin{aligned} \sqrt{n}\left\{ R_{1n}(u)+R_{2n}(u)-R(u)\right\}= & {} \frac{2}{\sqrt{n}}\sum _{i=1}^n\left[ \{X_i-F^{-1}(u)\}^+ f(X_i) -R(u)\right] \\{} & {} +r_{1n}(u), \quad u \in (0,1),\end{aligned}
with $$\int _0^1r_{1n}(u)^2w(u)du {\mathop {\longrightarrow }\limits ^{P}} 0$$.
As for $$R_{3n}(u)$$, we consider decomposition (36). In the proof of Theorem 6 it was shown that $$T_{1n}(u) {\mathop {\longrightarrow }\limits ^{P}} 0$$, for each $$u \in (0,1)$$. We have that
\begin{aligned} 0 \leqslant T_{1n}(u) \leqslant \left| \sqrt{n}\left\{ F^{-1}_n(u)- F^{-1}(u) \right\} \right| \,\frac{1}{n}\sum _{i=1}^n {\hat{f}}_n(X_i). \end{aligned}
From Theorems 2.1, 3.1.1 and 3.2.1 in Csörgő (1983), it follows that
\begin{aligned} n\int _0^{(n-1)/n}\left\{ F^{-1}_n(u)- F^{-1}(u) \right\} ^2w(u)du=O_P(1). \end{aligned}
\begin{aligned} 0 \leqslant \frac{1}{n}\sum _{i=1}^n {\hat{f}}_n(X_i) {\mathop {\longrightarrow }\limits ^{P}} {\mathbb {E}}\{f(X)\}<\infty . \end{aligned}
Thus, by dominated convergence theorem, it follows that
\begin{aligned} \int _0^{(n-1)/n} T_{1n}(u)^2w(u)du {\mathop {\longrightarrow }\limits ^{P}} 0. \end{aligned}
Analogously it can be seen that
\begin{aligned} \int _0^{(n-1)/n} T_{2n}(u)^2w(u)du {\mathop {\longrightarrow }\limits ^{P}} 0. \end{aligned}
From Theorems 2.1 and 3.2.1 in Csörgő (1983), it follows that
\begin{aligned} \sup _{0<u\leqslant \frac{n-1}{n}}\left| \sqrt{n}\left\{ F^{-1}_n(u)- F^{-1}(u) \right\} f(F^{-1}(u))- \frac{1}{\sqrt{n}} \sum _{i=1}^n \Big [ \textbf{1}\{X_i \leqslant F^{-1}(u)\}-u \Big ] \right| {\mathop {\longrightarrow }\limits ^{a.s.}}0. \end{aligned}
Notice that the convergence in (37) holds for each $$u \in [0,1]$$. Since $$\mu (u)$$ is a continuous function, it follows that such convergence is uniform on the interval [0, 1]. Therefore,
\begin{aligned} \int _0^{(n-1)/n} \left\{ T_{3n}(u)-\frac{1}{\sqrt{n}} \sum _{i=1}^n \frac{\textbf{1}\{X_i \leqslant F^{-1}(u)\}-u }{f(F^{-1}(u))} \mu (u) \right\} ^2w(u)du {\mathop {\longrightarrow }\limits ^{P}} 0. \end{aligned}
Summarizing, it has been shown that
\begin{aligned} \sqrt{n}\{R_{1n}(u)+{} & {} R_{2n}(u)+R_{3n}(u)-R(u)\}=\frac{1}{\sqrt{n}} \sum _{i=1}^n Y_i(u)+r_{2n}(u), \\{} & {} \int _0^{(n-1)/n}r_{2n}(u)^2w(u)du {\mathop {\longrightarrow }\limits ^{P}} 0. \end{aligned}
The same steps given in the proof of Theorem 6 to show that $$\sqrt{n}R_{4n}(u)=o_P(1)$$ can be used to see that $$\int _0^{(n-1)/n}n R_{4n}(u)^2w(u)du {\mathop {\longrightarrow }\limits ^{P}} 0$$. This completes the proof. $$\square$$
### Proof of Corollary 2
Define
\begin{aligned} {\widetilde{Y}}_i(u)=\left\{ \begin{array}{ll} Y_i(u) &{} \displaystyle \text{ if } u \in {\mathbb {I}}_n=\left( 0, \, (n-1)/n \right] ,\\ 0 &{} \displaystyle \text{ if } u \in \left( (n-1)/n, \, 1 \right] , \end{array} \right. \end{aligned}
$$1 \leqslant i \leqslant n$$, and $$W_n(u)=\frac{1}{\sqrt{n}} \sum _{i=1}^n {\widetilde{Y}}_i(u)$$, $$u \in [0,1]$$.
From Theorem 7, we have that
\begin{aligned} n\int _0^1\left\{ R_n(u)-R(u)\right\} ^2w(u)du= & {} \int _0^1 W_n(u)^2w(u)du \nonumber \\{} & {} +n\int _{(n-1)/n}^1R(u)^2w(u)du+o_P(1). \end{aligned}
(41)
We first see that $${\widetilde{Y}}_1 \in L^2(w)$$. With this aim we write $${\widetilde{Y}}_1(u)={\widetilde{Y}}_{11}(u)-{\widetilde{Y}}_{12}(u)+{\widetilde{Y}}_{13}(u)$$, with
\begin{aligned} {\widetilde{Y}}_{11}(u)=2 \{X_1-F^{-1}(u)\}^+ f(X_1)\textbf{1}(u \in {\mathbb {I}}_n), \qquad {\widetilde{Y}}_{12}(u)=2R(u)\textbf{1}(u \in {\mathbb {I}}_n), \end{aligned}
and
\begin{aligned} {\widetilde{Y}}_{13}(u)=\frac{\textbf{1}\{X_i \leqslant F^{-1}(u)\}-u }{f(F^{-1}(u))} \mu (u) \textbf{1}(u \in {\mathbb {I}}_n). \end{aligned}
In the proof of Theorem 4 we saw that R, $$\{X_1-F^{-1}(u)\}^+ \in L^2$$; since f is bounded, we also have that $${\widetilde{Y}}_{11}, \, {\widetilde{Y}}_{12} \in L^2(w)$$. As for $${\widetilde{Y}}_{13}$$,
\begin{aligned} \int _0^1{\widetilde{Y}}_{13}^2(u)w(u)du=\int _{{\mathbb {I}}_n}\left[ \textbf{1}\{X_1 \leqslant F^{-1}(u)\}-u \right] ^2\mu ^2(u)du \leqslant \int _0^1\mu ^2(u)du<\infty , \end{aligned}
because $$\mu (u) \leqslant M$$, $$\forall u \in [0,1]$$, since f is bounded. Thus, $${\widetilde{Y}}_1 \in L^2(w)$$.
From the central limit theorem in Hilbert spaces and the continuous mapping theorem,
\begin{aligned} \int _0^1W_n(t)^2w(u)du {\mathop {\longrightarrow }\limits ^{{\mathcal {L}}}} \Vert Z\Vert ^2_w. \end{aligned}
(42)
Since R is a decreasing function with $$\displaystyle \lim _{u\uparrow 1} R(u)=0$$, and w is bounded
\begin{aligned} 0 \leqslant n\int _{(n-1)/n}^1R(u)^2w(u)du \leqslant M n\int _{(n-1)/n}^1R(u)^2du \leqslant M R^2((n-1)/n) \rightarrow 0. \nonumber \\ \end{aligned}
(43)
The result follows from (41)–(43). $$\square$$ | 23,384 | 63,950 | {"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": 2, "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-2024-38 | latest | en | 0.696637 |
https://programmingpraxis.com/2017/04/25/ | 1,679,481,716,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.22/warc/CC-MAIN-20230322082826-20230322112826-00412.warc.gz | 544,695,307 | 20,804 | Etude On A Binary Tree
April 25, 2017
We have today another in our occasional series of exercises on binary trees; the input tree need not necessarily be ordered or balanced:
Given a binary tree containing integers, find the sum of all nodes at an even distance from the root, less the sum of all nodes at an odd distance from the root.
For instance, given the binary tree shown below, the requested sum is 1 – 2 – 3 + 4 + 5 + 6 + 7 – 8 – 9 – 10 – 11 – 12 – 13 – 14 – 15 = -74:
``` 1
2 3
4 5 6 7
8 9 10 11 12 13 14 15```
(I’m not an artist; you’ll have to imagine the lines connecting the various levels.)
Your task is to write a program to compute alternate sums and differences of the nodes of a binary tree. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. | 248 | 902 | {"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-2023-14 | latest | en | 0.894402 |
http://lists.distributed.net/pipermail/rc5/1999-April/020384.html | 1,590,619,685,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396163.18/warc/CC-MAIN-20200527204212-20200527234212-00392.warc.gz | 74,622,253 | 1,947 | # [RC5] Thoughts on Stats & "Block Hoarders"
bullicante bullicante at yahoo.com
Tue Apr 20 17:14:03 EDT 1999
```Block hoarding doesn't matter. The only thing that matters is the
keyrate. Why? Because the correct key could be anywhere. Suppose
there are only 16 keys and two decoders that can test a key in 1
minute. One decoder picks 2 keys and sits on them. The other decoder
is left with the remaining 14 keys. How long on average will it take to
find the correct key? The odds are 2 out of 16 the hoarder will get the
correct key, in which case the other decoder will take 15 or 16 minutes
to solve it. But the odds are 14 out of 16 that it will be solved in 14
minutes or less. (2/16)*((15+16)/2) + (14/16)*((1+2+..+14)/14) = 8.5.
Suppose the hoarder takes 4 keys: (4/16)*((13+14+15+16)/4) +
(12/16)*((1+2+...12)/12) = 8.5.
Bill Broecker wrote:
>
> Here's my \$.02:
>
> If I understand things correctly (and probably don't), older clients
> (everything older than 440?) use(d) a LIFO buffer system. In this case, one
> of these "block hoarders" could be sitting on THE finished key, but may not
> turn it in for some lengthy period of time, because they want to see their
> name at the top of the stats.
>
--
To unsubscribe, send 'unsubscribe rc5' to majordomo at lists.distributed.net
rc5-digest subscribers replace rc5 with rc5-digest
``` | 426 | 1,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-24 | latest | en | 0.893014 |
http://engineeredsoftware.com/rma_tutorial/example4.htm | 1,660,669,468,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572408.31/warc/CC-MAIN-20220816151008-20220816181008-00255.warc.gz | 14,130,695 | 1,645 | Example 4
Home
Life Data Analysis
A new product is required to demonstrate 99% reliability at 1.5 million cycles with 80% confidence. Given a Weibull shape parameter of 2.8, how long must 8 units be tested to demonstrate the required reliability assuming no failures?
### Solution
1. Select "Testing" from the menu, and the following screen appears.
2. Enter the shape parameter of 2.8, a sample size of 8, a confidence level of 0.8, reliability of 0.99 and a time of 1.5. Click the "Test Duration" radio button, and the test duration will be computed. To demonstrate 99% reliability with 80% confidence at 1.5 million cycles requires testing 8 units for 4 million cycles with no failures. This is shown in the figure below.
3. Change the Test Duration to 1.5, and click the "Sample Size" radio button. To demonstrate 99% reliability with 80% confidence at 1.5 million cycles requires testing 161 units for 1.5 million cycles with no failures. By testing a sample size of 8 for 4 million cycles ( a total of 32 million test cycles) instead of 161 samples for 1.5 million cycles ( a total of 241.5 million test cycles), the total number of test cycles is reduces by 80%. If the shape parameter is greater than 1, the total number of test cycles decreases as the test duration increases. | 329 | 1,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-33 | latest | en | 0.803236 |
https://www.softmath.com/math-book-answers/perfect-square-trinomial/how-do-you-solve-two-step.html | 1,718,552,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00283.warc.gz | 875,748,320 | 9,623 | ## What our customers say...
Thousands of users are using our software to conquer their algebra homework. Here are some of their experiences:
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• simplifying terms worksheet | 1,063 | 4,869 | {"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-2024-26 | latest | en | 0.912872 |
http://bigwww.epfl.ch/teaching/projects/abstracts/hugi/ | 1,679,672,732,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00444.warc.gz | 6,366,077 | 2,766 | Student Projects: Niklaus Hugi English only BIG > Teaching > Student Projects > Niklaus Hugi
CONTENTS
Home page Events Members Publications Tutorials and Reviews Research Demos Download Algorithms Teaching
Multi-dimensional Interpolation and Approximation for Robot Calibration Niklaus Hugi Section Microtechnique, EPFL Semester projectJune 2005 Abstract The Laboratory of Robotics Systems (LSRO) has developed two high precision robots of 3 and 6 degrees of freedom. In order to calibrate the systems, a large amount of input-output pairs have been measured. Input: position of the motors, Output: position of the end-effector. At these "training points" the input-output relation is now known exactly. At all other points the relation is found by interpolation (or, in the case of noisy measurements, by approximation). We approached the problem from a theoretical point of view and considered the general case of non-uniform sampling in m dimensions, where the solution can be expressed through radial basis functions (RBF). For the case of uniform sampling we derived a solution for the approximation and the interpolation problem expressed in the basis of cubic B-splines. In contrast to radial basis functions, B-spines have the advantage of a finite support which allows fast reconstruction. Three algorithms have been implemented in MATLAB: multi-dimensional interpolation based on the existing MATLAB function "interpn" three-dimensional interpolation and approximation for non-uniform sampling based on radial basis functions multi-dimensional interpolation and approximation based on cubic B-splines The LSRO has provided two experimental data sets of the "Delta Cube" robot. The first set of 15^3 training points was used to compute the RBF and B-spline coefficients. The second set of 10^3 reference points was then used to validate the proposed algorithms. All three algorithms produced good results. The goal of errors below 100 nm has been surpassed with errors around 17 nm. It has also been shown, that better results can be achieved if interpolation is replaced by approximation (accuracy improvement of roughly 10%). A graphical user interface has been created in order to facilitate further experiments. | 427 | 2,226 | {"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-2023-14 | latest | en | 0.909004 |
http://www.blogher.com/supersisters | 1,444,087,097,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736677950.64/warc/CC-MAIN-20151001215757-00119-ip-10-137-6-227.ec2.internal.warc.gz | 32,119,490 | 18,894 | ## Supersisters!
