url
stringlengths 6
1.61k
| fetch_time
int64 1,368,856,904B
1,726,893,854B
| content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
| warc_record_offset
int32 9.6k
1.74B
| warc_record_length
int32 664
793k
| text
stringlengths 45
1.04M
| token_count
int32 22
711k
| char_count
int32 45
1.04M
| metadata
stringlengths 439
443
| score
float64 2.52
5.09
| int_score
int64 3
5
| crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.06
1
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.gradesaver.com/textbooks/math/applied-mathematics/elementary-technical-mathematics/chapter-1-section-1-10-addition-and-subtraction-of-decimal-fractions-exercises-page-61/43
| 1,537,341,110,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-39/segments/1537267155942.15/warc/CC-MAIN-20180919063526-20180919083526-00016.warc.gz
| 752,043,464
| 13,905
|
# Chapter 1 - Section 1.10 - Addition and Subtraction of Decimal Fractions - Exercises - Page 61: 43
$8.68$
#### Work Step by Step
Perform the subtraction first: \begin{array}{ccc} \require{cancel} & & &\space 7\space3 \\&&&1\cancel{8}.\cancel{4}0 \\&- & &\underline{13.72} \\&&&04.68 \end{array} Add $4$ to the result to obtain: $=4.68 + 4 \\=8.68$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
| 163
| 513
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2018-39
|
longest
|
en
| 0.681578
|
https://www.unitsconverters.com/en/Apostilb-To-Lumenpersquaremeterpersteradian/Unittounit-4092-4084
| 1,721,330,839,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00505.warc.gz
| 884,287,058
| 39,945
|
Formula Used
1 Candela per Square Meter = 3.14159265359 Apostilb
1 Candela per Square Meter = 1 Lumen per Square Meter per Steradian
1 Apostilb = 0.31830988618377 Lumen per Square Meter per Steradian
## Apostilbs to Lumen per Square Meter per Steradians Conversion
asb stands for apostilbs and lm/m²*sr stands for lumen per square meter per steradians. The formula used in apostilbs to lumen per square meter per steradians conversion is 1 Apostilb = 0.31830988618377 Lumen per Square Meter per Steradian. In other words, 1 apostilb is 4 times smaller than a lumen per square meter per steradian. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
## Convert Apostilb to Lumen per Square Meter per Steradian
How to convert apostilb to lumen per square meter per steradian? In the luminance measurement, first choose apostilb from the left dropdown and lumen per square meter per steradian from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from lumen per square meter per steradian to apostilb? You can check our lumen per square meter per steradian to apostilb converter.
How to convert Apostilb to Lumen per Square Meter per Steradian?
The formula to convert Apostilb to Lumen per Square Meter per Steradian is 1 Apostilb = 0.31830988618377 Lumen per Square Meter per Steradian. Apostilb is 3.14169022934339 times Smaller than Lumen per Square Meter per Steradian. Enter the value of Apostilb and hit Convert to get value in Lumen per Square Meter per Steradian. Check our Apostilb to Lumen per Square Meter per Steradian converter. Need a reverse calculation from Lumen per Square Meter per Steradian to Apostilb? You can check our Lumen per Square Meter per Steradian to Apostilb Converter.
How many Candela per Square Meter is 1 Apostilb?
1 Apostilb is equal to 0.3183 Candela per Square Meter. 1 Apostilb is 3.14169022934339 times Smaller than 1 Candela per Square Meter.
How many Candela per Square Foot is 1 Apostilb?
1 Apostilb is equal to 0.3183 Candela per Square Foot. 1 Apostilb is 3.14169022934339 times Smaller than 1 Candela per Square Foot.
How many Kilocandela per Square Meter is 1 Apostilb?
1 Apostilb is equal to 0.3183 Kilocandela per Square Meter. 1 Apostilb is 3.14169022934339 times Smaller than 1 Kilocandela per Square Meter.
How many Lambert is 1 Apostilb?
1 Apostilb is equal to 0.3183 Lambert. 1 Apostilb is 3.14169022934339 times Smaller than 1 Lambert.
## Apostilbs to Lumen per Square Meter per Steradians Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like luminance finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like asb to lm/m²*sr through multiplicative conversion factors. When you are converting luminance, you need a Apostilbs to Lumen per Square Meter per Steradians converter that is elaborate and still easy to use. Converting Apostilb to Lumen per Square Meter per Steradian is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Apostilb to Lumen per Square Meter per Steradian conversion along with a table representing the entire conversion.
Let Others Know
| 915
| 3,671
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.579378
|
http://boymamateachermama.com/2016/04/18/teacher-mama-free-numbers-battleship-games/
| 1,540,104,905,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583513760.4/warc/CC-MAIN-20181021052235-20181021073735-00265.warc.gz
| 56,723,352
| 14,710
|
# FREE Numbers Battleship Games
I saw this amazing version of Battleship somewhere on Pinterest. I have tried and tried to find the original poster, but it seems that it is quite popular and many people have shared it. So this is my version of the game. I played it with my first graders this morning and they loved it. I have to admit, I kinda loved it too. 🙂
### Materials
templates
card stock
bingo chips
laminator
vis-a-vis marker
cardboard for privacy
### Preparation
Choose the version you would like to play. Version one (left) uses numbers while version two (right) uses coordinates.
Print the templates on card stock and laminate. If you do not have access to a laminator, papers can simply be copied and recycled when done or you can put the papers in a plastic sleeve.
### How to Play (Version 1)
Give each player a game board and a vis-a-vis marker.
Have each player find a private spot and mark off five “ships” as follows:
two squares
three squares
four squares
five squares
six squares
Put students in pairs with a privacy shield of some kind between them.
Give each pair some bingo chips or pennies.
Player one names a number to see if s/he has “hit” a battleship on his/her partner’s board. If s/he has, the partner puts a chip on that space on his/her board. If not, the partner does nothing.
The player making the guess, should put a dot on the numbers s/he has already guess so s/he does not repeat a number. If s/he gets a “hit,” have him/her put an x on the spot.
Play continues until one player has “sunk” all his/her partner’s battleships.
### How to Play (Version 2)
The game is set up and played the same way, only instead of guessing numbers, students use coordinates to guess where the battleships lie.
## This FREE game is available HERE.
The After School Link Up is a great place to share ideas and to find new ideas to do with your children after school or in your homeschool.  The After School Link Up goes live every Monday. So, if you are a blogger, an educator or just some one looking for some good ideas, be sure to stop by Boy Mama Teacher Mama (or the other co-hosts) on Mondays and see what others have to share.
We would love to have you link up your School-Age Post (Ages 5 and up) about your learning week after school including Crafts, Activities, Playtime and Adventures that you are doing to enrich your children’s lives after their day at school, homeschool or on the weekend! When linking up, please take a moment to comment on at least one post linked up before yours! By linking up you’re giving permission for us to share on our After School Pinterest Board or Feature on our After School Party in the upcoming weeks!
2016 After School Party HOSTS
The Educators’ Spin On It * Planet Smarty Pants * Boy Mama Teacher Mama * Mama Smiles * Relentlessly Fun, Deceptively Educational * The Measured Mom * This Reading Mama * Creative Family Fun
If this is your first time visiting Boy Mama Teacher Mama, welcome! Did you know that you can also find us on Facebook, Pinterest, Twitter and G+? Stop by to see what is happening!
© Boy Mama Teacher Mama 2016
| 752
| 3,161
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43
|
latest
|
en
| 0.950145
|
http://www.idealessaywriters.com/if-you-want-a-return-of-12-percent-how-much-will-you-pay-for-the-stock-what-if-you-want-a-return-of-8-percent/
| 1,519,580,835,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891816841.86/warc/CC-MAIN-20180225170106-20180225190106-00774.warc.gz
| 444,958,130
| 13,968
|
## If you want a return of 12 percent, how much will you pay for the stock? What if you want a return of 8 percent?
6.16
Interest Rate Risk. Both Bond Bill and Bond Ted have 7 percent coupons, make semiannual payments, and are priced at par value. Bond Bill has 3 years to maturity, whereas Bond Ted has 20 years to maturity. If interest rates suddenly rise by 2 percent, what is the percentage change in the price of Bond Bill? Of Bond Ted? If rates were to suddenly fall by 2 percent instead, what would the percentage change in the price of Bond Bill be then? Of Bond Ted? Illustrate your answers by graphing bond prices versus YTM. What does this problem tell you about the interest rate risk of longer-term bonds?
7.11
Valuing Preferred Stock. E-Eyes.com has a new issue of preferred stock it calls 20/20 preferred. The stock will pay a \$20 dividend per year, but the first dividend will not be paid until 20 years from today. If you require a return of 8 percent on this stock, how much should you pay today?
7.12
Stock Valuation. Alexander Corp. will pay a dividend of \$2.72 next year. The company has stated that it will maintain a constant growth rate of 4.5 percent a year forever. If you want a return of 12 percent, how much will you pay for the stock? What if you want a return of 8 percent? What does this tell you about the relationship between the required return and the stock price?
