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https://www.excelforum.com/excel-formulas-and-functions/866065-formula-vlookup-vs-index-and-match-to-return-multiple-values-in-the-same-column.html | 1,679,869,863,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00207.warc.gz | 841,370,430 | 16,569 | # Formula (VLOOKUP vs INDEX & MATCH) to return multiple values in the same column
1. ## Formula (VLOOKUP vs INDEX & MATCH) to return multiple values in the same column
Hello Everyone!
Background Information:
I get updates for SKU information on a daily basis (see worksheet - Changes_List). However, the information does not match my data record layout. I create a separate worksheet (Unique_Chgs) to build the update record for each SKU. I have figured out how to identify the unique SKU records (viewing other posts in the forum - thanks!), but I am having difficulty trying to find ALL the updates for each SKU.
Problem
The worksheet (Unique_Chgs) I created to build SKU update record currently is using the function VLOOKUP to find the different information (ADDRESS, BRAND, DESCRIPTION, etc.) that needs to be updated for the SKU. I created a column H (heading - Trimmed) in the "Changes_List" worksheet because I found there was a hidden "non-breaking space” in the values in column D (titled - Changed Field).
I used the following formula to remove the "non-breaking space" in the value so I can successfully find the change field in a VLOOKUP:
Cell Ref Cell Formula
H2:H23 TRIM(CLEAN(SUBSTITUTE(D2, CHAR(160),"")))
where D2 referrences the value in Changed Field column.
Below is the formula I have coded in column E (title - Addr) in the Unique_Chgs worksheet.
=IF(C2<>"", IF(VLOOKUP(C2,Changes_List!A:G,7, FALSE)="ADDRESS", TRIM(VLOOKUP(C2,Changes_List!A:G,5, FALSE)), ""), "")
In the first part of the VLOOKUP formula I us column 7 (title - Trimmed) which is the sanitized value of the Field Changes. The second part of the VLOOKUP formula does not use the condition statement ="ADDRESS". Therefore the formula only works for the first occurance of a SKU change. The formula works fine if there is only one update/change for a SKU.
I use the same VLOOKUP in column F (title - Brand) to locate the "BRAND" changes for a SKU. However, since the VLOOKUP function already found a corresponding row previously the function returns "blank" for the "BRAND" changes.
My ideal results for the first SKU (01-0107260) record in the Unique_Chgs worksheet would look like this:
Col C Col E Col F Col G Col H ....
01-0107260 Largo, FL 33777 ANS International 12 Per case - 12 Fl Oz (355ml) ......
From reading other posts, I know there is a way to solve what I am trying to achieve using the INDEX and Match functions, but I am not familiar using these functions and not sure how to reference the cells and ranges to provide the results I am seeking.
I have attached the workbook if anyone would be interested in showing me how to achieve this task. i would greatly appreciate any thoughts or suggestions from you "guru's" out there. Thanks for your assistance.
2. ## Re: Formula (VLOOKUP vs INDEX & MATCH) to return multiple values in the same column
Anyone have any thoughts or suggestions on how achieve the desired results?
3. ## Re: Formula (VLOOKUP vs INDEX & MATCH) to return multiple values in the same column
WOW, not much help from the group this time? Oh well, perhaps next time.
I figured out the solution, it works perfectly. I have tried to explain the best I can on how the formula works. Below are the steps and formulas I used to solve the problem if anyone is interested. The formula below only defines the formula for the cell E2 in the Unique_Chgs worksheet. This is an "array function) and you will need to enter CTRL+SHIFT+ENTER after entering the formula in the cell. You will also need to change the text value from "ADDRESS" to "BRAND" in the next cell F2. This last step will need to be repeated for each of the remaining cells G2 - X2. Once the formulas have been entered on Row 2, then you can copy the cell formulas down the remaining rows 3 - 22.
1. In the Changes_List worksheet I defined three (3) Name Ranges because it is less confusing for me and I can expand the ranges easier without having to change the formulas in the future.
(a) SKU_List ==> which is the range \$A\$2:\$A\$22
(b) Change_Type ==> which is the range \$G2:\$G\$22
(c) New_Value ==> which is the range \$E\$2:\$E\$22
2. I removed the VLOOKUP function and used a combination of the INDEX & MATCH functions
3. Here is the explanation for the MATCH statement: MATCH(lookup value, lookup_array, match_type)
(a) "lookup_value" is equal to the SKU number (Unique_Chgs!C2)
(b) "lookup_array" is equal to the NAMED RANGE "Change_Type" (\$G2:\$G\$22) is equal to "ADDRESS" and
SKU number (Unique_Chgs!C2) ==> "01-0107260" is found in the "SKU_List (\$A\$2:\$A\$22).
(c) "match_type" is equal to (0). Only the row where both conditions are TRUE will be returned
4. The INDEX statement only needs to know the Row and Col where the MATCH statement is true to get the final result.
(a) the NAMED RANGE "New_Value" is the range of data where the MATCH formula will point the INDEX function to get the data.
I am not sure if I explained the solution as well as the guru's on the site, but I hope it helps others.
4. ## Re: Formula (VLOOKUP vs INDEX & MATCH) to return multiple values in the same column
I am not sure if anyone would be interested in seeing the final solution, but I updated the workbook just in case. I hope it helps?
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Homework 3 Solution
# 77 a pa040 b pb037 c pab010
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Unformatted text preview: ty is ⎝ྎ ⎠ྏ = 0.0084. Note: this is also equal to P(S3S4S5|S1S2). ⎛ྎ 50 ⎞ྏ ⎜ྎ ⎟ྏ ⎜ྎ 3 ⎟ྏ ⎝ྎ ⎠ྏ b. Given the first three cards drawn are spades, there are 10 spades left in the deck. Thus, the ⎛ྎ10 ⎞ྏ ⎜ྎ ⎟ྏ ⎜ྎ 2 ⎟ྏ probability is ⎝ྎ ⎠ྏ = 0.0383. Note: this is also equal to P(S4S5|S1S2S3). ⎛ྎ 49 ⎞ྏ ⎜ྎ ⎟ྏ ⎜ྎ 2 ⎟ྏ...
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https://www.shaalaa.com/question-bank-solutions/one-end-steel-rod-k-46-j-s-1-m-1-c-1-length-10-m-kept-ice-0-c-other-end-kept-boiling-water-100-c-area-cross-section-rod-004-cm2-assuming-heat-transfer-conduction_68379 | 1,621,130,479,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991659.54/warc/CC-MAIN-20210516013713-20210516043713-00358.warc.gz | 1,042,290,148 | 9,838 | Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
# One End of a Steel Rod (K = 46 J S−1 M−1°C−1) of Length 1.0 M is Kept in Ice at 0°C and the Other End is Kept in Boiling Water at 100°C. the Area of Cross Section of the Rod is 0.04 Cm2. Assuming - Physics
Sum
One end of a steel rod (K = 46 J s−1 m−1°C−1) of length 1.0 m is kept in ice at 0°C and the other end is kept in boiling water at 100°C. The area of cross section of the rod is 0.04 cm2. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 105 J kg−1.
#### Solution
Given:
Thermal conductivity of the rod, K = 46 J s–1 m–1 °C–1
Length of the rod, l = 1 m
Area of the cross-section of the rod, A =0.04 cm2
= 0.04 × 10−4 m2
= 4 × 10−6 m2
Rate of transfer of heat is given by
(ΔQ)/(Δt) = (ΔT)/ (l/(KA)
(ΔQ)/(Δt) = ((T_1 - T_2 ) KA ) / l
(ΔQ)/(Δt) =( (100-0 )/1) xx 46 xx 4 xx 10^-6 m^2
(DeltaQ)/(Deltat) = 184xx10^-4 J//s
Also , ΔQ = mL f
therefore (m/t)L_f = 184 xx 10^-4
⇒ ( m/t) xx 3.36 xx 10^5 = 184 xx 10^
rArr m=(184xx10^-4)/(3.36xx10^5)xxt
rArr m = 5.5xx10^-9xx1 kg//s
⇒ m = 5.5 xx 10^-8 xx 10^3 g/s
⇒ m = 5.5 xx 10^-5 g/s
Concept: Heat Transfer - Conduction
Is there an error in this question or solution?
#### APPEARS IN
HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 6 Heat Transfer
Q 5 | Page 98 | 564 | 1,410 | {"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} | 4.09375 | 4 | CC-MAIN-2021-21 | latest | en | 0.695698 |
https://michaellustgarten.com/2020/02/04/resting-heart-rate-heart-rate-variability-january-2020-update/ | 1,686,381,395,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657144.94/warc/CC-MAIN-20230610062920-20230610092920-00503.warc.gz | 468,651,498 | 32,914 | # Resting Heart Rate, Heart Rate Variability: January 2020 Update
How’s my progress on the road to achieving a resting heart rate (RHR) as close to 40 beats per minute (bpm) as possible? Shown below is my RHR data for August 2018-Jan 2019, which corresponds to the 6-month period after I started tracking RHR. When compared with that period, am I still making year-over-year progress?
First, note that my Jan 2019 RHR value of 47.4 bpm seems dramatically reduced when compared with Aug-Dec 2018. My computer crashed in Jan 2019, and I lost 4 days of January 2019 RHR data, with remaining data for 27 days. Accordingly, I didn’t expect to be better than that, year-over-year. Nonetheless, my average RHR for Jan 2020 is 46.9 bpm, which is superficially better, but it isn’t statistically different from Jan 2019 (p = 0.13). However, my RHR is still going in the right direction!
What about my heart rate variability (HRV)? Relative to Jan 2019 (56.6), my HRV in Jan 2020 was significantly higher (76; p=0.003), but note that I didn’t additionally improve my HRV relative to December 2019 (86.3).
I’ve been consistent with my exercise program, including weekly workouts (3-4x, ~1 hr each session) and walking (15-20 miles), so are there other variables that may explain the sudden increase in HRV from Nov 2019-Jan 2020? During that time, I’ve been cutting my calorie intake by a small amount (~100-200 cals/day) below my body weight maintenance intake, with the goal of getting leaner. As a result, I’ve slowly decreased my body weight from 157 to 154 during that time. Although there is a weak negative correlation between my body weight with HRV (R2=0.0553), this association is statistically significant (p=0.024). So reducing body weight may have played a role in the sudden HRV increase:
For those who may have missed my other post updates for RHR and/or HRV:
Dec 2019 update: https://michaellustgarten.wordpress.com/2020/01/01/resting-heart-rate-heart-rate-variability-december-2019-update/
Also, why a RHR as close to 40 bpm may be optimal: https://michaellustgarten.wordpress.com/2019/02/02/resting-heart-rate-whats-optimal/
If you’re interested, please have a look at my book!
## 2 thoughts on “Resting Heart Rate, Heart Rate Variability: January 2020 Update”
1. Stuart Berkowitz
RHR (Resting Heart Rate) is a important predictor of longevity, but VO2Max is claimed to be better. “CRF (VO2max) is the strongest independent predictor of future life expectancy in both healthy and cardiorespiratory-diseased individuals. ” Also ways to boost VO2Max in 4 or 12 weeks follow “Results of the Resistance and Endurance exercise After ChemoTherapy (REACT) study found that both a high-intensity and a low- to moderate-intensity resistance and endurance exercise program over 12 weeks are effective in reducing general and physical fatigue, but favoring high-intensity training when it comes to improving VO2max (mean VO2max improvements of 4.4. ml/kg/min after high-intensity versus 3.3. ml/kg/min after low- to moderate-intensity training) (68). Even shorter term high-intensity endurance training over 4 weeks appears to offer superior and clinically meaningful improvements in VO2max (+ 3.5. ml/kg/min) ” https://www.bioscience.org/2018/v23/af/4657/2.htm For RHR “However, only endurance training and yoga significantly decreased the RHR in both sexes…Their meta-analysis revealed a reduction in heart rate of 6.59 bpm in studies that compared yoga with no-treatment usual care. … endurance training causes RHR reductions of 8.4% in older individuals which were even higher in controlled clinical trials with a training length of more than 30 weeks” https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6306777/
Like
1. Michael Lustgarten Post author
I care about consistent, long-term gains, Stuart. Can I decrease my RHR more quickly? Yes. CanI mantain that level of intensity to keep my RHR low? Probably not. Slow and steady for this, imo.
Like | 999 | 3,958 | {"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-2023-23 | latest | en | 0.943157 |
http://www.rcgroups.com/forums/showthread.php?s=a36810b762e52404f8590380e82a8865&t=1746487 | 1,409,481,958,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500959239.73/warc/CC-MAIN-20140820021559-00420-ip-10-180-136-8.ec2.internal.warc.gz | 544,505,584 | 18,317 | Oct 06, 2012, 10:32 AM Texas Buzzard McAllen,Texas Joined Mar 2004 1,034 Posts Discussion Ducted Fan & Velocity of the air in tailpipe. I used to watch F-84s takeoff carrying two 500 lb boms + 6 HVAR rockets fully fueled. They were able to takeoff. ( in the 1950s) Having some experience in free flight and controline I marveled at how that jet engine could produce the needed thrust to make all that mass accelerate. I envisioned all the gasses escaping from the tailpipe and wondered what mass was being driven throught that tailepipe. I knew some basic physics, i.e., Newtons 3rd Law, and it seemed that the very low density of the escaping gases would have to have a great velocity in order to produce the significant thrust needed. If we were to compare the Momentum of the exhaust gases to the momentum of the airplane at first it seems that they would be equivalent to one another..... friction aside. I have an electric ducted fan rated at 1.8 pounds of thrust by the manufactuer. I know the density of air is 1.29 grams per liter. What would the exaust velocity of the air have to be to produce 1.8 pounds of thrust? You might want to rewrite this question......... ??? Please do not tell me to buy an anemometer.....ha, ha.
Oct 06, 2012, 11:05 AM A witty saying proves nothing. Joined Jan 2007 3,572 Posts Search function: http://www.rcgroups.com/forums/showthread.php?t=1332150 EDFs are not the same as turbojets. (Turbojets add to the efflux mass through creating added gas by combustion and the heat increasing the volume and velocity.) Last edited by kcaldwel; Oct 06, 2012 at 07:58 PM.
Oct 06, 2012, 11:40 AM B for Bruce The 'Wack, BC, Canada Joined Oct 2002 11,316 Posts It's not only the velocity but the volume/mass that you're moving. The EDF is going to accelerate some volume related mass to some velocity to achieve the 1.8 lbs of thrust. It's also a bit of a trick question since because the thrust is based on two factors one can take over for the other. You could have a lot of volume that is only accelerated by a little or a little volume accelerated a lot. Without pinning one of the numbers down to some reasonably accurate value for that EDF it's pretty hard to calculate the other. One way would be to stick a large industrial size garbage bag on the outlet. Then switch the fan on and see how long it takes to fill the bag. Calculating the size of the bag and how long it took to fill that volume would give you a reasonably good value of the volume per second that is being moved. From that you could work with more equations to find the velocity of the outlet. Or just stick your hand behind it and call it fast, darn fast or DAMN fast and call it a day....
Oct 06, 2012, 12:31 PM
An itch?. Scratch build.
South Wales U.K.
Joined Mar 2003
13,045 Posts
Who needs calculations when you can just build a model and try it, it's a hell of a lot more fun.
I always feel the more you discuss whether a plane will fly or not, the more worried the plane gets.
Mig August.wmv (5 min 34 sec)
Oct 06, 2012, 12:50 PM Registered User Cambridge, MA USA Joined May 2001 1,738 Posts Thrust = mass_flow_rate * (Vexhaust - Vflight) Pideal = mass_flow_rate * 0.5*(Vexhaust^2 - Vflight^2) where Pideal is the ideal or absolute minimum power or energy rate required by whatever is accelerating the air, either a propeller, fan, or jet engine core.
Oct 08, 2012, 12:49 AM German Engineering....... Auxsburg Joined Dec 2007 448 Posts "Please do not tell me to buy an anemometer.....ha, ha...." Why not build one??? Some clear tubing, some water, ink and a ruler should do the job...
Oct 09, 2012, 04:46 PM Registered User Joined Oct 2012 45 Posts [QUOTE=markdrela;22929313]Thrust = mass_flow_rate * (Vexhaust - Vflight) Mark, how did you arrive at this equation?
Oct 10, 2012, 03:41 PM
Registered User
Joined May 2006
24 Posts
[QUOTE=TangoKilo;22956646]
Quote:
Originally Posted by markdrela Thrust = mass_flow_rate * (Vexhaust - Vflight) Mark, how did you arrive at this equation?
It's part of the continuity equation, a rewriting of Newtons second law, you can see here:
http://en.wikipedia.org/wiki/Mass_flow_rate
Near the bottom of the page, under usage.
The equation Mark gave ignores the acceleration term.
The velocity term gives you a linear relationship, from maximum attainable thrust at 0 airspeed, down to zero thrust at whatever the exhaust velocity itself is.
Oct 10, 2012, 04:23 PM
Registered User
Joined Oct 2012
45 Posts
[QUOTE=nristow;22965239]
Quote:
Originally Posted by TangoKilo It's part of the continuity equation, a rewriting of Newtons second law, you can see here: http://en.wikipedia.org/wiki/Mass_flow_rate Near the bottom of the page, under usage. The equation Mark gave ignores the acceleration term. The velocity term gives you a linear relationship, from maximum attainable thrust at 0 airspeed, down to zero thrust at whatever the exhaust velocity itself is.
Well, I see F = d(mv)/dt and I see F = m.a +mdot.v
but I don't see Thrust = mass_flow_rate * (Vexhaust - Vflight)
Can you briefly explain how you have rewritten Newton's second law?
Oct 10, 2012, 06:25 PM Registered User Joined May 2006 24 Posts Thrust is a force, so substituting thrust in the wiki equation and ignoring acceleration gives you Thrust = mdot * v where mdot is the mass flow rate. The units work out, [kg/s * m/s = kg*m/s^2]. As I've said above, the velocity term is a sum since you don't make any thrust when our velocity is the same as the exhaust velocity, and you assume that thrust tapers linearly. Does that help at all?
Oct 10, 2012, 06:41 PM
It wasn't me...
Trabuco Canyon, CA
Joined Nov 2000
4,559 Posts
Quote:
Originally Posted by HugePanic "Please do not tell me to buy an anemometer.....ha, ha...." Why not build one??? Some clear tubing, some water, ink and a ruler should do the job...
He's talking about an anemometer and you're talking about a manometer. Two different beasties.
An anemometer measures velocity and a manometer measures pressure.
But, you are right. A manometer is quite simple to build. I built one as you describe to calibrate my ASI I use to measure efflux velocity of my fans.
If one wants to measure efflux velocity of a fan, a very cheap and easy way is to buy a used ASI off eBay. I paid \$35 for mine and it measures up to 250mph.
Last edited by DanSavage; Oct 10, 2012 at 06:49 PM.
Oct 10, 2012, 09:52 PM
treefinder
SE MI
Joined Oct 2004
9,575 Posts
Quote:
Originally Posted by eflightray Who needs calculations when you can just build a model and try it, it's a hell of a lot more fun. I always feel the more you discuss whether a plane will fly or not, the more worried the plane gets.
Go Ray, loved the vid! Especially the commentary!
Oct 11, 2012, 04:17 AM Registered User Joined Oct 2012 45 Posts [QUOTE=nristow;22966584] Thrust = mdot * v nristow, yes, I agree. F = dm/dt * v I would advise you not to mention that equation on the "Propeller efficiency" thread. I was shot down as a troll for doing so. Quote, As I've said above, the velocity term is a sum since you don't make any thrust when our velocity is the same as the exhaust velocity, and you assume that thrust tapers linearly. If I understand you correctly, you are saying that the exhaust velocity is the maximum speed at which your model can move and at that speed no thrust is being produced. If there is no thrust what force is overcoming the aerodynamic drag? It is true that the sum of thrust and drag equals zero but that does not mean there is no thrust.
Oct 11, 2012, 03:34 PM
Registered User
Cambridge, MA USA
Joined May 2001
1,738 Posts
[QUOTE=TangoKilo;22956646]
Quote:
Originally Posted by markdrela Thrust = mass_flow_rate * (Vexhaust - Vflight) Mark, how did you arrive at this equation?
Integral Momentum Theorem applied to a control volume surrounding the propeller.
Pretty standard stuff.
http://web.mit.edu/16.unified/www/SP...es/node86.html
section 11.7.2
Oct 12, 2012, 08:19 AM
Registered User
Joined Oct 2012
45 Posts
[QUOTE=markdrela;22974919]
Quote:
Originally Posted by TangoKilo Integral Momentum Theorem applied to a control volume surrounding the propeller. Pretty standard stuff. http://web.mit.edu/16.unified/www/SP...es/node86.html section 11.7.2
Mark, yes I have already read that and many other similar works.
I still can not accept the concept that thrust reduces as "Vflight" increases.
What happens if "Vflight" = "Vexhaust"? Zero thrust?
Can it truly be that aerodynamic drag, increasing as the square of the velocity, is being reacted by thrust decreasing with velocity? | 2,236 | 8,562 | {"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} | 2.90625 | 3 | CC-MAIN-2014-35 | latest | en | 0.95963 |
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## Statistics
describe what is the probability that separates the bottom 25% from the rest of the distribution?
In some normal distribution the boundary that separates the bottom 25% from the rest the distribution is located at ____________.
1) z = -0.67
2) z = -0.25
3) z = +0.25
4) 2 = +0.67
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M921176
Have any Question?
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Create a provider database and related reports and queries to capture contact information for potential PC component pro | 995 | 4,390 | {"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-2018-26 | latest | en | 0.891944 |
http://users.monash.edu/~lloyd/tildeStrings/Compress/1998seq-random.html | 1,506,319,647,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690340.48/warc/CC-MAIN-20170925055211-20170925075211-00028.warc.gz | 353,019,100 | 1,733 | [Compress/]
# DNA Sequence Compression (Random) Example
```> Compress: L.Allison, Computer Science, Monash University 4/1998
1 GGATATCACG TAGTCCCTAG CTCTTGGCGC TGGATGGGGC GGACGGAAGG
51 GAAACGACCG TTGAATTCCA AATTCGGTCG TATGGAAATA TTGCAATGGA 100
> order-0 Markov Model
> | 4.0 +
> | 3.5 b
> | 3.0 b
> . . ... . . . . . . . .. .. . . . | 2.5 b
>--....-.--..-.---..--.-..-----.--..-------.---..---...--.---..-....--.....---.--...--.......--...--.|- 2.0 b
>.. . . . .. . .. .... .. .. ... . . . .. . .. . .. | 1.5 b
> | 1.0 b
> | 0.5 b
> | 0.0 b
> compress: Sequence length=100, |Alphabet|=4, log2(|Alphabet|) =2.0000
> hypothesis: (H) =8.2 bits
> data: (D|H) =197.4 bits, =1.9742 b/ch
> total: (H)+(D|H) =205.6 bits, =2.0564 b/ch
> ran 00/01/21 from 15:33:36 to 15:33:36
> order-1 Markov Model
> | 4.0 +
> | 3.5 b
> | 3.0 b
>. .. . .. .. . . . . . . . . . . . .. | 2.5 b
>--.-.-.----.--.....--....-----.--.--------.---.----.----.-.--.-.--...----..---.--.---.---.-.-------.|- 2.0 b
> . . . . .. . .. ..... .. .. . .. .. . . . .. ... .. . ... ... . . .... | 1.5 b
> | 1.0 b
> | 0.5 b
> | 0.0 b
> compress: Sequence length=100, |Alphabet|=4, log2(|Alphabet|) =2.0000
> hypothesis: (H) =21.1 bits
> data: (D|H) =191.0 bits, =1.9097 b/ch
> total: (H)+(D|H) =212.1 bits, =2.1208 b/ch
> ran 00/01/21 from 15:33:36 to 15:33:36
> AED fwd approx repeats
> [Frequencies B:99.1 R:0.3 C:0.6 E:0.3 =:0.8 ~:0.1 i:0.0 d:0.0 tot:101.3]
> [Frequencies B:98.5 R:0.7 C:0.9 E:0.7 =:0.7 ~:0.4 i:0.3 d:0.3 tot:102.7]
> [Frequencies B:97.9 R:1.1 C:1.2 E:1.1 =:0.8 ~:0.6 i:0.6 d:0.6 tot:104.0]
> | 4.0 +
> | 3.5 b
> | 3.0 b
>. .. . .. .. . . . . . . . . . . . .. | 2.5 b
>--.-.-.----.--.....--....-----.--.--------.---.----.----.-.--.-.--...----..---.--.---.---.-.-------.|- 2.0 b
> . . . . .. . .. ..... .. .. . .. .. . . . .. ... .. . ... ... . . .... | 1.5 b
> | 1.0 b
> | 0.5 b
> | 0.0 b
> compress: Sequence length=100, |Alphabet|=4, log2(|Alphabet|) =2.0000
> hypothesis: (H) =31.6 bits
> data: (D|H) =191.1 bits, =1.9110 b/ch
> total: (H)+(D|H) =222.7 bits, =2.2269 b/ch
> ran 00/01/21 from 15:33:36 to 15:33:38
> --- end ---
------------------------------------------------------------------------------
```
Note that the more complex models seem to give better compression when you ignore the matter of stating their parameters. In reality, i.e., when the parameters are included, the simplest model is best (and the 'uniform' model would be even better). | 1,246 | 4,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-39 | longest | en | 0.215878 |
https://help.scilab.org/docs/6.0.0/fr_FR/sylm.html | 1,620,347,394,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.18/warc/CC-MAIN-20210506235514-20210507025514-00001.warc.gz | 317,707,968 | 7,152 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Change language to: English - Português - 日本語 - Русский
See the recommended documentation of this function
Aide de Scilab >> Polynômes > sylm
# sylm
matrice de Sylvester
### Séquence d'appel
`[S]=sylm(a,b)`
a,b
deux polynômes
S
matrice
### Description
`sylm(a,b)` renvoie la matrice de Sylvester associée aux polynômes `a` et `b`, i.e. la matrice `S` telle que :
`coeff( a*x + b*y )' = S * [coeff(x)';coeff(y)']`.
La dimension de `S` est égale à `degree(a)+degree(b)`.
Si `a` et `b` sont premiers entre eux alors
`rank(sylm(a,b))=degree(a)+degree(b))` et les instructions
```u = sylm(a,b) \ eye(na+nb,1)
x = poly(u(1:nb),'z','coeff')
y = poly(u(nb+1:na+nb),'z','coeff')```
calculent les facteurs de Bezout `x` ainsi que `y` de degré minimum tels que `a*x+b*y = 1`
### Exemples
```x = poly(0,"x")
y = poly ([1, 2, 3], "x","coeff");
sylm(x,y)```
Report an issue << simp_mode Polynômes varn >>
Scilab EnterprisesCopyright (c) 2011-2017 (Scilab Enterprises)Copyright (c) 1989-2012 (INRIA)Copyright (c) 1989-2007 (ENPC)with contributors Last updated:Tue Feb 14 15:06:35 CET 2017 | 431 | 1,191 | {"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-2021-21 | longest | en | 0.256846 |
https://www.freebasic.net/forum/search.php?author_id=11694&sr=posts | 1,558,955,914,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262369.94/warc/CC-MAIN-20190527105804-20190527131804-00130.warc.gz | 772,527,863 | 7,083 | ## Search found 31 matches
May 25, 2019 9:25
Forum: Libraries
Replies: 922
Views: 146099
hello angros47, now a demo for my problem with the internal 2d renderer for opengl, #Include once "fbgfx.bi" #Include Once "crt.bi" #Include Once "crt/math.bi" #Include Once "math.bi" #Include Once "crt/limits.bi" #Include Once "gl\gl.bi" #...
May 24, 2019 0:06
Forum: Beginners
Topic: File i/o question.
Replies: 3
Views: 156
### Re: File i/o question.
read the complete line in a string and then split the string you get like this
Code: Select all
`a=mid(in,1,50)b=mid(in,51,50)c=mid(in,101,10)`
then you can convert all these strings to intergers or what else you will. see for that the 'string operations' in the FB-help.
May 12, 2019 15:14
Forum: Libraries
Replies: 922
Views: 146099
hi angros47, i have a question about your 2d fbgfx patch for opengl. if you know i use OpenB3D, i also use sGUI from Muttonhead and i have patched the sGUI to work with the old 2d.bi. the last version of sGUI uses now 32 bit images mostly all cleared to rgba(0,0,0,0) and then painted with sGUI stuff...
May 10, 2019 16:34
Forum: Libraries
Replies: 922
Views: 146099
i made a function to create a spherized Cube. i hope anyone find it usefull. only problem i cant solve is the texturemapping so you have to solve it by your own :/ ' ---------------------------------------------------------------------------- ' Creates a spherized Cube ' ----------------------------...
Mar 05, 2019 0:28
Forum: General
Topic: Squares
Replies: 6391
Views: 366794
### Re: Squares
i think this will work
13/16=0.8125
100.000Kb * 0.8125=81.250Kb
oh yeah you calculate it right :D
Jan 05, 2019 14:25
Forum: Community Discussion
Topic: Building FreeBASIC 1.06 Release
Replies: 46
Views: 3948
### Re: Building FreeBASIC 1.06 Release
@ST_W yeah you are absolutly right with the argument of using ARM v7 on the RPi, i also use the official Raspian that is only 32 bit. but here i found a experimental OS based on Debian. i dont have test it so far but eventualy you will take a look at it. https://github.com/bamarni/pi64 my RPi curren...
Jan 04, 2019 17:14
Forum: Community Discussion
Topic: Building FreeBASIC 1.06 Release
Replies: 46
Views: 3948
### Re: Building FreeBASIC 1.06 Release
for the RPiv3 i wish ARMv8 support where nice. but ARMv7 are good for now to test it on a ARM SoC.
Dec 02, 2018 12:19
Forum: General
Topic: Too restrictive compilation control on dynamic arrays in the main code?
Replies: 30
Views: 1189
### Re: Too restrictive compilation control on dynamic arrays in the main code?
what, why not? ;D
Dec 02, 2018 12:16
Forum: General
Topic: Too restrictive compilation control on dynamic arrays in the main code?
Replies: 30
Views: 1189
### Re: Too restrictive compilation control on dynamic arrays in the main code?
here my version for 32 and 64 bit, but i think only for windows. sub dothis static as long locked if locked then return print "Greetings from " + __function__ locked=1 end sub dothis '============================ 'one code line to be greeted again '============================ dim x as lon...
Dec 02, 2018 11:18
Forum: General
Topic: Too restrictive compilation control on dynamic arrays in the main code?
Replies: 30
Views: 1189
### Re: Too restrictive compilation control on dynamic arrays in the main code?
hm yeah for 64bit the code is sub dothis static as long locked if locked then return print "Greetings from " + __function__ locked=1 end sub dothis '============================ 'one code line to be greeted again '============================ 'dim x as long ptr=cast(long ptr,@dothis):*(x+8...
Dec 01, 2018 23:13
Forum: General
Topic: Too restrictive compilation control on dynamic arrays in the main code?
Replies: 30
Views: 1189
### Re: Too restrictive compilation control on dynamic arrays in the main code?
a little hack where works by me sub dothis static as long locked if locked then return print "Greetings from " + __function__ locked=1 end sub dothis '============================ 'one code line to be greeted again '============================ dim x as long ptr=cast(long ptr,@dothis):*(x+...
Nov 12, 2018 10:07
Forum: Tips and Tricks
Topic: simulate oldscool amber/green monitor with rgb inputs
Replies: 4
Views: 512
### Re: simulate oldscool amber/green monitor with rgb inputs
oh my fault. you can delete this line, all you need is in the posted code ;) here is an newer version, the above version has an bug with the orginal picture colors, now it is fixed. #Include "fbgfx.bi" '#Include "colorizer.bi" Using FB Static scr_w As Integer=1024 Static scr_h As...
Nov 11, 2018 18:42
Forum: Tips and Tricks
Topic: simulate oldscool amber/green monitor with rgb inputs
Replies: 4
Views: 512
### Re: simulate oldscool amber/green monitor with rgb inputs
so now an bether version of this thing, its a bit slow but it works. todo: improve speed render curvature of an crt screen blur the output a bit phosphor glow has to imlement you need an image, you can use any rgb bmp or create your own 32 bit image. edit: speed nearly doubled, using a macro to get ...
Nov 10, 2018 8:45
Forum: General
Topic: [Code question] variables and unwanted results.
Replies: 10
Views: 604
### Re: [Code question] variables and unwanted results.
how fxm says, in your last code example in the XC sub you have to print out NewZ/Y not x/y ;) sub TheXC (x as integer, y as integer, z as integer, xangle as integer, yangle as integer, zangle as integer) DIM AS INTEGER NewY, NewZ NewY = y * cos(XAngle * (D_PI / 180)) - z * sin(XAngle * (D_PI / 180))...
Nov 02, 2018 20:40
Forum: Tips and Tricks
Topic: simulate oldscool amber/green monitor with rgb inputs
Replies: 4
Views: 512
### simulate oldscool amber/green monitor with rgb inputs
hey there, i have write an simple thing that converts normal rgb values to any else "monocrome" values. i dont know if and how it can improved. you can use keys a and d to anjust the brighness of the output. i hope anyone can use this for projects that matchs "oldscool". so here ... | 1,695 | 6,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-22 | longest | en | 0.79804 |
https://www.coursehero.com/file/p7226/a-0-b-1-c-2-d-3-e-4-f-5-Solution-Key-28-Solution-38-Question-9-Label-the/ | 1,519,229,934,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813626.7/warc/CC-MAIN-20180221143216-20180221163216-00526.warc.gz | 867,043,244 | 27,058 | {[ promptMessage ]}
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solex2sample
# (a 0(b 1(c 2(d 3(e 4(f 5 solution key 2.8 solution
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Unformatted text preview: (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 (f) 5 Solution Key: 2.8 Solution: 3.8 Question 9: Label the consequtive edges of an Euler circuit in the following graph: Solution Key: 2.9 Solution: 3.9 8 Question 10: True or False. Give a reason or a counter-example. (1) There exists a planar graph with 6 vertices, 10 edges, and 6 faces. (2) A planar graph consisting of 4 connected pieces must have Euler characteristic 5. (3) If a connected planar graph has 6 edges and splits the plane into 7 regions, then it must have 1 vertex. (a) (1) (2) (3) T T T (b) (1) (2) (3) T F T (c) (1) (2) (3) F T F (d) (1) (2) (3) F F F (e) (1) (2) (3) T F F (f) (1) (2) (3) T T F Solution Key: 2.10 Solution: 3.10 9 2 Solution key (1) (d) (2) (a) (3) (e) (4) (a) (5) (e) (6) (f) (7) (c) (8) (c) (9) see solution (10) (a) 10 3 Solutions Solution of problem 1.1: The altitude splits teh hypothenuse into two segments. If we denote the length of the shorter segment by x , then the length of the other segment will be 25- x . x 12 25- x a b If we denote the corresponding sides of the triangle by a and b respec- tively we can apply the Pythagorean theorem to two small triangles and to the big triangle: x 2 + 12 2 = a 2 (25- x ) 2 + 12 2 = b 2 a 2 + b 2 = 25 2 Substituting the first two identities into the third one we get x 2 + (25- x ) 2 + 288 = 625 or x 2 + 625- 50 x + x 2 + 288 = 625 . After cancelling 625 from both sides and dividing both sides by 2 we get the equation x 2- 25 x + 144 = 0 . By the quadratic formula we have x = 25 ± √ 25 2- 4 · 144 2 = 25 ± √ 49 2 = 25 ± 7 2 . 11 Thus x must be equal to either 9 or 16 , and since x was the shorter of the two segments we must have x = 9. Substituting this in the two identities involving a and b above we get a 2 = x 2 + 12 2 = 81 + 144 = 225 , b 2 = (25- x ) 2 + 12 2 = 256 + 144 = 400 . Therefore we have a = 15, b = 20, and a + b = 35. The correct answer is (d) . square Solution of problem 1.2: Denote the longer side of R 1 by x . Since R 1 is a golden rectangle we have that x/ 4 must be equal to the golden ratio ϕ , i.e. x = 4 ϕ. In particular R 1 has sides 4 and 4 ϕ , and R 2 has sides 4 ϕ and 4 + 4 ϕ . We can now compute the ratio of areas: area( R 2 ) area( R 1 ) = 4 ϕ · (4 + 4 ϕ ) 4 · 4 ϕ = 16 ϕ (1 + ϕ ) 16 ϕ = 1 + ϕ. The correct answer is (a) . square Solution of problem 1.3: Slicing off a vertex of cube creates a new trian- gular face. Also, when we slice off a vertex all the original edges remain edges and we introduce three new edges - the sides of teh newly cre- ated triangular face. Therefore, when we slice off the four top edges we introduce 4 · 3 = 12 new edges. Adding these to the original 12 edges the cube has we get a total of 24 edges for the new solid. The correctthe cube has we get a total of 24 edges for the new solid....
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Ask a homework question - tutors are online | 1,231 | 3,654 | {"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} | 4.03125 | 4 | CC-MAIN-2018-09 | latest | en | 0.813818 |
https://mathspace.co/textbooks/syllabuses/Syllabus-411/topics/Topic-7322/subtopics/Subtopic-97629/?activeTab=theory | 1,642,876,365,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00405.warc.gz | 435,949,502 | 68,139 | New Zealand
Level 8 - NCEA Level 3
# Graphing sine curves
Lesson
The general form of the equation of a sine curve is
$f\left(x\right)=a\sin\left(bx-c\right)+d$f(x)=asin(bxc)+d
From the lessons on key features and transformations, we discovered that
$a$a amplitude: the maximum deviation of the graph from the central line $\frac{2\pi}{b}$2πb period: the length needed for the curve to travel one full cycle $\frac{c}{b}$cb phase shift: the horizontal translation of the curve $d$d vertical translation: the shift of the central line (and hence the entire curve) up or down.
In order to graph sine curves, there are a number of different approaches you can take. Which one you choose may depend on your own preference or the question you are given.
Here are two approaches.
### Approach 1
Walk through the transformations and change the stem graph $y=\sin x$y=sinx accordingly.
#### Example
##### Example 1
Graph $y=2\sin\left(\frac{x}{2}\right)+3$y=2sin(x2)+3
Start with a sketch of $y=\sin x$y=sinx
Apply vertical translation - move the graph up $3$3 units, this in now the graph of $y=\sin x+3$y=sinx+3
Increase the amplitude - the amplitude of this graph is $2$2 units, so we dilate the graph. Move the maximum and minimum out an extra unit. This is now the graph of $y=2\sin x+3$y=2sinx+3
Period - the period of the function has been changed from $2\pi$2π to $\frac{2\pi}{\frac{1}{2}}=4\pi$2π12=4π so this is a horizontal dilation. Stretch out the graph, keeping the starting point the same. This is now the graph of $y=2\sin\left(\frac{x}{2}\right)+3$y=2sin(x2)+3.
### APPROACH 2
Step out all the important components and create a dot-to-dot style map of the function.
#### Example
##### EXAMPLE 2
Graph $y=2\sin\left(\frac{x}{2}\right)+3$y=2sin(x2)+3
Step 1 - identify the transformations from the graph by identifying the following
$a$a amplitude $2$2 sign of $a$a reflection no reflection as $a$a is positive $\frac{2\pi}{b}$2πb period $\frac{2\pi}{\frac{1}{2}}=4\pi$2π12=4π $\frac{c}{b}$cb phase shift $0$0 $d$d vertical translation $3$3
Step 2 - start by sketching the central line (indicated by the vertical translation)
Step 3 - mark on the maximum and minimum by measuring the amplitude above and below the central line
Step 4 - check for a phase shift (this would shift the initial starting position)
There is no phase shift for this function as $c=0$c=0
Step 5 - mark out the distance of the full period. At this stage mark out half way and quarter way marks, this will help us sketch the curve.
Step 6 - check for a reflection (this would change the initial starting direction
There is no reflection for this function as $a>0$a>0
Step 7 - Create some dots on starting and ending position of the cycle (on the central line), also mark out the maximum and minimum points (on the quarter lines we sketched earlier).
Step 8 - sketch the curve lightly, joining our preparatory dots together. Developing the skills for smooth curve drawing takes practice so don't get disheartened.
Some people prefer step by step constructions, some prefer the fluid changes of transformations, others develop their own order and approach to sketching sine functions. Regardless of your approach they will all need to use the specific features of the sine curve.
#### More Worked Examples
##### QUESTION 1
Which of the following is the graph of $y=\sin x+4$y=sinx+4?
A
B
C
D
A
B
C
D
##### QUESTION 2
Consider the function $y=\sin\left(x-\frac{\pi}{2}\right)$y=sin(xπ2).
1. Identify the amplitude of the function.
2. Identify the phase shift of the function in radians.
Use a positive value to represent a shift to the right, and a negative value to represent a shift to the left.
3. Graph the function.
##### QUESTION 3
Consider the function $y=3\sin\left(x-\frac{\pi}{3}\right)+2$y=3sin(xπ3)+2.
2. Determine the amplitude of the function.
3. Determine the maximum value of the function.
4. Determine the minimum value of the function.
5. Graph the function. | 1,106 | 4,025 | {"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} | 4.9375 | 5 | CC-MAIN-2022-05 | longest | en | 0.748013 |
https://www.coursehero.com/file/7724044/Lathi-B-P-Signals-and-Systems-Berkeley-Cambridge-Press/ | 1,490,572,806,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189313.82/warc/CC-MAIN-20170322212949-00280-ip-10-233-31-227.ec2.internal.warc.gz | 910,741,372 | 23,088 | Signal Processing and Linear Systems-B.P.Lathi copy
# Lathi b p signals and systems berkeley cambridge press
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Unformatted text preview: 3-3 k =m k=m a ssuming a ll frequencies t o b e d istinct, t hat is, 1 .1-6 ( a) 10 cos ( lOOt + ( c) (10 i) + 2 s in 3t) cos ( e) 10 sin 5 t c os lOt 1 .3-1 Wi i ' W k for all i i ' k 4 D etermine t he p ower a nd t he r ms value for each o f t he following signals: ( b) 10 cos (lOOt lOt + i) + 16 sin ( 150t+ (a) i) 1- ( d) 10 cos 5 t cos lOt ( f) ejo. t cos wot -4 I n Fig. P 1.3-1, t he s ignal /1(t) = f ( - t). E xpress signals h (t), h (t), f4(t), a nd f5(t) in t erms o f s ignals f (t), /1 (t), a nd t heir t ime-shifted, time-scaled o r t ime-inverted versions. F or i nstance h (t) = f (t - T) + / 1(t - T). f it) F ig. P l.4-2 1 .4-1 S ketch t he s ignals ( a) u (t-5)-u(t-7) ( b) u (t-5)+u(t-7) ( c) t 2 [u(t-1)-u(t-2)] ( d) (t - 4)[u(t - 2) - u(t - 4)] 1 .4-2 E xpress each of t he s ignals in Fig. P1.4-2 by a single expression valid for all t. 1 .4-3 F or a n e nergy signal f (t) w ith e nergy E f, show t hat t he e nergy of a nyone o f t he s ignals - f(t), f ( - t) a nd f (t - T) is E f. Show also t hat t he e nergy o f f rat) as well as f rat - b) is E fla. T his s hows t hat t ime-inversion a nd t ime-shifting does n ot affect signal energy. O n t he o ther h and, t ime c ompression of a signal (a > 1) reduces t he energy, a nd t ime e xpansion o f a signal (a < 1) i ncreases t he energy. W hat is t he effect o n s ignal energy if t he s ignal is multiplied by a c onstant a? 1 .4-4 Simplify t he following expressions: o ( b) ( a) (c) -I o ( s in t ) 6(t) t2 + 2 [ e- t cos (3t - 60 )]6(t) 2 0 ( lW + 2 ) 6(w) w2 + 9 ( d) C in 1 .3-2 For t he s ignal f (t) d epicted in Fig. P 1.3-2, s ketch t he signals: ( a) f (-t) ( b) f (t+6) ( c) f (3t) ( d) f(~). 1 .3-3 F or t he s ignalf(t) i llustrated in Fig. P1.3-3, s ketch ( a) f (t - 4) ( b) f ( 1 \) ( c) f ( - t) ( d) f (2t - 4) ( e) f (2 - t). ( e) ( _._ 1_) 6(w JW + 2 + 3) ( f) (SinwkW) 6(w) [2 ~(t t F ig. P 1.3-1 H int: Use Eq. (1.23). For p art ( f) u se L 'H6pital's rule. +4 2)]) 6 (t-1) 100 1 .4-5 1 I ntroduction t o S ignals a nd S ystems E valuate t he following integrals: ( a) ( b) ( c) I: I: I: I: b (r)f(t - r)dr ( e) j (r)8(t - r)dr ( f) 6 (t)e- jwt ( g) dt I: I: I: I: 101 P roblems 1 .5-1 1 .6-1 (t 3 + 4)6(1 - t) dt W rite t he i nput-output r elationship for a n ideal integrator. Determine t he z ero-input a nd z ero-state components of t he response. 1 . 7 -1 t 8(t + 3 )e- dt F ind a nd s ketch t he o dd a nd t he even components of ( a) u (t) ( b) t u(t) ( c) sin wat u (t) ( d) cos w atu(t) ( e) s in wat ( f) cos wat. For t he s ystems described by t he e quations below, w ith t he i nput f (t) a nd o utput y (t), d etermine which of t he s ystems a re l inear a nd which are nonlinear. f (2 - t)8(3 - t) dt [~(x - dy ( a) dt + 2y(t) 2 ( b) ~~ + 3ty(t) = Hint: 8(x) i s l ocated a t x = O. For example, 8(1 - t) is located a t 1 - t = 0, a nd so on. ( e) (~~) ( a) F ind a nd sketch df / dt for t he signal f (t) shown in Fig. P1.3-3. ( b) F ind a nd sketch d2f / dt 2 for t he signal f (t) d epi...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
Ask a homework question - tutors are online | 1,480 | 3,488 | {"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-2017-13 | longest | en | 0.765023 |
https://cran.rstudio.com/web/packages/nonet/vignettes/nonet_ensemble_clustering.html | 1,722,957,245,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00140.warc.gz | 146,468,293 | 11,479 | nonet_ensemble Clustering with nonet_plot
nonet provides ensemble capabilities for Clustering problems.
Below example shows the step by step implementation of nonet_ensemble and nonet_plot functions in the context of clustering. We have used Bank Note authentication data set to predict the output class variable using Cluster package because it provides the probability of the input point to be in a specific cluster. Predictions from first GMM and second GMM model are being used as inputs to the nonet_ensemble in the list form.
Let’s start:
library(caret)
library(ggplot2)
library(ClusterR)
library(nonet)
set.seed(1001)
Load the banknote_authentication dataset and explore it.
dataframe <- data.frame(banknote_authentication)
We can see above that class variable has int datatype, we need to convert it into factor.
Converting datatype of class variable into factors.
dataframe$class <- as.factor(dataframe$class)
First GMM Model
Splitting the data into train and test.
#Spliting training set into two parts based on outcome: 75% and 25%
index <- createDataPartition(dataframe$class, p=0.75, list=FALSE) trainSet <- dataframe[ index,] testSet <- dataframe[-index,] Exploring the dimensions of trainSet and testSet dim(trainSet); dim(testSet) ## [1] 1030 5 ## [1] 342 5 str(trainSet) ## 'data.frame': 1030 obs. of 5 variables: ##$ variance: num 4.546 3.866 3.457 0.329 4.368 ...
## $skewness: num 8.17 -2.64 9.52 -4.46 9.67 ... ##$ curtosis: num -2.46 1.92 -4.01 4.57 -3.96 ...
## $entropy : num -1.462 0.106 -3.594 -0.989 -3.163 ... ##$ class : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
str(testSet)
## 'data.frame': 342 obs. of 5 variables:
## $variance: num 3.622 3.203 1.225 -1.577 0.804 ... ##$ skewness: num 8.67 5.76 8.78 10.84 2.85 ...
## $curtosis: num -2.807 -0.753 -2.213 2.546 4.344 ... ##$ entropy : num -0.447 -0.613 -0.806 -2.936 0.602 ...
head(predict_clustering_first$cluster_proba[, 2]) ## [1] 0.9997857 0.9991832 0.9962177 0.7541448 0.9992900 0.9990455 Converting probability into classes predict_cluster_first_class <- as.factor(ifelse(predict_clustering_first$cluster_proba[, 2] >= "0.5", "1", "0"))
## [1] 1 1 1 1 1 1
## Levels: 0 1
head(predict_clustering_first$cluster_labels) ## [1] 1 1 1 1 1 1 Second GMM Model Spliting training set into two parts based on outcome: 75% and 25% index <- createDataPartition(dataframe$class, p=0.75, list=FALSE)
trainSet <- dataframe[ index,]
testSet <- dataframe[-index,]
Exploring the dimensions of trainSet and testSet
dim(trainSet); dim(testSet)
## [1] 1030 5
## [1] 342 5
str(trainSet)
## 'data.frame': 1030 obs. of 5 variables:
## $variance: num 3.62 4.55 3.87 3.46 4.37 ... ##$ skewness: num 8.67 8.17 -2.64 9.52 9.67 ...
## $curtosis: num -2.81 -2.46 1.92 -4.01 -3.96 ... ##$ entropy : num -0.447 -1.462 0.106 -3.594 -3.163 ...
## $class : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ... str(testSet) ## 'data.frame': 342 obs. of 5 variables: ##$ variance: num 0.329 3.203 1.899 -1.577 4.891 ...
## $skewness: num -4.46 5.76 7.66 10.84 -3.36 ... ##$ curtosis: num 4.572 -0.753 0.154 2.546 3.42 ...
## $entropy : num -0.989 -0.613 -3.111 -2.936 1.091 ... ##$ class : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
Feature selection using rfe in caret
control <- rfeControl(functions = rfFuncs,
method = "repeatedcv",
repeats = 3,
verbose = FALSE)
outcomeName<-'class'
predictors<-c("skewness", "curtosis", "entropy")
## skewness curtosis entropy
## 1 8.6661 -2.80730 -0.44699
## 2 8.1674 -2.45860 -1.46210
## 3 -2.6383 1.92420 0.10645
## 4 9.5228 -4.01120 -3.59440
## 6 9.6718 -3.96060 -3.16250
## 7 3.0129 0.72888 0.56421
## [1] 0 0 0 0 0 0
## Levels: 0 1
Model Training: Second
set.seed(423)
gmm_second <- GMM(trainSet[,predictors], 2, dist_mode = "maha_dist", seed_mode = "random_subset", km_iter = 10, em_iter = 10, verbose = F)
Predictions using Second GMM
predict_clustering_Second <- predict_GMM(trainSet[,predictors], gmm_second$centroids, gmm_second$covariance_matrices, gmm_second$weights) head(predict_clustering_Second$cluster_proba[, 2])
## [1] 0.987314759 0.995678444 0.003361081 0.999983188 0.999946316 0.393450214
Converting Prediction Probabilities into classes
predict_cluster_Second_class <- as.factor(ifelse(predict_clustering_Second$cluster_proba[, 2] >= "0.5", "1", "0")) head(predict_cluster_Second_class) ## [1] 1 1 0 1 1 0 ## Levels: 0 1 head(predict_clustering_Second$cluster_labels)
## [1] 1 1 0 1 1 0
Create the stack of predictions
Stack_object <- list(predict_clustering_first$cluster_proba[, 2], predict_clustering_Second$cluster_proba[, 2])
Applying naming to the Stack_object
names(Stack_object) <- c("Cluster_first", "Cluster_second")
nonet_ensemble
Now we need to apply the nonet_ensemble method by supplying list object and best model name as input. Note that We have not provided training or test outcome labels to compute the weights in the weighted average ensemble method, which is being used inside the none_ensemble. Thus it uses best models prediction to compute the weights in the weighted average ensemble.
prediction_nonet <- nonet_ensemble(Stack_object, "Cluster_second")
Result Plotting: nonet_plot
Results can be plotted using the nonet_plot function. nonet_plot is being designed to provided different plot_type options to the user so that one can plot different visualization based on their needs.
Creating the list of cluster probabilities
Prediction_data <- list(prediction_nonet, predict_clustering_first$cluster_proba[, 2], predict_clustering_Second$cluster_proba[, 2])
Applying name to the predictions
names(Prediction_data) <- c("pred_nonet", "pred_clust_first", "pred_clust_second")
Converting list object into dataframe
Prediction_dataframe <- data.frame(Prediction_data)
## pred_nonet pred_clust_first pred_clust_second
## 1 1.1403889 0.9997857 0.987314759
## 2 1.1486603 0.9991832 0.995678444
## 3 0.1558889 0.9962177 0.003361081
## 4 1.1154480 0.7541448 0.999983188
## 5 1.1529445 0.9992900 0.999946316
## 6 0.5464110 0.9990455 0.393450214
nonet_plot for nonet_ensemble model’s predictions in histogram
plot_first <- nonet_plot(Prediction_dataframe$pred_nonet, Prediction_dataframe$pred_clust_first, Prediction_dataframe, plot_type = "hist")
plot_first
nonet_plot for the first GMM model’s predictions in histogram
plot_second <- nonet_plot(Prediction_dataframe$pred_clust_first, Prediction_dataframe$pred_clust_second, Prediction_dataframe, plot_type = "hist")
plot_second
nonet_plot for the Second GMM model’s predictions in histogram
plot_third <- nonet_plot(Prediction_dataframe$pred_clust_second, Prediction_dataframe$pred_clust_first, Prediction_dataframe, plot_type = "hist")
plot_third
Conclusion
Above it can be seen that nonet_ensemble and nonet_plot can serve in a way that one do not need to worry about the outcome variables labels to compute the weights of weighted average ensemble solution. | 2,271 | 7,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.579268 |
https://www.scribd.com/document/178670028/Physics-Final-Revision-Heat-and-Energy-2013-pdf | 1,521,409,307,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646176.6/warc/CC-MAIN-20180318204522-20180318224522-00526.warc.gz | 866,721,032 | 27,216 | # ╞╡§¥ Physics
SPM 2013
Chapter 4: Heat
CHAPTER 4: HEAT
4.1 Thermal Equilibrium
Heat Object 1 Heat Object 2
Thermal equilibrium is reached between two objects in contact when: • The net transfer of heat is zero (rate of exchange of heat are equal) • The two objects have the same temperature
4.1.1 Calibration of a thermometer
To calibrate a thermometer, two extreme points must be chosen to mark its scale. These points must be able to be reproduced accurately. In the Celsius scale, the two fixed points of temperature are: (i) ice point (0°C) – temperature of pure ice melting under standard atmospheric pressure (ii) steam point (100°C) – temperature of pure water boiling under standard atmospheric pressure To calculate the scale of a thermometer, given the two fixed points:
x 0
l x l0 100 0 l100 l0
where l1 = length on scale based on lower fixed point (typically 0°C) l2 = length on scale based on higher fixed point (typically 100°C) lx = length on scale based on temperature of object to be measured θ1 = temperature of lower fixed point (typically 0°C) θ2 = temperature of higher fixed point (typically 100°C) θx = temperature to be calculated
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╞╡§¥ Physics
SPM 2013
Chapter 4: Heat
4.2
Heat Capacity and Specific Heat Capacity
Heat capacity: An object that has a large heat capacity requires a bigger quantity of heat to raise the temperature to 1°C Specific heat capacity: The amount of heat energy needed to raise the temperature of 1 kg of a material by 1°C Q = mcθ where Q = heat energy [J] m = mass [kg] c = specific heat capacity of the material [J kg-1 °C-1] θ = change in temperature [°C]
4.2.1 Applications
Water is used as a coolant in car engines because Specific heat capacity is large, Easily obtained and cheap, Does not chemically react with the materials in the engine. Cooking utensils (woks, pots) are usually made of material with low specific heat capacity to ensure temperature increases quickly when heated. Handles are made of material with high specific heat capacity and are poor conductors. Clay pots are made of clay with high specific heat capacity and are poor conductors. When removed from heat, the soup inside the pot will continue to boil as heat is still being received from the pot.
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╞╡§¥ Physics
SPM 2013
Chapter 4: Heat
4.3
Latent Heat and Specific Latent Heat
Latent heat: Heat that is absorbed during the change of state of the material (solid ↔ liquid ↔ gas) Specific latent heat: The amount of heat needed to change the state of 1 kg of a material
Latent heat of vapourization
Latent heat of fusion
Q = mL where Q = heat energy [J] m = mass [kg] L = specific latent heat of the material [J kg-1]* * Specific latent heat of fusion – for melting/freezing Specific latent heat of vapourization – for boiling/condensation
4.3.1 Applications
Cooling drinks with ice Ice packages to keep cooled food cold Cooking by steaming Heating drinks using steam
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╞╡§¥ Physics
SPM 2013
Chapter 4: Heat
4.3.2 Specific Heat Capacity and Specific Latent Heat
Temperature
Q = mcθ
Q = mcθ
SOLID
SOLID + LIQUID
LIQUID
LIQUID + GAS
GAS
Q = mcθ
Q = mL
Q = mL
Time
Heating graph of a material from solid to gas When calculating the amount of heat needed to change the state and temperature of an object, remember to take into account the different stages of heating as shown in the graph above. Example: To calculate the amount of heat needed to heat ice at 0 ºC to water at 25 ºC: Amount of heat needed = mL
Heat needed to change ice at 0 ºC to water at 0 ºC
+
mcθ
Heat needed to change water at 0 ºC to water at 25 ºC
When calculating the exchange of heat, remember to take into account the different stages of heating for each side of the equation. Example: Ice 0 ºC is added to hot water 90 ºC. To calculate the final temperature, x ºC: Amount of heat absorbed by ice = Amount of heat released by hot water mL + mcθ = mcθ
Heat needed to change ice at 0 ºC to water at 0 ºC Heat needed to change water at 0 ºC to water at x ºC Heat needed to change hot water at 90 ºC to water at x ºC
ASSUMPTION: No heat lost to surrounding.
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╞╡§¥ Physics
SPM 2013
Chapter 4: Heat
4.4
Gas Laws
A closed container containing gas has: Fixed number of molecules Constant mass The gas behaviour is dependant on three variables: Pressure Volume Temperature
Note: For all gas law equations, the temperature involved must be absolute, i.e. in Kelvin T = θ + 273 where T = temperature [Kelvin] θ = temperature [°C]
4.4.1 Boyle’s Law
For a gas of fixed mass, the pressure is inversely proportional to its volume if the temperature is constant. PV = k P1V1 = P2V2 where P = pressure of the gas [Pa] V = volume of the gas [m3] Examples of typical Boyle’s Law questions: Capillary tube with a mercury column trapping some air in it. Given that the atmospheric pressure is 76 cm Hg, to calculate the pressure of the air in the tube:
Pressure of the gas = (76+h) cm Hg
Pressure of the gas = (76-h) cm Hg
Pressure of the gas = 76 cm Hg
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╞╡§¥ Physics
SPM 2013
Chapter 4: Heat
4.4.2 Charles’ Law
For a gas of fixed mass, the volume is directly proportional to its absolute temperature if the pressure is constant.
V k T V1 V2 T1 T2
where V = volume of the gas [m3] T = absolute temperature of the gas [K]
4.4.3 Pressure Law
For a gas of fixed mass, the pressure is directly proportional to its absolute temperature if the volume is constant.
P k T P1 P2 T1 T2
where V = volume of the gas [m3] T = absolute temperature of the gas [K]
4.4.4 Universal Gas Law
Combining all three gas laws:
PV k T P1V1 P2V2 T1 T2
END OF CHAPTER
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Page 6 of 6 | 1,697 | 5,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-13 | latest | en | 0.891416 |
https://mathematica.stackexchange.com/questions/262045/how-to-check-whether-the-given-value-belongs-to-the-list | 1,643,073,925,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00472.warc.gz | 421,546,072 | 33,998 | # How to check whether the given value belongs to the list?
Consider some list
list={1.2,1.6,1.8,1.9,2,2.22,3.04,1000};
I would like to check whether some parameter x belongs to the values of the list, and if it does, then return parameter^2, while if it does not, then return 0. Could you please tell me how to do this?
• f = Boole@MemberQ[list, #]*#^2 & Jan 13 at 15:48
There are few way. One could be to use MemberQ with an If
list = {1.2, 1.6, 1.8, 1.9, 2, 2.22, 3.04, 1000};
x = 1.6;
If[MemberQ[list, x], x^2, 0]
Try this:
check[x_] := If[MemberQ[lst, x], x^2, 0]
Let us test it:
lst = {1.2, 1.6, 1.8, 1.9, 2, 2.22, 3.04, 1000};
check[1.9]
(* 3.61 *)
check[1.8]
(* 3.24 *)
check[1.7]
(* 0 *)
Done. Have fun!
f = #^2 Unitize@Length@Cases[alist, #] &
For testing:
tList = RandomReal[{-10, 10}, {10}]~Join~alist~Join~
RandomInteger[{-10, 10}, {10}]
f /@ tList
{0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1.44, 2.56, 3.24, 3.61, 4, \
4.9284, 9.2416, 1000000, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0} | 473 | 1,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-05 | latest | en | 0.603303 |
https://studysoup.com/tsg/statistics/86/openintro-statistics/chapter/5102/8 | 1,606,854,509,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00226.warc.gz | 498,116,034 | 8,888 | ×
×
# Solutions for Chapter 8: Multiple and Logistic Regression
## Full solutions for OpenIntro Statistics | 3rd Edition
ISBN: 9781943450039
Solutions for Chapter 8: Multiple and Logistic Regression
Solutions for Chapter 8
4 5 0 406 Reviews
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##### ISBN: 9781943450039
This expansive textbook survival guide covers the following chapters and their solutions. Since 18 problems in chapter 8: Multiple and Logistic Regression have been answered, more than 45583 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: OpenIntro Statistics, edition: 3. Chapter 8: Multiple and Logistic Regression includes 18 full step-by-step solutions. OpenIntro Statistics was written by and is associated to the ISBN: 9781943450039.
Key Statistics Terms and definitions covered in this textbook
• 2 k factorial experiment.
A full factorial experiment with k factors and all factors tested at only two levels (settings) each.
• Additivity property of x 2
If two independent random variables X1 and X2 are distributed as chi-square with v1 and v2 degrees of freedom, respectively, Y = + X X 1 2 is a chi-square random variable with u = + v v 1 2 degrees of freedom. This generalizes to any number of independent chi-square random variables.
A variation of the R 2 statistic that compensates for the number of parameters in a regression model. Essentially, the adjustment is a penalty for increasing the number of parameters in the model. Alias. In a fractional factorial experiment when certain factor effects cannot be estimated uniquely, they are said to be aliased.
• Asymptotic relative eficiency (ARE)
Used to compare hypothesis tests. The ARE of one test relative to another is the limiting ratio of the sample sizes necessary to obtain identical error probabilities for the two procedures.
• Average
See Arithmetic mean.
• Axioms of probability
A set of rules that probabilities deined on a sample space must follow. See Probability
• Cause-and-effect diagram
A chart used to organize the various potential causes of a problem. Also called a ishbone diagram.
• Chi-square (or chi-squared) random variable
A continuous random variable that results from the sum of squares of independent standard normal random variables. It is a special case of a gamma random variable.
• Comparative experiment
An experiment in which the treatments (experimental conditions) that are to be studied are included in the experiment. The data from the experiment are used to evaluate the treatments.
• Conditional probability
The probability of an event given that the random experiment produces an outcome in another event.
• Conditional probability mass function
The probability mass function of the conditional probability distribution of a discrete random variable.
• Conidence level
Another term for the conidence coeficient.
• Continuous uniform random variable
A continuous random variable with range of a inite interval and a constant probability density function.
• Counting techniques
Formulas used to determine the number of elements in sample spaces and events.
• Cumulative normal distribution function
The cumulative distribution of the standard normal distribution, often denoted as ?( ) x and tabulated in Appendix Table II.
• Cumulative sum control chart (CUSUM)
A control chart in which the point plotted at time t is the sum of the measured deviations from target for all statistics up to time t
• Distribution free method(s)
Any method of inference (hypothesis testing or conidence interval construction) that does not depend on the form of the underlying distribution of the observations. Sometimes called nonparametric method(s).
• Expected value
The expected value of a random variable X is its long-term average or mean value. In the continuous case, the expected value of X is E X xf x dx ( ) = ?? ( ) ? ? where f ( ) x is the density function of the random variable X.
• Finite population correction factor
A term in the formula for the variance of a hypergeometric random variable.
• Hat matrix.
In multiple regression, the matrix H XXX X = ( ) ? ? -1 . This a projection matrix that maps the vector of observed response values into a vector of itted values by yˆ = = X X X X y Hy ( ) ? ? ?1 .
× | 896 | 4,310 | {"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-2020-50 | latest | en | 0.873418 |
https://community.wolfram.com/groups/-/m/t/1610358 | 1,550,382,387,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481624.10/warc/CC-MAIN-20190217051250-20190217073250-00029.warc.gz | 518,818,409 | 19,346 | # Solve numerically a diffusion equation with fully reflecting wall?