BlogHer Original Post
Sister Math is very different from regular run-of-the-mill math. It's sort of like string theory is to algebra--it just operates on a very different set of rules. For instance, in Sister Math, the amount of times your sibling has done something is directly proportional to how much trouble they will be in when caught. They may have only done it two times if no one is ever going to know but that number can swell well into the double digits if mum or dad notices.
2 + 2 really can equal "ten times today alone" if the transgression is something like raiding the liquor cabinet.
Sister Math is a shifty sort of thing where the length of time you dated your loser boyfriend changes depending on the who is hearing the story and the number of times you have borrowed your sister's pink sweater shifts depending on what she needs from you.
It is rumoured that a mathematician on the shortlist for the next Fields Medal has made studying Sister Math her life's work and well-covered in her research are Patience, Jen, and Kristen, the writers of PBS's new Supersister blog. These three sisters create a blog that is six times as fun, nine times as interesting, and twelve times as amusing as your regular blog. Which is entirely possible under Sister Math.
More importantly, the advice of a sister is somehow twenty times more helpful than advice given by even the most revered of parenting experts. Building on that unique element of Sister Math, the Supersister blog, which discusses raising children, is also twenty times more helpful to you.
It's a blog where chocolate is necessary, capes are encouraged, and children are many. These three sisters dish on parenting--raising girls, raising boys, and bringing the whole family together with projects such as book making and documenting acts of kindness. I figuratively sat down with Supersister, Jen Lemen, in this online interview to learn more about growing up with sisters and raising their broods together.
Mel: So, tell me, what is the best part about growing up with sisters?
Jen: Definitely the clothes. You have your own wardrobe times at least two or three, as long as you follow the unspoken rules of sharing: new items are off limits and you probably should not lend your sister's best sweater to your best friend's mother.
Also, there's this fierce loyalty we had growing up. We could fight like cats and dogs but if someone was mean to one of us, there was always hell to pay from the others.
Mel: A threesome can sometimes lead to divide and conquer techniques by ganging up on one sister. Did this ever happen in your youth and what would you do if you noticed siblings or cousins in the younger generation doing it now?
Jen: There are (and were!) times when two of us are getting along better than we are with the others because of stage of life or current interests, but it moves around with time and circumstance. The same is true for our kids. We went through a stage where all the kids were pining away for Josiah who was pining away for Madeleine. Now Carter is the main attraction and Madeleine (as the only school-aged girl) is kind of on the outs. We let things be for the most part and redirect if we think someone's feelings are getting hurt.
Mel: What was the best thing you got away with in your youth?
Jen: I had an incredibly boring youth—all episodes of rebellion reserved for my mid-thirties instead. But I did skip school once and I used to drive my Chevy Nova at 100 mph to school everyday (really). It's a miracle I did not kill myself or someone else. I'm still notoriously bad at following the rules of the road.
Mel: Damn, I didn't know Chevy Novas made it to 100 mph. Will your fourth sister make an appearance on the blog?
Jen: I don't know, but that would be very fun if she did. She is a brilliant writer and extremely witty. I'm sure her commentary on our collective parenting would be very revealing and insightful.
Mel: Did all three of you begin blogging at the same time or did one sister rope the others into the writing world?
Jen: I don't know the exact chronology of who started when, but I can say that we've all been doing this for so long now it's hard to remember. We are all writers in our own right with the compressed time of mothers, so it was just a matter of time for each of us to discover the web and make it our medium.
Jen: In our family, Valentine's Day was as big a deal as Christmas. My father brought us each bunches of flowers, boxes of chocolate and my mother made heart shaped meatloaf. I love that kind of magic, and so do my kids.
Mel: What is your best tip for staying calm when everything descends into chaos with the kids?
Jen: I say surrender is the way to go. I've learned over the years that most chaos springs from me clinging to a definition of normal that is simply not realistic. Every child in America falls apart when they are tired or hungry or over-stimulated—why not mine? I do much better when I'm empathetic and try to solve problems bit by bit instead of resisting the inevitable storms.
Melissa is the author of the infertility and pregnancy loss blog, Stirrup Queens and Sperm Palace Jesters. She keeps a categorized blogroll of over 1400 infertility blogs and writes the daily Lost and Found and Connections Abound, a news source for the infertility blogosphere. Her infertility book, The Land of If, is forthcoming from Seal Press in Spring 2009. She is also an editor at Bridges, the awareness consortium that is currently seeking writers for its 100 Words Project. | 1,203 | 5,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-40 | latest | en | 0.956549 |
https://oeis.org/A295969 | 1,627,680,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153980.55/warc/CC-MAIN-20210730185206-20210730215206-00502.warc.gz | 451,730,609 | 4,058 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A295969 Numbers k such that (5*10^k - 473)/9 is prime. 0
2, 3, 4, 9, 10, 16, 27, 46, 97, 103, 214, 229, 241, 285, 643, 999, 3154, 14115, 43012, 85855, 87622, 87712, 90142, 168379 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS For k > 1, numbers such that k-2 occurrences of the digit 5 followed by the digits 03 is prime (see Example section). a(25) > 2*10^5. LINKS Makoto Kamada, Factorization of near-repdigit-related numbers. Makoto Kamada, Search for 5w03 EXAMPLE 3 is in this sequence because (5*10^3 - 473)/9 = 503 is prime. Initial terms and primes associated: a(1) = 2, 3; a(2) = 3, 503; a(3) = 4, 5503; a(4) = 9, 555555503; a(5) = 10, 5555555503; etc. MATHEMATICA Select[Range[2, 100000], PrimeQ[(5*10^# - 473)/9] &] PROG (PARI) isok(k) = isprime((5*10^k - 473)/9); \\ Michel Marcus, Dec 01 2017 CROSSREFS Cf. A056654, A268448, A269303, A270339, A270613, A270831, A270890, A270929, A271269. Sequence in context: A249943 A251620 A128944 * A174437 A127150 A256189 Adjacent sequences: A295966 A295967 A295968 * A295970 A295971 A295972 KEYWORD nonn,more,hard AUTHOR Robert Price, Nov 30 2017 EXTENSIONS a(24) from Robert Price, Jan 21 2018 STATUS approved
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Last modified July 30 17:21 EDT 2021. Contains 346359 sequences. (Running on oeis4.) | 608 | 1,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.683099 |
https://discuss.leetcode.com/topic/104868/java-straight-forward-no-searching | 1,513,581,480,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948609934.85/warc/CC-MAIN-20171218063927-20171218085927-00383.warc.gz | 545,531,406 | 8,789 | # Java, straight forward, no searching
• The key is:
From last digit to first digit:
Find one number from four digits that can make it larger, if found, set the following digits to min, if not found, continue this on the digit before it.
When reach -1 th digit, means there is no solution bigger than input, just set all digits to min and return.
``````char min, max;
char[] time;
public String nextClosestTime(String s) {
time = s.toCharArray();
min=(char)Math.min(Math.min(time[0],Math.min(time[1],time[3])),time[4]);
max=(char)Math.max(Math.max(time[0],Math.max(time[1],time[3])),time[4]);
solve(4);//try looking for solution from last digits to first digits
if(time[0]=='#'){ //no solution found, turn to another day
Arrays.fill(time,min);
time[2]=':';
}
return new String(time);
}
public void solve(int i){//from the ith position, look for smallest value that is bigger than it
if(i<0){//didn't find any, mark no solution, need to turn to another day
time[0]='#';
return;
}
if(i==2){//':' sign, jump cross it
solve(1);
return;
}
char ans = max;
for(int j=0;j<5;j++){//try to find a suitable number to replace ith position
if(j!=2&&time[j]>time[i]) ans=(char)Math.min(ans,time[j]);
}
if(ans==time[i]||!isvalid(i,ans)){//didn't find a suitable one or the one found is not valid, go to i-1
solve(i-1);
return;
}
time[i]=ans;//set the number
for(int j=i+1;j<5;j++){//set the following number to smallest value possible
if(j!=2)time[j]=min;
}
}
public boolean isvalid(int i,char ans){
if(i==3)return ans<'6';
if(i==1)return time[i-1]<'2'||(time[i-1]=='2'&&ans<'4');
if(i==0)return ans<'2'||(ans=='2'&&time[i+1]<'4');
return true;
}``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 501 | 1,736 | {"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} | 2.890625 | 3 | CC-MAIN-2017-51 | latest | en | 0.56492 |
https://pypi.org/project/empyrical-reloaded/ | 1,716,278,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058385.38/warc/CC-MAIN-20240521060250-20240521090250-00203.warc.gz | 416,742,549 | 13,837 | empyrical computes performance and risk statistics commonly used in quantitative finance
## Project description
Common financial return and risk metrics in Python.
## Installation
empyrical requires Python 3.9+. You can install it using pip:
pip install empyrical-reloaded
or conda from the conda-forge channel
conda install empyrical-reloaded -c conda-forge
empyrical requires and installs the following packages while executing the above commands:
• numpy>=1.9.2
• pandas>=1.0.0
• scipy>=0.15.1
Optional dependencies include yfinance to download price data from Yahoo! Finance and pandas-datareader to access Fama-French risk factors and FRED treasury yields.
Note that pandas-datareader is not compatible with Python>=3.12.
To install the optional dependencies, use:
pip install empyrical-reloaded[yfinance]
or
pip install empyrical-reloaded[datreader]
or
pip install empyrical-reloaded[yfinance,datreader]
## Usage
### Simple Statistics
Empyrical computes basic metrics from returns and volatility to alpha and beta, Value at Risk, and Shorpe or Sortino ratios.
import numpy as np
from empyrical import max_drawdown, alpha_beta
returns = np.array([.01, .02, .03, -.4, -.06, -.02])
benchmark_returns = np.array([.02, .02, .03, -.35, -.05, -.01])
# calculate the max drawdown
max_drawdown(returns)
# calculate alpha and beta
alpha, beta = alpha_beta(returns, benchmark_returns)
### Rolling Measures
Empyrical also aggregates return and risk metrics for rolling windows:
import numpy as np
from empyrical import roll_max_drawdown
returns = np.array([.01, .02, .03, -.4, -.06, -.02])
# calculate the rolling max drawdown
roll_max_drawdown(returns, window=3)
### Pandas Support
Empyrical also works with both NumPy arrays and Pandas data structures:
import pandas as pd
from empyrical import roll_up_capture, capture
returns = pd.Series([.01, .02, .03, -.4, -.06, -.02])
factor_returns = pd.Series([.02, .01, .03, -.01, -.02, .02])
# calculate a capture ratio
capture(returns, factor_returns)
-0.147387712263491
### Fama-French Risk Factors
Note: requires optional dependency pandas-datareader - see installation instructions above.gst
import pandas as pd
import empyrical as emp
risk_factors = emp.utils.get_fama_french()
Mkt-RF SMB HML RF Mom
Date
1970-01-02 00:00:00+00:00 0.0118 0.0129 0.0101 0.00029 -0.0340
1970-01-05 00:00:00+00:00 0.0059 0.0067 0.0072 0.00029 -0.0153
1970-01-06 00:00:00+00:00 -0.0074 0.0010 0.0021 0.00029 0.0038
1970-01-07 00:00:00+00:00 -0.0015 0.0040 -0.0033 0.00029 0.0011
1970-01-08 00:00:00+00:00 0.0004 0.0018 -0.0017 0.00029 0.0033
2024-03-22 00:00:00+00:00 -0.0023 -0.0087 -0.0053 0.00021 0.0043
2024-03-25 00:00:00+00:00 -0.0026 -0.0024 0.0088 0.00021 -0.0034
2024-03-26 00:00:00+00:00 -0.0026 0.0009 -0.0013 0.00021 0.0009
2024-03-27 00:00:00+00:00 0.0088 0.0104 0.0091 0.00021 -0.0134
2024-03-28 00:00:00+00:00 0.0010 0.0029 0.0048 0.00021 -0.0044
### Asset Prices and Benchmark Returns
Empyrical use yfinance to download price data from Yahoo! Finance. To obtain the S&P returns since 1950, use:
Note: requires optional dependency yfinance - see installation instructions above.
import empyrical as emp
symbol = '^GSPC'
returns = emp.utils.get_symbol_returns_from_yahoo(symbol,
start='1950-01-01')
import seaborn as sns # requires separate installation
import matplotlib.pyplot as plt # requires separate installation
fig, axes = plt.subplots(ncols=2, figsize=(14, 5))
with sns.axes_style('whitegrid'):
returns.plot(ax=axes[0], rot=0, title='Time Series', legend=False)
sns.histplot(returns, ax=axes[1], legend=False)
axes[1].set_title('Histogram')
sns.despine()
plt.tight_layout()
plt.suptitle('Daily S&P 500 Returns')
### Documentation
See the documentation for details on the API.
## Support
Please open an issue for support.
## Contributing
Please contribute using Github Flow. Create a branch, add commits, and open a pull request.
## Testing
• install requirements
• "pytest>=6.2.0",
pytest tests
## Project details
Uploaded Source
Uploaded Python 3 | 1,312 | 4,115 | {"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} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.676855 |
https://www.jiskha.com/members/profile/posts.cgi?name=PsyDAG | 1,500,628,090,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423764.11/warc/CC-MAIN-20170721082219-20170721102219-00448.warc.gz | 790,123,020 | 8,243 | # Posts by PsyDAG
Total # Posts: 23,458
Algebra 1
5 gallons = 220 miles Let x = additional miles traveled 220/5 = x/10 I'll leave the explanations up to you.
English
science
If you had diploid cells instead of haploid cells forming the zygote, each generation would have double the number of chromosomes.
maths
Let x = second son's share. .65x + x + 1.35x = 3873 Solve for x.
Stats
P(A) is right. P(C) = 2/38 = .053 Only two choices are green. Why did you add 18 twice?
maths (?)
Cannot see which face is shaded.