Type of paper Academic level Subject area Essay Term Paper Research Paper Coursework Book Report Book Review Movie Review Dissertation Thesis Thesis Proposal Research Proposal Dissertation Chapter - Abstract Dissertation Chapter - Introduction Chapter Dissertation Chapter - Literature Review Dissertation Chapter - Methodology Dissertation Chapter - Results Dissertation Chapter - Discussion Dissertation Services - Editing Dissertation Services - Proofreading Formatting Admission Services - Admission Essay Admission Services - Scholarship Essay Admission Services - Personal Statement Admission Services - Editing Editing Proofreading Case Study Lab Report Speech Presentation Math Problem Article Article Critique Annotated Bibliography Reaction Paper PowerPoint Presentation Statistics Project Multiple Choice Questions (None-Time-Framed) Other (Not listed) High School Undergraduate Master Ph. D. Art Architecture Dance Design Analysis Drama Movies Music Paintings Theatre Biology Business Chemistry Communications and Media Advertising Communication Strategies Journalism Public Relations Creative writing Economics Accounting Case Study Company Analysis E-Commerce Finance Investment Logistics Trade Education Application Essay Education Theories Pedagogy Teacher's Career Engineering English Ethics History African-American Studies American History Asian Studies Canadian Studies East European Studies Holocaust Latin-American Studies Native-American Studies West European Studies Law Criminology Legal Issues Linguistics Literature American Literature Antique Literature Asian Literature English Literature Shakespeare Studies Management Marketing Mathematics Medicine and Health Alternative Medicine Healthcare Nursing Nutrition Pharmacology Sport Nature Agricultural Studies Anthropology Astronomy Environmental Issues Geography Geology Philosophy Physics Political Science Psychology Religion and Theology Sociology Technology Aeronautics Aviation Computer Science Internet IT Management Web Design Tourism Number of pages Paper urgency Cost per page: 1 pages/275 words 2 pages/550 words 3 pages/825 words 4 pages/1100 words 5 pages/1375 words 6 pages/1650 words 7 pages/1925 words 8 pages/2200 words 9 pages/2475 words 10 pages/2750 words 11 pages/3025 words 12 pages/3300 words 13 pages/3575 words 14 pages/3850 words 15 pages/4125 words 16 pages/4400 words 17 pages/4675 words 18 pages/4950 words 19 pages/5225 words 20 pages/5500 words 21 pages/5775 words 22 pages/6050 words 23 pages/6325 words 24 pages/6600 words 25 pages/6875 words 26 pages/7150 words 27 pages/7425 words 28 pages/7700 words 29 pages/7975 words 30 pages/8250 words 31 pages/8525 words 32 pages/8800 words 33 pages/9075 words 34 pages/9350 words 35 pages/9625 words 36 pages/9900 words 37 pages/10175 words 38 pages/10450 words 39 pages/10725 words 40 pages/11000 words 41 pages/11275 words 42 pages/11550 words 43 pages/11825 words 44 pages/12100 words 45 pages/12375 words 46 pages/12650 words 47 pages/12925 words 48 pages/13200 words 49 pages/13475 words 50 pages/13750 words 51 pages/14025 words 52 pages/14300 words 53 pages/14575 words 54 pages/14850 words 55 pages/15125 words 56 pages/15400 words 57 pages/15675 words 58 pages/15950 words 59 pages/16225 words 60 pages/16500 words 61 pages/16775 words 62 pages/17050 words 63 pages/17325 words 64 pages/17600 words 65 pages/17875 words 66 pages/18150 words 67 pages/18425 words 68 pages/18700 words 69 pages/18975 words 70 pages/19250 words 71 pages/19525 words 72 pages/19800 words 73 pages/20075 words 74 pages/20350 words 75 pages/20625 words 76 pages/20900 words 77 pages/21175 words 78 pages/21450 words 79 pages/21725 words 80 pages/22000 words 81 pages/22275 words 82 pages/22550 words 83 pages/22825 words 84 pages/23100 words 85 pages/23375 words 86 pages/23650 words 87 pages/23925 words 88 pages/24200 words 89 pages/24475 words 90 pages/24750 words 91 pages/25025 words 92 pages/25300 words 93 pages/25575 words 94 pages/25850 words 95 pages/26125 words 96 pages/26400 words 97 pages/26675 words 98 pages/26950 words 99 pages/27225 words 100 pages/27500 words 101 pages/27775 words 102 pages/28050 words 103 pages/28325 words 104 pages/28600 words 105 pages/28875 words 106 pages/29150 words 107 pages/29425 words 108 pages/29700 words 109 pages/29975 words 110 pages/30250 words 111 pages/30525 words 112 pages/30800 words 113 pages/31075 words 114 pages/31350 words 115 pages/31625 words 116 pages/31900 words 117 pages/32175 words 118 pages/32450 words 119 pages/32725 words 120 pages/33000 words 121 pages/33275 words 122 pages/33550 words 123 pages/33825 words 124 pages/34100 words 125 pages/34375 words 126 pages/34650 words 127 pages/34925 words 128 pages/35200 words 129 pages/35475 words 130 pages/35750 words 131 pages/36025 words 132 pages/36300 words 133 pages/36575 words 134 pages/36850 words 135 pages/37125 words 136 pages/37400 words 137 pages/37675 words 138 pages/37950 words 139 pages/38225 words 140 pages/38500 words 141 pages/38775 words 142 pages/39050 words 143 pages/39325 words 144 pages/39600 words 145 pages/39875 words 146 pages/40150 words 147 pages/40425 words 148 pages/40700 words 149 pages/40975 words 150 pages/41250 words 151 pages/41525 words 152 pages/41800 words 153 pages/42075 words 154 pages/42350 words 155 pages/42625 words 156 pages/42900 words 157 pages/43175 words 158 pages/43450 words 159 pages/43725 words 160 pages/44000 words 161 pages/44275 words 162 pages/44550 words 163 pages/44825 words 164 pages/45100 words 165 pages/45375 words 166 pages/45650 words 167 pages/45925 words 168 pages/46200 words 169 pages/46475 words 170 pages/46750 words 171 pages/47025 words 172 pages/47300 words 173 pages/47575 words 174 pages/47850 words 175 pages/48125 words 176 pages/48400 words 177 pages/48675 words 178 pages/48950 words 179 pages/49225 words 180 pages/49500 words 181 pages/49775 words 182 pages/50050 words 183 pages/50325 words 184 pages/50600 words 185 pages/50875 words 186 pages/51150 words 187 pages/51425 words 188 pages/51700 words 189 pages/51975 words 190 pages/52250 words 191 pages/52525 words 192 pages/52800 words 193 pages/53075 words 194 pages/53350 words 195 pages/53625 words 196 pages/53900 words 197 pages/54175 words 198 pages/54450 words 199 pages/54725 words 200 pages/55000 words 30 days 6 hours 12 hours 24 hours 48 hours 3 days 4 days 5 days 7 days 10 days 20 days USD GBP CAD AUD EUR Total:
| 2,188
| 7,960
|
{"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}
| 2.65625
| 3
|
CC-MAIN-2018-09
|
longest
|
en
| 0.890812
|
http://metamath.tirix.org/mpests/negmod
| 1,720,839,945,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00780.warc.gz
| 21,227,159
| 2,891
|
# Metamath Proof Explorer
## Theorem negmod
Description: The negation of a number modulo a positive number is equal to the difference of the modulus and the number modulo the modulus. (Contributed by AV, 5-Jul-2020)
Ref Expression
Assertion negmod ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to \left(-{A}\right)\mathrm{mod}{N}=\left({N}-{A}\right)\mathrm{mod}{N}$
### Proof
Step Hyp Ref Expression
1 rpcn ${⊢}{N}\in {ℝ}^{+}\to {N}\in ℂ$
2 recn ${⊢}{A}\in ℝ\to {A}\in ℂ$
3 negsub ${⊢}\left({N}\in ℂ\wedge {A}\in ℂ\right)\to {N}+\left(-{A}\right)={N}-{A}$
4 1 2 3 syl2anr ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to {N}+\left(-{A}\right)={N}-{A}$
5 4 eqcomd ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to {N}-{A}={N}+\left(-{A}\right)$
6 5 oveq1d ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to \left({N}-{A}\right)\mathrm{mod}{N}=\left({N}+\left(-{A}\right)\right)\mathrm{mod}{N}$
7 1 mulid2d ${⊢}{N}\in {ℝ}^{+}\to 1\cdot {N}={N}$
8 7 adantl ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to 1\cdot {N}={N}$
9 8 oveq1d ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to 1\cdot {N}+\left(-{A}\right)={N}+\left(-{A}\right)$
10 9 oveq1d ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to \left(1\cdot {N}+\left(-{A}\right)\right)\mathrm{mod}{N}=\left({N}+\left(-{A}\right)\right)\mathrm{mod}{N}$
11 1cnd ${⊢}{A}\in ℝ\to 1\in ℂ$
12 mulcl ${⊢}\left(1\in ℂ\wedge {N}\in ℂ\right)\to 1\cdot {N}\in ℂ$
13 11 1 12 syl2an ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to 1\cdot {N}\in ℂ$
14 renegcl ${⊢}{A}\in ℝ\to -{A}\in ℝ$
15 14 recnd ${⊢}{A}\in ℝ\to -{A}\in ℂ$
16 15 adantr ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to -{A}\in ℂ$
17 13 16 addcomd ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to 1\cdot {N}+\left(-{A}\right)=-{A}+1\cdot {N}$
18 17 oveq1d ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to \left(1\cdot {N}+\left(-{A}\right)\right)\mathrm{mod}{N}=\left(-{A}+1\cdot {N}\right)\mathrm{mod}{N}$
19 14 adantr ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to -{A}\in ℝ$
20 simpr ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to {N}\in {ℝ}^{+}$
21 1zzd ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to 1\in ℤ$
22 modcyc ${⊢}\left(-{A}\in ℝ\wedge {N}\in {ℝ}^{+}\wedge 1\in ℤ\right)\to \left(-{A}+1\cdot {N}\right)\mathrm{mod}{N}=\left(-{A}\right)\mathrm{mod}{N}$
23 19 20 21 22 syl3anc ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to \left(-{A}+1\cdot {N}\right)\mathrm{mod}{N}=\left(-{A}\right)\mathrm{mod}{N}$
24 18 23 eqtrd ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to \left(1\cdot {N}+\left(-{A}\right)\right)\mathrm{mod}{N}=\left(-{A}\right)\mathrm{mod}{N}$
25 6 10 24 3eqtr2rd ${⊢}\left({A}\in ℝ\wedge {N}\in {ℝ}^{+}\right)\to \left(-{A}\right)\mathrm{mod}{N}=\left({N}-{A}\right)\mathrm{mod}{N}$
| 1,369
| 2,737
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.765625
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.264005
|
http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities_ApplicationsActivities_BlackScholesOptionPricing&diff=prev&oldid=7931
| 1,600,683,978,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600400201601.26/warc/CC-MAIN-20200921081428-20200921111428-00534.warc.gz
| 144,683,328
| 5,490
|
# SOCR EduMaterials Activities ApplicationsActivities BlackScholesOptionPricing
(Difference between revisions)
Nchristo (Talk | contribs)
(New page: == Black-Scholes option pricing model - Convergence of binomial == * Black-Scholes option pricing formula: <br> The value $C[itex] of a European call option at time [itex]t=0$ ...)
## Black-Scholes option pricing model - Convergence of binomial
• Black-Scholes option pricing formula:
The value C < math > ofaEuropeancalloptionattime < math > t = 0 is: $C=S_0 \Phi (d_1) - \frac{E}{e^{rt}} \Phi(d_2)$
$d_1=\frac{ln(\frac{S_0}{E})+(r+\frac{1}{2} \sigma^2)t} {\sigma \sqrt{t}}$
$d_2=\frac{ln(\frac{S_0}{E})+(r-\frac{1}{2} \sigma^2)t} {\sigma \sqrt{t}}=d_1-\sigma \sqrt{t}$
Where,
S0 Price of the stock at time t = 0
E Exercise price at expiration
r Continuously compounded risk-free interest
σ Annual standard deviation of the returns of the stock
t < math > Timetoexpirationinyears < br > < math > Φ(di) Cumulative probability at di of the standard normal distribution N(0,1)
• Binomial convergence to Black-Scholes option pricing formula:
The binomial formula converges to the Black-Scholes formula when the number of periods n < math > islarge.Intheexamplebelowwevaluethecalloptionusingthebinomialformulafordifferentvaluesof < math > n and also using the Black-Scholes formula. We then plot the value of the call (from binomial) against the number of periods n < math > .ThevalueofthecallusingBlackScholesremainsthesameregardlessof < math > n. The data used for this example are: Failed to parse (lexing error): S_0=\$30, \ E=\$29,\ R_f=0.05, \sigma=0.30,\ \mbox{Days to expiration}=40 .
• For the binomial option pricing calculations we divided the 40 days into intervals from 1 to 100 (by 1).