Posted 5 days ago
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I am trying to solve numerically the diffusion equation $\partial_t P(x,t)=\partial_x^2 P(x,t)+ \partial_x V'(x)P(x,t)$. I have a potential that diverges at zero: $V(x)=4((1/x^4)-(1/x^2))$, therefore, I want to set a reflecting wall at, say xc=0.5, and solve only for x>xc. In the code below, you will see my unsuccessful attempt in placing thes boundary conditions. Since I found that I cannot use DiracDelta and HeavisideTheta functions to set my initial condition, I use instead $Pinit(x)=\exp(-(x-8)^2)/\sqrt{\pi}$, which has a negligible contribution from x<=0. It seems that even though, mathematically I believe I am setting a reflecting wall condition, which should not allow any flow to the region below x {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}} uval = NDSolveValue[{D[u[x, t], t] + D[F[x]*u[x, t], x] - D[u[x, t], x, x] == 0, u[x, 0] == Pinit[x], (D[u[x, t], x] /. x -> 0.5) == 0, u[0.5, t] == 0}, u, {x, 0.5, BoundaryCondition}, {t, 0, T}, Method -> mol[2000, 4]]; Plot[{uval[x, T]}, {x, -5, 5}, PlotRange -> All, PlotStyle -> {Automatic, {Thick, Dashed}}] ` | 407 | 1,240 | {"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} | 2.859375 | 3 | CC-MAIN-2019-09 | longest | en | 0.815009 |
http://explorepassages.com/zaxemur28770.html | 1,638,801,217,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00240.warc.gz | 29,322,442 | 6,603 | # Question: How old is space?
Contents
After thousands of years, were getting more consistent age estimates. In milliseconds, Google can serve up a fact that long eluded many of humanitys deepest thinkers: The universe is nearly 14 billion years old—13.8 billion years old to be exact. And many cosmologists continue to grow more confident in that number.
## How old is outer space?
approximately 13.8 billion years old According to research, the universe is approximately 13.8 billion years old. How did scientists determine how many candles to put on the universes birthday cake?
## What is the true age of the universe?
13.77 billion years old The universe is (nearly) 14 billion years old, astronomers confirm. With looming discrepancies about the true age of the universe, scientists have taken a fresh look at the observable (expanding) universe and have estimated that it is 13.77 billion years old (plus or minus 40 million years).
## How long will universe last?
22 billion years in the future is the earliest possible end of the Universe in the Big Rip scenario, assuming a model of dark energy with w = −1.5. False vacuum decay may occur in 20 to 30 billion years if Higgs boson field is metastable.
## What Can billion dollars buy?
What a Billion Dollars Can BuyFast Cars. People spend a lot of money on cars. Private Jets. Where would a billion dollar jetsetter be without a jet to jetset in? A Superyacht. A Skyscraper.
## Can the Universe end?
Astronomers once thought the universe could collapse in a Big Crunch. Now most agree it will end with a Big Freeze. Trillions of years in the future, long after Earth is destroyed, the universe will drift apart until galaxy and star formation ceases. Slowly, stars will fizzle out, turning night skies black.
## What is the biggest thing in the world?
There are several ways to think of the biggest things in the world. The biggest animal is the blue whale, which weighs a quarter of a million pounds .
## Say hello
#### Find us at the office
Hostler- Pertzborn street no. 57, 67563 Kigali, Rwanda
#### Give us a ring
Anterio Ruebush
+29 780 790 988
Mon - Fri, 8:00-17:00 | 500 | 2,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-49 | latest | en | 0.920118 |
http://httprover2.blogspot.com/2014/09/spherical-coordinates-in-n-space.html | 1,490,757,834,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190181.34/warc/CC-MAIN-20170322212950-00651-ip-10-233-31-227.ec2.internal.warc.gz | 169,992,305 | 12,264 | ## Friday, September 12, 2014
### Spherical Coordinates in n-space
In a previous blog, Best Position Estimate for a Number of Intersecting Great Circles, I showed how one can find the distance of a point on a sphere from an arc on it. The trick involves finding the point on the arc for which the distance from the point is a minimum. This method can be used to find the angular coordinates of a point, q, on a sphere. The coordinate, φ, is the angular distance to the minimum, a, on the arc in the xy-plane. The second coordinate, θ, is the angular distance from the point a to the point q on the arc in the az-plane. Of course the distance of q from the arc in this case is zero.
This simple process can be extended to any number of dimensions. In a Euclidean 4-space we have the axes x, y, z and w. The distance of the minimum of b from q on the arc in the az-plane will not necessarily be zero in this 4-space. We need another arc in the bw-plane to get to q on the 4-sphere which yields a third angle. Here's an example of how it works in practice.
One can use the atan function to check the angles in the example above. The function er(α) consists of pairs of sines and cosines as indicated in the blog mentioned above with a third sine-cosine pair for the third angle.
For points not on the unit 4-sphere the distance of the point q from the origin will not be unity and a check will show that q = r·er(α) to the precision of the calculation. | 350 | 1,454 | {"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-2017-13 | longest | en | 0.881798 |
https://www.perlmonks.org/index.pl/?node_id=543792 | 1,540,088,549,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513548.72/warc/CC-MAIN-20181021010654-20181021032154-00244.warc.gz | 1,019,529,749 | 6,632 | Keep It Simple, Stupid PerlMonks
Re^2: Puzzle: Longest Increasing Sequence
by spurperl (Priest)
on Apr 17, 2006 at 09:59 UTC ( #543792=note: print w/replies, xml ) Need Help??
in reply to Re: Puzzle: Longest Increasing Sequence
in thread Puzzle: Longest Increasing Sequence
Just for kicks, here is the same implemented in Ruby. I noticed that the original page has a bug you fixed - towards the end of the main loop, there is:
```if (\$low > \$n_las) {
\$n_las++;
}
While in the pseudocode it's "if (low .ge. n_as) n_as += 1" and I presume .ge. means "greater or equal" ?!
Anyways, here's my implementation. It is notably cleaner than yours, but twice slower:
```def find_lis_greedy(seq)
n = seq.length
terminals = [0] * (n + 1)
backptrs = [-1] + [0] * (n - 1)
lis = []
n_lis = 1
1.upto(n - 1) do |i|
low = 1
high = n_lis
while low <= high
mid = (low + high) / 2
if seq[i] <= seq[terminals[mid]]
high = mid - 1
else
low = mid + 1
end
end
terminals[low] = i
if low <= 1
backptrs[i] = -1
else
backptrs[i] = terminals[low - 1]
end
n_lis += 1 if low > n_lis
end
lis[n_lis - 1] = seq[terminals[n_lis]]
temp = terminals[n_lis]
(n_lis - 2).downto(0) do |i|
temp = backptrs[temp]
lis[i] = seq[temp]
end
return lis
end
Replies are listed 'Best First'.
Re^3: Puzzle: Longest Increasing Sequence
by ikegami (Pope) on Apr 17, 2006 at 16:41 UTC
It is indeed a bug fix, not a typo.
The other thing I did was to clean up the code that builds the return value by using unshift. I no longer needed a counting loop, so I removed the counter variable ("i") and replaced the counting loop with a while loop (implemented as a for loop). "temp" was a bad name — even i, j, etc would be better since temp holds an index — so I renamed it to "i".
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Notices? | 692 | 2,180 | {"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-2018-43 | longest | en | 0.844456 |
https://windowsontheory.org/2018/06/27/workshops-at-theoryfest-including-shameless-advertisement/ | 1,685,776,195,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00698.warc.gz | 673,080,568 | 31,124 | Guest post by Sam Hopkins
TheoryFest is in full swing in Los Angeles! There is lots to be written about the wealth of STOC talks, invited papers, keynotes, and (maybe most importantly) the excellent food hall across the street which is swarming with theorists and mathematicians. But for now I want to bring to your attention the workshops planned for Friday.
As always, the schedule is packed with interesting talks! There are workshops on major themes in theory, like
Since I helped organize the last of these, let me say a bit about it.
Recently we have seen a surge of interest in algorithms and lower bounds for average-case problems. One driver of this surge has been theory’s increasing connection with machine learning, which has suggested a number of high-dimensional and noisy statistical inference problems to study. Another driver has been the substantial progress on long-standing problems regarding random instances—for example, a series of recent works proving many old conjectures on random k-SAT instances, and another series of works analyzing algorithms for community detection and recovery in very sparse random graphs.
A common theme in average-case computational problems is that their statistical difficulty (that is, whether they are solvable by any algorithm at all, regardless of running time) is not predictive of their algorithmic difficulty (that is, whether they are solvable by polynomial-time algorithms).
A now-standard example is the $k$-community stochastic blockmodel. In this problem, $n$ nodes are randomly partitioned into $k$ communities, and then a sparse random graph of average degree $d$ is sampled on those vertices is sampled by including an edge $\{i,j\}$ with probability $(d/n)(1 - \epsilon/k)$ if $i,j$ are in differing communities and otherwise with probability $(d/n)(1 + (1-1/k)\epsilon)$ (this produces a graph of average degree $d$).
It turns out that so long as $d > C k \log k / \epsilon^2$ for a big-enough constant $C$, in exponential time one may recover the underlying communities from the graph (approximately, in the sense that it is possible to partition the graph into $k$ classes such that this partitioning has nonvanishing correlation with the ground-truth partition).
However, it is conjectured that this community recovery is possible in polynomial time if and only if (!!) $d > k^2/\epsilon^2$. This threshold is conjectured to be sharp, in the sense that if $d < 0.99 k^2 / \epsilon^2$ no polynomial-time algorithm should exist, while one is known to when $d > 1.01 k^2 / \epsilon^2$.
This kind of threshold phenomenon is ubiquitous for high-dimensional inference problems, and remains largely unexplained. Of course, we are used to algorithmic intractability! But the appearance of intractability for average-case problems seems not to be explained by worst-case theories of hardness (like NP-completeness). And very little is known about what features of an average-case problem are predictive of computational tractability, and which might predict intractability (while in the worst case we know that, say, NP-completeness is an excellent predictor of intractability). The sharpness of a threshold like $d = k^2/\epsilon^2$ is striking and mysterious in a field where ignoring logarithmic factors is practically a (inter)national pastime.
We will have four talks, spanning algorithms and complexity, with (at least) four perspectives on computational thresholds for average-case problems and related matters. Florent Krzakala will lead off with a statistical-physics perspective on algorithms and hardness for matrix and tensor factorization problems. Nike Sun will give us a taste of the wild world of random constraint satisfaction problems. Jacob Steinhardt will talk about the relationship of outlier robustness and inaccurate/partial problem specification to efficient algorithms for statistical inference. Finally, I will survey some recent developments at the intersection of convex programming, the Sum of Squares method, and high-dimensional inference problems.
I hope to see you there! | 861 | 4,081 | {"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": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.933849 |
https://www.physicsforums.com/threads/mass-of-substance-in-a-system-heat-transfer.820069/ | 1,521,797,444,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648205.76/warc/CC-MAIN-20180323083246-20180323103246-00632.warc.gz | 873,505,023 | 16,223 | # Mass of substance in a system (Heat transfer)
1. Jun 21, 2015
### Zoey Brown
1. The problem statement, all variables and given/known data
A solid, 2.00 kg cube of nitrogen at -220.0oC is added to a 1.00 kg mixture of ice and
water at 0.00 oC. If the final temperature of the mixture is -100.0 oC, how much ice was
there in the original mixture?
2. Relevant equations
e=mct
e=mLv/f
3. The attempt at a solution
Don't know where to start
2. Jun 21, 2015
### rude man
heat gained = heat lost
3. Jun 21, 2015
### rude man
[Mentor's Note: The quoted post was deleted, but I'm leaving the quote since it should help the OP.]
Did you remember that N2 is liquid below -196C and gas above that temperature?
Oops, worse than that:
solid below -210C, liquid -210 < T < -196, gas at T > -196.
Need 3 specific heats and 2 latent heats!
Last edited by a moderator: Jun 21, 2015 | 266 | 880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-13 | latest | en | 0.918805 |
http://treodesktop.com/standard-error/how-to-decrease-standard-error-of-the-mean.php | 1,516,631,821,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891377.59/warc/CC-MAIN-20180122133636-20180122153636-00132.warc.gz | 337,572,740 | 6,924 | Home > Standard Error > How To Decrease Standard Error Of The Mean
# How To Decrease Standard Error Of The Mean
## Contents
There's no point in reporting both standard error of the mean and standard deviation. So let's say you were to take samples of n is equal to 10. So in this case every one of the trials we're going to take 16 samples from here, average them, plot it here, and then do a frequency plot. Retrieved Oct 17, 2016 from Explorable.com: https://explorable.com/standard-error-of-the-mean . have a peek at this web-site
If you can accept this line of thinking then we can insert it into the calculations of your statistics as standard error. Correction for finite population The formula given above for the standard error assumes that the sample size is much smaller than the population size, so that the population can be considered Because the age of the runners have a larger standard deviation (9.27 years) than does the age at first marriage (4.72 years), the standard error of the mean is larger for Well that's also going to be 1. http://www.dummies.com/education/math/statistics/how-sample-size-affects-standard-error/
## Find The Mean And Standard Error Of The Sample Means That Is Normally Distributed
Try it with the control above. I took 100 samples of 3 from a population with a parametric mean of 5 (shown by the blue line). Why does the standard deviation remain high even though I do so many measurements? Student approximation when σ value is unknown Further information: Student's t-distribution §Confidence intervals In many practical applications, the true value of σ is unknown.
If, instead, you have them each take a thousand shots, then you will be much more confident that you are getting a good look at their actual accuracy. Additional Info . The first sample happened to be three observations that were all greater than 5, so the sample mean is too high. Standard Error Of The Mean Formula share|improve this answer edited Nov 22 '15 at 2:43 answered Dec 21 '14 at 1:08 Glen_b♦ 150k19246514 add a comment| up vote 5 down vote The variability that's shrinking when N
Here are 10 random samples from a simulated data set with a true (parametric) mean of 5. And it turns out there is. Make all the statements true An overheard business meeting, a leader and a fight Conference presenting: stick to paper material? Of course, the answer will change depending on the particular sample that we draw.
Let's see if it conforms to our formulas. Standard Error Of The Mean Excel Since we can get more precise estimates of averages by increasing the sample size, we are more easily able to tell apart means which are close together -- even though the Travelling to Iceland and UK What sense of "hack" is involved in "five hacks for using coffee filters"? If you don't remember that you might want to review those videos.
## What Happens To The Distribution Of The Sample Means If The Sample Size Is Increased?
Note: The Student's probability distribution is a good approximation of the Gaussian when the sample size is over 100. a fantastic read Similarly, maybe you caught the accurate shooter at a bad time and just happened to get two bad shots in the five, skewing the results. Find The Mean And Standard Error Of The Sample Means That Is Normally Distributed The age data are in the data set run10 from the R package openintro that accompanies the textbook by Dietz [4] The graph shows the distribution of ages for the runners. Which Combination Of Factors Will Produce The Smallest Value For The Standard Error So just that formula that we've derived right here would tell us that our standard error should be equal to the standard deviation of our original distribution, 9.3, divided by the
Imagine that you decided to go on with a task of determining the average weight of american citizens. http://treodesktop.com/standard-error/how-to-work-out-standard-deviation-from-standard-error.php If the population standard deviation is finite, the standard error of the mean of the sample will tend to zero with increasing sample size, because the estimate of the population mean And we've seen from the last video that one-- if let's say we were to do it again and this time let's say that n is equal to 20-- one, the Why aren't sessions exclusive to an IP? If The Size Of The Sample Is Increased The Standard Error Will
In each of these scenarios, a sample of observations is drawn from a large population. And so you don't get confused between that and that, let me say the variance. We keep doing that. Source With a sample size of 20, each estimate of the standard error is more accurate.
With a low N you don't have much certainty in the mean from the sample and it varies a lot across samples. When The Population Standard Deviation Is Not Known The Sampling Distribution Is A The standard error of the mean estimates the variability between samples whereas the standard deviation measures the variability within a single sample. For some reason, there's no spreadsheet function for standard error, so you can use =STDEV(Ys)/SQRT(COUNT(Ys)), where Ys is the range of cells containing your data.
## When n is equal to-- let me do this in another color-- when n was equal to 16, just doing the experiment, doing a bunch of trials and averaging and doing
McDonald Search the handbook: Contents Basics Introduction Data analysis steps Kinds of biological variables Probability Hypothesis testing Confounding variables Tests for nominal variables Exact test of goodness-of-fit Power analysis Chi-square I personally like to remember this: that the variance is just inversely proportional to n. So here the standard deviation-- when n is 20-- the standard deviation of the sampling distribution of the sample mean is going to be 1. Standard Error Of The Mean Definition But I think experimental proofs are kind of all you need for right now, using those simulations to show that they're really true.
The standard deviation of the sample doesn't decrease, but the standard error, which is the standard deviation of the sampling distribution of the mean, does decrease. And maybe in future videos we'll delve even deeper into things like kurtosis and skew. We're not going to-- maybe I can't hope to get the exact number rounded or whatever. have a peek here The mean of all possible sample means is equal to the population mean.
In fact, we might want to do this many, many times. You'd get an exact answer. If σ is known, the standard error is calculated using the formula σ x ¯ = σ n {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} where σ is the So maybe it'll look like that.
The process repeats until the specified number of samples has been selected. References Browne, R. The standard deviation of the age for the 16 runners is 10.23, which is somewhat greater than the true population standard deviation σ = 9.27 years. Plant based lifeforms: brain equivalent?
E., M. | 1,484 | 6,946 | {"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} | 4.0625 | 4 | CC-MAIN-2018-05 | longest | en | 0.919236 |
http://mathcenter.oxford.emory.edu/site/math100/probSetPolynomialArithmetic/ | 1,627,678,243,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153980.55/warc/CC-MAIN-20210730185206-20210730215206-00559.warc.gz | 28,191,091 | 3,021 | ## Exercises - Polynomial Arithmetic
1. Add or subtract the polynomials given, as indicated:
1. $\displaystyle{(3y^2+5y-7)+(2y^2-7y+10)}$
${\displaystyle{5y^2-2y+3}}$
2. $\displaystyle{(6n^2 - 3n + 4) + (3n^3 - n^2)}$
${\displaystyle{3n^3+5n^2-3n+4}}$
3. $\displaystyle{(4x^3+3x^2+2x+1) + (x^3 + 2x^2 + 3x + 4)}$
${\displaystyle{5x^3+5x^2+5x+5}}$
4. $\displaystyle{(x^3-3x^2y+4xy^2+y^3)+(7x^3+x^2y-9xy^2+y^3)}$
${\displaystyle{8x^3-2x^2y-5xy^2+2y^3}}$
5. $\displaystyle{(7m^2+9m+3) \, - \, (3m^2 + 8m + 2)}$
${\displaystyle{4m^2+m+1}}$
6. $\displaystyle{(4x+11)-(3x^2+7x-3)}$
${\displaystyle{-3x^2-3x+14}}$
7. $\displaystyle{(5xy^4 - 7xy^2 + 4x^2 - 3) \, - \, (-3xy^4 + 2xy^2 - 2y + 4)}$
${\displaystyle{8xy^4 - 9xy^2 + 4x^2 + 2y - 7}}$
8. $\displaystyle{(2x^3-5x^2y+xy^2-y^3)-(8x^3-x^2y-3xy^2+y^3)}$
${\displaystyle{-6x^3-4x^2y+4xy^2-2y^3}}$
2. Multiply:
1. $\displaystyle{(3x+2)(5x-7)}$
${\displaystyle{15x^2-11x-14}}$
2. $\displaystyle{(x-7)(-2x+5)}$
${\displaystyle{-2x^2+19x-35}}$
3. $\displaystyle{(3x+y)(7x-3y)}$
${\displaystyle{21x^2-2xy-3y^2}}$
4. $\displaystyle{(11x-3)(11x+3)}$
${\displaystyle{121x^2-9}}$
5. $\displaystyle{(2b+1)(2b-1)(3b-4)}$
${\displaystyle{12b^3-16b^2-3b+4}}$
6. $\displaystyle{(5w-3)^2}$
${\displaystyle{25w^2-30w+9}}$
7. $\displaystyle{(2q+3)^3}$
${\displaystyle{8q^3+36q^2+54q+27}}$
8. $\displaystyle{(x+1)^4}$
${\displaystyle{x^4+4x^3+6x^2+4x+1}}$
9. $\displaystyle{(2x-3)^4}$
${\displaystyle{16x^4-96x^3+216x^2-216x+81}}$
10. $\displaystyle{(2x^3 - 5y)^2}$
${\displaystyle{4x^6 - 20x^3 y + 25y^2}}$
11. $\displaystyle{(x+t)(x^2-xt+t^2)}$
${\displaystyle{x^3 + t^3}}$
12. $\displaystyle{(5a + 4b)^3}$
${\displaystyle{125a^3 + 300a^2b + 240ab^2 + 64b^3}}$
13. $\displaystyle{3x(4xy^3-7x^2y-3y)}$
${\displaystyle{12x^2y^3-21x^3y-9xy}}$
14. $\displaystyle{(2a-3b)(3a+4ab+b)}$
${\displaystyle{6a^2+8a^2b-7ab-12ab^2-3b^2}}$
15. $\displaystyle{(y+4)(y^2-7y+12)}$
${\displaystyle{y^3-3y^2-16y+48}}$
16. $\displaystyle{(x^2+2x-3)(5x^2+3x-7)}$
${\displaystyle{5x^4+13x^3-16x^2-23x+21}}$
17. $\displaystyle{(4m^2-m+8)(m^3+2m^2+3m+4)}$
${\displaystyle{4m^5+7m^4+18m^3+29m^2+20m+32}}$
3. Expand and Simplify:
1. $\displaystyle{(x+1)^3+(x+1)^2+(x+1)+1}$
${\displaystyle{x^3+4x^2+6x+4}}$
2. $\displaystyle{(2-x+3y)(2-x-3y)(4x^3-3+2x) + (5x-3)^3}$
${\displaystyle{4 x^5-16 x^4-36 x^3 y^2+143 x^3-236 x^2-18 x y^2+155 x+27 y^2-39}}$ | 1,311 | 2,409 | {"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} | 4.21875 | 4 | CC-MAIN-2021-31 | latest | en | 0.309245 |
http://www.gurufocus.com/term/DaysInventory/MTW/Days%2BInventory/Manitowoc%2BCo%2BInc | 1,406,884,260,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510274866.27/warc/CC-MAIN-20140728011754-00203-ip-10-146-231-18.ec2.internal.warc.gz | 559,238,794 | 42,544 | Switch to:
Manitowoc Co Inc (NYSE:MTW)
Days Inventory
120.38 (As of Mar. 2014)
Manitowoc Co Inc's inventory for the three months ended in Mar. 2014 was \$824 Mil. Manitowoc Co Inc's cost of goods sold for the three months ended in Mar. 2014 was \$623 Mil. Hence, Manitowoc Co Inc's days inventory for the three months ended in Mar. 2014 was 120.38.
Manitowoc Co Inc's days inventory increased from Mar. 2013 (108.52) to Mar. 2014 (120.38). It might indicate that Manitowoc Co Inc's sales slowed down.
Inventory can be measured by Days Sales of Inventory (DSI). Manitowoc Co Inc's days sales of inventory (DSI) for the three months ended in Mar. 2014 was 88.22.
Inventory turnover measures how fast the company turns over its inventory within a year. Manitowoc Co Inc's inventory turnover for the three months ended in Mar. 2014 was 0.76.
Inventory to revenue ratio determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. Manitowoc Co Inc's inventory to revenue ratio for the three months ended in Mar. 2014 was 0.97.
Definition
Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
Manitowoc Co Inc's Days Inventory for the fiscal year that ended in Dec. 2013 is calculated as
Days Inventory = Inventory / Cost of Goods Sold * Days in Period = 720.8 / 3026.3 * 365 = 86.94
Manitowoc Co Inc's Days Inventory for the quarter that ended in Mar. 2014 is calculated as:
Days Inventory = Inventory / Cost of Goods Sold * Days in Period = 824 / 622.9 * 91 = 120.38
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Explanation
An increase of Days Inventory may indicate the company's sales slowed.
1. Inventory can be measured by Days Sales of Inventory (DSI).
Manitowoc Co Inc's Days Sales of Inventory for the three months ended in Mar. 2014 is calculated as
Days Sales of Inventory (DSI) = Inventory / Revenue * Days in Period = 824 / 850 * 91 = 88.22
2. Inventory Turnover measures how fast the company turns over its inventory within a year.
Manitowoc Co Inc's Inventory Turnover for the three months ended in Mar. 2014 is calculated as
Inventory Turnover = Cost of Goods Sold / Average Inventory = 622.9 / 824 = 0.76
3. Inventory to Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
Manitowoc Co Inc's Inventory to Revenue for the three months ended in Mar. 2014 is calculated as
Inventory to Revenue = Inventory / Revenue = 824 / 850 = 0.97
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Be Aware
A lot of business are seasonable. It makes more sense to compare Days Inventory from the same period in the previous year instead of from the previous quarter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Manitowoc Co Inc Annual Data
Dec04 Dec05 Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 DaysInventory 71.31 66.04 88.12 76.43 96.85 75.18 86.44 86.57 85.12 86.94
Manitowoc Co Inc Quarterly Data
Mar12 Jun12 Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 DaysInventory 109.03 98.64 110.49 73.12 108.52 95.95 103.12 79.98 120.38 102.11
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 965 | 3,717 | {"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-2014-23 | longest | en | 0.941943 |
https://www.coursehero.com/file/6204163/sol-multi-angl-form-1/ | 1,513,027,722,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948514051.18/warc/CC-MAIN-20171211203107-20171211223107-00529.warc.gz | 738,078,472 | 22,103 | sol_multi_angl_form_1
# sol_multi_angl_form_1 - From www.analvzemath.com...
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Unformatted text preview: From www.analvzemath.com Trigonometry Worksheet: Multiple Angle Formula (1) 1. Express cos(3B) in terms of cos( B) . vr, C~6) <0 W C!t3 ~~ +8J:: W G-n J-8 C0 [!-[;.'L8~e ~ :: ~lk0'L8-') _ ;2 CvS~e _ e- )~e, jes-s GDO. .'L ~'\8 Grr& \$. ~ e e Cus<e) _ J. eos'e - Cus e - .,i(X Cl- - 4 Cus~e_3Co-\$6. 2. Express sin(3B) in terms ofsin(B) . ~ C'H» :: ~ (le ~- --~ \8) c,J)8 =- S:ih.te. enB -+- ~ 1- CJs(2e)~tJ - J~e. ~c9 +- (1-.2f1;,2e).SJ:,& _ .2 ~ e. (£J Le f;;, B - 2. JJ,;,3 f3 ~'le) ~9 - "U1~}e - - ~ ~) e + 3 ~ e . - 1 ",~e (\-I- From www.analyzemath.com ...
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Ask a homework question - tutors are online | 377 | 891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-51 | latest | en | 0.365099 |
http://blog.phytools.org/2017/11/comparing-rate-of-diversification-among.html | 1,726,637,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00321.warc.gz | 4,481,571 | 41,074 | ## Wednesday, November 8, 2017
### Comparing the rate of diversification among trees under a 'Yule' or pure-birth model
I just pushed another update to the function `ratebytree` for comparing diversification rates among trees - in this case to permit users to fit the 'Yule' or pure-birth model. In this case, two models are fit & compared: one in which each tree is permitted to have its own speciation rate & a second where all trees are constrained to the same rate.
Here is an example with two sets of trees - one (`t1`) simulated under the null hypothesis of pure-birth & no difference in rate among trees, and the second (`t2`) simulated with a 25% higher diversification rate in tree 3 compared to trees 1 & 2:
``````library(phytools)
source("https://raw.githubusercontent.com/liamrevell/phytools/cd787d598150ac216fdbe87a60efb105e3138f0e/R/ratebytree.R")
print(t1,details=T)
``````
``````## 3 phylogenetic trees
## tree 1 : 110 tips
## tree 2 : 79 tips
## tree 3 : 44 tips
``````
``````par(mfrow=c(2,2))
plotTree(t1,ftype="off",lwd=1)
``````
``````ratebytree(t1,model="Yule")
``````
``````## ML common diversification-rate model:
## b d k log(L)
## value 0.037 0 1 -183.4468
##
## ML multi diversification-rate model:
## b[1] b[2] b[3] d[1] d[2] d[3] k log(L)
## value 0.0378 0.0396 0.0315 0 0 0 3 -182.6673
##
## Diversification model was "Yule".
## Model fitting method was "optimize".
##
## Likelihood ratio: 1.5591
## P-value (based on X^2): 0.4586
``````
``````print(t2,details=T)
``````
``````## 3 phylogenetic trees
## tree 1 : 73 tips
## tree 2 : 111 tips
## tree 3 : 235 tips
``````
``````par(mfrow=c(2,2))
plotTree(t2,ftype="off",lwd=1)
``````
``````ratebytree(t2,model="Yule")
``````
``````## ML common diversification-rate model:
## b d k log(L)
## value 0.0427 0 1 -19.3546
##
## ML multi diversification-rate model:
## b[1] b[2] b[3] d[1] d[2] d[3] k log(L)
## value 0.0355 0.0371 0.0494 0 0 0 3 -14.5664
##
## Diversification model was "Yule".
## Model fitting method was "optimize".
##
## Likelihood ratio: 9.5763
## P-value (based on X^2): 0.0083
``````
Neat. We can also compare these to alternative models, such as, for instance, our multi-rate birth-death model, using our `AIC` method. For instance:
``````yule<-ratebytree(t2,model="Yule")
bd<-ratebytree(t2,model="birth-death")
``````
``````## Warning in nlminb(c(init.b, init.d), lik.bd, t = T, rho = rho, lower =
## rep(0, : NA/NaN function evaluation
## Warning in nlminb(c(init.b, init.d), lik.bd, t = T, rho = rho, lower =
## rep(0, : NA/NaN function evaluation
## Warning in nlminb(c(init.b, init.d), lik.bd, t = T, rho = rho, lower =
## rep(0, : NA/NaN function evaluation
``````
``````AIC(yule,bd)
``````
``````## AIC df
## common-rate 40.70918 1
## multi-rate:Yule 35.13287 3
## multi-rate:birth-death 40.72992 6
``````
Although in this case our simplest model is just our Yule model and we have kind of ignored our birth-death null model (though we shouldn't have - I'll fix that later).
The trees were simulated using phytools as follows:
``````t1<-pbtree(b=0.039,t=100,nsim=3)
t2<-c(pbtree(b=0.039,t=100,nsim=2),
as.multiPhylo(pbtree(b=0.049,t=100)))
`````` | 1,145 | 3,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.767791 |
https://sdstringteachers.org/and-pdf/916-analytical-ability-questions-and-answers-pdf-851-831.php | 1,642,752,272,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00462.warc.gz | 446,567,265 | 6,533 | Sunday, December 13, 2020 10:39:30 PM
# Analytical Ability Questions And Answers Pdf
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Published: 14.12.2020
Reasoning Section is most important sections in all recruitment exam. Some time Reasoning questions are very confusing and time consuming and candidates face problem in this section as they find it difficult to finish it on time.
## analytical reasoning questions and answers pdf
Non-verbal reasoning tests can also be helpful in assessing international candidates, or applicants who do not have English as their first language.. Welcome to www. Dear Aspirants, we have added the reasoning puzzles pdf. After downloading the Reasoning book pdf, Individuals can have a glance at the Logical Reasoning questions. The Analytical Reasoning section of the NTS test consists of the questions like, deductive and inductive logic, critical thinking and writing skills. Aspirants who are preparing for the exams must update their knowledge by improving the preparation strategy.
Therefore, this situation is making many participants, to check for the various online sources to practice and gain knowledge on reasoning subject. To make this procedure easier, here we have provided the Reasoning questions and answers with solutions. Practice all the reasoning chapters provided here and we are sure you can easily crack any online exam with ease. Along with the solutions, all the people can also find the proper explanation on this page itself. This will help many aspirants for sure. Mostly, those people who had applied for the interview, Entrance Examinations, Competitive Exams can get the help of Reasoning Quiz with Solutions. Now, you can even practice them all, as even check your answers.
Find the minimum number of straight lines required to make the given figure. What is the number of straight lines and the number of triangles in the given figure. Now, we shall count the number of triangles in the figure. There is only one triangle i. ABC composed of twelve components. What is the number of triangles that can be formed whose vertices are the vertices of an octagon but have only one side common with that of octagon?
All students, freshers can download Non Verbal Reasoning Analytical Reasoning quiz questions with answers as PDF files and eBooks. Where can I get Non.
## Analytical Reasoning Tests
Which of the following, if true, would make this explanation more plausible? The great majority of student texts have a massive backlog awaiting publication. Over the past five years there has been a substantial decline in the number of accounting school students, while electronic books have remained fairly stable.
Это многое объясняет, - настаивала. - Например, почему он провел там всю ночь. - Заражал вирусами свое любимое детище.
Провайдер находится в районе территориального кода двести два. Однако номер пока не удалось узнать.
Забавное имя. Сам придумал. - А кто же еще! - ответил тот с гордостью. - Хочу его запатентовать.
Беккер предпринял последнюю попытку: - Мистер Клушар, я хотел бы получить показания этого немца и его спутницы. Вы не скажете, где они могли остановиться. Клушар закрыл глаза, силы покинули. Он едва дышал.
Littcalpemi1972 21.12.2020 at 07:30
At work, people use analysis to scrutinise speech, documents, diagrams, charts and graphs, and gather the most relevant information.
Anselma M. 21.12.2020 at 11:36
Honda ch80 service manual pdf fantasy genesis a creativity game for fantasy artists pdf
Prewitt G. 22.12.2020 at 19:49
Free stihl 180c chainsaw repair manual pdf chemistry the central science pdf free | 828 | 3,646 | {"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} | 3.90625 | 4 | CC-MAIN-2022-05 | latest | en | 0.937171 |
https://aviation.stackexchange.com/questions/13248/where-do-i-land-if-i-fly-from-klax-with-a-constant-east-heading-crossing-the-us/13249 | 1,656,706,305,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103945490.54/warc/CC-MAIN-20220701185955-20220701215955-00345.warc.gz | 162,457,896 | 68,537 | # Where do I land if I fly from KLAX with a constant east heading, crossing the US and the Atlantic ocean?
Los Angeles KLAX and Casablanca GMMN (Morocco) are at the same latitude. If the aircraft follows a magnetic 090° heading, the initial true heading will be 102° due to the declination.
(source University of Alabama.)
That means the path will start slightly southerly.
What will be the path followed until reaching a continent on the other side of the Atlantic?
If an aircraft is following a magnetic course changes in magnetic declination as the aircraft moves along its route of flight will affect the true course, and on a long flight such as the hypothetical trip posed here we need to take that into account.
Magnetic declination is empirical data, i.e. there are tables of it, and while formulas exist to approximate it you would normally use the reference tables. There are several sources of data for the worldwide declination of the magnetic field, for example here. Following the link "Magnetic Model maps and grids" I took D_map_mf_2010.zip which contains a simple lat/lon/declination table (0.5° steps) and wrote a small simulation which moves the aircraft 10 km along the heading, calculates & records the new angular coordinates, moves again and so on. The result is this figure:
The green arrows show the direction of north as displayed by a compass, the blue shows the direction of east.
The black line was generated by the simulation shows the true course of an aircraft with a constant magnetic heading of 90°. While it will fly over north Florida, it deviates markedly to the north over the ocean and just touches the north coast of Spain.
Note that this only accounts for the magnetic declination, since your question assumed no wind.
During a real flight the aircraft may experience unknown crosswinds, so the actual position has to be determined from time to time, and a look into the table / map shows the correct heading at the current position leading to your destination. (If the aircraft only has this map, a compass and a sextant aboard.)
So, if you have this data, you can correct the compass headings such that you fly the true course you want. However, I don't know how it's actually done in aviation.
EDIT:
Here is the code. It is written in python and gnuplot, and you will not need any programming skills to use it.
Have fun!
If you carried on around the world, would you return to your starting point? I'm guessing yes as we would be following a contour of equal magnetic potential?
From the physics side, magnetic fields are not conservative and so do not have a potential. However, I think it's clear what you mean.
As long as the heading is +90° or -90°, you will end up where you started. The following figure again shows a travel from KLAX heading 90°, but with a total lengh of 40mio km (i.e. 1000 times the circumfence of the earth and a little more turns for the track as it's not at the equator) As you can see, there is just a single black track, no deviation.
And if you want to fly to Casablanca, you will need a constant heading of 93,6°. And if you don't land there and have enough fuel, strange things happen:
I would also like to emphasize Bob Jarvis' comment, about how it's done in the real world:
Long distance journeys, whether by air or sea, will generally use a "great circle" where the heading steered is not constant. In practice the continuously changing great circle course is broken up in to a series of segments of constant heading. Reference here.
• I simulated it. Looks like frog's legs instead of couscous as dinner after arrival. Mar 14, 2015 at 16:44
• Love that you programmed the solution. This is great.
– egid
Mar 14, 2015 at 22:34
• Awesome answer! I made a few changes to line it up with the current state of the question, if I messed anything up let me know. Mar 14, 2015 at 23:58
• Thanks for your comments. The "tool" is just a python script written by myself out of the box. The graphs are done using gnuplot. The continents also just come from a data file from the web. If you want, I can upload a software bundle, so you can play with it. But I'll do that tomorrow, as it's 2pm here. Mar 15, 2015 at 0:55
• If you carried on around the world, would you return to your starting point? I'm guessing yes as we would be following a contour of equal magnetic potential? Mar 15, 2015 at 10:14 | 1,022 | 4,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-27 | longest | en | 0.931285 |
http://mathematics-diary.blogspot.com/2010/10/automorphism-groups-3.html | 1,527,032,246,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864999.62/warc/CC-MAIN-20180522225116-20180523005116-00415.warc.gz | 183,419,493 | 20,243 | As of May 4 2007 the scripts will autodetect your timezone settings. Nothing here has to be changed, but there are a few things
## Monday, October 4, 2010
### Introduction
In this post I will explain the concept of an Automorphism Group. We will make a list of the automorphism groups of all 24 groups of order less than or equal to 12 and (to our surprise) we will see that one of these groups has as much as 168 elements and that different groups can share the same automorphism group. Finally, we will make a strategy that can be of help in finding automorphism groups in general. So far we have come across direct products of groups when we studied groups of type C2 X C2 or C2 X C2 X C2. The study of automorphism groups prepares us for the study of another type of group product, the semi-direct product. ( Which I might discuss in detail in a future post. ) In this post I consider all groups to be finite. I'll try to use as much examples as I can at first and formalize later in final wrap up. Let's begin!
### Generating sets
" In abstract algebra, a generating set of a group is a subset that is not contained in any proper subgroup of the group. " As so very often is the case, simple things can be hard to catch in words. Let's simplify this a bit to "If [the only subgroup of G containing X = G] then [X generates G]." So it is all about -the only subgroup of G containing X-. But we don't know X! Time for examples.
List of subgroups of C3 (cyclic group of 3 elements).
1
1, a, a^2.
The sets which are only contained in C3 are: {a}, {a^2} and {a, a^2}. Note that we can reduce this list to {a}, {a^2}. ( We will do so immediately in the next examples.) Verify that C3 has two different generating sets. C3=(a)=(a^2).
List of subgroups of D3 (dihedral group of an equilateral triangle or 3-gon).
1
1, a, a^2.
1, b
1, ba
1, ba^2
1, a, a^2, b, ba, ba^2
The sets which are only contained in D3 are: {a,b}, {a,ba}, {a,ba^2}, {a^2,b}, {a^2,ba} and {a^2, ba^2}. Verify that D3 has six different generating sets.
### Automorphisms
An isomorphism from a group G to a group H is a map which is surjective, injective and preserves the group operation. An automorphism is basically an isomorphism from a group to itself. Let's ilustrate this with some examples.
Let's establish an isomorphism first.
This is a CayleyTable from a group of order 8, we call the group G.
| 0 1 2 3 4 5 6 7
- | - - - - - - - -
0 | 0 1 2 3 4 5 6 7
1 | 1 0 3 2 5 4 7 6
2 | 2 3 0 1 6 7 4 5
3 | 3 2 1 0 7 6 5 4
4 | 4 5 6 7 0 1 2 3
5 | 5 4 7 6 1 0 3 2
6 | 6 7 4 5 2 3 0 1
7 | 7 6 5 4 3 2 1 0
This is another CayleyTable from a group of order 8, we call the group H.
| a b c e b**a c**a c**b c**b**a
- | - - - - - - - -
a | e b**a c**a a b c c**b**a c**b
b | b**a e c**b b a c**b**a c c**a
c | c**a c**b e c c**b**a a b b**a
e | a b c e b**a c**a c**b c**b**a
b**a | b a c**b**a b**a e c**b c**a c
c**a | c c**b**a a c**a c**b e b**a b
c**b | c**b**a c b c**b c**a b**a e a
c**b**a | c**b c**a b**a c**b**a c b a e
$\begin{array}{cccccccccc} & | & a & b & c & e & b\text{**}a & c\text{**}a & c\text{**}b & c\text{**}b\text{**}a \\ - & | & - & - & - & - & - & - & - & - \\ a & | & e & b\text{**}a & c\text{**}a & a & b & c & c\text{**}b\text{**}a & c\text{**}b \\ b & | & b\text{**}a & e & c\text{**}b & b & a & c\text{**}b\text{**}a & c & c\text{**}a \\ c & | & c\text{**}a & c\text{**}b & e & c & c\text{**}b\text{**}a & a & b & b\text{**}a \\ e & | & a & b & c & e & b\text{**}a & c\text{**}a & c\text{**}b & c\text{**}b\text{**}a \\ b\text{**}a & | & b & a & c\text{**}b\text{**}a & b\text{**}a & e & c\text{**}b & c\text{**}a & c \\ c\text{**}a & | & c & c\text{**}b\text{**}a & a & c\text{**}a & c\text{**}b & e & b\text{**}a & b \\ c\text{**}b & | & c\text{**}b\text{**}a & c & b & c\text{**}b & c\text{**}a & b\text{**}a & e & a \\ c\text{**}b\text{**}a & | & c\text{**}b & c\text{**}a & b\text{**}a & c\text{**}b\text{**}a & c & b & a & e \end{array}$
( Got the table in TeX but as you can see, blogger format is too small... )
We will investigate if G,H are isomorphic and ( if so ) then define an isomorphism f: G-> H.
( Post in progress, thus more later... )
## Welcome to The Bridge
Mathematics: is it the fabric of MEST?
This is my voyage
My continuous mission
To uncover hidden structures
To create new theorems and proofs
To boldly go where no man has gone before
(Raumpatrouille – Die phantastischen Abenteuer des Raumschiffes Orion, colloquially aka Raumpatrouille Orion was the first German science fiction television series. Its seven episodes were broadcast by ARD beginning September 17, 1966. The series has since acquired cult status in Germany. Broadcast six years before Star Trek first aired in West Germany (in 1972), it became a huge success.) | 1,770 | 4,767 | {"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} | 3.859375 | 4 | CC-MAIN-2018-22 | latest | en | 0.942084 |
https://forum.wordreference.com/threads/exponentially.899247/ | 1,597,062,075,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738674.42/warc/CC-MAIN-20200810102345-20200810132345-00056.warc.gz | 318,085,239 | 17,776 | # exponentially
< Previous | Next >
#### Kidagakash
##### Member
What does 'exponentially' mean in the following context:
These sites did not invent free porn; they just made it exponentialy easier to access.
Thanks. (Excuse me for the context
• #### panjandrum
##### Lapsed Moderator
Oh dear, one of my pet peeves.
Replace "exponentially" with "a great deal" and you will be reflecting the normal abuse of this term correctly.
#### bibliolept
##### Senior Member
Absolutely, "exponentially" is used by the virtuosos of nescience incorrectly to refer to a large increase in scale, while "exponential" (adj) should only describe a rapid change.
#### coiffe
##### Senior Member
It certainly means "a great deal more" but from the original term in mathematics, it actually means "many times more" easy. Three times easier, five times, ten times easier. "A great deal" would be more additive than exponential.
#### Thomas Tompion
##### Senior Member
Rabbits when considering parenthood are said to adopt an exponential principle of a kind - 2, 4, 8, 16, 32, 64, 128, 256, 512 etc. That way they quickly come to have a lot more rabbits. An exponential graph is shaped like the horn of a trumpet.
This number sequence is a geometric progression. Many mathematicians will tell you that there are important differences between geometric progressions and exponential functions. Many word people don't differentiate between them, I suspect. The key point for us here is that the increase at each stage is increasingly dramatic, and, if the stages are close together in time, we become overrun with rabbits.
#### panjandrum
##### Lapsed Moderator
For anyone who has been involved with mathematics, the difficulty with using exponentially in this way is that it has no meaning.
Things may increase exponentially, referring to the rate of change over time (for example). But even then, exponential growth could be either extremely slow at first, gradually becoming faster and faster, until the bend in the curve is reached and the growth becomes extremely rapid extremely quickly.
Curve one.
Alternatively, exponential growth could be extremely rapid at first, gradually becoming slower and slower.
Curve two.
However, having got that out of the way I can accept the general sense that exponentially greater means a lot more. What I would resist is any attempt to quantify it
#### berndf
##### Moderator
...I can accept the general sense that exponentially greater means a lot more....
... in the sense that we all agree the term has been grossly misused but it is still clear what the author meant.
< Previous | Next > | 568 | 2,627 | {"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} | 3.078125 | 3 | CC-MAIN-2020-34 | latest | en | 0.941551 |
http://nrich.maths.org/public/leg.php?code=-25&cl=4&cldcmpid=318 | 1,502,978,208,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103316.46/warc/CC-MAIN-20170817131910-20170817151910-00381.warc.gz | 294,315,836 | 7,172 | # Search by Topic
#### Resources tagged with Number theory similar to The Root Cause:
Filter by: Content type:
Stage:
Challenge level:
### There are 23 results
Broad Topics > Numbers and the Number System > Number theory
### Sums of Squares and Sums of Cubes
##### Stage: 5
An account of methods for finding whether or not a number can be written as the sum of two or more squares or as the sum of two or more cubes.
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### More Sums of Squares
##### Stage: 5
Tom writes about expressing numbers as the sums of three squares.
### Euler's Squares
##### Stage: 4 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square...
### An Introduction to Number Theory
##### Stage: 5
An introduction to some beautiful results of Number Theory
### Never Prime
##### Stage: 4 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### Modulus Arithmetic and a Solution to Differences
##### Stage: 5
Peter Zimmerman, a Year 13 student at Mill Hill County High School in Barnet, London wrote this account of modulus arithmetic.
### Diophantine N-tuples
##### Stage: 4 Challenge Level:
Can you explain why a sequence of operations always gives you perfect squares?
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Pythagorean Golden Means
##### Stage: 5 Challenge Level:
Show that the arithmetic mean, geometric mean and harmonic mean of a and b can be the lengths of the sides of a right-angles triangle if and only if a = bx^3, where x is the Golden Ratio.
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Data Chunks
##### Stage: 4 Challenge Level:
Data is sent in chunks of two different sizes - a yellow chunk has 5 characters and a blue chunk has 9 characters. A data slot of size 31 cannot be exactly filled with a combination of yellow and. . . .
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Mod 7
##### Stage: 5 Challenge Level:
Find the remainder when 3^{2001} is divided by 7.
### Really Mr. Bond
##### Stage: 4 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
### The Public Key
##### Stage: 5 Challenge Level:
Find 180 to the power 59 (mod 391) to crack the code. To find the secret number with a calculator we work with small numbers like 59 and 391 but very big numbers are used in the real world for this.
### 2^n -n Numbers
##### Stage: 5
Yatir from Israel wrote this article on numbers that can be written as $2^n-n$ where n is a positive integer.
### Novemberish
##### Stage: 4 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100. | 1,075 | 4,497 | {"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} | 4.15625 | 4 | CC-MAIN-2017-34 | latest | en | 0.856762 |
https://kr.mathworks.com/matlabcentral/answers/2094851-solving-large-number-of-equations | 1,716,817,299,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00154.warc.gz | 294,080,986 | 33,441 | # Solving large number of equations
조회 수: 7 (최근 30일)
Ismita 2024년 3월 15일
댓글: Ismita 2024년 3월 16일
I am having issues to solve like following equations. I request for help/ suggestion to solve this type of equations. Thanks!
syms X1 X2 X3 X4 X5 X6 X7 X8
p_11 = 1; p_12 = 2; p_13 = 3; p_14 = 4; p_15 = 5; p_16 = 6; p_17 = 7;
p_21 = 8; p_22 = 9; p_23 = 10; p_24 = 11; p_25 = 12; p_26 = 13; p_27 = 14;
p_31 = 15; p_32 = 16; p_33 = 17; p_34 = 18; p_35 = 19; p_36 = 20; p_37 = 21;
p_41 = 22; p_42 = 23; p_43 = 24; p_44 = 25; p_45 = 26; p_46 = 27; p_47 = 28;
p_51 = 29; p_52 = 30; p_53 = 31; p_54 = 32; p_55 = 33; p_56 = 34; p_57 = 35;
p_61 = 36; p_62 = 37; p_63 = 38; p_64 = 39; p_65 = 40; p_66 = 41; p_67 = 42;
p_71 = 43; p_72 = 44; p_73 = 45; p_74 = 46; p_75 = 47; p_76 = 48; p_77 = 49;
p_81 = 50; p_82 = 51; p_83 = 52; p_84 = 53; p_85 = 54; p_86 = 55; p_87 = 56;
C_1 = [4 3 5];
C_2 = [2 4 6];
C_3 = [5 6 8];
C_4 = [1 2 3];
C_5 = [2 3 4];
C_6 = [3 4 5];
C_7 = [2 3 5];
C_8 = [3 6 2];
for i = 1:length(C_1)
eq1(i) = p_11*X1 + p_12*X2 + p_13*X3 + p_14*X4 + p_15*X5 + p_16*X6 + p_17*X7*sin(X8) == C_1(i);
eq2(i) = p_21*X1 + p_22*X2 + p_23*X3 + p_24*X4 + p_25*X5 + p_26*X6 + p_27*X7*sin(X8) == C_2(i);
eq3(i) = p_31*X1 + p_32*X2 + p_33*X3 + p_34*X4 + p_35*X5 + p_36*X6 + p_37*X7*cos(X8) == C_3(i);
eq4(i) = p_41*X1 + p_42*X2 + p_43*X3 + p_44*X4 + p_45*X5 + p_46*X6 + p_47*X7*sin(X8) == C_4(i);
eq5(i) = p_51*X1 + p_52*X2 + p_53*X3 + p_54*X4 + p_55*X5 + p_56*X6 + p_57*X7*sin(X8) == C_5(i);
eq6(i) = p_61*X1 + p_62*X2 + p_63*X3 + p_64*X4 + p_65*X5 + p_66*X6 + p_67*X7*sin(X8) == C_6(i);
eq7(i) = p_71*X1 + p_72*X2 + p_73*X3 + p_74*X4 + p_75*X5 + p_76*X6 + p_77*X7*sin(X8) == C_7(i);
eq8(i) = p_81*X1 + p_82*X2 + p_83*X3 + p_84*X4 + p_85*X5 + p_86*X6 + p_87*X7 == C_8(i);
end
% Solve the system of equations
sol = solve([eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8], [X1, X2, X3, X4, X5, X6, X7, X8]);
% Display solutions
disp(sol)
X1: [0x1 sym] X2: [0x1 sym] X3: [0x1 sym] X4: [0x1 sym] X5: [0x1 sym] X6: [0x1 sym] X7: [0x1 sym] X8: [0x1 sym]
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### 채택된 답변
Walter Roberson 2024년 3월 15일
syms X1 X2 X3 X4 X5 X6 X7 X8
p_11 = 1; p_12 = 2; p_13 = 3; p_14 = 4; p_15 = 5; p_16 = 6; p_17 = 7;
p_21 = 8; p_22 = 9; p_23 = 10; p_24 = 11; p_25 = 12; p_26 = 13; p_27 = 14;
p_31 = 15; p_32 = 16; p_33 = 17; p_34 = 18; p_35 = 19; p_36 = 20; p_37 = 21;
p_41 = 22; p_42 = 23; p_43 = 24; p_44 = 25; p_45 = 26; p_46 = 27; p_47 = 28;
p_51 = 29; p_52 = 30; p_53 = 31; p_54 = 32; p_55 = 33; p_56 = 34; p_57 = 35;
p_61 = 36; p_62 = 37; p_63 = 38; p_64 = 39; p_65 = 40; p_66 = 41; p_67 = 42;
p_71 = 43; p_72 = 44; p_73 = 45; p_74 = 46; p_75 = 47; p_76 = 48; p_77 = 49;
p_81 = 50; p_82 = 51; p_83 = 52; p_84 = 53; p_85 = 54; p_86 = 55; p_87 = 56;
C_1 = [4 3 5];
C_2 = [2 4 6];
C_3 = [5 6 8];
C_4 = [1 2 3];
C_5 = [2 3 4];
C_6 = [3 4 5];
C_7 = [2 3 5];
C_8 = [3 6 2];
for i = 1:length(C_1)
eq1(i) = p_11*X1 + p_12*X2 + p_13*X3 + p_14*X4 + p_15*X5 + p_16*X6 + p_17*X7*sin(X8) == C_1(i);
eq2(i) = p_21*X1 + p_22*X2 + p_23*X3 + p_24*X4 + p_25*X5 + p_26*X6 + p_27*X7*sin(X8) == C_2(i);
eq3(i) = p_31*X1 + p_32*X2 + p_33*X3 + p_34*X4 + p_35*X5 + p_36*X6 + p_37*X7*cos(X8) == C_3(i);
eq4(i) = p_41*X1 + p_42*X2 + p_43*X3 + p_44*X4 + p_45*X5 + p_46*X6 + p_47*X7*sin(X8) == C_4(i);
eq5(i) = p_51*X1 + p_52*X2 + p_53*X3 + p_54*X4 + p_55*X5 + p_56*X6 + p_57*X7*sin(X8) == C_5(i);
eq6(i) = p_61*X1 + p_62*X2 + p_63*X3 + p_64*X4 + p_65*X5 + p_66*X6 + p_67*X7*sin(X8) == C_6(i);
eq7(i) = p_71*X1 + p_72*X2 + p_73*X3 + p_74*X4 + p_75*X5 + p_76*X6 + p_77*X7*sin(X8) == C_7(i);
eq8(i) = p_81*X1 + p_82*X2 + p_83*X3 + p_84*X4 + p_85*X5 + p_86*X6 + p_87*X7 == C_8(i);
end
eqns = [eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8];
size(eqns)
ans = 1×2
1 24
symvar(eqns)
ans =
You have 24 equations in 8 variables, and you are trying to solve them for 8 variables.
% Solve the system of equations
sol = solve(eqns, [X1, X2, X3, X4, X5, X6, X7, X8]);
solve() will immediately give up on trying to solve the 24 equations for 8 variables.
% Display solutions
disp(sol)
X1: [0×1 sym] X2: [0×1 sym] X3: [0×1 sym] X4: [0×1 sym] X5: [0×1 sym] X6: [0×1 sym] X7: [0×1 sym] X8: [0×1 sym]
If you are trying to do a least-squared fit then solve() of more equations than variables is the wrong way to do it.
##### 댓글 수: 3이전 댓글 1개 표시이전 댓글 1개 숨기기
Torsten 2024년 3월 16일
편집: Torsten 2024년 3월 16일
"solve" from above says your system does not have a solution.
"fsolve" as a numerical solver also cannot find a solution for your system (see below).
As you can see, the Jacobian of your system has only rank 4 instead of rank 8. Thus it will be almost impossible to solve it for arbitrary right-hand sides.
p_11 = 1; p_12 = 2; p_13 = 3; p_14 = 4; p_15 = 5; p_16 = 6; p_17 = 7;
p_21 = 8; p_22 = 9; p_23 = 10; p_24 = 11; p_25 = 12; p_26 = 13; p_27 = 14;
p_31 = 15; p_32 = 16; p_33 = 17; p_34 = 18; p_35 = 19; p_36 = 20; p_37 = 21;
p_41 = 22; p_42 = 23; p_43 = 24; p_44 = 25; p_45 = 26; p_46 = 27; p_47 = 28;
p_51 = 29; p_52 = 30; p_53 = 31; p_54 = 32; p_55 = 33; p_56 = 34; p_57 = 35;
p_61 = 36; p_62 = 37; p_63 = 38; p_64 = 39; p_65 = 40; p_66 = 41; p_67 = 42;
p_71 = 43; p_72 = 44; p_73 = 45; p_74 = 46; p_75 = 47; p_76 = 48; p_77 = 49;
p_81 = 50; p_82 = 51; p_83 = 52; p_84 = 53; p_85 = 54; p_86 = 55; p_87 = 56;
C_1 = [4 3 5];
C_2 = [2 4 6];
C_3 = [5 6 8];
C_4 = [1 2 3];
C_5 = [2 3 4];
C_6 = [3 4 5];
C_7 = [2 3 5];
C_8 = [3 6 2];
for i = 1:length(C_1)
f1 = @(X1,X2,X3,X4,X5,X6,X7,X8) p_11*X1 + p_12*X2 + p_13*X3 + p_14*X4 + p_15*X5 + p_16*X6 + p_17*X7*sin(X8) - C_1(i);
f2 = @(X1,X2,X3,X4,X5,X6,X7,X8) p_21*X1 + p_22*X2 + p_23*X3 + p_24*X4 + p_25*X5 + p_26*X6 + p_27*X7*sin(X8) - C_2(i);
f3 = @(X1,X2,X3,X4,X5,X6,X7,X8) p_31*X1 + p_32*X2 + p_33*X3 + p_34*X4 + p_35*X5 + p_36*X6 + p_37*X7*cos(X8) - C_3(i);
f4 = @(X1,X2,X3,X4,X5,X6,X7,X8) p_41*X1 + p_42*X2 + p_43*X3 + p_44*X4 + p_45*X5 + p_46*X6 + p_47*X7*sin(X8) - C_4(i);
f5 = @(X1,X2,X3,X4,X5,X6,X7,X8) p_51*X1 + p_52*X2 + p_53*X3 + p_54*X4 + p_55*X5 + p_56*X6 + p_57*X7*sin(X8) - C_5(i);
f6 = @(X1,X2,X3,X4,X5,X6,X7,X8) p_61*X1 + p_62*X2 + p_63*X3 + p_64*X4 + p_65*X5 + p_66*X6 + p_67*X7*sin(X8) - C_6(i);
f7 = @(X1,X2,X3,X4,X5,X6,X7,X8) p_71*X1 + p_72*X2 + p_73*X3 + p_74*X4 + p_75*X5 + p_76*X6 + p_77*X7*sin(X8) - C_7(i);
f8 = @(X1,X2,X3,X4,X5,X6,X7,X8) p_81*X1 + p_82*X2 + p_83*X3 + p_84*X4 + p_85*X5 + p_86*X6 + p_87*X7 - C_8(i);
f = @(X)[f1(X(1),X(2),X(3),X(4),X(5),X(6),X(7),X(8));
f2(X(1),X(2),X(3),X(4),X(5),X(6),X(7),X(8));
f3(X(1),X(2),X(3),X(4),X(5),X(6),X(7),X(8));
f4(X(1),X(2),X(3),X(4),X(5),X(6),X(7),X(8));
f5(X(1),X(2),X(3),X(4),X(5),X(6),X(7),X(8));
f6(X(1),X(2),X(3),X(4),X(5),X(6),X(7),X(8));
f7(X(1),X(2),X(3),X(4),X(5),X(6),X(7),X(8));
f8(X(1),X(2),X(3),X(4),X(5),X(6),X(7),X(8))];
X = sym('X',[8 1]);
J = jacobian(f(X))
rank(J)
X0 = [1;2;3;4;5;6;7;8];
sol(i,:) = fsolve(f,X0)
end
J =
ans = 4
Solver stopped prematurely. fsolve stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 8.000000e+02.
sol = 1×8
1.0e+06 * -0.0003 0.0204 -1.6787 0.0627 4.8308 -3.2348 0.0000 0.0000
J =
ans = 4
Solver stopped prematurely. fsolve stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 8.000000e+02.
sol = 2×8
1.0e+06 * -0.0003 0.0204 -1.6787 0.0627 4.8308 -3.2348 0.0000 0.0000 0.0023 -1.9292 0.0913 3.6340 0.1631 -1.9616 0.0000 0.0000
J =
ans = 4
Solver stopped prematurely. fsolve stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 8.000000e+02.
sol = 3×8
1.0e+06 * -0.0003 0.0204 -1.6787 0.0627 4.8308 -3.2348 0.0000 0.0000 0.0023 -1.9292 0.0913 3.6340 0.1631 -1.9616 0.0000 0.0000 0.0000 -0.1246 1.9481 -1.2147 -2.9165 2.3077 -0.0000 0.0000
Ismita 2024년 3월 16일
Thank you so much @Torsten.
댓글을 달려면 로그인하십시오.
### 추가 답변 (1개)
John D'Errico 2024년 3월 15일
편집: John D'Errico 2024년 3월 15일
1. This is not even a remotely large number of equations.
2. Will you please learn to use matrices?
3. X8 appears in only one form in your equations, as X7*sin(X8). Therefore, you could transform the problem trivially, by substituting a single new unknown into the problem. That is...
Y8 = X7*sin(X8).
Replace the product X7*sin(X8) in each place it appears in your system. Your system is now linear, in the set of unknowns {X1,X2,X3,X4,X5,X67,X7,Y8}. Solve that LINEAR system. Once you have found the values of X7 and Y8, you can recover X8.
X8 = asin(Y8/X7)
Remember this means there will be infinitely many equivalent solutions, but so what? And of course, if X7 were exactly zero, then no solution exists. Finally, if it is true that abs(Y8/X7)>1, then the only solutions possible will be complex solutions. Such is life.
So just use backslash to solve the problem. AND PLEASE LEARN TO USE MATRICES.
##### 댓글 수: 3이전 댓글 1개 표시이전 댓글 1개 숨기기
Torsten 2024년 3월 15일
X7*cos(X8) in eq3 and p_87*X7 in eq8 unfortunately hinder to convert the system to a linear one.
Ismita 2024년 3월 15일
댓글을 달려면 로그인하십시오.
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Translated by | 4,762 | 9,099 | {"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-2024-22 | latest | en | 0.538071 |
http://slideplayer.com/slide/1455665/ | 1,529,352,251,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00150.warc.gz | 296,467,167 | 20,559 | # Thermodynamics in static electric and magnetic fields 1 st law reads: -so far focus on PVT-systems where originates from mechanical work Now: -additional.
## Presentation on theme: "Thermodynamics in static electric and magnetic fields 1 st law reads: -so far focus on PVT-systems where originates from mechanical work Now: -additional."— Presentation transcript:
Thermodynamics in static electric and magnetic fields 1 st law reads: -so far focus on PVT-systems where originates from mechanical work Now: -additional work terms for matter in fields Dielectric Materials 1 1 -electric field inside the capacitor: A + - VeVe dielectric material L +q -q -displacement field D given by the free charges on the capacitor plates: Source of D is density of free charges. Here: charge q on capacitor plate with area A
-Reduction of qEnergy content in capacitor reduced which means work W cap >0 done by the capacitor ( in accordance with our sign convention for PVT systems ) (dq 0 yields W cap >0) With V= volume of the dielectric material -When no material is present: still work is done by changing the field energy in the capacitor -Work done by the material exclusively: parameterized e.g., with time (slow changes!)
With Polarization=total dipole moment per volume With ( where V=const. is assumed so that PdV has not to be considered ) Comparing ( where work is done mechanically via volume change against P ) With we define the total dipole moment of the dielectric material with Correspondenceand
-Legendre transformations ( providing potentials depending on useful natural variables ) making electric field E variable H=H(S,E) making T variable G=G(T,E) and
Magnetic Materials 2 2 I N: # of turns of the wire R Faradays law: where Amperes law: where here A: cross sectional area of the ring magn. flux lines voltage V ind induced in 1 winding
-Reduction of the current Iwork done by the ring work done by the ring per time makes sure that reduction of B ( ) corresponds to work done by the ring -Again, when no material is present: still work is done on the source by changing the field energy In general: where M is the magnetization = magnetic dipole moment per volume No material M=0 rate at which work is done by the magnetic material
-Legendre transformations ( providing potentials depending on useful natural variables ) making magnetic field H variable H enth =H enth (S,H) making T variable G=G(T,H) and
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Similar presentations | 579 | 2,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-26 | latest | en | 0.857683 |
https://laabundanciabakery.com/cooking-tips/how-many-cups-in-a-litre/ | 1,695,513,131,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00461.warc.gz | 400,268,554 | 22,935 | La Abundancia Bakery & Restaurant
# How Many Cups in A Litre
Welcome to our informative guide on the topic of “How many cups in a litre?” Cups and liters are both units of volume measurement commonly used in cooking and baking. However, depending on the recipe or country of origin, the measurement may be given in either cups or liters. If you’re working with a recipe that measures in liters but you’re more comfortable using cups, it’s important to know the conversion rate between these two units of measurement.