Maths
5 students scored more than 70. 40% of 50 (20) students scored below 55. That leaves (50-25) students scoring between 55-70. What percentage is that?
statistics check my work..thank you
Same here.
statistics check my work
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance. Didn't check the math, but if this is what you did, this and the rest is on point.
math
To get total of 21, the next card must be a 4. (4-1)/49 = ? Sorry, if there are no Aces, (4-1)/45. Without an ace, the player can draw a 5(4 chances), 6 (4 chances)... K,Q or Jack (11 chances). Fill in the remaining chances and divide that sum by 49 (or if the case may be, 45.
math
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 13/52 * (13-1)/52-1) = ?
math
6^3 = 6 * 6 * 6
Math
(7*13) - 87 = ?
Math
4x + x = 60 4x = ?
Math
Since the percentages add to 90%, the 90 students are in the remaining 10%, meaning that there were a total of 900 students. .4 * 900 = ? Cannot draw diagrams here.
math
Find the mean first = sum of scores/number of scores 55*1 + 65*309 + 75*1433.... Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance. Standard deviation = square root of ...
math
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
Stats in need of help
35/650 = ?
stat
Do not have table A-2. Lack data.
Help me with math
Find the mean first = sum of scores/number of scores Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Stats need help
Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores. Multiply by 100.
Stats Help Help
Find the mean first = sum of scores/number of scores Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
STATISTICS
What do you consider are "the four levels of measurement"?
Stats
Disagree with second and third answers.
math
Z = (mean1 - mean2)/standard error (SE) of difference between means = 246/standard error (SE) of difference between means SEdiff = ?(SEmean1^2 + SEmean2^2) SEm = SD/?n If only one SD is provided, you can use just that to determine SEdiff. Find table in the back of your ...
bio
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
math
Let x = original sum This is assuming that the 1/5 was of the original sum rather than after the 1/4 was spent. x - .25x - .2x - 78 = 21 Solve for x.
math
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability of Z = ±3. Or you can use this: http://davidmlane.com/hyperstat/z_table.html
Math
Assuming I interpreted your equation correctly: 5(2n+3) - (n+8) = 13 Combine terms. 10n + 15 - n - 8 = 13 9n = 6 Take it from there.
math
Is 26.7 seconds less than 30?
math
40m/1.5sec = 26.7 sec 30m/1sec = 30 sec
Mah
Online "^" is used to indicate an exponent, e.g., x^2 = x squared. Area = ? r^2 = 3.14 * 8 * 8 You do the math.
College Math
Cannot draw on these posts.
Math
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
Science
Agree
algebra
9230 = x/2 + 30 Subtract 30 from both sides, then multiply both sides by 2.
MATHS
First, please do not use all capitals. Online it is like SHOUTING. Not only is it rude, but it is harder to understand. Thank you. Second, we cannot diagram on these posts.
Maths
It would help if you proofread your questions before you posted them. Although you don't indicate your units, I assume that the ratio is 1 cm/500,000 Km. 500,000/1 = x/7 Solve for x. If the 500,000 is actually in cm, divide x by 10,000. (1000 m in a Km and 10 cm in a m.)
maths
(5/8)*(8/5) = 40/40 = 1 1 to any power = 1
statistic
What "two variables"? I only see the memory test. Z = (score-mean)/SEm SEm = SD/?n Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability related to the Z score.
Science (Biology)(?)
Although this is not my area of expertise, I would assume that anything "linked" cannot be separated.
Science
With half the mass, he would weigh half as much as on earth.
English
1 and 3 are advisable to do. 2 and 4 are your preference.
English
#3
STATISTICS
90% = mean ± Z (SEm) Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability (± .05) and its Z score. 90% = mean ± 1.645 SEm .5 = 1.645 SEm SEm = SD/?...
Math
4215
Math
Rain = (1-.3)^3 Claudia = 1/9 = .11 Ally = 2/17 = .12 I'll let you explain.
Math
If they are in groups of 4: 700 * n/4 = ? However, that is only in groups of 4. If there are 1, 2 or 3 extra, there still would be an extra charge of 700 pesos, any fraction would be another 700 pesos.
Math
let x = total number of students (1-4/9)x = 755 x = 1359 1359 - 755 = ?
algebra
.8C + .2G = 1 Punch
algebra
Rate * Time = Work done
math
S = 2G S-10 = 3(G-10) Substitute 2G for S in the second equation and solve for G. Insert that value into the first equation to solve for S. Check by putting both values into the second equation.
Math (?)
What were the other choices?
Math
It would help if you proofread your questions before you posted them. Is the length of the pole 11.2 meters or 11 halves (11/2) of a meter? 1 - 1/3 - 1/4 = 1 - 4/12 - 3/12 = 5/12 So the blue portion is 5/12 of whatever the length of the pole is.
math
"not selecting a consonant" = selecting a vowel. There are 3 vowels in the 8 letters. Take it from here.
Algebra (?)
Lacking data.
Algebra (?)
Unfamiliar with your notation \$\$\$\$\$
Science
Since this is not my area of expertise, I searched Google under the key words "silent mutation" to get th1s possible source: https://en.wikipedia.org/wiki/Silent_mutation In the future, you can find the information you desire more quickly, if you use appropriate key ...
Math (?)
What shaded area? You cannot copy and paste here.
English - Writing
Previous sentence needs a comma. In order to better assist you, select the best retirement benefit package to fit your personal needs. Select the best retirement benefit package to fit your personal needs.
family
The father of your half brother can be your father, if you have different mothers. If you both have the same mother, then he is your step father.
science
Density = mass + volume Volume = 2^3 Take it from here.
math (?)
No following statements. Also it would be easier to respond to if the data were shown as 9,12,8,7....
statistics
The probability of the heart in your example would be 13/50 if the king or ace did not include a heart. You have still not included probability that either/both a king or ace will be a heart. Thus the probability of the heart in your next equations would be either 12/50 or 11/...
Science
An independent variable is the potential stimulus or cause, usually directly manipulated by the experimenter, so it could also be called a manipulative variable. A dependent variable is the response or measure of results. Extraneous variables — other than the independent ...
Math
3t = 5c
Math
The field is 30 by 10. Figure it from there.
science
From Google: Eukaryotic organisms, such as algae, fungi, and higher plants, have multilayered cell walls composed in large part of either cellulose or chitin . Cellulose and chitin are polysaccharides , meaning they are composed of many linked sugar molecules.
science
agree
Math
You figured the area. The athlete runs the perimeter = 50 * 4 = 200 Kilometer = 1000 meters. (3*1000)/200 = ?
STATISTICS
.95 = mean ± Z(SD) = 1050 ± 290Z Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability (±.025) and its Z score. Insert the Z score into the equation ...
math (?)
By what process?
Statistics
Have you considered the fact the the king and/or the 10 could be a heart? Consider those probabilities. Either-or probabilities are found by adding the individual probabilities.
Math
A is right.
Math
Disagree. Watch out for "always" and "never" answers.
math
2/3 * 3/4 = ?
Math
Cannot diagram here. Distance A-C, use Pythagorean theorem. https://en.wikipedia.org/wiki/Pythagorean_theorem Total distance = 9 + 12 = ?
Health (Ms. Sue Check Please)
Not knowing WHY the doctor is worried about Kori's cardiovascular health, I would tend to agree.
Biology
Since this is not my area of expertise, I searched Google under the key words "process of digestion steps" to get these possible sources: https://www.google.com/search?client=safari&rls=en&q=process+of+digestion+steps&ie=UTF-8&oe=UTF-8 In the future, you can find the...
life sciences (?)
Lack data. No leaves.
English
1. 3 * 300ml = 3 * .3 liters = .9 liters 2 and 3 are even less, but 2 = .005 liters, 3 = .015 liters 4. is a possibility, but that is too much water. It would be better to express all in liters, but you would want them to drink more than .0015 liters a day.
English
All are okay, but 2 is better than 1. 3. would be like using weights that are too heavy. 4 would probably relate to frequency or duration of exercise. 5 would relate to frequency. 6. would relate to duration.
Maths (?)
No data. Cannot draw a flow chart here.
Math
However, I will give you a start. Convert all to cubic yards. Even so, there are several problems in making a comparison. Here are some examples: Knowing the size of the yard = (80/3)^2 sq yards is not really helpful, because you don't know the depth of the soil that will ...
Math
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
Biology-Genes
Use a Punnett square. https://en.wikipedia.org/wiki/Punnett_square Red fish are all rr. Green fish are either GG or Gr.
Math (?)
No tables shown. Cannot copy and paste here.
mathematics
Cannot draw on these posts.
algebra
Z = (mean1 - mean2)/standard error (SE) of difference between means SEdiff = ?(SEmean1^2 + SEmean2^2) SEm = SD/?n If only one SD is provided, you can use just that to determine SEdiff. Find table in the back of your statistics text labeled something like "areas under ...
Mathematics Algebra
value ? (b^2-a^2)
Math
Welcome
Math
you got it!
Math
1) 8/500 = x/20,000 2) 15/50 = x/2000
MMATH08
First, please do not use all capitals. Online it is like SHOUTING. Not only is it rude, but it is harder to understand. Thank you. Second, it would help if you proofread your questions before you posted them. If you are using only "10 aquare" tiles of gigantic size: ...
Algebra
5) Multiply first equation by 5. 5x - 20y = 60 6) Multiply first equation by 4. 4y - 24x = -12 Subtracting first equation from the second, what does that tell you?
Science (?)
Lack data.
Science
agree
Algebra
x/3x = 1/3x = 1/3(100) = .333
Science
It has to pump blood into the whole body.
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Edexcel computer science 08/06/16 watch
1. (Original post by stormblazer999)
I think I got something similar in terms of the code - not sure though.
For the logical operations, there were 4 boxes - I got AND OR NOT AND
What did you put for computational methods (parallel and sequential)
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2. (Original post by damndaniel2)
i got select (id, description)
from tblname
where id = 'g'
order by (id) i thought it said to order by the id or something haha
also order by automatically means ascending so asc is not needed.
For the question on the back i did so bad. I did something like:
R1 = 0
r2 = 0
r3 = 0
rr1 = r1 + 1
rr2 = r2 + 2
rr3 = r3 + 4
if r5 = 0
print (int(r1*r2*r3))
else
print (int(r5 - 1))
something like that and i know this is defo wrong!
For the logical operations one i think i put:
And
or
not
i think lol wbu?
posted from tsr mobile
or or and not
3. (Original post by DamnDaniel2)
Oh yh I put AND at the end as well haha. How many marks was that?
Posted from TSR Mobile
4 delicious marks. Good job.
4. (Original post by Obi_xtra)
or or and not
Some people put that but the majority put AND OR NOT AND because:
How can the first box be OR? It needs to be a value that's larger than or equal to 1 AS WELL AS less than or equal to 5 in order to become a weekday.
And the third box is definitely NOT.
5. (Original post by stormblazer999)
4 delicious marks. Good job.
Hahaha thank you. For the truth table thing what did u say for the reason why people use it? I said it's used to find logical errors as they are hard to detect and fix because the error in the code occurred without the program realising it
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6. (Original post by DamnDaniel2)
What did you put for computational methods (parallel and sequential)
Posted from TSR Mobile
Oh that was pretty okay.
I just said that sequential models execute line by line, from start to finish, in a linear fashion. Parallel models execute in parallel and simultaneously on multiple processing devices. Parallel models are more productive/efficient/faster than linear models because more than one task is being performed at the same time.
7. (Original post by DamnDaniel2)
Hahaha thank you. For the truth table thing what did u say for the reason why people use it? I said it's used to find logical errors as they are hard to detect and fix because the error in the code occurred without the program realising it
Posted from TSR Mobile
Erm, it didn't ask why a TRACE TABLE is used. It asked the PURPOSE OF THE ALGORITHM. I forgot what I put in as an answer though.
8. (Original post by stormblazer999)
Erm, it didn't ask why a TRACE TABLE is used. It asked the PURPOSE OF THE ALGORITHM. I forgot what I put in as an answer though.
Yh sorry that's what I meant. And loool I always get mixed up with a trace table and truth table
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9. (Original post by stormblazer999)
Oh that was pretty okay.
I just said that sequential models execute line by line, from start to finish, in a linear fashion. Parallel models execute in parallel and simultaneously on multiple processing devices. Parallel models are more productive/efficient/faster than linear models because more than one task is being performed at the same time.
Yup that's what I put, however, I didn't talk about multiprocessing devices
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10. (Original post by stormblazer999)
Erm, it didn't ask why a TRACE TABLE is used. It asked the PURPOSE OF THE ALGORITHM. I forgot what I put in as an answer though.
Also for the pseudocode one where you had to change bits of the pseudocode to reach the requirements what did you put? I created a constant variable and made two other variables called Area and Area2 and set them to 0. I added an input as well cos we want it to work with every number
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11. (Original post by DamnDaniel2)
Yh sorry that's what I meant. And loool I always get mixed up with a trace table and truth table
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Well, either way tbh, I don't think you'll get the mark because I don't think finding logical errors had to do with the algorithm. Sorry
12. (Original post by stormblazer999)
Well, either way tbh, I don't think you'll get the mark because I don't think finding logical errors had to do with the algorithm. Sorry
I just checked the internet and on various sites it says that I got it correct lol
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13. (Original post by DamnDaniel2)
I just checked the internet and on various sites it says that I got it correct lol
Posted from TSR Mobile
Oh really? Well done man. Sorry for reducing your confidence.
14. (Original post by stormblazer999)
Oh really? Well done man. Sorry for reducing your confidence.
Hahaha no it's fine Also for the codes do you think I got like nearly full marks for them or even full marks?
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15. (Original post by DamnDaniel2)
Hahaha no it's fine Also for the codes do you think I got like nearly full marks for them or even full marks?
Posted from TSR Mobile
I would say full marks IMO. They seem like they covered everything.
16. (Original post by stormblazer999)
I would say full marks IMO. They seem like they covered everything.
Oh hopefully! I really want that A* hahaha. I really need someone to create an unofficial mark scheme but it seems like barely anyone does edexcel computer science!