• The snapshot below from the SOCR Black Scholes Option Pricing model applet shows the path of the stock.
| 548
| 1,857
|
{"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.796875
| 4
|
CC-MAIN-2020-40
|
longest
|
en
| 0.65702
|
http://mathhelpforum.com/calculus/11980-calc-problem-differencial-equation-print.html
| 1,529,661,700,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864387.54/warc/CC-MAIN-20180622084714-20180622104714-00294.warc.gz
| 196,811,983
| 2,900
|
# calc problem differencial equation
Printable View
• Feb 26th 2007, 05:38 PM
clockingly
calc problem differencial equation
hi, i was wondering if i did this problem right. i had to solve the following differential equation for y as a function of t using separation of variables:
y' = yt^2
so this is what i did.
i set dy/dt = yt^2
then from that i got dy/y = t^2 dt, which is the same as 1/y dy = t^2 dt.
i took the integral of both sides.
the integral of 1/y dy is ln(y)
the integral of t^2 dt is t^3/3 + C
so then i got that y = e^(ln(t^3/3)+C)
this is the same as y = e^(ln(t^3/3) * e^C
then i got y = (t^3 * e^c)/3 for a final answer.
• Feb 26th 2007, 06:47 PM
ThePerfectHacker
Quote:
Originally Posted by clockingly
hi, i was wondering if i did this problem right. i had to solve the following differential equation for y as a function of t using separation of variables:
y' = yt^2
y=0 is a trivial solution.
Assume y not = 0 (in fact it is zero everywhere or nowhere).
Now divide,
y'/y=t^2
Thus,
INT y'/y dt = INT t^2 dt
ln |y| = (1/3)t^3+C
y=exp[(1/3)t^3 +C]=e^c * exp[1/3t^3]=C*e^{1/3t^3} for C>0.
Note, y=0 was also a solution. Corresponding for C=0.
Thus,
y=C*e^{1/3t^3} for C>=0.
Is the solution on this non-empty open interval.
| 430
| 1,257
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.8125
| 4
|
CC-MAIN-2018-26
|
latest
|
en
| 0.926987
|
https://sites.uclouvain.be/archives-portail/cdc2019/en-cours-2019-LSTAT2200
| 1,627,129,299,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046150264.90/warc/CC-MAIN-20210724094631-20210724124631-00545.warc.gz
| 525,422,938
| 7,489
|
# Survey and Sampling
lstat2200 2019-2020 Louvain-la-Neuve
Survey and Sampling
Note from June 29, 2020
Although we do not yet know how long the social distancing related to the Covid-19 pandemic will last, and regardless of the changes that had to be made in the evaluation of the June 2020 session in relation to what is provided for in this learning unit description, new learnig unit evaluation methods may still be adopted by the teachers; details of these methods have been - or will be - communicated to the students by the teachers, as soon as possible.
4 credits
15.0 h + 5.0 h
Q2
Teacher(s)
Kestemont Marie-Paule;
Language
French
Main themes
Topics to be treated - General framework of inference in finite population; population, sampling, statistics for the inference based on experimental data, linear homogenous estimation: elementary units, complex units. - Sampling with unequal probabilities: Hansen-Hurwitz and Horvitz-Thompson estimators, for the particular case of simple random sampling. - Estimators improvement through auxiliary information: ratio estimator, regression estimator - Sampling from complex units: stratified sampling, cluster sampling, two stages sampling. - Sampling from biological populations: basic issues in sampling, estimation of the population size.
Aims
At the end of this learning unit, the student is able to : 1 Objective (in terms of abilities and knowledge) This course aims at providing the student the basic knowledges on the sampling methods, with a particular, but not exclusive, emphasis on sampling from (finite) human populations. At the end of the course, the student should be able to correctly designing a simple survey and analysing the results.
The contribution of this Teaching Unit to the development and command of the skills and learning outcomes of the programme(s) can be accessed at the end of this sheet, in the section entitled “Programmes/courses offering this Teaching Unit”.
Content
General framework of inference in finite population :
• Techniques of random samplings and estimators properties.
• Simple random sampling
• Stratified random sampling
• Uneven probability sampling
• Cluster sampling
• Multi-level sampling
Estimation improvement by use of auxiliary information.
Teaching methods
8 x 2 hours of masterful presentations and 2 x 2 hours of practical exercices on computer.
Evaluation methods
Written examination in session : 14 points on 20.
Individual project delivered for the beginning of the first session : 6 points on 20.
Online resources
MOODLEUCL : lecture LSTAT2200.
Bibliography
Tillé, Y. (2001). Théorie des sondages : échantillonnage et estimation en populations finies, (Cours et exercices avec solutions), Dunod, Paris.
Mouchart M. et J.-M. Rolin (1981), Enquêtes et Sondages, Série " Recyclage en Statistique ", Vol.5, , Louvain : U.C.L., Comité de Statistique.
Sharon Lohr (1999), Sampling : Design and Analysis, Duxbury Press Rao Poduri S.R.S. (2000), Sampling Methodologies with Applications, London : Chapman and Hall.
Teaching materials
• transparents sur moodle
Faculty or entity
LSBA
#### Programmes / formations proposant cette unité d'enseignement (UE)
Title of the programme
Sigle
Credits
Prerequisites
Aims
Mineure en statistique et science des données
Approfondissement en statistique et sciences des données
Minor in Statistics, Actuarial Sciences and Data Sciences
Master [120] in Data Science : Statistic
Master [120] in Data Science Engineering
Certificat d'université : Statistique et sciences des données (15/30 crédits)
Master [120] in Data Science: Information Technology
Master [120] in Economics: General
Master [120] in Statistic: General
| 846
| 3,680
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2021-31
|
latest
|
en
| 0.893885
|
https://community.powerbi.com/t5/Desktop/Moving-average-of-a-percentage-Measure/m-p/260878
| 1,553,311,769,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-13/segments/1552912202711.3/warc/CC-MAIN-20190323020538-20190323042538-00225.warc.gz
| 441,963,772
| 106,185
|
cancel
Showing results for
Did you mean:
Frequent Visitor
## Moving average of a percentage Measure
Hi everyone,
I need to create a moving average over 6 months of percentage.
For example, in september my moving average should be efficiency of september + august + july + june + may + april /6
In August, it should be august + july + june + may + april + march /6...
I tried the solution found by Preetish_1 :
but it didnt work for me.
I have two tables :
- Table_1 contains my Efficiency in % and id_Time
- Table_2 contains my dimTime with year col, month col and date col
Could you help me guys ?
Thank you
Valentin
1 ACCEPTED SOLUTION
Accepted Solutions
Super User
## Re: Moving average of a percentage Measure
Hi,
It's no surprise that your formula is not working. There is no date column in your FACT_SRL table. You must have a date column there and that date column should bear a relationship with the Caldate column of the SUM_Temps table. Only then will the Time Intelligence functions such as DATESBETWEEN() or DATESINPREIOD() work.
9 REPLIES 9
Moderator
## Re: Moving average of a percentage Measure
You can create a measure below:
Moving_Average_6_Months =
CALCULATE (
AVERAGEX ( 'Table1', 'Table1'[Efficiency]) ,
DATESINPERIOD (
'Table2'[Date],
LASTDATE ( 'Table2'[Date]),
-6,
MONTH
)
)
Best Regards,
Qiuyun Yu
Community Support Team _ Qiuyun Yu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super User
## Re: Moving average of a percentage Measure
Based on the dataset shared by @v-qiuyu-msft, this will also work
`=CALCULATE(AVERAGEX ('Table1','Table1'[Efficiency]),DATESBETWEEN('Table2'[Date],EDATE(MIN('Table2'[Date]),-5),max(Table2[Date])))`
Hope this helps.
Frequent Visitor
## Re: Moving average of a percentage Measure
Thank you for your replies @v-qiuyu-msft & @Ashish_Mathur
But I try both of yours solutions and both didn't work for me :/
I forgot to precise, my "Efficiency" is a calculated measure from a rate between two values ("allocated time"/"passed time")
Another thing which could be a problem : when I calculate LASTDATE(CalDate), it returns (2017-12-01) because it's my last date in my dimTime.. But my last values are in september !
Anyway it seems to ignore the time period when it calculates.
My dimTime & relation between two tables :
Super User
## Re: Moving average of a percentage Measure
Hi,
Ensure that you drag the year and month from the Dim_Temps Table.
Frequent Visitor
## Re: Moving average of a percentage Measure
Yes, it's the year and month from Dim_Temps Table.
How can I attach my pbix file ?
Super User
Frequent Visitor
Thank's !
It's here
Super User
## Re: Moving average of a percentage Measure
Hi,
It's no surprise that your formula is not working. There is no date column in your FACT_SRL table. You must have a date column there and that date column should bear a relationship with the Caldate column of the SUM_Temps table. Only then will the Time Intelligence functions such as DATESBETWEEN() or DATESINPREIOD() work.
Highlighted
Frequent Visitor
Hi,
| 800
| 3,139
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.6875
| 3
|
CC-MAIN-2019-13
|
latest
|
en
| 0.860838
|
https://www.analyzemath.com/intermediate_algebra/questions_5_sol.html
| 1,580,244,043,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579251783000.84/warc/CC-MAIN-20200128184745-20200128214745-00483.warc.gz
| 748,963,952
| 34,287
|
# Solutions and Explanations to Intermediate Algebra Questions in Sample 5
Detailed solutions and full explanations to the multiple choice intermediate algebra questions in sample 5 are presented.
1. If f(x) = 4x3 - 4x2 + 10, then f(-2) =
Solution
Substitute x by -2 in f(x) as follows
f(-2) = 4(-2)3 - 4(-2)2 + 10
= 4(-8) - 4(4) + 10 = - 32 - 16 + 10 = - 38
2. Which of these values of x satisfies the inequality -7x + 6 ≤ -8
A. -2 B. 0 C. -7 D. 2
Solution
Solve the inequality
-7x + 6 ≤ -8 , given
-7x + 6 - 6 ≤ -8 - 6 , add - 6 to both sides
-7x ≤ - 14 , simplify
-7x / -7 ≥ -14 / -7 , divide by - 7 and CHANGE symbol of inequality
x ≥ 2 , solution set
The answer to the above question is D since 2.
3. The domain of the function f(x) = √(6 - 2x) is given by
Solution
f(x) is real if the expression under the radical is positive or equal to zero. Hence to find the domain of we need to solve the following inequality.
(6 - 2x) ≥ 0
x ≤ 3 , domain of f
4. The lines y = 2x and 2y = - x are
A. parallel B. perpendicular C. horizontal D. vertical
Solution
Horizontal lines are of the form y = constant and vertical lines are of the from x = constant and therefore the two lines are neither horizontal nor vertical. Let us find the slopes of the two given lines
y = 2x has a slope equal to 2
2y = - x is equivalent to y = -(1/2) x and its slope is equal to -(1/2)
Since the slopes are not equal, the two lines are not parallel. The product of the two slopes is given by
2(-1/2) = - 1
and hence the two lines are perpendicular.
5. The equation |-2x - 5| - 3 = k has no solution if k =
A. -5 B. -3 C. 7 D. 0
Solution
We first rewrite the given equation in the form
|-2x - 5| = k + 3
The term |-2x - 5| is either positive or equal to zero. Therefore the above equation has no solutions whenever the expression k + 3 is negative. The values of k for which the above equation has no solutions are solutions of the inequality
k + 3 < 0 or k < - 3
The answer is A since - 5 is less than - 3.