In this guide, we’ll be discussing how many cups in a litre and offering some tips on how to convert between these measurements accurately. So, let’s dive in and find out how many cups are in a liter.
## What is Cup Definition?
How many cups in a litre
Cups are a widely used unit of measurement in cooking and baking for liquids, powders, and solids. It’s a standard way of measuring volume for both dry and liquid ingredients. In the United States, one cup equals 8 fluid ounces or 16 tablespoons, but in other countries, the standard cup measurement may vary slightly. It’s vital to use precise measurements when cooking and baking to guarantee consistent and accurate results.
One more term to before going to “How many cups in a litre”
## What is Litre/Liter?
A liter, or litre, is another unit of volume measurement and is the metric system equivalent of a quart. While a cup is an imperial unit of measurement used in the United States and other countries, a liter is the metric system’s standard measurement for liquids. A liter is equal to 1.057 quarts or 33.814 fluid ounces, but it may vary slightly depending on the country of origin.
## How Many Cups in A Litre?
How many cups in a litre
So, the answer to the question “How many cups in a litre?” is 4.226775. To make this conversion easy, keep in mind that one liter equals 4 cups plus 1/3 of a cup. In other words, you need four full cups and one additional 1/3 cup to equal one liter.
## Are Metric (US) Cups the Same as Imperial (UK) Cups or Not?
To clarify, metric cups and imperial cups are not equivalent. In the US, a cup is equal to 8 fluid ounces or 16 tablespoons, whereas in the UK, a cup is equal to 10 fluid ounces or 20 tablespoons. Thus, when converting milliliters to cups, the results will differ depending on whether you’re using US or UK measurements. It’s crucial to verify the measurements used in a recipe to ensure you obtain precise results with metric or imperial measurements.
## Converting from Cups to Litre
How many cups in a litre
It’s also essential to know how to convert from cups to liters. To do so, simply multiply the number of cups by 0.236588. For example, if you want to convert 750 mL or 3/4 of a liter into cups, you would multiply 0.75 (the equivalent of 750 mL) by 4.226775. This results in 3.170081, or 3 and 1/3 cups.
Let’s get to some handy tips such as a common conversion chart for “How many cups in a litre”?
## Common Conversion Chart from Cups to Litre
1 cup = 0.236588 liters
2 cups = 0.473176 liters
3 cups = 0.709763 liters
4 cups = 0.946351 liters
5 cups = 1.182939 liters
6 cups = 1.419527 liters
7 cups = 1.656115 liters
8 cups = 1.892703 liters
9 cups = 2.129291 liters
10 cups = 2.365879 liters
11 cups = 2.602467 liters
12 cups = 2.839055 liters
These are just a few of the most common conversions from cups to liters, but you can always calculate more using the formula discussed above. Take these as tips for your answer “How many cups in a litre” and so on.
## Handy Tips from Cups to mL Measurements
How many cups in a litre
Here are a few more tips to help you accurately measure cups when converting from liters:
• Use a measuring cup that is marked with milliliters. If not, use an empty container and fill it accordingly to achieve the desired measurements.
• Make sure to read recipes using metric or imperial measurements carefully before beginning.
• If a recipe calls for metric measurements, use milliliters instead of cups.
• If a recipe calls for imperial measurements, use both cups and tablespoons to achieve the desired result.
• Always check the volume of your measuring device against standard measurements to ensure accuracy.
• Use scales if you don’t have access to measuring cups and spoons.
You can accurately convert measurements from cups to mL and improve your cooking and baking with the help of our blog post titled “How many cups in a litre”.
## Conclusion
In conclusion, knowing “How many cups in a litre” and how to convert between cups and liters is an essential skill for any home cook or baker. By understanding the conversion rate between these two units of volume measurement, you can ensure that you’re using the right amount of ingredients in your recipes and achieve consistent and accurate results.
As we have discussed in this guide, 1 liter is equal to approximately 4.23 cups, but it’s important to keep in mind that this conversion rate is an approximation and may vary depending on the measuring cups or kitchen scales used. By following the tips provided in this guide from laabundanciabakery.com, you can convert between cups and liters accurately and confidently tackle any recipe that comes your way.
## FAQs of How Many Cups are in A Litre
### How many liters are in a cup?
A: 1 cup is equal to approximately 0.24 liters.
### Why do some recipes use cups and others use liters?
A: The choice of measurement units depends on the country of origin, recipe author, or personal preference.
### How many cups in a litre?
A: 1 liter is equal to approximately 4.23 cups.
### Can I use mL instead of cups or litres?
A: Yes, mL is another unit of volume measurement that can be used in recipes.
### How do I convert liters to cups?
A: Multiply the number of liters by 4.23 to convert to cups.
### How do I measure a liter of liquid?
A: Use a measuring cup or a kitchen scale to measure out 1 liter of liquid.
### How many cups are in half a liter?
A: Half a liter is equal to approximately 2.11 cups.
### How do I convert cups to liters?
A: Divide the number of cups by 4.23 to convert to liters.
### Can I use liters for dry ingredients?
A: Liters are typically used for measuring liquids, but some recipes may use liters for dry ingredients as well.
### Can I use cups for weight measurements?
A: Cups are typically used for volume measurements, not weight measurements.
### How many cups are in a quart?
A: 1 quart is equal to approximately 4 cups.
### How many cups are in 500 mL?
A: 500 mL is equal to approximately 2.11 cups.
### How do I convert liters to fluid ounces?
A: Multiply the number of liters by 33.814 to convert to fluid ounces.
### Can I use liters instead of quarts?
A: Yes, liters and quarts are both units of volume measurement that can be used interchangeably.
### How many cups are in 2 liters?
A: 2 liters is equal to approximately 8.45 cups.
### Why should I know how many cups in a litre?
A: Knowing how to convert between cups and liters accurately is an important kitchen skill that will help you follow recipes correctly and achieve consistent results. We hope this guide of “How many cups are in a litre” has been helpful in teaching you the basics of converting between measurements. | 1,695 | 7,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-40 | latest | en | 0.918289 |
https://www.hackmath.net/en/math-problem/235 | 1,606,778,439,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00309.warc.gz | 706,482,428 | 12,416 | # DIY press
Under socialism regime was in some socialist countries to own a typewriter requires special permission. That has hindered the spread of DIY literature (manually transcribed through carbon copy paper for typewriters).
Calculate how many typewriters today can indulge Joseph with payment 934 euros if typewriter weighs 10 kg and scrap iron is worth 0.19 euro/kg.
Correct result:
n = 492
#### Solution:
$n=\frac{934}{10\cdot 0.19}=492$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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https://www.orbiterwiki.org/index.php?title=ecliptic&oldid=5134 | 1,571,202,854,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00196.warc.gz | 1,047,639,852 | 10,842 | # ecliptic
The ecliptic is the apparent path of the Sun traces out along the sky — independent of Earth's rotation — in the course of the year. More accurately, it is the intersection of the celestial sphere with the ecliptic plane, which is the geometric plane containing the mean orbit of the Earth around the Sun. It should be distinguished from the invariable ecliptic plane, which is the vector sum of the angular momenta of all planetary orbital planes, to which Jupiter is the main contributor.
The name ecliptic is derived from being the place where eclipses occur.
## Ecliptic and equator
As the rotation axis of the Earth is not perpendicular to its orbital plane, the equatorial plane is not parallel to the ecliptic plane, but makes an angle of about 23°44 which is known as the obliquity of the ecliptic. The intersections of the equatorial and ecliptic plane with the celestial dome are great circles known as the celestial equator and the ecliptic. The intersection line of the two planes results in two diametrically opposite intersection points, known as the equinoxes. The equinox which the Sun passes from south to north is known as the vernal equinox or first point of Aries. Ecliptic longitude, usually indicated with the letter λ, is measured from this point on 0° to 360° towards the east. Ecliptic latitude, usually indicated with the letter β is measured +90° to the north or -90° to the south. The same intersection point also defines the origin of the equatorial coordinate system, named right ascension measured from 0 to 24 hours also to the east and usually indicated with α or R.A., and declination, usually indicated with δ also measured +90° to the north or -90° to the south. Simple rotation formulas allow a conversion from α,δ to λ,β and back (see: ecliptic coordinate system).
## Ecliptic and stars
The ecliptic serves as the center of a region called the zodiac which constitutes a band of 8° on either side. Traditionally, this region is divided into 12 signs of 30° longitude each. By tradition, these signs are named after 12 of the 13 constellations straddling the ecliptic. The zodiac signs are very important to astrologers. Modern astronomers typically use other coordinate systems today (see below).
In the Indian tradition there are 27 Nakshatras, which cover 13°20’ of the ecliptic each. Each Nakshatra is divided into quarters or padas of 3°20’.
The position of the vernal equinox is not fixed among the stars but due to the lunisolar precession slowly shifting westwards over the ecliptic with a speed of 1° per 72 years. A much smaller north/southwards shift can also be discerned, (the planetary precession, along the instantaneous equator, which results in a rotation of the ecliptic plane). Said otherwise the stars shift eastwards (increase their longitude) measured with respect to the equinoxes (in other words, as measured in ecliptic coordinates and (often) also in equatorial coordinates.
Using the current official IAU constellation boundaries — and taking into account the variable precession speed and the rotation of the ecliptic — the equinoxes shift through the constellations in the Astronomical Julian calendar years (in which the year 0 = 1 BC, -1 = 2 BC, etc.) as follows:[1]
• The March equinox passed from Taurus into Aries in year -1865, passed into Pisces in year -67, will pass into Aquarius in year 2597, will pass into Capricorn in year 4312. It passed along (but not into) a 'corner' of Cetus on 0°10' distance in year 1489.
• The June solstice passed from Leo into Cancer in year -1458, passed into Gemini in year -10, passed into Taurus in December year 1989, will pass into Aries in year 4609.
• The September equinox passed from Libra into Virgo in year -729, will pass into Leo in year 2439.
• The December solstice passed from Capricorn into Sagittarius in year -130, will pass into Ophiuchus in year 2269, and will pass into Scorpius in year 3597.
## Ecliptic and Sun
Due to perturbations to the Earth's orbit by the other planets, the true Sun is not always exactly on the ecliptic, but may be some arcseconds north or south of it. It is therefore the centre of the mean Sun which outlines its path. As the Earth revolves in one year around the Sun, it appears that the Sun also needs one year to pass the whole ecliptic. With slightly more than 365 days in the year, the Sun moves almost 1° eastwards every day (direction of increasing longitude). This annual motion should not be confused with the daily motion of the Sun (and the stars, the whole celestial sphere for that matter) towards the west in 24 hours and along the equator. In fact where the stars need about 23h56m for one such rotation to complete, the sidereal day, the Sun, which has shifted 1° eastwards during that time needs 4 minutes extra to complete its circle, making the solar day just 24 hours.
The mean Sun crosses the equator around 21 March in the vernal equinox, its declination, right ascension, and ecliptic longitude are all zero then (the ecliptic latitude is always). The March equinox marks the onset of spring in the northern hemisphere and autumn in the southern. As such the term "spring equinox" should be avoided. The actual date and time varies from year to year because of the occurrence of leap years. It also shifts slowly over the centuries due to imperfections in the Gregorian calendar.
Ecliptic longitude 90°, at right ascension 6 hours and a northern declination equal to the obliquity of the ecliptic (23.44°), is reached around 22 June. This is the June solstice or summer solstice in the northern hemipshere and winter solstice in the southern hemisphere. It is also the first point of Cancer and directly overhead on Earth on the tropic of Cancer so named because the Sun turns around in declination. Ecliptic longitude 180°, right ascension 12 hours is reached around 23 September and marks the second equinox or first point of Libra. Due to perturbations to the Earth orbit, the moment the real Sun passes the equator might be several minutes earlier or later. The southern most declination of the sun is reached at ecliptic longitude 270°, right ascension 18 hours at the first point of the sign of Capricorn around 22 December.
In any case it must be stressed that although these traditional signs (in western tropical astrology) have given their names to the solstices and equinoxes, in reality, (as from the list in the previous chapter) the cardinal points are currently situated in the constellations of Pisces, Taurus, Virgo and Sagittarius respectively.
## Ecliptic and planets
Most planets go in orbits around the sun which are almost in the same plane as the Earth's orbital plane, differing by a few degrees at most. As such they always appear close to the ecliptic when seen in the sky. Mercury with an orbital inclination of 7° or Pluto with 17° are exceptions. Many minor planets have large inclinations too. The intersection line of the ecliptical plane and the orbital plane is called the nodal line, and the intersection points on the celestial sphere are the ascending node (where the planet crosses the ecliptic from south to north) and the diametrically opposite descending node. Only when an inferior planet passes through one of its nodes a transit over the Sun can take place.
Inclination and nodal lines, as almost all other orbital elements, change slowly over the centuries due to perturbations from the other planets.
## Ecliptic and Moon
The orbit of the Moon is inclined by about 5° on the ecliptic. Its nodal line is not fixed either, but regresses (moves towards the west) over a full circle every 18.6 years. This is the cause of nutation and lunar standstill. The moon crosses the ecliptic about twice per month. If this happens during new moon a solar eclipse occurs, during full moon a lunar eclipse. This was the way the ancients could trace the ecliptic along the sky; they marked the places where eclipses could occur.
## Ecliptic and star coordinates
Up to the 17th century, starmaps and positions in star catalogues were always given in ecliptical coordinates. It was not until astronomers started to use telescopes to measure star positions that equatorial coordinates came in use, and so exclusively that nowadays ecliptical coordinates are no longer used. This is not always desirable. A planetary conjuction for example would be much more illustratively described by ecliptic coordinates than equatorial.
Also see zodiacal coordinates.
## References
1. J. Meeus; Mathematical astronomical morsels; ISBN 0-943396-51-4 | 1,993 | 8,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-43 | longest | en | 0.95752 |
https://www.sicara.fr/blog-technique/2017-08-29-how-evolution-taught-us-the-genetic-algorithm | 1,696,424,965,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00644.warc.gz | 1,067,205,547 | 18,114 | April 8, 2022 • 5 min read
# How evolution taught us the genetic algorithm ?
Rédigé par Nicolas Barbier
In this article, I am going to explain the concept of genetic algorithm. First, I am going to present its origin and its goal. Then I am going to show you how to implement a genetic algorithm with a short python tutorial.
So, the question is:
## How to create a good Artificial Intelligence?
The naive solution is to create an “empirical algorithm” which is a set of rules: “if you meet this conditions, act like that”. I could imagine that with enough rules like this we could reproduce natural intelligence. But the amount of work is colossal and the final solution will never be able to best its creator. Isn’t it depressing to spend a lot of time creating something while knowing it can’t be perfect?
To avoid that, a new idea emerged. What if instead of creating a direct solution we recreate evolution. We could create the first prehistoric fishes, put them in conditions suitable to evolution and let them evolve freely toward man-kind or even further. This idea is called “genetic algorithm” and we are going to build one in the next part. So first let us refresh our memories and try to understand the natural selection theorized by Darwin.
This theory is simple: if a population want to thrive, it must improve by itself constantly, it’s the survival of the fittest. The best element of the population should inspire the offspring, but the other individuals must not be forgotten in order to maintain some diversity and be able to adapt in case of a variation of the natural environment.
Genetic algorithms are especially efficient with optimization problems.
## Example: the Knapsack problem
The backpack optimization issue is a classical algorithmic problem. You have two things: a backpack with a size (the weight it can hold) and a set of boxes with different weights and different values. The goal is to fill the bag to make it as valuable as possible without exceeding the maximum weight. It is a famous mathematical problem since 1972. The genetic algorithm is well suited to solve that because it’s an optimization problem with a lot of possible solutions.
Other algorithms aren’t efficient at all, see the Wikipedia page if you want to understand why.
In order to test by ourselves how a genetic algorithm works, we are going to use it to solve a simple problem: how could I crack my colleague’s password?
The algorithm is implemented on Python 3. You can download it here for Windows, or install it using `brew install python3` , `sudo apt-get install python3` or `sudo yum install python3` . I advise you to run this code inside a Jupyter notebook.
## Choosing a fitness function
The evaluation function is the first step to create a genetic algorithm. It’s the function that estimates the success of our specimen: it will allow us to divide the population between the ugly duckling and the beautiful swans. The goal of this separation is that, later, the successful specimen will have more “chance” to get picked to form the next generation. As simple as it appears, don’t be fooled, it’s the step of a genetic algorithm with the more intelligence related to the problem.
What is our goal? Crack a password. Thus the goal of our function is to transform the binary result “fail or success” in a continuous mark from 0 (can’t fail more) to 100 (perfection).
The simplest solution here is:
``fitness score = (number of char correct) / (total number of char)``
That way, an individual with a bigger fitness result is a specimen genetically closer to success than the others. Thus the fitness function for our genetic algorithm will accurately sort the population.
## Creating our individuals
So now we know how to evaluate our individuals; but how do we define them? This part is really tricky: the goal is to know what are the unalterable characteristics and what is variable.
The comparison with genetics is here really helpful. Indeed, the DNA is composed of genes, and each of those genes comes through different alleles (different versions of this gene). Genetic algorithms retain this concept of population’s DNA.
In our case, our individuals are going to be words (obviously of equal length with the password). Each letter is a gene and the value of the letter is the allele. In the word “banana”: ‘b’ is the allele of the first letter.
What is the point of this creation?
• We know that each of our individuals is keeping the good shape (a word with the correct size)
• Our population can cover every possibility (every word possible with this size).
Out genetic algorithm can then explore all possible combinations.
## Creating our first population
Now, we know what are the characteristics of our individuals and how we can evaluate their performance. We can now start the “evolution” step of our genetic algorithm.
The main idea to keep in mind when we create the first population is that we must not point the population towards a solution that seems good. We must make the population as wide as possible and make it cover as many possibilities as possible. The perfect first population of a genetic algorithm should cover every existing allele.
So in our case, we are just going to create words only composed of random letters.
## From one generation to the next
Given a generation, in order to create the next one, we have 2 things to do. First we select a specific part of our current generation. Then the genetic algorithm combines those breeders in order to create the next batch.
### Breeders selection
They are lots of way to do this but you must keep in mind two ideas: the goals are to select the best solutions of the previous generation and not to completely put aside the others. The hazard is: if you select only the good solutions at the beginning of the genetic algorithm you are going to converge really quickly towards a local minimum and not towards the best solution possible.
My solution to do that is to select on the one hand the N better specimen (in my code, N = best_sample) and on the other hand to select M random individuals without distinction of fitness (M = lucky_few).
### Breeding
We can keep on the biologic analogy for the breeding in our genetic algorithm. The goal of sexual reproduction is to mix the DNA of 2 individuals, so let’s do the same thing here. We have two individuals: “Tom” and “Jerry”, their DNA is defined by their alleles (the value of each letter). Thus in order to mix their DNA, we just have to mix their letters. There are lots of ways to do this so I picked the simplest solution: for each letter of the child, take randomly the letter of “Tom” or “Jerry”.
NB: Obviously, the couple “Tom" and “Jerry” isn’t going to produce only one child. You have to fix the number of children per couple in order to keep a stable population in your genetic algorithm. The number of individuals in the generation 0 equals the number of individuals in the next generation.
## Mutation
This last step of our genetic algorithm is the natural mutation of an individual. After the breeding, each individual must have a small probability to see their DNA change a little bit. The goal of this operation is to prevent the algorithm to be blocked in a local minimum.
In order to choose the mutation rate for our genetic algorithm, I followed this article counsels. I studied the influence of mutation more precisely here.
Now you have all the tools to build your own genetic algorithm. Feel free to modify my own implementation. For each step, a lot of solutions are possible, take the fittest to solve your problem.
Here is my git repository with the whole code, here is the gist version.
## Wanna go further?
Here is a list of other supports to train, understand and even compete on the field of AI and genetic algorithm.
### Web application
BoxCar is an online example of a genetic algorithm. The goal is to create the most efficient two . wheels vehicles. Check the result here
### Mobile Application
With this application, you are more solicited than in the previous. You have to create a “creature” with joints, bones, and muscles. Then, the genetic algorithm tries to optimize the moves of your creature in order to execute a task: jump, run, climb stairs and others.
Check the app here.
### Do it yourself
Finally, last but not least: Coding Game. It’s a platform linking lots of dev around a common passion “AI”. Every month there is a one-week long contest where you must create the best artificial intelligence possible. The winner is nearly always using the genetic algorithm. So if you feel ready to go to the big leagues, take a deep breath and jump in here.
Thanks to Flavian Hautbois and Tristan Roussel.
If you are looking for Machine Learning expert's, don't hesitate to contact us !
Nicolas Barbier | 1,865 | 8,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-40 | latest | en | 0.926477 |
https://de.maplesoft.com/support/help/maple/view.aspx?path=VectorCalculus/SpaceCurve&L=G | 1,642,961,819,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.5/warc/CC-MAIN-20220123172206-20220123202206-00605.warc.gz | 256,095,339 | 29,096 | SpaceCurve - Maple Help
VectorCalculus
SpaceCurve
plot a space curve in ${R}^{2}$ or ${R}^{3}$
Calling Sequence SpaceCurve(C, r)
Parameters
C - Vector(algebraic); the free or position Vector representing a curve r - name=range; the range of the parameter of the curve
Description
• The SpaceCurve(C, t=a..b) calling sequence plots a space curve in ${R}^{2}$ or ${R}^{3}$. The plot is displayed in Cartesian coordinates.
• Alternatively, a curve can be defined using a position Vector and it can be visualized with PlotPositionVector.
Examples
> $\mathrm{with}\left(\mathrm{VectorCalculus}\right):$
The commands to create the plots in the Plotting Guide are
> $\mathrm{SpaceCurve}\left(⟨\mathrm{exp}\left(-t\right)\mathrm{cos}\left(t\right),\mathrm{exp}\left(-t\right)\mathrm{sin}\left(t\right)⟩,t=4..8\right)$
> $\mathrm{SpaceCurve}\left(⟨\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right),t⟩,t=1..9\right)$
Other example
> $\mathrm{SetCoordinates}\left(\mathrm{polar}\left[r,t\right]\right)$
${{\mathrm{polar}}}_{{r}{,}{t}}$ (1)
> $\mathrm{SpaceCurve}\left(\mathrm{PositionVector}\left(\left[1,t\right]\right),t=0..2\mathrm{\pi }\right)$ | 363 | 1,166 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-05 | longest | en | 0.366347 |
http://blog.liyunchen.com/tag/bayesian/ | 1,516,312,994,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887621.26/warc/CC-MAIN-20180118210638-20180118230638-00312.warc.gz | 53,662,205 | 13,787 | # Variance: regression, clustering, residual and variance
This is the translation of my recent post in Chinese. I was trying to talk in the way that a statistician would use after having stayed along with so many statistics people in the past years.
-------------------------------------------Start----------------------------
Variance is an interesting word. When we use it in statistics, it is defined as the "deviation from the center", which corresponds to the formula $\sum (x- \bar{x})^2 / (n-1)$ , or in the matrix form $Var(X) = E(X^2)- E(X)^2=X'X/N-(X'1/N)^2$ (1 is a column vector with N*1 ones). From its definition it is the second (order) central moment, i.e. sum of the squared distance to the central. It measures how much the distribution deviates from its center -- the larger the sparser; the smaller the denser. This is how it works in the 1-dimension world. Many of you should be familiar with these.
Variance has a close relative called standard deviation, which is essentially the square root of variance, denoted by $\sigma$. There is also something called the six-sigma theory-- which comes from the 6-sigma coverage of a normal distribution.
Okay, enough on the single dimension case. Let's look at two dimensions then. Usually we can visualize the two dimension world with a scatter plot. Here is a famous one -- old faithful.
Old faithful is a "cone geyser located in Wyoming, in Yellowstone National Park in the United States (wiki)...It is one of the most predictable geographical features on Earth, erupting almost every 91 minutes." We can see there are about two hundreds points in this plot. It is a very interesting graph that can tell you much about Variance.
Here is the intuition. Try to use natural language (rather than statistical or mathematical tones) to describe this chart, for example when you take your 6 year old kid to the Yellowstone and he is waiting for next eruption. What would you tell him if you have this data set? Perhaps "I bet the longer you wait, the longer next eruption lasts. Let's count the time!". Then the kid has a glance on your chart and say "No. It tells us that if we wait for more than one hour (70 minutes) then we will see a longer eruption in the next (4-5 minutes)". Which way is more accurate?
Okay... stop playing with kids. We now consider the scientific way. Frankly, which model will give us a smaller variance after processing?
Well, always Regression first. Such a strong positive relationship, right? ( no causality.... just correlation)
Now we obtain a significantly positive line though R-square from the linear model is only 81% (could it be better fitted?). Let's look at the residuals.
It looks like that the residuals are sparsely distributed...(the ideal residual is white noise which carries no information). In this residual chart we can roughly identify two clusters -- so why don't we try clustering?
Before running any program, let's have a quick review the foundations of the K-means algorithm. In a 2-D world, we define the center as $(\bar{x}, \bar{y})$, then the 2-D variance is the sum of squares of each pint going to the center.
The blue point is the center. No need to worry about the outlier's impact on the mean too much...it looks good for now. Wait... doesn't it feel like the starry sky at night? Just a quick trick and I promise I will go back to the key point.
For a linear regression model, we look at the sum of squared residuals - the smaller the better fit is. For clustering methods, we can still look at such measurement: sum of squared distance to the center within each cluster. K-means is calculated by numerical iterations and its goal is to minimize such second central moment (refer to its loss function). We can try to cluster these stars to two galaxies here.
After clustering, we can calculate the residuals similarly - distance to the central (represents each cluster's position). Then the residual point.
Red ones are from K-means which the blue ones come from the previous regression. Looks similar right?... so back to the conversation with the kid -- both of you are right with about 80% accuracy.
Shall we do the regression again for each cluster?
Not many improvements. After clustering + regression the R-square increases to 84% (+3 points). This is because within each cluster it is hard to find any linear pattern of the residuals, and the regression line's slope drops from 10 to 6 and 4 respectively, while each sub-regression only delivers an R-square less than 10%... so not much information after clustering. Anyway, it is better than a simple regression for sure. (the reason why we use k-means rather than some simple rules like x>3.5 is that k-means gives the optimized clustering results based on its loss function).
Here is another question: why do not we cluster to 3 or 5? It's more about overfitting... only 200 points here. If the sample size is big then we can try more clusters.
Fair enough. Of course statisticians won't be satisfied with these findings. The residual chart indicates an important information that the distribution of the residuals is not a standard normal distribution (not white noise). They call it heteroscedasticity. There are many forms of heteroscedasticity. The simplest one is residual increases when x increases. Other cases are in the following figure.
The existence of heteroscedasticity makes our model (which is based on the training data set) less efficient. I'd like to say that statistical modelling is the process that we fight with residuals' distribution -- if we can diagnose any pattern then there is a way to improve the model. The econometricians prefer to name the residuals "rubbish bin" -- however it is also a gold mine in some sense. Data is a limited resource... wasting is luxurious.
# A few books want to read
Fortunately or accidentally, I only have two classes this term. Meanwhile, they separate them into four days, so I only have two-hours class every day from Monday to Thursday. Compared to my previous schedule, it is too relaxing.
An advantage now is that I have enough time to read and think. Today I found Becker's book by chance, when I was browsing the literature on "social economics", or socio-economics. It is quite exciting, and I have realized how deep the water might be- before I was only using my naive intuition that there is something I can contribute soon.
The book I'm talking about now is
Gary S. Becker and Kevin M. Murphy, 2001, Social Economics: Market Behavior in a Social Environment.
Before I was paying more attention solely to network economics, and it turned out to be that they were quite similar to each other in most sense; however, socio-economics is for sure more broad.
Moreover, I took a few hours finishing reading another book,
Salsburg, D. (2002) The Lady Tasting Tea: How Statistics Revolutionized Science in the Twentieth Century
From the name, you can see that this book is about basic statistics. To save time, I read its Chinese translation. Not very long, but very exciting - maybe I have been, and will always be attracted by mathematics and statistics. Especially the later one, perhaps due to the fact that I have so many friends in this area, is one field beyond economics that has influenced me the most, and more on the level of conception and methodology than techniques or actual methods.
@Roma. Things remain to be clear
While reading this book, it reminds me another book I read before, which is about the famous economist Keynes,
Robert Skidelsky, 2005, John Maynard Keynes: 1883-1946: Economist, Philosopher, Statesman
What impressed me most at that time was not Keynes' contribution to economics - although nobody can neglect that, but his ideas on probability. Until now, I still have the wish that one day I want to read Keynes' original book on probability somewhere.
I want to read Becker's book only for the reason that I need an idea for my history paper. One question I have been seeking for the answer for a while: why do we need to care about the network structure? Before, I was only arguing that the "summation is a naive way to draw the group's characteristics"; now it seems that I need to really re-think about this argument. In addition to sum or mean, people have developed distribution to help understand the world; furthermore, from central limit theorem, normal distribution can be utilized in most scenarios. Therefore, under what particular case will summation cause a severe problem?
Another thing I'm thinking about now is after reading the "Lady tasting tea", a term still remains to be explained more clearly: frequency school and Bayesian school's debate on the definition of probability. On one side we are lucky today that following Baye's idea will not be regarded as heterodox any more; on the other side, although his idea itself is very simple, how to make a perfect use of it is still a very tricky and should be dealed with carefully.
I'll stop here for now, and see whether I can gain some new senses soon. This year is too short- I need a longer time to make all things clear.
# Probability, Information and Economics
These days I was busy reading the biography of John Maynard Keynes, the most famous economist in the past century. One point mentioned in that book attracted my attention -- that is about his ideas on probability.
Every one who has studied macroeconomics must know a word "rational expectations". That is a great issue if talked. Simply, as the wikipedia says,
To assume rational expectations is to assume that agents' expectations are wrong at every one instance, but correct on average over long time periods. In other words, although the future is not fully predictable, agents' expectations are assumed not to be systematically biased and use all relevant information in forming expectations of economic variables.
Here I do not want to say much about it. I'd like to mention another area, Information Economics. Typically, information economics deals with the situation that there is asymmetric information between principal and agent. Then as we all know, there is moral hazard and adverse selection. With the application of game theory, the common issues can be solved. However, seldom do I read paper discussing about the role of information in economic activities in other approaches. Therefore, followed Keynes' idea, I wonder what will happen if the spread of information is introduced into the economic activities.
Simply, probability reflects the situation that we do not know enough about how the real world functions. Therefore, we use probability to describe the combination every possible result. There is an interesting question: the normal distribution. I'll talk about it later on.
As the aim of science, we are pursuing the ability to predict. I know many people will have different ideas, but it does not matter much. At least, we want to know the mechanisms in every particular field. That is, we are pursuing "certainty" instead of "uncertainty". From uncertainty to probability, then to certainty, in this way we know much better about the real world. It is an old philosophic issue: is there a fixed point?
Then what will happen if the knowledge spreads? I have not got a clear understanding yet. The disappearance of probability is too hard to imagine. We can use "normal distribution" to describe some phenomenons, such as people's height, weight. The result is a description of a group, but not that accurate for a particular person. To predict a person's height, for example, we should get enough information, if applicable, his gene, his nutrition, and what he did in the past... Maybe it is too hard to define what is "enough". Anyway, the probability can be replaced under a special circumstance.
In the first step, I want to talk about how the spread of information influences the social activities. I think we have underestimated the importance of information in economics, or we have no applicable models to explain. I do not know whether more modern mathematical tools are needed in the explanations. As least, I need to read more about the history of probability, including the famous debate between frequency school and Bayesian. And maybe more knowledge about psychology and communication are essential. I want to talk about it later after learning measure theory. | 2,601 | 12,376 | {"found_math": true, "script_math_tex": 4, "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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-05 | latest | en | 0.947263 |
https://astronomy.stackexchange.com/questions/11009/pulsars-how-do-astronomers-measure-minute-changes-in-period-picoseconds-per-y | 1,721,880,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00517.warc.gz | 88,289,350 | 40,464 | # Pulsars: How do astronomers measure minute changes in period (~picoseconds per year)?
I've been to some talks that mention how stable the period of a millisecond pulsar is over long periods of time. Recently, it was mentioned that astronomers have calculated the change in period over time to be less than 10^-12 seconds per year for several pulsars. No one I've talked to seems to know any details of this calculation. How do we calculate such small differences in period? How much data must be collected and what are the exposure times for imaging such rapid phenomena? A source/paper would be excellent. I apologize that I don't have a citation for the 10^-12s figure, but the lack of citation is mostly my reason for posting this question.
• I'm not sure of the details of the calculation, either, and would be interested to see some details. I know the basic idea: the pulsar is emitting a lot of electromagnetic radiation, and that constitutes an energy loss. The energy has to have come from something. If it doesn't have an accretion disk or the intense magnetic field of a magnetar, then the most likely source is the conversion of angular momentum. As its angular momentum is lost, the rotation speed goes down. I don't know the conversion mechanism or calculations right now, though. Commented Jun 15, 2015 at 23:25
Let us suppose that the pulsar is spinning down at a uniform rate. So it has a period $P$ and a rate of change of period $dP/dt$ that is positive and constant (in practice there are also second, third, fourth etc. derivatives to worry about, but this doesn't change the principle of my answer).
Now let's assume you can measure the period very accurately - say you look at the pulsar today and measure its radio signals for a few hours, do a Fourier transform of the signal and get a nice big peak with a period of 0.1 seconds (for example).
With that period, you can "fold" the data to create an average pulse profile. This pulse profile can then be cross-correlated with subsequent measurements of the pulse to determine an offset between the predicted time of "phase zero" in the profile, calculated using the 0.1 s period, and the actual time of phase zero. This is often called an "O-C" curve or a residuals curve.
If you have the correct period and $dP/dt=0$, then the residuals will scatter randomly around zero with no trend as you perform later and later observations (see plot (a) from Lorimer & Kramer 2005, The Handbook of Pulsar Astronomy). If the initial period was in error, then the residuals would immediately begin to depart from zero on a linear trend.
If however, you have the period correct, but $dP/dt$ is positive, then the residuals curve will be in the form of a parabola (see plot (b)).
If you have second, third etc. derivatives in the period, then this will affect the shape of the residuals curve correspondingly.
The residuals curve is modelled to estimate the size of the derivatives of $P$. The reason that $dP/dt$ can be measured so precisely is that pulsars spin fast and have repeatable pulse shapes, so changes in the phase of the pulse quickly become apparent and can be tracked over many years.
Mathematically it works something like this. The phase $\phi(t)$ is given by $$\phi(t) \simeq \phi_0 + 2\pi \frac{\Delta t}{nP} - \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$ where $\phi_0$ is an arbitrary phase zero, $\Delta t$ is the time between the first and last observation and $n$ is the integer number of full turns the pulsar has made during that time. If the period is approximately correct, then $n = int(\Delta t/P)$.
The "residual curve" would be given by $$\phi_0 - 2\pi\frac{\Delta t}{nP} -\phi(t) \simeq \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$
For example, if the period of a $P \sim 0.01$ second pulsar changed by a picosecond in a year, then there would be an accumulated residual of almost $10^{-4}$ seconds after 1 year of observation. Depending on how "sharp" the pulse is, then this shift of about 1% in the phase of the pulse might be detectable.
Perhaps needless to say, but there are a host of small effects and corrections to make in order to get this very high precision timing. You need to know exactly how the Earth is moving in its orbit. The proper motion of the pulsar on the sky also has an effect. These and more can be found in Lorimer and Kramer's book, but there is also a summary here.
My comment notwithstanding, here's a solution to a homework problem that does the calculation. It doesn't specify the exact mechanism that converts angular momentum (aka rotational energy) into electromagnetic radiation. In this case, it is just an assertion of the problem (partially justified with what I said in my comment: the energy must come from somewhere, and if there doesn't seem to be any sources other than angular momentum, then it must be coming from the angular momentum).
Slightly rephrasing that link's content for the sake of future accessibility:
The pulsar is radiating energy (which we observe as radio waves). Since the total energy in the universe must be conserved, this radio energy must come from somewhere. In this case, it is taken out of the rotational kinetic energy of the pulsar: thus, it gradually slows down. We're interested in a relation between the pulsar luminosity and its rotational period. In general, the kinetic energy of a rotating body is given by $$E=\frac{1}{2} I \omega^2 = 2\pi^2 I P^2.$$ Since Luminosity is the time derivative of energy, we are now in a position to relate the quantities we are interested in: $$L= \frac{\partial E}{\partial t} = -4\pi^2 I P^{-3} \frac{\partial P}{\partial t}.$$ Rearranging this in terms of the quantity we want – the rate of change of the period – gives: $$\frac{\partial P}{\partial t} = \frac{-L P^3}{4\pi^2 I}.$$ If we assume that this neutron star is a homogeneous sphere (not really true, but a simple approximation), then its moment of inertia is just: $$I_{\text{sphere}}=\frac{2}{5}M R^2,$$ and so the final rate of change of period we get is: $$\frac{\partial P}{\partial t} = -\frac{5}{8\pi ^2} \frac{L P^3}{MR^2}.$$
So as long as we have measurements of the quantities on the right-hand side, we can just plug them in to get a value for the rate of change in the period. The hardest to measure is usually the moment of inertia (where the mass, $M$, and radius, $R$, terms come from). These are easier to get from eclipsing binary systems, since then there are nice relations between their orbital paths and masses.
• Going by this and this, the mechanism comes for a magnetic dipole radiation. Which, if I understand correctly, means that the magnetic field accelerates surface (or nearby) electrons, and applies torque to them. Acceleration means they must radiate away energy, torque means angular momentum is transferred. It's hard to get an explanation that really spells it out in full, it seems. Commented Jun 16, 2015 at 4:34
• But we don't have the quantities on the RHS. The mass, radius and moment of intertia are essentially unknowns. The pulsar "luminosity" is also rather hard to estimate in the context of this calculation. Commented Jun 16, 2015 at 9:12
• @RobJeffries Thanks for pointing that out. I was fairly sure that most of those quantities were going to give a lot of compounding sources of measurement problems, and so there had to be another way around the issue. Glad to see you posted about such a way. Commented Jun 16, 2015 at 10:08 | 1,828 | 7,503 | {"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} | 3.015625 | 3 | CC-MAIN-2024-30 | latest | en | 0.962389 |
http://mathhelpforum.com/discrete-math/87538-simple-graphs-paths-print.html | 1,529,494,864,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00434.warc.gz | 207,214,447 | 3,040 | Simple Graphs and Paths
• May 4th 2009, 10:50 PM
pupnat
Simple Graphs and Paths
This might seem easy, but im having trouble putting my proofs into words.
Let G = (V,E) be simple graph,
I need to prove for all vertices (x,y,z) of G, the follwoing are true:
1) There is a path from x to x.
2) If there is a path from x to y then there is path y -x.
3)If there is a path from x- y, and y-z then there is path x-z.
Now im not sure whats expected when proofing this.
Do i just say a path from x to x exists since its a path to itself?
and...
if a path x-y exists then there is a path y- x because x and y are two connected vertices by a edge?
and..
if there is a path from x-y, and y-z then there is a path x-z since vertices x and y are connected by an edge, and y-z is connected by an edge, this provides a path from vertices x to z through vertices y.
Id like some feedback if possible thanks.
• May 9th 2009, 09:10 AM
TiRune
Just work from the definitions and find a path that satisfies the conditions.
1: You can take the vertice sequence {x} which starts and stops at x and exists by axiom.
2: if F: (a,b,...,z) is the path from x to y, then G: (z,...,b,a) is the path from y to x, which exists because every edge q in G exists by assumption since it exists in F.
3: if F: (a,b,...,m), G: (n,o...,z) are the respective paths from x to y and from y to z, then the path H:(a,b,...,m,n,o,...,z) is the path from x to z, and all edges exist because they existed in F and G.
I couldn't be arsed with indices of the vertices, which you should use if you do this for homework, but using the alphabet is clearer.
General rule of thumb when solving these things: don't think too deeply and just construct the answers straight from the definition and question. | 488 | 1,761 | {"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-2018-26 | latest | en | 0.948786 |
https://douglasyao.github.io/notes/stats/2020/11/05/Johnstone-and-Paul-2018-Proceedings-of-the-IEEE-PCA-in-High-Dimensions-An-orientation.html | 1,713,651,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00482.warc.gz | 203,792,584 | 4,666 | # Johnstone and Paul 2018 Proceedings of the IEEE - PCA in High Dimensions - An orientation
Authors: Iain M. Johnstone and Debashis Paul
## Synopsis
This paper provides a review of the properties of eigenvalues and eigenvectors of the random covariance matrices in which the data is “high dimensional” (or the number of features is comparable to the number of samples). It turns out that the eigenvalues of such matrices exhibit some really interesting properties. This information is pertinent for answering questions like: “which eigenvectors/eigenvalues of a matrix are significant?” or “what proportion of variance in a matrix can be explained by low-rank signal vs. independent noise?”
## Summary
### Background
When performing PCA, we start with a $$p \times n$$ matrix $$X$$, where $$n$$ is the number of samples and $$p$$ is the number of features. We then find the eigenvalues and eigenvectors of the $$p \times p$$ sample covariance matrix $$\frac{1}{n} X X^T$$. Although PCA is simply a procedure applied to a matrix of fixed values, it can also be conceptualized in terms of random matrices, in which each sample is drawn from a multivariate distribution with some mean vector and covariance matrix.
As $$n$$ goes to infinity and $$p$$ is fixed, the sample covariance matrix will converge to the population covariance matrix (a.k.a $$E[XX^T]$$), and the sample eigenvalues will also converge to the population eigenvalues. However, what happens if $$n$$ and $$p$$ _both _go to infinity and the ratio $$\frac{p}{n}$$ stays constant? What will the eigenvalues of the sample covariance matrix look like? This is the boundary case in which the paper focuses on.
### Figure 2
This figure illustrates the phenomenon known as “eigenvalue spreading,” in which sample eigenvalues are more spread out than population eigenvalues. Both figures denote a scenario in which the matrix has $$p$$ = 100 features. On the left, all 100 eigenvalues of the population covariance matrix are equal to 1, while the eigenvalues of the sample covariance matrix (where $$n$$ = 200) are spread out. The same phenomenon is observed when the eigenvalues of the population covariance matrix are spread evenly between 5 and 25 (on the right).
### Figure 3
Given a random matrix (features mean centered and variance standardized, samples iid), when $$n$$ and $$p$$ go to infinity and their ratio stays constant, the distribution of eigenvalues of the sample covariance matrix will converge to a Marchenko-Pastur distribution (which depends only on the ratio between $$n$$ and $$p$$). This figures shows what this distribution looks like for two values $$\frac{n}{p} = 1$$ and $$\frac{n}{p} = 4$$.
### Figure 4
This figure describes the distribution of the largest eigenvalue in random covariance matrices. A very interesting fact is that even if the population covariance matrix has eigenvalues that are larger than 1, if these eigenvalues are smaller than some critical point indicated in the figure, then the distribution of sample eigenvalues will be indistinguishable from that arising from a population covariance matrix in which all eigenvalues are equal to 1.
First, the way we model this is through the “spiked covariance” model, which is a way to model noise on top of low-rank structure. The model involves a “base” covariance matrix $$\Sigma_0$$ and a low-rank perturbation:
$\Sigma = \Sigma_0 + \sum_k^K h_k \mathbf{u}_k \mathbf{u}_k^T$
In the simplest case, $$\Sigma_0$$ is the identity matrix, while $$h_k$$ and $$\mathbf{u}_k$$ are the eigenvalues and eigenvectors respectively of some low-rank structure in the covariance matrix. This model is called the “spiked” covariance model because it results in a population eigenvalue distribution with many points at 1 (corresponding to noise) and spikes at each $$h_k$$ (corresponding to signal). This can also be conceptualized in terms of the probabilistic PCA model $$\mathbf{x = Wz + e}$$, where $$\mathbf{x}$$ is our sample, $$\mathbf{W}$$ are all the eigenvectors of the sample covariance matrix, $$\mathbf{z}$$ are the projected points in PC space (assumed to be randomly drawn from $$N(\mathbf{0}, \mathbf{I})$$), and $$\mathbf{e}$$ is isotropic noise (drawn from $$N(\mathbf{0}, \mathbf{I})$$). The covariance matrix of $$\mathbf{x}$$ is defined as $$\mathbf{W W}^T + \mathbf{I}$$.
Continuing on: let $$l_1$$ represent an eigenvalue of our matrix greater than 1, with the remaining eigenvalues equal to 1. If $$l_1$$ is smaller than $$1 + \sqrt{\gamma}$$ (where $$\gamma$$ is $$\frac{p}{n}$$), then the distribution of eigenvalues will be exactly the same as if $$l_1 = 1$$ (a.k.a. it will look as if the matrix has no low-rank structure at all. This fact is pretty cool). Moreover, the corresponding eigenvector for $$l_1$$ will be completely uncorrelated with the true eigenvector. The top eigenvalue will be distributed according to a Tracy-Widom distribution around the value $$(1 + \sqrt{\gamma})^2$$.
### Figure 5
This figure describes the scenario in which the population covariance matrix has eigenvalues that are larger than the critical point. If $$l_1$$ is greater than $$1 + \sqrt{\gamma}$$, then the top eigenvalue will now be distributed according to a Gaussian distribution while being significantly upwardly biased relative to $$l_1$$ (interesting!). Meanwhile, the corresponding eigenvector will now be correlated with the true eigenvector, where the degree of correlation will depend on $$l_1$$ and $$\gamma$$:
### Figure 6
This figure shows the relationship between upward bias of $$\lambda(l_1)$$ (a.k.a. the sample value of $$l_1$$) relative to its population value. The upward bias will converge to $$\gamma$$ as $$l_1$$ increases. More precisely, we have the following relationship between $$\lambda(l_1)$$ and $$l_1$$:
$\lambda(l_1) - l_1 = \gamma \frac{l_1}{l_1 - 1}$
*****
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January 22, 2012, 12:30
#1
Senior Member
Mostafa Mahmoudi
Join Date: Jan 2012
Posts: 319
Rep Power: 8
hi,
I have a big problem.
I have a cubic that have three prependicular solid pipes in it and them are related to the suroundings that have the temperature of 320K except pipes in inlet and outlet. I create the mesh file in Gambit.
due to pipes temerature my fluid heated. fluid comes in at 0.1m/s and 300K and exit from opposite face.
I want to find the temperature of end of pipes but in exit face I don't know what BC shoud I make.
fluent automatically seperate the pipes into a wall and four circle that made from intersecting by the cubic face. I dont know what BC should I make.
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January 22, 2012, 13:15
#2
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Sijal
Join Date: Mar 2009
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Quote:
fluent automatically seperate the pipes into a wall and four circle that made from intersecting by the cubic face. I dont know what BC should I make.
What type of boundary conditions you want to specify? and where?
January 23, 2012, 04:12
#3
Senior Member
Mostafa Mahmoudi
Join Date: Jan 2012
Posts: 319
Rep Power: 8
Quote:
Originally Posted by Far What type of boundary conditions you want to specify? and where?
hi,
in enterance T=300K and velocity is 0.1m/s
in walls we have T=320K that cause the pipes heated
Exit face is just an outlet that the ambient temperature is 300K.
problem is in the automatically seperation of walls. Fluent consider the intersection of pipe and surroundings as a wall!!!
what can I do?
January 23, 2012, 14:21
#4
Member
Join Date: Jun 2011
Posts: 83
Rep Power: 8
Quote:
Originally Posted by adambarfi hi, in enterance T=300K and velocity is 0.1m/s in walls we have T=320K that cause the pipes heated Exit face is just an outlet that the ambient temperature is 300K. problem is in the automatically seperation of walls. Fluent consider the intersection of pipe and surroundings as a wall!!! what can I do?
Please check your mesh in Gambit. You have the wall imposed on the outlet of the pipe.
If that is the case you may have to subtract the volume of the pipes from the volume of the container in which you have the pipes.
January 23, 2012, 17:50 #5 Super Moderator Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,358 Blog Entries: 6 Rep Power: 45 Check that all volumes have mesh
January 26, 2012, 03:00 #6 Senior Member Mostafa Mahmoudi Join Date: Jan 2012 Posts: 319 Rep Power: 8 hi, I cheched it agian, that was allright. I could solve the problem, I assume that at inlet and out let we have first fluid that Enter to the body and exit from it. this was responsed. but still I have a problem I want my pipes that related to the walls have the same temperature with walls. my surrounding walls the temperature of 320K that due to the fluid flow it changes. I want the intersection of pipes and walls that fluent assume it as a wall have tempereture same the walls. here is anyone can help me? thanks
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# 5. In Fig. 6.17, is a line. Ray is perpendicular to line . OS is another ray lying between rays and . Prove that 6. It is given that and is produced to point P. Draw a figure from the given information. If ray YQ bisects , find Fig. 6.17 and reflex .
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Question Text 5. In Fig. 6.17, is a line. Ray is perpendicular to line . OS is another ray lying between rays and . Prove that 6. It is given that and is produced to point P. Draw a figure from the given information. If ray YQ bisects , find Fig. 6.17 and reflex . Updated On Feb 3, 2023 Topic All topics Subject Mathematics Class Class 9 Answer Type Video solution: 1 Upvotes 59 Avg. Video Duration 10 min | 355 | 1,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.866782 |
https://activelyshare.com/cost-behavior/ | 1,657,010,406,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00477.warc.gz | 126,722,739 | 17,777 | # Recognizing Cost Behavior, Its Nature And How To Calculate It
Cost behavior is part of cost accounting. Maybe you still don’t understand in detail the characteristics of cost behavior and even are confused about what the difference between expenses and costs is.
Therefore, the purpose of this cost behavior is useful for analyzing these costing activities, so that later it does not have a negative impact on company operations.
Of course, as a businessman, you will find out how the cost behavior changes in the company, especially for manufacturing. So this influence becomes important for making decisions, estimating future costs, and evaluating.
Cost behavior is a description of the company’s activities, where the activity shows an up and down activity in company’s operations. So when the costs incurred are too much than the activity, this will be a loss for the company in the future.
Cost behavior is a change in the volume of activity which is divided into three namely fixed costs, variable costs and semi-variable costs.
## Understanding the Definition of Cost Behavior?
Basically, cost behavior is a relationship between total and changes in the volume of company activity. Usually this behavior has 3 types of changes in the volume of cost activities, such as fixed costs, variable costs, and semi-variable costs.
In planning and controlling these 3 types of cost behavior, fixed costs and variable costs are divided into two groups, namely:
1. Fixed costs are divided into two costs, namely Committed Fixed Costs and Discretionary Fixed Costs.
2. Variable costs are divided into two, namely Engineered Variable Costs and Discretionary Variable Costs.
## What are the Types of Cost Behavior?
As in the previous discussion, the types of cost behavior are divided into 3 types, namely:
### 1. Fixed Cost Behavior (Fixed Cost)
Fixed costs are costs that have a fixed total amount even though there are changes in the volume of certain business activities. However, the fixed costs per unit will change due to changes in the volume of activity.
Usually the size of fixed costs will be influenced by cost behavior objectives that affect the company’s long term, technology, management strategies and methods. Therefore, these fixed costs are also divided into two types, namely:
a. Committed Fixed Costs
Committed fixed costs represent most of the fixed costs arising from ownership of plant, basic organization and equipment.
In the cost behavior of committed fixed costs, all fixed costs will be incurred and cannot be reduced, in order to maintain the company’s ability to meet cost behavior objectives on a long-term basis. For example the cost of depreciation, property taxes, rent, salaries, and insurance.
b. Discretionary Fixed Costs
Discretionary fixed costs are also usually called managed or programmed costs, namely costs that arise from decisions to provide budgets on a regular basis or usually annually. Where directly reflects the policy of top management, which relates to the maximum amount of fees.
This cost behavior can describe the optimum relationship either input or output costs, but can be measured through the volume of sales, products, or services. Examples of discretionary fixed costs include research and development costs, consultant fees, advertising costs, employee training program costs, sales promotion costs.
### 2. Variable Costing Behavior (Variable Costing)
The term variable costing or variable cost is a cost whose total amount can change, but is proportional to changes in the volume of company activities, such as the cost of raw materials. Thus, the types of variable costs are also divided into 2, namely:
a. Engineered Variable Costs
Engineered cost variable is a cost associated with a specific physical relationship to a measure of activity. Usually all variable costs are engineered costs, so these costs change according to input or output, for example the use of raw material costs.
b. Discretionary Variable Costs
The previous explanation has said that all variable costs are engineered costs, but some of these variable costs can also be said to be discretionary variable costs.
The existence of discretionary variable costs depends on the company’s management decisions, so that the policy between input and output costs has a close relationship. When the output costs change, the input costs also change. For example, the cost of advertising.
### 3. Semi Variable Cost
In semi-variable costs are costs that have both fixed and variable elements, where semi-variable costs in fixed costs can be said to be the minimum amount of costs to provide services. In addition, variable costs are influenced by changes in the volume of activity.
## How to Define the Characteristics of a Cost Behavior Pattern?
There are 3 factors that influence in determining the nature of cost behavior and must be considered, as follows:
1. Choose a cost and investigate its behavior pattern, so that this cost is said to be a dependent variable and is represented by the symbol y.
2. Choosing the nature of the independent variable, can cause these costs to fluctuate. So the calculation function is stated as y = f(x).
3. Choose the relevant activities, where the independent and dependent variable costs are expressed in the applicable cost function where the cost formula is y = a + bx.
## What is the Cost Behavior Method and How is It Calculated?
In the cost method, it can be seen that there are two approaches to cost behavior, namely:
1. Analytical approach
2. Historical approach
Following are the functions of the 3 historical approach methods, namely:
### a. High and Low Point Method
In estimating the cost function, analyze these costs by means of the highest level of activity. Where this way compares the lowest activity level in the past. For this reason, this cost difference will be included in the variable cost.
• Sample Cost Behavior Problem – High and Low Point Method
The activities and costs for the repair and maintenance of ABCDEF,LLC in 2020 are presented in the following table data:
Month Repair And Maintenance Cost Machine Clock 1 750.000 6.000 2 715.000 5.500 3 800.000 4.250 4 600.000 4.500 5 600.000 4.500 6 550.000 7.000 7 600.000 6.000 8 875.000 4.500 9 800.000 6.000 10 750.000 6.000 11 530.000 4.000 12 1.000.000 8.000 Amount 8.570.000 66.250
So it can be seen that activity in November 2020 is the lowest level of activity, but in December 2020 activity is the highest.
In the number of machine hours in the repair and maintenance costs, also have these two levels of activity. By comparing and calculating the difference:
Information Highest Lowest Difference Number of Machine Hours 8.000 4.000 4.000 Repair and Maintenance Cost \$1.000.000 \$530.000 \$470.000
### b. Guard Fee Method
This method uses the calculation of costs that must still be incurred, even though the company is temporarily closed, the product will be equal to zero.
• Example of Cost Behavior Problem – Cost Precautionary Method
For example, for the repair and maintenance rate of 8,000 machine hours per month, the costs incurred are \$ 1,000,000. However, in the calculation, if the company does not produce, the repair costs will be \$ 470,000 per month.
In determining variable and fixed costs can be calculated as follows:
Variable Costs = Repair and Maintenance Costs – Fixed Costs (Maintenance Costs)
Variable Cost = \$ 1,000,000 – \$ 470,000
Variable Cost = \$ 530,000
Variable Cost per hour = \$ 530,000 / 8000 hours = \$ 66 per machine hour
### c. Least Square Method
In this method, it is related through the cost to the volume of activity, where it is in the form of a straight line with the regression line equation y = a + bx. So that at y is the dependent variable whose changes are determined by changes in the variable x, namely the independent variable.
The estimation of the example of the cost behavior can be done by using the calculation of the estimated interval number. Where this calculation estimates the actual cost.
This is a complete explanation of how you can understand cost behavior. In understanding these costs, a businessman or accountant can estimate which potential changes in costs in the future.
Therefore, it is a good step in running your business, as well as having difficulty calculating business expenses or even if you cannot control finances and bookkeeping.
Trending in Accounting | 1,794 | 8,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-27 | latest | en | 0.949119 |
https://www.geeksforgeeks.org/sum-of-bitwise-and-of-sum-of-pairs-and-their-bitwise-and-from-a-given-array/?ref=rp | 1,674,780,942,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00490.warc.gz | 804,406,397 | 29,926 | # Sum of Bitwise AND of sum of pairs and their Bitwise AND from a given array
• Last Updated : 19 Jan, 2023
Given an array arr[] consisting of N integers, the task is to find the sum of Bitwise AND of (arr[i] + arr[j]) and Bitwise AND of arr[i] and arr[j] for each pair of elements (arr[i], arr[j]) from the given array. Since the sum can be very large, print it modulo (109 + 7).
Examples:
Input: arr[] = {8, 9}
Output: 0
Explanation: The only pair from the array is (8, 9). Sum of the pair = (8 + 9) = 17. Bitwise AND of the pairs = (8 & 9) = 8. Therefore required Bitwise AND = (17 & 8) = 0.
Input: arr[] = {1, 3, 3}
Output: 2
Explanation:
Pair (1, 3): Required Bitwise AND = (1 + 3) & (1 & 3) = (4 & 1) = 0.
Pair (3, 3): Required Bitwise AND = (3 + 3) & (3 & 3) = (6 & 3) = 2.
Therefore, total sum = 0 + 0 + 2 = 2.
Naive Approach: The simplest approach is to generate all possible pairs from the given array and check if there exist any such pair that satisfy the given condition or not. If found to be true, then count this pair. After checking for all the pairs, print the resultant count
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
• Considering the pair (X, Y), the Bitwise AND at a bit, say bth is 1, then (X + Y) must also have the bth bit as a set bit to contribute at the bth bit.
• Considering only the last b bits of both the numbers of such pairs, it can be observed that to satisfy the above condition, bth bit and (b + 1)th bit, both will have to be set.
• Therefore, from the above steps, it can be derived that to satisfy the above conditions, the following condition should satisfy:
• Therefore, the task reduces to find the count of pairs satisfying the above condition for each bit position and increment the result by the value cnt*2b for 0 ≤ b < 31.
Follow the steps below to solve the problem:
• Initialize a variable, say ans, to store the resultant sum.
• Iterate over the range [0, 30] and perform the following operations:
• Initialize a vector V that stores the pairs satisfying the above conditions.
• Traverse the given array arr[] and insert the value arr[j] – (arr[j] >> (i + 1))*(1 << (i + 1)) modulo M in the vector V, if ((arr[j] >> (i + 1)) &1) is true.
• Sort the vector V in ascending order.
• Traverse the vector V and perform the following:
• Calculate the value 2(i + 1) + 2i – V[j] and store it in a variable say Y.
• Find count of points in V which is greater than or equal to Y in V and store it in a variable say cnt i.e., cnt = V.size() – (lower_bound(V.begin() + j + 1, V.end(), Y) – V.begin()).
• Update the ans as ans = (ans+ cnt* 2i)%M.
• After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
## C++14
`// C++ program for the above approach` `#include ``#define M 1000000007``using` `namespace` `std;` `// Function to find the sum of Bitwise AND``// of sum of pairs and their Bitwise AND``// from a given array``void` `findSum(``int` `A[], ``int` `N)``{`` ``// Stores the total sum`` ``long` `long` `ans = 0;` ` ``for` `(``int` `i = 0; i < 30; i++) {` ` ``vector<``long` `long``> vec;` ` ``for` `(``int` `j = 0; j < N; j++) {` ` ``// Check if jth bit is set`` ``if` `((A[j] >> i) & 1) {` ` ``// Stores the right shifted`` ``// element by(i+1)`` ``long` `long` `X = (A[j] >> (i + 1));` ` ``// Update the value`` ``// of X`` ``X = X * (1 << (i + 1));`` ``X %= M;` ` ``// Push in vector vec`` ``vec.push_back(A[j] - X);`` ``}`` ``}`` ``// Sort the vector in`` ``// ascending order`` ``sort(vec.begin(), vec.end());` ` ``// Traverse the vector vec`` ``for` `(``int` `j = 0; j < vec.size(); j++) {` ` ``// Stores the value`` ``// 2^(i+1)- 2^(i)- vec[j]`` ``int` `Y = (1 << (i + 1))`` ``+ (1 << i)`` ``- vec[j];` ` ``// Stores count of numbers`` ``// whose value > Y`` ``int` `idx = lower_bound(`` ``vec.begin() + j + 1,`` ``vec.end(), Y)`` ``- vec.begin();` ` ``// Update the ans`` ``ans += (vec.size() - idx)`` ``* 1ll * (1 << i);` ` ``ans %= M;`` ``}`` ``}` ` ``// Return the ans`` ``cout << ans % M << endl;``}` `// Driver Code``int` `main()``{`` ``int` `arr[] = { 1, 3, 3 };`` ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);`` ``findSum(arr, N);` ` ``return` `0;``}`
## Java
`import` `java.util.ArrayList;``import` `java.util.Collections;` `class` `Main {`` ``public` `static` `final` `long` `M = ``1000000007``;` ` ``public` `static` `void` `findSum(``int``[] A, ``int` `N)`` ``{` ` ``// Stores the total sum`` ``long` `ans = ``0``;` ` ``for` `(``int` `i = ``0``; i < ``32``; i++) {`` ``ArrayList vec = ``new` `ArrayList<>();` ` ``for` `(``int` `j = ``0``; j < N; j++)`` ``{` ` ``// Check if jth bit is set`` ``if` `((A[j] >> i & ``1``) == ``1``)`` ``{` ` ``// Stores the right shifted element by(i+1)`` ``long` `X = (A[j] >> (i + ``1``)) * (``1` `<< (i + ``1``)) % M;` ` ``// Push in vector vec`` ``vec.add(A[j] - X);`` ``}`` ``}` ` ``// Sort the vector in ascending order`` ``Collections.sort(vec);` ` ``// Traverse the vector vec`` ``for` `(``int` `j = ``0``; j < vec.size(); j++)`` ``{` ` ``// Stores the value 2^(i+1)- 2^(i)- vec[j]`` ``long` `Y = (``1` `<< (i + ``1``)) + (``1` `<< i) - vec.get(j);` ` ``// Stores count of numbers whose value > Y`` ``long` `idx = Collections.binarySearch(vec.subList(j + ``1``, vec.size()), (``long``) Y);` ` ``// Update the ans`` ``ans += (vec.size() - (j + ``1``) - idx) * 1L * (``1` `<< i)-``2``;`` ``ans %= M;`` ``}`` ``}` ` ``// Return the ans`` ``System.out.println(ans % M -``1``);`` ``}` ` ``public` `static` `void` `main(String[] args) {`` ``int``[] arr = {``1``, ``3``, ``3``};`` ``int` `N = arr.length;`` ``findSum(arr, N);`` ``}``}` `// This code is contributed by aadityaburujwale.`
## Python3
`# Python 3 program for the above approach``M ``=` `1000000007``from` `bisect ``import` `bisect_left` `# Function to find the sum of Bitwise AND``# of sum of pairs and their Bitwise AND``# from a given array``def` `findSum(A, N):`` ` ` ``# Stores the total sum`` ``ans ``=` `0` ` ``for` `i ``in` `range``(``30``):`` ``vec ``=` `[]` ` ``for` `j ``in` `range``(N):`` ``# Check if jth bit is set`` ``if` `((A[j] >> i) & ``1``):`` ``# Stores the right shifted`` ``# element by(i+1)`` ``X ``=` `(A[j] >> (i ``+` `1``))` ` ``# Update the value`` ``# of X`` ``X ``=` `X ``*` `(``1` `<< (i ``+` `1``))`` ``X ``%``=` `M` ` ``# Push in vector vec`` ``vec.append(A[j] ``-` `X)` ` ``# Sort the vector in`` ``# ascending order`` ``vec.sort(reverse``=``False``)` ` ``# Traverse the vector vec`` ``for` `j ``in` `range``(``len``(vec)):`` ` ` ``# Stores the value`` ``# 2^(i+1)- 2^(i)- vec[j]`` ``Y ``=` `(``1` `<< (i ``+` `1``)) ``+` `(``1` `<< i) ``-` `vec[j]` ` ``# Stores count of numbers`` ``# whose value > Y`` ``temp ``=` `vec[j``+``1``:]`` ``idx ``=` `int``(bisect_left(temp,Y))` ` ``# Update the ans`` ``ans ``+``=` `((``len``(vec) ``-` `idx) ``*` `(``1` `<< i))` ` ``ans ``%``=` `M`` ``ans ``/``=` `7` ` ``# Return the ans`` ``print``(``int``(ans ``%` `M))` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``arr ``=` `[``1``, ``3``, ``3``]`` ``N ``=` `len``(arr)`` ``findSum(arr, N)` ` ``# This code is contributed by SURENNDRA_GANGWAR.`
## Javascript
`// JavaScript program for the above approach``const M = 1000000007;` `// Function to find the sum of Bitwise AND``// of sum of pairs and their Bitwise AND``// from a given array``function` `findSum(A, N) {`` ``let ans = 0;` ` ``for` `(let i = 0; i < 30; i++) {`` ``const vec = [];` ` ``for` `(let j = 0; j < N; j++) {`` ``if` `((A[j] >> i) & 1) {`` ``let X = (A[j] >> (i + 1)) % M;`` ``X = X * (1 << (i + 1)) % M;`` ``vec.push(A[j] - X);`` ``}`` ``}` ` ``vec.sort((a, b) => a - b);` ` ``for` `(let j = 0; j < vec.length; j++) {`` ``const Y = (1 << (i + 1)) + (1 << i) - vec[j];`` ``const temp = vec.slice(j + 1);`` ``const idx = temp.findIndex(el => el >= Y);`` ``ans += (temp.length - idx) * (1 << i);`` ``ans %= M;`` ``}`` ``}` ` ``ans /= 7;`` ``console.log(parseInt(ans % M)+1);``}` `const arr = [1, 3, 3];``const N = arr.length;``findSum(arr, N);`
Output:
`2`
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
My Personal Notes arrow_drop_up | 3,281 | 9,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-06 | latest | en | 0.844818 |
www.beltbrake.com | 1,386,459,009,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163056101/warc/CC-MAIN-20131204131736-00045-ip-10-33-133-15.ec2.internal.warc.gz | 248,363,732 | 6,153 | ## TECHNICAL INFORMATION
1. Introduction
The Purpose of this document is to give a practical example of how to calculate the number of brake rollers that need to be installed on a specific configuration of conveyor belt. In order to successfully understand the terminology used in this document the following basic concepts regarding conveyor belt layouts and friction need to be understood.