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17. What did you guys get for the trace table question? I think i did it right but Im not quite sure
18. (Original post by DamnDaniel2)
Oh hopefully! I really want that A* hahaha. I really need someone to create an unofficial mark scheme but it seems like barely anyone does edexcel computer science!
Posted from TSR Mobile
Yeah, last year only about 540 people in the entire country did edexcel computer science :/
19. (Original post by DamnDaniel2)
Oh hopefully! I really want that A* hahaha. I really need someone to create an unofficial mark scheme but it seems like barely anyone does edexcel computer science!
Posted from TSR Mobile
Yeah, last year only about 540 people in the entire country did edexcel computer science :/
Subject Statistics from last year: http://qualifications.pearson.com/co...ng_with_1).pdf
First page where it specifies Comp Science
20. (Original post by Synx-b)
What did you guys get for the trace table question? I think i did it right but Im not quite sure
I got like 1, 2, 4 and 5? Then 2, 3... Something like that? Not sure. What did you get?
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Directions: Study the following information carefully and answer the questions given beside:
Important for :
1
Five persons – Ashi, Bala, Diya, Neha and Rupa were standing in a row facing north direction, but not necessarily in the same order. The distances between two adjacent persons are successive multiples of four (i.e. if the distance between the 1st and the 2nd person is 4 m, 1st and the 3rd person is 8m and between 1st and 4th person is 12 m and so on.)
(i.e. suppose Ashi stand on the left end of the line then the remaining people will stand at a distance as follows – 4 m, 8 m, 12 m and so on, from the end)
Ashi was standing at the left end of the line. Two persons were standing between Ashi and Bala. Diya stand second to the left of Bala. Rupa was an immediate neighbor of Diya. Only one person stands between Rupa and Neha. Joya starts walking from point X which was 8 m in north from Rupa. Joya moves 17 m in north-east direction to reach point Y. Point Y and Neha were inline vertically. Neha moves 6 m in the north direction and stops at point Z. Person Ashi moves in the direction to point Z. (Set 1 Q. 1)
What is the shortest distance between the initial position of Neha and final position of Joya?
» Explain it
D
Following the figure we made using the given information we can say that the shortest distance between the initial position of Neha and final position of Joya = (15 + 2 + 6) m = 23 m.
Hence, the correct answer is option D.
Common Explanation:
Reference:
Ashi was standing at the left end of the line.
Two persons were standing between Ashi and Bala.
Diya stand second to the left of Bala.
Rupa was an immediate neighbor of Diya.
Only one person stands between Rupa and Neha.
Inference:
Using the given hints we can fix the positions of all the five persons standing in the row.
Reference:
Joya starts walking from point X which was 8 m in north from Rupa.
Joya moves 17 m in north-east direction to reach point Y.
Point Y and Neha were inline vertically.
Inference:
Here, in the figure below the path of Joya is denoted by Red colour, and I.P. and F.P. means initial position and final position of the person.
Reference:
Neha moves 6 m in the north direction and stops at point Z.
Person Ashi moves in the direction to point Z.
Inference:
In the figure below, the path of Neha and Ashi is denoted by Blue and green colour respectively, and I.P. and F.P. means initial position and final position of the person.
2
Five persons – Ashi, Bala, Diya, Neha and Rupa were standing in a row facing north direction, but not necessarily in the same order. The distances between two adjacent persons are successive multiples of four (i.e. if the distance between the 1st and the 2nd person is 4 m, 1st and the 3rd person is 8m and between 1st and 4th person is 12 m and so on.)
(i.e. suppose Ashi stand on the left end of the line then the remaining people will stand at a distance as follows – 4 m, 8 m, 12 m and so on, from the end)
Ashi was standing at the left end of the line. Two persons were standing between Ashi and Bala. Diya stand second to the left of Bala. Rupa was an immediate neighbor of Diya. Only one person stands between Rupa and Neha. Joya starts walking from point X which was 8 m in north from Rupa. Joya moves 17 m in north-east direction to reach point Y. Point Y and Neha were inline vertically. Neha moves 6 m in the north direction and stops at point Z. Person Ashi moves in the direction to point Z. (Set 1 Q. 2)
What is the direction of initial position of Joya with respect to final position of Ashi?
» Explain it
C
Following the figure we made using the given information we can say that initial position of Joya was in North – West direction with respect to final position of Ashi.
Hence, the correct answer is option C.
Common Explanation:
Reference:
Ashi was standing at the left end of the line.
Two persons were standing between Ashi and Bala.
Diya stand second to the left of Bala.
Rupa was an immediate neighbor of Diya.
Only one person stands between Rupa and Neha.
Inference:
Using the given hints we can fix the positions of all the five persons standing in the row.
Reference:
Joya starts walking from point X which was 8 m in north from Rupa.
Joya moves 17 m in north-east direction to reach point Y.
Point Y and Neha were inline vertically.
Inference:
Here, in the figure below the path of Joya is denoted by Red colour, and I.P. and F.P. means initial position and final position of the person.
Reference:
Neha moves 6 m in the north direction and stops at point Z.
Person Ashi moves in the direction to point Z.
Inference:
In the figure below, the path of Neha and Ashi is denoted by Blue and green colour respectively, and I.P. and F.P. means initial position and final position of the person.
3
Five persons – Ashi, Bala, Diya, Neha and Rupa were standing in a row facing north direction, but not necessarily in the same order. The distances between two adjacent persons are successive multiples of four (i.e. if the distance between the 1st and the 2nd person is 4 m, 1st and the 3rd person is 8m and between 1st and 4th person is 12 m and so on.)
(i.e. suppose Ashi stand on the left end of the line then the remaining people will stand at a distance as follows – 4 m, 8 m, 12 m and so on, from the end)
Ashi was standing at the left end of the line. Two persons were standing between Ashi and Bala. Diya stand second to the left of Bala. Rupa was an immediate neighbor of Diya. Only one person stands between Rupa and Neha. Joya starts walking from point X which was 8 m in north from Rupa. Joya moves 17 m in north-east direction to reach point Y. Point Y and Neha were inline vertically. Neha moves 6 m in the north direction and stops at point Z. Person Ashi moves in the direction to point Z. (Set 1 Q. 3)
What is the shortest distance between the final position of Ashi and initial position of Joya?
» Explain it
A
Following the figure we made using the given information we can say that the shortest distance between final position of Ashi and initial position of Joya = (82 + 22)1/2 m = 2√17 m
Hence, the correct answer is option A.
Common Explanation:
Reference:
Ashi was standing at the left end of the line.
Two persons were standing between Ashi and Bala.
Diya stand second to the left of Bala.
Rupa was an immediate neighbor of Diya.
Only one person stands between Rupa and Neha.
Inference:
Using the given hints we can fix the positions of all the five persons standing in the row.
Reference:
Joya starts walking from point X which was 8 m in north from Rupa.
Joya moves 17 m in north-east direction to reach point Y.
Point Y and Neha were inline vertically.
Inference:
Here, in the figure below the path of Joya is denoted by Red colour, and I.P. and F.P. means initial position and final position of the person.
Reference:
Neha moves 6 m in the north direction and stops at point Z.
Person Ashi moves in the direction to point Z.
Inference:
In the figure below, the path of Neha and Ashi is denoted by Blue and green colour respectively, and I.P. and F.P. means initial position and final position of the person.
4
Five persons – Ashi, Bala, Diya, Neha and Rupa were standing in a row facing north direction, but not necessarily in the same order. The distances between two adjacent persons are successive multiples of four (i.e. if the distance between the 1st and the 2nd person is 4 m, 1st and the 3rd person is 8m and between 1st and 4th person is 12 m and so on.)
(i.e. suppose Ashi stand on the left end of the line then the remaining people will stand at a distance as follows – 4 m, 8 m, 12 m and so on, from the end)
Ashi was standing at the left end of the line. Two persons were standing between Ashi and Bala. Diya stand second to the left of Bala. Rupa was an immediate neighbor of Diya. Only one person stands between Rupa and Neha. Joya starts walking from point X which was 8 m in north from Rupa. Joya moves 17 m in north-east direction to reach point Y. Point Y and Neha were inline vertically. Neha moves 6 m in the north direction and stops at point Z. Person Ashi moves in the direction to point Z. (Set 1 Q. 4)
If Ashi had stopped at point T which is 5m north from Bala instead of point Z, then what would be the shortest distance between initial and final position of Ashi?
» Explain it
C
Following the figure we can say that the shortest distance between initial and final position of Ashi is 13 m.
Note- T is the final position of Ashi as per the question.
= (12)2 + (5)2 = 169
= 13
Hence option C is correct.
Common Explanation:
Reference:
Ashi was standing at the left end of the line.
Two persons were standing between Ashi and Bala.
Diya stand second to the left of Bala.
Rupa was an immediate neighbor of Diya.
Only one person stands between Rupa and Neha.
Inference:
Using the given hints we can fix the positions of all the five persons standing in the row.
Reference:
Joya starts walking from point X which was 8 m in north from Rupa.
Joya moves 17 m in north-east direction to reach point Y.
Point Y and Neha were inline vertically.
Inference:
Here, in the figure below the path of Joya is denoted by Red colour, and I.P. and F.P. means initial position and final position of the person.
Reference:
Neha moves 6 m in the north direction and stops at point Z.
Person Ashi moves in the direction to point Z.
Inference:
In the figure below, the path of Neha and Ashi is denoted by Blue and green colour respectively, and I.P. and F.P. means initial position and final position of the person.
5
There is PQ axis in such a way that P is in north and Q is in south direction. There is RS axis in such a way that R is in west direction and S is in east direction. PQ axis and RS axis intersect at a point X in such a way that PX is 20 m, XQ is 23 m, XR is 19 m, XS is 31 m.
Car A starts from point R and travels 28 m in south direction and then it takes a left turn and travels 45 m. Car B starts from point P and travels 26 m in east direction. Car C starts from point S and travels 7 m in north direction and then it takes a left turn and travel 5 m and again it takes a left turn and travels 30 m. (Set 2 Q. 1)
What is the shortest distance between the final positions of car A and car B?
» Explain it
C
Following the figure we made using the given information we can say that the shortest distance between the final positions of car A and car B = (5 + 23 + 20) m = 48 m.
Hence, the correct answer is option C.
Common Explanation:
Reference:
There is PQ axis in such a way that P is in north and Q is in south direction. There is RS axis in such a way that R is in west direction and S is in east direction. PQ axis and RS axis intersect at a point X in such a way that PX is 20 m,
XQ is 23 m, XR is 19 m, XS is 31 m.
Inference:
Using the given hints we can create a following figure:
Reference:
Car A starts from point R and travels 28 m in south direction and then it takes a left turn and travels 45 m. Car B starts from point P and travels 26 m in east direction. Car C starts from point S and travels 7 m in north direction and then it takes a left turn and travel 5 m and again it takes a left turn and travels 30 m.
Inference:
Here, the paths of car A, car B and car C will be denoted by red, blue and green colour respectively.
In the above figure I.P. means initial position and F.P. means final position.
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Team Smartkeeda | 5,348 | 21,226 | {"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} | 4.28125 | 4 | CC-MAIN-2020-34 | latest | en | 0.961297 |
https://www.numwords.com/words-to-number/en/2505 | 1,563,880,610,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529276.65/warc/CC-MAIN-20190723105707-20190723131707-00100.warc.gz | 775,751,640 | 1,759 | NumWords.com
# How to write Two thousand five hundred five in numbers in English?
We can write Two thousand five hundred five equal to 2505 in numbers in English
< Two thousand five hundred four :||: Two thousand five hundred six >
Five thousand ten = 5010 = 2505 × 2
Seven thousand five hundred fifteen = 7515 = 2505 × 3
Ten thousand twenty = 10020 = 2505 × 4
Twelve thousand five hundred twenty-five = 12525 = 2505 × 5
Fifteen thousand thirty = 15030 = 2505 × 6
Seventeen thousand five hundred thirty-five = 17535 = 2505 × 7
Twenty thousand forty = 20040 = 2505 × 8
Twenty-two thousand five hundred forty-five = 22545 = 2505 × 9
Twenty-five thousand fifty = 25050 = 2505 × 10
Twenty-seven thousand five hundred fifty-five = 27555 = 2505 × 11
Thirty thousand sixty = 30060 = 2505 × 12
Thirty-two thousand five hundred sixty-five = 32565 = 2505 × 13
Thirty-five thousand seventy = 35070 = 2505 × 14
Thirty-seven thousand five hundred seventy-five = 37575 = 2505 × 15
Forty thousand eighty = 40080 = 2505 × 16
Forty-two thousand five hundred eighty-five = 42585 = 2505 × 17
Forty-five thousand ninety = 45090 = 2505 × 18
Forty-seven thousand five hundred ninety-five = 47595 = 2505 × 19
Fifty thousand one hundred = 50100 = 2505 × 20
Sitemap | 385 | 1,245 | {"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} | 3.171875 | 3 | CC-MAIN-2019-30 | latest | en | 0.82308 |
https://www.physicsforums.com/threads/making-sense-of-differentials.790298/ | 1,696,417,361,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00881.warc.gz | 1,002,688,161 | 17,217 | # Making Sense of Differentials
• PFuser1232
#### PFuser1232
When I first came across differentials, I was told that they could be thought of as infinitesimal changes. However, I can't get my head around how they're actually used to model physical problems. For example, if ##x## is the x-coordinate of a moving body, then ##dx## is an infinitesimally small change in position. More generally, if ##\vec{r}## is the position vector of a body, then ##d\vec{r}## is an infinitesimally small change in position. What I don't understand is how we think of things like distance, mass, and area in terms of differentials. For example, we think of ##dA## as an infinitesimally small area; ##dm## as an infinitesimally small mass; and ##ds## as an infinitesimally small distance. Why do we avoid the concept of "change" when talking about mass, area, and distance (to name a few)?