6. The inequality corresponding to the statement:"the price is no less than 100 Dollars" is
Solution
If the price is no less than 100 Dollars, then the price is either equal to or greater than 100 Dollars.
x ≥ 100
7. Which of these relations DOES NOT represent a function?
A. {(2,3),(-4,3),(7,3)} B. {(0,0),(-1,-1),(2,2)} C. {(2,3),(-5,3),(2,7)} D. {(-1,3),(-5,3),(-9,0)}
Solution
For the relation in C, when x = 2, there are two possible values of y: 3 or 7 and therefore the relation in C is not a function.
8. Which of these points DOES NOT lie on the graph of y = -x + 3?
A. (9,- 6) B. (3,0) C. (-2,5) D. (2,2)
Solution
Substitute the coordinates of the given points in the given equation and check which one gives a false statement.
Point (9,- 6) : - 6 = -(9) + 3 , - 6 = - 6 , true , point lies on the line
Point (3,0) : 0 = - (3) + 3 , 0 = 0 , true , point lies on the line
Point (-2,5) : 5 = - (-2) + 3 , 5 = 5 , true , point lies on the line
Point (2,2) : 2 = - (2) + 3 , 2 = 1 , false , point DOES NOT lie on the line
9. What is the slope of the line perpendicular to the line y = -5x + 9?
Solution
The slope of the given (in slope intercept form) line is equal to - 5. Let m be the slope of the line perpendicular to the given line. Two lines are perpendicular if the product of their slopes is equal to -1. Hence
m*(-5) = - 1
Solve for m. Hence
m = 1/5 is the slope of a line perpendicular to the given line.
10. Which property is used to write:3(x y) = (3 x)y?
A. Commutative property of multiplication B. Multiplicative inverse property C. Distributive property D. Associative property of multiplication
Solution
We may use the Associative property of multiplication to write
3(x y) = (3 x)y
11. In which quadrant do the lines x = 3 and y = - 4 intersect?
Solution
The two lines intersect at the point (3 , -4) which is in quadrant IV.
12. The value of 2 - | - 2 | is
Solution
2 - | - 2 | = 2 - 2 , since | - 2 | = 2
= 1 / 2 2 , since a -n = 1 / an
= 1/4 = 0.25
13. If a and b are positive real numbers, then (a0 - 3b0)5 =
Solution
Simplify.
(a0 - 3b0)5 = (1 - 3*1)5 = (- 2)5 = - 32
14. Which inequality describes the situation:"length L is at most 45 cm".
Solution
C
15. The equation m x - 8 = 6 - 7(x + 3) DOES NOT have any solution if m =
Solution
Solve for x.
m x - 8 = 6 - 7(x + 3)
m x + 7x = 6 - 21 + 8
x(m + 7) = -7
x = - 7 / (m + 7)
m cannot be equal to - 7 otherwise the denominator will be zero.
16. The equation - m x + 1 = 13 - 4(x + 3) is an identity if m =
Solution
Expand both the right side.
- m x + 1 = - 4 x + 1
The above is an identity if m = 4.
17. Which of the following is ALWAYS true?
Solution
Every function is a relation
18. Which of these inequalities has NO solutions?
Solution
The absolute value of any expression is positive or equal to zero. Hence the inequality .
|x + 3| < -2
has no solutions
19. The lines y = (a - 5)x + 5 and y = -2x + 7 are parallel if a =
Solution
Two lines are parallel if their slopes are equal. Hence
a - 5 = - 2
Solve for a
a = 3
20. The lines y = (a - 5)x + 5 and y = -2 x + 7 are perpendicular if a =
Solution
Two lines are perpendicular if the product of their slopes is equal to -1. Hence
-2(a - 5) = - 1
Solve for a
a = 11/2
1. B
2. D
3. C
4. B
5. A
6. B
7. C
8. D
9. C
10. D
11. C
12. B
13. C
14. D
15. C
16. A
17. B
18. C
19. B
20. A
Algebra Questions and problems
More ACT, SAT and Compass practice
| 1,840
| 5,377
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05
|
longest
|
en
| 0.904476
|
https://www.got-it.ai/solutions/excel-chat/excel-tutorial/date-and-time/get-most-recent-day-of-week
| 1,713,617,589,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00754.warc.gz
| 714,591,948
| 19,035
|
Get instant live expert help with Excel or Google Sheets
“My Excelchat expert helped me in less than 20 minutes, saving me what would have been 5 hours of work!”
#### Post your problem and you'll get expert help in seconds
Your message must be at least 40 characters
Our professional experts are available now. Your privacy is guaranteed.
# How to Get the Most Recent Day of the Week in Excel
In most calculations involving dates and day of the week, Excel provides a solution and saves us enormous amount of time from performing calculations on our own. This step by step tutorial will assist all levels of Excel users in getting the most recent day of week using the MOD function.
Figure 1. Final result: Get most recent day of week
Final formula: `=B3-MOD(B3-2,7)`
## Syntax of MOD Function
MOD function returns the remainder after dividing a number by a divisor; the result of MOD always follows the sign of the divisor
`=MOD(number, divisor)`
• number – the number for which we want to divide by a divisor and find the remainder
• divisor – the number by which we want to divide number
## Syntax of TODAY Function
`=TODAY()`
• TODAY function does not have any arguments
• TODAY returns the current date
• It can be used in combination with other functions and mathematical operations to obtain the desired results
## Setting up our Data
Our table consists of two columns: Date (column B) and Most Recent Monday (column C). We have a list of dates in column B, formatted such that the day of the week is displayed for easier reference. We want to determine the most recent Monday and record the results in column C.
Figure 2. Sample data to get most recent day of week
By default, each day of the week is assigned a number by Excel where “1” refers to Sunday and “7” refers to Saturday. For this example, we want to get the most recent Monday. Monday has the day of value of “2”.
## Get most recent Monday
In order to get the most recent Monday, we will be using a formula with the MOD function. Let us follow these steps:
Step 1. Select cell C3
Step 2. Enter the formula: `=B3-MOD(B3-2,7)`
Step 3. Press ENTER
Step 4. Copy the formula in C3 to cells C4:C9 by clicking the “+” icon at the bottom-right corner of cell C3 and dragging it down
Figure 3. Get most recent Monday by using the MOD function
Our formula subtracts a value to the date in column B such that the result is the most recent Monday. This part of the formula MOD(B3-2,7) subtracts the day of the week, in this case 2, from the date then divides the result by 7. The remainder from the division operation is then subtracted from the date.
For our first example, B3 is a Thursday so its value is 5. The formula then becomes:
` = B3-MOD(5-2,7)`
` = B3-MOD(3,7)`
` = B3-3`
The most recent Monday is three days before the date in B3, or three days before January 24, 2019. Finally, our formula returns the date January 21, 2019, which is a Monday.
The table below shows the results for the most recent Monday in column C. Note that for row 7, the date “January 28, 2019” is already a Monday. Hence, the date in C7 is the same date as in B7.
Figure 4. Output: Get most recent Monday
## Note
We can determine the most recent day of the week from any current date by using the TODAY and MOD function. Let’s say we want to get the most recent Monday as of today, we use the formula:
`=TODAY()-MOD(TODAY()-2,7)`
If we want a different day of the week other than Monday, we simply replace the number “2” with the appropriate day of the week.
Most of the time, the problem you will need to solve will be more complex than a simple application of a formula or function. If you want to save hours of research and frustration, try our live Excelchat service! Our Excel Experts are available 24/7 to answer any Excel question you may have. We guarantee a connection within 30 seconds and a customized solution within 20 minutes.
| 991
| 3,942
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.765625
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.892813
|
https://community.fabric.microsoft.com/t5/Desktop/Sort-by-Day-of-Week/m-p/595197
| 1,718,763,547,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861796.49/warc/CC-MAIN-20240618234039-20240619024039-00173.warc.gz
| 147,216,372
| 115,413
|
cancel
Showing results for
Did you mean:
Find everything you need to get certified on Fabric—skills challenges, live sessions, exam prep, role guidance, and more. Get started
Anonymous
Not applicable
## Sort by Day of Week
Hello,
I have a column that list a range of dates from 2013 - present. I was asked to analyze the data by each day of the week, for example, Monday would display specific information that occured just on every monday and the same for tuesday, etc.
People have been pointing me to
`WeekDay = WEEKDAY(Table[Date],2)`
But it is not working. We are attempting to analyze the amount of tickets we recieve per day of the week and would like vizualize our busiest days.
Thanks
1 ACCEPTED SOLUTION
Employee
Hi @Anonymous,
Suppose your data table that lists a range of dates from 2013 to today is called "Table". Then, new a calculated column and type into this DAX formula: WeekDay = WEEKDAY(Table[Date],2)
The numer "2" determine the start weekday is Monday in a week. To help you understand the WEEKDAY function, please see: https://docs.microsoft.com/en-us/dax/weekday-function-dax
Best regards,
Yuliana Gu
Community Support Team _ Yuliana Gu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
3 REPLIES 3
Employee
Hi @Anonymous,
After adding a calculated column WeekDay = WEEKDAY(Table[Date],2), you should place [WeekDay] column and [tickets amount] column into a table visual. Choose "Sum" as its aggregation function, it will automatically sum up the ticket amounts sorted by each weekday.
Best regards,
Yuliana Gu
Community Support Team _ Yuliana Gu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable
@v-yulgu-msftThank you! WeekDay = WEEKDAY(Table[Date],2), Do I enter this as is? It seems to be having a problem with "TABLE". I am brand new to DAX. What is table refering to? Is the "2" refering to the number of a day of the week, "Tues = 2, Wed = 3 etc."?
Thank you and sorry for the beginner questions.
kyle
Employee
Hi @Anonymous,
Suppose your data table that lists a range of dates from 2013 to today is called "Table". Then, new a calculated column and type into this DAX formula: WeekDay = WEEKDAY(Table[Date],2)
The numer "2" determine the start weekday is Monday in a week. To help you understand the WEEKDAY function, please see: https://docs.microsoft.com/en-us/dax/weekday-function-dax
Best regards,
Yuliana Gu
Community Support Team _ Yuliana Gu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Announcements
#### Europe’s largest Microsoft Fabric Community Conference
Join the community in Stockholm for expert Microsoft Fabric learning including a very exciting keynote from Arun Ulag, Corporate Vice President, Azure Data.
#### Power BI Monthly Update - June 2024
Check out the June 2024 Power BI update to learn about new features.
#### New forum boards available in Real-Time Intelligence.
Ask questions in Eventhouse and KQL, Eventstream, and Reflex.
Top Solution Authors
Top Kudoed Authors
| 764
| 3,192
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.671875
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.926205
|
https://www.shaalaa.com/question-bank-solutions/arithmetic-progression-finding-their-general-term-which-following-are-arithmetic-progression-1-2-1-3-1-4-1-5_30857
| 1,575,777,301,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00128.warc.gz
| 873,221,525
| 10,323
|
Share
Which of the Following Are in Arithmetic Progression 1/2, 1/3 , 1/4 , 1/5 - ICSE Class 10 - Mathematics
ConceptArithmetic Progression - Finding Their General Term
Question
Which of the following are in arithmetic progression
1/2, 1/3 , 1/4 , 1/5
Solution
1/2, 1/3 , 1/4 , 1/5
d_1 = 1/3 - 1/2 = (2 - 3)/6 = -1/6
d_2 = 1/4 - 1/3 = (3 - 4)/12 = -1/12
Since d_1 != d_2, the given sequence is not in arithematic progression
Is there an error in this question or solution?