1.1 Conveyor Belt Angle
The correct angle at which a conveyor belt operates is important in determining the correct number of BELT-BRAKETM to install. Both the height of the conveyor as well as the height difference between the head pulley and the tail pulley is required as seen in Figure 1 to determine the conveyor belt angle. The angle can be calculated from normal trigonometric relationships.
Sinθ = (H/L) therefore θ = Sin-1(H/L) (1)
1.2 Normal Force
A normal force is a force that acts perpendicularly to a surface. When the angle at which a surface is orientated is at a zero angle (θ=0°) the normal force is equal to the weight of the object that is lying on the surface. However as soon as the surface is raised at a certain angle the normal force is fractionally less depending on the angle at which the surface is inclined. The normal force can be calculated for the following four vector force combinations and layouts as seen in Figure 2.
The configuration that is applicable to the mass being distributed on a conveyor belt at a certain angle theta (θ) is
N = mgCosθ [N] (2)
Hence only a portion of the mass acts perpendicular to the surface of the conveyor belt and the other portion want to roll back down the incline.
1.3 Coefficient of Friction
Friction is the phenomena of the interlocking forces between the irregularities between two surfaces. These forces increase when trying to move the surfaces relative to each other until just prior to motion occurring when the forces are at there greatest. After there is relative motion the friction forces are found to be slightly less than the amount just prior to the relevant movement between the two surfaces.
The experimental Static coefficient of friction can be determined by placing a mass at an incline and determining the tangent of the threshold angle at which the mass starts sliding down the incline.
The experimental Kinetic coefficient of friction is determined by using the tangent of the minimum angle at which the mass will maintain a constant velocity down the incline when initially assisted to overcome the static friction threshold and set into motion. The Kinetic coefficient of friction is found to be constant for various velocities and is always less that the Static coefficient of friction.
It is important to note that coefficient of kinetic and static friction is a function of the two materials which are being used and hence there is not a coefficient of friction for say wood or for glass. There is however a unique coefficient between say tar and a tyre. It is also important to note that the surface finish plays a big role in the coefficient.
1.4 Weight distribution
When there is certain average mass distributed over a load-carrying member, which in this case is the conveyor belt we need to know what the weight distribution in N/m is. The symbol q is used for the weight distribution. Both the mass being transported per hour and the weight of the conveyor belt in kg/m needs to be used in order to accurately determine the weight distribution.
To give an example of how we calculate the weight distribution we use the following example.
Example 1:
1) Velocity of Conveyor Belt (V) = 2 m/s
2) Tonnage per hour (T) = 1000 tons per hour
3) Mass of belt per meter (mB) = 30 kg/m
The mass distribution is calculated as follows:
1 Ton = 1000 kg therefore 1000 ton = 1.0x106 kg (1000 000 kg)
Mass transported per second = (1.103)/3600 = 277.78 kg/s
Mass distribution = Mass per second/ belt velocity (V) + Belt mass (mB)
= 277.78/2 + 30kg/m = 169 kg/m>
Weight distribution = Mass Distribution* Gravitational Constant
Therefore q = 1657N/m (g = 9.81 m/s2)
Mass Distribution (q) =g((T.1000/3600.V) + MB)) [N/m] (3)
This value is needed to determine the normal and horizontal force components that react through the BELT-BRAKETM rollers and hence its structure.
1.5 Configuration Factor
A trough will typically be a three or a five roll configuration and therefore a different percentage of the mass that is distributed on a section of the conveyor will act on the idler or on the installed BELT-BRAKETM. On average the percentage of mass that acts on the centre idler of a three roll configuration is 64% and on a 5 roll configuration approximately 38 %. The remaining percentage on a three roll configuration is then 18% per wing idler.
Figure 4 illustrates the approximate factor for both the three and five roller through. Therefore we need to incorporate a factor which will be used to determine correctly the normal force that will act on the BELT-BRAKETM. We will call this factor the configuration factor CF. The configuration is designate the subscript 3 to the three roll and 5 to the 5 roll configuration.
Configuration factor: CF3 = 0.64; 0.18 or CF5 = 0.38; 0.20; 0.11 (4)
2. Calculating the number of BELT-BRAKETM rollers to be installed.
The number of BELT-BRAKETM rollers that needs to be installed can be calculated as follows.
Firstly the amount of weight that needs to be stopped will be formulated as a function of the angel of the conveyor. This amount will be multiplied by a safety factor. The amount of braking force that can be realised by the BELT-BRAKETM will then be formulated as a function of the angle of the conveyor, the friction coefficient and the Configuration Factor. These two can then be equated to determine how many BELT-BRAKETM rollers need to be installed for a specific configuration.
2.1 Total mass that needs to be stopped.
The amount of friction force that is required to stop the mass that is lying on the conveyor belt is the component of mass that want to roll back down the specific incline. In the same way that the component of the mass that is acting normal to the conveyor belt is given by m.Cosθ so the component that wants to roll back is given by m.Sinθ
This total mass that want to roll down the conveyor is given by the total length of the belt (L) multiplied by the mass distribution (q). The total mass is also multiplied by the safety factor (SF). This total quantity is called the force required to stop (FRS) and is given in Equation 5.
FRS =(SF)LqSinθ [N] (5)
2.2 Friction force that can be realised by a BELT-BRAKETM
To differentiate between the friction force that is required and the friction force that can be realised we allocate FFB to the friction force that a BELT-BRAKETM is capable of producing. We assume that the total weight that acts on a trough is equal to the weight distribution that lies between two troughs. This distance will be designated the symbol LR. Figure 5 indicates LR more clearly.
Therefore the friction force that can be realised per BELT-BRAKETM is the normal force m.Cosθ multiplied by the friction coefficient µ multiplied by the Configuration factor given Equation 4 (CF). The mass that acts on any roller is the mass distribution q multiplied by LR.
FFB = CF.µ.q.LR.cosθ [N] (6)
The total amount of friction force that is required is the total number of BELT-BRAKETM rollers that will be installed (NBR) multiplied by the amount that each can stop or the friction force that can be realised by each BELT-BRAKETM (FFB). Therefore the friction force required is given the designation FFR.
FFR = NBR FFB [N] (7)
All that is still required is to determine the number of rollers to install. This is done by noting that the friction force that is required FFR in Equation 7 should at least be greater or equal to the Force that we require to stop FRS in Equation 5.
FFR ≥ FRS (8)
Therefore
CF.NBRLR.µCosθ ≥ q.L.(SF).Sinθ
NBR ≥ L.Tanθ.SF / LR.µ.CF (9)
3. Symbols
Symbol Description µ [Dimensionless]> Static Friction Coefficient θ [Degrees] Angle of Conveyor belt N [Newton] Normal Force NBR [Dimensionless] Number BELT-BRAKETM Rollers L [m] Length of Conveyor Belt g [m.s-2] Gravitational Constant q [N.m-1] Weight Distribution V [m.s-1] Velocity NBR [Dimensionless] No of Brake Rollers to be Installed FFR [N] Stopping force realised by all BELT-BRAKETM installed FRS [N] Stopping Friction force required to stop FFB [N] Stopping Force realised by a BELT-BRAKETM LR [m] Distance between brake rollers T Kg/hour Tonnage per hour SF [Dimensionless] Safety Factor normally 2 LR [m] Distance between Idler Rollers mB [kg/m] Belt Mass
4. References
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https://www.efinancialmodels.com/downloads/monte-carlo-analysis-in-excel-without-macros-190803/ | 1,632,543,347,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00505.warc.gz | 764,558,589 | 59,436 | # Monte Carlo Analysis in Excel without Macros
This is downloadable for FREE! Traditional sensitivity analysis involves testing a limited number of scenarios (e.g. base, upside, and downside). The Monte Carlo method tests a large number (several hundred or thousands) of āscenariosā in which the inputs are drawn as random numbers.
Traditional sensitivity analysis involves testing a limited number of scenarios (e.g. base, upside, and downside). Each scenario is a set of predefined inputs. This approach shows the outcomes of the model from various perspectives but does not give a precise likelihood of a particular result to happen.
In contrast, the Monte Carlo method tests a large number (several hundreds or thousands) of āscenariosā in which the inputs are drawn as random numbers. The results of the model (gross profit, IRR, etc.) are also represented by ranges of numbers. Analyzing statistical patterns of those ranges the analyst can determine mathematically the chances of output being within a specific range or being higher or lower than a certain threshold.
In this publication, I am sharing a technique of Monte Carlo analysis in Excel. My approach is based on standard Excel functions and data tables without macros. The accompanying model performs essential Monte Carlo simulation: drawing random numbers of certain distribution types, making correlations, interpreting the outcomes.
You will get two downloadable files: 1 Full Excel Template and 1 PDF file with Instructions and Explanations.
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https://handcrafted-gold-jewelry.com/qa/what-change-you-will-observe-while-heating-ice.html | 1,604,121,129,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107912807.78/warc/CC-MAIN-20201031032847-20201031062847-00674.warc.gz | 319,992,666 | 9,461 | What Change You Will Observe While Heating Ice?
How do you keep ice from melting for 24 hours?
Wrap the icu cube in an insulating material that has been cooled below the freezing temperature.
This could be a thick cloth.
Then put this in something that has very low heat conduction such as a vacuuum bottle, a styrofoam container, or other insulated container.
Some will melt over 24 hours..
Which liquids melt the fastest when frozen?
On average, water melted in 145 minutes; sweet tea in 119 minutes; Coke in 118 minutes; PowerAde in 115 minutes and milk melted in 102 minutes. My results were that milk melted the fastest over-all out of all the liquids.
Does freezing gain or lose energy?
In freezing does the liquid lose or gain thermal energy and does the movement of particles increase or decrease? It loses thermal energy and the movement of particles decreases. … Thermal energy gains and the movement of particles increase.
Does heat increase mass?
2 Answers. All internal energy such as thermal, rotational, and internal potential energy contributes to the rest mass of an object. … So, yes, a hot object has greater rest mass and would weigh more when measured, if a scale were sensitive enough.
What happens when heat is taken away from ice?
If you remove heat from a solid system, that wiggling will reduce. So, basically, lowering the temperature of an ice cube will simply result in less movement at the molecular level. The practical limit of this process is absolute zero, -273.15 Kelvin, the lowest possible temperature.
How fast do ice cubes melt?
A standard 1 ounce cube (30 grams) will take 90 to 120 minutes to melt at the same temperature. The same 1oz (30g) ice cube submerged in a cup of hot water of 185° F (85° C) will take about 60-70 seconds to melt.
What causes the ice to melt?
Ice melts when heat energy causes the molecules to move faster, breaking the hydrogen bonds between molecules to form liquid water. In the melting process, the water molecules actually absorb energy.
How does temperature affect matter?
When heat is added to a substance, the molecules and atoms vibrate faster. As atoms vibrate faster, the space between atoms increases. The motion and spacing of the particles determines the state of matter of the substance. … Solids, liquids and gases all expand when heat is added.
Why does boiling require more energy than melting?
Because the heat of vaporization is higher than the heat of fusion. It takes more energy to break the attraction between liquid phase molecules into gas then it does between solid phase molecules into liquid.
When you heat a solid what happens?
In a solid the strong attractions between the particles hold them tightly packed together. Even though they are vibrating this is not enough to disrupt the structure. When a solid is heated the particles gain energy and start to vibrate faster and faster.
What is the process of changing water into ice called?
Water is a liquid and its solid form is known as ice. When the water converts from the liquid phase (that is, water) to the solid phase (that is, ice), the process is known as freezing. … Heat is lost to the surroundings as the water freezes. The ethalapy of fusion of water or heat of fusion of water is 6 kJ/mol.
Which ice cube will melt the fastest?
A spherical ice cube with the same volume has a surface area of approximately 43.5 square centimeters. Therefore, the cubical ice cube melts faster than either of the other two shapes since it has the greatest surface area.
Is energy added or removed in melting?
Adding Energy: When a solid is at its melting point, any energy added to it is used to overcome the attractions that hold the particles in place.
What kind of change happens when ice melts?
When you melt an ice cube (H2O), you have a physical change because you add energy. You added enough energy to create a phase change from solid to liquid. Physical actions, such as changing temperature or pressure, can cause physical changes. No chemical changes took place when you melted the ice.
What happens to ice as it absorbs energy?
The solid ice particles absorb heat energy from the warmer air, giving the particles energy and enabling them to move away from one another. … This is what happens when the ice cube (a solid) turns into water (a liquid).
What changes do you observe when you heat ice cubes initially?
If I continued to heat up an ice cube, it will melt into the liquid state of water. This phase change is called melting. … When the molecules are sped up even more, the phase change of evaporation (water taking the form of a gas) occurs. The molecules can be sped up by heat.
Why do you have to warm the water above room temperature before adding the ice?
Experiment #2 – Latent heat of fusion of ice. Energy is required to change water from a solid to a liquid, i.e. to melt ice. … The needed energy will come from a cup of warm water. The amount of water and its temperature will be measured before adding some ice and then again after the ice has been melted. | 1,069 | 5,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-45 | latest | en | 0.909872 |
www.easyreadsystem.com | 1,490,737,134,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189903.83/warc/CC-MAIN-20170322212949-00179-ip-10-233-31-227.ec2.internal.warc.gz | 518,207,570 | 113,264 | # Reading Disabilities Can Impact Math Too: Strategies for Adding and Subtracting Numbers
Many students that struggle with reading can also experience trouble with some mathematical concepts. In particular, difficulty with directionality can impact both reading and math as it can manifest as confusion when discriminating left from right, East and West, as well as letter, word and number reversals. There are numerous mathematical concepts which require directionality, and compensatory strategies can help these students navigate around this hurtle. Because number lines tend to be displayed horizontally from left to right, adding and subtracting numbers can be difficult to understand and execute for some students. So what can we do to help?
Compensatory Strategies for Number Lines
Instead of teaching a number line that spans from left to right, I like to use a ladder that travels up and down. Students that have difficulty with directionality find this much easier to conceptualize. Whenever you add, you go up the ladder. Whenever you subtract, you go down the ladder. What’s more, when introducing negative numbers, utilizing a ladder that travels both “above ground” and below ground,” as illustrated in the initial blog image, will continue to make adding and subtracting a breeze, as the same rule applies -adding results in traveling up the ladder and subtracting down the ladder.
Stair Hop: A Game for Adding and Subtracting Whole Numbers and Integers
If students need a concrete method for grasping and practicing adding and subtracting whole numbers and integers, I love to use a staircase. Label each stair using a sequence of numbers with masking or duct tape. Break the class into teams of two players. For practicing adding and subtracting negative and positive integers, have one member of the team stay at the bottom of the staircase with you and the other member stand on the middle step of the staircase. For this example, let’s say you have a staircase that has forty steps. Players start on the middle stair. The middle stair is labeled zero and all the steps above it are labeled with the counting numbers and the steps below are labeled with negative numbers. One at a time the players at the bottom of the stairs flip a coin. Heads means add, and tails means subtract. They will also roll a die which will indicate the number of steps to navigate. Say the first student flips a head and rolls a three. They would instruct their team member to add three, and therefore climb 3 stairs. Then the other players continue. Play ends when one of the players reaches the top of the stairs or the bottom of the stairs. If you would like to play with the counting numbers, label the stairs 1-40. As in the prior example, have the players begin on the middle step, in this case, the 20th step. You can also play this game in the classroom by creating a ladder like the image displayed at the top of this blog and giving each pair of students a coin, a die and two different colored paperclips. They can play against each other by starting at zero with their selected paperclip and taking turns to see who can win by being the first to the top or bottom of the ladder.
I hope you found this strategy helpful. I would love to hear your thoughts! Also, if you play the game, I would love to see some images!!
Dr. Erica Warren is a reading specialist, educational therapist and author of multisensory, and mindful educational materials. She resides in New York, where she works one on one with students as a “personal trainer for the brain” and an educational consultant/teacher trainer. Dr. Warren offers her own materials at Good Sensory Learning and Teachers Pay Teachers. You can also get free advice and resources by following her blog here.
## 2 thoughts on “Reading Disabilities Can Impact Math Too: Strategies for Adding and Subtracting Numbers”
1. Great idea for helping visual learners understand the difference between positive and negative integers. I can definitely see this being a beneficial study tool for students when manipulating the addition and subtraction of positive and negative integers.
Thanks for sharing!
Jennifer Smith-Sloane
4mulaFun
• Hi Jennifer
So glad you like it! Erica produces some fantastic materials over on her blog.
Sarah | 887 | 4,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-13 | longest | en | 0.901752 |
https://math.answers.com/other-math/What_is_1_fifth_plus_4_fifthteens | 1,726,161,261,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00653.warc.gz | 365,662,402 | 47,008 | 0
# What is 1 fifth plus 4 fifthteens?
Updated: 4/28/2022
Wiki User
11y ago
1/5 + 4/15 = 9/15. In simplest form, this would be either 3/5 as a fraction or 0.6 as a decimal number.
Wiki User
11y ago | 78 | 204 | {"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-2024-38 | latest | en | 0.911449 |
https://netlib.org/lapack/explore-html/d9/d6b/sdrvsy__aa_8f_aa35f165614d9237b27c591803135b1fb.html | 1,696,426,100,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00337.warc.gz | 453,357,598 | 10,450 | LAPACK 3.11.0 LAPACK: Linear Algebra PACKage
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## ◆ sdrvsy_aa()
subroutine sdrvsy_aa ( logical, dimension( * ) DOTYPE, integer NN, integer, dimension( * ) NVAL, integer NRHS, real THRESH, logical TSTERR, integer NMAX, real, dimension( * ) A, real, dimension( * ) AFAC, real, dimension( * ) AINV, real, dimension( * ) B, real, dimension( * ) X, real, dimension( * ) XACT, real, dimension( * ) WORK, real, dimension( * ) RWORK, integer, dimension( * ) IWORK, integer NOUT )
SDRVSY_AA
Purpose:
` SDRVSY_AA tests the driver routine SSYSV_AA.`
Parameters
[in] DOTYPE ``` DOTYPE is LOGICAL array, dimension (NTYPES) The matrix types to be used for testing. Matrices of type j (for 1 <= j <= NTYPES) are used for testing if DOTYPE(j) = .TRUE.; if DOTYPE(j) = .FALSE., then type j is not used.``` [in] NN ``` NN is INTEGER The number of values of N contained in the vector NVAL.``` [in] NVAL ``` NVAL is INTEGER array, dimension (NN) The values of the matrix dimension N.``` [in] NRHS ``` NRHS is INTEGER The number of right hand side vectors to be generated for each linear system.``` [in] THRESH ``` THRESH is REAL The threshold value for the test ratios. A result is included in the output file if RESULT >= THRESH. To have every test ratio printed, use THRESH = 0.``` [in] TSTERR ``` TSTERR is LOGICAL Flag that indicates whether error exits are to be tested.``` [in] NMAX ``` NMAX is INTEGER The maximum value permitted for N, used in dimensioning the work arrays.``` [out] A ` A is REAL array, dimension (NMAX*NMAX)` [out] AFAC ` AFAC is REAL array, dimension (NMAX*NMAX)` [out] AINV ` AINV is REAL array, dimension (NMAX*NMAX)` [out] B ` B is REAL array, dimension (NMAX*NRHS)` [out] X ` X is REAL array, dimension (NMAX*NRHS)` [out] XACT ` XACT is REAL array, dimension (NMAX*NRHS)` [out] WORK ` WORK is REAL array, dimension (NMAX*max(2,NRHS))` [out] RWORK ` RWORK is REAL array, dimension (NMAX+2*NRHS)` [out] IWORK ` IWORK is INTEGER array, dimension (2*NMAX)` [in] NOUT ``` NOUT is INTEGER The unit number for output.```
Definition at line 149 of file sdrvsy_aa.f.
152*
153* -- LAPACK test routine --
154* -- LAPACK is a software package provided by Univ. of Tennessee, --
155* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
156*
157* .. Scalar Arguments ..
158 LOGICAL TSTERR
159 INTEGER NMAX, NN, NOUT, NRHS
160 REAL THRESH
161* ..
162* .. Array Arguments ..
163 LOGICAL DOTYPE( * )
164 INTEGER IWORK( * ), NVAL( * )
165 REAL A( * ), AFAC( * ), AINV( * ), B( * ),
166 \$ RWORK( * ), WORK( * ), X( * ), XACT( * )
167* ..
168*
169* =====================================================================
170*
171* .. Parameters ..
172 REAL ONE, ZERO
173 parameter( one = 1.0e+0, zero = 0.0e+0 )
174 INTEGER NTYPES, NTESTS
175 parameter( ntypes = 10, ntests = 3 )
176 INTEGER NFACT
177 parameter( nfact = 2 )
178* ..
179* .. Local Scalars ..
180 LOGICAL ZEROT
181 CHARACTER DIST, FACT, TYPE, UPLO, XTYPE
182 CHARACTER*3 MATPATH, PATH
183 INTEGER I, I1, I2, IFACT, IMAT, IN, INFO, IOFF, IUPLO,
184 \$ IZERO, J, K, KL, KU, LDA, LWORK, MODE, N,
185 \$ NB, NBMIN, NERRS, NFAIL, NIMAT, NRUN, NT
186 REAL ANORM, CNDNUM
187* ..
188* .. Local Arrays ..
189 CHARACTER FACTS( NFACT ), UPLOS( 2 )
190 INTEGER ISEED( 4 ), ISEEDY( 4 )
191 REAL RESULT( NTESTS )
192* ..
193* .. External Functions ..
194 REAL DGET06, SLANSY
195 EXTERNAL dget06, slansy
196* ..
197* .. External Subroutines ..
198 EXTERNAL aladhd, alaerh, alasvm, serrvx, sget04, slacpy,
201* ..
202* .. Scalars in Common ..
203 LOGICAL LERR, OK
204 CHARACTER*32 SRNAMT
205 INTEGER INFOT, NUNIT
206* ..
207* .. Common blocks ..
208 COMMON / infoc / infot, nunit, ok, lerr
209 COMMON / srnamc / srnamt
210* ..
211* .. Intrinsic Functions ..
212 INTRINSIC max, min
213* ..
214* .. Data statements ..
215 DATA iseedy / 1988, 1989, 1990, 1991 /
216 DATA uplos / 'U', 'L' / , facts / 'F', 'N' /
217* ..
218* .. Executable Statements ..
219*
220* Initialize constants and the random number seed.
221*
222* Test path
223*
224 path( 1: 1 ) = 'Single precision'
225 path( 2: 3 ) = 'SA'
226*
227* Path to generate matrices
228*
229 matpath( 1: 1 ) = 'Single precision'
230 matpath( 2: 3 ) = 'SY'
231*
232 nrun = 0
233 nfail = 0
234 nerrs = 0
235 DO 10 i = 1, 4
236 iseed( i ) = iseedy( i )
237 10 CONTINUE
238*
239* Test the error exits
240*
241 IF( tsterr )
242 \$ CALL serrvx( path, nout )
243 infot = 0
244*
245* Set the block size and minimum block size for testing.
246*
247 nb = 1
248 nbmin = 2
249 CALL xlaenv( 1, nb )
250 CALL xlaenv( 2, nbmin )
251*
252* Do for each value of N in NVAL
253*
254 DO 180 in = 1, nn
255 n = nval( in )
256 lwork = max( 3*n-2, n*(1+nb) )
257 lwork = max( lwork, 1 )
258 lda = max( n, 1 )
259 xtype = 'N'
260 nimat = ntypes
261 IF( n.LE.0 )
262 \$ nimat = 1
263*
264 DO 170 imat = 1, nimat
265*
266* Do the tests only if DOTYPE( IMAT ) is true.
267*
268 IF( .NOT.dotype( imat ) )
269 \$ GO TO 170
270*
271* Skip types 3, 4, 5, or 6 if the matrix size is too small.
272*
273 zerot = imat.GE.3 .AND. imat.LE.6
274 IF( zerot .AND. n.LT.imat-2 )
275 \$ GO TO 170
276*
277* Do first for UPLO = 'U', then for UPLO = 'L'
278*
279 DO 160 iuplo = 1, 2
280 uplo = uplos( iuplo )
281*
282* Set up parameters with SLATB4 and generate a test matrix
283* with SLATMS.
284*
285 CALL slatb4( matpath, imat, n, n, TYPE, KL, KU, ANORM,
286 \$ MODE, CNDNUM, DIST )
287*
288 srnamt = 'SLATMS'
289 CALL slatms( n, n, dist, iseed, TYPE, RWORK, MODE,
290 \$ CNDNUM, ANORM, KL, KU, UPLO, A, LDA, WORK,
291 \$ INFO )
292*
293* Check error code from SLATMS.
294*
295 IF( info.NE.0 ) THEN
296 CALL alaerh( path, 'SLATMS', info, 0, uplo, n, n, -1,
297 \$ -1, -1, imat, nfail, nerrs, nout )
298 GO TO 160
299 END IF
300*
301* For types 3-6, zero one or more rows and columns of the
302* matrix to test that INFO is returned correctly.
303*
304 IF( zerot ) THEN
305 IF( imat.EQ.3 ) THEN
306 izero = 1
307 ELSE IF( imat.EQ.4 ) THEN
308 izero = n
309 ELSE
310 izero = n / 2 + 1
311 END IF
312*
313 IF( imat.LT.6 ) THEN
314*
315* Set row and column IZERO to zero.
316*
317 IF( iuplo.EQ.1 ) THEN
318 ioff = ( izero-1 )*lda
319 DO 20 i = 1, izero - 1
320 a( ioff+i ) = zero
321 20 CONTINUE
322 ioff = ioff + izero
323 DO 30 i = izero, n
324 a( ioff ) = zero
325 ioff = ioff + lda
326 30 CONTINUE
327 ELSE
328 ioff = izero
329 DO 40 i = 1, izero - 1
330 a( ioff ) = zero
331 ioff = ioff + lda
332 40 CONTINUE
333 ioff = ioff - izero
334 DO 50 i = izero, n
335 a( ioff+i ) = zero
336 50 CONTINUE
337 END IF
338 ELSE
339 ioff = 0
340 IF( iuplo.EQ.1 ) THEN
341*
342* Set the first IZERO rows and columns to zero.
343*
344 DO 70 j = 1, n
345 i2 = min( j, izero )
346 DO 60 i = 1, i2
347 a( ioff+i ) = zero
348 60 CONTINUE
349 ioff = ioff + lda
350 70 CONTINUE
351 izero = 1
352 ELSE
353*
354* Set the last IZERO rows and columns to zero.
355*
356 DO 90 j = 1, n
357 i1 = max( j, izero )
358 DO 80 i = i1, n
359 a( ioff+i ) = zero
360 80 CONTINUE
361 ioff = ioff + lda
362 90 CONTINUE
363 END IF
364 END IF
365 ELSE
366 izero = 0
367 END IF
368*
369 DO 150 ifact = 1, nfact
370*
371* Do first for FACT = 'F', then for other values.
372*
373 fact = facts( ifact )
374*
375* Form an exact solution and set the right hand side.
376*
377 srnamt = 'SLARHS'
378 CALL slarhs( matpath, xtype, uplo, ' ', n, n, kl, ku,
379 \$ nrhs, a, lda, xact, lda, b, lda, iseed,
380 \$ info )
381 xtype = 'C'
382*
383* --- Test SSYSV_AA ---
384*
385 IF( ifact.EQ.2 ) THEN
386 CALL slacpy( uplo, n, n, a, lda, afac, lda )
387 CALL slacpy( 'Full', n, nrhs, b, lda, x, lda )
388*
389* Factor the matrix and solve the system using SSYSV_AA.
390*
391 srnamt = 'SSYSV_AA'
392 CALL ssysv_aa( uplo, n, nrhs, afac, lda, iwork,
393 \$ x, lda, work, lwork, info )
394*
395* Adjust the expected value of INFO to account for
396* pivoting.
397*
398 IF( izero.GT.0 ) THEN
399 j = 1
400 k = izero
401 100 CONTINUE
402 IF( j.EQ.k ) THEN
403 k = iwork( j )
404 ELSE IF( iwork( j ).EQ.k ) THEN
405 k = j
406 END IF
407 IF( j.LT.k ) THEN
408 j = j + 1
409 GO TO 100
410 END IF
411 ELSE
412 k = 0
413 END IF
414*
415* Check error code from SSYSV_AA .
416*
417 IF( info.NE.k ) THEN
418 CALL alaerh( path, 'SSYSV_AA ', info, k,
419 \$ uplo, n, n, -1, -1, nrhs,
420 \$ imat, nfail, nerrs, nout )
421 GO TO 120
422 ELSE IF( info.NE.0 ) THEN
423 GO TO 120
424 END IF
425*
426* Reconstruct matrix from factors and compute
427* residual.
428*
429 CALL ssyt01_aa( uplo, n, a, lda, afac, lda,
430 \$ iwork, ainv, lda, rwork,
431 \$ result( 1 ) )
432*
433* Compute residual of the computed solution.
434*
435 CALL slacpy( 'Full', n, nrhs, b, lda, work, lda )
436 CALL spot02( uplo, n, nrhs, a, lda, x, lda, work,
437 \$ lda, rwork, result( 2 ) )
438 nt = 2
439*
440* Print information about the tests that did not pass
441* the threshold.
442*
443 DO 110 k = 1, nt
444 IF( result( k ).GE.thresh ) THEN
445 IF( nfail.EQ.0 .AND. nerrs.EQ.0 )
446 \$ CALL aladhd( nout, path )
447 WRITE( nout, fmt = 9999 )'SSYSV_AA ',
448 \$ uplo, n, imat, k, result( k )
449 nfail = nfail + 1
450 END IF
451 110 CONTINUE
452 nrun = nrun + nt
453 120 CONTINUE
454 END IF
455*
456 150 CONTINUE
457*
458 160 CONTINUE
459 170 CONTINUE
460 180 CONTINUE
461*
462* Print a summary of the results.
463*
464 CALL alasvm( path, nout, nfail, nrun, nerrs )
465*
466 9999 FORMAT( 1x, a, ', UPLO=''', a1, ''', N =', i5, ', type ', i2,
467 \$ ', test ', i2, ', ratio =', g12.5 )
468 RETURN
469*
470* End of SDRVSY_AA
471*
subroutine slaset(UPLO, M, N, ALPHA, BETA, A, LDA)
SLASET initializes the off-diagonal elements and the diagonal elements of a matrix to given values.
Definition: slaset.f:110
subroutine slacpy(UPLO, M, N, A, LDA, B, LDB)
SLACPY copies all or part of one two-dimensional array to another.
Definition: slacpy.f:103
subroutine alasvm(TYPE, NOUT, NFAIL, NRUN, NERRS)
ALASVM
Definition: alasvm.f:73
subroutine xlaenv(ISPEC, NVALUE)
XLAENV
Definition: xlaenv.f:81
subroutine alaerh(PATH, SUBNAM, INFO, INFOE, OPTS, M, N, KL, KU, N5, IMAT, NFAIL, NERRS, NOUT)
ALAERH
Definition: alaerh.f:147
double precision function dget06(RCOND, RCONDC)
DGET06
Definition: dget06.f:55
subroutine slatms(M, N, DIST, ISEED, SYM, D, MODE, COND, DMAX, KL, KU, PACK, A, LDA, WORK, INFO)
SLATMS
Definition: slatms.f:321
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 ssytrf_aa(UPLO, N, A, LDA, IPIV, WORK, LWORK, INFO)
SSYTRF_AA
Definition: ssytrf_aa.f:132
subroutine ssysv_aa(UPLO, N, NRHS, A, LDA, IPIV, B, LDB, WORK, LWORK, INFO)
SSYSV_AA computes the solution to system of linear equations A * X = B for SY matrices
Definition: ssysv_aa.f:162
subroutine slarhs(PATH, XTYPE, UPLO, TRANS, M, N, KL, KU, NRHS, A, LDA, X, LDX, B, LDB, ISEED, INFO)
SLARHS
Definition: slarhs.f:205
subroutine slatb4(PATH, IMAT, M, N, TYPE, KL, KU, ANORM, MODE, CNDNUM, DIST)
SLATB4
Definition: slatb4.f:120
subroutine serrvx(PATH, NUNIT)
SERRVX
Definition: serrvx.f:55
subroutine sget04(N, NRHS, X, LDX, XACT, LDXACT, RCOND, RESID)
SGET04
Definition: sget04.f:102
subroutine spot02(UPLO, N, NRHS, A, LDA, X, LDX, B, LDB, RWORK, RESID)
SPOT02
Definition: spot02.f:127
subroutine ssyt01_aa(UPLO, N, A, LDA, AFAC, LDAFAC, IPIV, C, LDC, RWORK, RESID)
SSYT01_AA
Definition: ssyt01_aa.f:124
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### sample cuda problems in finance
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Last edited by Cuchulainn on April 24th, 2015, 10:00 pm, edited 1 time in total.
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### sample cuda problems in finance
QuoteOriginally posted by: outrunparallel binomial tree works fine.Do you have the paper? Have you read it?BM prices at a _single_ S with a speedup of 285 is not spectacular. A real FDM can price a whole matrix of values in one go.
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Jean Piaget
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### sample cuda problems in finance
QuoteOriginally posted by: outrunQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: outrunparallel binomial tree works fine.Do you have the paper? Have you read it?BM prices at a _single_ S with a speedup of 285 is not spectacular. A real FDM can price a whole matrix of values in one go...a tree ..a grid: that's not the point... The point is that explicit is more efficient parallelizable. You can have cores work on various parts of the tree (or grid)Well, I take it then that you have _not_ read the article.There authors are talking about a speedup factor of 2! here They have conclusions why the speedup is so low.Even more worrying is they use floats!(??) The book by Sandors and Kandrot also use float. Is there no support for double.QuoteThe point is that explicit is more efficient parallelizable.Not necessarily. Example: FTCS(aka explicit Euler) scheme for heat PDE.
Last edited by Cuchulainn on April 25th, 2015, 10:00 pm, edited 1 time in total.
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### sample cuda problems in finance
QuoteWe learned from the experiments that the lack of explicit mechanism in CUDA for synchronising CPU and GPU executions is a major obstacle for the hybrid processing to achieve high performance.N. Zhang et al.Really?
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### sample cuda problems in finance
QuoteOriginally posted by: outrunparallel FFT is yet another option ..I think this is good case.QuoteThe Cooley-Tukey algorithm that FFTW uses to perform one-dimensional transforms works by breaking a problem of size N=N1N2 into N2 problems of size N1 and N1 problems of size N2. The way in which we parallelize this transform, then, is simply to divide these sub-problems equally among different threads. Ray tracing of course .. that's why GPU was incarnated.
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Jean Piaget
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### sample cuda problems in finance
Here is another approach Quote* Only very old GPUs don't have doubles, that was the first generation.So I can throw that CUDA book you recommended in the bin; it uses float.
Last edited by Cuchulainn on April 26th, 2015, 10:00 pm, edited 1 time in total.
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### sample cuda problems in finance
QuoteOriginally posted by: outrunQuoteOriginally posted by: CuchulainnHere is another approach Quote* Only very old GPUs don't have doubles, that was the first generation.So I can throw that CUDA book you recommended in the bin; it uses float.Yes, you can put it on a blanked outside your home if you can't come up with the obvious solution yourself. I've said it before: I'm an expert in spoon feeding a toddler Brussels sprouts -one of the most difficult tasks on earth-, and you are acting like one.Indeed; communication is not your strong point.
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### sample cuda problems in finance
Some food for thought: Stepanov's litmus test in a scary, heterogeneous world: does C++ still pass?Slides: http://accu.org/content/conf2015/AlexVo ... s_Test.pdf
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28. If a firm is a price taker, then the demand curve for the firm's product is: A. Equal to the total revenue curve B. Perfectly inelastic C. Perfectly elastic D. Unit elastic 29. In the short run, a purely competitive firm that seeks to maximize profit will produce: 30. Assume the price of a product sold by a purely competitive firm is \$5. Given the data in the accompanying table, at what output level is total profit highest in the short run? 31. Which is necessarily true for a purely competitive firm in short-run equilibrium?
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32. A purely competitive firm will be willing to produce even at a loss in the short run, as long as: A. The loss is smaller than its total variable costs B. The loss is smaller than its marginal costs C. The loss is smaller than its total fixed costs D. Price exceeds marginal costs 33. Refer to the diagram. At the profit-maximizing output, the firm will realize: 34. Answer the question on the basis of the following cost data for a firm that is selling in a purely competitive market: Refer to the data. If the market price for the firm's product is \$28, the competitive firm will:
35. Refer to the cost table above. If price of the product were \$30 per unit, the firm would: 36. Refer to the above graph. It shows short-run cost curves for a competitive firm. At what price would the firm break even? A. P1 B. P2 C. P3 37. The short-run supply curve for a competitive firm is the:
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38. The wage rate increases in a purely competitive industry. This change will result in a(n): 39. The following table applies to a purely competitive industry composed of 100 identical firms. Refer to the table. At the equilibrium price, each of the 100 firms in this industry will produce: 40. The short-run supply curve for a purely competitive industry can be found by: A. multiplying the AVC curve of the representative firm by the number of firms in the industry. B. adding horizontally the AVC curves of all firms. C. summing horizontally the segments of the MC curves lying above the AVC curve for all firms. D. adding horizontally the immediate market period supply curves of each firm.
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Open Journal of Applied Sciences
Vol.09 No.05(2019), Article ID:92741,6 pages
10.4236/ojapps.2019.95034
An NC Algorithm for Sorting Real Numbers in $O\left(\frac{nlogn}{\sqrt{loglogn}}\right)$ Operations
Yijie Han, Sneha Mishra, Md Usman Gani Syed
School of Computing and Engineering, University of Missouri at Kansas City, Kansas City, MO, USA
Received: April 8, 2019; Accepted: May 27, 2019; Published: May 30, 2019
ABSTRACT
We apply the recent important result of serial sorting of n real numbers in $O\left(n\sqrt{\mathrm{log}n}\right)$ time to the design of a parallel algorithm for sorting real numbers in $O\left({\mathrm{log}}^{1+\epsilon }n\right)$ time and $O\left(\frac{n\mathrm{log}n}{\sqrt{\mathrm{log}\mathrm{log}n}}\right)$ operations. This is the first NC algorithm known to take $o\left(n\mathrm{log}n\right)$ operations for sorting real numbers.
Keywords:
Parallel Algorithms, Sorting, Sort Real Numbers, Complexity
1. Introduction
It is known widely that serial comparison sorting takes $\theta \left(n\mathrm{log}n\right)$ time for sorting n numbers [1] . Although integer sorting can outperform the $\Omega \left(n\mathrm{log}n\right)$ lower bound for sorting n integers [2] [3], these algorithms generally do not apply to the problem of sorting real numbers. It has been known that n integers can be sorted in $O\left(n\mathrm{log}\mathrm{log}n\right)$ time and linear space [2] [3] . The $O\left(n\mathrm{log}n\right)$ time bound remains for sorting real numbers ever since. Only very recently Han showed that real numbers can be converted to integers for the sorting purpose in $O\left(n\sqrt{\mathrm{log}n}\right)$ time [4], thus enabling the serial sorting of real numbers in $O\left(n\sqrt{\mathrm{log}n}\right)$ time.
Parallel sorting algorithms for sorting real numbers run on the PRAM (Parallel Random Access Machine) model are known [5] [6] . The AKS sorting network [5] can be transformed into an EREW (Exclusive Read Exclusive Write) PRAM algorithm with $O\left(\mathrm{log}n\right)$ time and $O\left(n\mathrm{log}n\right)$ operations. Cole’s parallel merge sort [6] sorts n numbers in $O\left(\mathrm{log}n\right)$ time using n processors on the EREW PRAM. On the CRCW (Concurrent Read Concurrent Write) PRAM Cole showed [7] that his parallel merge sort can run in $O\left(\mathrm{log}n/\mathrm{log}\mathrm{log}\left(2p/n\right)\right)$ time using p processors. Also see [7] .
There are also parallel algorithms for integer sorting [8] [9] [10] [11] . In the case of integer sorting, the operation bound can be improved to below $O\left(n\mathrm{log}n\right)$. In particular, [10] presents a CRCW PRAM integer sorting algorithm with $O\left(\mathrm{log}n\right)$ time and $O\left(n\mathrm{log}\mathrm{log}n\right)$ operations and [11] presents an EREW PRAM integer sorting algorithm with $O\left(\mathrm{log}n\right)$ time and $O\left(n\sqrt{\mathrm{log}n}\right)$ operations.
For sorting real numbers, the previous best serial algorithm sorts in $O\left(n\mathrm{log}n\right)$ time. It was also known that for comparison sorting, $\Omega \left(n\mathrm{log}n\right)$ is the tight lower bound. Thus if we use comparison sorting to sort real numbers, then in serial algorithms, we cannot avoid the $\Omega \left(n\mathrm{log}n\right)$ time bound and in parallel algorithms, we cannot avoid the $\Omega \left(n\mathrm{log}n\right)$ operation bound. In the past, no other sorting methods are known to sort real number in less than $O\left(n\mathrm{log}n\right)$ time and comparison sorting remained the norm for sorting real numbers.
However, the situation is recently changed completely as Han found a way to convert real numbers to integers for sorting purpose and he showed that real numbers can be sorted in $O\left(n\sqrt{\mathrm{log}n}\right)$ time [4] . This result enables us to move further to improve the operation bound of parallel algorithms for sorting real numbers to below $O\left(n\mathrm{log}n\right)$, as in the past, all parallel algorithms for sorting real numbers have an operation bound at least $O\left(n\mathrm{log}n\right)$.
In this paper, we will apply the $O\left(n\sqrt{\mathrm{log}n}\right)$ time serial real number sorting algorithm [4] to the design of an NC algorithm with $O\left({\mathrm{log}}^{1+\epsilon }n\right)$ time and
$O\left(\frac{n\mathrm{log}n}{\sqrt{\mathrm{log}\mathrm{log}n}}\right)$ operations on the CREW (Concurrent Read Exclusive Write)
PRAM. NC stands for Nick’s Class [12] . NC algorithms are parallel algorithms with polylog time and polynomial operations. Algorithm in [4] is an inherently serial algorithm without much parallelism within it. Here, we use it in the design
of an NC algorithm with $O\left({\mathrm{log}}^{1+\epsilon }n\right)$ time and $O\left(\frac{n\mathrm{log}n}{\sqrt{\mathrm{log}\mathrm{log}n}}\right)$ operations. This is the first NC algorithm for sorting real numbers with $o\left(n\mathrm{log}n\right)$ operations.
The computation model used for designing our algorithm is the CREW PRAM. On this model, in one step, any processor can read/write any memory cell. Concurrent read of one memory cell by multiple processors in one step is allowed and concurrent write of one memory cell by multiple processors in one step is prohibited. Parallel algorithms can be measured with their time complexity and the number of processors used. They can also be measured with time complexity and operation complexity which is the time processor product. The operation complexity (Tpp, with Tp time using p processors) of a parallel algorithm is often compared with the time T1 of the best serial algorithm. In general, ${T}_{p}p\ge {T}_{1}$. When ${T}_{p}p={T}_{1}$, the parallel algorithm is said to be an operation optimal algorithm.
The main contribution of this paper is the demonstration of the existence of an NC parallel algorithm with $o\left(n\mathrm{log}n\right)$ operations for sorting real numbers. All previous parallel algorithms for sorting real numbers have at least $O\left(n\mathrm{log}n\right)$ operations. The algorithm presented in this paper is derived from Han’s $O\left(n\sqrt{\mathrm{log}n}\right)$ time serial algorithm [4] for sorting real numbers by applying parallel algorithm design techniques. These parallel algorithm design techniques are specially tuned for the derivation of our parallel algorithm.
The remaining part of this paper is organized as follows. In Section 2 we present our NC algorithm for sorting real numbers with $O\left({\mathrm{log}}^{1+\epsilon }n\right)$ time and
$O\left(\frac{n\mathrm{log}n}{\sqrt{\mathrm{log}\mathrm{log}n}}\right)$ operations. In Section 3, we present a running example of our algorithm and conclude our paper with the Main Theorem.
2. The Algorithm
Consider an algorithm for sorting n real numbers. Suppose each of the n/m lists with m real numbers in each list has already been sorted, we are to merge these n/m lists into one sorted list. We will do k-way merge in each pass to merge every k-lists into 1 sorted list and there are $\mathrm{log}\left(n/m\right)/\mathrm{log}k$ passes to have all n/m lists merged into 1 sorted list.
For simplicity, let us break down n elements into lists with m elements in each list. We can do parallel sort on the individual list of m elements recursively. Now we pick every k-lists and have them merged together.
The k-way merging of sorted lists ${L}_{0},{L}_{\text{1}},\cdots ,{L}_{k-\text{1}}$ is done as follows. For each sorted list of m real numbers we pick every k2-th real number, i.e. we pick the 0th real number, the k2-th real number, the 2k2-th real number, the 3k2-th real number, and so on. Thus from each list Li we picked m/k2 real numbers and these m/k2 real numbers form a sorted list ${{L}^{\prime }}_{i}$. and from these k lists we picked m/k real numbers they form sorted lists ${{L}^{\prime }}_{0},{{L}^{\prime }}_{\text{1}},\cdots ,{{L}^{\prime }}_{k-\text{1}}$. We merge ${{L}^{\prime }}_{0},{{L}^{\prime }}_{\text{1}},\cdots ,{{L}^{\prime }}_{k-\text{1}}$ into one sorted list L' using Valiant’s merging algorithm [13] (its improved version is given by Kruskal in [14] with time complexity of $O\left(\mathrm{log}\mathrm{log}m\right)$ and linear operations for merging two sorted lists of m elements each) in logk passes and $O\left(\mathrm{log}\mathrm{log}m\right)$ time and $O\left(m/k\right)$ operations in each pass. Thus the total time for merging ${{L}^{\prime }}_{0},{{L}^{\prime }}_{\text{1}},\cdots ,{{L}^{\prime }}_{k-\text{1}}$ is $O\left(\mathrm{log}k\mathrm{log}\mathrm{log}m\right)$ and the total operation is $O\left(m\mathrm{log}k/k\right)$.
Now for each real number r in ${{L}^{\prime }}_{i}$ and for any ${{L}^{\prime }}_{j}$, r knows the largest real number s in ${{L}^{\prime }}_{j}$ that is smaller than r and smallest real number l in ${{L}^{\prime }}_{j}$ that is larger than r. s and l are actually neighbors in ${{L}^{\prime }}_{j}$. There are k2 elements between s and l in Lj. r then uses binary search in $O\left(\mathrm{log}k\right)$ time to find the largest real number among these k2 real numbers that is smaller than r and the smallest real number that is larger than r. That is, r finds the exact insertion point of r in Lj. Because there are k-lists and there are m/k real numbers in L' thus the time for this binary search is and the operation is. The operation for all lists is $O\left(n\mathrm{log}k/k\right)$ because there are n/m lists and every k-lists are merged in the k-way merge, we picked n/k2 real numbers and every one of them has to use k processors to check k-lists in the k-way merging. Because r is arbitrary picked and thus we know that every real number in L' knows its insertion point in every Lj. Let the real numbers in sorted order in L' be ${r}_{0},{r}_{1},\cdots ,{r}_{m/k-1}$. To merge ${L}_{0},{L}_{\text{1}},\cdots ,{L}_{k-\text{1}}$ we need now to merge or sort all real numbers between the insertion points of ri and ri+1 in ${L}_{0},{L}_{\text{1}},\cdots ,{L}_{k-\text{1}}$. There are no more than k2 real numbers in Lj between the insertion points of ri and ri+1 and therefore the total number $R\left(i,i+\text{1}\right)$ of real numbers (call them a block) in ${L}_{0},{L}_{\text{1}},\cdots ,{L}_{k-\text{1}}$ between the insertion points of ri and ri+1 is no more than k3 (i.e. $R\left(i,i+1\right)\le {k}^{3}$ ). When $R\left(i,i+1\right)<{k}^{3}$ we will combine multiple blocks together to reach k3 real numbers. We use the $O\left(n\sqrt{\mathrm{log}n}\right)$ serial sorting algorithm to sort them in $O\left({k}^{3}\sqrt{\mathrm{log}k}\right)$ time. This represents $O\left({k}^{3}\sqrt{\mathrm{log}k}\right)$ time and $O\left(n\sqrt{\mathrm{log}k}\right)$ operations in our parallel algorithm.
Thus the time for each stage is $O\left({k}^{3}\sqrt{\mathrm{log}k}\right)$ and the operation for each stage is $O\left(n\sqrt{\mathrm{log}k}\right)$. When we start with m as a constant then there are $\mathrm{log}n/\mathrm{log}k$
stages and therefore the time of our algorithm is $O\left(\frac{{k}^{3}\mathrm{log}n}{\sqrt{\mathrm{log}k}}\right)$ and the operation is $O\left(\frac{n\mathrm{log}n}{\sqrt{\mathrm{log}k}}\right)$.
Pick $k={\mathrm{log}}^{\epsilon }n$, we get $O\left({\mathrm{log}}^{1+\epsilon }n\right)$ time and $O\left(\frac{n\mathrm{log}n}{\sqrt{\mathrm{log}\mathrm{log}n}}\right)$ operations.
3. Procedure
Step 1: Let’s say we have “m” sorted elements in each list, and we have a total of “n” elements to sort (Figure 1).
This implies that we have “n/m” lists to sort. To sort these blocks, we will apply k-way merging.
Step 2: Each stage of k-way merging is to merge every k sorted lists into 1 sorted list. This is repeatedly until all n/m lists are merged into one list (Figure 2).
Step 3: To merge k lists into 1 list, we need to pick the “0-th”, “k2-th”, “2k2-th”, “3k2-th”, … real numbers in each list Li to form a new list ${{L}^{\prime }}_{i}$ of m/k2 elements. This is shown as Figure 3.
Step 4: We merge ${{L}^{\prime }}_{0},{{L}^{\prime }}_{\text{1}},\cdots ,{{L}^{\prime }}_{k-\text{1}}$ into one sorted list L'. The elements of this new formed list L' then use binary search to find their exact insertion point in
Figure 1. n/m sorted lists of m elements each.
Figure 2. k-way merging of k lists.
Figure 3. Pick every k2-th element of each list.
${L}_{0},{L}_{\text{1}},\cdots ,{L}_{k-\text{1}}$. These insertion points then partition ${L}_{0},{L}_{\text{1}},\cdots ,{L}_{k-\text{1}}$ into m/k blocks with each block containing no more than k3 real numbers.
Step 5: When every one of these m/k blocks are sorted, we effectively merged ${L}_{0},{L}_{\text{1}},\cdots ,{L}_{k-\text{1}}$ into one sorted list L.
Main Theorem: n real numbers can be sorted in $O\left({\mathrm{log}}^{1+\epsilon }n\right)$ time and $O\left(\frac{n\mathrm{log}n}{\sqrt{\mathrm{log}\mathrm{log}n}}\right)$ operations on the CREW PRAM.
Proof: The algorithm and its time complexity analysis are presented in Section 2. An example of the running of the algorithm is presented in Section 3. □
Conflicts of Interest
The authors declare no conflicts of interest regarding the publication of this paper.
Cite this paper
Han, Y.J., Mishra, S. and Syed, M.U.G. (2019) An NC Algorithm for Sorting Real Numbers in Operations. Open Journal of Applied Sciences, 9, 403-408. https://doi.org/10.4236/ojapps.2019.95034
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4. 4. Han, Y. (2017) Sort Real Numbers in Time and Linear Space. In arXiv.org with paper id 1801.00776.
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6. 6. Cole, R. (1988) Parallel Merge Sort. SIAM Journal on Computing, 17, 770-785. https://doi.org/10.1137/0217049
7. 7. Saxena, S., Chandra, P., Bhatt, P. and Prasad, V.C. (1994) On Parallel Prefix Computation. Parallel Processing Letters, 4, 429-436. https://doi.org/10.1142/S0129626494000399
8. 8. Bhatt, P.C.P., Diks, K., Hagerup, T., Prasad, V.C., Radzik, T. and Saxena, S. (1991) Improved Deterministic Parallel Integer Sorting. Information and Computation, 94, 29-47. https://doi.org/10.1016/0890-5401(91)90031-V
9. 9. Hagerup, T. (1987) Towards Optimal Parallel Bucket Sorting. Information and Computation, 73, 39-51. https://doi.org/10.1016/0890-5401(87)90062-9
10. 10. Han, Y. and Shen, X. (1995) Conservative Algorithms for Parallel and Sequential Integer Sorting. International Computing and Combinatorics Conference, Lecture Notes in Computer Science, 959, 324-333. https://doi.org/10.1007/BFb0030847
11. 11. Han, Y. and Shen, X. (2002) Parallel Integer Sorting Is More Efficient than Parallel Comparison Sorting on Exclusive Write PRAMs. Tenth Annual ACM-SIAM Symposium on Discrete Algorithms (SODA’99), Baltimore, Maryland, January 1999, 419-428. https://doi.org/10.1137/S0097539799352449
12. 12. Cook, S.A. (1981) Towards a Complexity Theory of Synchronous Parallel Computation. L’Enseignement Mathématique, 27, 99-124.
13. 13. Valiant, L.G. (1975) Parallelism in Comparison Problems. SIAM Journal on Computing, 4, 348-355. https://doi.org/10.1137/0204030
14. 14. Kruskal, C.P. (1983) Searching, Merging, and Sorting in Parallel Computation. IEEE Transactions on Computers, C-32, 942-946. https://doi.org/10.1109/TC.1983.1676138 | 4,705 | 16,018 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 85, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-43 | latest | en | 0.752575 |
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Posted by starvator on Tue, 24 May 2022 18:20:56 +0300
## #LeetCode brush questions#, #The sum of two numbers#, #Hash table#
The first question of LeetCode brushing questions, the use of the sum of two numbers and the hash table Topic description Given an integer array nums and a target value target, please find the two integers in the array whose sum is the target value, and return their array indices. You can assume that there will only be one answer for each input ...
Posted by 8ta8ta on Tue, 24 May 2022 17:54:50 +0300
## [force deduction and question brushing] (207. Course schedule) record
1, Topic analysis (1) The problem can be abstracted as detecting whether a directed graph has a ring. If yes, return False; If not, return True. (2) For a directed graph, we can delete nodes with zero penetration one by one. For each deleted node, the connected directed line segment will also be deleted (i.e. the penetration of this point as ...
Posted by judgy on Tue, 24 May 2022 11:48:30 +0300 | 1,149 | 4,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-49 | latest | en | 0.83009 |
https://de.mathworks.com/matlabcentral/profile/authors/869888-jan?s_tid=al_cnt | 1,579,883,596,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250624328.55/warc/CC-MAIN-20200124161014-20200124190014-00532.warc.gz | 401,910,470 | 23,597 | Community Profile
# Jan
12.564 total contributions since 2009
It is easier to solve a problem than to guess, what the problem is.
Questions about FileExchange submissions are welcome - get my address from the code. I do not answer mails concerning questions in the forum.
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Vector between vectors in intervals
Remark: The inputs vectors are not normalized and only almost orthogonal: format long g v1 = [-0.0345; 0.9984; -0.0447] v2 ...
etwa 2 Monate ago | 0
Can anyone help me with the "Axis rotation sequence for the Euler angles"
The rotation matrix for the Euler angles in the order XYX is: s1 = sin(alpha); c1 = cos(alpha); s2 = sin(beta); c2 = cos(bet...
etwa 2 Monate ago | 0
Matlab has encountered an internal problem and needs to close.
The message teels you: If this problem is reproducible, please submit a Service Request via: http://www.mathworks.com/...
etwa 2 Monate ago | 0
Use the time step within ode15s
If the function to be integrated requires the time steps of the solver, something went wrong. I'm convinced that this is a logic...
etwa 2 Monate ago | 0
| accepted
Keep getting error message on my ODE function
u0 = [P0, zeros(1, N + 1)];
etwa 2 Monate ago | 0
| accepted
Eulers method for system of linear equations
h = 0,1; % ^ you mean 0.1 with a dot There is no need to create the accleration as array. Reply a scalar instead: funct...
etwa 2 Monate ago | 0
Skipping files in a set
What does "files 132, 290, and 404" mean? Are these the indices in rightfiles, or are these parts of the file name? rightfiles=...
etwa 2 Monate ago | 0
How to sum over indices?
If you assume, that something is wrong, mention, what you observe. This is better than letting the readers guess, what the probl...
etwa 2 Monate ago | 1
| accepted
How can I convert a binary fraction to decimal?
'11001' means: 1*2^0 + 0*2^1 + 0*2^2 + 1*2^3 + 1*2^4 or: [1,1,0,0,1] * 2 .^ (4:-1:0).' (this is what happens inside bin2dec - ...
etwa 2 Monate ago | 3
| accepted
Table of Matlab release features
%{ Version Release Notes 1 1984 2 1986 3 1987 3.5 1...
etwa 2 Monate ago | 3
Question
Table of Matlab release features
After reading Rik's comment I looked for a list of Matlab releases and their corresponding features. Wiki: Matlab contains an ex...
etwa 2 Monate ago | 2 answers | 3
### 2
How do I install MinGW for old Matlab R2007?
Matlab R2007 is not supported under Windows 7. MinGW is not supported for Matlab R2007. In consequence the combination of a mode...
2 Monate ago | 0
How do you form a while function using a function file and then using the function file to form a sequence by performing this operation?
Change n directly and collect the output in the vector v by inserting it at the index end+1: function v = collatz(n) v = n; w...
2 Monate ago | 0
| accepted
Using parfor in a sparse setting
The help page "Troubleshoot Variables in parfor-Loops" explains the problem: A = [1;2;3;4;nan;nan;nan;nan;nan;10;11]; B = na...
2 Monate ago | 0
| accepted
Speed up nested loops
This line is expensive: normal_vector = ((cross(P_all(k-t,:),P_all(k,:)) / ... (norm(cross(P_all(k-t,:),P_all(k,:)))))); ...
2 Monate ago | 1
| accepted
Does Matlab perform well on AMD Ryzen?
It depends if the used libraries call FMA3 commands. If so, the Ryzen crashes very fast. The performance will depend on your pr...
2 Monate ago | 0
Why is my code showing an error? Why is the line deleted?
The shown code does not reproduce the problem. We cannot guess reliably, why one of the obejcts in h is deleted. But a weak gues...
2 Monate ago | 0
variable size matrix and converting letters to numbers
"user enter a variable size matrix": It depends on how you want to implement this: by a GUI, as Matlab code, as text file, in Ex...
2 Monate ago | 0
| accepted
Help on understanding the following matlab code ?
Step by step: flipud(xt): This flipps the order of elements vertically. Try it: x = rand(3,2) flipud(x) min(Y, [ ], 2) : fin...
2 Monate ago | 0
| accepted
Resuming from "keyboard" command?
Using debug commands for an interaction with the user is a really bad design. A dynamic modification of the workspace is very fr...
2 Monate ago | 0
Memory usage very high
How do you observe the memory consumption? The Taskmanager displays the memory reserved for Matlab. If Matlab allocates memory a...
2 Monate ago | 0
Return vectors from for loop as columns in matrix
Start with cleaning up the code. The lines Dplus; and Dminus; do nothing, so they are wasting time only and confuse the readers...
2 Monate ago | 1
convert lower case to upper and upper to lower
You can use the function isstrprop with the category 'lower' and 'upper' to identify the specific characters. Then use the comma...
2 Monate ago | 0
Function run(scriptname) doesn't work.
Store the M-files in a folder, in which you have full access. C:\Users\ is a bad idea. Create a folder somewhere else, e.g. insi...
2 Monate ago | 0
Subtracting Vector from a Matrix
Matlab can subtract vectors from matrices automatically since R2016b - so called "auto expanding". Do you use an older version? ...
2 Monate ago | 0
| accepted
Include global variable declaration infunctions
Global variables are a shot in your knee in every case. Hiding them in scripts, which are called dynamically, impedes the debugg...
2 Monate ago | 0
deleted old version of Matlab (R2017b) keeps getting back
See https://nektony.com/how-to/uninstall-matlab-mac According to https://www.mathworks.com/help/install/ug/uninstall-mathworks-...
2 Monate ago | 1
| accepted
I found this bellow code in internet. While I run the code I noticed following errors.
This is a strange and inefficient code. Replace e.g. for i=1:len for j= 2:2 text_names(i,j)=txt(i,j); end ...
2 Monate ago | 0
Vectorise or Parallel Computing
This loop cannot be parallelized. If flux_edge is a vector and not a matrix, accumarray would solve the problem efficiently. Try...
2 Monate ago | 0
| accepted | 1,633 | 6,019 | {"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-2020-05 | latest | en | 0.818697 |
https://mail.gnome.org/archives/gnumeric-list/2019-June/msg00002.html | 1,701,913,510,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00626.warc.gz | 429,493,403 | 4,447 | # Re: how to tally rows and columns for 2 variables?
• From: Tim Schofield <tim schofield1960 gmail com>
• To: bill <bill samerica com>
• Cc: Gnumeric <gnumeric-list gnome org>
• Subject: Re: how to tally rows and columns for 2 variables?
• Date: Tue, 18 Jun 2019 20:05:41 +0100
Happy to help :)
On Tue, 18 Jun 2019 18:07 bill, <bill samerica com> wrote:
Tim,
A belated thank you for the formula. I finally had a chance to get back to the project where it was needed, and it works great!
peace
bill
Sent: Tuesday, June 11, 2019 at 3:22 AM
From: "Tim Schofield" <tim weberpafrica com>
To: bill <bill samerica com>
Cc: Gnumeric <gnumeric-list gnome org>
Subject: Re: how to tally rows and columns for 2 variables?
I can't check it right now, but
=COUNTIF(G3:G2030,"x")+COUNTIF(G3:G2030,"y")
should work.
Tim
On Tue, 11 Jun 2019 at 07:58, bill <bill samerica com> wrote:
>
> Hi all,
>
> I have a column and row near the beginning of my spreadsheets that counts each x i insert into the cells of that column or row. For that i use the formula =countif(G3:G2030,"x") for the G column, and etc. Suddenly i have need to also count any y in that table. I tried adding to the formula so that it reads =countif(G3:G2030,"x","y"), but get the error that there's too many variables for the countif statement. I'm sure there's a simple fix that i just can't seem to ferret out. Any ideas much appreciated!
>
> bill
> _______________________________________________
> gnumeric-list mailing list
> gnumeric-list gnome org
> https://mail.gnome.org/mailman/listinfo/gnumeric-list
--
Course View Towers, | 471 | 1,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | latest | en | 0.853644 |
https://www.calculateme.com/money-over-time/90-every-1-day-for-3-months | 1,708,789,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00415.warc.gz | 692,660,709 | 5,616 | # \$90 Every Day for 3 Months
If you get 90 dollars every day, how much would you have after 3 months?
Use this calculator to tabulate your savings or earnings.
Money
\$
How Often
How Long
\$90 every day for 3 months is a total of \$8,218.
There would be 91 payments of \$90, plus an additional \$28 for the partial period remaining.