I was first introduced to the concept of differentiation at the age of 16 when taking my first calculus class. We were not told what ##\frac {dy}{dx}## actually meant, we were just told 'tricks and tips' so to speak for dealing with differentiation and integration. i.e. 'When you differentiate a polynomial, reduce the powers of x by 1, and multiply by the old power.' was the first thing we were taught. It wasn't until I was introduced to differentiation during mechanics and physics modules that the concept was explained in the terms you have described it. It may help to think of ##dA##, ##dm##, ##ds## as infinitesimally small 'amounts' of something, rather than changes. Can you provide an example of where you believe the concept of 'change' has been abandoned?
That's the problem with taking Calculus so young- you don't have the 'maturity' yet to grasp the theory behind the math (or at least your teachers don't think you do) so you just learn "rules". The derivative is defined in terms of a limit: $\frac{dy}{dx}= \lim_{h\to 0} \frac{f(x+h)- f(x)}{h}$. One result of that is that, while the derivative is NOT a fraction, it can be "treated like one". You cannot, for example, prove the "chain rule", $\frac{dy}{dx}\frac{dx}{dt}= \frac{dy}{dt}$ just by saying "the dx terms cancel" but you can proving by going back before the limit, cancelling in the "difference quotient" then taking the limit again.
Because the derivative can be "treated like a quotient" we introduce the "differentials", dx and dy, separately to make use of that property. You say you were "told that they could be thought of as infinitesimal changes". That's fine if you were taught "non-standard" Calculus where you actually introduce the notion of "infinitesmals" rigously. But that requires some very deep "symbolic logic" showing that one can introduce "infinitesmals" into the number system (and what you wind up with is NOT the standard real number system). Lacking that, you need to consider "differentials" as just a convenient notation that happens to work.
Last edited by a moderator:
. Why do we avoid the concept of "change" when talking about mass, area, and distance (to name a few)?
I'm not sure what you are asking. "Change" of things is often dealt with in physics. Is the question: "Why do we use expressions where symbols like $dy$ appear as ordinary variables instead of only as part of a fraction like $\frac{dy}{dx}$ ?"
If you browse contributions to old threads concerning differentials, you won't find a unanimous point of view - even from the experts who generally agree on other aspects of mathematics. I think differentials have a standard definition in differential geometry, but there is no standard definition in elementary calculus. The treatment of differentials in calculus varies from textbook to textbook. Physics texts often reason with differentials without establishing any rigorous definition for them.
Last edited:
That's the problem with taking Calculus so young- you don't have the 'maturity' yet to grasp the theory behind the math (or at least your teachers don't think you do) so you just learn "rules". The derivative is defined in terms of a limit: $\frac{dy}{dx}= \lim_{h\to 0} \frac{f(x+h)- f(x)}{h}$. One result of that is that, while the derivative is NOT a fraction, it can be "treated like one". You cannot, for example, prove the "chain rule", $\frac{dy}{dx}\frac{dx}{dt}= \frac{dy}{dt}$ just by saying "the dx terms cancel" but you can proving by going back before the limit, cancelling in the "difference quotient" then taking the limit again.
Because the derivative can be "treated like a quotient" we introduce the "differentials", dx and dy, separately to make use of that property. You say you were "told that they could be thought of as infinitesimal changes". That's fine if you were taught "non-standard" Calculus where you actually introduce the notion of "infinitesmals" rigously. But that requires some very deep "symbolic logic" showing that one can introduce "infinitesmals" into the number system (and what you wind up with is NOT the standard real number system). Lacking that, you need to consider "differentials" as just a convenient notation that happens to work.
I write the following assuming this post was a reply to my own.
I had calculus classes for two years before university, using the 'tricks' we had been taught. I had trouble with the concept of differentiation and integration for that time as it had not been properly explained. I suppose the teaching of differentiation followed the order it was developed by Newton and the concept of the limit was not mentioned until I reached university. It was then that we learned differentiation from First Principles in our first week of calculus. After that first week the differentiation and integration that had been a slight problem for me before, was now 'simple', so to speak. I felt like I had a much deeper understanding of the calculus I had learned over the past two years and more confident in my ability to differentiate functions more complicated than polynomials. I feel as though the concept of a limit should have been introduced at the earliest point, and then once we had understood the groundwork for what we were doing we could move onto learning the 'tricks' that help differentiate and integrate more easily. Although, you could argue that the two years of study before university gave me the foundation to appreciate the further insight gained from my introduction to the limit notation and the explanation of differentiation from First Principles.
Anyway, back to the topic at hand. OP I suggest reading 'Zero, the biography of a dangerous idea' by Charles Seife. The book focuses around the concepts of 0 and ∞, building up to an explanation of the development of calculus, and the different notation used. The passage also explains how Newton developed the idea (and his notation) and how Newton's ideas differed slightly from those of Leibniz. It then goes on to explain a little about the 0/0 problem and l'Hôpital's rule. The book is popular science and as such is easy to digest, if a little slow at getting to what it's trying to convey. | 1,608 | 7,066 | {"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} | 3.75 | 4 | CC-MAIN-2023-40 | latest | en | 0.975938 |
https://ninjasquad.org/find-middle-element-of-linked-list-in-java/ | 1,679,728,465,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00776.warc.gz | 491,259,903 | 21,860 | # Find Middle Element of Linked List in Java
In this article, we’ll learn how to find middle element of a linked list using multiple approach in Java.
### Approach 1: Keep track of the LinkedList size
In first approach, we can keep track of size of the linked list. We can have a size counter initialized as zero. Increase or decrease the counter by 1, on addition or deletion of nodes from linked list respectively.
In this case, middle element’s index will be (size-1)/2
``````class LinkedList<T> {
int size = 0;
public Node<T> get(int index) {
// get by index
return node;
}
size++;
}
public void delete(int index) {
// delete element
size--;
}
public void findMiddleElement() {
get((size - 1) / 2);
}
}
``````
### Approach 2: Traverse the LinkedList to find size
Sometime we have given only the head node of the linked list and no information about the size.
In simplest approach, we can traverse through whole linked list starting from head till end to find the size of the linked list.
In this case, middle element’s index will be (size()-1)/2
``````class LinkedList<T> {
public Node<T> get(int index) {
// get by index
return node;
}
public int size() {
int size = 1;
while (node.getNext() != null) {
node = node.getNext();
size++;
}
return size;
}
public void findMiddleElement() {
get((size() - 1) / 2);
}
}
``````
If linked list is having n elements, then this approach requires n iteration to get size and n/2 iteration to get middle elements.
Total iteration is (n + n/2)
### Approach 3: Fast and Slow pointers
This approach is also applicable when size of the linked list is unknown and only head node is given.
In this approach, we iterate through the linked list using two pointers. Fast pointer jumps 2 nodes in each iteration, and the slow pointer jumps only one node per iteration.
When the fast pointer reaches the end of the list, the slow pointer will be at the middle element.
``````class LinkedList<T> {
public Node<T> findMiddleElement() {
while (fast != null && fast.getNext() != null) {
fast = fast.getNext().getNext();
slow = slow.getNext();
}
return slow;
}
}
``````
This is best approach when size is unknown as we are able to find the middle elements in just n/2 iterations.
### Summary
We learned that it is always good to keep track of linked list size to find the middle element. Moreover, when size of the linked list is unknown then Fast and slow pointer approach is the way to go.
Source: Internet
We are offering free coding tuts
X | 594 | 2,499 | {"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 | 3 | CC-MAIN-2023-14 | latest | en | 0.717567 |
https://katinalindaa.com/skincare/how-many-moles-of-oxygen-are-present-in-one-mole-of-kclo3.html | 1,660,009,021,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00530.warc.gz | 333,413,895 | 18,334 | How many moles of oxygen are present in one mole of KClO3?
Contents
How many moles of O2 are in one mole of KClO3?
2- moles of KClO3 on heating produces 3- moles of oxygen gas.
How many moles are in KClO3?
1 grams KClO3 is equal to 0.0081599680129254 mole.
How many moles of O2 are made for every 2 mol of KClO3?
2KClO3 –> 2KCl + 3O2
The balanced equation tells us that we get 3 moles of O2 for every 2 moles of KClO3.
When 1 mole of ammonia and 1 mole of O2 are allowed to react then?
while 1 mole of O2 requires = 5/4 = 0.8 mole of NH3. As there is 1 mole of NH3 and 1 mole of O2 , so all oxygen will be consumed.
How many moles of O2 are formed from 1.65 moles of KClO3?
1.65 mol KClO33mol O2= 2.475 mol O22 mol KClO32. How many moles of KClO3 are needed to make 3.50 moles of KCl?
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What is the mole of O2?
One mole of oxygen gas, which has the formula O2, has a mass of 32 g and contains 6.02 X 1023 molecules of oxygen but 12.04 X 1023 (2 X 6.02 X 1023) atoms, because each molecule of oxygen contains two oxygen atoms.
How many moles of KClO3 are used to produce 10 moles O2?
3. How many moles of KClO3 is used to produce 10 moles of O2? x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3 (3 mol O2) Page 2 4.
How many moles of O2 can be produced from 12 moles KClO3?
How many moles of O2 can be produced by letting 12.00 moles of KClO3 react? The KClO3 / O2 molar ratio is 2/3. 2 mol KClO3 / 3 mol. O2 = 12.00 mol KClO3 / x = 18.00 mol.
How many moles of KClO3 are required for producing 2.4 moles of oxygen?
Answer: The amount of KClO3 required 196 g. Given: 196 g of is requires to produce 2.4 moles of oxygen.
How many moles of oxygen are produced when 26.5 mol of aluminum oxide are decomposed 17.7 mol?
∴ 26.5 mole of the Aluminium oxide produces (3/2) × 26.5 mole of the Oxygen. = 39.75 moles. Hence, the numbers of the moles of the oxygen produced is 39.75 moles. | 693 | 1,951 | {"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} | 4.15625 | 4 | CC-MAIN-2022-33 | latest | en | 0.911409 |
https://discuss.leetcode.com/topic/64154/very-intesresting-que-given-dishes-and-its-ingredients-group-dishes-that-has-common-ingredients | 1,521,706,380,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647782.95/warc/CC-MAIN-20180322073140-20180322093140-00518.warc.gz | 570,008,639 | 14,168 | # Very Intesresting Que : Given dishes and its ingredients. Group dishes that has common ingredients
• E.g:
Input:
"Pasta" -> ["Tomato Sauce", "Onions", "Garlic"]
"Chicken Curry" --> ["Chicken", "Curry Sauce"]
"Fried Rice" --> ["Rice", "Onions", "Nuts"]
Follow up: What is the time and space complexity?
• This is an interesting yet simple question. I was trying to use 'sets' to compare intersections. I'm guessing the input is a hashmap.
EDIT: I guess using sets was not the best idea. I create additional dictionary:
``````def group_items(mydict):
d2 = collections.defaultdict(list)
for k, v in mydict.iteritems():
for name in v:
d2[name].append(k)
out = [i for i in map(tuple, d2.values()) if len(i) > 1]
return out
mydict = {"Pasta":["Tomato Sauce", "Onions", "Garlic"], "Chicken Curry":["Chicken", "Curry Sauce"], \
"Fried Rice":["Rice", "Onions", "Nuts"], "Salad":["Spinach", "Nuts"],
print group_items(mydict)
``````
output:
``````[('Sandwich', 'Quesadilla'), ('Salad', 'Fried Rice'), ('Chicken Curry', 'Quesadilla'), ('Fried Rice', 'Pasta')]
``````
• Assuming the input is map, following java code solves the problem. Complexity will be O(n^2) as each item needs to iterated over at least once. I simply invert the map from a dish->List<Ingredients> to Ingredient -> List<Dishes> .
``````static void printDishesWithCommonIngredients(Map<String, String[]> dishIngredientsMap)
{
Map<String, Set<String>> invertedMap = new HashMap<String, Set<String>>();
for (String dish : dishIngredientsMap.keySet())
{
String[] ingredients = dishIngredientsMap.get(dish);
for (String ing : ingredients)
{
if (invertedMap.containsKey(ing))
{
}
else
{
Set<String> dishSet = new HashSet<String>();
invertedMap.put(ing, dishSet);
}
}
}
for (Set<String> dishes : invertedMap.values())
{
if (dishes.size() > 1)
System.out.println(Arrays.toString(dishes.toArray()));
}
}
``````
• Similar to modqhx, basically creating an inverse dictionary and printing out values from it which are greater than 1 ingredient. So running complexity should be O(n) since we are running through each value (of each key) only once while creating a dictionary of equivalent size. So space will be O(n) as well?
``````d = { "pasta": ["Tomato Sauce", "Onions", "Garlic"],
"Chicken Curry": ["Chicken", "Curry Sauce"],
"Fried Rice" : ["Rice", "Onions", "Nuts"],
def group_by_ingredients(d):
d2 = {}
for k,v in d.iteritems():
for x in v:
d2.setdefault(x,[]).append(k)
for k,v in d2.iteritems():
if len(v) > 1:
print v
``````
=============
``````>>> d
{'Chicken Curry': ['Chicken', 'Curry Sauce'], 'Sandwich': ['Cheese', 'Bread'], 'Salad': ['Spinach', 'Nuts'], 'Quesadilla': ['Chicken', 'Cheese'], 'Fried Rice': ['Rice', 'Onions', 'Nuts'], 'pasta': ['Tomato Sauce', 'Onions', 'Garlic']}
>>> group_by_ingredients(d)
['Fried Rice', 'pasta']
``````
• @dhawalverma got something similar:
``````public void group(Map<String, String[]> map) {
Map<String, Set<String>> invertedMap = new HashMap<>();
for (Map.Entry<String, String[]> entry : map.entrySet()) {
for (String ingredient : entry.getValue()) { | 858 | 3,088 | {"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-2018-13 | latest | en | 0.708853 |
https://www.electro-tech-online.com/threads/logic-divider-question.158800/ | 1,701,355,031,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00026.warc.gz | 861,425,368 | 22,110 | Continue to Site
# Logic divider question
Status
Not open for further replies.
#### throbscottle
##### Well-Known Member
I made an ajustable interval timer using a 4060 oscillator/divider, which works well, but quite naturally gives a maximum useful interval of t/2^13. Since it's a 14 stage divider, is there a 'neat trick' to get an interval using the full cycle of the output instead of just the high or just the low portion?