APPEARS IN
Solution Which of the Following Are in Arithmetic Progression 1/2, 1/3 , 1/4 , 1/5 Concept: Arithmetic Progression - Finding Their General Term.
S
| 232
| 643
|
{"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}
| 3.4375
| 3
|
CC-MAIN-2019-51
|
latest
|
en
| 0.785417
|
https://www.coursehero.com/file/6047859/a9/
| 1,521,504,593,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257647244.44/warc/CC-MAIN-20180319234034-20180320014034-00532.warc.gz
| 791,835,613
| 48,120
|
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
# a9 - F-1 explicitly b Find the derivative matrices DF x,y...
This preview shows page 1. Sign up to view the full content.
Math 237 Assignment 9 Due: Friday, Nov 26th 1. Invent an invertible transformation that transforms the ellipse 3 x 2 + 6 xy + 4 y 2 = 4 onto the unit circle and determine the inverse map. 2. Invent an invertible transformation f : R 3 R 3 that maps the ellipsoid x 2 +8 y 2 +6 z 2 + 4 xy - 2 xz + 4 yz = 9 onto the unit sphere. 3. Consider the maps F : R 2 R 2 defined by ( u, v ) = F ( x, y ) = ( y + xy, y - xy ). a) Show that F has an inverse map by finding
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: F-1 explicitly. b) Find the derivative matrices DF ( x,y ) and DF-1 ( u,v ) and verify that DF ( x,y ) DF-1 ( u,v ) = I . c) Verify that the Jacobians satisfy ∂ ( x,y ) ∂ ( u,v ) = h ∂ ( u,v ) ∂ ( x,y ) i-1 . 4. Find the Jacobian of ( u,v ) = T ( x,y ) = ( x 2 + y 2 ,x 2-y 2 )....
View Full Document
{[ snackBarMessage ]}
Ask a homework question - tutors are online
| 366
| 1,113
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.40625
| 3
|
CC-MAIN-2018-13
|
latest
|
en
| 0.683278
|
https://coolconversion.com/weight/2100-drachme-to-grain
| 1,534,284,905,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-34/segments/1534221209585.18/warc/CC-MAIN-20180814205439-20180814225439-00650.warc.gz
| 672,272,890
| 23,008
|
Please get in touch with us if you:
1. Have any suggestions
2. Have any questions
3. Have found an error/bug
4. Anything else ...
# 2100 Drachmes to Grains (dr to gr) Conversion
## How many grains in 2100 drachmes?
There are 123150.21965647 grains in 2100 drachmes
To convert any value in drachmes to grains, just multiply the value in drachmes by the conversion factor 58.642961741177. So, 2100 drachmes times 58.642961741177 is equal to 123150.21965647 grains. See details below and use our calculator to convert any value in drachmes to grains.
To use this drachmes to grains, converter simply type the dr value in the box at left (input). The conversion result in gr will immediately appear in the box at right.
### Drachmes to grains Converter
Enter values here: Results here:
=
Detailed result here
To calculate a drachme value to the corresponding value in grain, just multiply the quantity in drachmes by 58.642961741177 (the conversion factor). Here is the drachmes to grains conversion formula: Value in grains = value in drachmes * 58.642961741177 Supose you want to convert 2100 drachmes into grains. In this case you will have: Value in grains = 2100 * 58.642961741177 = 123150.21965647 (grains)
Using this converter you can get answers to questions like:
1. How many drachmes are in 2100 grains?
2. 2100 drachmes are equal to how many grains?
3. how much are 2100 drachme in grains?
4. How to convert drachmes to grains?
5. What is the conversion factor to convert from drachmes to grains?
6. How to transform drachmes in grains?
7. What is the drachmes to grains conversion formula? Among others.
## Values Near 2100 drachme in grain
drachmegrain
130076235.850263531
140082100.146437648
150087964.442611766
160093828.738785884
170099693.034960002
1800105557.33113412
1900111421.62730824
2000117285.92348235
2100123150.21965647
2200129014.51583059
2300134878.81200471
2400140743.10817883
2500146607.40435294
2600152471.70052706
2700158335.99670118
2800164200.2928753
2900170064.58904941
Note: some values may be rounded.
## Sample Weight / Mass Conversions
### Disclaimer
While every effort is made to ensure the accuracy of the information provided on this website, we offer no warranties in relation to these informations.
| 650
| 2,258
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.296875
| 3
|
CC-MAIN-2018-34
|
latest
|
en
| 0.708563
|
https://casadimenotti.com/math-aid-worksheets/math-aids-integers-worksheets-download-them-and-try-to-solve/
| 1,550,862,537,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-09/segments/1550247522457.72/warc/CC-MAIN-20190222180107-20190222202107-00396.warc.gz
| 517,534,137
| 16,472
|
# Math Aid Worksheets Division Aids Com Adding And Subtracting Rationalers Answer Key Fractions Mixed
By Claudette Lesperance at December 28 2018 19:52:38
So what kinds of worksheets should you get? Anything where you feel that your child needs further drill. We often have this notion that worksheets are just for math. This, of course, is not true. While they are excellent tools for reviewing math facts such as the multiplication tables and division facts, they are just as useful for reviewing parts of speech or the states in the union. When you're teaching your student to write, there are a whole host of worksheets online that you can use. Many of these include clipart that will help the students learn the sounds of letters and letter combinations. There are other sheets that help the student learn to write his or her numbers.
What sacrifices do I need to make to reach this goal? Any substantial goal requires sacrifice. The more significant the goal, likely the more substantial the sacrifice. This is a reality check: Are you prepared to make the sacrifices necessary to reach your goal? What information or skills do I need if I am to achieve this goal? Most big goals require us to grow personally in knowledge or skill. If you can figure out where the gaps are from the outset, and begin to fill them, you will progress toward your goal very rapidly.
## Gallery of Math Aid Worksheets
We are all aware that in academics, Math is one of the toughest subjects since it involves numbers and a lot of solving. It makes you think and rationalize every detail of your solution. Distinguishing Story Structure: Story structure can be defined by the way in which the text is arranged or organized into a plot. For example, by understanding characters, setting, problem/conflict, climax, and validation, students increase comprehension. By understanding headings, subheadings, picture graphs and bold words, students also increase comprehension.
The basic skill you require in order to successfully assist your child with its 3rd grade math worksheets problems is to be able to identify the difficulty. Does it lack the requisite skills that it should have already possessed for the work at hand? If that is the case then it is best that you take the child back to the missing link, so that your it learns what it is missing and move forward. Graphic Organizers - Graphic organizers are visual diagrams to aid in organizing information. Others names many include maps, webs, graphs, charts, etc. These organizers assist students in determining main ideas and supporting details. They help students visually summarize a reading selection.
| 524
| 2,653
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2019-09
|
longest
|
en
| 0.952081
|
https://yetanothermathblog.com/2022/05/26/rankines-the-mathematician-in-love/
| 1,685,609,104,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00456.warc.gz
| 1,165,536,512
| 24,627
|
# Rankine’s “The Mathematician in Love”
The 1874 poem “The Mathematician in Love” by Scottish mechanical engineer William Rankine (in the book From Songs and Fables) has been published in many places (e.g., poetry.com, New Scientist and the scanned version is available at the internet archive. However, the mathematical equations Rankine presented at the end of his poem are only available in the scanned versions. As WordPress can render LaTeX, the poem quoted below includes those last few lines.
The Mathematician in Love
William J. M. Rankine
I.
A mathematician fell madly in love
With a lady, young, handsome, and charming:
By angles and ratios harmonic he strove
Her curves and proportions all faultless to prove,
As he scrawled hieroglyphics alarming.
II.
He measured with care, from the ends of a base.
The arcs which her features subtended:
Then he framed transcendental equations, to trace
The flowing outlines of her figure and face.
And thought the result very splendid.
III.
He studied (since music has charms for the fair)
The theory of fiddles and whistles, —
Then composed, by acoustic equations, an air,
Which, when ’twas performed, made the lady’s long hair
Stand on end, like a porcupine’s bristles.
IV.
The lady loved dancing: – he therefore applied.
To the polka and waltz, an equation;
But when to rotate on his axis he tried.
His centre of gravity swayed to one side.
And he fell, by the earth’s gravitation.
V.
No doubts of the fate of his suit made him pause.
For he proved, to his own satisfaction.
That the fair one returned his affection; – “because,
As every one knows, by mechanical laws,
Re-action is equal to action.”
VI.
“Let x denote beauty, – y manners well-bred, –
x, Fortune, – (this last is essential), –
Let L stand for love” – our philosopher said, –
“Then z is a function of x, y and 0,
Of the kind which is known as potential.”
VII.
“Now integrate L with respect to dt,
(t Standing for time and persuasion);
Then, between proper limits, ’tis easy to see,
The definite integral Marriage must be: —
(A very concise demonstration).”
VIII.
Said he – “If the wandering course of the moon
By Algebra can be predicted,
The female affections must yield to it soon” –
But the lady ran off with a dashing dragoon,
And left him amazed and afflicted.
End notes:
Equation referred to in Stanza VI.–
$L=\phi(x,y,z)= \int\int\int \frac{f(x,y,z)}{\sqrt{(x-\xi)^2+(y-\nu)^2+(z-\zeta)^2}} \, d\xi d\nu d\zeta$
Equation referred to in Stanza VII.–
$\int_{-\infty}^\infty L\, dt = M$
## One thought on “Rankine’s “The Mathematician in Love””
1. What a unique piece. I enjoyed this a lot and will be reading more of your blog.
| 682
| 2,669
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.71875
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.95067
|
http://www.talkstats.com/tags/expected-value/
| 1,670,277,510,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00414.warc.gz
| 96,456,261
| 12,556
|
# expected value
1. ### Help required with statistics course
Hi, I need to complete a statistics course to complete my degree but I am struggling with it quite a bit. I have attempted some of the questions on my current assignment but overall I'm not sure how to go about it at all. I was wondering would anyone be able to help me to understand how to do...
2. ### Expected value of chi-square distribution
The chi-square distribution is defined as the distribution of Y = X_1**2 + X_2**2 + ... + X_k**2 with X_i independent standard normally distributed (mean 0, std 1). The expected value of a sum of variables is E(X_1 + X_2) = E(X_1) + E(X_2) The expected value of a product of independent...
2
4. ### joint probability distributions
Hi, I am very puzzled as to how they achieved μ=2/5 in the example below. I understand that μ is just expected value which would mean summing x or y with its possible realizations. I'm thinking this is very obvious, however, I just can't see how they arrived at 2/5, which just proves I...
5. ### Coin probability problem
We throw the coin 1 000 000 times. How many times on average will make 13 successful heads? Now the problem with the naive: 1 000 000/(2^13) is that once it made 13 heads the 14 head will happen with 1/2 probability , but it will count as 2 13 successful heads. 15 will happen 4xtimes less, but...
6. ### Word Problem - Student Who Misbubbles Exam
What I have done so far: If the student was expected to make a 75%, we basically expected he would get 15 of the 20 questions correct. Now that he has misbubbled his answers, we expect those 15 he should've gotten right to now be wrong. For the remaining questions he now has a 1/10 chance of...