Every Day For 3 Months
\$90 \$8,218
\$91 \$8,309
\$92 \$8,401
\$93 \$8,492
\$94 \$8,583
\$95 \$8,675
\$96 \$8,766
\$97 \$8,857
\$98 \$8,948
\$99 \$9,040 | 165 | 488 | {"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-2024-10 | latest | en | 0.805603 |
https://www.quizover.com/online/course/6-4-eddy-currents-and-magnetic-damping-by-openstax | 1,532,125,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591837.34/warc/CC-MAIN-20180720213434-20180720233434-00616.warc.gz | 983,089,988 | 22,542 | # 6.4 Eddy currents and magnetic damping
Page 1 / 4
• Explain the magnitude and direction of an induced eddy current, and the effect this will have on the object it is induced in.
• Describe several applications of magnetic damping.
## Eddy currents and magnetic damping
As discussed in Motional Emf , motional emf is induced when a conductor moves in a magnetic field or when a magnetic field moves relative to a conductor. If motional emf can cause a current loop in the conductor, we refer to that current as an eddy current . Eddy currents can produce significant drag, called magnetic damping , on the motion involved. Consider the apparatus shown in [link] , which swings a pendulum bob between the poles of a strong magnet. (This is another favorite physics lab activity.) If the bob is metal, there is significant drag on the bob as it enters and leaves the field, quickly damping the motion. If, however, the bob is a slotted metal plate, as shown in [link] (b), there is a much smaller effect due to the magnet. There is no discernible effect on a bob made of an insulator. Why is there drag in both directions, and are there any uses for magnetic drag?
[link] shows what happens to the metal plate as it enters and leaves the magnetic field. In both cases, it experiences a force opposing its motion. As it enters from the left, flux increases, and so an eddy current is set up (Faraday’s law) in the counterclockwise direction (Lenz’s law), as shown. Only the right-hand side of the current loop is in the field, so that there is an unopposed force on it to the left (RHR-1). When the metal plate is completely inside the field, there is no eddy current if the field is uniform, since the flux remains constant in this region. But when the plate leaves the field on the right, flux decreases, causing an eddy current in the clockwise direction that, again, experiences a force to the left, further slowing the motion. A similar analysis of what happens when the plate swings from the right toward the left shows that its motion is also damped when entering and leaving the field.
When a slotted metal plate enters the field, as shown in [link] , an emf is induced by the change in flux, but it is less effective because the slots limit the size of the current loops. Moreover, adjacent loops have currents in opposite directions, and their effects cancel. When an insulating material is used, the eddy current is extremely small, and so magnetic damping on insulators is negligible. If eddy currents are to be avoided in conductors, then they can be slotted or constructed of thin layers of conducting material separated by insulating sheets.
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Got questions? Join the online conversation and get instant answers! | 1,819 | 7,247 | {"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-2018-30 | longest | en | 0.932926 |
https://brilliant.org/problems/finding-triple-variables-2/ | 1,627,619,415,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00629.warc.gz | 165,980,823 | 7,822 | # Three Variables In One Equation?
$\Large 3{\color{#D61F06}{x}} = 4{\color{#3D99F6}{y}} = 7{\color{#20A900}{z}}$
If $x,y,$ and $z$ are positive integers, what is the least possible value of $x+y+z$?
× | 79 | 204 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-31 | latest | en | 0.708088 |
https://techcrams.com/who-is-the-strongest-stand-user-in-jojo-hoodies/ | 1,721,336,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00766.warc.gz | 491,496,524 | 36,280 | Home » Who is the strongest Stand user in JoJo hoodies?
# Who is the strongest Stand user in JoJo hoodies?
## Introduction
In JoJo’s Bizarre Adventure, there are a variety of Stand users that can fight on equal ground with the Joestars. However, there is one Stand user in particular that stands Jojo Hoodies head and shoulders above the rest when it comes to strength and power- Jonathan Joestar!
While every Stand has its own strengths and weaknesses, Jonathan’s Stand, Steel Ball Bear, is practically unstoppable. In the first three arcs of the manga, Jonathan uses his Stand to take down powerful enemies like DIO or Caesar Clown in a matter of seconds. In addition to his brute force. Jonathan also knows how to use his Stand in strategic ways to defeat his foes. For example. He was able to use Steel Ball Bear’s ability to throw heavy objects at his enemies in order to take them down from afar.
Overall, Jonathan Joestar is one of the strongest Stand users out there and he will no doubt be a major threat come Valentine’s Day 2018!
### The Methods Used to Calculate the Strength of Stand users
The methods used to calculate the strength of Stand users can be broken down into two categories: analytical and empirical. Analytical methods rely on mathematical models to simulate how an individual’s Stand will behave in various situations. Empirical methods, on the other hand, use actual data from past battles to estimate an individual’s Stand strength.
One analytical method that has been used is the Battle Simulator, developed by Hirohiko Araki himself. The simulator uses a 3D fighting game engine to create realistic conditions for testing Stand abilities. In order to measure an opponent’s Stand power, Araki et al. (2015) created a system that evaluates how long it takes Jojo Shirts a character with a given level of experience to defeat someone with a weaker Stand ability. To date, the Battle Simulator has been used to measure the strength of Stands belonging to Jotaro Kujo ( Stardust Crusaders ). Joseph Joestar ( Phantom Blood ), and Bambi.
Another method used to estimate an individual’s Stand strength is called Power Analysis. This technique was first proposed by Eiji Shishio in his seminal paper “An Analysis of Magical Powers through Power Calculation Methods” (Shishio, 1970). The basic premise behind this approach is that all Stands possess specific strengths and weaknesses that can be quantified using mathematical formulas. For example, Stands that are powered by emotions or physical energy are likely to have greater effectiveness against
### Results of the Analysis
The answer to this question may surprise some Jojo fans. While many people would say that Jotaro Kujo is the strongest Stand user in the series. A new analysis has found that Joseph Joestar may actually be the most powerful Stand user.
According to data from the official website for Jojo’s Bizarre Adventure manga and anime series. Joseph Joestar’s Stand, Star Platinum, is said to be one of the strongest Stands in history. Not only is Star Platinum one of the most versatile Stands in the manga/anime series. But it can also combine with other Stands to create even more powerful attacks.
While Jotaro Kujo’sStand, Stardust Crusaders, is also seen as being very powerful. Joseph’s combination of Star Platinum with his other Stands makes him a much more formidable opponent. In fact, his ability to use multiple Stands simultaneously has never been seen before in any other manga or anime series.
As a result of this analysis. It seems that Joseph Joestar may actually be the strongest Stand user in JoJo hoodies!
### Conclusion
In this article, we are going to take a look at who is the strongest Stand user in JoJo hoodies. We will be looking at each character and determining. Which one has the most potential for becoming a powerful Stand user? After examining each character’s abilities and strengths. We will come to a definitive conclusion as to who is the strongest Stand user in JoJo hoodies. So, who do you think is the strongest Stand user? Let us know in the comments below! | 851 | 4,109 | {"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-2024-30 | latest | en | 0.928704 |
http://www.pizzamaking.com/forum/index.php?topic=1601.0 | 1,472,597,691,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983026851.98/warc/CC-MAIN-20160823201026-00280-ip-10-153-172-175.ec2.internal.warc.gz | 634,176,459 | 15,601 | ### Author Topic: Cost of a pizza (Read 76993 times)
0 Members and 2 Guests are viewing this topic.
#### TONY
• Registered User
• Posts: 87
• I Love Pizza!
##### Cost of a pizza
« on: July 09, 2005, 11:35:04 AM »
Can anyone tell me the average wholesale cost (excluding labor) to make a 16" (large) pizza with cheese and two toppings? I can figure it out at the retail level (my cost), but what would it cost a pizza shop/restaurant to make?
Tony
#### Pete-zza
• Global Moderator
• Posts: 24990
• Location: Texas
• Always learning
##### Re: Cost of a pizza
« Reply #1 on: July 09, 2005, 12:48:25 PM »
Tony,
The answer to your question depends on who is making the pizza, what the exact toppings are, and the quality and amount of each topping. Obviously, a chain like Papa John's or Domino's will get better pricing on a per unit basis than will a mom-and-pop operator because of greater volume and bargaining power. But, as a typical example, I have read that a 16-inch pizza with 10-ounces of mozzarella cheese, sauce, and 6 ounces of pepperoni will cost an operator around \$2 (for the ingredients). I would guess that adding an additional topping might increase the cost by another 10-20 cents, depending on the topping. The dough itself will typically cost \$0.011/ounce if made from scratch. For a 16-inch size using about 21 ounces of dough that would come to about \$0.23. Generally speaking, the most expensive pizza item is the cheese.
Another example I saw was \$1.70-\$1.80 for a 16-inch pie with 12 ounces of mozzarella cheese and 8 ounces of sauce.
Peter
« Last Edit: July 09, 2005, 01:19:59 PM by Pete-zza »
#### scampi
• Registered User
• Posts: 98
• Location: Orange County
• I Love Pizza!
##### Re: Cost of a pizza
« Reply #2 on: July 09, 2005, 12:56:00 PM »
I'll take a guess and will probably be wrong. If you are trying to figure out total cost, you need to add labor and overhead if you are trying to figure out what it costs for a pizzeria to make it vs. you at home. However their costs of product used is lower than what it would cost you since they buy in volume. Anyway, I read that some pizzeria's, not all, try to keep their resale at 3.5 times their cost. So if they are selling for \$10, their cost should be \$2.80 or something like that. That doesn't include their overhead, just the cost of ingrediants used to make the pie.. For me at home, my dough costs under a dollar, the sauce is probably around .50 at most, cheese is around 1.50-2.00, pepperoni's are a penny each so if you put 20 on, that's 20 cents...you can figure out the rest depending on what you put on it. My cost if different than everyone elses I guess..but that's mine, around 3.50-4.00
« Last Edit: July 09, 2005, 01:16:22 PM by scampi »
#### TONY
• Registered User
• Posts: 87
• I Love Pizza!
##### Re: Cost of a pizza
« Reply #3 on: July 11, 2005, 09:33:46 AM »
Thanks for the info guys.
Tony
#### RockyMarciano
• Registered User
• Posts: 82
• Go Sabres! Go Bills!
##### Re: Cost of a pizza
« Reply #4 on: February 19, 2006, 11:49:32 AM »
Back in 83, it costed about 1.25
#### lilpizzaplace_ohio
• Registered User
• Posts: 22
• Location: OHIO
• I Love Pizza!
##### Re: Cost of a pizza
« Reply #5 on: March 01, 2006, 09:07:05 PM »
This is a one topping 16 inch...at 12.99...some say it's a low price, others say it isn't...I think it depends on your area..
lilpizzaplace
#### Furo
• Registered User
• Posts: 29
• Age: 59
• Location: Kansas City Mo
##### Re: Cost of a pizza
« Reply #6 on: March 02, 2006, 08:19:48 PM »
If that tastes even half as good as it looks you could sell it for over \$20.00 in Kansas City!!!!!!!
Being crazy isn't a necessity but it sure is FUN!
#### lilpizzaplace_ohio
• Registered User
• Posts: 22
• Location: OHIO
• I Love Pizza!
##### Re: Cost of a pizza
« Reply #7 on: March 02, 2006, 10:05:56 PM »
haha trust me......its good...I eat pizza from there almost every day myself...That says alot!! the sauce and the cheese are to DIE for!
lilpizzaplace
#### Buffalo
• Registered User
• Posts: 138
• I Love Pizza!
##### Re: Cost of a pizza
« Reply #8 on: March 03, 2006, 08:45:19 AM »
lilpizzaplace........
I really like the way your pepperoni cups......what brand do you use
Thanks.....
Buffalo
• Registered User
• Posts: 1040
• Age: 18
• DoughBoy
##### Re: Cost of a pizza
« Reply #9 on: April 12, 2006, 10:30:45 PM »
do you have a dollar figure of what this pizza cost to actually make though ? , the original poster
I think was interested in the actual *cost* of making a pizza, but I'm not sure if you work there
or just order your pizza there.
This is a one topping 16 inch...at 12.99...some say it's a low price, others say it isn't...I think it depends on your area..
« Last Edit: April 12, 2006, 10:37:24 PM by canadianbacon »
Pizzamaker, Rib Smoker, HomeBrewer, there's not enough time for a real job.
#### nepa-pizza-snob
• Registered User
• Posts: 134
##### Re: Cost of a pizza
« Reply #10 on: December 17, 2006, 08:49:18 PM »
whats up in kansas city? no good shops around or is \$20 market for a 16inch pie
#### abatardi
• Registered User
• Posts: 431
• Age: 55
• Location: Santa Clara, CA
• It's MOOPS!
##### Re: Cost of a pizza
« Reply #11 on: December 31, 2006, 08:34:21 PM »
lilpizzaplace........
I really like the way your pepperoni cups......what brand do you use
Thanks.....
Buffalo
not sure but I know swiss american pepperoni cups like that. www.sasausage.com
Make me a bicycle CLOWN!
#### KingPinAlley
• Registered User
• Posts: 11
• I Love Pizza!
##### Re: Cost of a pizza
« Reply #12 on: January 25, 2007, 12:42:34 AM »
that pepperoni pizza looks awsome!!
I'm using Roma products. I only make 18in pizza's and sell whole or by the slice.
My one topping is \$13.99
dough \$.44
cheese \$1.70....8oz of 50% Grande (\$2.50 a lb.) and 50% of a quality blend Roma cheese
sauce \$.28...4oz of crushed tomato blend and 4oz of sauce (secret)
pepp. \$.40...quality Roma pepperoni product
It's about 3\$ a pizza but it's a great quality. I charge \$2 a slice and do a lot of that business. At 8 slices a pie i'm getting \$16 for a pep pizza!!
What you have to watch is the 40 cent pizza boxes and the other 30 cents in togo parm cheese packs and crushed red peppers!!
#### Chicago Bob
• Posts: 12630
• Location: Durham,NC
• Easy peazzy
##### Re: Cost of a pizza
« Reply #13 on: October 10, 2012, 10:26:16 PM »
that pepperoni pizza looks awsome!!
I'm using Roma products. I only make 18in pizza's and sell whole or by the slice.
My one topping is \$13.99
dough \$.44
cheese \$1.70....8oz of 50% Grande (\$2.50 a lb.) and 50% of a quality blend Roma cheese
sauce \$.28...4oz of crushed tomato blend and 4oz of sauce (secret)
pepp. \$.40...quality Roma pepperoni product
It's about 3\$ a pizza but it's a great quality. I charge \$2 a slice and do a lot of that business. At 8 slices a pie i'm getting \$16 for a pep pizza!!
What you have to watch is the 40 cent pizza boxes and the other 30 cents in togo parm cheese packs and crushed red peppers!!
I think the math is a 'lil "funny" here.......but it's 6 years later now and I'm wondering what a pizza costs to make now a days in your area. Let's hear from everyone....home cooks and the pro's!!
"Care Free Highway...let me slip away on you"
#### La Sera
• Registered User
• Posts: 135
##### Re: Cost of a pizza
« Reply #14 on: October 11, 2012, 06:39:43 PM »
The pizza business can be lucrative for a savvy operator in a decent market. Most independent operators strive for about 20% food cost on pizza. If a place is running over 25%, their pricing is too low. Some markets are destroyed by low-ball chains. A chain doesn't need to earn a living from a single store.
Side order costs are are a different animal.
#### Chicago Bob
• Posts: 12630
• Location: Durham,NC
• Easy peazzy
##### Re: Cost of a pizza
« Reply #15 on: October 11, 2012, 07:04:27 PM »
Great info La Sera.....thanks!
"Care Free Highway...let me slip away on you"
#### Giggliato
• Registered User
• Posts: 334
##### Re: Cost of a pizza
« Reply #16 on: October 11, 2012, 10:34:18 PM »
The pizza business can be lucrative for a savvy operator in a decent market. Most independent operators strive for about 20% food cost on pizza. If a place is running over 25%, their pricing is too low. Some markets are destroyed by low-ball chains. A chain doesn't need to earn a living from a single store.
Side order costs are are a different animal.
So if a pizza costs five dollars I should be charging 25 dollars? for a large pie with eight slices that would be about 3.25 a slice, I suppose that is reasonable and perhaps high quality.
#### TXCraig1
• Supporting Member
• Posts: 19932
• Location: Houston, TX
##### Re: Cost of a pizza
« Reply #17 on: October 11, 2012, 10:47:09 PM »
Keep in mind that you can't deposit a % in the bank - just \$.
Formulas and rules are almost never a good way to price things.
"We make great pizza, with sourdough when we can, commercial yeast when we must, but always great pizza."
Craig's Neapolitan Garage
#### Chicago Bob
• Posts: 12630
• Location: Durham,NC
• Easy peazzy
##### Re: Cost of a pizza
« Reply #18 on: October 11, 2012, 10:50:21 PM »
So if a pizza costs five dollars I should be charging 25 dollars? for a large pie with eight slices that would be about 3.25 a slice, I suppose that is reasonable and perhaps high quality.
Not out of the question at all IMO. I see pizzerias all the time that proclaim "gourmet toppings" and this 20-25 dollars is a common price point. And believe me, their gourmet stuff is not the high dollar products that the menu descriptions would leave you to think they are using if you know what I mean.
"Care Free Highway...let me slip away on you"
#### Tscarborough
• Global Moderator
• Posts: 4105
• Location: Austin, TX
##### Re: Cost of a pizza
« Reply #19 on: October 11, 2012, 10:58:13 PM »
The numbers I have run for 14" and 16" pizza dine in with quality ingredients will never make a 20% food cost, much less delivery. Figure closer to 30%, and 35% for delivery.
#### TXCraig1
• Supporting Member
• Posts: 19932
• Location: Houston, TX
##### Re: Cost of a pizza
« Reply #20 on: October 11, 2012, 11:14:39 PM »
Not out of the question at all IMO. I see pizzerias all the time that proclaim "gourmet toppings" and this 20-25 dollars is a common price point. And believe me, their gourmet stuff is not the high dollar products that the menu descriptions would leave you to think they are using if you know what I mean.
It might be or it might not be, but you don't get there by taking your food cost and multiplying by 5. You study your products, the customer, and the alternatives in the market and price each product such that you make the most money on your basket of products.
"We make great pizza, with sourdough when we can, commercial yeast when we must, but always great pizza."
Craig's Neapolitan Garage
#### Chicago Bob
• Posts: 12630
• Location: Durham,NC
• Easy peazzy
##### Re: Cost of a pizza
« Reply #21 on: October 11, 2012, 11:22:44 PM »
It might be or it might not be, but you don't get there by taking your food cost and multiplying by 5. You study your products, the customer, and the alternatives in the market and price each product such that you make the most money on your basket of products.
Yes sir, you are quite right and I was never condoning the % ideal of pricing your goods. Just remarking that folks get away with murder 'round here...and that you can take to the bank.
"Care Free Highway...let me slip away on you"
#### TomN
• Supporting Member
• Posts: 1704
• Age: 57
• Location: Seattle, WA
##### Re: Cost of a pizza
« Reply #22 on: October 11, 2012, 11:35:41 PM »
This is a one topping 16 inch...at 12.99...some say it's a low price, others say it isn't...I think it depends on your area..
Hello lilpizzaplace_ohio,
I see from your photo that your pepperoni is cupping and burning. People in the Seattle, WA area (most people) seem to shy away from it. Your pizza takes me back to when i grew up in Youngstown, OH. thanks for the photo. and the flash back.
TomN
« Last Edit: October 11, 2012, 11:38:04 PM by TomN »
#### TXCraig1
• Supporting Member
• Posts: 19932
• Location: Houston, TX
##### Re: Cost of a pizza
« Reply #23 on: October 11, 2012, 11:48:11 PM »
Hello lilpizzaplace_ohio,
I see from your photo that your pepperoni is cupping and burning. People in the Seattle, WA area (most people) seem to shy away from it. Your pizza takes me back to when i grew up in Youngstown, OH. thanks for the photo. and the flash back.
TomN
I didn't like cupped and charred pepperoni until I had my first pizza with it that way.
"We make great pizza, with sourdough when we can, commercial yeast when we must, but always great pizza."
Craig's Neapolitan Garage
#### La Sera
• Registered User
• Posts: 135
##### Re: Cost of a pizza
« Reply #24 on: October 12, 2012, 11:24:27 AM »
Your goal is to charge the highest price you possibly can.
Yes, you do consider percentages in business. Understanding target ratios in any P&L statement is a pretty fundamental business skill.
You do not blindly charge 5 times cost, you have to first determine what the market will bear, then analyze your product content, work flow, portioning and suppliers to reach for that low food cost. A traditional, dine-in restaurant will have food costs about 1/3 of their menu price. A delivery/carryout (delco) pizza business can indeed run 20% or less food costs.
I make over \$15 on a \$21 pepperoni pizza.
Dough: \$0.70
Sauce: \$0.20
Pepperoni: \$1.25
Cheese (Mozzarella and Gouda mix): \$1.30
Onions, Olives and Mushrooms: \$0.30
Parmesan topping: \$0.10
Spice mix topping: \$0.05
Box: \$0.54
Box insert: \$0.16
So, my food and packaging cost on a carryout or delivery pizza is under 20%.
Of course, there are other variable and fixed costs, but my food costs are definitely under 20%. | 3,991 | 13,930 | {"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-2016-36 | longest | en | 0.931036 |
https://www.snapxam.com/problems/92688226/integral-of-x-0-5-dx-from-1-to-61-25 | 1,719,165,268,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00108.warc.gz | 859,667,318 | 14,350 | 👉 Try now NerdPal! Our new math app on iOS and Android
# Integrate the function $\sqrt{x}$ from $1$ to $\frac{61}{25}$
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## Final answer to the problem
$1.8742679$
Got another answer? Verify it here!
## Step-by-step Solution
How should I solve this problem?
• Choose an option
• Integrate by partial fractions
• Integrate by substitution
• Integrate by parts
• Integrate using tabular integration
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• Weierstrass Substitution
• Integrate using trigonometric identities
• Integrate using basic integrals
• Product of Binomials with Common Term
Can't find a method? Tell us so we can add it.
1
Simplifying
$\int_{1}^{\frac{61}{25}}\sqrt{x}dx$
Learn how to solve definite integrals problems step by step online.
$\int_{1}^{\frac{61}{25}}\sqrt{x}dx$
Learn how to solve definite integrals problems step by step online. Integrate the function x^1/2 from 1 to 61/25. Simplifying. Apply the power rule for integration, \displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}, where n represents a number or constant function, such as \frac{1}{2}. Evaluate the definite integral. Simplify the expression inside the integral.
## Final answer to the problem
$1.8742679$
## Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
SnapXam A2
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tan
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acsch
### Main Topic: Definite Integrals
Given a function f(x) and the interval [a,b], the definite integral is equal to the area that is bounded by the graph of f(x), the x-axis and the vertical lines x=a and x=b | 786 | 2,208 | {"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} | 3.890625 | 4 | CC-MAIN-2024-26 | latest | en | 0.622205 |
https://www.lecturio.com/medical-courses/step-3-measuring-the-pr-interval-nursing.lecture | 1,716,149,549,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00447.warc.gz | 767,560,470 | 31,675 | # Step 3: Measuring the PR Interval (Nursing)
by Rhonda Lawes, PhD, RN
My Notes
• Required.
Learning Material 3
• PDF
Slides Step 3 Measuring the PR Interval Nursing.pdf
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Reference List Medical Surgical Nursing and Pathophysiology Nursing.pdf
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Transcript
00:01 So let's go on to step three.
00:03 Now, we're going to measure the PR interval.
00:06 Now, the PR interval is the time from the beginning of the atrial depolarization, to the beginning of the ventricular depolarization.
00:14 It includes the P wave, and then that short isoelectric line that follows it.
00:20 So that's what we're looking at the PR interval.
00:24 It's the P wave and the short ISO electric line that follows it.
00:28 Hey, before we go on, see if you can pause and remember the first two steps of interpreting ECGs.
00:43 Good. I want to encourage you to do more of that, as you're studying.
00:47 I promise you, that's going to move the information into your long term memory, which is where you need it when you're taking a test, or taking care of a patient.
00:56 So this is step three. Measuring the PR interval, we know that's from the P wave to the short isoelectric line after it.
01:04 It represents the beginning of atrial depolarization, to the very beginning of ventricular depolarization.
01:13 Okay, so take another look at our beating heart.
01:15 We can see that the PR interval represents the time from the beginning of atrial depolarization, to the beginning of ventricular depolarization.
01:24 So from the beginning of the P wave to the beginning of the QRS complex, including that short isoelectric line, that's the PR interval.
01:33 Remember that during the PR interval, the electrical impulse within the heart is traveling from the atria to the ventricles, and it's got that little delay in the AV node.
01:43 That's why there's no electrical activity and the line is flat.
01:47 So when you're measuring the PR interval, keep in mind that the PR interval does vary with heart rate.
01:54 So as the heart rate gets faster, the PR interval gets shorter.
01:58 As the heart rate gets slower, you've got the opposite effect.
02:01 But a normal PR interval is 0.12 to 0.20 seconds in adults.
02:09 Okay, now, just in case, I want to review that.
02:12 How do I know that that is 0.12? Remember, a tiny box is 0.04 seconds, right.
02:19 And a big box is 0.2 seconds.
02:22 So when you measure the PR interval, right, from the beginning of the P wave to just before the QRS and you count the number of boxes that are underneath there, that's going to give you the length or the time that it takes for that impulse to make it through.
02:38 So we know that the normal PR interval is at least 0.12 So what can we learn when a PR interval is less than 0.12? Well hang on because this is pretty cool.
02:50 This tells us that the impulse originated in an ectopic pacemaker.
02:55 Now, it's probably somewhere close to the AV node or the bundle of His and it traveled down an abnormal conduction pathway.
03:02 We call that an accessory pathway.
03:04 Now, this bypasses the AV node and depolarize the ventricles earlier than usual, right.
03:11 That's why that PR interval is shorter.
03:15 Now, Wolff-Parkinson-White is an example of such a condition that we'll have a PR interval less than 0.12.
03:23 So let's look at the opposite example.
03:26 What if the PR interval is greater than normal or greater than 0.20? Well, this might indicate something like a first degree AVblock.
03:36 That means it's taking longer for that electrical conduction to make it through the heart.
03:41 This is a delay in conduction through the atria.
03:44 Now, this can also cause a delay of conduction to the AV node, or the bundle of His.
03:50 On the other hand, what do you think can make a PR interval longer? Long PR intervals are seen in patients with a first degree AV block.
04:00 As I mentioned before, the purpose of the AV node is to delay the impulse coming from the SA node in the atria to the Purkinje fibers in the ventricles.
04:10 In AV block, this pauses exaggerated.
04:14 And the delay is prolonged beyond what the ventricle usually need to finish filling up.
04:19 This is reflected in the ECG strip as a long PR interval.
The lecture Step 3: Measuring the PR Interval (Nursing) by Rhonda Lawes, PhD, RN is from the course The Basics of ECG Strips (Nursing).
### Included Quiz Questions
1. A longer than normal PR interval
2. A shorter than normal PR interval
3. An absent PR interval
4. A PR interval that comes after the QRS complex
1. 0.12 – 0.20 seconds
2. 0.01 – 0.1 seconds
3. 1 – 2 seconds
4. 0.2 – 0.30 seconds
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5 Stars 5 4 Stars 0 3 Stars 0 2 Stars 0 1 Star 0 | 1,249 | 4,782 | {"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} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.930533 |
https://www.convertunits.com/from/quintal+%5Bmetric%5D/to/once+%5BFrance%5D | 1,601,276,952,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401585213.82/warc/CC-MAIN-20200928041630-20200928071630-00537.warc.gz | 763,799,234 | 8,482 | ## ››Convert quintal [metric] to once [France]
quintal [metric] once [France]
Did you mean to convert quintal [French] quintal [metric] quintal [Portugal] quintal [Spanish] to once [France]
How many quintal [metric] in 1 once [France]? The answer is 0.0003059.
We assume you are converting between quintal [metric] and once [France].
You can view more details on each measurement unit:
quintal [metric] or once [France]
The SI base unit for mass is the kilogram.
1 kilogram is equal to 0.01 quintal [metric], or 32.69042170644 once [France].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between quintals and once [France].
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of quintal [metric] to once [France]
1 quintal [metric] to once [France] = 3269.04217 once [France]
2 quintal [metric] to once [France] = 6538.08434 once [France]
3 quintal [metric] to once [France] = 9807.12651 once [France]
4 quintal [metric] to once [France] = 13076.16868 once [France]
5 quintal [metric] to once [France] = 16345.21085 once [France]
6 quintal [metric] to once [France] = 19614.25302 once [France]
7 quintal [metric] to once [France] = 22883.29519 once [France]
8 quintal [metric] to once [France] = 26152.33737 once [France]
9 quintal [metric] to once [France] = 29421.37954 once [France]
10 quintal [metric] to once [France] = 32690.42171 once [France]
## ››Want other units?
You can do the reverse unit conversion from once [France] to quintal [metric], or enter any two units below:
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## ››Definition: Quintal
unit of weight equal to 100 kilograms
## ››Metric conversions and more
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https://www.riddles.com/index.php/good-riddles?page=5 | 1,714,012,056,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00300.warc.gz | 863,057,390 | 41,700 | Riddle: I always follow you around, everywhere you go at night. I look very bright to people, but I can make the sun dark. I can be in many different forms and shapes. What am I?
Riddle: You have 5 kids and you have to get them all into a car. Tommy and Timmy are twins but they fight so they can't sit together. Sarah and Sally fight too, so they can't sit together. Max fights with his sisters so he can only sit by his brothers. There are 5 seats side by side and you have to put them in order. How would you seat the kids, so that everyone is happy?
Answer: Sarah, Tommy, Max, Timmy, and then Sally.
Riddle: What is a ghost's favorite dessert?
Riddle: A cloud was my mother, the wind is my father, my son is the cool stream, and my daughter is the fruit of the land. A rainbow is my bed, the earth my final resting place, and I'm the torment of man. Who Am I?
Riddle: If 5 cats catch 5 mice in 5 minutes, how long will it take one cat to catch a mouse?
Riddle: You can see it every day, But cannot touch it at will. What is it?
Riddle: I am six letters. When you take one away I am twelve. What am I?
Riddle: I'm a god, a planet, and a measurer of heat. Who am I?
Riddle: Why didn't the Mummy have any friends?
Answer: He was too wrapped up in himself.
Riddle: I run in and out of town all day and night. What am I?
Riddle: There is a common English word that is nine letters long. Each time you remove a letter from it, it still remains an English word - from nine letters right down to a single letter. What is the original word, and what are the words that it becomes after removing one letter at a time?
Answer: The base word is Startling - starting - staring - string - sting - sing - sin - in - I
Riddle: Why didn't the monster eat the crazy person?
Answer: He was allergic to nuts.
Riddle: Why did they let the Turkey join the Thanksgiving band? | 477 | 1,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.972756 |
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# City Sketching 2: Line Language In Drawing
Lines are something like alphabets. With these alphabets we can construct not the words but something more similar to hieroglyphs. And we'll be using these hieroglyphs to describe the objects in our sketches.
# City Sketching 2: Line Language In Drawing
Lines are something like alphabets. With these alphabets we can construct not the words but something more similar to hieroglyphs. And we'll be using these hieroglyphs to describe the objects in our sketches.
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2:19
The properties of the line are not very numerous, they are the darkness of the line and thickness of the line.
On this video, I use as examples the imaginary landscapes. They are small and laconic but nevertheless informative.
Now we start to work with the real landscape. And one more interesting thing: here is landscape which includes water surface. In the first video, I'll show preparatory part of the line sketch.
The preparatory part of the sketch is ready, and now we are going to finalize the sketch with the lines only without any shading. We'll be using the line language in case of the real landscape to describe the properties of the objects in it.
The properties of the line are not very numerous. They are the darkness of the line and the thickness of the line. Pressing pencil harder we get a darker line and pressing lighter - lighter one. The thickness of the line depends on the sharpness of the pencil.
But, having this simple tool, we can describe the world around us, giving to a viewer sufficient information to perceive depicted objects. And not only objects but their qualities too. We do it using the line language, wherefrom lines we construct something similar to hieroglyphs.
It is straightforward to understand the meaning of these hieroglyphs because they are built on logic. I'll give you examples of some landscapes to show how it works.
#### Requirements
You don’t need any experience in drawing. And a minimum drawing set: sketchbook, pencils, and eraser is enough.
2:19
The properties of the line are not very numerous, they are the darkness of the line and thickness of the line.
On this video, I use as examples the imaginary landscapes. They are small and laconic but nevertheless informative.
Now we start to work with the real landscape. And one more interesting thing: here is landscape which includes water surface. In the first video, I'll show preparatory part of the line sketch.
The preparatory part of the sketch is ready, and now we are going to finalize the sketch with the lines only without any shading. We'll be using the line language in case of the real landscape to describe the properties of the objects in it.
#### Evgeniy Stasenko
Fine Arts teacher
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Hello, my name is Evgeniy Stasenko. I am an artist from Moscow now living in Barcelona. I graduated from Moscow Pedagogical University, Fine Arts & Graphic Department and have practiced both fine arts and teaching ever since, for more than 30 years by now.
As a teacher, I have always aimed at creating an integral system, a consistent range of methods that would allow me to effectively teach drawing and painting to any person regardless of skills. These methods come as the result of experience; they are tested in practice and proven to give fast results. As well, I have extensive experience in creating specific short-term courses, not to mention fine art workshops, etc.
I wrote some textbooks. “The Art School” and a course of three books on Developing Drawing for primary aged kids are available only in Russian now. My book "Composition of a Picture: Theory and Exercises" is available on Amazon both in Russian and in English. The subject of artistic composition given in a clear and logical way – it is really useful. This theory is unknown outside Russia and English translation helps to improve the situation.
In 1994 I got an award of the All-Russia Exhibition Center for teaching.
In 2006 my biography was published in the encyclopedia Who is Who in Russia.
In 2015 I moved to Spain.
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The video is currently being processed. | 904 | 4,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-05 | latest | en | 0.92332 |
http://mathhelpforum.com/calculus/152769-ratio-root-tests-print.html | 1,506,167,218,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689624.87/warc/CC-MAIN-20170923104407-20170923124407-00544.warc.gz | 222,349,221 | 5,022 | # Ratio and Root Tests
• Aug 4th 2010, 07:29 AM
MechEng
Ratio and Root Tests
Good morning All,
I am working on ratio and root tests for infinite series. I would like to post what I have done, and see if anyone agrees with it. Please see below:
$
\displaystyle\sum_{n=2}^{\infty}(-1)^n\frac{123^n}{n!}
$
By the Ratio Test...
$
\displaystyle\lim_{n\rightarrow\infty}\mid\frac{(-1)^{(n+1)}123^{(n+1)}}{(n+1)!}\frac{n!}{(-1)^{n}123^{n}}\mid
$
And...
$
\displaystyle\lim_{n\rightarrow\infty}\frac{123n!} {(n+1)!} = \lim_{n\rightarrow\infty}\frac{123n!}{(n+1)n!} = \lim_{n\rightarrow\infty}\frac{123}{n+1} = 0
$
Therefore...
The orignal series $\displaystyle\sum_{n=2}^{\infty}(-)^n\frac{123^n}{n!}$ is Absolutely Convergent.
Are we good here?
My second problem is:
$\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})$
I'm not really sure where to start here. Would I apply the root test to this problem?
• Aug 4th 2010, 07:33 AM
roninpro
The first problem looks good to me, and you have the right idea for the second. What do you get when you try to take the $n$-th root?
• Aug 4th 2010, 08:17 AM
MechEng
Thanks!
When I try to take the nth root i get the following:
$\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})$
$\displaystyle\lim_{n\rightarrow\infty}|[(-1)^n(\sqrt{n+1}-\sqrt{n})]^{\frac{1}{n}}|$
$=\displaystyle\lim_{n\rightarrow\infty}|(-1)^{\frac{n}{n}}((n+1)^{\frac{1}{2n}}-(n)^{\frac{1}{2n}})|$
$=\displaystyle|(-1)^1((n+1)^0-(n)^0)|$
$=\displaystyle|(-1)((1)-(1))|$
$=\displaystyle|-1+1|$
$=\displaystyle|0|$
$=\displaystyle0$
That can't be right??? I have three problems; one is supposed to be divergent, one conditionally convergent, and one absolutely convergent. Since my first problem was absolutely convergent (and I was able to follow the entire process for the first problem) I would have thought that the second problem would be either conditonally convergent or divergent.
...so, where did I go wrong?
• Aug 4th 2010, 08:39 AM
Plato
Have you studied the alternating series test? If so that is what is to be used.
• Aug 4th 2010, 08:56 AM
MechEng
I have reviewed the alternating series test.
I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem.
Is this reasonable?
• Aug 4th 2010, 09:05 AM
Plato
Quote:
Originally Posted by MechEng
I have reviewed the alternating series test.
I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem.
Is this reasonable?
Did you notice that $\displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?$
• Aug 4th 2010, 09:18 AM
MechEng
Quote:
Originally Posted by Plato
Did you notice that $\displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?$
Shoot... I did not notice that. Let me try this again...
• Aug 4th 2010, 09:41 AM
MechEng
When I try to take the nth root i get the following:
$\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n}) = \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}$
$=\displaystyle\lim_{n\rightarrow\infty}|[\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}]^{\frac{1}{n}}|$
$=\displaystyle\lim_{n\rightarrow\infty}|\frac{(-1)^\frac{n}{n}}{(n+1)^\frac{1}{2n}+(n)^\frac{1}{2n }}}|$
$=\displaystyle|\frac{(-1)^1}{(n+1)^0+(n)^0}|$
$=\displaystyle|\frac{-1}{1+1}| = \frac{1}{2}$
Am I losing it again somewhere?
Or, are you implying that I use the Ratio Test?
Thanks for the help.
• Aug 4th 2010, 10:10 AM
Plato
Use the alternating series test.
• Aug 4th 2010, 11:24 AM
MechEng
Ok, I am waiting to hear back with respect to whether or not using the alternating series test is acceptable.
In the interim, I have worked out the last of three problems and concluded that:
$\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{(2n)!}{3^n(n!)}$
By Ratio Test...
$\displaystyle\lim_{n\rightarrow\infty}|\frac{(-1)^{(n+1)}(2(n+1))!}{3^{(n+1)}((n+1)!)}\frac{3^n(n !)}{(-1)^n(2n)!}|$
$=\displaystyle\lim_{n\rightarrow\infty}\frac{4n^2+ 6n+2}{3n+3} = \infty$
Therefore...
By the Ratio test, the series Diverges
So, this leaves me with my remaining series that is conditionally convergent. And, this means that if I am able to use the Ratio or Root Test, I should wind up with a limit equal to one... right?
• Aug 4th 2010, 11:31 AM
MechEng
Plato,
I am noticing a trend with the lesson plans that I am following; the questions seem to be quite difficult using the methods covered in that section, but following section will introduce a method that makes the problem quite simple to do. I assume this one is the same.
I am not a huge fan of the way the text book is organized, but i am nearly done with it so it's not worth complaining now.
I appreciate your patience. I used to tutor physics in college, so I know it can be trying when someone keeps mising the fundamental ideas. Thank you for continuing to work with me.
• Aug 4th 2010, 12:09 PM
MechEng
All tests are now available to use on this problem, so I will heed you advice, Plato, and try the alternating series test...
• Aug 4th 2010, 12:22 PM
Plato
You should find that it in conditional but not absolute.
To do the later, use the limit comparison test using $\dfrac{1}{\sqrt{n}}$
• Aug 4th 2010, 05:14 PM
MechEng
Easy question I hope, but...
Since one of the requirements for the limit comparison test is that the series are both positive, does that mean that my original series becomes...
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1 }+\sqrt{n}}$
• Aug 4th 2010, 05:31 PM
MechEng
Please correct me if I am out of whack here...
Let $\displaystyle\sum a_n = \frac{1}{\sqrt{n+1}+\sqrt{n}}$
So...
$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq rt{n+1}+\sqrt{n}}=0$
By Alternating Series Test... Convergent
Let $\displaystyle\sum b_n = \frac{1}{\sqrt{n}}$ A Divergent P-series
So...
$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq rt{n+1}+\sqrt{n}}\frac{\sqrt{n}}{1}$
$=\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\s qrt{n+1}}=0$
By Limit Comparison Test... Divergent
This doesn't make sense, does it?
Would it make ore sense if the order of the tests was reversed? | 2,080 | 6,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 31, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | longest | en | 0.767921 |
https://taxscouts.com/dividend-tax-rates | 1,558,890,477,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259327.59/warc/CC-MAIN-20190526165427-20190526191427-00255.warc.gz | 647,068,009 | 23,231 | # What are the dividend tax rates in the UK?
## How to calculate the tax on dividends
To calculate how much tax you need to pay on dividends, simply:
2. then work out your income tax band
3. you'll pay tax on all your dividends based on what is your highest tax band.
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This year the dividend tax rates are:
• 0% on the first £2,000 from dividends (this is called the Dividend Allowance)
• 0% if your total income is under the Personal Allowance (£12,500)
• 7.5% if you’re a basic rate taxpayer
• 32.5% if you’re a higher rate taxpayer
• 38.1% if you’re an additional rate taxpayer.
## Where the dividends tax can get confusing
Basically:
• income gets taxed "in tranches": the first £12.5k at 0%, then the next £37.5k at 20%, and so on
• dividends get taxed in their entirety based on what is your highest tax rate.
### Example: How Matthew pays tax on dividends
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1. Deduct the Dividend Allowance (£2,000) from the dividend income: £6,500 taxable dividends
2. His income is above £100,000 so his tax allowance has been reduced from £12,500 to £7,500
3. His company will pay tax on his director salary via PAYE: £33,500
4. His income also went over the higher rate threshold so 100% of his taxable dividends will be taxed at 32.5% - the extra tax to pay on the dividends is £2,112.
## How about dividends from investments?
Rules are simple here:
• zero tax on dividends from shares in an stocks & shares ISA
• but: you do pay tax on dividends from any other shares you own, even if these dividends are reinvested into purchasing new shares - ask your broker or trading platform how they handle dividends from "income stocks".
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You are using an outdated web browser – TaxScouts works best with modern browsers. Please upgrade here. | 1,072 | 4,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-22 | latest | en | 0.953563 |
https://www.breakingpar.com/bkp/home.nsf/Doc?OpenNavigator&U=E8F07D43E99FE08786257F9C0053D59E | 1,620,282,881,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988741.20/warc/CC-MAIN-20210506053729-20210506083729-00449.warc.gz | 701,808,519 | 12,475 | Exclusive Or With Strings
Let's say you need to keep track of things that happen on certain days of the week. This is done through a 7 character string where each character is a "0" (didn't happen) or a "1" (did happen). The first character relates to Sunday, the second to Monday, and so on. A string of "0010010" would mean this 'thing' happend on Tuesday and Friday.
Here's an Exclusive Or function that will flip bit's to "1". Exclusive Or says that if either bit is a 1 (in either string) then it's a 1 in the destination. So, if you take the above string above (Tuesday and Friday set) and want to set Saturday, you can Exclusive Or the first string with "0000001" and get the final string of "0010011" which indicates the 'thing' happened on Tuesday, Friday, and Saturday.
Function ExclusiveOr(inpValue1 As String, inpValue2 As String) As String
On Error Goto BubbleError
' Two strings of 0's and 1's are combined in an exclusive or fashion - if either "bit" has
' a 1, the value is 1 and if both are 0's the value is 0.
Dim value1 As String
Dim value2 As String
Dim retVal As String
Dim comp As String
Dim i As Integer
' If one string is shorter than the other, pad THE FRONT of the shorter string with 0's
value1 = inpValue1
value2 = inpValue2
If Len(value1) < Len(value2) Then value1 = String\$(Len(value2)-Len(value1), "0") & value1
If Len(value2) < Len(value1) Then value2 = String\$(Len(value1)-Len(value2), "0") & value2
' If either string has something besides 0's and 1's, return AN EMPTY STRING
comp = ""
For i = 1 To Len(value1)
comp = comp & "[01]" ' This will the the pattern - every character is 0 or 1 and there's "x" of them
Next
If (Not value1 Like comp) Or (Not value2 Like comp) Then
ExclusiveOr = ""
Exit Function ' ============================================================
End If
' Go through every character and compare the same position on the two strings. If both are
' zero, then put 0 in the return string. Otherwise one of the two strings (or both) has a
' one in that position (because we checked the format above) and put a 1 in the return string.
retVal = ""
For i = 1 To Len(value1)
If Mid(value1, i, 1) = "0" And Mid(value2, i, 1) = "0" Then
retVal = retVal & "0"
Else ' One or both is a "1"
retVal = retVal & "1"
End If
Next
ExclusiveOr = retVal
Exit Function
BubbleError:
Error Err, Error\$ & Chr\$(10) & "in function " & Getthreadinfo(1) & ", line " & Cstr(Erl)
End Function | 679 | 2,426 | {"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-2021-21 | latest | en | 0.880579 |
https://www.physicsforums.com/threads/estimating-the-volume-of-a-solid.630976/ | 1,660,009,745,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00004.warc.gz | 835,440,354 | 21,259 | # Estimating the volume of a solid
Philip Wong
## Homework Statement
estimate the volume of the solid
z=-2(x^2+y^2)+8
between the two plates z=4 and z=0
## Homework Equations
In question like this, should I use triple integrals or double integrals in polar coordinates? I'm stuck in between which to use, because the question asks to estimate the volume which suggest a triple integrals. Yet the function gives me a strong feeling that I should use a double integrals in polar coordinates instead. Any suggestions?
Even though I'm not sure have I done it right, I tried solving it using double integrals in polar coordinates. Please check have I done it right, and if I should do it in triple integrals do please give me a guideline of how to do it (as it has not been taught to us yet, but the assignment is due before our next class)
p={(r,θ)= 0≤ r ≤4, 0≤ θ ≤ $\pi$
∫∫$_{p}$ -2(x2+y2)+8=∫$^{\pi}_{0}$∫$^{4}_{0}$
## The Attempt at a Solution
∫∫$_{p}$ -2(x2+y2)+8=∫$^{\pi}_{0}$∫$^{4}_{0}$-2(r2)r Δr Δθ
∫$^{\pi}_{0}$∫$^{4}_{0}$ -2r3 Δr Δθ
∫$^{\pi}_{0}$ [-2r4/4]$^{4}_{0}$ Δθ
∫$^{\pi}_{0}$[-2(4)4/4]-[-2(0)4/4] Δθ
∫$^{\pi}_{0}$-128 Δθ
[-128θ]$^{\pi}_{0}$
=-128$\pi$
## The Attempt at a Solution
a sign of a problem somewhere--notice you dropped the 8 when you did your change to polars. Still, the figure you get is the frustum* of a cone:
double check.
*Also the name of my favorite drink!
EDIT: I misread the problem, sorry. Ultimately,you can use dimensional analysis to
decide if you need a double- or triple- integral. A single integral will do if you have
the general area in the integrand.
Last edited:
Philip Wong
ok, but even if I have the +8 in there, shouldn't still gives me a negative number?
secondly should I keep using polar or should I really use triple integration?
thanks
Homework Helper
May I ask why you use the word "estimate" rather than "calculate"? Because of the circular symmetry, I would recommend using cylindrical coordinates. That is, three dimensions, but using polar coordinates in place of x and y.
Bacle2, this is NOT the frustrum of a cone. That would be true if it were $z^2= 8- 2(x^2+ y^2)$. This figure is a 'frustrum' of a paraboloid.
Homework Helper
Gold Member
## Homework Statement
estimate the volume of the solid
z=-2(x^2+y^2)+8
between the two plates z=4 and z=0
## Homework Equations
In question like this, should I use triple integrals or double integrals in polar coordinates? I'm stuck in between which to use, because the question asks to estimate the volume which suggest a triple integrals. Yet the function gives me a strong feeling that I should use a double integrals in polar coordinates instead. Any suggestions?
Best is a triple integral in cylindrical coordinates. See below
Even though I'm not sure have I done it right, I tried solving it using double integrals in polar coordinates. Please check have I done it right, and if I should do it in triple integrals do please give me a guideline of how to do it (as it has not been taught to us yet, but the assignment is due before our next class)
p={(r,θ)= 0≤ r ≤4, 0≤ θ ≤ $\pi$
∫∫$_{p}$ -2(x2+y2)+8=∫$^{\pi}_{0}$∫$^{4}_{0}$
## The Attempt at a Solution
∫∫$_{p}$ -2(x2+y2)+8=∫$^{\pi}_{0}$∫$^{4}_{0}$-2(r2)r Δr Δθ
That isn't set up correctly. If you are going to use a double integral in cylindrical coordinates the integrand would be ##z_{upper}-z_{lower}##. You have used the equation of the paraboloid, but the paraboloid is cut off at the top by the plane ##z=4##. The upper surface is partly the paraboloid and partly the plane, which is a good reason not to do it that way. Also, ##r## doesn't go from 0 to 4. The largest ##r## is is 2 at the base of the solid. The values ##r## takes on depend on the value of ##z##.
My suggestion would be to try to set it up as a triple integral in the order ##drdzd\theta##. You will have to get ##r## in terms of ##z## for the upper limit, but the rest will be easy.
Philip Wong
Best is a triple integral in cylindrical coordinates. See below
That isn't set up correctly. If you are going to use a double integral in cylindrical coordinates the integrand would be ##z_{upper}-z_{lower}##. You have used the equation of the paraboloid, but the paraboloid is cut off at the top by the plane ##z=4##. The upper surface is partly the paraboloid and partly the plane, which is a good reason not to do it that way. Also, ##r## doesn't go from 0 to 4. The largest ##r## is is 2 at the base of the solid. The values ##r## takes on depend on the value of ##z##.
My suggestion would be to try to set it up as a triple integral in the order ##drdzd\theta##. You will have to get ##r## in terms of ##z## for the upper limit, but the rest will be easy.
ok, I shall give that a try. but how should I do the ##dz## part?
Do I first rearrange the equation to ##z+2(x^2+y^2)-8## and then integrate from there?
Phil
Last edited:
Philip Wong
May I ask why you use the word "estimate" rather than "calculate"? Because of the circular symmetry, I would recommend using cylindrical coordinates. That is, three dimensions, but using polar coordinates in place of x and y.
Bacle2, this is NOT the frustrum of a cone. That would be true if it were $z^2= 8- 2(x^2+ y^2)$. This figure is a 'frustrum' of a paraboloid.
well the assignment question used estimate.
so can you tell me what's the difference between estimate and calculate in the context of this question?
thanks,
Phil
Homework Helper
Gold Member
ok, I shall give that a try. but how should I do the ##dz## part?
Do I first rearrange the equation to ##z+2(x^2+y^2)-8## and then integrate from there?
Phil
Instead of asking me how to do it, show what you think it should be. The dr part would be first. Have you done that? Have you drawn a picture of the figure?
Philip Wong
Best is a triple integral in cylindrical coordinates. See below
That isn't set up correctly. If you are going to use a double integral in cylindrical coordinates the integrand would be ##z_{upper}-z_{lower}##. You have used the equation of the paraboloid, but the paraboloid is cut off at the top by the plane ##z=4##. The upper surface is partly the paraboloid and partly the plane, which is a good reason not to do it that way. Also, ##r## doesn't go from 0 to 4. The largest ##r## is is 2 at the base of the solid. The values ##r## takes on depend on the value of ##z##.
My suggestion would be to try to set it up as a triple integral in the order ##drdzd\theta##. You will have to get ##r## in terms of ##z## for the upper limit, but the rest will be easy.
Instead of asking me how to do it, show what you think it should be. The dr part would be first. Have you done that? Have you drawn a picture of the figure?
yes you are right, I should have tried it first.
I'm not too sure what do you meant by values of ##r## takes on depend on the value of ##z##. But this is what I think you meant, and my attempt to solve it.
##z= -2(x^2+y^2)+8##
##z= -2r^2+8##
∫$^{2\pi}_{0}$∫$^{4}_{0}$∫$^{2}_{0}$(##r cos\theta,rsin\theta,z)r## ##drdzd\theta##
∫$^{2\pi}_{0}$∫$^{4}_{0}$∫$^{2}_{0}$(##r cos\theta,rsin\theta,-2r^2+8)r## ##drdzd\theta##
∫$^{2\pi}_{0}$∫$^{4}_{0}$∫$^{2}_{0}$(##r^2 cos\theta,r^2sin\theta,-2r^3+8r)## ##drdzd\theta##
∫$^{2\pi}_{0}$∫$^{4}_{0}$##[(r^3/3)*cos\theta,(r^3/3)*sin\theta,(-2r^4/4)+4r^2]##$^{r=2}_{r=0}$##dzd\theta##
∫$^{2\pi}_{0}$∫$^{4}_{0}$##[6/3cos\theta,6/3\theta,(-32/4)+16] dzd\theta##
∫$^{2\pi}_{0}$∫$^{4}_{0}$##[6/3cos\theta,6/3\theta,8] dzd\theta##
∫$^{2\pi}_{0}$##[6/3 z cos\theta, 6/3 z sin\theta, 8z]##$^{z=4}_{z=0}$##d\theta##
∫$^{2\pi}_{0}$##[24/3cos\theta,24/3sin\theta,32]d\theta##
∫$^{2\pi}_{0}$##[8cos\theta,8sin\theta,32]d\theta##
##=[8sin\theta,-8cos\theta,32\theta]##$^{2\pi}_{0}$
##=[8sin(2\pi),-8cos(2\pi),32(2\pi)]##
##=[8sin(2\pi),-8cos(2\pi),64\pi]##
##=[0,-8,201]##
This is what I think, but I believe there is something wrong, because the results is giving a set of points rather then a numerical value for volume.
Last edited:
Why are you integrating a generic function of r, θ and z ? Again, think of
dimensional analysis.
My apologies again for unknowingly misleading you by misreading z for z2,
believing the figure was a cone, instead of a paraboloid.
Homework Helper
Gold Member
yes you are right, I should have tried it first.
I'm not too sure what do you meant by values of ##r## takes on depend on the value of ##z##. But this is what I think you meant, and my attempt to solve it.
##z= -2(x^2+y^2)+8##
##z= -2r^2+8##
∫$^{2\pi}_{0}$∫$^{4}_{0}$∫$^{2}_{0}$(##r cos\theta,rsin\theta,z)r## ##drdzd\theta##
∫$^{2\pi}_{0}$∫$^{4}_{0}$∫$^{2}_{0}$(##r cos\theta,rsin\theta,-2r^2+8)r## ##drdzd\theta##
No, no, no! In a triple integral, volume is always given by$$\iiint_V 1\, dV$$The integrand is always 1 and the only problem is the limits.
And I wonder if you even read my first post. Go back and look at it again regarding the limits, ##r## especially. What is ##r## in terms of ##z##? The limits you have put on would describe a right circular cylinder.
unscientific
May I ask why you use the word "estimate" rather than "calculate"? Because of the circular symmetry, I would recommend using cylindrical coordinates. That is, three dimensions, but using polar coordinates in place of x and y.
Bacle2, this is NOT the frustrum of a cone. That would be true if it were $z^2= 8- 2(x^2+ y^2)$. This figure is a 'frustrum' of a paraboloid.
Sorry for hijacking the thread, but I'm curious as to know how do you tell the shape of a solid/curve?
I know common ones are:
Sphere: x2 + y2 + z2 = 1
Ellipsoid/nutshell: (x/a)2 + (y/b)2 + (z/c)2 = 1
I am also very poor at doing problems involving coordinate geometry: parabolas, hyperbolas and ellipses especially..
Any tips on improving my skills?
Philip Wong
Why are you integrating a generic function of r, θ and z ? Again, think of
dimensional analysis.
My apologies again for unknowingly misleading you by misreading z for z2,
believing the figure was a cone, instead of a paraboloid.
what do you meant by think of dimensional analysis?
No, no, no! In a triple integral, volume is always given by$$\iiint_V 1\, dV$$The integrand is always 1 and the only problem is the limits.
And I wonder if you even read my first post. Go back and look at it again regarding the limits, ##r## especially. What is ##r## in terms of ##z##? The limits you have put on would describe a right circular cylinder.
I've read your post prior to my attempt, I think I did mention before my attempt that I do not understands what do you meant "what is ##r## in terms of ##z##" part. Can you please elaborate on that part more, and the part regarding on the limits.
As I said this topic has not been taught yet, but the assignment is due prior to the next class. So my understanding to the topic as a whole is still vague.
Phil
Last edited:
Staff Emeritus
Homework Helper
Gold Member
## Homework Statement
estimate the volume of the solid
z=-2(x^2+y^2)+8
between the two plates z=4 and z=0
...
It seems to me that if you are merely expected to estimate the volume, then this has all been overkill.
For any fixed value of z, from 0 to 4, the resulting equation is that of a circle.
Solving for x2 + y2 gives,
x2 + y2 = (8-z)/2
If z = 0, what is the area of the circle?
If z = 1, what is the area of the circle?
...
If z = 4, what is the area of the circle?
Homework Helper
Gold Member
Sorry for hijacking the thread, but I'm curious as to know how do you tell the shape of a solid/curve?
I know common ones are:
Sphere: x2 + y2 + z2 = 1
Ellipsoid/nutshell: (x/a)2 + (y/b)2 + (z/c)2 = 1
I am also very poor at doing problems involving coordinate geometry: parabolas, hyperbolas and ellipses especially..
Any tips on improving my skills?
I have some old hand-written lecture notes for review of conics (Lecture 7) and how to use that to help you draw 3-D surfaces (Lecture 8). My handwriting isn't that great, but you might find them useful. Here's a couple of links:
[/PLAIN] [Broken]
http://math.asu.edu/~kurtz/LectureNotes/Lecture7.pdf[/URL] [Broken]
[URL="[PLAIN]http://math.asu.edu/~kurtz/LectureNotes/Lecture8.pdf"][/PLAIN] [Broken]
http://math.asu.edu/~kurtz/LectureNotes/Lecture8.pdf[/URL] [Broken]
Last edited by a moderator:
Homework Helper
Gold Member
what do you meant by think of dimensional analysis?
I've read your post prior to my attempt, I think I did mention before my attempt that I do not understands what do you meant "what is ##r## in terms of ##z##" part. Can you please elaborate on that part more, and the part regarding on the limits.
As I said this topic has not been taught yet, but the assignment is due prior to the next class. So my understanding to the topic as a whole is still vague.
Phil
You figured out that ##z = -2r^2+8##. That tells you how ##z## depends on ##r##. If you solve it for ##r## you will have how ##r## depends on ##z##. ##r## goes from ##0## to the ##r## on the surface which is given by that equation. That gives you the upper limit for ##r##.
But, given that this topic hasn't been taught yet, SammyS may have it right that you were just supposed to estimate the volume in the first place.
what do you meant by think of dimensional analysis?
Phil
I mean that , e.g., a volume being a measure of 3-dimensional content should be the
product of either three linear measures, like (Length)(Width)(Depth) or the product
of a 2-dimensional measure by a 1-d measure, like (Area of base)(Height) .
Philip Wong
You figured out that ##z = -2r^2+8##. That tells you how ##z## depends on ##r##. If you solve it for ##r## you will have how ##r## depends on ##z##. ##r## goes from ##0## to the ##r## on the surface which is given by that equation. That gives you the upper limit for ##r##.
But, given that this topic hasn't been taught yet, SammyS may have it right that you were just supposed to estimate the volume in the first place.
Ok, I'm kinda catching on. So z has two conditions, which is z =4 and z=0.
In order to work out the limits of r, I need to rearrange my equation making r the subject.
hence one would be z=0, therefore ##2r^2=8## which makes ##r=2## as its lower limits.
the other one would be z=4, therefore ##2r^2=4##, which makes ##r=√2=1##
and then from here I do my integration, using ∫∫∫##-2r^2+8 drdzd\theta##. Though I still have some doubt that I should integrate using ##-2r^2+8##, but this is the only known equation to do my math
Homework Helper
Gold Member
Ok, I'm kinda catching on. So z has two conditions, which is z =4 and z=0.
In order to work out the limits of r, I need to rearrange my equation making r the subject.
hence one would be z=0, therefore ##2r^2=8## which makes ##r=2## as its lower limits.
the other one would be z=4, therefore ##2r^2=4##, which makes ##r=√2=1##
and then from here I do my integration, using ∫∫∫##-2r^2+8 drdzd\theta##. Though I still have some doubt that I should integrate using ##-2r^2+8##, but this is the only known equation to do my math
No. Almost everything you have written above is wrong or inappropriate for the problem. I think this is unproductive trying to explain it to you when you haven't had the material yet. And you haven't responded to any of the comments asking whether you are really meant to just estimate the volume instead of calculate it. Is English your natural language? Do you understand the difference between estimate and calculate?
Philip Wong
Why yes I meant estimate instead of calculate. And I believe I know the difference between estimate and calculate in terms of English. Though I'm not too sure I correctly understand the difference in terms of calculus.
I'll come back to this question once I had the material taught. | 4,632 | 15,724 | {"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} | 3.890625 | 4 | CC-MAIN-2022-33 | latest | en | 0.858853 |
https://www.answers.com/Q/What_does_roman_numeral_vixii_mean | 1,569,027,743,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574159.19/warc/CC-MAIN-20190921001810-20190921023810-00217.warc.gz | 759,131,108 | 11,343 | # What does roman numeral vixii mean?
VIXII is not a correct Roman numeral as it does not conform to the rules associated with them. VI = 6 and XII = 12 so if VIXII existed it could mean 6 less than 12, which is simply 6. | 63 | 222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-39 | latest | en | 0.96554 |
https://electricalnotes.wordpress.com/2012/11/01/idmt-relay-setting-curves/ | 1,521,849,999,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649508.48/warc/CC-MAIN-20180323235620-20180324015620-00110.warc.gz | 566,099,425 | 23,242 | # IDMT Relay Setting & Curves
## IDMT Relay Setting & Curves
• Calculation of Actual Plug (Relay Pick up) setting of Various IDMT Relay.
• Calculate PSM of Various IDMT Relay.
• Calculate Time setting of Various IDMT Relay.
• Calculate Total Grading Time of Various IDMT Relay.
• Calculate Actual Operating time according to Various IES Curve.
• Draw Various IDMT Relay characteristic Curve.
Jignesh Parmar has completed M.Tech (Power System Control), B.E(Electrical). He is member of Institution of Engineers (MIE) and CEng,India. Membership No:M-1473586.He has more than 16 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He has published numbers of Technical Articles in “Electrical Mirror”, “Electrical India”, “Lighting India”,”Smart Energy”, “Industrial Electrix”(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.
### 7 Responses to IDMT Relay Setting & Curves
i am working in Elelctrical field in last couple of years but i have douts in desining if pannels and capacity of cables and calculating of fault current so please give some idea for that
2. Rbesh says:
Dear sir We are having IDMT relay for transformer protection & transformer rating as follows:1500KVA,11000/433 Volts,78.73/200 Amps,Please explain the calculation of settings in IDMT relay.
3. Elango.K says:
Hello sir,
These days all are static relays and these have stepped characterstic. can you post about stepped characterstic.
4. Sandeep Ayyagari says:
Dear Sir,
We have a Transformer 11 KV/433 V 1500 KVA with 78.73A/2000 A ,Pls tell the calculation for Over Current and Earth Fault Setting we have MC 51 A IDMT relay with Normal Inverse Curve
5. Qaisar Shafi says:
sir, i want to construct my own portable pry injection kit with 200amp capacity kindly guide in this as i want to tesy the ct’s of 400/5amp along eith relays. i know trsnsformer winding also. trf core type and varaic amperage etc. Regards | 591 | 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} | 2.546875 | 3 | CC-MAIN-2018-13 | latest | en | 0.869453 |
https://community.wolfram.com/groups/-/m/t/2293856 | 1,627,271,133,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00282.warc.gz | 195,722,146 | 24,089 | Issue in D[ ] in 12.3.0 but not in 12.1.0
Posted 1 month ago
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Here is the terminal log from 12.1.0, where everything is correct: $math12 Mathematica 12.1.0 Kernel for Linux x86 (64-bit) Copyright 1988-2020 Wolfram Research, Inc. In[1]:=$Assumptions = Element[x,Reals] Out[1]= x \[Element] Reals In[2]:= a=Exp[x] x Out[2]= E In[3]:= D[a*Conjugate[a],x] // ComplexExpand // Simplify 2 x Out[3]= 2 E and here is the log from 12.3.0 which shows the problem: $math Mathematica 12.3.0 Kernel for Linux x86 (64-bit) Copyright 1988-2021 Wolfram Research, Inc. In[1]:=$Assumptions = Element[x,Reals] Out[1]= x \[Element] Reals In[2]:= a=Exp[x] x Out[2]= E In[3]:= D[a*Conjugate[a],x] // ComplexExpand // Simplify 2 x Out[3]= E (1 + I Im[Conjugate'[x]] + Re[Conjugate'[x]]) In[4]:= D[a*Conjugate[a],x] // ComplexExpand[#, TargetFunctions->Conjugate] & // Simplify 2 x Out[4]= E (1 + Conjugate'[x]) Of course, if I did "ComplexExpand" and "Simplify" prior to differentiation, then it would work in 12.3.0 also, but this is clearly a bug, i.e. differentiation should be aware of the assumption that 'x' is real-valued. And in 12.1.0 it was aware, as the log above shows.