Sorry, you need to explain that better as I've no idea what you mean. Mind you, could be the wine, will look again in the morning, may make sense then.
Mike.
You can make the counter programmable by passing 0's through diodes to the reset input, after inverting it.
However that only reduces the max interval, sounds like you want to increase it.
I'm not stating but I dont think you change the max cycle count, as the clock signal between stages isnt accessible.
One 'trick' you might be able to do if you want a multiple of 2 division, maybe even /4 is to run the 4060's osc at 1/2 or 1/4 your incoming clock freq, and use said clock freq signal to synchronise the 4060's osc.
Ah, I've led everybody up the garden path, and I wasn't even trying!
It's being used as a stand alone 1-shot timer. So if the oscillator is running at 2kHz, for example, the 4060 is a 14 stage divider, so the 14th stage gives me an output of 0.122Hz, the period of which is 8.192 seconds.
But as a simple timer I can only use the high part or the low part of the output, so the oscillator needs to run at half the speed. So I was wondering if there is some way to use both half cycles and then have the timed interval end. OR'ing all the outputs together gets me most of the way. I was hoping there might be a "well known method" to use for this?
Nope. I've used the 4060 in a lot of timing applications, and if you want the convenience of the Q14 output driving the downstream circuit or device, then you get only the half-period.
BUT - if you need only an edge to indicate the end of the period (as opposed to a level throughout the period), then a series capacitor and resistor to Vdd on the Q14 output will produce a negative edge at the end of the full cycle. This is a negative-edge differentiator.
ak
Thanks AK. I was trying to get my head around doing that. For some reason I can't remember, I'd discounted it, but I'll re-visit the idea.
I think it was because the output keeps a transistor on for the interval being timed and I didn't want to change the circuit...
You could use a couple of sections of NAND gate to make a flip-flop.
Set it at power on or with the 4060 reset, then reset when the appropriate counter stage goes high or low, directly or via an C-R network to get an edge?
That allow any stage of the counter to be used, either high or low state.
That's a good idea - but also defeats the idea of a 1 chip circuit!
That's a good idea - but also defeats the idea of a 1 chip circuit!
Then chuck the 4060 away, and replace it with a PIC
Then you make it do whatever you want, and as long as you want.
There is another oscillator counter chip in the cmos range, it has loads of functions, cant remember the no. but you might be able to find it with google.
Otherwise might as well go with 2 4060's, or how about a 4040 series'd up to it.
The CD4521 is a 24 bit oscillator/divider. For the same oscillator components, the longest output period is 1K times longer than the 4060. However, the outputs start at Q18, so if you want something only 2x or 4x of your 4060 circuit, at least one osc component will have to change.
ak
#### Attachments
• CD4521.pdf
284.6 KB · Views: 67
The CD4521 is a 24 bit oscillator/divider. For the same oscillator components, the longest output period is 1K times longer than the 4060. However, the outputs start at Q18, so if you want something only 2x or 4x of your 4060 circuit, at least one osc component will have to change.
ak
I just used a cd4521 to make a clock from a stepper motor with a cd4017 as a divider. Clicks once every 18 seconds to move the minute hand around once per hour for a 200-step motor. I stumbled on the cd4521 last week in my junk box as I was struggling with the cd4060. I didn't realize it had the onboard oscillator circuitry like the 4060. Just add crystal and two caps.
The CD4521 has such a long chain of dividers, that you can still employ common 32.762Khz crystals and still have a precise 512 second period.
For longer periods of many hours, you can use the powerline frequency.
The CD4521 has such a long chain of dividers, that you can still employ common 32.762Khz crystals and still have a precise 512 second period.
For longer periods of many hours, you can use the powerline frequency.
or you can cascade a second cd4521 and get a delay of more than 6 months. Or anything between by cascading a cd4060.
Status
Not open for further replies. | 1,214 | 4,873 | {"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-2023-50 | latest | en | 0.950725 |
https://stats.stackexchange.com/questions/29389/multivariate-time-series-model-evaluation-with-conditional-moments | 1,723,595,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00118.warc.gz | 421,186,691 | 41,262 | # Multivariate time series model evaluation with conditional moments
Consider multivariate time series models that estimate potentially time-varying conditional means, variances, and correlations (one type of model might be a VAR(p)+Garch(1,1)+DCC Gaussian Copula model). I will simulate from one "true" distribution and compare other potential models to it. I don't particularly care about RMSE because I care just as much about the conditional covariance matrix as the conditional means. For the sake of simplicity, assume that the conditional mean and conditional covariance are sufficient to explain "true" distribution.
I can calculate the conditional mean vector and conditional covariance matrix of each model and the true model at any given point in time. Alternately, I can calculate the forecasted conditional means and covariances on a rolling basis.
1. Is there any reason to prefer the Kullback-Leibler divergence or Bhattacharyya distance as comparison methodologies or should I just do both? Alternately, is there a different method that is even more appropriate.
2. In the constant multivariate Gaussian case, is it possible to make statistical inferences based on these comparison methodologies, such as that the divergence is not statistically bigger than zero or that the divergence of one is bigger than another?
3. If I calculate the conditional means and variances at any given point in time, I could calculate the comparison statistic in each period. This would produce a time series of divergences/distances. Is it reasonable to make statistical inferences in this case by just assuming it is a typical time series?
• I don't get why you are throwing away essentially all the reputation you have accumulated so far to hope for an answer to this. The reason it is not getting attention is probably because it is so specicalized that no one who has seen it thus far has the expertise to answer it. If you get no answer for the next six days you lose 50 reputation points and get no benefit. Commented Jun 1, 2012 at 19:26
• I got +100 reputation for joining since I have more reputation on another stackexchange site. Since the question is important, figured it couldn't hurt.
– John
Commented Jun 4, 2012 at 16:23
• Why would you compare means and covariances between models, where one is such that you know it's true? Or are you trying to say you're knowingly fitting 'wrong' models on (real or simulated) data, and trying to come up with a routine that flags when something is wrong or when the model is dangerous or unsuitable or fits poorly? Is this for a trading strategy? Commented Nov 2, 2016 at 3:43
• @Taylor The idea was that I cared about the density as much as the mean. This question was from a few years ago. Nowadays I would probably use WAIC or Leave one out cross validation.
– John
Commented Nov 2, 2016 at 12:40
• @John my fault I didn't see the datestamp Commented Nov 2, 2016 at 15:32 | 647 | 2,931 | {"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-2024-33 | latest | en | 0.948841 |
https://www.mathsglow.com/page/3/ | 1,657,215,284,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00145.warc.gz | 912,481,829 | 12,416 | ## Intermediate Maths Solutions for Limits and Continuity Exercise 8(b)
Intermediate Maths solutions for Limits and Continuity Exercise 8(b) Inter Maths 1B Solutions for Exercise 8(b) Limits and Continuity are given. These solutions are very easy to understand. Please study the text book very well. Observe the example problems and solutions given in the text book. Try them. Observe the given solutions and try them [...]
## Intermediate Maths solutions for Limits and Continuity Exercise 8(a)
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## Inverse Trigonometric Functions,Intermediate first year Mathematics 1A chapter 8 problems with solutions,junior inter maths solutions for Trigonometry
INVERSE TRIGONOMETRIC FUNCTIONS, INTERMEDIATE FIRST YEAR MATHEMATICS 1A CHAPTER 8 PROBLEMS WITH SOLUTIONS Mathematics intermediate first year 1A and 1B solutions for some problems. These solutions are very easy to understand. First year 1A : Solutions for functions, mathematical induction, functions, addition of vectors, trigonometric ratios upto transformations, trigonometric equations, hyperbolic functions, inverse trigonometric functions [...]
## Trigonometric Equations,Intermediate Mathematics first year 1 A chapter 7 problems with solutions,Junior inter Trigonometry
TRIGONOMETRIC EQUATIONS, INTERMEDIATE FIRST YEAR 1 A CHAPTER 7 PROBLEMS WITH SOLUTIONS Mathematics intermediate 1A and 1B solutions for some problems. These solutions are very simple to understand. Inter first year 1A : Solutions for functions, mathematical induction, addition of vectors, trigonometric ratios upto transformations, trigonometric equations, hyperbolic functions, inverse trigonometric functions and properties of [...]
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Inter Maths Solutions for Limits and Continuity Exercise 8(d) intermediate maths solutions for Limits and Continuity exercise 8(d) textbook.. Thses are very easy to understand. Observe the example problems and solutions given in the textbook. Study the textbook lesson very well. M 1b solutions for textbook Inter Locus Exercise 1(a) Transformation of axes Exercise 2(a) [...]
## MATHEMATICAL INDUCTION, Intermediate 1st year problems with solutions,solutions for Mathematical Induction Inter first year 1A
MATHEMATICAL INDUCTION, INTERMEDIATE FIRST YEAR PROBLEMS WITH SOLUTIONS Mathematics intermediate first year 1A and 1B solutions for some problems. These solutions are very simple to understand. Junior inter 1A : Functions, mathematical induction, functions, addition of vectors, trigonometric ratios upto transformations, trigonometric equations, hyperbolic functions, inverse trigonometric functions and properties of triangles. Junior inter 1B : [...]
## Intermediate Maths Solutions for Exercise 10(a) Properties of Triangles
Intermediate Maths Solutions for Exercise 10(a) Properties of Triangles Inter Maths 1A text book Exercise 10(a) Properties of Triangles solutions are given. These solutions are very easy to understand. Please study the text book lesson Properties of Triangles very well. Observe the example problems and solutions given in the text book. Observe the given solutions [...] | 1,022 | 4,838 | {"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-2022-27 | latest | en | 0.877358 |
https://39thstreetgourmet.com/qa/how-much-does-2-cups-of-flour-weigh.html | 1,627,253,301,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151866.98/warc/CC-MAIN-20210725205752-20210725235752-00491.warc.gz | 89,942,560 | 9,178 | # How Much Does 2 Cups Of Flour Weigh?
## How many pounds is 2 cups?
Conversion Tablecups to poundscuplb10.521621.043231.564817 more rows.
## How many cups of flour are in a 2 lb bag?
6 2/3 cupsCups per pound:There are 3 1/3 cups of flour per pound and about 6 2/3 cups per 2-lb bag.
## What can I use to measure 1 cup?
measuring cup = standard coffee mug. measuring tablespoon = dinner spoon. measuring teaspoon = coffee spoon.
## How much yeast do I use for 3 cups of flour?
A packet of active dry yeast = 2 1/2 teaspoons. If using instant yeast, you can cut that down to 2 teaspoons. Assuming you are not using preferments, one packet is enough yeast for a standard recipe, which generally means 3-4 cups flour.
## How much does cup of flour weigh?
125 grams1 cup of flour weighs 125 grams. The volume is the same, but the weight is different (remember: lead and feathers). One other benefit to using metric measurements is accuracy: scales often only show ounces to the quarter or eighth of an ounce, so 4 1/4 ounces or 10 1/8 ounces.
## How much does 2 cups of sugar weigh?
400 grams2 US cups of granulated sugar weighs 400 grams. (or precisely 399.834119685 grams.
## How many ounces is 2 cups of sugar?
Conversion Tableounces to cupsozcup10.140620.281330.421917 more rows
## How can I measure 1 cup of flour without a measuring cup?
Pour more flour into the mug or use a spoon to scoop some out until you get it at the right level. If you need less than 1 cup (120 g) of flour, just fill the cup up less according to the amount you need. For example, if you need 1/2 cup (60 g) of flour, fill it up to about 7-8 mm below halfway.
## How much is 2 cups in grams?
Packed Brown SugarCupsGramsOunces1/4 cup55 g1.9 oz1/3 cup73 g2.58 oz1/2 cup110 g3.88 oz1 cup220 g7.75 ozNov 19, 2020
## How many cups is 250g flour?
1½ cupsWhite flour – plain, all-purpose, self-raising, speltWHITE FLOUR – GRAMS TO CUPSGramsCups100g½ cup + 2 tbsp200g1¼ cups250g1½ cups + 1 tbsp4 more rows•Sep 20, 2018
## How many grams is 2 cups all purpose flour?
240 grams2 US cups of all purpose flour weighs 240 grams.
## How do I measure a cup of rice without a measuring cup?
Press your finger through the rice and see how much rice you’ve added to the pot. Then place your finger tip on top of the layer of the rice. The same measure you just noted is the same amount of water to add.
## Which flour absorbs more water?
Here are a few simple things to keep in mind concerning flour absorption. Higher protein flour absorbs more water than lower protein flour. This means that a recipe that calls for bread flour may require more water than one that uses all-purpose flour.
## How much does 2 cups self raising flour weigh?
Self-Rising Flour: 1 cup = 4 ounces = 113 grams. Baking powder: 1 teaspoon = 4 grams. Baking soda: 1/2 teaspoon = 3 grams. Butter: 1/2 cup =1 stick = 4 ounces = 113 grams.
## What does 2 cups weigh?
Begin converting cups to pounds by understanding a few basic conversion points. 16 ounces equals one pound or two cups. Another way to look at the equivalent is that one cup weighs eight ounces and therefore two cups equal 16 ounces and this is the same weight of one pound–16 ounces.
## How much water do I use for 2 cups of flour?
1 cup waterThe ideal ratio is 1 cup water to 2 cups flour. The trick however is to add water in several stages. Add one third water at the beginning to the flour and try and consolidate everything in to a ball. Add another one third and try to smooth the whole ball at much as possible.
## How can I measure 2 cups without a measuring cup?
Use an object as a reference point.A teaspoon is about the size of the tip of your finger.A tablespoon is about the size of an ice cube.1/4 cup is about the size of a large egg.1/2 cup is about the size of a tennis ball.A full cup is about the size of a baseball, an apple or a fist. X Research source
## How much water do I use for 3 cups of flour?
Divide the weight of the water by the weight of the flour and then multiply the result by 100. For example, a recipe containing 1 1/4 cups of water (10 ounces) and 3 cups of all-purpose flour (15 ounces) will have a 67 percent (10/15 x 100 = 67) hydration level, indicating a moderately airy crumb.