7. ### Die With Multiple Same Sides
Hi everyone - see my problem below. With a regular die I know my denominator is 36 for each probability. What I'm wondering here is if that changes since multiple sides of the die have the same number... Right now I have the probability of a sum of 2 is 1/36, a sum of 3 as 2/36, sum of 4 as...
8. ### Expected Value + Variance
Hi guys just a quick question I am told i have two values which are 100 with probability of 0.8 and 1000 with probability of 0.2. Now, I know the E(X) is just 100(.8) + 1000 (.2). I know my var(x) = e(x^2) - e(x)^2 so is my var(x) just 100^2(.8) and 1000^(.2) ?
9. ### Sudokua problem using related binomial distribution
It is given: If you succeed to solve a Sudoku with probability 0.4 and you try many times, how likely is it that you succeed twice before you fail 5 times? What is the expected number of failures before two successes? I don't understand which one between the binomial, geometric, negative...
10. ### Resulting expected value of a bounded exponential pdf
I need to determine the distribution function of an exponential random variable with mean 2, given that its outcome is larger than 2 and give the resulting expected value? I know that the mean is 1/lambda, so lambda should be 1/2. I get that way 0.5*e^(-0.5x) >= 2. How can I deal with the...
llll
12. ### Expected Value of Y = max (X1, X2, X3, .... , Xn)
I'm having a hard time figuring out what the expected value of Y = max(X1, X2, X3, ... , Xn) with pdf (x;θ) = 2x/θ^2. So it is the nth value of order statistics Y1<Y2<Y3.....<Yn. Does anyone know?
13. ### Simple but quite confusing Problem
Hi all, I'm solving the problems in probability but, the question below looks simple but quite confusing to me. Regarding question 2-(c), any tips or advice would be appreciated to me. 2. Suppose that we have N balls numbered 1 to N. If we let Xi be the number on the ith drawn ball so...
14. ### Probability that X restaurants will be visited if Y people choose independently
I am trying to determine for Y people selecting from X restaurants (with an equal probability) what the expected value of the number of restaurants chosen would be. All decisions are independent. Ex: 2 people and 2 restaurants - E(X) = 1(.5) + 2(.5) = 1.5 restaurants will be visited on...
15. ### Help with the statistics solving
Violations 0 0 1 2 3 4 5 >5 Fine A$1000 0 0 10 10 10 20 50 (Please note from the table that 0 violations = 0 fines, 1 violation = 0 fines, 2 violations =$10,000 fine, 3 violations = $10,000 fine, 4 violations =$10,000 fine, 5...
16. ### 'Alternate' proof that the expected value of the sample mean is the population mean
It would be appreciated if someone could verify that this makes sense. By definition \bar{x} = \frac{\sum x_i}{n} So taking its expectation we get \bar{x} = \frac{1}{n} E[\sum x_i] Now, as we have a population of size N and a sample size of size n, we have {N\choose n} different samples and...
17. ### Expected value in a triangular distribution
Hi All, I'm struggling to understand following problem. It says, Demand follows a triangular distribution. Minimum 500 Mode 6,000 Maximum 24,000 Expected demand is given as 10,167. To me, expected value should be the mode i.e. 6000. I'm struggling to understand the...
18. ### Expected value (mean)
Hi All I'm looking for a metric that shows a close trend (good correlation) to the mean (expected value) of a distribution, so I can investigate the trend of that metric instead of mean when I don't have access to mean. Any suggestion? Thank you
19. ### Combining expected values and variances
Hi! First post here so cut me a bit of slack :) Hopefully an easy topic too (I'm new to stats). If I have some values for E(X) (e.g. 5min) and Var(X) (e.g. 2min^2) for a uniform distribution for say, the waiting time at a bank, how can I calculate, for example, the E(X) and Var(X) over a week...
20. ### Expected Value of Convolution
I am interested in estimating E[g(h(X))] where... - X ~ Normal(mu,sigma) - h(.) = CDF of Standard Normal distribution - g(.) = inverse CDF of Beta distribution with known parameters alpha and beta Any suggestions as to how to proceed? Taylor approximation? Any help appreciated. Jesse
| 1,578
| 5,948
|
{"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}
| 3.59375
| 4
|
CC-MAIN-2022-49
|
latest
|
en
| 0.956977
|
https://learn.careers360.com/school/question-explain-solution-rd-sharma-class-12-chapter-differentiation-exercise-10-point-2-question-16-maths/?question_number=16.0
| 1,716,402,796,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00768.warc.gz
| 303,747,727
| 39,610
|
#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 16 maths
Answer:$\sec x(\sec x+\tan x)$
Hint: You must know the rules of solving derivative of trigonometric function.
Given: $\sqrt{\frac{1+\sin x}{1-\sin x}}$
Solution:
Let $y=\sqrt{\frac{1+\sin x}{1-\sin x}}$
Differentiating with respect to x
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1+\sin x}{1-\sin x}\right)^{\frac{1}{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{\frac{1}{2}-1} \times \frac{d}{d x}\left(\frac{1+\sin x}{1-\sin x}\right)$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{1}{2}} \times\left\{\frac{(1-\sin x)(\cos x)-(1+\sin x)(-\cos x)}{(1-\sin x)^{2}}\right\}$$\ldots \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{1}{2}}\left\{\frac{(\cos x)(1-\sin x)-(1+\sin x)(-\cos x)}{(1-\sin x)^{2}}\right\}$
$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{1}{2}}\left\{\frac{2 \cos x}{(1-\sin x)^{2}}\right\}$
$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1+\sin x}(1-\sin x)^\frac{3}{2}}$
$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1+\sin x} \sqrt{1-\sin x}(1-\sin x)}$
$\frac{d y}{d x}=\frac{\cos x}{\sqrt{1-\sin ^{2} x} \times(1-\sin x)}$
$\frac{d y}{d x}=\frac{\cos x}{\cos x(1-\sin x)}$
\begin{aligned} &\frac{d y}{d x}=\frac{1}{(1-\sin x)} \times\left(\frac{1+\sin x}{1-\sin x}\right) \\ &\frac{d y}{d x}=\frac{(1+\sin x)}{\left(1-\sin ^{2} x\right)} \end{aligned}
$\frac{d y}{d x}=\frac{(1+\sin x)}{\left(\cos ^{2} x\right)}$
$\frac{d y}{d x}=\frac{1}{(\cos x)}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)$
$\frac{d y}{d x}=\sec x(\sec x+\tan x)$
| 783
| 1,763
|
{"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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.40625
| 4
|
CC-MAIN-2024-22
|
latest
|
en
| 0.308437
|
http://www.physicsforums.com/showthread.php?s=3f5dee1c992ecdc0b526b996c8fec056&p=4479078
| 1,386,531,788,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-48/segments/1386163800358/warc/CC-MAIN-20131204133000-00079-ip-10-33-133-15.ec2.internal.warc.gz
| 484,443,679
| 7,027
|
# Quick notation question
by strokebow
Tags: notation
P: 101 Hi, I want to be able to say the following with mathematical symbols; 1) A is any integer. 2) A is a non-integer (not a whole number). Please help. Thanks
PF Patron HW Helper Sci Advisor Thanks P: 11,935 For example: 1) $$A\in\mathcal{Z}$$ 2) $$A\notin\mathcal{Z}$$
P: 170 $A$ is an integer: $A\in\mathbb Z$. $A$ is not an integer: $A\notin\mathbb Z$. $A$ is a real number which is not an integer: $A\in\mathbb R\backslash\mathbb Z$. $A$ is not a whole number: $A\notin\mathbb Z_+$.
Related Discussions General Physics 2 General Math 2 Introductory Physics Homework 2 Special & General Relativity 4 General Math 4
| 222
| 680
|
{"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}
| 2.515625
| 3
|
CC-MAIN-2013-48
|
latest
|
en
| 0.704148
|
jamiehalesblog.blogspot.nl
| 1,500,992,865,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-30/segments/1500549425254.88/warc/CC-MAIN-20170725142515-20170725162515-00483.warc.gz
| 162,032,111
| 31,203
|
## Wednesday, May 24, 2017
### Don't Forget About The Stats: Quantitative Research
In the context of everyday language statistics (numbers, quantitative representations) are used to represent basketball player's free throw average, death rates, life spans and so on. In science statistics are tools used in describing, organizing, summarizing and analyzing data. Learning about stats will help you think in terms of probabilities, and allow you to gain a better understanding of research data. Statistics are not easy, but with some effort the basics can be learned by most people. Research methods and statistics are often taught together in college courses. Quantitative research uses stats, and these stats are essential in an effort to represent the data; stats are required for making sense of the research.
Descriptive statistics are numerical measures that describe a population by providing information on the central tendency of the distribution, the width of distribution (dispersion, or variability), the shape of distribution (Jackson, 2009). Inferential statistics are procedures that allow us to make an inference from a sample to the population. That is, we are able to make generalizations about a population based on the information derived from the sample.
A key reason we need statistics is to be able to effectively interpret research. Without statistics it would be very difficult to analyze the collected data and make decisions based on the data. Statistics give us an overview of the data and allow us to make sense of what is going on. Without statistics, in many cases, it would be extremely difficult to find meaning in the data. Statistics provides us with a tool to make an educated inference.
Most scientific and technical journals contain some form of statistics. Without an understanding of statistics, the statistical information contained in the journal will be meaningless. An understanding of basic statistics will provide you with the fundamental skills necessary to read and evaluate most results sections. The ability to extract meaning from journal articles, and the ability to evaluate research from a statistical perspective are basic skills that will increase your knowledge and understanding of the article of interest. To reiterate, quantitative research uses stats, and to assess statistical validity, at least a basic understanding of stats is essential.
When researchers question a study’s statistical validity they are questioning issues relevant to how well the conclusions coincide with the results, represented as statistics. Interrogating statistical validity may include some of the following questions: If the study found a difference what is the probability that the conclusion was a false alarm? If the study’s finding found no difference what is the probability that a real relationship went unnoticed? What is the effect size? Is the difference between groups statistically significant? Are the finding practically significant? What type of inferential stats were used to assess predictions? Could different statistical procedures have been used?
Gaining knowledge in the area of statistics will help you become a better-informed consumer. Statistics are difficult for many people. Students often cringe when they hear the word - statistics. Learning about statistics requires the same strategies as learning about other topics (strategies to improve learning and memory). Once an individual learns theoretical aspects and calculations used for basic statistical procedures the learning of more complex statistics become much easier. Everyone benefits from learning the basics of statistics. Statistics is not an easy subject compared to many other subjects, but the subject is much easier when one doesn't have negative expectations and realizes that with the appropriate cognitive effort and understanding of some rather basic mathematical principles the subject is learnable. Being knowledgeable in the area of statistics will be beneficial across domains of scholarly and everyday life.
Recently I asked Dr. Jonathan Gore (from Eastern Kentucky University) the following question- Why is a basic understanding of stats important for the public? He gave the following answer:
"My answer to why stats is important is that pretty much everything operates based on probability. Even some of the "hard" sciences are starting to realize that phenomena that used to only require a basic equation are now having to factor in probability to account for all that they observe."