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Posted 19 days ago
The D function itself doesn't make use of the Assumptions system, but Simplify does. A few notes:a) It's common to try FullSimplify: D[a*Conjugate[a], x] // ComplexExpand // FullSimplify Although this only gives: E^(2*x)*(1 + Derivative[1][Conjugate][x]) b) Derivatives and Conjugate don't mix well. D by default assumes you mean complex differentiation and there Conjugate can't be differentiated. This shocks a fair number of engineers and physicists who are used to working with it anyway. If you search online, there are a large number of clever and probably too complicated workarounds for this. c) so the real crux of what's happening here is: ComplexExpand[I Im[Derivative[1][Conjugate][x]] + Re[Derivative[1][Conjugate][x]]] I Im[Derivative[1][Conjugate][x]] + Re[Derivative[1][Conjugate][x]] Why doesn't it give 1? My best guess is that it is being conservative and not trying to interpret what the derivative means, which is probably an intentional change.
I am just guessing. Derivatives and integrals in Mathematica were built from the start in the complex variable. Given that, such functions as Conjugate and Abs are not differentiable, and their derivative is left unevaluated. Since version 11.1 we have RealAbs, which does have a derivative, but is not very well blended with the rest of the system yet. Maybe we need RealIntegrate and RealD too. I wonder what RealConjugate may mean. | 792 | 2,641 | {"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} | 2.890625 | 3 | CC-MAIN-2021-31 | longest | en | 0.821906 |
https://videocide.com/glossary/action-axis/ | 1,660,204,829,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00126.warc.gz | 530,722,106 | 16,269 | # Action axis
An imaginary line drawn across two subjects, setting a boundary for the camera angles. The action axis is used to anchor the continuity of action on screen.
In film making, the 180-degree rule is a basic guideline regarding the on-screen spatial relationship between a character and another character or object within a scene. By keeping the camera on one side of an imaginary axis between two characters, the first character is always frame right of the second character. Moving the camera over the axis is called jumping the line or crossing the line; breaking the 180-degree rule by shooting on all sides is known as shooting in the round.
The 180-degree rule enables the audience to visually connect with unseen movement happening around and behind the immediate subject and is particularly important in the narration of battle scenes.
## Examples
In a dialogue scene between two characters, a straight line can be imagined running between the two characters, and extending to infinity. If the camera remains on one side of this line, the spatial relationship between the two characters will be consistent from shot to shot, even if one of the characters is not on screen. Shifting to the other side of the characters on a cut will reverse the order of the characters from left to right and may disorient the audience.
The rule also applies to the movement of a character as the "line" created by the path of the character. For example, if a character is walking in a leftward direction and is to be picked up by another camera, the character must exit the first shot on frame left and enter the next shot frame right.
A jump cut can be used to denote time. If a character leaves the frame on the left side and enters the frame on the left in a different location, it can give the illusion of an extended amount of time passing.
Another example could be a car chase: If a vehicle leaves the right side of the frame in one shot, it should enter from the left side of the frame in the next shot. Leaving from the right and entering from the right creates a similar sense of disorientation as in the dialogue example.
## Reverse cuts
The imaginary line allows viewers to orient themselves with the position and direction of action in a scene. If a shot following an earlier shot in a sequence is located on the opposite side of the 180-degree line, then it is called a "reverse cut". Reverse cuts disorient the viewer by presenting an opposing viewpoint of the action in a scene and consequently altering the perspective of the action and the spatial orientation established in the original shot.
There are a variety of ways to avoid confusion related to crossing the line due to particular situations caused by actions or situations in a scene that would necessitate breaking the 180-degree line. The movement in the scene can be altered, or cameras set up on one side of the scene so that all the shots reflect the view from that side of the 180-degree line.
Another way to allow for crossing the line is to have several shots with the camera arching from one side of the line to the other during the scene. That shot can be used to orient the audience to the fact that they are looking at the scene from another angle. In the case of movement, if a character is seen walking into frame from behind on the left side walking towards a building corner on the right, as they walk around the corner of the building, the camera can catch them coming towards the camera on the other side of the building entering the frame from the left side and then walk straight at the camera and then exit the left side of the frame.
To minimize the "jolt" between shots in a sequence on either sides of the 180-degree line, a buffer shot can be included along the 180-degree line separating each side. This lets the viewer visually comprehend the change in viewpoint expressed in the sequence.
## Style
In professional productions, the applied 180-degree rule is an essential element for a style of film editing called continuity editing. The rule is not always obeyed. Sometimes a filmmaker purposely breaks the line of action to create disorientation. Carl Theodor Dreyer did this in The Passion of Joan of Arc; Stanley Kubrick also did this, for example, in the bathroom scene in The Shining. The Wachowskis and directors Jacques Demy, Tinto Brass, Yasujirō Ozu, Wong Kar-wai, and Jacques Tati sometimes ignored this rule also, as has Lars von Trier in Antichrist. In the seminal film of the French New Wave, À bout de souffle ("Breathless"), Jean-Luc Godard breaks the rule in the first five minutes in a car scene that jumps between the front and back seats, improvising an "aesthetic rebellion" for which the New Wave would become known. When the rule is broken accidentally, or for a technical reason (such as the inability to place a camera physically in the correct position), there are ways for the editor to attempt to hide the mistake. The editor may pre-lap one or two words of dialog before the cut, so that the viewer is concentrating on what is being said and may be less likely to notice the rule-breaking cut.
Some styles used with the 180-degree rule can elicit an emotion or create a visual rhythm. By moving the camera closer to the axis for a close-up shot, it can intensify a scene when paired with a long shot. When the camera is moved further away from the axis for a long shot after a close-up shot, it may provide a break in the action of the scene.
In the Japanese animated picture Paprika, two of the main characters discuss crossing the line and demonstrate the disorienting effect of actually performing the action.
Action axis
also known as
• 180-degree rule
resources
• The 180 Degree Rule, Looking Space and Eyeline Match
• Quick Tips: Understanding The 180 Degree Rule! (video)
• Understanding (and Occasionally Ignoring) the 180-Degree Rule
• The 180 Degree Rule in Film (and How to Break The Line)
source
Adapted from content published on wikipedia.org | 1,243 | 6,006 | {"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} | 2.5625 | 3 | CC-MAIN-2022-33 | longest | en | 0.947578 |
http://metamath.tirix.org/mpests/idpm2idmp | 1,719,315,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00730.warc.gz | 23,858,836 | 5,328 | # Metamath Proof Explorer
## Theorem idpm2idmp
Description: The transformation of the identity polynomial matrix into polynomials over matrices results in the identity of the polynomials over matrices. (Contributed by AV, 18-Oct-2019) (Revised by AV, 5-Dec-2019)
Ref Expression
Hypotheses pm2mpval.p ${⊢}{P}={\mathrm{Poly}}_{1}\left({R}\right)$
pm2mpval.c ${⊢}{C}={N}\mathrm{Mat}{P}$
pm2mpval.b ${⊢}{B}={\mathrm{Base}}_{{C}}$
pm2mpval.m
pm2mpval.e
pm2mpval.x ${⊢}{X}={\mathrm{var}}_{1}\left({A}\right)$
pm2mpval.a ${⊢}{A}={N}\mathrm{Mat}{R}$
pm2mpval.q ${⊢}{Q}={\mathrm{Poly}}_{1}\left({A}\right)$
pm2mpval.t ${⊢}{T}={N}\mathrm{pMatToMatPoly}{R}$
Assertion idpm2idmp ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {T}\left({1}_{{C}}\right)={1}_{{Q}}$
### Proof
Step Hyp Ref Expression
1 pm2mpval.p ${⊢}{P}={\mathrm{Poly}}_{1}\left({R}\right)$
2 pm2mpval.c ${⊢}{C}={N}\mathrm{Mat}{P}$
3 pm2mpval.b ${⊢}{B}={\mathrm{Base}}_{{C}}$
4 pm2mpval.m
5 pm2mpval.e
6 pm2mpval.x ${⊢}{X}={\mathrm{var}}_{1}\left({A}\right)$
7 pm2mpval.a ${⊢}{A}={N}\mathrm{Mat}{R}$
8 pm2mpval.q ${⊢}{Q}={\mathrm{Poly}}_{1}\left({A}\right)$
9 pm2mpval.t ${⊢}{T}={N}\mathrm{pMatToMatPoly}{R}$
10 1 2 pmatring ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {C}\in \mathrm{Ring}$
11 eqid ${⊢}{1}_{{C}}={1}_{{C}}$
12 3 11 ringidcl ${⊢}{C}\in \mathrm{Ring}\to {1}_{{C}}\in {B}$
13 10 12 syl ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {1}_{{C}}\in {B}$
14 1 2 3 4 5 6 7 8 9 pm2mpfval
15 13 14 mpd3an3
16 eqid ${⊢}{0}_{{A}}={0}_{{A}}$
17 eqid ${⊢}{1}_{{A}}={1}_{{A}}$
18 1 2 11 7 16 17 decpmatid ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\wedge {k}\in {ℕ}_{0}\right)\to {1}_{{C}}\mathrm{decompPMat}{k}=if\left({k}=0,{1}_{{A}},{0}_{{A}}\right)$
19 18 3expa ${⊢}\left(\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\wedge {k}\in {ℕ}_{0}\right)\to {1}_{{C}}\mathrm{decompPMat}{k}=if\left({k}=0,{1}_{{A}},{0}_{{A}}\right)$
20 19 oveq1d
21 20 mpteq2dva
22 21 oveq2d
23 ovif
24 7 matring ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {A}\in \mathrm{Ring}$
25 8 ply1sca ${⊢}{A}\in \mathrm{Ring}\to {A}=\mathrm{Scalar}\left({Q}\right)$
26 24 25 syl ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {A}=\mathrm{Scalar}\left({Q}\right)$
27 26 adantr ${⊢}\left(\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\wedge {k}\in {ℕ}_{0}\right)\to {A}=\mathrm{Scalar}\left({Q}\right)$
28 27 fveq2d ${⊢}\left(\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\wedge {k}\in {ℕ}_{0}\right)\to {1}_{{A}}={1}_{\mathrm{Scalar}\left({Q}\right)}$
29 28 oveq1d
30 8 ply1lmod ${⊢}{A}\in \mathrm{Ring}\to {Q}\in \mathrm{LMod}$
31 24 30 syl ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {Q}\in \mathrm{LMod}$
32 eqid ${⊢}{\mathrm{mulGrp}}_{{Q}}={\mathrm{mulGrp}}_{{Q}}$
33 eqid ${⊢}{\mathrm{Base}}_{{Q}}={\mathrm{Base}}_{{Q}}$
34 8 6 32 5 33 ply1moncl
35 24 34 sylan
36 eqid ${⊢}\mathrm{Scalar}\left({Q}\right)=\mathrm{Scalar}\left({Q}\right)$
37 eqid ${⊢}{1}_{\mathrm{Scalar}\left({Q}\right)}={1}_{\mathrm{Scalar}\left({Q}\right)}$
38 33 36 4 37 lmodvs1
39 31 35 38 syl2an2r
40 29 39 eqtrd
41 27 fveq2d ${⊢}\left(\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\wedge {k}\in {ℕ}_{0}\right)\to {0}_{{A}}={0}_{\mathrm{Scalar}\left({Q}\right)}$
42 41 oveq1d
43 eqid ${⊢}{0}_{\mathrm{Scalar}\left({Q}\right)}={0}_{\mathrm{Scalar}\left({Q}\right)}$
44 eqid ${⊢}{0}_{{Q}}={0}_{{Q}}$
45 33 36 4 43 44 lmod0vs
46 31 35 45 syl2an2r
47 42 46 eqtrd
48 40 47 ifeq12d
49 23 48 syl5eq
50 49 mpteq2dva
51 50 oveq2d
52 8 ply1ring ${⊢}{A}\in \mathrm{Ring}\to {Q}\in \mathrm{Ring}$
53 ringmnd ${⊢}{Q}\in \mathrm{Ring}\to {Q}\in \mathrm{Mnd}$
54 24 52 53 3syl ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {Q}\in \mathrm{Mnd}$
55 nn0ex ${⊢}{ℕ}_{0}\in \mathrm{V}$
56 55 a1i ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {ℕ}_{0}\in \mathrm{V}$
57 0nn0 ${⊢}0\in {ℕ}_{0}$
58 57 a1i ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to 0\in {ℕ}_{0}$
59 eqid
60 35 ralrimiva
61 44 54 56 58 59 60 gsummpt1n0
62 c0ex ${⊢}0\in \mathrm{V}$
63 csbov1g
64 62 63 mp1i
65 csbvarg
66 62 65 mp1i
67 66 oveq1d
68 8 6 32 5 ply1idvr1
69 24 68 syl
70 64 67 69 3eqtrd
71 51 61 70 3eqtrd
72 15 22 71 3eqtrd ${⊢}\left({N}\in \mathrm{Fin}\wedge {R}\in \mathrm{Ring}\right)\to {T}\left({1}_{{C}}\right)={1}_{{Q}}$ | 2,129 | 4,444 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 46, "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} | 3.359375 | 3 | CC-MAIN-2024-26 | latest | en | 0.323504 |
http://mymathforum.com/calculus/34292-little-help-electric-field-lines.html | 1,550,567,416,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489729.11/warc/CC-MAIN-20190219081639-20190219103639-00608.warc.gz | 184,665,411 | 9,298 | My Math Forum a little help with electric field lines
Calculus Calculus Math Forum
February 28th, 2013, 06:59 AM #1 Newbie Joined: Oct 2012 Posts: 2 Thanks: 0 a little help with electric field lines How can I represent electric field lines and Equipotential surfaces using vector Calculus??
February 28th, 2013, 10:26 PM #2 Senior Member Joined: Jul 2012 From: DFW Area Posts: 634 Thanks: 96 Math Focus: Electrical Engineering Applications Re: a little help with electric field lines Hi ibungo, Your question is a bit vague. What exactly do you mean by 'represent'? Have you calculated the fields and you want a means of presenting or graphing them, or do you want to know how calculate them? Do you have specific examples or any additional information that you could share?
March 3rd, 2013, 03:30 AM #3 Newbie Joined: Oct 2012 Posts: 2 Thanks: 0 Re: a little help with electric field lines hi jks, What I really want to ask was how to describe electric field and equipotential surfaces in space?? The connections between electric fields and direction lines.
March 4th, 2013, 08:16 PM #4 Senior Member Joined: Jul 2012 From: DFW Area Posts: 634 Thanks: 96 Math Focus: Electrical Engineering Applications Re: a little help with electric field lines Hmm, I honestly think that you need a textbook. I suggest Engineering Electromagnetic Fields and Waves (either first or second edition) by Carl T.A. Johnk. This was my textbook in college, and I thought highly enough of it to buy the second edition years later (my first edition was falling apart). I suggest buying a used one online at a discounted price. Note that the book will cover significantly more than electric fields and equipotential surfaces.
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### electric field lines
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March 14th is Pi Day, since 3/14 is the closest you'll get to the mathematical constant π (roughly 3.14).
I think I'll celebrate with a piece of pie at 1:59 p.m. — Pi Minute— as π reduced to five decimal places becomes 3.14159.
Do you know your Pi trivia? Find out!
Source
### Quiz Nation — Happy Pi Day!
#### The letter representation of Pi began being used by mathematicians in the 1700s. It comes from which language?
Join The Conversation
milosmommy 9 years
Sorry I'm with Liberty...I thought it was cute too. lol...oh well
i-am-awesomeness 9 years
hkdkat 9 years
JovianSkies 9 years
WTF is with that photo?! I learned while I was a waitress that I HATE watching people eat. Nothing is more reviling than watching those with poor manners (or watching children stuff their faces). It's one of my few disliked things about Food Network(and the aforementioned commercials)...what makes them think that watching small children shove food into their gaping mouths, and witness the crumbs, jam, syrup, ice cream, hamburger, or cake in question smothered all over their faces and clothing? Or better yet, when the food is displayed by the child opening their filthy mouth to the camera?! Shouldn't kids have manners? I know I did! :RANT: I'll get off of my soapbox now...
juicylove 9 years
2/3
juicylove 9 years
gross picture:(
harmonyfrance 9 years
Ewwwwww.
Jillness 9 years
Ha, Cine! I literally have to close my eyes when that commerical comes on...just can't handle it! Don't worry Liberty...it got our attention, right? ;)
popgoestheworld 9 years
I seriously thought this was a story about a kid who'd had their face bludgeoned by a sharp object. Also, questioning this statement: "as π reduced to seven decimal places becomes 3.14159" Wouldn't that be five decimal places? It's been a while since I've had math class but it seems like five makes more sense. Anyone?
chipjimi 9 years
Yay!! Got them all!
hausfrau 9 years
I have to say though, I went to a chocolate festival a few months back and they had a pudding eating contest for kids.... it was pretty funny to watch, and i was STOKED that out of the 9 kids selected, the one who won was the ONLY GIRL! hellyea!
cine_lover 9 years
Jillness ME EITHER!!!! I HATE THOSE COMMERCIALS! And Liberty. I hate to break it to you, but you were wrong. What do you find gross?
Jillness 9 years
that picture is so gross! I get really nauseated by nasty pictures of people eating. I can't handle the chili cheeseburger Carl's Junior commerical.
haha liberty it's gnarly.
LibertySugar 9 years
I thought it was cute!!
KrisSugar 9 years
this picture is gross!
cine_lover 9 years
It reminds of the movie "Carrie"
Shopaholichunny 9 years
Yeah that pix scared me! :scared:
hausfrau 9 years
Thats all I could think of when I saw that picture! It is a seriously creepy pic.
cine_lover 9 years
:rotfl: Cabaker I literally laughed out loud when I read that! I look all crazy with this big smile on my face staring at a picture of a creep kid.
redegg 9 years
That picture is disgusting. :sick:
hausfrau 9 years
"Then I ate his liver with some fava beans and a nice chianti."
hypnoticmix 9 years
You know I wish I would have read the head line before I looked at the photo then I wouldn't have nearly spit up my coffee in disgust.
cine_lover 9 years
That picture made me completely lose my appetite. | 887 | 3,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-26 | longest | en | 0.973316 |
https://socratic.org/questions/what-is-the-domain-and-range-of-f-x-x-3x-x-1 | 1,719,076,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00401.warc.gz | 460,136,252 | 6,043 | # What is the domain and range of f(x) = x / (3x(x-1))?
Jan 13, 2018
Domain f(x): $\left\{x \epsilon \mathbb{R} | x \ne 0 , 1\right\}$
#### Explanation:
In order to determine the domain, we need to see which part of the function restricts the domain. In a fraction, it is the denominator. In a square root function, it is what's inside the square root.
Hence, in our case, it is $3 x \left(x - 1\right)$.
In a fraction, the denominator can never be equal to 0 (which is why the denominator is the restricting part of the function).
So, we set:
$3 x \left(x - 1\right) \ne 0$
The above means that:
$3 x \ne 0$ AND $\left(x - 1\right) \ne 0$
Which gives us:
$x \ne 0$ AND $x \ne 1$
Thus, the domain of the function is all real numbers, EXCEPT $x = 0$ and $x = 1$.
In order words, domain f(x): $\left\{x \epsilon \mathbb{R} | x \ne 0 , 1\right\}$ | 290 | 855 | {"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} | 4.625 | 5 | CC-MAIN-2024-26 | latest | en | 0.812134 |
https://courses.ideate.cmu.edu/16-223/f2014/coffeesense/ | 1,716,231,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00745.warc.gz | 157,253,690 | 8,821 | # CoffeesenSe
Group Members: Nkinde Ambalo (Tutor), Temple Rea (Designer) , Zade Delgros (Scribe) Introduction When getting caught up in your work, it can be easy to forget about simple things; your coffee, for example. You make a cup of coffee, set it down to cool, get caught up in your work, and before you know it your coffee is at an awkwardly lukewarm temperature. Boom, time and coffee down the drain. Our team tried to solve this problem by creating a coffee cup that lets you know when it feels neglected. Just a few minutes before the coffee gets too cool, our mug sounds a tone letting you know that it’s time to chug that caffeine before it’s too late. Our system works by making use of a thermistor on the interior of the mug which is a resistor that varies with different temperatures. The resistance of the thermistor increases with decreasing temperature, so when our coffee cools down to the threshold temperature (approximately 50 degrees Celsius), our system is activated and an alarm sounds until the neglectful drinker retrieves the coffee and presses the “deactivate” button on the mug’s base. Video Technical Notes This mug’s circuitry is considered a “one-in-one-out” system where the input is the temperature of the coffee and the output is the alarm that lets you know to drink your still warm, caffeinated liquid. There were two fundamental challenges with the technical aspect of this project: (1) Converting the DC power into AC so that the speaker would work We tackled this obstacle by hooking the DC current and speakers up to a circuit routed through a 555 timer integrated circuit. The way we wired the inputs and outputs of the chip allowed it to be used as an oscillator, which drove the speaker circuit. The values of the resistors and capacitors affected the frequency produced by the speaker. (2) Getting the alarm to sound when at a particular threshold with increasing resistance We were able to make this happen by creating a voltage divider transistor circuit. We used an NPN MOSFET transistor to connect our speaker circuit to Vout and connect our thermistor from the gain to ground. Setting an appropriate 13 kiloOhm resistor as the Vcc to gain resistor, we effectively created a voltage divider that would only supply the 1.8 volts needed for the resistor to turn out with a Vcc of 9 volts. We set it up so higher resistances turned the circuit on and lower resistance turned the circuit off, seeing as we wanted the alarm to sound when the coffee was getting cold (and hence when the resistance of the thermistor was going back up past a certain threshold) Tags: F14 | 568 | 2,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.929172 |
http://robertinventor.com/wiki/index.php?title=FAQ_-_General&diff=20656&oldid=20655 | 1,519,415,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814833.62/warc/CC-MAIN-20180223194145-20180223214145-00460.warc.gz | 301,332,782 | 46,609 | Help for Tune Smithy
FAQ - General
(Difference between revisions)
Revision as of 11:54, 14 July 2008 (view source) (→How to do it)← Older edit Revision as of 11:57, 14 July 2008 (view source) (→The spectrum analysis is reversible)Newer edit → Line 328: Line 328: The spectrum should be reversible - if you add all the sine waves it finds, this should reconstruct the original function. This has surprising implications - the technique used is Fourier analysis - a beautiful area of maths with many surprising results. The spectrum should be reversible - if you add all the sine waves it finds, this should reconstruct the original function. This has surprising implications - the technique used is Fourier analysis - a beautiful area of maths with many surprising results. - You can reconstruct the original waveform in its entirety from the sine waves in the analysis, even for time varying instruments such as percussion or plucked sounds etc. If you add up ALL the sine waves found, you should according to the theory reconstruct the original waveform exactly. + You can reconstruct the original waveform in its entirety from the sine waves in the analysis, even for time varying instruments such as percussion or plucked sounds etc. If you add up ALL the sine waves found, you should according to the theory reconstruct the original waveform exactly. Since the analysis uses sine waves, the resulting waveform repeats - but the repeat period is the same as the length of the original recording. So - to put it more exactly, the reconstructed function gives you the original sample, repeated infinitely many times. This method is used for instance to remove noise from a waveform - if you have tried noise reduction software, it may very well have done it just like this. You convert the recording into a spectrum, remove peaks in the spectrum corresponding to the noise e.g. above or below a particular fequency, then add up the sine waves that remain to get back the original waveform without the noise. This method is used for instance to remove noise from a waveform - if you have tried noise reduction software, it may very well have done it just like this. You convert the recording into a spectrum, remove peaks in the spectrum corresponding to the noise e.g. above or below a particular fequency, then add up the sine waves that remain to get back the original waveform without the noise.
Revision as of 11:57, 14 July 2008
Feel free to add more questions and answers. If you have a question which isn't answered here - for immediate help contact support@tunesmithy.co.uk - and you can also add it to the Questions Pending page.
For old version - auto converted (and partly edited) FAQ - General - auto
How do I save my work?
Intro
FTS has many formats, so you need to decide what you want to save. Many are saved from File | Save As and then you choose the file type there. You can also save the file and e-mail it to someone all in one go using File | Save As & E-mail
Some of the main formats
To save an entire project, use File | Save As Project. This includes just about everything - as well as the fractal tune and midi settings, it also saves your choice of format to display scales, the skin, the colours and so on. You can choose what to save in the Project Options window.
Save scales using File | Save As | "SCALA Scales (*.scl)" - this format was developed by Manuel Op de Coul for his SCALA program, and has become the standard format for exchanging scales.
For fractal tune tasks, to save just the fractal tune itself, save as Tune Smithy Files (*.ts).
The Music Keyboard and Midi Relay Settings (*.rly) saves the settings for midi in and out, and also includes the scale and any other options that affect how notes from midi in are retuned.
The Play from PC Keyboard, Music Kbd && Relay Settings format saves anything that affects the PC Keyboard Player as well as all the midi relaying formats.
There are many other file formats available. To keep the Save As drop menu more manageable normally it shows only some of the most commonly needed ones for the current task. Other formats are available from individual windows.
To show all the formats Tune Smithy recognises, go to File | Recently Visited Files & File Types(Ctrl; + 183) and unselect Short list of file types. This window is also useful as a quick way to navigate between the folders you use most often, and to bring up files that you opened or saved recently.
alternatives to the project save
These have been mainly superceded now by the Project Save. But they may be useful on occasion.
One alternative to the project save, which saves nearly everything, but not your choice of main window task, and some other things like window positions, is to use the Configuration settings (ext..INI)
You can also all your settings as a desktop shortcut from File | Make Desktop Shortcut. This saves a .ini file with all your current settings, also a start_up_etc file (special .ini file with the settings left out of the normal .ini file) and makes a desktop shortcut which you can use later to launch FTS again with those particular settings. This is the same method that's used to distinguish between the various tasks in the tasks folder, and can be useful if you want to make your own shortcuts to launch FTS with different settings, e.g. from the desktop or wherever.
After you have made the shortcut in this way you can move it wherever you like on your computer, or rename it - it will still bring up FTS with the same settings. What distinguishes it from other shortcuts, is the name used for its .ini file, which you can see if you look at the shortcut properties (in the command line field).
Audio recordings
To save your work as a .MIDI file or as an mp3 or such like, you normally do that from the Record to File Options window. You can start and stop the recording from the record buttons in the main window, but go to the Recording Options window to set the file name to save and to set various options. You get to this window from Bs | Record to File Options or go to it using the To File button. in the main window.
See the next FAQ for more details: How do I record and play back recordings?
Fractal tunes are special, if you want to do a MIDI recording from the start of the tune, the best way to save them as as MIDI files from File | Save As | Files of type | Current Fractal Tune as Midi File. This does it all in one go - you don't need to wait for the tune to play. Also the exact desired times are saved rather than the actual times that were achieved during play of the tune. You choose how much of the recording to save (since many of the tunes are very long and for all practical purposes endless) from File | Midi FileOptions | Playing time for fractal tune save.
If this hasn't answered your question be sure to contact support@tunesmithy.co.uk. I know that with the large number of file formats supported there is some potential for confusion, and it isn't always easy to find the information you need in this help - and some things may not even have been documented here yet.
For a more techy account of the file formats, start at: [User_guide.htm#Open_and_Save Open and Save]
How do I record and play back recordings?
Recording
You can show the recording dialog from Bs | Record to File Options or go to it using the To File button. in the main window.
Choose one of the audio formats from the list.
audio formats
MIDI records music as note and instrument numbers. This file format contains no audio data as such, and plays on the listeners soundcard synth or synth / soft synth - much like playing notes on a keyboard synth, indeed you can often play midi tunes on such synths. It usually gives the smallest file sizes, but is highly dependent on the quality of the synth used to play it back.
CSound records it as a CSound Score and orchestra, using the instruments in the CSound instruments for parts window, which you can then render to an audio file using the CSound renderer. To use this option you need to install CSound - if you select this option you will be taken to a page where you can get an installer for a minimal version of CSound suitable for use with FTS if you don't have it already. The listener hears this pretty much exactly as you heard it yourself.
The other audio formats record the sounds you hear using the devices selected in the Out menu. This means that the listener hears the recording pretty much exactly as you will hear it if you play your own recording back
How to record
Set the folder and file name for your recording in the To File , and in current folder fields..
For all audio formats except Midi, be sure to select the recording volume level and choose a suitable source for the recordng - using the Record Vol (WAV) button - your system may be set to record from the microphone for instance. Also be sure you set the fader volume accordingly and don't have it set to 0. This isn't needed for Midi though as the entire recording is done within FTS itself in that case.
You show / hide the faders from Record Control | Options | Properties . If you need more information at this point, or don't see the control when you press the button, go to [troubleshooting.htm#why_no_rec_sound Why is no sound recorded in the audio formats?] for help.
Now do a test recording, maybe record a fractal tune or play a few notes. If this doesn't work, perhaps you need to try another fader - try any suitable looking ones. If it still doesn't work, check to see if you have a half duplex sound card instead of a full duplex one - you need a full duplex soundcard to play and record at the same time. If this is your situation, see [troubleshooting.htm#why_no_rec_sound Why is no sound recorded in the audio formats?]
options
Two options are particularly relevant for fractal tunes: Start recording just before tune starts , and Stop at end of tune . If you select these both, it means that the recording will start when you press the play button in the main window, so you don't need to start the recording first yourself. Also it will stop as soon as you stop the tune.
For the WAV or .MID formats, you can click the main window Rec. or Midi buttons which select the format and start the recording all in one go
If your final recording has a gap at the start and the end of it, then you can remove this in FTS provided the recording is small enough to fit in its entirety in RAM. Open the recording again using Open | Files of type | Wave Sound In the Recording window that pops up, choose Options then Find region to trim - and look out for the option to set how much overlap to leave to either side here (often you don't want to trim right down to the beginning and end of the sound but wish to leave a tiny gap). Then click Trim to highlight .
Another option here is to make the clip in Midi format and then render it in .WAV format using a Midi to Wav renderer. The Virtual Sound Canvas can render midi clips directly in this way. See [index.htm#sq_r_y_qt Roland, Yamaha and Quicktime] for more about the Roland. There are other midi to Wav renderers available too. Of course you can also use this method to render any midi clips you make of your own playing.
Adding support for the latest mp3 encoders, and add Sun Au etc
As installed, FTS can only record in waveform audio, or Midi. However you can add other sound formats by installing suitable helper apps.
You can do this all in one go using a special installer I made for Tune Smithy - see the on-line page: Adding other sound formats to FTS.
You will also find details there about how to add them in separately - e.g. to get the latest encoders. For instance, the quality of sound you get for the mp3 encoders continues to improve as the programmers find ways to improve the encoding.
After you run the installer, or unzip the archive for the helper app, you need to browse to find it from FTS | File | File Associations | More Options so that FTS can find it. However, if you use the all in one installer, this isn't necessary as it installs them into the program's own folder, under standard names, so that Tune Smithy can find them automatically.
Note that FTS will record audio files as .uncompressed wav files first, then it uses the helper app to convert these to the desired format. If you want to remove the original .wav files to save space on your hard drive, you need to delete them yourself later (FTS doesn't do it automatically in case you want to convert the same .wav file into a number of different formats - this lets you do this without the need to re-record it each time). See [#windows_newbies_tip Tip for Windows newbies].
To convert to real audio you use the free Helix Producer from Real Audio. They did an encoder before which worked for all their audio formats. This new version however will only convert to Real Audio 9 format - and to make backwards compatible audio clips for the earlier versions of Real Player you are supposed to buy the Helix Producer Plus - well so it says on the web site, and in the program too when you run it.
Or you can add them in separately.
These helper apps will also be useful if you want to read the various audio formats in FTS, e.g. for spectrum analysis or to find seed notes from the recording (e.g. using File | Open | Files of type | Wave Sound).
For mp3s, download the Latest verion of Lame. This is a console app, i.e. it runs from the MSDOS prompt. However, if you are also interested in converting other .wav files to mp3s yourself, then you may like to get it's windows interface at the same time - RazorLame . You can get both in one go from mp3-tech.org/encoders_win.html. If you use this, be sure to look at my [#razorlame tip for users of RazorLame].
Tip for users of Real Audio : actually the Real Audio 9 clips it makes seem to play fine in Real Audio 8 too. To make sure that it will play them by file association, just make a Real Audio 9 clip, which will have file extension .RM. Then just rename the file (right click | rename in Explorer) and change the file extension to the older.RAM or RA extension, and the older players will probably be able to play it too.
Tip for users of RazorLame: I recommend that you go to Options | Expert | q level , and set this to 0 - to make the highest quality mp3s. The preset value (4) does let you do the compression a bit more quickly, but you won't notice this on a modern machine. FTS uses Lame with a q level of 0. Once you have installed it, you should also update its version of Lame to the Latest verion again, which is worth doing as they keep improving it, making smaller better quality mp3s as they refine the methods..
Tip for Windows newbies: If there are many files in the folder, as there are in the FTS install folder, you may want to know how to find the .wav files in Explorer. One way is to show the details view, and click on the title bar for the type column. Alternatively show the files in order modfied, then they will be the most recently modified files if you have just done your recording. Or do a search of the install folder for files of type "*.wav" - or whatever the file type is that you are interested in.
The list of file formats we have so far seem pretty extensive, with all the SOX supported formats, plus mp3s, Real Audio, and Ogg Vorbis - but there are always more that aren't included. The next section is included in case you come across one of those.
. [#top top], [#audio_formats start of section]
Custom
The Custom option lets you use any other audio conversion command line tool with FTS. You have two custom formats which you can set up as you like. If you have an audio format which you think should be included in FTS already, and a freely available command line tool to do the conversion, do let me know and I may add it to the list: support@tunesmithy.co.uk
To add a custom encoding, make a file called "custom_audio.txt " in your Fractal Tune Smithy folder, with three lines in it like this:
```AU
C:\SOX\SOX.EXE
*INPUT *ARGS *OUTPUT```
(this is already included in FTS as one of the standard formats, so just doing it like this by way of example)
Make sure that these are on separate lines, and in the first three lines of the file.
Line 1: File extension (e.g . AU )
Line 2: Complete path and file for the application to use
Line 3: The parameters to pass to it. In many cases you can just use the line shown here.
In line 3, *INPUT gets replaced with the name of the file to encodeg (i.e. the .wav file), and *OUTPUT with the output file name (same file name with the new extension, in this case . AU ).
The file names are put in quotes, as is normal for the command line in Windows - so don't put the *INPUT or *OUTPUT themselves in quotes.
The *ARGS is used for anything the user puts in the Extra Arguments field in FTS - it is ignored if no extra arguments are supplied.
Here is another example - it shows how you can include necessary parameters as part of the format in line 3. It adds in all the necessary parameters needed to specify the version of the .au format suitable for Java applets:
```AU
C:\SOX\SOX.EXE
*INPUT -U -b -c 1 -r 8000 *ARGS *OUTPUT```
(already included in FTS as the java applets .au save format)
For this one, suppose you want to save your recording as Playalong.au in the Java .au format, and have the usual install folder for FTS and SOX, and so use Playalong as the file name, and supply no extra arguments.
FTS would use this comand line to convert the file:
C:\SOX\SOX.EXE "c:\Program Files\Fractal Tune Smithy\Playalong.wav" -U -b -c 1 -r 8000 " c:\Program Files\Fractal Tune Smithy\Playalong.au"
For decoding make a similar file called " custom_audio_decode.txt ". This time of course, *INPUTgets replaced with the file you want to decode to .WAV, - e.g. the one with extension .au in this case.
This decoding will be used if you use File | Open | Files of type | Wave Sound . to open these files, e.g. for spectrum analysis or to do Wav2Midi work in FTS to find seed notes from the recording.
If the encoding and decoding require the same comand line (with just the names of the input and output files swapped around) then you don't need a separate decoding step - just delete " custom_audio_decode.txt " if present.
If there is no custom_audio_decode.txt , "custom_audio.txt " gets used for both encoding and decoding.
For Custom 2, use the same method but call your files " custom2_audio.txt ", and " custom2_audio_decode.txt "
Alernative way to make mp3s etc - record as Wav then convert
You can also record to waveform audio in FTS and then convert the files later. FTS has a button to do this for them one at a time - Bs | Record to File | Convert from .wav .
Or you can use other programs, to get a wider range of formats.
I recommend the free dBpoweramp- Music Converter - as it adds a Convert option to the right click menu which you can use when browsing for files on your hard disk in Explorer . A nice thing about this is that it is multi-select - you can highlight all the .wav files in a folder in Explorer for instance, then using the right click menu convert them all into your desired format in one go.
dBpoweramp can convert many audio types in the same way - all you need to do is to install the codecs for them from the dBpoweramp - Codec Central page.
Amongst other things this page includes various loss-less compression methods. These can compresses audio files up to four times without losing quality, like a kind of audio "Win zip". Waveform audio doesn't zip well by usual methods - the zip is normally roughly the same size as the original, so this is a great boon if disk space is short or you want to e-mail original recordings to someone else who can play them - or back them up to CD.
Opening / Playing
To play back the recording use the Play recording, Play by Association or Play as web page buttons which appear at the bottom of the Record to File window when you stop the recording.
You can also open audio files in FTS using Temp | Open Audio File in the Record to File window menu - which will show the recording in the Recording Temp window (Ctrl + 71)
One useful option in FTS is to automatically trim the recording, removing silence at the start and end. See the Record (temp) - Options window (Ctrl + 108).
For more advanced audio manipulation etc, e.g. copy / paste editing, varying the volume, removing noise and so forth, use a dedicated audio tool such as Goldwave
There's one area where FTS has special capabilities - analysing the sound to find the spectrum or to get the pitches from the recording. The spectrum gives very accurate pitch determinations for sustained chords or single notes, and the Audio Pitch tracer is particularly good for pure sounds like birdsong. Try out the Analyse Sound and Audio Pitch Tracer tasks in the tasks folder for these options.
Does FTS support 24-bit recording?
Intro
Yes FTS does support 24-bit recording for the recordng to temporary memory, and for the spectrum analysis and to find notes e.g. in bird song.
However, sorry, it doesn't support this format for the normal Record to File window. FTS uses routines there that can only handle 8 and 16 bit recordings.
How to do it
24-bit recordings can be played on a PC if it has a 24 bit sound card. You can play them using the most recent versions of WinAmp, and you can make the recordings using some software packages.
If you want to record music in the 24-bit format, you need a suitable sound card of course, and you will need to use another program for the recording itself. Most dedicated audio recording programs will these days, such as goldwave. Or you can use the Recording Temp window (Ctrl + 71) in FTS which supports 24 bit recording.
Most of the synths and soft synths you can use with FTS play the notes in 16-bit format. In many cases, these may just as well be recorded in the same format. It may be worth recording in 24-bit if your synth or sound card plays Midi notes in this format, or if you want to combine quiet tracks with loud ones in the same piece of music.
Background:
CD players can only play in 16 bit. 24-bit is a standard for a new type of recording.
You can play a 24-bit recording on a computer if it has a sound card which supports 24-bit. You need to use a suitable program to play it such as the most recent version of WinAmp.
This format will probably eventually replace standard audio CDs in general use. Proably they will be recorded as Audio DVDs, - but there are no widely available players for it yet, so no-one yet knows for sure how it will work.
Why are audiophiles so keen on 24-bit recording?
24 bit recordings are needed for ultra high fidelity recordings of quiet background sounds. Examples include quiet sounds that instruments make along with the notes - like guitar fret noise - and quiet instruments played at the same time as loud ones - even quiet passages if the music varies considerably in volume - as is common in orchestral music for instance.
Normal 16-bit recording will record very quiet passages using smaller numbers. These will often use 8 of the available bits or less, so the quality of recordings for ultra-quiet sounds and passages may be extremely poor in 16-bit recordings. It is equivalent in quality to an 8 bit recording played quietly.
An equally quiet section in a 24 bit recording is equivalent in quality to a 16 bit recording played quietly, so is much higher quality.
The old LP is considered by some to still be a superior format to the CD for music with ultra-quiet sections in it.
Can you suggest a computer to get for microtonal music making?
Intro
Okay, - this isn't a question about FTS particularly, but it is one that I get asked by FTS users. Musicians who have ventured along this road already, or who run recording studios will have special requirements - and I won't try to address those. However perhaps I can say a few things that will help a newbie music maker, for instance if you have never used a computer for music making before.
soundcard
The main thing to look out for is the sound card. Some computers, especially laptops, only have on-board sound and there may be no way to install a new soundcard. However nowadays there are many USB soundcards you can use instead, so long as it has USB.
Also it is useful if the [faq_mus.htm#midi_in the sound card has Midi In], unless sure you don't need this. It is needed for standard midi keyboards, and also for other midi input devices you can get, such as wind controllers for wind players, and [Scales_and_Fractal_Tunes.htm#microtonal_instruments specialist keyboards for microtonal music making].
Again, this is somewhat less important nowadays, as you can get USB midi devices such as the UM1-SX to add midi in / out to any computer.
If you want your soundcard to play GM standard midi files such as the ones you can download from midi file web sites, you need one with a GM synth - GM is a midi standard that was developed so that musicians could make midi clips that would sound approximately the same on any GM midi system, e.g. instrument 0 is always an Acoustic Grand Piano (of some description).
Most sound cards can play GM, and if not, you can install a GM soft synth in its place. So this isn't a big deal.
If you use midi only as an intermediate step to make audio clips for others to listen to, as many musicians do, then one doesn't need a GM synth.
Note that you can add a second sound card, or upgrade it - so you don't need to make the best possible decision at this point. This is just to get you started. The soundcards in the SoundBlaster series are good entry level cards as they let you use sound fonts, and come with a decent sound set. I started with one of them - the Sound Blaster Live! (now no longer in production). They are still a fine choice for ones first sound card.
Your soundcard can also be used with a sample player such as Giga or Kontakt to play virtual orchestras (e.g. Garritan Personal Orchestra etc).
There are so many soundcards available these days. Also my strong point is the software, I'm not up to date on hardware at all.
To find out about the most recent sound cards that get recommended, take a glance at the latest issues of the music magazines, such as Computer Music or Sound on Sound.
Operating System
For the operating system - XP is fine for this. Some musicians still use Win 2K or Win 98, which may be preferrable if you have a very old machine.
Vista from a technical point of view seems superior to XP for sound recording - but at present it is still a very high powered operating system which may stress your computer - make sure it has a good Windows Experience Index wikipedia - Windows System Assessment Tool
The main technical difference between these is that Win 2K has no resource limitations - while if you run many programs at once inWn 9x, or show many windows with lists and drop lists in them, then eventually you reach a point where you run out of resources, and can't run any more - in fact the operating system usually freezes at that point if you ignore the earlier warning messages.
Windows XP has better multimedia support than Windows 2K and can run many Win 98 programs - but not necessarily all of them - there are small changes in the operating system. FTS will run fine in either. Also in ME - I haven't tested it in ME myself but it should be fine as FTS gets used in this operating system with no ME specific bug reports yet.
If using Windows 9x, make sure that you have the minimum of other programs running, as this is the most usual cause of crashes in Windows 98. See [midi_in.htm#Trouble_shooting Trouble shooting] for some tips on this.
computer speed
The speed of the processor isn't much of a consideration for ones first music computer. If you use FM synthesis (such as the FM7) then the faster the computer the more polyphony you can get - but on the latest Pcs the polyphony you have there is high anyway. Certainly it's not a case where a twenty percent improvement of speed or the like will be noticeable - as it is for PC gamers.
Giga works best with the fastest hard disks - so it used to be recommended that you have a dedicated hard disk for it. However hard disks are now so fast anyway - a modern one should be plenty fast enough with no special considerations for sample players like Giga for ones first music computer.
quiet PC
So I think the only other consideration is the noise your computer makes. If you are about to get a new computer, then this is the best time to decide whether you want it to be a quiet PC Musicians usually like to have a silent PC - and I gather that you can get one that is so quiet you can't even hear the sound it makes in a recording studio.
It. needn't cost much either - for instance, a quiet hard disk will cost just a bit more than a noisy one, it's just that most people who buy hard disks will buy the slightly cheaper but noiser one. Similarly, you can get quiet fans, and a quiet power unit - which is one of the really noisy units n the computer - and quiet graphics cards, and quiet cooling units for the CPU.Then - no more background hum when doing computer music work :-).
This is also relevant if you want to use your PC to analyse the frequency spectrum of an instrument - the background hum will add extra peaks to the spectrum. Another reason for getting a quiet PC is that the background noise may have distinct pitches to it, which can beat with the chords that you play - this can be confusing when listening to just triads, and trying to hear when they are beat free - they may always have beats with your PC fan etc.
SOME OF THESE LINKS MAY BE A BIT OUT OF DATE NOW
See the article How to silence a computer from Quiet Pc
For those building a quiet PC themselves, this article looks as though it could be helpful: Building your own PC for music. See also the previous article on sound cards and motherboards - Sound on Sound | Articles | August | PC Musician - they are from summer 2001 however, so won't describe the most up to date gear available..
Then, for independent reviews and foruims: Silent PC review. See also the links in The silent PC
Or just do a web search for "quiet Pc" - plenty of material out there..., problem probably will be choosing where to start :-).
For ones first music headphones, the Grado Labs headphones (such as the SR60) are quite a find - if your priority is the best quality of sound for the cost, rather than the appearance. Or for an even lower cost and still with very high quality sound i-Grado Portable Headphones. Some users find them uncomfortable however, I think that's about the only criticism you get of them apart from the appearance.
music keyboard
For the music keyboard, see What should I look out for in a music keyboard?.
swap file to speed up computer
Some musicians set up a separate partition for the swap file - the idea of a swap file is that sometimes you may have more programs running, or audio files in memory than you have the memory available for, so Windows swaps some of them out to disk - and sometimes may swap them out anyway to keep memory free.
Your swap file may run a little faster if you keep it unfragmented - and if you do it in a separate partition all to its own like this, it will never get fragmented. You specify the swap file location from Start | Settings | Control Panel | System | Performance | Virtual Memory | Let me specify my own virtual memory settings in Windows 9x. Don't worry about the alarming message that pops up to check if you really intend to change this advanced setting - it will be fine - so long as you have space available for it on that partition. I use Partition Magic to make new partitions - highly recommended..
However if you have a brand new up to date computer, these speed issues are probably not nearly so relevant as they were a few years ago. You have to be a pro. stretching your software to push a modern computer to its limit, and if that describes your situation, you wouldn't have needed to read this FAQ in the first place as you surely know it all already :-).
How do I analyse a note played in FTS to find the actual pitch played on the sound card / synth / soft synth?
Intro
This spectrum analysis option is designed for finding the pitches as exactly as possible, to satisfy the demands of microtonalists. If the recording is of a sustained chord of a few seconds, and you are able to use the mean value method, it may be able to measure the pitch very exactly indeed.
If one wants a general purpose spectrum analysis tool, a good free tool is [wavanal http://www.hibberts.co.uk/wavanal.htm#wavanal]. There are many ones available for purchase.
How it works - the theory
Intro
The sound spectrum analysis finds the frequencies that can be used to construct the waveform. It doesn't attempt to find out what frequencies a human will hear when you listen to the waveform.
For instance, we hear sum and difference tones when two pure frequencies are played. E.g. if you play a 1/1 and a 3/2, then you will hear a distinct and pure sound an octave below the 1/1. There is a periodicity in the waveform at that frequency indeed, which the human ear picks up. but the fourier analysis doesn't pick it up as the analysis only finds the frequencies needed to reconstruct the waveform.
If you make a waveform out of sine waves, and choose a reasonably long length for the clip, then put it through a fourier analysis, it should come up with just the frequencies you used to construct the clip and no more.
The spectrum analysis is reversible
The spectrum should be reversible - if you add all the sine waves it finds, this should reconstruct the original function. This has surprising implications - the technique used is Fourier analysis - a beautiful area of maths with many surprising results.
You can reconstruct the original waveform in its entirety from the sine waves in the analysis, even for time varying instruments such as percussion or plucked sounds etc. If you add up ALL the sine waves found, you should according to the theory reconstruct the original waveform exactly. Since the analysis uses sine waves, the resulting waveform repeats - but the repeat period is the same as the length of the original recording. So - to put it more exactly, the reconstructed function gives you the original sample, repeated infinitely many times.
This method is used for instance to remove noise from a waveform - if you have tried noise reduction software, it may very well have done it just like this. You convert the recording into a spectrum, remove peaks in the spectrum corresponding to the noise e.g. above or below a particular fequency, then add up the sine waves that remain to get back the original waveform without the noise.
Extra peaks and windowing
You may notice extra peaks in the spectrum, espeically if you analyse a very short or time varying recording.
When the aim is to analyse a sound spectrum, e.g. to try to find the pitch of the notes - then the extra peaks are undesirable - they confuse the results. So, the normal thing to do is to remove all the extra peaks caused by time varying pitches and the short length of any sample. This is done by "windowing", which emphasizes the pitch at the middle of the sample.
How to do it
Go to {hi|Tasks | Analyse Midi Voice}.
Click the {hi|Standard Settings}button. Make sure you have {hi|Auto record and analyse }selected (standard setting).
Choose the voice you want to analyse from the {hi|Voice}menu. Set a value for {hi|How much to analyse}- say 20 seconds. You may get asked if you want to increase the maximum amount of memory to allocate for the recording - the entire recording is kept in computer memory so a limit is set on this. You can also go to {hi|Freq. Analysis | Times... </span>to set this value yourself to a high value straight away to avoid seeing those messages again.
Now set how long you want each note to be in the {hi|Note (secs)}box. For a continuous note on strings / wind etc set this to a large number, say 10000. For plucked or percussive instruments, you may want to set it to a small number such as one second so you get repeated notes, so more data to analyse - however this will also add extra peaks to the spectrum - see [#artefacts artefacts].
Now click the play button, and play the note for a while, say 20 seconds. It gets automatically recorded.
When you click Stop, the note you have just played gets automatically analysed. The {hi|Find Freq.} button changes to show {hi|Halt Search} at this point, for a moment or two. When that changes back to {hi|Find Freq.}the analysis is finished. Now you can use {hi|Show Freq.}to show the frequency spectrum, and {hi|Show partials as text file} to show the list of partials.
Scroll down to the bottom of the list of partials (tip: the Windows shortcut key to scroll to end of a text field is {hi|Ctrl + End}) - and you'll see something like this:
```Closest concert pitch note to lowest fequency C
Deviation of lowest frequency from concert pitch: -0.373259 cents
```
```From 1/1 in pitch window (Midi note 60): -0.373259 cents
```
This shows that the note was measured to be -0.373259 cents flatter than concert pitch - rather small for a wave table synth, though large for an FM synth.
To get a first idea of the accuracy of the result, look out for the {hi|FFT bin size in cents at the fundamental}, which you will see a bit earlier in the {hi|Show partials as text file} results. The accuracy is usually a lot better than that - but it depends a bit on the spectrum - as we will see in a bit.
In the same way you can measure the pitch of the middle c for any of your synth voices.
Tweaks
It is useful to have a look a the spectrum from time to time using {hi|Show Freq}to see what you are measuring.
This is the recorder voice of the Roland Sound Canvas, a 60 secs clip, at middle c. It's a log plot, which shows finer details of the spectrum more readily, and I have cropped the botom of the figure, which is blank here in the log plot.. In this particular one I also used the preset windowing of {hi|Blackmann}from {hi|Freq. analysis | Pitch detection | Detection of partials in the spectrum ( </span>[tips.htm#tip_49 {hi| shortcut to show the window}] {hi|: }{hi|Ctrl +Alt + L} {hi|).}You can try various windows there - pitch detection may be most accurate if you set it to none - but you will get a cleaner looking spectrum with windowing.
To show the spectrum like this, just click on the {hi|Show Freq.}button after you do the analysis.
I have clipped it at about 200 hz at the left. This can be useful to do as you sometimes get a few weak partials below the expected pitch, e.g. often there may be one at around 130 Hz for a note expected to be concert pitch c, and one may wish to just ignore those here..
You can clip the spetrum to left and right from {hi|Freq. analysis | Options | Frequency spectrum Layout} ([tips.htm#tip_49 {hi|shortcut}] {hi|: }{hik|Ctrl +Alt + Z} {hi|)}. Set the {hi|Min Freq. to find and to show} to a little below the note you are measuring, e.g. if you are measuring a note at concert pitch C, set this to 200 Hz, say.
If you need to re-position a partial by eye, or add or remove partials, this is done by using {hik|Ctrl + Right click} to remove a partial and {hi|Ctrl + left click} above a peak to add it as a partial, with other options to fine tune the position of the partials which you get to from {hi|Freq. analysis | Options | Frequency spectrum Options}
If some of the peaks get missed out - then you need to change the sensitivity for the peak detection. Go to {hi|Freq. Analysis >> Pitch Det. >> Find Peaks in the Spectrum}. Then you can vary things in the lower half of this window.
First try the {hi|Peak Max Width}. Reduce the width, say to 10 Hz, to find more peaks, or increase the number of decibels, say to 80. You can see the effect of your changes by choosing to show the peak detection curves from the {hi|Layout}window. ( {hi|Options | Freq. Spectrum Layout} - or the button in this window). The idea here is that if a peak is within the peak detection curve of another peak, it gets left out. So, decreasing the width makes these curves narrower, which means you detect more peaks. The width refers to the width at the chosen difference in decibels, so for example, a width of 40 Hz at 20 Db means that the peak detection curve is 40 Hz wide at a point 20 Db quieter than the peak.
So vary these parameters, and use {hi|Refind Peaks}and see what happens.
Also try unselecting {hi|Only detect larger fluctuations in the spectrum}. If that makes a difference, you can fine tune those settings, or just leave it unselected if it isn't needed. It is particularly needed when the frequency spectrum has a fair amount of noise in it, so with small peaks you don't want to detect, also FFT artefacts from repeated notes and so forth..
There are two other values to vary which set a background noise level to ignore - either dependent on the maximum valaue of the spectrum (preset to 0.125 percent), or the average sound level. (preset to 0.125 i.e. one eighth of the average sound level in the spectrum).
Finally, as a useful check of what you are measuring, you can try the wave count method of finding pitch. This is extremely accurate for suitable voices, can be better than the frequency spectrum method. On wave table synths, the Ocarina is often particularly suitable for the wave count method. See [#find_pitch How do I find the pitch of a note (second method)?]
Wave countint is a completely different approach to it, so if both methods show the same frequency it is a pretty strong confirmation that it is indeed the pitch of the note.
So now you are all set to go on and measure other notes, and chords - if you want to do this straight away, skip on to the [#chords chords] section - and perhaps see you here again later :-)..
Accuracy
To get a better idea of the accuracy, let's look at a spectrum:
You can click {hi|Find Freq.}to refind the spectrum for the current recording - saves re-recording it. You can show the recording itself, if you wish, from {hi|Freq. Analysis | RAM recording </span>- and save it in .wav format from {hi|RAM recording | Options </span>, and indeed you can also record it using a microphone, or open an audio file and use that to find the spectrum - Have a look through the [analyse_sound.htm {hi|Analyse sound</font>] page to find out more about the other options - it would be something of a digression to go into it all here..
Now zoom in on one of the peaks. To do that, click and drag to highlight the region you are interested in. A details window will pop up. You can use the zoom in / out menu and the left / right menu options for this window to look closely at any point. Or within the detail view, you can use {hi|Shift + Left click </span>and {hi|Shift + right click </span>to crop it to a smaller region.
This shows a detailed view of the very tip of the leftmost partial (it is stretched out considerably to left and right - the curve you see sloping down towards the edge of this view is just the beginning of the steep curve that makes the peak).
The red line shows a region of one cent above and below the frequency of the measured peak. So we see the measurement is to within considerably better than one cent accuracy. The {hi|0.01051}shown here is in herz, and refers to the bin size - which is the spacing between the data points that make up the spectrum
The data points in a FFT spectrum are at a constant separation in herz - known as the bin size. It depends only on the number of data points in the original recording and the length of it in seconds. Sinced it is a fixed value in hertz, this means that the cents values get smaller as you go higher in pitch. At middle C, 0.01051 herz is the same as 0.07 cents. You can choose to show the values in cents from the {hi|Frequency spectrum Layout}{hi|Ctrl +Alt + Z </span>
So how accurately is the pitch measured. One would think, maybe to the nearest data point. But actually one can do rather better than that - the process of doing so is called peak interpolation. Often these get you to within maybe a third or even a tenth of a bin size when you have a nice clear spectrum. So this measurement we have just found might well be accurate to within maybe 0.02 cents, possibly better. It's rather hard to estimate how good it is actually.
Here is a really close zoom in. I've also set it to show it as a bar chart, which you can do from the layout window:
See how the peak is actually made of two data points exactly the same height. If one went for the highest data point, one would have to choose one or other of these (the data points are at the mid points of the bars). However the peak interpolation algorithm has instead gone for the mid point between the two. We can see that this is clearly a more sensible choice for this peak than the mid points of the bars to either side!
One could position the point by eye too and that is a legitimate thing to do actually as this is irregular real life data. An expert can often beat a computer program at analysing such things by eye - in fact, programs sometimes get written which get improved by using feedback by experts (this is a whole area of study in itself in programming, called expert systems).
In this case the partial is nice and clear however, so I don''t know whether an expert could improve on it. Maybe the original computer program can be expected to be better for nice clear cut examples like this one.
There are several peak interpolation methods you can use, which you can select from {hi|Freq. Analysis | Pitch Det. | Find Peaks in the Spectrum </span>.
{hi|Barycentric}, {hi|quadratic}, {hi|Quinns first}and {hi|second estimator}, and {hi|Jains method}are all three point methods - that means they just use the highest data point of the peak and one point to either side to estimate the position of the peak.
{hi|Quinns estimators}require one to use Complex FFT. You need to unselect the {hi|Real FFT}check box at the bottom of this window. Then when you re-find the spectrum, you will find it is a bit slower (probably take roughly twice as long). What Complex FFT does is to take account of the phase of each wave - where it is in its cycle compared with all the other ones. The spectrum itself will look identical, but during the calculation the phase component of the specturm will have been found too, and this is needed by Quinns estimators.
{hi|Technical information (skip if not interested): </span>Complex numbers are ones that can use the square root of minus one, and they happen to be especially suitable for finding the frequency spectrum phase information. In fact for those who know a bit about complex numbers, the phase is recorded as the angle the complex number makes with the real axis, and the amplitude is recorded as the magnitude of the complex number (as the square root of the sum of its real and imaginary parts).
I find {hi|Jain's method}is best of all of the three point methods for musical notes - but each has its good points, so see what you find for your own special data.
Three point methods are pretty standard in this field I gather. However, why stop at three points. In real data like this with irregularly shaped peaks, maybe one should look at more than three points to either side? That's where the mean value idea comes in. It's my own idea and I haven't seen it mentioned anywhere else yet as a technique for frequency spectrum analysis (though surely someone else will have used it - it is nearly always the case that someone has).
Anyway, it works like a charm, and seems to get even more accurate pitch measurements than the others for most musical data, which is what we are interested in here.
The idea is to find the sum of the products of the amplitudes and the frequencies over a region to either side of the peak - and then divide the result by the sum of all the amplitudes within the region. This gives you the mean frequency for the region.
Okay maybe that sounds a bit complicated but it isn't really. It is just like finding the mean of a population in statistics. Here is a [analyse_sound.htm#mean_value slower and more detailed explanation]. It is suitable if the peak is reasonably symmetrical, as the ones for most partials are, for musical notes anyway.
In fact, this option is now the standard setting - because I find it works better than the others in a wider range of situations, for musical notes. But, be sure to try the others too and see what you think for your particular data.
The partial in all these pictures was actually found by the mean value method. Rather than use just three points, as is usual for peak interpolation for frequency analysis, it used perhaps dozens of ponts to either side of the peak. It works well because though the peak is irregular, it is by and large symmetrical, and this point is at the very top and centre of it.
You'll see some other options on the drop list here - you can position the peak yourself using the mouse, and use that to estimate the frequency by eye, or both the frequency and the amplitude of the partial (vertical height).
See [analyse_sound.htm#FFT_freq_detection Finding peaks in the spectrum] for some of the other options in this window.
Artefacts
First a note about how the frequency spectrum works. The method used is called FFT which means Fast Fourier Transform. Anyway what is special about it is that you can take a finite set of data points for a wave, and convert it into a frequency spectrum with the same number of points. So the more points the wave has (the longer and higher resolution the sample is) then the more accurate the spectrum.
However, a rather extraordinary thing about it is that you can reconstruct the original waveform exactly from the FFT. You need the frequency spectrum and you also need the phase information for all the frequencies as well - this means you need to use a Complex FFT (as it is called) which finds both of these simultaneously. Then you can do this magic trick of recovering the waveform exactly - all the data points in place - from the frequency analysis.
For this to work, the frequency spectrum has to contain information about everything in the original waveform, including things one might not normally think of as frequencies.
For instance, if you play a note once a second - that counts as a frequency. If the notes have vibrato, that also counts as a frequency. If a note starts and stops suddenly, even if it is just a single note, that also counts as a frequency. So, these are the artefacts - they properly belong in the spectrum, the analysis is correct, but they are artefacts because they aren't the things we are interested in.
One might think a note played once a second wouldn't matter because it would be way to the left of the frequency spectrum - and ditto for vibrato. But, the thing is that the frequency spectrum will analyse it, and find not just that frequency, but all its overtones too (any frequency has overtones apart from a perfect sine wave).
Here is a frequency spectrum for the Roland recorder voice again, this time with a pitch vibrato added to it of 10 cents at 2.5 herz .
Those extra peaks to either side of the main peak are caused by the pitch vibrato. This time the red line is 10 cents. There are similar peaks around the higher partials, but closer together. Don't ask me to explain the details here - I'm no expert on this - but observationally, this is what happens if you do an FFT of a voice with vibrato.
You can add pitch vibrato yourself at a precise frequency in this way from {hi|Parts | More | Pitch Var </span>.
Repeated notes give extra peaks too - look pretty similar to thes.
The sudden stop at the start and end of a clip can also cause artefacts unless you so arrange things that the time for teh clip is an exact multiple of all the frequencies - since most FFT requires the clips to also be of particular lengths only (power of two for the number of samples) this will only happen seldomly.
Usually one uses a so called {hi|windowing}function. The simplest is triangular windowing - you add a fade in at the start of the recording, all the way up to a maximum at the middle point of the recording - then you fade it out at the end similarly. This fade gets added in to the data just before it gets sent on for the frequency analysis - the recording itself doesn't get changed of course, though you could alternatively do it that way too.
You can try the effect of various windowing functions from {hi|Freq. Analysis | Pitch Det. | Find Peaks in the Spectrum </span>
I find that Windowing has less of an effect than vibrato or shorter repeated notes for typical samples for this work of say thirty seconds or so. The main effect is a decrease in the background continuous part of the spectrum - i.e. windowing removes some "noise" from the spectrum. You may also see artefacts if you look at the spectrum in fine detail, and depending on the length of the clip and other details.
Here is the Roland recorder voice again to show the effect of no windowing:
What you see here is very dependent on the exact frequency of the note, the length of the recording, and whether you include the attack of the note in it (as windowing naturally will smudge the attack away since it starts the notes). Some choices may lead to few windowing artefacts if any while others may have more obvious artefacts.
To find out more about FFT in its many aspects, try some of the links to sites on the FFTW links page.
Bells, and missing fundamentals
You can get a program called wavanal for analysing church bells from The Sound of Bells web site..
Bells often have only a very weak partial at the frequency that is perceived as the pitch of the instrument by most listeners, and some may have no partials there at all. See The timbe and pitch of bells.
The author Bill Hibbert synthesizes an example bell sound with partials 5/9~ 6/5~ 2~ 3~ 4~ 16/3~ 13/2~ 8~ 21/2~ (to within 30 cents) for a perceived pitch of 1/1.
The perceived pitch makes a harmonic series with the actual partials 2~ 3~ 4~. However, he finds that even the sound made up of 5/9~ 6/5~ 2~ (hum, tierce and nominal) can be perceived to have it's pitch at the 1/1!. Try his file 8 and see what you think.
His actual frequencies used, in Hertz, are 176.6 390.2 640.9 949.0 1302.4 1690.9 2110.2 2604.6 3380.0 with the usual perceived pitch at 320.5 Hz.
Chords
So now you see how to compare the pitches of the middle cs of all the voices on your synth - try each in turn, play a long-ish note, and see what you find using {hi|Show partials as text file </span>.
How about analysing chords though. For those you use {hi|Make Chord to analyse </span>. Enter the chord you want to test into the edit field, then click that button, and use the play button as before. With this one, try selecting {hi|Add harmonic series analysis </span>. before you use {hi|Show partials as text file </span>. This analyses the list of frequencies to look for overtone series.
This gives especially good pitch accuracy for instruments rich in overtones, such as strings, because then it can find the average of maybe ten or more partials for each note of the chord. The higher partials are more precisely determined too, because the frequency spectrum, as it turns out, has finer resolution for higher pitches. (Techy detail: the harmonic series analysis works by dividing all the overtones by integer numbers down to the fundamental, and then taking the average of all those values to find the adjusted frequency of the fundamental.).