## How many ounces is 2 cups of flour?
8.812 US cups of flour weighs 8.81 ( ~ 8 3/4) ounces.
## How do you measure 2 cups of flour?
How to Measure Flour with Measuring CupsFirst, fluff up the flour in the bag or canister. Flour settles easily, becoming tightly packed inside a bag or jar. … Second, spoon the flour into the measuring cup. … Then, scrape a knife across the top of the measuring cup to level the flour.
## How do you convert cups to weight?
Please be careful as a cup of water weighs more than a cup of flour for example….How to convert weight to cups in recipes.US cupsMetricImperial1/4 cup30g1 oz1/3 cup40g1 1/2 oz1/2 cup65g2 1/4 oz2/3 cup85g3 oz3 more rows•Jan 31, 2013 | 1,328 | 4,950 | {"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} | 3.328125 | 3 | CC-MAIN-2021-31 | latest | en | 0.893155 |
https://www.bionicturtle.com/forum/tags/modified-duration/ | 1,611,442,632,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538741.56/warc/CC-MAIN-20210123222657-20210124012657-00646.warc.gz | 678,899,158 | 13,774 | What's new
# modified-duration
1. ### Calculation of macaulay duration and Modified duration
Hi, I am confused on the difference in how John Hull (John Hull Example 4.6) and Tuckman (Tuckman Table 4.6) calculates their Macaulay duration in order to determine the modified duration. In John Hull's example, he uses continuous compounding to determine present value of the cash flows and...
2. ### YouTube T4-36: Fixed Income: Simple bond illustrating all three durations (effective, mod, Mac)
Macaulay duration is the bond's weighted average maturity (where the weights are each cash flow's present value as a percent of the bond's price; in this example, the bond's Macaulay duration is 2.8543 years. Modified duration is the true (best) measure of interest rate risk; in this example...
3. ### YouTube T4-35: Fixed Income: Modified and Macaulay Duration
Using my rebuild of Bruce Tuckman's Table 4-6, this video illustrates the calculation of Macaulay and modified duration. Macaulay duration is the bond's weighted average maturity. Modified duration is the best measure of the bond's interest rate risk.
4. ### Hull, Instructional video , ch4 -Duration
Dear David, Thanks a lot for video lectures they are much inspiring Still I was little bit confused with all these different names duration, modified duration, Macauly duration,.. etc...I will shortly examine mine view of this and kindly ask you to comment ( but without laughing:)) According to...
5. ### Derivation of modified duration formula for par bond
In which study guide or video can i find the derivation of the formula of modified duration of par bond? I understand MD formula for par bond is: MD=[1-(1+y)˄-maturity] / y But still searching for derivation..
6. ### P1.T3.714. Duration, modified duration and dollar duration (Hull Chapter 4)
Learning objectives: Calculate the duration, modified duration, and dollar duration of a bond. Evaluate the limitations of duration and explain how convexity addresses some of them. Questions 714.1. A very risky two-year bond with a face value of \$100.00 pays a semi-annual coupon of 18.0% and... | 487 | 2,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-04 | latest | en | 0.867473 |
http://mathmastersnyc.com/section-1.2-straight-lines.html | 1,516,664,537,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891546.92/warc/CC-MAIN-20180122232843-20180123012843-00369.warc.gz | 212,405,995 | 13,762 | Try the Free Math Solver or Scroll down to Tutorials!
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# Section 1.2: Straight lines
(Terminology: All lines are “straight” lines. Every line is a curve. Not every curve is a line.
A line is a curve that is not curved.)
We will review some familiar facts about lines very quickly with very few examples:
The line has constant “steepness”, not the other curve. The line rises at a constant rate, not the
other curve.
Picture of line, with two distinguished points
Our measure of steepness: slope,
(Question: Why do we refer to slope with the letter, “m”; why not, say, “s”?)
Note that it does not matter to which of the two points we affix the subscripts, “1” and “2”.
In fact, it does not matter which two points on the line we choose in order to compute the slope.
Our measure of steepness is a signed measure.
Line is rising Line is falling Horizontal line Vertical line m>0 m>0 m=0 m: undefined (Don’t say: “No slope”).
Convince yourself for each of the above pictures, that the claimed sign of the slope is correct.
In each case, pick any two points along the line, one with x1 and one with x2; choose one as the
starting point, one as the final point. Note the sign of the run, note the sign of the rise,
What is the sign of the resulting ratio,
What does the sign of the slope tell us about the relationship between x and y?
If the slope is positive, we say that x and y have a “positive” relationship. In this case, x and y
move in the same direction: if x increases, so does y; if x decreases, so does y.
If the slope is negative, we say that x and y have a “negative” relationship. In this case, x and y
move in opposite directions: if x increases, then y decreases; if x decreases, then y increases.
The steeper the line, the greater the magnitude of the slope.
Equation of a line: The equation of a line represents a condition which is satisfied by every
point on that line and only by the points on that line.
Different forms for the equation of a line (This is not an exhaustive list.):
1. Equation of a horizontal line: A horizontal line consists of all of the points with the same
y-coordinate, b. Therefore the equation of a horizontal line can be written y = b.
2. Equation of a vertical line: A vertical line consists of all of the points with the same
x-coordinate, a. Therefore the equation of a vertical line can be written x = a.
3. Point-slope form: A line with slope m which passes through the point (x0, y0 ) satisfies the
equation y − y0 = m(x − x0 ). (Notice that this is just a restatement of the definition of slope:
namely, rise = slope & run.).
4. Slope-intercept form: A line with slope m and y-intercept b satisfies the equation y = mx + b.
(Remember, the y-intercept of a line is the y-coordinate of the point where the line intersects the
y-axis, i.e., the y-coordinate of the point on the line with x-coordinate 0. The x-intercept is
defined similarly.)
Question:
True or false? Every line has exactly one y-intercept.
If this statement is false, describe the exceptional lines and the number of y-intercepts they
possess.
Answer: False. There are lines which lack a y-intercept. (Which lines are they?) There is one line
which has infinitely many y-intercepts. (Which line is it?)
Example: Find the equation of the line which passes through the points (3, 2) and (1, 5).
First, we find the slope of the line:
The point-slope form of the line is therefore, .
Solving for y, we obtain the slope-intercept form: (y-intercept = 13/2)
(To check your answer, you should plug in the two specified points which must lie on the line.
For example, when x = 3, It checks! Now check the point, (1, 5).)
5. General form: The equation of every line can be expressed in so-called “general form”,
Ax + By = C, where A, B, and C are constants, and A and B cannot both be zero.
Question: Why was this last restriction (“A and B cannot both be 0”) added? If both A and B are
equal to zero, what is the graph of the resulting equation.
Hint: Consider two cases: 0x + 0y = 0 and 0x + 0y = C, where C ≠ 0
There are infinitely many ways to write the equation of a given line in general form. After all,
multiplying the equation by any non-zero number does not change the set of points which satisfy
the equation.
So which numbers should we use as our A, B, C?
Some like to insist that A, B, and C be integers (if possible), with no common factors.
We may sometimes impose this restriction, but a far more sensible restriction would be to insist
that (if possible) A, B, and C possess meaningful interpretations.
Example: Refer to the last example. Write the equation of the line in general form.
Move the x-term to the left by addition to obtain, .
Since there is no context for meaningful interpretation, we may as well pretty up the equation:
multiply both sides by 2 to obtain 3x + 2y = 13.
Once past Chapter 1, we will work almost exclusively with the general form. We will refer to
Ax + By = C, as a linear equation in two variables. The obvious extensions are Ax + By + Cz = D,
a linear equation in three variables,...etc.,... , a linear equation
in n variables.
(Question: What is so general about the general form ? Why isn’t y = mx + b sufficiently
“general”?.
Hint: Try to write the equation of a vertical line in the form y = mx + b.)
The slope-intercept form, y = mx + b, is often used when we are studying the relationship
between two variables x and y, and it seems natural to us to think of y “depending” on x.
The general form is often used when we are studying the relationship between two variables x
and y, and do not see any basis for arbitrarily picking out one of the variables, say y, and calling it
the “dependent” variable, and isolating this dependent variable on the left hand side of the
equation. (See the widget example below. Is there any natural choice for “dependent” variable?))
Example: Let x and y represent the number of Deluxe and Regular widgets produced per day,
respectively. Each Regular widget requires 2.5 hours of labor; each Deluxe widget requires 3.2
hours of labor; there are 18 available hours of labor per day. If all of the available labor is to be
used each day, then x and y must satisfy the relationship 2.5x + 3.2y = 18.
Anticipatory note: The constraint we have imposed on production of widgets does not seem all
that reasonable. If we have 18 available hours of labor, then we cannot use more than 18 hours,
but it should certainly be possible to use less than 18 hours. When we get to “linear
programming” in Chapter 3, we will impose a much more realistic constraint on the production
of widgets, namely the inequality 2.5x + 3.2y ≤ 18.
To graph a line:
It is clear to all of you that two points determine a line, so that once you have plotted any two
points on the line, the graph of the line can be constructed by using a straightedge to draw the
line passing through the two specified points.
Recommended method for sketching (non-vertical, non-horizontal) lines:
Set x = 0, solve for y to obtain the y-intercept. Plot it.
Set y = 0, solve for x to obtain the x-intercept. Plot it.
Sketch the line passing through these two intercepts.
This method has many advantages for us:
1. Since we will be working with equations in general form, it is especially easy to solve for
the intercepts. To find the y-intercept, just cover up the x-term, and solve for y. To find the
x-intercept, just cover up the y-term, and solve for x.
2. Since the points are on the axes, not in the interior of the quadrants, they are easier to plot.
3. As we will see in Chapters 3 and 4, the intercepts often have great significance.
Example: Sketch the line, 3x + 2y = 13. (The x-intercept is 13/3 ; the y-intercept is 13/2.)
Note: This method will not work on lines which pass through the origin. If the line passes
through the origin, you will need to plot a 2nd point in addition to the origin. You could plug any
value of x (or y) into the equation, solve for the other coordinate, then plot the resulting point.
Example: Sketch the line, 2x - y = 0. (One point is (0,0). To get a second point, I set x equal to
(say) 1, solve for y. My second point is (1,2).)
Assignment: Work the problems in Section 1.2. These should be familiar. If they are not
familiar, then you should certainly work these problems. | 2,243 | 8,782 | {"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} | 4.40625 | 4 | CC-MAIN-2018-05 | latest | en | 0.940644 |
https://www.basculasbalanzas.com/understanding-measures/ | 1,718,367,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00470.warc.gz | 592,699,330 | 13,211 | # Understanding Measures
Measures are an important concept in mathematics, physics and other disciplines. These mathematical objects allow a comparison of the properties of physical objects. They are used in a variety of contexts, including probability theory and integration theory.
In mathematics, a measure is a countably additive set function with values in the real numbers or infinity. The foundations of modern measure theory were laid by such mathematicians as Emile Borel, Henri Lebesgue, Nikolai Luzin, and Johann Radon.
## Units
A unit is a standard measurement that can be used to describe the size of an object or amount of something. It can be a number, symbol or abbreviation. There are two major systems of units that are commonly used: the metric system and the U.S customary system. In physics, there are seven fundamental physical quantities that can be measured in base units, which are the meter, kilogram, second, ampere, Kelvin, mole and candela (Table 1.1). Other physical quantities are described by mathematically combining these base units.
When performing calculations, it is important to know the units that are being used. For example, if a measurement is given in gallons and cups, the conversion factor must be used to convert from one unit to the other. This will make the calculation make sense. For example, 1 gallons equals 8 fluid ounces.
## Uncertainty
If three different people measure the length of a piece of string, each will get slightly different results. This variation is due to uncertainty in the measurement process. This uncertainty can be reduced by using a more precise measurement technique. However, there is no way to eliminate it completely.
The most realistic interpretation of a measured value is that it represents a dispersion of possible values. This is sometimes described as a’most probable’ or ‘true’ value, but this is arbitrary and at the whim of the metrologist who uses the estimation method.
The combined standard uncertainty is the product of the standard uncertainties of all input quantities, including any corrections for systematic errors. The combined standard uncertainty is often multiplied by a coverage factor to obtain an expanded measurement uncertainty which indicates the range of values that could reasonably represent the true quantity value within a specified level of confidence. This coverage factor is typically a Type A evaluation, but it may also include a Type B component.
## Scales
Scales are a fundamental part of musical theory and one of the most important concepts to understand if you want to play music. They are the building blocks of chords and harmonic progressions, and knowing them can help you play songs in any key. Scales are also useful for improvising and songwriting.
A scale is a set of notes that belong together and are ordered by pitch. They are a basis for melodies and harmony, and create various distinctive moods and atmospheres. There are many different scales, including major, minor and church modes.
A scale is a sequence of notes, and the intervals between them are what determine its quality. Intervals can be either tones or semitones. A tone is the distance between two adjacent frets, and a semitone is the distance between a note and its next higher or lower note. These intervals are called scale steps, and they are used to define the pattern of the scale.
## Measures of a set
Measures of a set are a fundamental concept in mathematical analysis, probability theory, and more. A measure is a function that assigns a length or area to a set. Its value is the sum of all the elements in the set. It is called a finite measure if its sum is a real number, or s-finite if it can be decomposed into a countable union of measurable sets with finite measure.
The concept of measures is also used in physics to describe the distribution of mass or other conserved properties. Negative values are often seen as signs, resulting in signed measures. The study of the geometry of measures is one of the main goals of geometric measure theory. A core result in this area is the class of rectifiable measures. Other important results include the characterization of non-rectifiable measures and a generalization of the Riemann integrable functions.
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Understanding Cardinality in E-R Diagrams
Navigate the Article
Key Takeaways
– Cardinality refers to the number of times one entity can be associated with another entity in an E-R diagram.
– Participation refers to whether an entity must participate in a relationship with another entity to exist.
– Cardinality can be represented using different notations, such as crowsfeet or min/max notation.
– Consistency in notation is important when representing cardinality constraints in E-R diagrams.
– There are two approaches to specifying cardinality and participation constraints: the “Look-Here” method and the “Look-Across” method.