If the objective is to thoroughly analyze the study, don't skip over the "Results" section when reading the paper. A key guideline for the Results section is a presentation of numerical findings that should be stated clearly, concisely and accurately. The methodology provides detailed information regarding processes used in the collection of data, while statistical procedures provide information on detecting meaningful signals among the noise: making sense of the data collected.
The book contains 76 questions and answers regarding scientific research methods and stats. It also contains practice problems involving statistical procedures.
References are available upon request
## Saturday, January 14, 2017
### Better Study Strategies
Studying should be cognitively challenging, persistent and structured. Strong memory is not built easily or overnight. This article consists of links to articles relevant to learning / memory and key points from my seminar- Strategies To Maximize Learning.
How To Study
The effort required to form strong memory is often intense for students. Students often spend hours trying to master new information. Of course, methods to enhance memory are important for everyone, not just students. For example, when a friend recommends a new shoe store we want to remember the name of of it, or when going to the grocery it is important to remember the items we need to pick up. What are some strategies that can be used to strengthen memory?
Trying To Remember
In one study researchers investigated the role of intentional-encoding instructions and task relevance at study on visual memory performance (Varakin & Hale, 2014). Task relevance was manipulated by having participants keep a running tally of either the objects they were attempting to remember or an irrelevant category of objects during study. Half of the participants within each level of task relevance were further instructed to remember one category of objects for a subsequent recognition memory test (intentional memory group) , and the other half of the participants were not informed of a memory test (incidental memory group). Intentional-encoding instructions improved recognition discrimination only when participants were not already keeping a tally of the to-be-remembered objects. This result suggests that intentional-encoding instructions may improve visual memory due
Building a Better Memory
Are learning and memory completely distinct? No; both are experienced based. “[M]emory is the consequence of learning from an experience- that is, the consequence of acquiring new information” , asserts James McGaugh (memory researcher, author of Memory and Emotion). Learning is a process of memory formation. There are 2 general categories of memory: explicit and implicit.
Key Points from- Strategies to Maximize Learning (Hale, 2014):
Memory is the product of learning
Memory formation = brain change
All cognition, emotion, feeling, perception and learning emanate from the brain
Healthy brain is imperative to maximize learning / memory
Mind- body is a unit- not separate
All cognition, emotion, feeling, perception and learning emanate from the brain Healthy brain is imperative to maximize learning / memory Mind- body is a unit- not separate
Foundations of memory include: brain health, focused attention, elaborative encoding, spaced rehearsal and testing
Understanding is imperative for strong memory
Studying should be structured: progressive, organized, spaced over multiple sessions and involve accurate evaluation
## Wednesday, November 16, 2016
### Rationality Quotient: Comprehensive Assessment of Rational Thinking
Stanovich and colleagues recently developed a prototype of the first comprehensive assessment of rational thinking. The test is discussed, and presented in detail in the new book, titled- The Rationality Quotient.
Up until publication of- The Rationality Quotient - components of rational thinking had been tested using various tasks, but a comprehensive test was not available. I first discussed the development of such a test with Stanovich, in 2013- interview here.
In the following interview (conducted in November, 2016) Stanovich provides detailed answers to important questions about the test.
What are some of the initial reactions, regarding the RQ, from academics?
Uniformly positive so far, and I believe that is because we were careful in the book to be explicit about two things. First, we were clear about what our goals were and the goals were circumscribed. Secondly, we included an entire chapter contextualizing our test (the Comprehensive Assessment of Rational Thinking, CART) and discussing caveats regarding its use as a research instrument or otherwise. In fact, I think we have already entirely achieved our aims. We have a prototype test that is a pretty comprehensive measure of the rational thinking construct and that is grounded in extant work in cognitive science. Now, this is not to deny that there is still much work to be done in turning the CART into a standardized instrument that could be used for practical purposes. But of course a finished test was not our goal in this book. Our goal was to show a demonstration of concept, and we have done that. We have definitively shown that a comprehensive test of rational thinking was possible given existing work in cognitive science. This is something that I have claimed in previous books but had not empirically demonstrated with the comprehensiveness that we have here by introducing the CART. As I said, there are more steps left in turning the CART into an “in the box” standardized measure, but that is a larger goal than we had for this book.
I think that, at least so far, most academics have understood our goals and the feedback has been good. We wrote a summary article on the CART in a 2016 issue of the journal Educational Psychologist (51, 23-34) and the feedback from that community has been good.
Are there components of the RQ that can be expected to show a strong positive correlation with intelligence?
The CART has 20 subtests and four thinking dispositions scales (the latter are not part of the total score). Collectively they tap both instrumental rationality and epistemic rationality. In cognitive science, instrumental rationality means behaving in the world so that you get exactly what you most want, given the resources (physical and mental) available to you. Epistemic rationality concerns how well beliefs map onto the actual structure of the world. The two types of rationality are related. In order to take actions that fulfill our goals, we need to base those actions on beliefs that are properly calibrated to the world.
The CART assesses epistemic thinking errors such as: the tendency to show incoherent probability assessments; the tendency toward overconfidence in knowledge judgments; the tendency to ignore base rates; the tendency not to seek falsification of hypotheses; the tendency to try to explain chance events; the tendency to evaluate evidence with a myside bias; and the tendency to ignore the alternative hypothesis.
Additionally, CART assesses instrumental thinking errors such as: the inability to display disjunctive reasoning in decision making; the tendency to show inconsistent preferences because of framing effects; the tendency to substitute affect for difficult evaluations; the tendency to over-weight short-term rewards at the expense of long-term well-being; the tendency to have choices affected by vivid stimuli; and the tendency for decisions to be affected by irrelevant context.
Importantly, the test also taps what we call contaminated mindware. This category of thinking problem arises because suboptimal thinking is potentially caused by two different types of mindware problems. Missing mindware, or mindware gaps, reflect the most common type—where Type 2 processing does not have access to adequately compiled declarative knowledge from which to synthesize a normative response to use in the override of Type 1 processing. However, in the book, we discuss how not all mindware is helpful or useful in fostering rationality. Indeed, the presence of certain kinds of mindware is often precisely the problem. We coined the category label contaminated mindware for the presence of declarative knowledge bases that foster irrational rather than rational thinking. Four of the 20 subtests assess contaminated mindware.
My purpose in digressing here to describe the CART is to point out that given the number and complexity of rational thinking skills, it is likely that the subtests will have correlations with intelligence that are quite variable. The four subtests with the highest correlations are: the Probabilistic Reasoning Subtest; the Scientific Reasoning Subtest; the Reflection Versus Intuition Subtest; and the Financial Literacy Subtest. Correlations with these subtests tend to .50or higher. Most of the subtests of the CART correlate with intelligence in the range of .25 to .50 (a few have even lower correlations). Some very important components of rational thinking do show considerable dissociation from intelligence. Overconfidence (measured by the Knowledge Calibration Subtest of the CART) shows only a .38 correlation with intelligence. This represents a substantial amount of dissociation for a key component of rational thinking. Kahneman, for example, devoted substantial portions of his best-selling book to this component of rational thinking. Myside bias (measured by our Argument Evaluation Subtest) likewise shows a correlation of .38, indicating a substantial dissociation. This thinking bias is at the center of many discussions of what it means to be rational. Some of the subtests that most directly measure the components of the axiomatic approach to utility maximization show relatively mild correlations with intelligence. For example, the Framing Subtest shows a fairly low .28 correlation. Framing measures a foundational aspect of rational thinking according to the axiomatic approach.
Finally, some subtests of immense practical importance show very low correlations with intelligence in the CART. The skill of assessing numerical expected value shows a correlation of only .21, and the ability to delay for greater monetary reward shows a correlation of only .06. The tendency to believe in conspiracies shows a modest correlation of .34.
Do you think rationality will acquire the same high level status as intelligence in the near future?
Not in the near future, no. Our goal with the book was more modest—to simply raise awareness of the importance of rational thinking and the ability of modern cognitive psychology to measure it. The result of our efforts will, we hope, redress the imbalance between our tendency to value intelligence versus rationality. In our society, what gets measured gets valued. Our aim in developing the CART was to draw attention to the skills of rational thought by measuring them systematically. In the book, we are careful to point out that we operationalized the construct of rational thinking without making reference to any other construct in psychology, most notably intelligence. Thus, we are not trying to make a better intelligence test. Nor are we trying to make a test with incremental validity over and above IQ tests. Instead, we are trying to show how one would go about measuring rational thinking as a psychological construct in its own right. We wish to accentuate the importance of a domain of thinking that has been obscured because of the prominence of intelligence tests and their proxies. It is long overdue that we had more systematic ways of measuring these components of cognition, that are important in their own right, but that are missing from IQ tests. Rational thinking has a unique history grounded in philosophy and psychology, and several of its subcomponents are firmly identified with well-studied paradigms. The story we tell in the book is of how we have turned this literature into the first comprehensive device for the assessment of rational thinking (the CART).
Why does society need a comprehensive assessment of rational thinking?
To be globally rational in our modern society you must have the behavioral tendencies and knowledge bases that are assessed on the CART to a sufficient degree. Our society is sometimes benign, and maximal rationality is not always necessary, but sometimes, in important situations, our society is hostile. In such hostile situations, to achieve adequate degrees of instrumental rationality in our present society the skills assessed by the CART are essential. In Chapter 15 of The Rationality Quotient we include a table showing that rational thinking tendencies are linked to real life decision making. In that table, for each of the paradigms and subtests of the CART, an association with a real-life outcome is indicated. The associations are of two types. Some studies represent investigations where a laboratory measure of a bias was used as a predictor of a real-world outcome. Others are reports of real-world analogues of biases that were originally discovered in the lab. Clearly more work remains to be done on tracing the exact nature of the connections—that is, whether they are causal. The sheer number of real-world connections, however, serves to highlight the importance of the rational thinking skills in our framework. Now that we have the CART, we could, in theory, begin to assess rationality as systematically as we do IQ. If not for professional inertia and psychologists’ investment in the IQ concept, we could choose tomorrow to more formally assess rational thinking skills, focus more on teaching them, and redesign our environment so that irrational thinking is not so costly. Whereas just thirty years ago we knew vastly more about intelligence than we knew about rational thinking, this imbalance has been redressed in the last few decades because of some remarkable work in behavioral decision theory, cognitive science, and related areas of psychology. In the past two decades cognitive scientists have developed laboratory tasks and real-life performance indicators to measure rational thinking tendencies such as sensible goal prioritization, reflectivity, and the proper calibration of evidence. People have been found to differ from each other on these indicators. These indicators are structured differently from the items used on intelligence tests. We have brought this work together by producing here the first comprehensive assessment measure for rational thinking, the CART.
## Tuesday, October 25, 2016
### Science and Rationality in Modern Society
Science and rationality are important in modern, technologically advanced, industrial societies. Science is a large enterprise consisting of multiple components. Science, although fallible, is the great reality detector. Rationality, in this context, refers to rationality as it is conceptualized in cognitive science. Rationality is concerned with judgment and decision making. Rationality consists of two main categories- instrumental and epistemic. Instrumental rationality reflects goal optimization, and epistemic reflects evidence based beliefs. There is overlap between the two categories of rationality. In my most recent book- In Evidence We Trust: The need for science, rationality & statistics- I provide information on various aspects of science, rationality and mathematical procedures (statistics) used in describing and making inferences in the context of scientific research.