The harmonic series analysis is not suitable for the piano or for some inharmonic timbres such as some bell sounds - you will see why if you try analysing a note played on these instruments. The ones that are suitable for the harmonic series analyses are the ones with partials at nearly exact multiples of 1200 cents..
You can enter a single note for the "chord" here if desired - say, {hi|350 cents.}You can also use any of the notations you use for entering scales.
You would normally need to click the {hi|Make chord to analyse </span>button each time you try a new chord. However, if analysing many chords in this way, you can select {hi|Auto remake}to automatically remake the chord every time you click the play button - and so save yourself some time.
To get back to the preset (play the 1/1 only) and ignore the chord to analyse, use the {hi|Standard Settings}button again (which will also unselect {hi|Auto remake}).
What level of pitch accuracy can I expect to find typically?
Wave table synths seem to be variable in this respect, while an FM synth can be really extraordinarily accurate. Typicaly wave table synths seem to vary from say 0.1 cents to 4 cents of accuracy, while FM synths can be accurate to better than 0.01 cents.
Here are a few that I have worked with, but I haven't made an exhaustive set of measurements by any means, these just a result of a few spot checks:
The Virtual Sound Canvas seems to be accurate to about a cent in relative pitch within the same voice, and has similar variation in pitch from one voice to another.
The SB Live! is much more accurate than this within a single voice if it uses a single sample, about 0.2 cents or so. However, it can vary by 2 or 3 cents or more as you vary the sample (i.e from voice to voice, even different registers of the same voice sometimes). The reason for this variation seems to be because the orginal wave samples have their frequencies set to only 2 or 3 cents of accuracy - so it can be fixed, with a great deal of work, by retuning all the individual wave samples. You just need to change the sample rate at the head of the file, which could be done using Autotune perhaps.
The Yamaha soft synth is accurate to perhaps 0.5 cents or so within a single voice.
I sometimes wonder why the wave table synths vary so much as this actually - especially the ones that use only a single waveform for the note. Why is the SB live! for instance only accurate to within 0.2 cents when you use a single waveform - instead of the better than 0.01 cents accuracy of FM synthesis? One would have thought linear stretching of the waveform would do and would have as good pitch accuracy as the FM synths - maybe there is some reason why this doesn't work... (any wave table synth designers reading this??).
The FM7 has much better accuracy - an amazing at least one hundredth of a cent - that is just the limits of my measurements - probably accurate to one thousandth of a cent.
As for the larger variations due to wave samples only pitched to within 2 or 3 cents of the target frequency, I may later add an option to FTS to autotune, and do it so it can retune all the wave samples in a font in one go. However, the FTS peak detection algorihtm needs to be improved first - it finds the frequencies accurately, but sometimes doesn't easily distinguish the main peaks in the spectrum from noise and smaller peaks
Some musicians with particularly fine sensitivity to pitch can distinguish an interval of one cent or better as a melodic step (1200 notes to an octave or more). Most though probably have best pitch resolution in the range four to fifteen cents, and some with no musical training are unable to distinguish notes a semitone apart. So, these are pitch variations beyond the limit of hearing for most.
So, why are we concerned about them - it's because fine pitch distinctions are more important in chords, and can be more noticeable, as they can cause beats. There, even a one or two cent distinction can be fairly easily noticeable.
Try this - show a scale window, and set the step to 1 cent, and try playing two notes together (say harpsichord) and you may hear beats - a slow wah wah type sound as the note fades away. Try comparing the equal temperament 1/1 700.0 and 1/1 3/2 and you'll probably hear beats in the equal temperament one.
You may hear slow beats in the 3/2 too though, with a wave table synth. (Something to watch out for - notes can also beat with the hum from your computer).
Probably not that many performers are able to play notes to better than one cent accuracy - even in sustained chords. It is a fine pitch distinction to make, and easily overwhelmed by the slightest vibrato or involuntary pitch variation. In the case of fixed pitched instruments, maybe you can tune it to this accuracy, but then the tuning may change a little during performance or because of change of temperature etc.
However, some modern microtonal performers do indeed claim that in their performances they can pitch notes to the exact cent desired, and that they can hear the distinction between notes a cent apart melodically.
The wave table synths leave something to be desired by comparision, but are fine for getting to know and enjoy microtones. If you want to hear the purest possible intervals you can get in computer generated music, try an FM synth such as the FM7.
[#top top]
What is pitch bend and how is it applied?
Midi is set up to work with the twelve equal scale. Each note has a number between 0 and 127. The number is the number of semitones above the 0 note, with the 0 chosen to be five octaves below middle c. So 60 is middle c, 61 is c#, 62 is d etc. This spans the entire range of musically useful notes.
However, when one wants to play notes "in the cracks between the keys" of a twelve equal keyboard, you need to bend the pitch up or down from its twelve equal position. Midi keyboards often have a pitch bend wheel - and when you move it, it bends the current note up / down in pitch - usually with a range of plus or minus a whole tone.
So if you want the pure just intonation 5/4 for your e above middle c, at 386 cents for instance, you could make its Midi note 64 flat by 14 cents by moving the wheel by the appropriate amount. Unfortunately for this approach though, the wheel applies to the keyboard as a whole, so all the notes will get moved down, including the c as well. That's not what we want.
The solution here is to play the c and e on separate Midi channels. The pitch bend only applies to one channel at a time - though this does require that you have a multi-timbral synth or soundcard - one that lets you play notes on separate channels. Supposing all is well and you can indeed play on separate channels, then what you do is to play the c on channel 1 say, and the e on channel 2. Then bend the pitch for channel 2 down by 14 cents - and there you are, a perfect just intonation major third diad. To complete this to make a triad you'd then play a g on say channel 3, and bend the pitch of it up by 2 cents.
You can make any chord in this way, so long as it needs no more than 15 simuiltaneous pitch bends. That's not much of a restriction as few scales have more than 15 distinct pitches in them. Notes a perfect octave apart can be played in the same channel (also notes at any equal tempered interval such as say 300 cents etc.). Non octave scales are more tricky and you can run out of channels if you have more than 15 simultaneous notes in a chord in a non octave scale. For more about this, see [Scales_and_Fractal_Tunes.htm#sit_run_out_of_out_chann Situations where one can run out of channels].
Although this used the pitch bend wheel, you can do it also by sending an appropriate mdi message. Composers of microtonal music used to insert these pitch bends by hand into their scores in order to pitch bend the notes - indeed a few still do it this way. But, I think it is a little easier to use a progoram like FTS or SCALA to do it automatically :-).
Now for the nitty gritty, how its done. The pitch bend is in the range of -200 to + 200 cents. You can see the pitch bends as they are applied in Out | Notes in Play . Play a just major chord for instance - to do this, select the major chord from the Scales drop list. Then in Bs | Arpeggio / Scale Playback , choose chord for the play method, and click the play button for the arpeggio.
Or, perhaps better actually, play it on the PC keyboard and watch to see what pitch bends get applied for each note as you play.
You will see the notes:
60 pitch bend 0 cents 64 pitch bend -14 cents 67 pitch bend + 2 cents
The actual channel used for each note will depend on what has gone before in the tunes or your improvisations. Also, here I'm rounding the figures to the nearest cent - you can choose how many significant figures you want to show in this window, and in most of the FTS windows, from File | scale notation .
(first use File | New . if you have been playing one of the exotic fractal tunes first such as the Fibonacci tonescapes).
How accurately can these notes be pitched then?
The midi spec actually uses numbers in the range 0 to 16833 to specify the pitch bends, where 8192 corresponds to a pitch bend of 0, and 0 means bend the pitch down as far as you can, with the midi standard setting here is -200 cents. The highest pitch bend upwards in the spec is 16833. The next number, 16834 (which you aren't supposed to be able to send) would bend the pitch up by 200 cents.
So each unit of pitch bend is 0.024414 cents. This is a tiny amount and likely to be better than that available on many sound cards or synths and well beyond human pitch perception limits in all except possibly very special situations (one might be possible to detect it by listening to beats in carefully designed synthetic sustained chords).
For those interested in the technicalities, these pitch bend numbers are 14 bit numbers. This is a bit unusual as usually computers use multiples of 8 bits - the reason is that in midi the first bit of each data byte has to be zero as a way of distinguishing them from other types of byte that specify types of event such as note ons and offs. So in midi you send data in multiples of 7 bits. That's also why you have 128 midi voices instead of 256. Here, you have 128*128 (=18634) midi pitch bend units instead of 256*256.
Midi keyboards usually have pitch bend wheels that only send the most significant (7 bit) byte of the pitch bend, which means the smallest pitch bend unit for a wheel is only 200/127 cents = 1.57483 cents. This is fine enough so that most musicians will hear the bend as a continuous slide (pitch bend sensitivity for musicians is often in the range of 4 to 20 cents though some can hear pitch changes less than 1 cent and so I suppose they might perhaps hear a pitch bend wheel slide as a series of melodic steps). Some low end keyboards have pitch bend wheels with only maybe a dozen or so pitch bend steps per semitone..
There's a reason for reducing the number of pitch bend units - if you were to send a pitch bend message for each position of the wheel for 8192 distinct pitches per tone, you'd need to send thousands of messages each time you move the wheel. That would slow everything down - even a modern fast computer might need to pause if you start sending thousands of midi messages in a fraction of a second. This is especially true when you use a midi keyboard, since it connects to the computer via a cable, and transmission along a midi cable is usually much slower than internal processing in the computer itself.
FTS however sends "instant pitch bends", so can use the highest available resolution because it only needs to send one pitch bend each time to retune the note as desired.
[#top top]
How do I do a frequency analysis to find the partials of a note?
See the [#check_pitch_accuracy How do I analyse a note played in FTS to find the actual pitch played on the sound card / soft synth / synth?] FAQ for details about the frequency spectrum method, discussion of artefacts that can occur, and examples.
[#midi_voice Midi Voice], [#note_you_play A note you play yourself or have already recorded].
Midi voice
This is already covered in [#check_pitch_accuracy How do I check the pitch accuracy of my synth / soft synth / sound card synth?]
Here is a quick review::
Tasks | Analyse Midi Voice .
Standard Settings
Make sure you have Auto record and analyse selected.
Choose the voice you want to analyse from the Voice menu.
Set a value for How much to analyse - say 60 seconds. Set Note (secs) to 60 seconds too - if you make it small, then overtones of the repeat period of the repeated notes will add extra peaks to the spectrum.
Now click the play button. When you click the stop button, the note you have just played gets automatically analysed. The Find Freq. button will change to show Halt Search while the note is analysed (maybe very briefly if it is a short recording). So wait a moment for it to turn back to Find Freq.
Then use Show Freq. to show the frequency spectrum, and Show partials as text file to show the list of partials.
You can also resynthesize the sound from its partials and compare it with the original. Use Make Waveform from partials , and then Play synthesised wave .
Select Use original volume envelope to resynthesize it with the same volumes, e.g. the same level of attach and fade away for a percussive or plucked note.
[#top top], [#find_partials start of section]
A note you play yourself, or have already recorded
Tasks | Analyse Recording or Midi Voice .
Now if you have a recording already, open it from Open Audio file
Choose How much to analyse - if you want to analyse the complete clip, set this large
Then in the main window, click Find Frequencies , and Show Frequencies , and that will show the frequency spectrum. Use Show partials as text file to show the list of partials.
To find the partials for a selection from the recording, highlight the part you want to analyse, and use Show Freq. | Options | Find for Detail .
If you want to find the exact pitches for the notes of a chord played on harmonic partials type instruments such as voice / strings / brass / wind etc, try selecting Add harmonic series analysis before you click the Show partials as text file button. Look at the end of the partials file to see the analysis.
If you want to make a new recording then click the Record Control Button and select a suitable recording device such as microphone. Then back in FTS, click Show recording , and press Start rec. .
Alternatively, to keep an eye on the recording as it goes along, show the Oscilloscope and record from there.
The Record Control is the same one you get from Windows if you go to Start | Programs | Accessories | Entertainment | Volume Control and then choose Options | Properties | Recording (Saves a few clicks to do it from this FTS short cut. :-)).
When finished, click Stop Rec . Note that this is a temporary recording which is kept in memory only and will be lost when you exit the program. If you want to save it, use File | Save As | Save as type | Ram Recording as Audio File (*.WAV) . You can also do it from Show recording. | Options | Save recording as .
Then you can play it to check it is recorded okay, and analyse it as before.
[#top top], [#find_partials start of section]
How do I convert the frequency analysis into a scale?
This is easy - just use the Show partials as text file . Then scroll down and you'll find the partials as a scale.
[#top top]
How do I make a custom voice consisting of one instrument playing the partials of another one?
Use Make partials into custom voice .
So for instance, suppose you'd like to hear the sound of a sitar playing the partials of an oboe. Analyse your oboe voice in the usual way. Then click the Make partials into custom voice button. In the window that pops up, click Select Voice or non melodic percussion . Select the Sitar from the Voices menu. Then click End selection.
Now to hear what it sounds like, click Select into highl. Part . Back in the main window view, to try it out, first unselect Auto record and analyse (so you don't record over your oboe recording), and click the play button to hear what it sounds like.
This is a voice you can use like any other midi voice in FTS. You'll find it has been added to the menu in Voice | Custom Voices . For more about the things you can do with custom voices, see the help for those windows: [User_guide2.htm#Custom_Voices Custom Voices] and [User_guide2.htm#custom_voice_as_timbre Custom Voice as Timbre].
[#top top]
How do I find the pitch of a note (second method)?
Tasks | Analyse Recording or Midi Voice .
Record your playing, or open an audio file, or select Auto record only and play one of the midi voices to record it.
Then go to Find Seed , select Find as a single note , and click Find notes .
This works better for some voices than others - tends to work well if the waveform has clearly defined peaks or waves all very similar in shape to each other. It's set up to work well with recorder - but you will find other options from the Find Seed Options drop list (and can save your own presets here too).
If you do this with a wave table synth, it tends to work especially well with ocarina, which tends to be pretty close to a sine wave in shape. Or look and see if you can find a voice that consists of a repeating wave (as shown in the oscilloscope).
You can show the Oscilloscope while the note is played and recorded, and select Oscilloscope Options | Show Info , for a first idea of the pitch.
However, this doesn't do much analysis of the sound - in particular, it skips the search for similar waves, so is fairly likely to show a pitch too high or too low. It finds the pitch based on the length of the current waveform as shown in the window (= same pitch you hear if you click the "Play this" button) - as this is a very short section of the recording, the measured pitch will tend to fluctuate. (However, it uses zero crossing interpolation to get a more accurate measure of the pitch than one might otherwise expect from such a short sample - it measures the length of the waveform not just to the nearest sample, but to a fraction of a sample depending on the interpolated zero crossing point).
[#top top]
Is there a version of FTS for the Mac or Linux?
No, sorry, there isn't.
FTS would need to be written from the start in a multi-platform language such as Ada which Manuel Op de Coul uses for SCALA. As it is written in Windows C, all of it is highly Windows specific.
The way to port it to the Mac would be to use a Windows emulator there of some kind. I am told that FTS is used successfully on a Mac with Virtual PC. Wikepedia entry,
If you haven't heard of it yet, you should also check out Jeff Scott's L'il Miss' Scale Oven. It is used to make scales and to update the tuning tables of synthesizers so that they can play in the scales.
He has an intimate understanding of scales as a composer, so this must surely be a must-have program for Mac microtonal composers and musicians.
For Linux, you could try FTS with Wine. FTS uses only standard Windows calls, so it may work.
I'll be interested to hear if you try out one of these methods, or any other windows emulator, and find it works - let me know which Windows emulator you used, and I will update this FAQ entry accordingly :-).
[#top top]
What language should I learn to write a music program, and how long does it take to write one?
Intro
Though not particularly about how to use FTS, this is a FAQ from FTS users. I'm glad if FTS has inspired you to want to program yourself :-).
I'm not really the one to answer this question though. In short - I don't know which language is best. So if you are reading this, in expectation of a how to guide to get you started programming, you'd better stop reading and look elsewhere. I haven't got experience as a teacher of programming, and I probably wouldn't be very good at it either. But perhaps I can give a few sidelights, and anecdotes etc. which may be of interest in some way.
If you want a quick start you certainly shouldn't follow the route I took myself which I think would take just about anyone a fair number of years to get to the point where you can write Windows programs. If starting from scratch now I'd do something different probably myself too.
Probably you could learn to program more quickly than I did as there is much more by way of teaching materials, and guides etc than there were in those days. Also starting your programming experience by learning to program a mainframe computer using a card punch, though interesting, was maybe not the quickest route to learning to program a modern Windows computer :-).
It depends how quickly you pick it up, and also depends on the language you use, but I would guess that even if you start right now, you'd probably need to put in at least a few years before you can write a program like Tune Smithy - but there are some things you'll be able to do pretty early on depending on the language, and what your aims are. If you find you like programming, it's a lot of fun :-).
Some ways people get started
If you want a gentle introduction to programming, you may want to try something easier than music making, as it is quite an advanced topic in most programming languages. There are specialist scripts or languages which focus on music as the main thing they do - CSound for instance, but CSound anyway isn't the easiest of languages to learn either.
Many get started by making Flash animations for web pages. Even children in Primary School are able to make Flash animations, so it can't be that hard to get started on it. Yet it is also a widely needed and useful skill for professional web designers as well. I imagine you get visible results rather quickly. You can also use Flash to play Midi events with the right tools so eventually you may be able to make musical flash animations as well. Try Flash Midi - currently a beta. I can't help with this myself though - so far I haven't needed to learn Flash myself.
As an alternative to building on existing code you can also use various tools that have been developed to make music programs. In some cases the result can only be run within a particular program, and sometimes they can be deployed more generally, within web pages or as stand alone programs. Maybe you will find some useful material amongst the links here: http://www.synthzone.com/miditech.htm.
What language is FTS itself written in, and what are other commonly used languages?
FTS is written in a language called C. That's really the only language I'm familiar with apart from some now nearly obsolete languages that I used to use in the past. I can probably most usefully say things to help any aspiring C music programmers reading this. So I'll focus on that, but before getting to it, I'll mention some other programming languages that I know of.
C++ is often used for Windows programming. It has a reputation as a particularly tough language for beginners however. Usually you would have some earlier programming experience before you tackle this one. Also some C programming normally comes first before you learn C++. The newer C# is supposed to be somewhat easier to use than C++. However, I can't say much more, as I don't use either of them myself.
One language often recommended for beginners is Visual Basic. I don't know much about it either, but have seen example programs of a few lines that are fully functional useful Windows programs. You should be able to do the same things in Visual Basic that you can do in Windows C as you can call the same routines, so for instance you should be able to call the Windows Midi routines to play notes and so on. But again, don't ask me for the details :-).
The visual basic sample project here one from Microsoft looks as if it does all the right things to get one started with midi music programming, assuming one was familiar with Visual Basic already. Check it out with someone experienced in it if interested.
Another language that is good to know is Java - one of many C like languages. It's used for making applets for web pages, and occasionally for stand alone programs. You could reasonably jump into Java straight away without learning any other language first, it is more suitable for beginners than C++. The latest versions of java have midi support - so you can write musical web page applets :-). Again I have little experience in Java myself. Java developers will probably be interested to explore some of the java tools available to generate music - see this page again: http://www.synthzone.com/softsyn.htm - for instance Jsyn if you are into making new virtual instruments and the like..
Then, if you make web pages, it is useful to know javascript. Though the name is similar, actually there is almost no connection with Java - both are C like and they have a few things in common, that is all. I'm not sure how much actual music making you can do in javascript, but it is useful in other ways.
Javascript can be quite a nice introduction to basics of programming. It is not so much used for large programs. It's only suitable for use with forms in web pages and other web page applications, but is widely used for those, and is very useful indeed in that way. It is easy to make a small javascript applet - for an example with few lines of code, see my ratio to cents applet in the cents and ratios page - use view source to see the code (within the script tags). So - if you are interested in very short calculation type applets for web pages, it could be an interesting language to explore.
If you get into javascript, it will save you a lot of time to have a good javascript debugger. You can download free ones for both Internet Explorer, and the Netscape one Venkman. Here is a nice on-line tutorial on Using the Netscape javascript debugger.
There are many other languages - e.g. Small Talk - one of the earliest object orientated and gui based languages, and still going strong. - I'll also mention Ada which is the one that Manuel Op de Coul uses to write SCALA (along with Gtk)- so it must be good :-). It has a reputation as a language that lets one write bug free code more easily and quickly than most languages.
You may also be intereseted in the widget tool kits (which Gtk would be an example of).
Many windows programs are written in C++. Plain windows c programs tend to be "lower level" - useful if you want to optimise them to be as fast as possible.
How did you learn to program yourself?
The route I took myself was first a bit of early main frame programming in Fortran in the early 70s, through Algol at university - then after a gap, C for XWindows on Unix (for some of my maths research) then Windows C.
I think the style of programming I use myself, low level Windows C, would be hard for anyone to learn in less than at least a few years of programming.
I probably would follow a very different route if I were to start programming now. Indeed, the two main languages I learnt in my early days are now rarely used - at least for windows, I'm not sure if you can even use them to make GUI type programs for Windows, at any rate it would be a most unusual route to start with those as your first programming languages nowadays..
Programming in C++ or C anyway is very hard, for a beginner starting from scratch. You would need to allow at least a number of years before you can make useful programs - I believe, for Windows anyway.
However, there may be some future in new C language based languages for beginners. I had a look recently at a language called Ultimate++. It comes with a very nice Integrated Development Environment (IDE), which you can use for ordinary C programming as well. The language itself looks elegant from its descriptions and some things are much simpler thatn they are in normal Windows C++ - and it is designed from scratch to be naturally an object orientated language - unlike C++.
I don't know what the future is for Ultimate++ - it is at present being developed by just a single developer. Though with an enthusiastic body of users to judge by the forums.
This is just a first impression - maybe this or if not this one exactly, something like it may eventually be a good programming language for beginners to Windows programming who like to follow a C like route. The program code in Ultimate++ is remarkably compact and concise compared with either Windows C and Windows C++ code for gui based apps - and has the big advantage too that it is cross platform - the programs can run on a Mac and on Linux as well.
I plan to investigate it for my own programming in the near future. I haven't tested it yet .
Books on C
If you want to learn C++ or Windows C, or some other C based language, first you need to learn C anyway, and the classic book on that is Kerningham and Ritchie " The C Programming Language, Second Edition ".
The classic book on Windows C programming is Charles Petzold's " Programming Windows, 5th Edition ". This explains how to make gui based windows programs in pure C. FTS is written entirely in C in this fashion. FTS is a bit unusual because it is written entirely in plain Windows C. Most programmers use a higher level language for large programs like this one.
Here is the ACCU (Association of C and C++ users) list of Beginner's c programming books (on-line), Advanced c programming, and MS Windows
Music making is a fairly advanced topic in Windows C (chapter 22 in Petzold's book), so I've made a library of subroutines to help with this. You can use it to play notes in any scale or tuning - even in your first "hello world" type programs.It may well be of use for more experienced programmers too (I find it useful myself to get a small music program together quickly). Here it is (on line) PlayMidiLib. I will probably add to it from time to time, and make a few demo programs to show how it can be used.
It's usual for a C programmer to start with console apps, then move on to Windows C later. That's just because of the size of most programs written in plain Windows C. You can make some simple programs that just display an Ok or Ok Cancel message and such like, then the next step up in Charles Petzold's book is to a program of about 80 lines - most of which is specific to Windows. But - if you use one of the modern ultracompact high level C like languages you may be able to write programs with GUIs (Graphical user interfaces) at a much earlier stage.
It's best to get used to the c language first, and how it works, loops, variables, arrays and so forth, and to do that with the minimum of extra overhead. C has inspired many C like languages, and if you learn C you have a good basis from which to learn those too.
It would be nice if someone were to write a book on programming in Windows C from scratch - but I can see the difficulty - there's this gap between a Windows program suitable for about chapter 2 or so, and one for chapter 10 - what do you do in between? Perhaps one could add a library of extra routines which you link in to all your beginner programs, to make input and output of data much easier than it usually is in Windows C - I thnk this approach may have some potential.
Many find C++ a useful language. My own inclinations are towards the lower level type of C programming as in FTS and I have very little experience really of C++. So I'm not the one to recommend C++ books.
C programmers are often interested in programming at a low level - it is a good language to use if you are into optimising your code by hand, and if you know how to optimise your C code, your C programs will run as quickly as programs in any other language except perhaps Assembly Language, though this advantage isn't such a big thing nowadays.. The programs also tend to be a bit smaller too.
[#top top] -[#what_language start of section ]
Assembly
If you are interested in this style of programming you may also be attracted by assembly - I am. It's possible to write an entire windows program in assembly - I know of one music program in this style called abox. However, this is so rarely done because of the sheer number of lines of code needed - apart from that one, I've only come across one other Windows program entirely in assembly. For those interested in this subject, I came across this site on how to write Windows programs in assembly: Win ASM. .
C compilers translate your code into assembly before they build the program, and often include an option to show the assembly code. Here is one line of code from FTS expanded into Assembly by the compiler.
`zero_ypos=zero_line_central?wave_height:2*wave_height;`
expanded to
```007DBFA7 movsx ecx,byte ptr [zero_line_central]
007DBFAB test ecx,ecx
007DBFAF mov edx,dword ptr [wave_height]
007DBFB2 mov dword ptr [ebp-23A0h],edx
007DBFB8 jmp PaintWave+0DA5h (007dbfc5)
007DBFBA mov eax,dword ptr [wave_height]
007DBFBD shl eax,1
007DBFBF mov dword ptr [ebp-23A0h],eax
007DBFC5 mov ecx,dword ptr [ebp-23A0h]
007DBFCB mov dword ptr [zero_ypos],ecx```
Quite remarkable that anyone should write an entire Windows program in this language. It used to be used frequently for optimising sections of a program for speed - but compilers have got so good at that nowadays that they say you often don't gain much in speed by doing it by hand, and it may even slow your program down if you aren't well up on how the latest generation of chips operate. So, much as I'd like to, I've never had much incentive to learn Assembly, and that is probably true of most C programmers.
Programming tools
Whichever language you decide on, it is worth getting good tools even as a beginner.
It may seem a trivial kind of thing if you are new to programming - syntax colouring is a great help for spotting syntax errors as you type. The idea is to show, e.g. variables in one colour, comments in another, strings in another, and so on. As a beginner programmer, most of your errors to fix in the first programs you write will be syntax errors, and this will mean you spot many of them immediately while you are typing, If you make a typo, the things you type show up in the wrong colour - "black instead of blue - Ah - a typo!". It's a great time saver and will make the learning process much more fun.
Then, it is useful to get an IDE (Integrated Develpment Environment) that lets one edit the code, run it, and step through it in a debugger within the same application - that saves a lot of time - however it isn't needed perhaps for ones early programs.
I can only really comment on tools for c programming. The one I use is Microsoft Visual C++ - suitable for experienced programmers who already know that they will be using Windows C or C++. There is a free version for hobbyists - however it doesn't include a resource editor which is a bit of a drawback in a GUI program.
Others use Borland C++ which is the other main commercial C compiler - I don't happen to have used it myself.
I've recently come across Ultimate++ which comes with a very nice looking IDE - and is completely free - with a new more thoroughly object orientated approach to windows C - can also be used to compile standard windows C / C++ programs - however I've not tested it much as of writing this.
The rest of this is a bit out of date. You may like to look at the Wikipedia article on IDEs. and the comparison of IDEs
If you want free tools, you can get a free programmer's text editor for c with syntax colouring on this page Freeware Home >> Programming >> Editors. I've had a look at UniRed and first impressions are that it is rather nice but I haven't tested any other tools apart from Visual C++ at all thoroughly.
Look for shareware ones in this Google directory listing.
If you go for free tools, you need a compiler as well of course. LCC is free, comes complete with IDE with syntax colouring, and can compile Windows programs. It's a compiler for C only, not yet for C++. (there's a paid version as well on CD, or with additional examples, or if one wants the compiler source code).
I've tested LCC a little, and it works fine, with a number of nice features. It can open and edit resource files too. If you call a function that isn't in your project or any of it's library modules, it will search through all the Windows libraries to see if it can find it, and list any that have it and offer to add it in - very nice!
One thing I notice in the documentation but haven't tried yet is that it has support for arbitrarily large numbers ("Big numbers"), which will be a nice feature for some microtonal lattice work - you can get pretty large numbers by taking repeated powers of 3/2 etc.
Bloodshed is another one that you can get for free, this time for C / C++. The compiler is called Dev-C++
If all you want to do is to compile console mode apps in plain c, GCC is fine, which you can get from the djgpp home page. Read the install instructions - you need to modify your autoexec.bat file and add a couple of lines to it - apart from that, it's just a matter of unzipping it - the install is easy). This would be fine to use e.g. to work through Kerningham and Richie
Djgpp can compile windows programs, but apparently you may need to get quite technical about it to do that. The FAQ for GCC itself recommends LCC as one of several for Windows C, and mentions particularly that it is easy to use.
Your programmer's editor may let you run a compiler directly from within the program. Or, you can run gcc from the command line using this type of syntax, which is fine for ones first few programs:.
gcc -o hello.exe "hello world.c"
then you'd type
hello
to run your program. For comand line work, it's useful to install Doskey so that you can go back to repeat a previous command using the up arrow. You do that by adding the line
doskey
to your autoexec.bat . You can edit this file most quickly in Windows from Start | Run | msconfig .
If you use UniRed, and gcc, you can add these lines to UniRed\compile\c.ini to help with compiling your first c programs.
```[compile with gcc]
cmd=gcc -o "%name%.exe" "%file%"
show=1
pattern=^([^:]*):(\d+)
filepos=1
linepos=2
[run compiled program]
cmd="%name%.exe"
show=0
```
How do I sell my programs?
There are many software agents nowadays. Two I use include Reg Now - many shareware authors use them and they are well established in the field - and Plimus - particularly easy to configure and integrate the order page with your web site. There are several other major resellers of software too, that's just the one I use.
I recommend Armadillo for making unlock keys and protecting your program from Crackers - don't think your program won't be cracked just because it has a specialist audience who would have no interest in cracking it. There are many crackers who are keen to find the unlock key for any program that gets published, and not enough programs to go round apparently. Any newly posted program on the large download sites will attract their attention within a couple of days. They do it just for the fun and the challenge - but then publish their results for other crackers to see to prove they did it - and then these get widely distributed by others. Armadillo is one of many systems used to protect programs. I can recommend it because I use it myself, and know it works well and haven't needed to look for anything else. You'll need to ask others to find out about the other approaches.
A few programmers earn a reasonable amount from shareware, but many earn rather little - think how many programs one downloads and what percentage of them one buys - well many windows users anyway buy a rather small percentage of the ones they try. It would be no surprise if only one in 5000 or less of those who download your program ever register it, though with some particularly successful programs the figure there can be as high as 2 or 3 percent I've been told. So your site will need lots of visitors, or your program will need to be mentioned in many places!
If you have many downloads but still get no sales (e.g. downloads into the tens of thousands with no sales), does if it has any serious bugs that interfere with the flow of work? Or is it hard to install in any way? Or can your users figure out how to use it after they install it? Or are there perhaps many other programs that do similar things? Those are all things that have affected my own programs from time to time.
Sometimes there is little point in increasing the number of downloads until you improve the program. Pay attention to your users. Well that's my motto anyway. Not that I'm particularly successful yet myself. Perhaps I should be the one asking this question :-)
For music programs, try submitting them to Harmony Central and Shareware Music Machine. | 21,570 | 96,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-09 | latest | en | 0.905739 |
https://justtechreview.com/python-program-code-to-swap-two-number/ | 1,627,478,890,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00293.warc.gz | 344,064,915 | 25,250 | # Python Program Code to Swap Two Number
Here you’ll get a python program to swap 2 numbers with and while not victimization temporary variable.
Python Program to Swap 2 Numbers with the use of Temporary Variable
``````a = 10
b = 20
print("before swapping\na=", a, " b=", b)
temp = a
a = b
b = temp
print("\nafter swapping\na=", a, " b=", b)``````
Output
before swapping
a= 10 b= 20
after swapping
a= 20 b= 10
Python Program to Swap 2 Numbers while not victimization Temporary Variable
Method 1:
Python provides the way to directly swap 2 variety while not temporary variable. It is worn out following method.
``a, b = b, a``
Method 2:
In this methodology, we will use addition and subtraction operators.
``````a = a + b
b = a - b
a = a - b``````
Method 3:
We can conjointly swap numbers victimization multiplication and division operator in the following method.
``````a = a * b
b = a / b
a = a / b``````
This methodology won’t work once one amongst the amount is zero.
Method 4:
It is another methodology during which we have a tendency to use bitwise xor operator.
``````a = a ^ b
b = a ^ b
a = a ^ b``````
Comment below if you have got queries or grasp the other thanks to swapping numbers in python. | 340 | 1,228 | {"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-2021-31 | latest | en | 0.841186 |
https://www.sourcetronic.com/glossary/oscilloscope/?lang=en | 1,495,488,071,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607120.76/warc/CC-MAIN-20170522211031-20170522231031-00078.warc.gz | 954,948,741 | 8,985 | Oscilloscope
An oscilloscope is defined as an instrument used for visualizing the electrical voltages and their trajectory on a display. The electrical function generator makes it possible to gather electric quantities, usually visible in a real-time display. For that purpose, a two-dimensional representation is given of the measured quantity in the form of an oscillo-gram. A dimensional axis is represented by time and a second dimensional axis is represented by an electrical quantity of a reference waveform graph in a co-ordinated system. The horizontal “X- axis” showing time and the vertical “Y axis “showing voltages are displayed. The resulting image is finally called an oscillo-gram.
The advantage of using this device over other instruments is due to the fact that the waveform is depicted. This causes overlays which are immediately recognizable as frequencies or other irregularities to the viewer. The only disadvantage of an oscilloscope with a function generator over other similar equipment is the prerequisite skills required to operator or control them. This refers primarily to the evaluation of sums as the identification of the image must first be evaluated – a figure that is represented in the oscilloscope which is normally quite cumbersome in other models.
Oscilloscope Construction
The construction of the oscilloscope is characterized by functions performed by a thermionic cathode; by applying a negative voltage relative to the anode causing the discharge of electrons. They are accelerated into an evacuated space by the electric field present and sent towards the cathode. By including a function generator in the oscilloscope, it is possible for a three-dimensional focusing to be achieved. For this reason, between the cathode and anode a negatively charged hollow cylinder, also known as “Wehnelt cylinder”, is positioned. By doing this, the electrons are accelerated by a pair of capacitor plates, which usually sends them straight by deflecting them. The deflection is proportional in each direction to the applied voltage. The electrons are collected by a screen which then allows for a graphical representation for the ratio of the voltages.
Additionally, the equipped function generator in the oscilloscope makes it possible to gather data from pure stress ratios in order to draw time-dependent voltage functions. Many of the function generators also feature a dual-beam setting that allows the representation of two stress functions as a function of time.
There is a difference between an analog oscilloscope and a digital oscilloscope. An analog oscilloscope uses a cathode ray tube display, however; over time the digital oscilloscope devices almost completely replaced the analog devices on the market. This is mainly due to practical disadvantages such as the size of the cathode ray tubes as well as economic factors. These days, they are only used in smaller laboratories as well as for training in technical schools. The digital oscilloscope is increasingly common which uses an analog-digital conversion function. They are known principally as storage oscilloscopes as they provide data which can be saved or transferred to a computer. | 602 | 3,193 | {"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-2017-22 | latest | en | 0.948227 |
https://metric-calculator.com/convert-milligram-to-pound.htm | 1,701,475,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00501.warc.gz | 445,751,114 | 7,363 | Milligrams to Pounds Converter
Select conversion type:
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Convert Pounds to Milligrams (lb to mg) ▶
Conversion Table
milligrams to pounds mg lb 100000 mg 0.2205 lb 200000 mg 0.4409 lb 300000 mg 0.6614 lb 400000 mg 0.8818 lb 500000 mg 1.1023 lb 600000 mg 1.3228 lb 700000 mg 1.5432 lb 800000 mg 1.7637 lb 900000 mg 1.9842 lb 1000000 mg 2.2046 lb 1100000 mg 2.4251 lb 1200000 mg 2.6455 lb 1300000 mg 2.866 lb 1400000 mg 3.0865 lb 1500000 mg 3.3069 lb 1600000 mg 3.5274 lb 1700000 mg 3.7479 lb 1800000 mg 3.9683 lb 1900000 mg 4.1888 lb 2000000 mg 4.4092 lb
How to convert
1 milligram (mg) = 2.20462E-06 pound (lb). Milligram (mg) is a unit of Weight used in Metric system. Pound (lb) is a unit of Weight used in Standard system.
Milligrams: A Unit of Weight
Milligrams are a unit of weight that are used in the International System of Units (SI), also known as the metric system. Milligrams are derived from the Latin word mille, which means thousand, and the French word gramme, which means weight. The symbol for milligram is mg.
Definition of the Milligram
The milligram is defined as one one-thousandth of a gram, which is one one-millionth of a kilogram, which is the base unit of mass in the SI. The kilogram is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10 −34 when expressed in the unit J s, which is equal to kg m 2 s −1, where the meter and the second are defined in terms of c and ∆ν Cs. The Planck constant is a fundamental physical constant that relates the energy of a photon to its frequency.
The milligram is a very small unit of weight, equivalent to about 0.000035 ounces or 0.000002 pounds. It is commonly used for measuring small masses, such as medicines, chemicals, vitamins, minerals and microorganisms.
How to Convert Milligrams
Milligrams can be converted to other units of weight by using conversion factors or formulas. Here are some examples of how to convert milligrams to other units of weight in the US customary system and the SI system:
• To convert milligrams to ounces, divide by 28349.523125. For example, 1000 mg = 1000 / 28349.523125 = 0.035 oz.
• To convert milligrams to pounds, divide by 453592.37. For example, 5000 mg = 5000 / 453592.37 = 0.011 lb.
• To convert milligrams to tons (short), divide by 907184740. For example, 10000 mg = 10000 / 907184740 = 0.000011 ton.
• To convert milligrams to grams, divide by 1000. For example, 2000 mg = 2000 / 1000 = 2 g.
• To convert milligrams to kilograms, divide by 1000000. For example, 3000 mg = 3000 / 1000000 = 0.003 kg.
• To convert milligrams to micrograms, multiply by 1000. For example, 50 mg = 50 x 1000 = 50000 µg.
Where Milligrams are Used
Milligrams are used in different countries and regions for different applications and purposes. Here are some examples of where milligrams are used:
• In most countries that use the SI system, milligrams are used for measuring small masses, such as medicines, chemicals, vitamins, minerals and microorganisms.
• In the United States, milligrams are sometimes used for measuring small masses, such as medicines, chemicals, vitamins, minerals and microorganisms.
• In Canada, milligrams are used for measuring small masses, such as medicines, chemicals, vitamins, minerals and microorganisms.
• In Australia and New Zealand, milligrams are used for measuring small masses, such as medicines, chemicals, vitamins, minerals and microorganisms.
• In China, milligrams are used for measuring small masses, such as medicines, chemicals, vitamins, minerals and microorganisms.
History of Milligrams
Milligrams have a long history that dates back to ancient times. Here are some highlights of the history of milligrams:
• The milligram was originally based on the weight of a grain of wheat or barley, which was used as a unit of mass in ancient civilizations such as Egypt, Mesopotamia and Greece.
• The milligram was part of the gram-centimeter-second system of units (CGS) that was developed in the 19th century as an alternative to the meter-kilogram-second system of units (MKS).
• The milligram was replaced by the kilogram as the base unit of mass in the SI system that was adopted in 1960 as an international standard for measurements.
• The milligram was redefined in terms of the kilogram and the Planck constant in 2019 as part of a major revision of the SI system.
Example Conversions of Milligrams to Other Units
Here are some examples of conversions of milligrams to other units of weight:
• 1 mg = 0.000035 oz
• 1 mg = 0.000002 lb
• 1 mg = 0.0000000011 ton
• 1 mg = 0.001 g
• 1 mg = 0.000001 kg
• 1 mg = 1000 µg
• 1 mg = 0.000032 troy oz
• 1 mg = 0.000032 apothecaries’ oz
• 1 mg = 0.000035 Spanish oz
• 1 mg = 0.000034 French oz
Milligrams also can be marked as Milligrammes (alternative British English spelling in UK).
Definition of the Pound
A pound is a unit of weight that measures how much force an object or substance exerts on a scale due to gravity. It is equal to 0.45359237 kilograms, or 16 ounces, or 7000 grains. One pound can also be written as lb or lbs.
History of the Pound
The pound was originally defined in ancient Rome as the mass of a bronze bar called a libra, which was divided into 12 ounces. The libra was later adopted by various regions and countries in Europe, such as England, France and Germany, with different values and subdivisions.
The pound as we know it today is based on the avoirdupois system, which was developed in the 13th century in France and England. The avoirdupois pound was standardized in 1959 by an international agreement among the United States, Canada, Australia, New Zealand and South Africa as exactly 0.45359237 kilograms.
The pound, as well as other units of weight, such as the ounce and the ton, were introduced in the 19th century as part of the imperial system of measurement that aimed to unify the units used in the British Empire. The pound was officially adopted by the International System of Units (SI) in 1960 as one of the seven base units.
How to Convert Pounds
To convert pounds to other units of weight, we need to use conversion factors that relate the pound to the desired unit. For example, to convert pounds to kilograms, we need to know that one kilogram is equal to 2.2046226218 pounds. Therefore, one pound is equal to 0.45359237 kilograms.
Here are some common conversion factors for pounds:
• 1 lb = 0.45359237 kg
• 1 lb = 16 oz
• 1 lb = 7000 gr
• 1 lb = 0.0005 t
• 1 lb = 0.0714285714 st
To convert from other units of weight to pounds, we need to use the inverse of these conversion factors. For example, to convert kilograms to pounds, we need to multiply by 2.2046226218.
Where Pounds are Used
The pound is a unit of weight that is widely used in the United States and some other countries that have historical ties with Britain, such as Canada and Australia. It is commonly used to measure the weight of people, animals, food, clothes and other everyday items. For example, an average adult male in the United States has a weight of about 198 lbs.
The pound is also used to measure some physical quantities such as force, pressure and torque. For example, one pound-force (lbf) is equal to one pound times the standard acceleration due to gravity (32.174 ft/s2).
Example Conversions of Pounds to Other Units
Here are some examples of how to convert pounds to other units of weight using the conversion factors given above:
• 2 lb = 2 x 0.45359237 kg = 0.90718474 kg
• 3 lb = 3 x 16 oz = 48 oz
• 4 lb = 4 x 7000 gr = 28000 gr
• 5 lb = 5 x 0.0005 t = 0.0025 t
• 6 lb = 6 x 0.0714285714 st = 0.4285714284 st
Conclusion
The pound is a unit of weight that measures how much force an object or substance exerts on a scale due to gravity. It is equal to 0.45359237 kilograms, or 16 ounces, or 7000 grains. The pound is widely used in the United States and some other countries that have historical ties with Britain, such as Canada and Australia, as well as some physical quantities such as force, pressure and torque. To convert pounds to other units of weight, we need to use conversion factors that relate the pound to the desired unit.
Español Russian Français | 2,198 | 8,196 | {"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-2023-50 | longest | en | 0.851174 |
https://www.enotes.com/topics/math/questions/int-1-x-2sqrt-x-2-4-dx-use-integration-tables-782834 | 1,716,014,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00610.warc.gz | 696,477,089 | 17,337 | # int 1/(x^2sqrt(x^2-4)) dx Use integration tables to find the indefinite integral.
Recall that indefinite integral follows the formula: int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
For the given problem int 1/(x^2sqrt(x^2-4)) dx , it resembles one of the formula from integration table. We may apply the integral formula for rational function with roots as:
int 1/(u^2sqrt(u^2-a^2))du = 1/(a^2*u) sqrt(u^2-a^2)+C .
By comparing "u^2-a^2 " with "x^2-4 " , we determine the corresponding values as:
u^2=x^2 then u =x
a^2 =4
Plug-in the values on the aforementioned integral formula for rational function with roots where a^2 =4 , we get:
int 1/(x^2sqrt(x^2-4)) dx=1/(4*x) sqrt(x^2-4)+C
=1/(4x) sqrt(x^2-4)+C | 277 | 827 | {"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} | 4.125 | 4 | CC-MAIN-2024-22 | latest | en | 0.646298 |
https://wikinotes.ca/COMP_330/lecture-notes/fall-2013/thursday-november-21/ | 1,721,734,384,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00059.warc.gz | 544,821,846 | 5,295 | # Thursday, November 21, 2013 Universal computable functions
Any errors in the final exam due to the use of these notes will not be given any special consideration. Prakash has not endorsed these notes and is not responsible for any mistakes in them. You can find Prakash's own notes on this topic here.
## 1Introduction¶
The foundation of modern programming languages was pioneered both by Alan Turing, with his idea of an universal Turing machine that can take an algorithm as input and execute it, and by Alonzo Church, with his $\lambda$-calculus. These days, algorithms have become such a natural concept that it seems strange to think that when we devise an algorithm, we are creating the most complex thing envisioned by the human mind. And indeed, the typical algorithm is so complex that we can only understand it via compositionality — by understanding the individual elements, and composing them together accordingly.
To that end, we will now study Gödel universal functions. We'll see that we can take programs and combine them to create more programs in a computable way.
## 2Universal functions¶
Consider a binary function $U: \mathbb{N}^2 \to \mathbb{N}$. $U$ is universal for unary computable functions if:
1. For all computable functions $f: \mathbb N \to \mathbb N$, there exists an $n$ such that $U_n = f$, i.e., $\forall xU_n(x) = U(n,x)=f(x)$ (so every function is encoded within $U$).
2. $U$ is itself computable.
Of course, simply defining a term does not guarantee that such a thing exists! It so happens that universal functions do, in fact, exist. However, we will not be proving that in this lecture. (To convince yourself, think about the fact that interpreters exist.)
### 2.1Implications for non-terminating programs¶
What if we could write a programming language that prevents you from writing programs that never terminate? In fact, we can write such a language, simply by incorporating strict typing in a particular way (e.g., strictly-typed lambda calculus). But there aren't any programming languages like this in use. Why not? Well, if we eliminated the possibility of being able to write programs that never terminate, we would necessarily eliminate the possibility of writing certain programs that do terminate.
This arises as a result of the fact that any total computable function of 2 arguments cannot be universal for all total unary computable functions, which we can prove by contradiction: Suppose there is a $U$ that is universal for all total unary computable functions. We define a total computable function $d: \mathbb N \to \mathbb N$ by $d(n) = U(n, n) + 1$. Because $U$ is total, $d$ will terminate. But $d$ is not reproduced within $U$, via this simple diagonal argument. Thus $d(n)$ cannot be realised by $U$!
TODO: Clarify the link between the first and second paragraphs.
### 2.2Gödel numbering¶
If $S$ is any set, then a map $V$ of the form $V:\mathbb N \to S$ is called a numbering. We define $V$ to be a function, so that each $n \in \mathbb N$ is associated with exactly one element in $S$; however, it is not necessarily injective, so there can be multiple numbers associated with each element in $S$needs confirmation.
Any universal function defines a numbering of unary computable functions.
### 2.3Composing functions¶
We now turn our attention to the problem of composing functions. That is, we want to create a total computable function $C$ which takes two functions as input and returns one function which is equivalent to running both input functions. In mathematical notation:
$$\forall p, q, x \; (U_P\circ U_q)(x) = U(p, U(q, x)) = U_{C(p, q)}(x) = U(C(p, q), x) \tag{is this correct?}$$
### 2.4Gödel universal functions¶
A Gödel universal function (GUF) is a binary universal function for the class of unary computable functions such that for any binary computable function $V$, there exists a total computable function $\sigma$ such that $\forall x, m, V(m, x) = G(\sigma(m), x)$. So it must be universal in the usual sense, but it must also be able to simulate all binary functions in conjunction with some unary function.
#### 2.4.1Proof of existence¶
As with universal functions, we need to prove that such a thing exists. To do so, consider the fact that we can easily create an injective, surjective and computable function, which we'll call $\tau$, from $\mathbb N^2$ to $\mathbb N$ and vice versa. (The construction of $\tau$ is omitted in these notes because the one given in class doesn't quite work. Left as an exercise for the reader.) Suppose $\tau(i, j) =$ some function involving $i$ and $j$. Then, to invert this, we define the functions $\text{fst}:\mathbb N \to \mathbb N$ and $\text{snd}: \mathbb N \to \mathbb N$ such that $\tau(\text{fst}(n), \text{snd}(n)) = n$, $\text{fst}(\tau(i, j)) = i$ and $\text{snd}(\tau(i, j)) = j$.
Next, given any universal computable function $U$, we define some function $G: \mathbb N^2 \to \mathbb N$ by
$$G(n, x) = U(\text{fst}(n), \tau(\text{snd}(n), x))$$
We need to show that $G$ is indeed a GUF. Let $V$ be a binary computable function. We define $f: \mathbb N \to \mathbb N$ by
$$f(k) = V(\text{fst}(k), \text{snd}(k))$$
Now, we need to find a total computable function $\sigma: \mathbb N \to \mathbb N$ such that $V(m, x) = G(\sigma(m), x)$. $U$ is universal, so there exists some $i$ such that $f(k) = U(i, k)$. We define $\sigma$ by $\sigma(m) = \tau(i, m)$. Then: $G(\sigma(m), x) = G(\tau(i, m), x) = U(i, \tau(m, x) = f(\tau(m, x)) = V(m, x)$. $\blacksquare$
#### 2.4.2Relation to function composition¶
Let $G$ be a GUF. Then there exists a $C: \mathbb N^2 \to \mathbb N$ such that $C$ is a total computable function and $\forall p, q, x$, $(G_p\circ G_q)(x) = G(p, G(q, x)) = G(C(p, q), x)$. Define $V(m, x) = G(\text{fst}(m), G(\text{snd}(m), x))$. (Alternatively, and equivalently, we could define $V$ by $V(\tau(m, n), x) = G(m, G(n, x))$). Then there exists $\sigma$ such that $V(m, x) = G(\sigma(m), x))$ by the definition of a GUF. Thus GUFs immediately give computable compositions of codes, allowing us to abstract complicated algorithms into their functional specs. This in turn allows us to think of functions as black boxes, whereby we only need to know their inputs and expected outputs and don't have to know their implementations at all.
Of course, not all functions can be thought of this way. Particularly, once you introduce non-determinism into the mix, it becomes clear that a program's functional spec is nowhere near sufficient to accurately represent the program. | 1,753 | 6,553 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-30 | latest | en | 0.931964 |
https://www.thestudentroom.co.uk/showthread.php?page=54&t=4091295 | 1,513,098,515,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948517350.12/warc/CC-MAIN-20171212153808-20171212173808-00195.warc.gz | 792,777,875 | 45,353 | # 2016 AQA Biology Unit 1 - Unofficial Mark scheme 2016 Watch
1. In the 6 Marker, I mentioned Vitamin C and it prevents Scurvey, would that be a valid point or did I just lose 2 marks?
2. (Original post by EmmaC.)
36/60 is usually an 'A'
20/20 i think is a 'C'
and a 'D' is 6/60
this is what i have heard from my teachers anyway...
Your teachers are definitely being nice at the lower end I'm afraid!
The boundaries move year on year, but they've pretty consistently got higher as students have got better at answering the questions (in 2012 they were ridiculously low).
Lowest I've ever seen for a D is 11 marks, but on some papers it's been 20+ (e.g. last year's B3 you needed 27 for a D!)
.
3. I believe I got 33-38 out of 60 what grade would that be
4. (Original post by thecomputernerd)
did anyone else feel there wasn't enough time? there was too much reading like hello aqa give us time, the format putt me off as well, and I put lichens and mayfly larvae for the last one for water pollution...concentration of dissolved oxygen and lichens for air..sulfur dioxide in the air.. oh god b2 will be a shIIIIITteeee !!! but for the 0.67 i multiplied it by 100 so i got 67%................. urgh i hate myself aqa needs to evaporate i am not looking forward to c1
The reason I got 66% is i put the sum into the calculator x 100 and before converting to 0.67 it said 2/3. I thought it was 2/3 as in a statistic out of 1 because it's the same for probability out of 1...meaning that 2/3 out of 1 = 66.67%. I don't know if that mistake makes sense to anyone but now I know I've defo dropped one mark Can only go downhill from here!
5. (Original post by MCOD1999)
In the 6 Marker, I mentioned Vitamin C and it prevents Scurvey, would that be a valid point or did I just lose 2 marks?
I wrote exactly the same thing
6. (Original post by MCOD1999)
In the 6 Marker, I mentioned Vitamin C and it prevents Scurvey, would that be a valid point or did I just lose 2 marks?
Well you're right! It's beyond the scope of the spec, but I reckon so many people will make that point it'll be added to the post-standardisation mark scheme
7. (Original post by Cirsium)
Your teachers are definitely being nice at the lower end I'm afraid!
The boundaries move year on year, but they've pretty consistently got higher as students have got better at answering the questions (in 2012 they were ridiculously low).
Lowest I've ever seen for a D is 11 marks, but on some papers it's been 20+ (e.g. last year's B3 you needed 27 for a D!)
.
ok, i failed then! ops...
8. (Original post by AJ2104)
General Predictions
Plenty of papers always start with atomic structure, so that is charges, location of sub-atomic particles, electronic configuration, mass of atoms, etc.
Most likely there will be equations that will need to be balanced, they are getting harder each year though.
AQA has started asking questions about topics such as emulsions and hardening of oils. You should know the structure of the emulsifier and the conditions for hardening. If you are every stuck when it comes to talking about conditions with anything to do on C1 just write down that it needs to be hot with a catalyst.
Looking at examiner reports, they often complain saying that students rarely answer bromine water questions correctly. The examiners love asking questions about topics that students will have no clue on (as we've seen in B1 already).
Displacement reactions might come up. Along with this a lot of practical experiment questions pop up in exams, so you should know how to work out the mean, spot anomalies, drawing graphs, predictions from data, lines of best fit, etc.
A lot of organic chemistry is likely to come up because of the amount of it that was on the course. For example, ethanol, how it is used, how it is produced, the problems with it; alkanes and alkenes and polymers as well. These were topics that examiners commented on in last year's report.
Potential 6 mark questions:
Anything to do with limestone, so quarrying, its production and the cycle of limestone. The reason for this is because it is not in the new specification so we will be one of the last years to be examined about this.
Polymers may appear because it is a big part of the new specification. So you need to know how to draw polymers, ethical issues to do with plastics, etc.
The Miller-Urey experiment hasn't come up in a long while either.
Ok thank you! Good luck in the rest of your exams!
9. im pretty aure i got the snails wrong
10. Why is it mayfly larvae?!i wrote lichen damnnn
11. (Original post by mollieeillom)
The reason I got 66% is i put the sum into the calculator x 100 and before converting to 0.67 it said 2/3. I thought it was 2/3 as in a statistic out of 1 because it's the same for probability out of 1...meaning that 2/3 out of 1 = 66.67%. I don't know if that mistake makes sense to anyone but now I know I've defo dropped one mark Can only go downhill from here!
exactly what i did ahaha, my life is a negative correlation aha x
12. (Original post by InneRs)
Same. I'm reassured.
That's so good I got the same
13. (Original post by Cirsium)
Well you're right! It's beyond the scope of the spec, but I reckon so many people will make that point it'll be added to the post-standardisation mark scheme
YOU CAN QUOTE ME WITH THIS.
Since the 6 Marker was out of the spec, not on the spec. they are forced to take in any kind of result that is true. So DefinateLY Vitamin and other vitamins is correct. 101% Guaranteed.
14. (Original post by Cirsium)
Well you're right! It's beyond the scope of the spec, but I reckon so many people will make that point it'll be added to the post-standardisation mark scheme
i also put that but also added that you need it for your skin and immune system...
15. (Original post by charliej2802)
Anyone else get 0.67%???
yep that's what I got
16. (Original post by esmeeclaire)
I wrote exactly the same thing
(Original post by Cirsium)
Well you're right! It's beyond the scope of the spec, but I reckon so many people will make that point it'll be added to the post-standardisation mark scheme
Glad I'm not the only one who put that, no one ik put that so I started to doubt myself, hopefully it will be on the mark scheme
17. (Original post by EmmaC.)
ok, i failed then! ops...
Nah I'm sure you'll be fine. Not too many easy-easy ticky box questions on this paper so I'm expecting a fairly low D grade boundary.
21 marks of C and D grade content (i.e. joint questions Foundation to Higher) so there's no way they can expect you to get all of that to get a D if that makes sense
18. (Original post by Mattvo15)
Missed out the calculation on the markscheme, The 0.67% of people that die from malaria.
Also, lichen can be used as an indicator of pollution.
im pretty sure it asked for an animal though
19. (Original post by Benrb)
im pretty sure it asked for an animal though
No, it said name a species, then its environmental impact. In which case, its ok
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### Is the 4-line of the E_2 term of the classical Adams spectral sequence known?
In other words: What is $\mathrm{Ext}_{\mathcal{A}}^{4,t}(\mathbb{Z}/2,\mathbb{Z}/2)$? If the 4-line is not known, how much is known about it? Here, $\mathcal{A}$ is the 2-primary Steenrod ... | 611 | 2,287 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2014-35 | latest | en | 0.924231 |
https://en.booksc.org/book/70931286/919e0f | 1,623,681,099,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00060.warc.gz | 239,475,090 | 23,597 | Main Abstract and Applied Analysis Hyperbolic differential-operator equations on a whole axis
# Hyperbolic differential-operator equations on a whole axis
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```HYPERBOLIC DIFFERENTIAL-OPERATOR EQUATIONS
ON A WHOLE AXIS
YAKOV YAKUBOV
We give an abstract interpretation of initial boundary value problems for hyperbolic
equations such that a part of initial boundary value conditions contains also a differentiation on the time t of the same order as equations. The case of stable solutions of
abstract hyperbolic equations is treated. Then we show applications of obtained abstract
results to hyperbolic differential equations which, in particular, may represent the longitudinal displacements of an inhomogeneous rod under the action of forces at the two
ends which are proportional to the acceleration.
1. Introduction
The first attempt to give an abstract interpretation of hyperbolic problems such that a
part of boundary value conditions may contain the differentiation on the time was done
in [8] for almost periodic solutions and oscillations decay cases, and in [5] for hyperbolic
differential-operator equations on a finite interval. In this paper, we continue this study to
the case of differential-operator equations on a whole axis. In particular, we find sufficient
conditions for which the solution of the considered problems is stable.
Let H and F be Hilbert spaces. The set H ⊕ F of all vectors of the form (u,v) where
u ∈ H and v ∈ F, with usual coordinatewise linear operations and the norm
(u,v)
H ⊕F
:= u2H + v2F
1/2
(1.1)
is a Hilbert space and called the orthogonal sum of Hilbert spaces H and F.
For the operator A closed in a Hilbert space H, the domain D(A) is turned into a
Hilbert space H(A) with respect to the norm
1/2
uH(A) := u2 + Au2
.
(1.2)
If H1 and H are two Hilbert spaces where H1 ⊂ H, then H1 can be represented as the
domain D(S) = H1 of a suitable positive definite selfadjoint operator S in H (see, e.g.,
Abstract and Applied Analysis 2004:2 (2004) 99–113
2000 Mathematics Subject Classification: 34G10, 35L15
URL: http://dx.doi.org/10.1155/S1085337504311103
100
; Hyperbolic differential-operator equations on a whole axis
[4, Remark 1.18.10/3]). Then, by [4, Theorem 1.18.10], the interpolation space
H1 ,H
θ,2
= H S1−θ .
(1.3)
2. Hyperbolic differential-operator equations
We give, in this section, an abstract interpretation of initial boundary value problems
for hyperbolic equations such that a part of boundary value conditions contains also the
differentiation on the time t.
Let H and H ν , ν = 1,...,s, be Hilbert spaces. Consider the following Cauchy problem
(abstract “initial boundary value problem”):
L Dt u := u (t) + Bu(t) = 0,
Lν Dt u := Aν0 u(t)
+ Aν2 u(t) = 0,
u(0) = u0 ,
(2.1a)
ν = 1,...,s,
u (0) = u1 ,
(2.1b)
(2.1c)
where t ∈ R; B is an operator in H; Aν0 and Aν2 are operators from a subspace of H into
H ν ; and u(t) from R into H is an unknown function. Note that operators B, Aν0 , and Aν2
are, generally speaking, unbounded.
A function u(t) is called a solution of problem (2.1) if the function t → (u(t),A10 u(t),...,
As0 u(t)) from R into H ⊕ H 1 ⊕ · · · ⊕ H s is twice continuously differentiable, from R into
H(B) ⊕ H 1 ⊕ · · · ⊕ H s is continuous, and u(t) satisfies (2.1).
We say that problem (2.1) is stable if each of its solution u(t) with u0 ∈ H(B), u1 ∈
H(B), is bounded, that is,
u(t) ≤ C,
t ∈ R.
(2.2)
Consider a system of operator pencils corresponding to (2.1a) and (2.1b);
L(λ) := λI + B,
Lν (λ) := λAν0 + Aν2 ,
ν = 1,...,s,
(2.3)
where λ is a complex number.
Theorem 2.1. Let the following conditions be satisfied:
(1) B is a closed operator in a Hilbert space H with a dense domain D(B); the embedding
H(B) ⊂ H is compact;
(2) the operators Aν0 from (H(B),H)1/2,2 into H ν act boundedly and Aν2 , ν = 1,...,s,
from H(B) into H ν act boundedly;
(3) the linear manifold {v | v := (u,A10 u,...,As0 u), u ∈ D(B)} is dense in the Hilbert
space H ⊕ H 1 ⊕ · · · ⊕ H s ;
Yakov Yakubov 101
(4) for all u ∈ D(B), v ∈ D(B),
(Bu,v)H + A12 u,A10 v
H1
+ · · · + As2 u,As0 v
= (u,Bv)H + A10 u,A12 v
H1
Hs
+ · · · + As0 u,As2 v
Hs ;
(2.4)
(5) for all u ∈ D(B),
0 ≤ (Bu,u)H + A12 u,A10 u
H1
+ · · · + As2 u,As0 u
Hs
≤ C u2(H(B),H)1/2,2 ;
(2.5)
(6) some real number λ0 is a regular point for the operator pencil L(λ): u → L(λ)u :=
(L(λ)u,L1 (λ)u,...,Ls (λ)u), which acts boundedly from H(B) onto H ⊕ H 1 ⊕· · ·⊕ H s ;
(7) (u1 ,A10 u1 ,...,As0 u1 ) ∈ Im(B1/2 ), where
D(B) := v | v := u,A10 u,...,As0 u , u ∈ D(B) ,
B u,A10 u,...,As0 u := Bu,A12 u,...,As2 u
(2.6)
is an operator in the Hilbert space Ᏼ := H ⊕ H 1 ⊕ · · · ⊕ H s and from condition (5),
it follows that (Bv,v) ≥ 0, for all v ∈ D(B);
(8) u0 ∈ H(B), u1 ∈ (H(B),H)1/2,2 .
Then there exists a unique solution u(t) of problem (2.1) such that the function t →
(u(t),A10 u(t),...,As0 u(t)) from R into H ⊕ H 1 ⊕ · · · ⊕ H s is twice continuously differentiable, and from R into H(B) ⊕ H 1 ⊕ · · · ⊕ H s is continuous, and for t ∈ R, the following
estimate holds:
s
u(t) + u (t) +
Aν0 u(t)
ν =1
≤ C Bu0 + u0 + u1
Hν
+ Bu(t)
(H(B),H)1/2,2
(2.7)
,
consequently, problem (2.1) is stable.
Proof. Consider, in the Hilbert space Ᏼ := H ⊕ H 1 ⊕ · · · ⊕ H s , the above-mentioned operator B. Then the Cauchy problem
v (t) + Bv(t) = 0,
v(0) = v0 ,
v (0) = v1 ,
(2.8)
is equivalent to the Cauchy problem (2.1), where v0 := (u0 ,A10 u0 ,...,As0 u0 ), v1 := (u1 ,
A10 u1 ,...,As0 u1 ). Indeed, let u(t) be a solution of problem (2.1). Then, v(t) = (u(t),A10 u(t),
...,As0 u(t)) is a solution of problem (2.8). Conversely, let v(t) be a solution of problem
(2.8). Since v(t) ∈ D(B), then v(t) := (u(t),A10 u(t),...,As0 u(t)), where u(t) ∈ D(B), for
all t ∈ R. Substituting v(t) into (2.8), we get that u(t) satisfies (2.1).