Introduction
E-R diagrams, also known as Entity-Relationship diagrams, are a visual representation of the relationships between entities in a database. These diagrams are widely used in database design to model the structure and behavior of a database system. One important aspect of E-R diagrams is cardinality, which determines the number of times one entity can be associated with another entity. In this article, we will explore the concept of cardinality in E-R diagrams and its significance in database design.
Understanding Cardinality
Cardinality is a fundamental concept in E-R diagrams that helps define the relationships between entities. It specifies the number of instances of one entity that can be associated with another entity. In simpler terms, cardinality tells us “how many” times one entity can be linked to another. For example, in a database representing a university, the cardinality between the “Student” entity and the “Course” entity could be one-to-many, indicating that one student can be enrolled in multiple courses, but each course can have multiple students.
Crowsfeet Notation
One popular method to represent cardinality in E-R diagrams is the crowsfeet notation. In this notation, cardinality is depicted using decorations on the ends of lines connecting entities. A straight line perpendicular to the relationship line represents a cardinality of one, indicating that one instance of an entity is associated with another. On the other hand, a three-pronged ‘crow-foot’ symbol represents a cardinality of many, indicating that multiple instances of an entity are associated with another. This notation provides a visual representation of the cardinality constraints between entities, making it easier to understand the relationships in the database.
IE Method
Another method to represent cardinality in E-R diagrams is the IE method. This method is similar to crowsfeet notation but does not show attributes related to a relationship. Instead, cardinality and participation constraints are combined into min/max notation, represented by bars and crowfoot symbols. The min/max notation specifies the minimum and maximum number of instances of an entity that can be associated with another entity. For example, a bar on one end of the relationship line indicates a minimum cardinality of zero, meaning that an instance of the entity is not required to be associated with another. On the other hand, a crowfoot symbol represents a maximum cardinality of many, indicating that multiple instances of the entity can be associated with another.
Specifying Cardinality Constraints
When creating E-R diagrams, it is important to choose a methodology and stay consistent with the notation used for cardinality constraints. There are two approaches to specifying cardinality and participation constraints: the “Look-Here” method and the “Look-Across” method. The Look-Here method specifies the constraints next to the entity, making it easier to understand the cardinality at a glance. For example, a cardinality of one-to-many can be represented as “1:M” next to the entity. On the other hand, the Look-Across method requires looking to the other side of the relationship for meaning. This method can be more intuitive for some users, as it represents the cardinality in relation to the other entity involved in the relationship.
Benefits of Understanding Cardinality
Understanding cardinality is crucial for accurately representing relationships between entities in E-R diagrams. It helps database designers and developers define the structure and behavior of a database system. By correctly specifying the cardinality constraints, it becomes easier to query and manipulate data in the database. Additionally, cardinality constraints play a vital role in ensuring data integrity and preventing inconsistencies in the database. By enforcing cardinality constraints, database systems can maintain the integrity of relationships between entities and avoid data anomalies.
Conclusion
In conclusion, cardinality is an essential concept in E-R diagrams that determines the number of times one entity can be associated with another entity. It helps define the relationships between entities and plays a crucial role in database design. By using notations such as crowsfeet or min/max notation, cardinality constraints can be visually represented in E-R diagrams. Consistency in notation and understanding the different approaches to specifying cardinality constraints are important for accurately representing relationships. By understanding and correctly specifying cardinality constraints, database designers can create robust and efficient database systems that maintain data integrity and support complex relationships between entities. | 991 | 5,412 | {"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-2024-18 | latest | en | 0.901155 |
http://www.dreamincode.net/forums/topic/138748-reverse-an-array/ | 1,524,681,789,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00365.warc.gz | 405,203,005 | 18,291 | # Reverse an Array
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## 2 Replies - 4540 Views - Last Post: 13 November 2009 - 09:11 AMRate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://www.dreamincode.net/forums/index.php?app=forums&module=ajax§ion=topics&do=rateTopic&t=138748&s=3f5cd751c656577a92e2696c8ce79afd&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>
### #1 Ibanez Gio
• New D.I.C Head
Reputation: 0
• Posts: 8
• Joined: 25-October 09
# Reverse an Array
Posted 12 November 2009 - 10:57 PM
Ok so my assignment is to create a purse class and stuff and then have the methods in it called reverseOrder() and transfer(Purse other). Reverse will just take the the array and put themin opisite order (IE quarters, dimes, nickles, pennies turns into pennies, nickels, dimes, quarters)
The transfer will have purse a and b and say a has quarter, dime and purse b has nickel, penny then a.transfer( a is Purse[quarter,dime,nickel,penny]
any help??
here is the code
```import java.util.ArrayList;
public class Purse
{
ArrayList<String> coins;
public Purse()
{
coins = new ArrayList<String>();
}
public void addCoins(String coinName)
{
}
public String toString()
{
String preamble = "Purse[";
String body = "";
String postamble = "]";
for(int i = 0; i < coins.size(); i++)
{
body = body + coins.get(i);
if (i < coins.size() -1)
{
body = body + ", ";
}
}
return preamble + body + postamble;
}
public String reverse()
{
String preamble = "Purse[";
String body = "";
String postamble = "]";
for(int i = (coins.size() - 1); i >= 0; i--)
{
body = body + coins.get(i);
if (i > 0)
{
body = body + ", ";
}
}
return preamble + body + postamble;
}
public void transfer(Purse other)
{
}
public void reverseCoins()
{
}
}
```
Is This A Good Question/Topic? 0
## Replies To: Reverse an Array
### #2 Dophert
Reputation: 10
• Posts: 60
• Joined: 19-October 09
## Re: Reverse an Array
Posted 13 November 2009 - 01:43 AM
Why not use the Collections.reverse(List<E>) method ?
### #3 macosxnerd101
• Games, Graphs, and Auctions
Reputation: 12278
• Posts: 45,364
• Joined: 27-December 08
## Re: Reverse an Array
Posted 13 November 2009 - 09:11 AM
Because of the resizable nature of ArrayList vs. the static nature of an array, it might be easier to use the Collections.swap() method to write your own reverse() method. In this way, you can insure that your algorithm takes n/2 time, where n is the number of elements.
You can implement something like this:
```0. ArrayList <-- get ArrayList
1. i <-- 0
2. k <-- ArrayList.size()
3. while i < k
4. swap(i, k) //swap elems at i and k
end while
end algorithm
```
This is easier and more efficient than moving n items to the end of the list.
Try implementing this solution and feel free to post if you need any more help. Good luck!
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http://www.mathscore.com/math/practice/Fractions%20to%20Decimals/ | 1,718,588,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00154.warc.gz | 44,542,067 | 3,478 | MathScore EduFighter is one of the best math games on the Internet today. You can start playing for free!
## Fractions to Decimals - Sample Math Practice Problems
The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.
See some of our other supported math practice problems.
### Complexity=1
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1.
182100
2.
8850
### Complexity=3
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1.
56100
2.
125
### Complexity=5
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1.
24
2.
2825
### Complexity=7
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1.
126100
2.
3825
### Complexity=5, Mode=multiple
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1.
164200
2.
147300
### Complexity=15, Mode=multiple
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1.
24400
2.
248400
### Complexity=1
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1
182100
Solution
With denominator=100, take 182.0 and move the decimal two places to the left to get 1.82.
2
8850
Solution
Make the denominator 100 by multiplying the numerator and denominator by 2.
Now you have this:
176100
With denominator=100, take 176.0 and move the decimal two places to the left to get 1.76.
### Complexity=3
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1
56100
Solution
With denominator=100, take 56.0 and move the decimal two places to the left to get 0.56.
2
125
Solution
Make the denominator 100 by multiplying the numerator and denominator by 4.
Now you have this:
4100
With denominator=100, take 4.0 and move the decimal two places to the left to get 0.04.
### Complexity=5
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1
24
Solution
Make the denominator 100 by multiplying the numerator and denominator by 25.
Now you have this:
50100
With denominator=100, take 50.0 and move the decimal two places to the left to get 0.5.
2
2825
Solution
Make the denominator 100 by multiplying the numerator and denominator by 4.
Now you have this:
112100
With denominator=100, take 112.0 and move the decimal two places to the left to get 1.12.
### Complexity=7
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1
126100
Solution
With denominator=100, take 126.0 and move the decimal two places to the left to get 1.26.
2
3825
Solution
Make the denominator 100 by multiplying the numerator and denominator by 4.
Now you have this:
152100
With denominator=100, take 152.0 and move the decimal two places to the left to get 1.52.
### Complexity=5, Mode=multiple
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1
164200
Solution
Make the denominator 100 by dividing the numerator and denominator by 2.
Now you have this:
82100
With denominator=100, take 82.0 and move the decimal two places to the left to get 0.82.
2
147300
Solution
Make the denominator 100 by dividing the numerator and denominator by 3.
Now you have this:
49100
With denominator=100, take 49.0 and move the decimal two places to the left to get 0.49.
### Complexity=15, Mode=multiple
Find the decimal equivalent to the fraction. Example: 1/2 = .5
1
24400
Solution
Make the denominator 100 by dividing the numerator and denominator by 4.
Now you have this:
6100
With denominator=100, take 6.0 and move the decimal two places to the left to get 0.06. | 1,057 | 3,854 | {"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} | 4.34375 | 4 | CC-MAIN-2024-26 | latest | en | 0.773146 |
https://www.doorsteptutor.com/Exams/CBSE/Class-12/Economics/Questions/Topic-Determination-of-Income-and-Employment-6/Subtopic-Propensity-to-Consume-and-Propensity-to-Save-1/Part-1.html | 1,527,408,296,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00083.warc.gz | 706,968,684 | 11,769 | Determination of Income and Employment-Propensity to Consume and Propensity to Save (CBSE (Central Board of Secondary Education- Board Exam) Class-12 Economics): Questions 1 - 6 of 22
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Question number: 1
» Determination of Income and Employment » Propensity to Consume and Propensity to Save
Essay Question▾
Describe in Detail
Draw a straight line consumption curve. From it derive a savings curve explaining the process. Show on this diagram:
(a) the level of income at which Average Propensity to Consume is equal to one.
(b) a level of income at which Average Propensity to Save is negative
Explanation
AC is the consumption curve and OA is the consumption expenditure at zero
level of income.
Income minus consumption is savings. When income
… (92 more words) …
Question number: 2
» Determination of Income and Employment » Propensity to Consume and Propensity to Save
One Liner Question▾
Write in Brief
What is the relationship between marginal propensity to consume and marginal propensity to save?
Question number: 3
» Determination of Income and Employment » Propensity to Consume and Propensity to Save
One Liner Question▾
Write in Brief
What is the relationship between marginal propensity to save and marginal propensity to consume?
Question number: 4
» Determination of Income and Employment » Propensity to Consume and Propensity to Save
Write in Short
Complete the following table:
Income Saving MPC APC 0 -20 - - 50 -10 - - 100 0 - - 150 30 - - 200 60 - -
Question number: 5
» Determination of Income and Employment » Propensity to Consume and Propensity to Save
Appeared in Year: 2005
Essay Question▾
Describe in Detail
Level of income ( Rs. ) Consumption expenditure Marginal Propensity to consume Marginal Propensity to save 400 240 - - 500 320 - - 600 395 - - 700 465 - -
Explanation
Level of income ( Rs. ) (Y) Consumption expenditure (C) Marginal Propensity to consume Marginal Propensity to save 400 240 0.60 0.40 500 320 0.80 0.20
… (19 more words) …
Question number: 6
» Determination of Income and Employment » Propensity to Consume and Propensity to Save
Appeared in Year: 2011
Essay Question▾
Describe in Detail
Explain the relationship between investment multiplier and marginal propensity to consume.
Explanation
Investment multiplier involve that any change in investment can cause change in income and output.
Investment multiplier K =
Investment Multiplier shares direct positive relationship with marginal propensity to consume.
The relationship is expressed as follows.
K=
Assume that value of MPC is 0.6 then
K= = 2.5
If val
… (34 more words) …
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A305407 Expansion of e.g.f. 1/(1 + log(1 - x)*exp(x)). 0
1, 1, 5, 32, 274, 2939, 37833, 568210, 9753280, 188342949, 4041170695, 95380234366, 2455830637412, 68501591450447, 2057726452045145, 66227424015265178, 2273614433910697920, 82932491842062712873, 3202994529476330549163, 130577628147690206429038, 5603479009890212632226756 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS FORMULA a(n) ~ n! / ((1 + exp(r)/r) * (1 - exp(-r))^(n+1)), where r = 0.62747017959751658496114808922921433658821962606026068561095... is the root of the equation r*exp(1 - exp(-r)) = 1. - Vaclav Kotesovec, Mar 26 2019 EXAMPLE E.g.f.: A(x) = 1 + x + 5*x^2/2! + 32*x^3/3! + 274*x^4/4! + 2939*x^5/5! + 37833*x^6/6! + ... MAPLE a:=series(1/(1+log(1-x)*exp(x)), x=0, 21): seq(n!*coeff(a, x, n), n=0..20); # Paolo P. Lava, Mar 26 2019 MATHEMATICA nmax = 20; CoefficientList[Series[1/(1 + Log[1 - x] Exp[x]), {x, 0, nmax}], x] Range[0, nmax]! a[0] = 1; a[n_] := a[n] = Sum[HypergeometricPFQ[{1, 1, 1 - k}, {2}, -1] a[n - k]/(k - 1)!, {k, 1, n}]; Table[n! a[n], {n, 0, 20}] CROSSREFS Cf. A002104, A006153, A007840, A009324. Sequence in context: A166993 A328055 A265130 * A320349 A001923 A257710 Adjacent sequences: A305404 A305405 A305406 * A305408 A305409 A305410 KEYWORD nonn AUTHOR Ilya Gutkovskiy, May 31 2018 STATUS approved
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Last modified January 17 23:37 EST 2020. Contains 330995 sequences. (Running on oeis4.) | 757 | 1,996 | {"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} | 3.421875 | 3 | CC-MAIN-2020-05 | longest | en | 0.547625 |
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