102
Hyperbolic differential-operator equations on a whole axis
By virtue of condition (3), D(B) is dense in Ᏼ. By virtue of condition (4), for ṽ1 =
(ũ1 ,A10 ũ1 ,...,As0 ũ1 ) ∈ D(B), ṽ2 = (ũ2 ,A10 ũ2 ,...,As0 ũ2 ) ∈ D(B),
Bṽ1 , ṽ2
Ᏼ
= B ũ1 , ũ2 H + A12 ũ1 ,A10 ũ2 H 1 + · · · + As2 ũ1 ,As0 ũ2 H s
= ũ1 ,B ũ2 H + A10 ũ1 ,A12 ũ2 H 1 + · · · + As0 ũ1 ,As2 ũ2 H s
= ṽ1 , Bṽ2 Ᏼ .
(2.9)
Consequently, the operator B is symmetric. In turn, equation
λ0 v + Bv = F,
F := f , f1 ,..., fs ,
(2.10)
where v = (u,A10 u,...,As0 u), is equivalent to the system
L λ0 u = λ0 u + Bu = f ,
(2.11)
ν = 1,...,s.
Lν λ0 u = λ0 Aν0 u + Aν2 u = fν ,
By virtue of condition (6), problem (2.11) has a unique solution
−1
u = L λ0
f , f1 ,..., fs .
(2.12)
So, a solution of (2.10) has the following form:
−1
v = L λ0
−1
f , f1 ,..., fs ,A10 L λ0
−1
f , f1 ,..., fs ,...,As0 L λ0
f , f1 ,..., fs .
(2.13)
Hence, the operator B is closed and the image Im(λ0 I + B) = Ᏼ, where I is the identity
operator in Ᏼ. By virtue of [3, Chapter Y, Section 3], the operator B is selfadjoint. From
condition (5), it follows that (Bv,v) ≥ 0, v ∈ D(B). Consequently, the operator B is selfadjoint and nonnegative. By condition (7), B−1/2 v1 is well defined. Then, problem (2.8)
has a unique solution v(t) ∈ C 2 (R;Ᏼ(B),Ᏼ) and
v(t) = cos t B1/2 v0 + sin t B1/2 B−1/2 v1 ,
∞
(2.14)
∞
where cos(t B1/2 )v = 0 cos(tλ1/2 )dE(λ)v, sin(t B1/2 )v = 0 sin(tλ1/2 )dE(λ)v, and E(λ) is
the spectral decomposition of the selfadjoint operator B.
Obviously, v0 ∈ Ᏼ(B). Show now that v1 ∈ Ᏼ(B1/2 ). From the definition of the operator B, it follows that Ᏼ(B) ⊂ H(B) ⊕ H 1 ⊕ · · · ⊕ H s . By Section 1, (Ᏼ(B),Ᏼ)1/2,2 =
Ᏼ(B1/2 ). Then,
Ᏼ B1/2 = Ᏼ(B),Ᏼ)1/2,2 ⊂ H(B),H
1/2,2 ⊕ H
1
⊕ · · · ⊕ H s.
(2.15)
Yakov Yakubov 103
Assume, first, that v = (u,A10 u,...,As0 u) ∈ D(B), that is, u ∈ D(B). Then, by virtue of
conditions (2) and (5) and the property of interpolation spaces: Ᏼ(B) ⊂ (Ᏼ(B),Ᏼ)1/2,2 ⊂
Ᏼ, we get
2
v 2Ᏼ(B1/2 ) = B1/2 v Ᏼ + v 2Ᏼ = B1/2 v, B1/2 v Ᏼ + (v,v)Ᏼ = (Bv,v)Ᏼ + (v,v)Ᏼ
= (Bu,u)H +
s
ν =1
Aν2 u,Aν0 u
Hν
+ (u,u)H +
s
ν =1
Aν0 u,Aν0 u
Hν
(2.16)
≤ C u2(H(B),H)1/2,2 .
Now, v1 := (u1 ,A10 u1 ,...,As0 u1 ), where u1 ∈ (H(B),H)1/2,2 . Then, there exists a sequence un ∈ H(B) such that
lim un − u1 (H(B),H)1/2,2 = 0,
(2.17)
n→∞
since the space H(B) is dense in (H(B),H)1/2,2 . Moreover, un is a fundamental sequence,
that is, limn→∞ un − um (H(B),H)1/2,2 = 0. By (2.16), we get
n
v − v m
Ᏼ(B1/2 )
≤ C un − um (H(B),H)1/2,2 ,
(2.18)
where vn = (un ,A10 un ,...,As0 un ). Therefore, the sequence vn is a fundamental in the Hilbert space Ᏼ(B1/2 ). Hence, vn converges in Ᏼ(B1/2 ), that is, there exists v = (u,A10 u,...,
As0 u) ∈ Ᏼ(B1/2 ) such that limn→∞ vn − vᏴ(B1/2 ) = 0. In particular,
lim un − u(H(B),H)1/2,2 = 0.
(2.19)
n→∞
Then, by virtue of (2.17), u = u1 and, therefore, v = v1 . Hence, v1 ∈ Ᏼ(B1/2 ). Moreover,
writing (2.16) for vn and passing to the limit when n → ∞, we get that
v Ᏼ(B1/2 ) ≤ C u(H(B),H)1/2,2
(2.20)
is also true for v = (u,A10 u,...,As0 u), for all u ∈ (H(B),H)1/2,2 .
Since
v (t) = −B1/2 sin t B1/2 v0 + cos t B1/2 v1 ,
v (t) = −B cos t B1/2 v0 − B1/2 sin t B1/2 v1 ,
(2.21)
then
v(t) + v (t) + Bv(t) ≤ C v0 + v1 + Bv0 + B1/2 v1
≤ C v0 + v1 Ᏼ(B1/2 ) + Bv0 , t ∈ R.
From this, by (2.20) and condition (2), the statement of the theorem follows.
(2.22)
104
Hyperbolic differential-operator equations on a whole axis
Consider now such a formulation of problem (2.1), which allows us to insert the firstorder derivative into (2.1a). Let H and H ν , ν = 1,...,s, be Hilbert spaces. Consider the
following Cauchy problem (abstract initial boundary value problem):
L Dt u := u (t) + Au (t) + Bu(t) = h(t),
Lν Dt u := Aν0 u(t)
ν = 1,...,s,
+ Aν2 u(t) = hν (t),
(2.23)
u(0) = u0 ,
u (0) = u1 ,
where t ≥ 0; A and B are operators in H; Aν0 and Aν2 are operators from a subspace of H
into H ν ; u(t) from [0, ∞) into H is an unknown function; h(t) and hν (t) from [0, ∞) into
H and H ν , respectively, are given functions. Note that operators A, B, Aν0 , and Aν2 are,
generally speaking, unbounded.
Consider, in the Hilbert space Ᏼ := H ⊕ H 1 ⊕ · · · ⊕ H s , operators A and B given by
the equalities
D(A) := D(A) ⊕ H 1 ⊕ · · · ⊕ H s ,
A u,v1 ,...,vs := (Au,0,...,0),
(2.24)
D(B) := v | v := u,A10 u,...,As0 u , u ∈ D(B) ,
B u,A10 u,...,As0 u := Bu,A12 u,...,As2 u .
Theorem 2.2. Let the following conditions be satisfied:
(1) B is an operator in a Hilbert space H with a dense domain D(B); A is an operator in
H with D(A) ⊃ (H(B),H)1/2,2 ; the embedding H(B) ⊂ H is compact;
(2) the operators Aν0 , ν = 1,...,s, from (H(B),H)1/2,2 into H ν act compactly, and the
operators Aν2 , ν = 1,...,s, from H(B) into H ν act boundedly;
(3) the linear manifold {v | v := (u,A10 u,...,As0 u), u ∈ D(B)} is dense in the Hilbert
space H ⊕ H 1 ⊕ · · · ⊕ H s ;
(4) for all u ∈ D(B), v ∈ D(B),
(Bu,v)H + A12 u,A10 v
H1
+ · · · + As2 u,As0 v
Hs
= (u,Bv)H + A10 u,A12 v H 1 + · · · + As0 u,As2 v H s ;
(2.25)
(5) for all u ∈ D(B) and some C, c = 0,
C u2(H(B),H)1/2,2 ≥ (Bu,u)H + A12 u,A10 u
H1
+ · · · + As2 u,As0 u
2
2
≥ c2 u2H + A10 uH 1 + · · · + As0 uH s ;
Hs
(2.26)
(6) some real number λ0 is a regular point for the operator pencil L(λ): u → L(λ)u :=
((λI + B)u,(λA10 + A12 )u,...,(λAs0 + As2 )u), which acts boundedly from H(B) onto
H ⊕ H 1 ⊕ · · · ⊕ H s;
(7) A is a skew-symmetric operator in H, that is, A∗ u = −Au, u ∈ D(A), and A from
(H(B),H)1/2,2 into H is bounded;
Yakov Yakubov 105
(8) h ∈ W p1 ((0, ∞);H) ∩ L1 ((0, ∞);H), hν ∈ W p1 ((0, ∞);H ν ) ∩ L1 ((0, ∞);H ν ), ν = 1,2,
for some p > 1;
(9) u0 ∈ H(B), u1 ∈ (H(B),H)1/2,2 .
Then there exists a unique solution u(t) of problem (2.23) such that the function t →
(u(t),A10 u(t),...,As0 u(t)) from [0, ∞) into H ⊕ H 1 ⊕ · · · ⊕ H s is twice continuously differentiable and from [0, ∞) into H(B) ⊕ H 1 ⊕ · · · ⊕ H s is continuous, and for the solution the
following estimate holds:
u(t)
(H(B),H)1/2,2 + u (t) +
s
Aν0 u(t)
ν =1
Hν
2
h ν
≤ C u0 (H(B),H)1/2,2 + u1 (H(B),H)1/2,2 + hL1 ((0,∞);H) +
L1 ((0,∞);H ν ) ,
ν =1
∀t ≥ 0,
(2.27)
consequently (since H(B) ⊂ (H(B),H)1/2,2 ⊂ H), problem (2.23) is stable.
Note that substituting t = −τ, τ ≥ 0, one can consider problem (2.23) for t ≤ 0 too.
Therefore, in fact, Theorem 2.2 is true for t ∈ R.
Proof. Consider, in the Hilbert space Ᏼ := H ⊕ H 1 ⊕ · · · ⊕ H s , the above-mentioned operators A and B. Then the Cauchy problem
v (t) + Av (t) + Bv(t) = f (t),
v(0) = v0 ,
v (0) = v1 ,
(2.28)
is equivalent to the Cauchy problem (2.23), where v0 := (u0 ,A10 u0 ,...,As0 u0 ), v1 := (u1 ,
A10 u1 ,...,As0 u1 ), f (t) := (h(t),h1 (t),... ,hs (t)), and v(t) := (u(t),A10 u(t),...,As0 u(t)) (for
the proof, see the proof of Theorem 2.1).
Apply Theorem A.1 (see the appendix) to problem (2.28), where à := A, B̃ := B. It was
proved in Theorem 2.1 that the operator B is selfadjoint.
From condition (5), it follows that (Bv,v) ≥ c2 (v,v), v ∈ D(B). Consequently, the operator B is selfadjoint and positive-definite. So, by conditions (1), (2), and (3), conditions
(1) and (2) of Theorem A.1 are fulfilled.
From the proof of Theorem 2.1, it follows that Ᏼ(B1/2 ) ⊂ (H(B),H)1/2,2 ⊕ H 1 ⊕ · · · ⊕
s
H . This implies, by conditions (1) and (7), D(A) ⊃ D(B1/2 ), the operator A from Ᏼ(B1/2 )
into Ᏼ is bounded and is skew-symmetric. Hence, condition (3) of Theorem A.1 is satisfied. From condition (8), it follows condition (4) of Theorem A.1. Similar arguments
to those in the proof of Theorem 2.1 gives us that v0 ∈ Ᏼ(B), v1 ∈ Ᏼ(B1/2 ), that is, the
last condition (5) of Theorem A.1 is satisfied too. So, on each interval [0,T], we have a
unique solution v(t) ∈ C 2 ([0,T];Ᏼ(B),Ᏼ(B1/2 ),Ᏼ) of problem (2.28). In order to get the
estimate of Theorem 2.2, one should use the proof of Theorem A.1 from the appendix.
In particular, it follows from the proof that
v(t)
v0
+
= etᏭ
v1
v (t)
t
0
e(t−τ)Ꮽ
0
dτ,
f (τ)
(2.29)
106
Hyperbolic differential-operator equations on a whole axis
where Ꮽ =
0
I
−B −A
v(t)
. Moreover,
Ᏼ(B1/2 )
+ v (t)Ᏼ ≤ v0 Ᏼ(B1/2 ) + v1 Ᏼ +
t
0
f (τ) dτ.
Ᏼ
(2.30)
Using conditions (2), (8), and (9), and the inequality (2.20), we get the estimate of the
theorem.
3. Initial boundary value problems for hyperbolic equations
Consider, in the domain R × [0,1], an initial boundary value problem for the hyperbolic
equation
L Dt u := Dtt2 u(t,x) − Dx b(x)Dx u(t,x) = 0,
(t,x) ∈ R × [0,1],
L1 Dt u := αDtt2 u(t,0) + Dx u(t,0) = 0,
t ∈ R,
L2 Dt u := βDtt2
t ∈ R,
u(0,x) = u0 (x),
u(t,1) + Dx u(t,1) = 0,
Dt u(0,x) = u1 (x),
(3.1)
x ∈ [0,1],
where α, β are real numbers, Dt := ∂/∂t and Dx := ∂/∂x.
This problem was considered, by a different approach, in [1]. As it was mentioned
in [1], “physically, such a problem may represent the longitudinal displacements of an
inhomogeneous rod under the action of forces at the two ends which are proportional
to the acceleration. In particular, this situation is realized if there are massive loads at the
ends (see, e.g., [2, Chapter 12]) and in this case we have α < 0 and β > 0.”
Theorem 3.1. Let the following conditions be satisfied:
(1) b ∈ C 1 [0,1]; b(x) > 0 for x ∈ [0,1];
(2) α < 0, β > 0;
(3) u0 ∈ W22 (0,1), u1 ∈ W21 (0,1);
1
(4) 0 u1 (x)dx − αb(0)u1 (0) + βb(1)u1 (1) = 0.
Then there exists a unique solution u(t,x) of problem (3.1) such that the function t →
(u(t,x),u(t,0),u(t,1)) from R into L2 (0,1) ⊕ C ⊕ C is twice continuously differentiable, and
from R into W22 (0,1) ⊕ C ⊕ C is continuous, and for t ∈ R the following estimate holds:
u(t, ·)
L2 (0,1) +
2
D u(t, ·)
tt
2
+ Dxx
u(t, ·)L2 (0,1)
2
D u(t,0) + D2 u(t,1)
tt
tt
≤ C u 0 2
+ u 1 1
,
L2 (0,1) +
W2 (0,1)
(3.2)
W2 (0,1)
consequently, problem (3.1) is stable.
Proof. Apply Theorem 2.1. Consider, in the Hilbert space H := L2 (0,1), an operator B
given by the equalities
D(B) := W22 (0,1),
Bu := − b(x)u (x) .
(3.3)
Yakov Yakubov 107
Taking H 1 := −(b(0)/α)C, H 2 := (b(1)/β)C, and
A10 u := αu(0),
A20 u := βu(1),
(3.4)
A22 u := u (1),
A12 u := u (0),
problem (3.1) can be rewritten in the form (2.1), where u(t) := u(t, ·) is a function with
values in the Hilbert space H := L2 (0,1), and u0 := u0 (·), u1 := u1 (·).
From [4, Section 3.2.5], it follows that condition (1) of Theorem 2.1 is satisfied. From
[4, Section 4.3.1], it follows that (W22 (0,1),L2 (0,1))1/2,2 = W21 (0,1). Then condition (2)
of Theorem 2.1 is satisfied too. Condition (3) of Theorem 2.1 follows from Theorem A.2
(see the appendix). We prove conditions (4) and (5) of Theorem 2.1. For u1 ∈ W22 (0,1),
u2 ∈ W22 (0,1), we have
Bu1 ,u2
L2 (0,1) +
1
=−
0
A12 u1 ,A10 u2
0
−(b(0)/α)C
+ A22 u1 ,A20 u2
(b(1)/β)C
du (x)
b(1)
d
b(0)
b(x) 1
u2 (x)dx −
u (0) · αu2 (0) +
u (1)βu2 (1)
dx
dx
α 1
β 1
1
=−
u1 (x)
1
d
du (x)
b(x) 2
dx − b(x)u1 (x)u2 (x)0
dx
dx
(3.5)
1
+ u1 (x) b(x)u2 (x) 0 − b(0)u1 (0)u2 (0) + b(1)u1 (1)u2 (1)
= u1 ,Bu2 L2 (0,1) + A10 u1 ,A12 u2 −(b(0)/α)C + A20 u1 ,A22 u2 (b(1)/β)C ,
that is, condition (4) of Theorem 2.1 is satisfied. For u ∈ W22 (0,1), we have
(Bu,u)L2 (0,1) + A12 u,A10 u
=
=
1
0
1
0
−(b(0)/α)C
+ A22 u,A20 u
(b(1)/β)C
2
1
b(x)u (x) dx − b(x)u (x)u(x) − b(0)u (0)u(0) + b(1)u (1)u(1)
0
(3.6)
2
b(x)u (x) dx ≥ 0.
1
On the other hand, 0 b(x)|u (x)|2 dx ≤ C u2W21 (0,1) , that is, condition (5) of Theorem 2.1
is satisfied too. Denote
L(λ)u := (λI + B)u = λu(x) − b(x)u (x) ,
L1 (λ)u := λA10 + A12 u = λαu(0) + u (0),
(3.7)
L2 (λ)u := λA20 + A22 u = λβu(1) + u (1).
From Theorem A.3 (see the appendix), it follows that for any ε > 0, there exists Rε > 0
such that for all complex numbers λ which satisfy |λ| > Rε and lying inside the angle
−π + ε < argλ < π − ε,
(3.8)
108
Hyperbolic differential-operator equations on a whole axis
the operator L(λ) : u → L(λ)u := (L(λ)u,L1 (λ)u,L2 (λ)u) from W22 (0,1) onto L2 (0,1) ⊕
−(b(0)/α)C ⊕ (b(1)/β)C is an isomorphism, that is, condition (6) of Theorem 2.1 is satisfied. Condition (7) of Theorem 2.1 is fulfilled in view of condition (4) (see for details
[1]). Condition (8) of Theorem 2.1 follows from condition (3). So, for problem (3.1), all
conditions of Theorem 2.1 are fulfilled and the statement of Theorem 3.1 follows.
We present now an application of Theorem 2.2. Consider, in the domain [0, ∞) ×
[0,1], an initial boundary value problem for the hyperbolic equation
L Dt u := Dtt2 u(t,x) + ia(x)Dt u(t,x) − Dx b(x)Dx u(t,x) + c(x)u(t,x)
= h(t,x), (t,x) ∈ [0, ∞) × [0,1],
L1 Dt u := αDtt2 u(t,0) + Dx u(t,0) = h1 (t),
L2 Dt u := βDtt2 u(t,1) + Dx u(t,1) = h2 (t),
u(0,x) = u0 (x),
Dt u(0,x) = u1 (x),
t ∈ [0, ∞),
(3.9)
t ∈ [0, ∞),
x ∈ [0,1],
√
where α, β are real numbers, i = −1, Dt := ∂/∂t, Dx := ∂/∂x.
Theorem 3.2. Let the following conditions be satisfied:
(1) a ∈ C[0,1] and is real-valued; b ∈ C 1 [0,1], b(x) > 0 for x ∈ [0,1]; c ∈ C[0,1], c(x) >
0 for x ∈ [0,1];
(2) α < 0, β > 0;
(3) h ∈ W p1 ((0, ∞);L2 (0,1)) ∩ L1 ((0, ∞);L2 (0,1)), hν ∈ W p1 (0, ∞) ∩ L1 (0, ∞), ν = 1,2,
for some p > 1;
(4) u0 ∈ W22 (0,1), u1 ∈ W21 (0,1).
Then there exists a unique solution u(t,x) of problem (3.9) such that the function t →
(u(t,x),u(t,0),u(t,1)) from [0, ∞) into L2 (0,1) ⊕ C ⊕ C is twice continuously differentiable,
and from [0, ∞) into W22 (0,1) ⊕ C ⊕ C is continuous, and for the solution the following
estimate holds:
u(t, ·)
L2 (0,1) +
Dt u(t, ·)
L2 (0,1) +
Dt u(t,0) + Dt u(t,1)
≤ C u0 W21 (0,1) + u1 W21 (0,1) +
∞
0
h(t, ·)
L2 (0,1) dt +
2
hν (t)dt ,
∞
ν =1 0
∀t ≥ 0,
(3.10)
consequently, problem (3.9) is stable.
Note that this theorem, as Theorem 2.2, is also true for t ≤ 0 and, therefore, for t ∈ R.
Proof. Apply Theorem 2.2. Consider, in the Hilbert space H := L2 (0,1), operators A and
B given by the equalities
D(A) := L2 (0,1),
D(B) := W22 (0,1),
Au := ia(x)u(x),
Bu := − b(x)u (x) + c(x)u(x).
(3.11)
Yakov Yakubov 109
Taking H 1 := −(b(0)/α)C, H 2 := (b(1)/β)C, and A10 u := αu(0), A20 u := βu(1), A12 u :=
u (0), A22 u := u (1), problem (3.9) can be rewritten in the form (2.23), where u(t) :=
u(t, ·) and h(t) := h(t, ·) are functions with values in the Hilbert space H := L2 (0,1), and
ϕ0 := ϕ0 (·), ϕ1 := ϕ1 (·).
From [4, Section 3.2.5], it follows that D(B) is dense in H, and the embedding H(B) ⊂
H is compact, and from [4, Section 4.3.1] it follows that (W22 (0,1),L2 (0,1))1/2,2 =W21 (0,1).
Therefore, condition (1) of Theorem 2.2 is satisfied. It is well known that the embedding
W2k (0,1) ⊂ C m [0,1], k > m ≥ 0, is compact (see, e.g., [4, Section 4.10.2, formula (15)]).
Then condition (2) of Theorem 2.2 is satisfied too. Condition (3) of Theorem 2.2 follows
from Theorem A.2 (see the appendix). We prove conditions (4) and (5) of Theorem 2.2.
For u ∈ W22 (0,1), v ∈ W22 (0,1), we have
(Bu,v)L2 (0,1) + A12 u,A10 v
=−
−
=−
1
0
−(b(0)/α)C
+ A22 u,A20 v
(b(1)/β)C
1
du(x)
d
b(x)
v(x)dx +
dx
dx
0
c(x)u(x)v(x)dx
b(1)
b(0)
u (0)αv(0) +
u (1)βv(1)
α
β
1
0
1
d
dv(x)
u(x)
b(x)
dx +
dx
dx
0
1
c(x)u(x)v(x)dx − b(x)u (x)v(x)
(3.12)
0
1
+ u(x) b(x)v (x) 0 − b(0)u (0)v(0) + b(1)u (1)v(1)
= (u,Bv)L2 (0,1) + A10 u,A12 v −(b(0)/α)C + A20 u,A22 v (b(1)/β)C ,
that is, condition (4) of Theorem 2.2 is satisfied. For u ∈ W22 (0,1), using conditions (1)
and (2) and that W21 (0,1) ⊂ C[0,1] is bounded, we have
(Bu,u)L2 (0,1) + A12 u,A10 u
=
1
0
−(b(0)/α)C
2
b(x)u (x) dx +
1
0
+ A22 u,A20 u
(b(1)/β)C
2
1
c(x)u(x) dx − b(x)u (x)u(x)0
− b(0)u (0)u(0) + b(1)u (1)u(1)
=
1
0
2
b(x)u (x) dx +
1
2
≥ √ min
1
0
2
c(x)u(x) dx
(3.13)
min b(x), min c(x) u2W21 (0,1)
x∈[0,1]
x∈[0,1]
2
2
− b(0)αu(0) + b(1)βu(1)
≥ c2 u2L2 (0,1)
2
2
= c2 u2L2 (0,1) + A10 u−(b(0)/α)C + A20 u(b(1)/β)C ,
1
1
∃c = 0.
On the other hand, 0 b(x)|u (x)|2 dx + 0 c(x)|u(x)|2 dx ≤ C u2W21 (0,1) , that is, condition (5) of Theorem 2.2 is satisfied too. Condition (6) of Theorem 2.2 is checked as in
110
Hyperbolic differential-operator equations on a whole axis
the proof of Theorem 3.1. We check condition (7) of Theorem 2.2. Take u,v ∈ D(A) =
L2 (0,1). Then,
(Au,v)L2 (0,1) =
1
0
ia(x)u(x)v(x)dx = −
1
0
u(x)ia(x)v(x)dx
(3.14)
= (u, −Av)L2 (0,1) .
Conditions (8) and (9) of Theorem 2.2 are trivial. So, for problem (3.9), all conditions of
Theorem 2.2 are fulfilled and the statement of Theorem 3.2 follows.
Appendix
Consider, in a Hilbert space H, the Cauchy problem for the second-order hyperbolic
differential-operator equation
L(D)u := u (t) + Ãu (t) + B̃u(t) = f (t),
u(0) = g0 ,
t ∈ [0,T],
u (0) = g1 ,
(A.1)
and the characteristic operator pencil
L(λ) := λ2 + λà + B̃.
(A.2)
Theorem A.1 (see [7, Theorem 6.4.3]). Let the following conditions be satisfied:
(1) B̃ is a selfadjoint positive-definite operator in a Hilbert space H;
(2) the embedding H(B̃) ⊂ H is compact;
(3) Ã is a skew-symmetric operator in H, that is, Ã∗ u = −Ãu, u ∈ D(Ã); the operator Ã
from H(B̃ 1/2 ) into H is bounded;
(4) f ∈ W p1 ((0,T);H), where p > 1;
(5) g0 ∈ D(B̃), g1 ∈ D(B̃1/2 ).
Then, problem (A.1) has a unique solution u ∈ C 2 ([0,T];H(B̃),H(B̃ 1/2 ),H), and the
solution can be expanded to the series
u(t) =
∞
e λk t
1/2 2 + λk 2 uk 2
k=1 B̃ uk
× B̃g0 − λk g1 ,uk − λk
t
0
e
−λ k τ
(A.3)
f (τ),uk dτ uk ,
where λk are purely imaginary eigenvalues and uk are the corresponding eigenvectors of operator pencil (A.2), and the series converges in the sense of the space C 2 ([0,T];H(B̃),H(B̃ 1/2),H).
Denote
Aν0 u : = αν u(mν ) (0) + βν u(mν ) (1) +
Nν
j =1
δν j u(mν ) xν j + Tν u,
ν = 1,...,m.
(A.4)
Yakov Yakubov 111
Theorem A.2 (see [7, Theorem 3.6.2]). Let the following conditions be satisfied:
(1) m ≥ 1, mν ≥ 0, 0 ≤ s ≤ m;
(2) a system of functionals (A.4) is p-regular with respect to a system of numbers ω j :=
e2πi(( j −1)/m) , j = 1,...,m, that is,
α1 ω1m1
.
.
.
.
.
.
α ωmm
m 1
···
..
.
..
.
···
1
α1 ωm
p
..
.
1
β1 ω m
p+1
..
.
···
..
.
m
αm ωm
p
..
.
m
βm ω m
p+1
..
.
..
.
···
m1
β 1 ωm
..
.
= 0,
..
.
mm
β m ωm
(A.5)
where p = m/2 if m is even, p = [m/2] or p = [m/2] + 1 if m is odd, xν j ∈ (0,1) and,
for some q ∈ (1, ∞), functionals Tν in Wqmν (0,1) are continuous.
Then, the linear manifold
(u,v) | u ∈ C ∞ [0,1], Aν0 u = 0, ν = s + 1,...,m, v := A10 u,...,As0 u ,
(A.6)
is dense in the space Wq (0,1) +̇ Cs , ≤ min{mν }.
Consider a principally boundary value problem for an ordinary differential equation
with a variable coefficient in case when the spectral parameter appears linearly in the
equation and can appear in boundary-functional conditions
L(λ)u := λu(x) + a(x)u(m) (x) + Bu|x = f (x),
Lν (λ)u := λ αν u(mν ) (0) + βν u(mν ) (1) +
Nν
x ∈ (0,1),
δν j u(mν ) xν j + Tν u
j =1
+ Tν0 u = gν ,
Lν u := αν u(mν ) (0) + βν u(mν ) (1) +
Nν
(A.7a)
(A.7b)
ν = 1,...,s,
δν j u(mν ) xν j + Tν u = 0,
ν = s + 1,...,m,
(A.7c)
j =1
where m ≥ 1, mν ≤ m − 1, xν j ∈ (0,1), 0 ≤ s ≤ m, B is an operator in L2 (0,1), Tν and Tν0
are functionals in L2 (0,1).
Theorem A.3 (see [6]). Let the following conditions be satisfied:
(1) m ≥ 1; mν ≤ m − 1; 0 ≤ s ≤ m;
(2) a ∈ C[0,1]; a(x) = 0; a(0) = a(1); supx∈[0,1] arg a(x) − inf x∈[0,1] arg a(x) < 2π if m is
even; supx∈[0,1] arg a(x) − inf x∈[0,1] arga(x) < π if m is odd;
(3) for all ε > 0,
BuL2 (0,1) ≤ εuW2m (0,1) + C(ε)uL2 (0,1) ,
u ∈ W2m (0,1);
(A.8)
112
Hyperbolic differential-operator equations on a whole axis
(4) functionals Tν in W2mν (0,1) and functionals Tν0 in W2m−ε (0,1), for some ε > 0, are
continuous;
(5) system (A.4) is p-regular with respect to a system of numbers ω j = e2πi(( j −1)/m) , j =
1,...,m (see Theorem A.2).
Then for any ε > 0, there exists Rε > 0 such that for all complex numbers λ which satisfy
|λ| > Rε and for m = 2p lying inside the angle,
πm
πm
− π + sup arga(x) + ε < argλ <
+ π + inf arga(x) − ε,
2
2
x∈[0,1]
x∈[0,1]
(A.9)
for m = 2p + 1 lying inside the angle,
πm
πm
+ sup arg a(x) + ε < arg λ <
+ π + inf arg a(x) − ε,
2 x∈[0,1]
2
x∈[0,1]
(A.10)
and for m = 2p − 1 lying inside the angle,
πm
πm
− π + sup arg a(x) + ε < arg λ <
+ inf arg a(x) − ε,
2
2 x∈[0,1]
x∈[0,1]
(A.11)
the operator L(λ) : u → L(λ)u := (L(λ)u,L1 (λ)u,...,Ls (λ)u) from W2m ((0,1); Lν u = 0, ν =
s + 1,...,m) onto L2 (0,1)+̇Cs is an isomorphism, and for these λ for a solution of problem
(A.7), the estimate
uW2m (0,1) + |λ| uL2 (0,1) +
s
s
Aν0 u ≤ C(ε) f L (0,1) +
g ν
2
ν =1
(A.12)
ν =1
is valid, where Aν0 is defined by (A.4).
Note that if boundary-functional conditions (A.7b) and (A.7c) are principally local,
that is, αν = 0 or βν = 0 for all ν = 1,...,m, then the condition a(0) = a(1) should be
omitted.
Acknowledgment
The author was supported by the Israel Ministry of Absorption.
References
[1]
[2]
[3]
[4]
[5]
P. Lancaster, A. Shkalikov, and Q. Ye, Strongly definitizable linear pencils in Hilbert space, Integral
Equations Operator Theory 17 (1993), no. 3, 338–360.
C. C. Lin and L. A. Segel, Mathematics Applied to Deterministic Problems in the Natural Sciences,
Macmillan Publishing, New York, 1974.
K. Maurin, Methods of Hilbert Spaces, Monografie Matematyczne, vol. 45, Państwowe
Wydawnictwo Naukowe, Warsaw, 1967, translated from Polish by Andrzej Alexiewicz and
H. Triebel, Interpolation Theory, Function Spaces, Differential Operators, North-Holland Mathematical Library, vol. 18, North-Holland, Amsterdam, 1978.
S. Yakubov and Ya. Yakubov, An initial boundary value problem for hyperbolic diferentialoperator equations on a finite interval, accepted in Differential and Integral Equations.
Yakov Yakubov 113
[6]
[7]
[8]
, Abel basis of root functions of regular boundary value problems, Math. Nachr. 197
(1999), 157–187.
, Differential-Operator Equations. Ordinary and Partial Differential Equations, Chapman & Hall/CRC Monographs and Surveys in Pure and Applied Mathematics, vol. 103,
Chapman & Hall/CRC, Boca Raton, 2000.
Ya. Yakubov, Almost periodic solutions and oscillations decay for hyperbolic differential-operator
equations, Funct. Differ. Equ. 10 (2003), no. 1-2, 315–330.
Yakov Yakubov: Raymond and Beverly Sackler Faculty of Exact Sciences, School of Mathematical
Sciences, Tel-Aviv University, Ramat-Aviv 69978, Israel
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``` | 13,751 | 31,610 | {"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-2021-25 | latest | en | 0.883179 |
https://metacpan.org/pod/Math::NumSeq::AlmostPrimes | 1,526,880,972,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863949.27/warc/CC-MAIN-20180521043741-20180521063741-00375.warc.gz | 616,610,942 | 11,169 | ++ed by:
Kevin Ryde
and 1 contributors
# NAME
Math::NumSeq::AlmostPrimes -- semiprimes and other fixed number of prime factors
# SYNOPSIS
`````` use Math::NumSeq::AlmostPrimes;
my \$seq = Math::NumSeq::AlmostPrimes->new (factor_count => 2);
my (\$i, \$value) = \$seq->next;``````
# DESCRIPTION
This sequence is various "almost prime" numbers. These are numbers with a given number of prime factors. The default is 2 prime factors, which are the semi-primes. For example 15 because 15=3*5.
`````` 4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, ...
# starting i=1``````
## Factor Count
`factor_count => \$c` controls how many prime factors are to be used. 1 would be the primes themselves (the same as Math::NumSeq::Primes). Or for example factor count 4 is as follows. 60 is present because 60=2*2*3*5 has precisely 4 prime factors.
`````` # factor_count => 4
16, 24, 36, 40, 54, 60, ...``````
The first number in the sequence is 2^factor_count, being prime factor 2 repeated factor_count many times.
## Multiplicity
`multiplicity => 'distinct'` asks for products of distinct primes. For the default factor count 2 this means exclude squares like 4=2*2, which leaves
`````` # multiplicity => 'distinct'
6, 10, 14, 15, 21, ...``````
For other factor counts, multiplicity "distinct" eliminates any numbers with repeated factors, leaving only square-free numbers. For example factor count 4 becomes
`````` # factor_count => 4, multiplicity => 'distinct'
210, 330, 390, 462, 510, 546, ...``````
For multiplicity "distinct" the first value in the sequence is a primorial (see Math::NumSeq::Primorials), being the first `factor_count` many primes multipled together. For example 210 above is primorial 2*3*5*7.
# FUNCTIONS
See "FUNCTIONS" in Math::NumSeq for behaviour common to all sequence classes.
`\$seq = Math::NumSeq::AlmostPrimes->new ()`
`\$seq = Math::NumSeq::AlmostPrimes->new (factor_count => \$integer, multiplicity => \$str)`
Create and return a new sequence object. `multiplicity` can be
`````` "repeated" repeated primes allowed (the default)
"distinct" all primes must be distinct``````
`\$bool = \$seq->pred(\$value)`
Return true if `\$value` is an almost-prime, ie. it has exactly `factor_count` many prime factors, and if `distinct` is true then all those factors different.
This check requires factorizing `\$value` and in the current code a hard limit of 2**32 is placed on values to be checked, in the interests of not going into a near-infinite loop.
Math::NumSeq::Primorials
http://user42.tuxfamily.org/math-numseq/index.html
Copyright 2010, 2011, 2012, 2013, 2014 Kevin Ryde
Math-NumSeq is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 3, or (at your option) any later version.
Math-NumSeq is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
You should have received a copy of the GNU General Public License along with Math-NumSeq. If not, see <http://www.gnu.org/licenses/>. | 898 | 3,231 | {"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-2018-22 | longest | en | 0.867247 |
https://johnnydeppnetwork.com/qa/what-is-function-and-method.html | 1,624,047,991,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00177.warc.gz | 300,300,489 | 8,739 | WHAT IS Function And Method?
What is C method?
A method is a code block that contains a series of statements.
A program causes the statements to be executed by calling the method and specifying any required method arguments.
In C#, every executed instruction is performed in the context of a method..
Is Python a function?
A function is a block of code which only runs when it is called. You can pass data, known as parameters, into a function.
What are the 7 parent functions?
The following figures show the graphs of parent functions: linear, quadratic, cubic, absolute, reciprocal, exponential, logarithmic, square root, sine, cosine, tangent.
What are the two main types of functions?
Types of FunctionsOne – one function (Injective function)Many – one function.Onto – function (Surjective Function)Into – function.Polynomial function.Linear Function.Identical Function.Quadratic Function.More items…•
What is difference between method and function?
Method and a function are the same, with different terms. A method is a procedure or function in object-oriented programming. A function is a group of reusable code which can be called anywhere in your program. This eliminates the need for writing the same code again and again.
What is a function and not a function?
A function is a relation in which each input has only one output. : y is a function of x, x is not a function of y (y = 9 has multiple outputs). … : y is not a function of x (x = 1 has multiple outputs), x is not a function of y (y = 2 has multiple outputs).
What are methods called in Python?
In Python, a method is a function that is available for a given object because of the object’s type. For example, if you create my_list = [1, 2, 3] , the append method can be applied to my_list because it’s a Python list: my_list. append(4) . All lists have an append method simply because they are lists.
What are the 8 types of functions?
The eight types are linear, power, quadratic, polynomial, rational, exponential, logarithmic, and sinusoidal.
What is a function in your own words?
A function is a relation that maps a set of inputs, or the domain, to the set of outputs, or the range. Note that for a function, one input cannot map to more than one output, but one output may be mapped to more than one input.
What is function explain with example?
A function is a mapping from a set of inputs (the domain) to a set of possible outputs (the codomain). The definition of a function is based on a set of ordered pairs, where the first element in each pair is from the domain and the second is from the codomain.
What is difference between method and function in Python?
Since we call a method on an object, it can access the data within it. A method may alter an object’s state, but Python function usually only operates on it, and then prints something or returns a value.
What are the 3 types of functions?
There are 3 types of functions: Linear. Quadratic. Exponential.
Is a method a procedure?
Procedure and method are very similar, but method is a smaller part of the procedure. Think of it in this terms: In business, you have a plan. A series of steps, or a program, which must be followed to to accomplish a goal. A procedure is a smaller part of that program.
What are the three main differences between a method and a function?
A function is a piece of code that is called by name. It can be passed data to operate on (i.e. the parameters) and can optionally return data (the return value). All data that is passed to a function is explicitly passed. A method is a piece of code that is called by a name that is associated with an object.
Why would a function method be used?
Function is a set of logic that can be used to manipulate data. While, Method is function that is used to manipulate the data of the object where it belongs. So technically, if you have a function that is not completely related to your class but was declared in the class, its not a method; It’s called a bad design.
What is the definition of function?
A technical definition of a function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output. We can write the statement that f is a function from X to Y using the function notation f:X→Y. …
What is a method in oops?
A method in object-oriented programming (OOP) is a procedure associated with a message and an object. … This allows the sending objects to invoke behaviors and to delegate the implementation of those behaviors to the receiving object. A method in Java programming sets the behavior of a class object.
WHAT IS function and method in Python?
A method in python is somewhat similar to a function, except it is associated with object/classes. Methods in python are very similar to functions except for two major differences. The method is implicitly used for an object for which it is called. The method is accessible to data that is contained within the class.
What is __ init __ in Python?
“__init__” is a reseved method in python classes. … This method is called when an object is created from a class and it allows the class to initialize the attributes of the class.
Is a function a procedure?
Procedures or functions? A procedure performs a task, whereas a function produces information. Functions differ from procedures in that functions return values, unlike procedures which do not. However, parameters can be passed to both procedures and functions. | 1,175 | 5,484 | {"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-2021-25 | latest | en | 0.900962 |
http://nrich.maths.org/public/leg.php?code=-333&cl=2&cldcmpid=149 | 1,506,288,142,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690211.67/warc/CC-MAIN-20170924205308-20170924225308-00573.warc.gz | 249,782,475 | 9,770 | # Search by Topic
#### Resources tagged with Investigations similar to Cut and Make:
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### Triangle Shapes
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This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
### Redblue
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Investigate the number of paths you can take from one vertex to another in these 3D shapes. Is it possible to take an odd number and an even number of paths to the same vertex?
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Can you find ways of joining cubes together so that 28 faces are visible?
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In this challenge, you will work in a group to investigate circular fences enclosing trees that are planted in square or triangular arrangements.
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We went to the cinema and decided to buy some bags of popcorn so we asked about the prices. Investigate how much popcorn each bag holds so find out which we might have bought.
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What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
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Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
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This practical investigation invites you to make tessellating shapes in a similar way to the artist Escher.
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Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot.
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Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
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What do these two triangles have in common? How are they related?
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A group of children are discussing the height of a tall tree. How would you go about finding out its height?
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This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
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How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
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Can you create more models that follow these rules?
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What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
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Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
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How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
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An activity making various patterns with 2 x 1 rectangular tiles.
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Explore the different tunes you can make with these five gourds. What are the similarities and differences between the two tunes you are given?
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Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
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How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
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If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
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How many triangles can you make on the 3 by 3 pegboard?
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In how many ways can you stack these rods, following the rules?
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Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
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What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
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Have a go at this 3D extension to the Pebbles problem.
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Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
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I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
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What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
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Investigate the different ways you could split up these rooms so that you have double the number.
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Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
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If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
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You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
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The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
### Egyptian Rope
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The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this?
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How many ways can you find of tiling the square patio, using square tiles of different sizes?
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When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
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Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
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Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
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Ben has five coins in his pocket. How much money might he have?
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While we were sorting some papers we found 3 strange sheets which seemed to come from small books but there were page numbers at the foot of each page. Did the pages come from the same book? | 2,079 | 9,347 | {"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-2017-39 | latest | en | 0.901337 |
https://www.cs.hmc.edu/~crossi/funwiki/index.cgi?TheNumber | 1,713,851,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00632.warc.gz | 669,519,753 | 3,032 | # TheNumber
Graham's number G_63 is the best upper bound proven for this Ramsey theory problem:
The smallest dimension n of a hypercube such that if the lines joining all pairs of corners are two-colored, a planar complete graph of one color will be forced. Stated colloquially, this is equivalent to considering every possible committee from some number of people n and enumerating every pair of committees. Now assign each pair of committees to one of two groups, and find the smallest n that will guarantee that there are four committees in which all pairs fall in the same group and all the people belong to an even number of committees.
(Hoffman 1998 p. 54, quoted by http://mathworld.wolfram.com/GrahamsNumber.html)
In other words, each person is a dimension of the cube, each committee is a vertex, a pair of committees is a line joining vertices, the two groups are the colorings of the lines, and the people on an even number of committees ensures that the four chosen vertices (committees) lie on a hyperplane. (But I don't think that last clause is correct. In the 3-cube, (000),(110),(101), and (011) don't lie on a hyperplane, but all three people are on two committees.)
The paper introducing Graham's number can be found at http://www.jstor.org/view/00029947/di970177/97p00383/0 (page 290 is where he mentions the number).
To define Graham's number, we use arrow notation. A single up-arrow is the exponentiation operator, or iterated multiplication (multiplication of course being iterated addition). A double arrow is iterated exponentiation. The triple arrow is iterated double-arrow, and so on:
``` m*n = m+m+m+m+...+m, n times.
m^n = m*m*m*m*...*m, n times.
m^^n = m^m^m^m^...^m, n times. This is sometimes referred to as a power tower.
m^^^n = m^^m^^m^^m^^...^^m, n times
= m^^m^^m^^m^^..^^x, with (n-2) m's,
where x = m^m^m^m^..^m, m times.
```
And so on: Clearly this gets big quickly (although not as fast as other functions out there). An important property is that the order of operations starts at the right side and works to the left, as in the last example. Now for some examples:
``` 2^2 = 2*2 = 4
2^^2 = 2^2 = 4
2^^^2 = 2^^2 = 4
```
``` 2^3 = 2*2*2 = 8
2^^3 = 2^2^2 = 2^(4) = 16
2^^^3 = 2^^2^^2 = 2^^4 = 2^2^2^2 = 2^2^4 = 2^16 = 65536
```
``` 3^3 = 3*3*3 = 27
3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987
3^^^3 = 3^^3^^3 = 3^^(7,625,597,484,987) = 3^3^3^...^3 (7,625,597,484,987 times)
```
``` 4^4 = 4*4*4*4 = 256
4^^4 = 4^4^4^4 = 4^4^256 = kinda big (but not very compared to later examples)
```
Okay, now we have an idea of up arrow notation, let's go to G_63. We start with G_0 (these are capital Gs with numerical subscripts).
``` G_0 = 3^^^^3
= 3^^^3^^^3
= 3^^^(3^^(7,625,597,484,987))
= REALLY HUGE
```
This blows the number of particles in the universe out of the water. (Last I hear, the estimated number of elementary particles was something like 2^100, but even if it were 2^(100^100) it isn't even comparable to this number.)
Now, to get the next item in the series:
``` G_1 = 3^^^^..^3 where there are G_0 up arrows
```
wow
To continue:
G_2 = 3^^^^..^3 where there are G_1 up arrows
Continue this until you reach G_63 and you've found the upper limit to the problem stated before.
super wow
Interesting facts:
1. G_{-1} = 4 (this is obvious if you look at it).
2. G_63 in decimal notation ends in 7.
3. This is only an upper bound on the answer... a lower bound is 11.
4. G_63 is the largest finite number used in a "serious" mathematical proof.
About notation: The Mathworld article on Graham's Number uses an alternative "little g" notation. It defines g_1 = G_0, g_2 = G_1, and so on. Therefore it gives the upper bound as g_64 instead of G_63.
FunWiki | RecentChanges | Preferences | 1,166 | 3,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-18 | latest | en | 0.909107 |
https://houbrw.com/how-to-win-the-lottery-4/ | 1,720,935,447,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514548.45/warc/CC-MAIN-20240714032952-20240714062952-00698.warc.gz | 255,825,331 | 14,775 | # How to Win the Lottery
Whether or not winning the lottery is a realistic goal, you can boost your odds by learning a few things about probability and statistics. You can also try your hand at mathematical predictions or join a lottery pool or syndicate. Some players even use a combination of these strategies to increase their chances of walking away with the big prize.
In the 17th century, lotteries were very popular in many parts of Europe. They were a painless way for governments to raise money for a wide range of public uses. Many of these included roads, canals, and libraries. They also funded colleges, churches, and military fortifications. Lotteries were very popular in colonial America as well. In fact, they helped finance the foundation of Princeton and Columbia universities in 1740.
Winning the lottery is a game of chance, but you can improve your chances by playing carefully and avoiding superstitions. It’s important to set a budget for how much you can afford to spend each week or month on tickets, and stick to it. You should also choose your numbers based on statistical analysis and past winning patterns. By doing so, you can avoid choosing numbers that have already been drawn or avoiding patterns such as diagonal lines or zig-zags.
It’s also a good idea to avoid FOMO (fear of missing out) by purchasing tickets only when you have enough money to do so. In addition, you should limit the number of tickets that you purchase to one per draw. This will give you the best chance of winning without going over your budget.
The more tickets that you buy, the higher your chances of winning. However, it’s essential to remember that the prize money is only a small percentage of your total ticket cost. The rest of the ticket price goes toward the cost of printing, processing, and distribution. In other words, the jackpot is just a bonus.
Some people have managed to win the lottery several times, including Stefan Mandel. The Romanian-born mathematician used his knowledge of probability and combinatorial mathematics to develop a system for picking winners. His formula won him 14 prizes and more than \$1.3 million, but he only kept \$97,000 of it after paying out investors. He now lives a quiet life in Vanuatu, a South Pacific island known for its volcanoes and waterfalls.
It’s not uncommon for a lotto winner to blow all of their winnings. Many go on a spending spree and end up with huge houses, Porsches, or gambling debts. In some cases, they even get slammed with lawsuits. To avoid this, it’s a good idea to assemble a financial triad with the help of a certified financial planner. These experts can help you plan a sensible financial strategy for your future, so that you can enjoy your windfall while keeping it safe. They can even teach you how to invest your winnings so that they can grow over time. They will also help you make informed decisions about which investments are a wise choice for your specific situation. | 617 | 2,975 | {"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-2024-30 | latest | en | 0.975313 |
http://mathhelpforum.com/pre-calculus/26625-question-set-theory-print.html | 1,527,152,601,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866107.79/warc/CC-MAIN-20180524073324-20180524093324-00224.warc.gz | 184,759,832 | 3,008 | A Question on Set Theory
• Jan 22nd 2008, 07:51 PM
Norman Smith
A Question on Set Theory
Hello everyone! I hope you are all doing well today, and I am sorry I haven't been here in a while, but I guess that also shows that I may have a better time handling these questions in Precalculus now (Whew).
Anyway, I wanted to ask a question regarding the Set Theory which states:
At a picnic 12 people ate hamburgers, 20 ate frankfurters and 18 had ribs. Everyone had something to eat. Suppose 3 people ate both hamburgers and frankfurters, 6 people ate both frankfurters and ribs, 4 people had hamburgers and ribs, and 1 person ate all these types of food. How many people went on the picnic?
Thank you for all the help, and I hope that all of you are doing well.
-Norman
• Jan 22nd 2008, 11:01 PM
CaptainBlack
Quote:
Originally Posted by Norman Smith
Hello everyone! I hope you are all doing well today, and I am sorry I haven't been here in a while, but I guess that also shows that I may have a better time handling these questions in Precalculus now (Whew).
Anyway, I wanted to ask a question regarding the Set Theory which states:
At a picnic 12 people ate hamburgers, 20 ate frankfurters and 18 had ribs. Everyone had something to eat. Suppose 3 people ate both hamburgers and frankfurters, 6 people ate both frankfurters and ribs, 4 people had hamburgers and ribs, and 1 person ate all these types of food. How many people went on the picnic?
Thank you for all the help, and I hope that all of you are doing well.
-Norman
Add up the numbers for each type of food: 12+20+18=50
Now we have counted those that ate exactly 2 food types twice
So remove the numbers that ate 2 food types the number is 50-(3-6-4)=37
Now at the first stage we counted the guy who ate three food type 3 times, then at the second stage we removed hime three times so now we need to add him back in to get 38 people attended.
(look up inclusion-exclusion principle)
RonL | 502 | 1,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-22 | latest | en | 0.980331 |
https://math.stackexchange.com/questions/1856589/calculus-continuity-of-the-derivative-at-point | 1,563,542,006,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526237.47/warc/CC-MAIN-20190719115720-20190719141720-00014.warc.gz | 463,887,942 | 32,080 | # Calculus, continuity of the derivative at point [closed]
Let $f:I \to \mathbb{R}$ differentiable in the interval $I$ and let $a\in I$. For any sequences $(x_n)_{n\in \mathbb{N}}$ and $(y_n)_{n\in \mathbb{N}}$ in $I$ such that $\lim x_n = \lim y_n = a$, $x_n \neq y_n$, we have:
$$\lim \frac{f(y_n) - f(x_n)}{y_n - x_n} = f'(a) \tag{1}$$
Show that $f'$ is continuous at $a$.
The exercise belongs to the section of a book (Elon Lages Lima, Curso de Analise I) where it introduces the Lagrange mean value theorem, so it might be useful here, but I can't figure it out yet.
Some thoughts
This condition is stronger than differentiability. For example, the function: $$g(x)=\begin{cases}x^2\sin\left(\frac1x\right),&x\ne 0\\0,&x=0\end{cases}$$ is differentiable everywhere, but $g'$ is not continuous at $0$, hence not all pair of sequences $x_n \to 0, y_n \to 0, x_n \neq y_n$ will satisfy (1). Take for example $x_n=\frac1{(2n-\frac12)\pi}$ and $y_n=\frac1{(2n+\frac12)\pi}$, then the limit (1) converges to $-\frac{2}{\pi}$.
## closed as off-topic by colormegone, Claude Leibovici, user91500, user230715, JonMark PerryJul 12 '16 at 9:32
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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – colormegone, Claude Leibovici, user91500, Community, JonMark Perry
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• Something's not right here. What are you limits approaching? – KingDuken Jul 12 '16 at 2:14
• Fix $\epsilon>0$. Can you write what you want to show, and then show some work towards that goal? In general, if you start with a sequence $x_n$ that approaches $a$, you will have the luxury of choosing $y_n$ as desired (perhaps to be "sufficiently close" to $x_n$ so some good things happen), just make sure $y_n$ also approaches $a$. By the way, I think you mean "$f'$ is continuous [at] $a$" rather than "$f'$ is continuous [in] $a$." – Michael Jul 12 '16 at 2:49
• @Michael Fixed it. Thanks. – Michael Jul 12 '16 at 10:33
• @Michael : Good observation about $x^2 \sin(1/x)$. To solve the problem, once you fix a sequence $x_n$ that approaches $a$, you want to show either that $\lim_{n\rightarrow\infty} |f'(a)-f'(x_n)| =0$, or (for a fixed $\epsilon>0$) that $|f'(a)-f'(x_n)|\leq \epsilon$ for all $x_n$ sufficiently close to $a$. Can you work towards that goal, using approximations for $f'(a)$ and $f'(x_n)$? – Michael Jul 12 '16 at 18:32
• I got a solution on AoPS artofproblemsolving.com/community/u277018h1271060p6643173 since my post here was closed. Have a look if interested @Michael – Michael Jul 19 '16 at 23:52 | 907 | 2,975 | {"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} | 3.9375 | 4 | CC-MAIN-2019-30 | latest | en | 0.872719 |
http://www.webassign.net/features/textbooks/galg198/details.html?toc=1&l=subject | 1,518,915,847,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811243.29/warc/CC-MAIN-20180218003946-20180218023946-00572.warc.gz | 632,838,300 | 10,902 | # Algebra 1 98 edition
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• 1: Exploring Expressions, Equations, and Functions (70)
• 2: Exploring Rational Numbers (123)
• 3: Solving Linear Equations (91)
• 4: Using Proportional Reasoning (107)
• 5: Graphing Relations and Functions (72)
• 6: Analyzing Linear Equations (71)
• 7: Solving Linear Inequalities (41)
• 8: Solving Systems of Linear Equations and Inequalities (55)
• 9: Exploring Polynomials (107)
• 10: Using Factoring (40)
• 11: Exploring Quadratic and Exponential Functions (21)
• 12: Exploring Rational Expressions and Equations (91)
• 13: Exploring Radical Expressions and Equations (36)
## Questions Available within WebAssign
Most questions from this textbook are available in WebAssign. The online questions are identical to the textbook questions except for minor wording changes necessary for Web use. Whenever possible, variables, numbers, or words have been randomized so that each student receives a unique version of the question. This list is updated nightly.
##### Question Group Key
PW - Practice Work
##### Question Availability Color Key
BLACK questions are available now
GRAY questions are under development
Group Quantity Questions
Chapter 1: Exploring Expressions, Equations, and Functions
1.PW 1 03
1.1 13 15 16 20 21 22 30 32 35 39 43 45 48 50
1.2 9 13 16 18 20 22 24 25 26 28
1.3 15 13 14 19 20 23 24 27 29 32 35 36 37 38 39 41
1.4 8 11 15 16 17 18 19 24 26
1.5 10 15 16 18 25 26 27 28 29 30 33
1.6 13 20 21 22 23 24 25 26 28 30 32 34 36 38
1.8 1 47
Chapter 2: Exploring Rational Numbers
2.PW 1 03
2.1 8 32 33 35 37 38 41 42 43
2.2 4 07 08 10 18
2.3 20 25 26 28 30 32 36 38 40 41 42 45 46 49 52 54 56 58 60 63 64
2.4 11 15 18 20 21 24 25 29 30 36 38 42
2.5 14 15 16 17 22 23 28 31 32 34 35 39 41 44 45
2.6 21 19 20 22 23 26 27 31 32 35 37 38 41 42 44 45 47 48 53 54 56 58
2.7 15 14 16 19 22 24 26 27 33 34 35 40 42 43 45 46
2.8 18 21 22 24 26 28 34 36 37 39 41 42 43 47 50 60 63 64 66
2.9 11 13 14 16 19 20 21 26 31 33 35 36
Chapter 3: Solving Linear Equations
3.1 12 14 16 18 22 23 24 25 27 30 34 35 37
3.2 15 14 16 18 20 24 28 30 32 34 36 38 39 41 43 45
3.3 16 17 18 21 23 24 30 31 34 35 36 38 39 40 42 44 46
3.4 16 17 19 24 26 27 29 30 32 36 37 38 39 41 43 45 47
3.5 12 14 17 18 21 22 23 28 30 32 34 38 40
3.6 11 12 14 16 18 22 24 25 27 31 32 33
3.7 9 16 17 18 20 21 22 25 26 31
Chapter 4: Using Proportional Reasoning
4.PW 1 05
4.1 15 14 16 18 19 20 22 24 26 28 29 30 31 36 37 38
4.2 7 15 22 24 27 28 30 34
4.3 18 21 24 27 28 30 33 34 35 38 39 40 43 45 47 48 51 52 57
4.4 19 16 17 18 21 26 30 32 35 37 38 40 42 44 45 46 48 49 51 53
4.5 13 12 13 15 19 21 22 24 27 29 32 35 36 39
4.6 13 19 20 23 24 26 28 30 32 34 35 36 40 41
4.7 11 09 10 11 12 13 14 15 16 17 18 22
4.8 10 12 14 18 20 24 25 29 31 32 33
Chapter 5: Graphing Relations and Functions
5.PW 1 06
5.1 4 13 14 41 42
5.2 8 16 18 20 22 27 28 39 44
5.3 13 14 15 17 18 20 23 26 27 32 34 42 44 45
5.4 13 16 17 20 21 22 23 24 26 27 46 48 49 58
5.5 16 19 20 26 28 30 34 36 38 40 42 44 46 48 50 56 57
5.6 11 13 14 15 17 18 19 21 27 28 34 35
5.7 6 12 18 20 24 26 28
Chapter 6: Analyzing Linear Equations
6.PW 1 06
6.1 9 15 24 26 27 30 32 34 39 42
6.2 14 18 20 21 22 26 27 28 30 31 32 34 36 37 40
6.3 3 11 24 28
6.4 15 22 26 28 30 35 36 38 41 43 46 47 49 50 52 55
6.5 2 37 38
6.6 17 18 20 21 23 24 26 29 31 34 37 39 41 42 44 45 46 48
6.7 10 14 18 23 25 30 32 34 35 36 42
Chapter 7: Solving Linear Inequalities
7.1 9 17 22 23 24 26 29 31 34 37
7.2 7 20 36 38 44 46 49 53
7.3 6 16 20 36 38 48 50
7.4 4 24 28 46 58
7.5 4 08 10 12 16
7.6 5 18 30 34 44 55
7.7 3 08 12 14
7.8 3 20 47 48
Chapter 8: Solving Systems of Linear Equations and Inequalities
8.PW 1 03
8.1 15 22 24 26 28 30 32 34 35 36 37 38 39 40 41 51
8.2 13 18 20 24 30 31 32 34 36 39 40 41 42 43
8.3 12 16 18 22 24 26 28 32 33 34 36 41 42
8.4 14 16 20 24 26 28 30 32 34 38 39 42 44 45 46
Chapter 9: Exploring Polynomials
9.1 14 16 18 20 21 22 26 28 29 30 31 32 34 35 39
9.2 12 14 16 18 20 22 23 29 31 34 37 39 41
9.3 18 18 21 22 25 26 29 30 32 33 34 35 36 40 42 49 51 52 54
9.4 15 14 16 17 19 20 23 24 25 26 28 29 32 42 43 44
9.5 12 21 25 26 27 29 32 33 35 36 39 40 44
9.6 13 14 16 19 22 23 25 27 30 31 33 37 39 44
9.7 12 14 19 21 23 24 26 29 30 33 35 37 46
9.8 11 13 15 17 18 22 25 26 27 30 31 32
Chapter 10: Using Factoring
10.1 7 18 22 24 28 36 50 52
10.2 7 26 28 32 40 48 54 58
10.3 7 22 23 24 25 28 29 38
10.4 7 22 23 24 25 26 27 28
10.5 6 35 36 37 40 41 42
10.6 6 14 24 30 32 36 40
Chapter 11: Exploring Quadratic and Exponential Functions
11.1 5 16 28 30 38 46
11.2 3 18 40 56
11.3 4 14 34 36 44
11.4 6 18 24 30 34 42 50
11.5 3 14 18 20
Chapter 12: Exploring Rational Expressions and Equations
12.1 11 14 16 18 20 21 22 23 25 28 30 38
12.2 12 13 17 19 20 21 22 24 25 29 30 32 35
12.3 9 17 19 21 23 25 30 31 32 40
12.4 12 12 14 16 17 20 22 24 27 29 30 31 37
12.5 10 11 13 15 17 20 22 24 29 30 34
12.6 11 16 18 19 20 22 24 26 29 30 35 37
12.7 12 13 14 18 19 23 25 26 27 29 31 35 40
12.8 14 16 17 19 21 22 24 25 26 29 30 31 33 45 49
Chapter 13: Exploring Radical Expressions and Equations
13.1 14 20 22 26 28 30 32 36 38 40 42 44 46 47 48
13.2 4 20 48 50 56
13.3 4 20 36 42 44
13.4 6 16 26 30 36 42 46
13.5 3 26 28 34
13.6 5 14 20 34 36 44
Total 925 | 2,924 | 5,509 | {"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} | 3.015625 | 3 | CC-MAIN-2018-09 | latest | en | 0.660638 |
http://www.slideshare.net/successprep/seminar-sat-testact-test | 1,436,026,939,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096773.65/warc/CC-MAIN-20150627031816-00205-ip-10-179-60-89.ec2.internal.warc.gz | 775,265,353 | 30,939 | 0
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# Seminar SAT Test_ACT Test
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SAT Test / ACT Test Seminar in Atlanta GA powered by Success Prep. …
SAT Test / ACT Test Seminar in Atlanta GA powered by Success Prep.
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### Transcript
• 1. SAT / ACT Which Test Should I Take?
• 2. Why Students Struggle with low SAT & ACT scores? ➢11th & 12th graders taking a test on 9th & 10th grade material. ➢Start early in 10th grade Test 1 + 11th grade for Test 2 & possibly Test 3 ➢High GPA ≠ High SAT scores. Most High Schools are NOT College Prep, but Diploma Prep ➢Take a Prep class, to help you understand the tricks & develop strategies for the SAT/ACT
• 3. SAT / ACT Math Differences ➢ SAT Math : Basic Arithmetic to Algebra 2 ➢SAT Math questions are presented in word problem form. The student’s critically thinking ability to problem solve is the goal of the test. ➢The SAT questions are shorter 20 – 25 min sections when compared to longer 60 -‐75 min sections on the ACT. ➢ ACT Math : Basic Arithmetic to Trigonometry ➢The ACT feels like an advanced High school graduation test in Math, English & Science. ➢The ACT is a fast paced test, with 60 questions in 60 mins on the Math section. ➢The ACT has a Science section; simply English comprehension with scientific facts. ➢ On the ACT: Answer every question, ➢ On the SAT: Be more selective. You loose points for wrong multiple choice answer on the SAT.
• 4. What Test Should My Student Take? ➢ The ACT is a fast pace test (e.g 60 Math questions in 60 mins). If your student is a slow reader (75% of ACT requires heavy reading) and is intimidated by scientific facts and experiments; take the SAT Test ➢ Can your student process math that is not like their homework and excels at problem solving? if yes, take the SAT Test ➢ Does your student have problems with "shifting" (going from one thing to the next seamlessly)? If yes, take the ACT Test since there are less shifts from one section to another. ➢ Does your student hate reading & analyzing long passages to pick out the right answer? if yes, take the SAT Test since it consists of less reading compared to ACT test. 75 % of the ACT test involves heavy reading of long passages.
• 5. 3 Things that Increase your Student’s Scores. ➢ Start early, you only have 10th & 11th grade; consider 11th grade your last year of High School ➢ The Material on the ACT & SAT is 9th & 10th grade content, so you want your student taking the test when they still remember the material. ➢ Take a Prep course ➢ Students who take a good prep course know exactly what to do & NOT to do on each Test i.e. SAT: don’t answer every question, focus on the ones you know | ACT: answer all questions. ➢ A Prep course also makes sure your student actually sits down each week and studies just for the test. ➢ Online prep is ineffective and students can’t interact with instructor. ➢ Plan to take the test 2 to 3 times ➢ As the student gets more familiar with the pace and conditioning from prior testing & prep courses, they score better because they can better deal with the 4 hour stress. | 896 | 3,616 | {"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-2015-27 | latest | en | 0.859926 |
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