In Evidence We Trust
It is often said we live in the information age, but we also live in the mis-information age. How do we decide what constitutes knowledge and what constitutes nonsense? Maybe there are no wrong or right answers, and just opinions? This notion is fallacious. There are facts and opinions, right and wrong answers. There is a reality that extends beyond personal comforts and opinions (Mitchell & Jolley, 2010). In the context of science facts are tentative. They are assertions that are supported by the preponderance of evidence. Facts in the context of science (primary concern in this book) are based on levels of certainty, but absolute certainty is never attained. Scientific findings are presented in terms of probabilities and data (e.g. laws, principles, theories, etc.) is revised in accordance to findings.
Testimonials, anecdotes, they-says, wishful thinking and so on do not count for evidence. If these types of claims and feelings are labeled as evidence then any discussion of evidence is vacuous. Testimonials exist for almost any claim you can imagine. That does not mean that claims of this sort have no value. Experiences are confounded (confused by alternative explanations). Experiences may be very important in some contexts, and they may serve as meaningful research questions. However, a meaningful question or a possible future finding is not synonymous with evidence. Scientific evidence is drastically different than evidence as it relates to everyday discourse. As Joy Victoria points out- it should be obvious from the book's title that the type of evidence I am referring to in the book is derived from scientific findings (paraphrased).
The content in chapter one includes short-articles (old, new & revised), a science discussion roundtable (featuring individuals from various fields) and a nonsense detection kit. Some of the short articles presented in chapter one have been published on various internet sites, and some of the same or similar information may be discussed in across different articles. There are at least two key benefits that can occur when presenting similar information across different articles (in different contexts): strengthening of memory connections, and each article can be read as a stand-alone article. In the science discussion roundtable participants are asked two questions. One) Do you have any tips for people that are interested in enhancing their ability to read scientific research? Two) What is the biggest (or at least one of the biggest misconceptions) misconception about science? The Nonsense Detection Kit is presented at the end of chapter one. The impetus for designing the Nonsense Detection Kit was similar kits devised by Sagan, Shermer, and Lilienfeld.
Chapter two features short articles on rationality. Some of the same or similar information is contained across different articles. There are at least a couple of advantages to presenting information in this manner (refer to previously mentioned advantages in chapter one). Many of the articles focus on the rationality intelligence dichotomy. Also included in this chapter are interviews with Keith Stanovich and the Stanovich Research Lab (Keith Stanovich, Richard West and Maggie Toplak). In the interview with Stanovich, he discusses the development of an RQ Test. In the interview with the Stanovich lab, rationality and intelligence are discussed. Since the publication of the book Stanovich, West and Toplak have designed the first comprehensive test for rational thinking
Chapter three features frequently asked questions about research methods and statistics. Many of the questions are questions I have received in the past from my students. Some of the questions address basic research and statistics problems, while other questions are more complex. At the end of the chapter recommended sources are provided for readers that are interested in furthering their studies on research methods and statistics.
The book ends with an appendices section. Practice problems, and guidelines regarding APA citations and reference lists are given.
The content in this book may be difficult for some to comprehend. However, with some effort and patience the content is learnable for most people. In the words of Albert Einstein “Things should be made as simple as possible, but not any simpler.” Science, rationality and statistics can be simplified to a degree, but relative to most other topics these topics are difficult. This book is not written for cognitive misers (the cognitively lazy). This book is written for individuals that are interested in separating knowledge and nonsense, and are willing to put forth at least a moderate level of cognitive effort. This book is not written in the format often used by pop science writers.
I would like to thank Joy Victoria, Kitty Mervine, Jason Silvernail and Coert Visser for the review articles of IEWT they have written.
| 4,779
| 25,146
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.625
| 4
|
CC-MAIN-2017-30
|
longest
|
en
| 0.9312
|
https://999calculator.net/force-to-velocity-calculator/
| 1,696,125,371,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510734.55/warc/CC-MAIN-20231001005750-20231001035750-00320.warc.gz
| 93,323,204
| 12,147
|
# Force to Velocity Calculator
## About Force to Velocity Calculator (Formula)
The Force to Velocity Calculator is a fundamental tool in the realm of physics and engineering, allowing users to explore and understand the relationship between force and velocity in various mechanical systems. It is an essential tool for engineers, physicists, and researchers working on projects involving motion and dynamics. The calculator relies on a specific formula designed to quantify the relationship between force and velocity, offering valuable insights into the behavior of objects subjected to external forces.
The formula for calculating velocity based on force is:
Velocity (v) = (Force (F) / Mass (m)) × Time (t)
In this formula:
• Velocity (v): Velocity represents the speed and direction at which an object moves. It is typically measured in meters per second (m/s) or another unit of velocity.
• Force (F): Force is an external influence or interaction that can cause an object to accelerate or change its velocity. It is measured in newtons (N).
• Mass (m): Mass is the measure of the amount of matter in an object and is typically measured in kilograms (kg).
• Time (t): Time represents the duration over which the force is applied and is measured in seconds (s).
The Force to Velocity Calculator applies this formula to provide valuable insights into mechanical systems and their behavior:
1. Mechanical Engineering: Engineers use this calculator to analyze the effects of forces on objects in motion, helping design and optimize various mechanical systems such as engines, vehicles, and machinery.
2. Physics Experiments: Researchers use it in physics experiments to understand the relationship between force and velocity, making it an invaluable tool in scientific investigations.
3. Vehicle Dynamics: In automotive and aerospace industries, engineers employ this calculator to study the impact of forces on vehicles and aircraft, aiding in vehicle performance analysis and design.
4. Biomechanics: In the field of biomechanics, scientists use it to explore the forces affecting human and animal movement, assisting in the design of prosthetics and sports equipment.
To use the Force to Velocity Calculator, users input values for force, mass, and time, which correspond to their specific application or experiment. The calculator then computes the velocity (v), providing a numeric value that quantifies the object’s speed and direction under the influence of the applied force.
In conclusion, the Force to Velocity Calculator, driven by its specialized formula, is an essential tool for understanding motion dynamics in the fields of physics and engineering. It facilitates the analysis of how external forces impact the velocity and behavior of objects in motion. Whether optimizing mechanical systems, conducting physics experiments, or designing vehicles and equipment, this calculator plays a pivotal role in unraveling the intricate relationship between force and velocity in the world of motion and dynamics.
| 555
| 3,031
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.71875
| 4
|
CC-MAIN-2023-40
|
longest
|
en
| 0.931648
|
eurasianhub.com
| 1,660,821,909,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00044.warc.gz
| 234,903,970
| 92,926
|
# Consumption Function
A consumption function is a mathematical approach to finding out the relation between consumption and gross national income. The consumption expenditure is an essential component of aggregate demand. It is related to income(Y). Aggregate demand refers to the entire amount of money that individuals want to spend on goods and services produced in the economy at various levels of income.
If the income level (Y) is high then C will also high and visa versa is also true. The income (Y) and consumption (C) is represented by a functional relation known as the consumption function. Therefore consumption function is a functional relationship between Y (income) and C (Consumption)
This relationship is helpful in understanding the behavior of consumption expenditure with respect to the income level. On observing the behavior of C and Y the experts of economics come up with the following essential points:
• There is some slightest level of C at the value of Y =0. This is known as autonomous consumption. This leads to the value of S in negative terms.
• Consumption is positively related to the value of income. The rise in the value of Y also increases the value of C and visa-versa is also true.
• Also, the whole increase in the value of Y is not converted in C. Some part of it is saved. It means the rate at which Consumption rises often lags behind the rate at which income increases. To completely understand the behavior of C with respect to Y consider the table mentioned below:
Y(Income) C(Consumption) 0 20 50 60 100 100 150 140
This above-mentioned table is the tabular presentation of the consumption function
i)The value of 20 is the minimum level of consumption. When income is zero, this demonstrates that even when Y=0, there is always some minimal amount of consumption. This minimum level of consumption during Y is zero is termed autonomous consumption.
ii)The value of C rises as the value of Y rises. It means that C is positively related to the value of Y.
iii) When the value of Y rises from 0 to 50. At that time the value of C is also rising but its value is lower than Y. This shows saving. It means at a certain point the entire rise in Y is not turned into C. A part of it is saved.
Also read : Scope of Macroeconomics
## Diagrammatic Reprenstion
In the above graph, Consumption is represented on the Y-axis and income on the X-axis. There is a line Y which is inclined at an angle 45 . C line represents the consumption function. This line helps in observing the behavior of C with respect to Y.Line C starts from 20 which is the minimum level of income.The C line has a positive slope which is moving in the upward direction. More is the value of Y more will be the value of C. In the initial state, the value of Cis is higher than Y, Finally, its value starts lagging behind the value of Y.
## Relation With Marginal Propensity To Consume
The slope of line C is the representation of the C function. Let us elaborate it. Basically, it refers to the increase in the rate of C to the increase of Y . There is some proportion of additional income which goes into the consumption. This can be measured in terms of the rate of change in C to the rate of change in Y . This is termed Marginal prosperity to consume.
Marginal propensity to consume is the ratio of ΔC to the ΔY .
### Graph
From the above graph, the value of MPC is calculated as ΔC / ΔY =40/50=0.8
This shows out of every additional rupee of income the 80% part is spent for consumption.
The consumption function can be linear when MPC is constant.
## APC and MPC
The APC stands for average prosperity to consumption and MPC as marginal prosperity to consume. The average propensity to consume (APC) is the ratio of total C(consumption) to the total Y ( income). The MPC, on the other hand, is the ratio of increased consumption( ΔC )to increased income (ΔY ).
### Mathematical representation of APC and MPC
APC=C/Y
MPC=ΔC/ΔY
as Y=150 and C=140 then
APC=140/150=0.933
as ΔY=50 and ΔC =40
MPC=40/50=0.8
In essence, APC shows consumption per unit of total income and MPC shows the consumption per unit of additional income.
## Algebraic Representation of C function
The consumption function formula is expressed as :
C=+bY
where C̅ is the minimum level of consumption. It is the value of consumption when Y is equal to 0. The b symbol is MPC.
The above algebraic equation helps us in calculating the different values of C corresponding to the different values of Y.
C=20+0.8Y(as given)
C when Y=0 =20+0.8(0)=20
C when Y=50 =20+0.8(50)=60
C when Y=100 =20+0.8(100)=100
C when Y=150 =20+0.8(150)=140
We come to the conclusion that the estimated values of consumption are exactly same as mentioned in table we mentioned above in the article. Thus we can calculate the values of consumption once we know the values
| 1,143
| 4,881
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33
|
longest
|
en
| 0.953955
|
https://thermopedia.com/jp/content/687/
| 1,719,266,564,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00045.warc.gz
| 513,744,736
| 5,469
|
ライブラリ登録: Guest
25341
The Dirac Delta Function δ(x - a) is an impulsive function defined as zero for every value of x, except for the point x ≠ a where it jumps to an infinitely large value. However, its graph encloses a unit area. It can be regarded as an idealization of a unit impulse. We define δ(x - a) through the following two properties:
This function has the following important property: for any continuous function f(x),
that is, δ(x - a) applied to f(x) detects its value at x = a.
We can heuristically show the validity of this property using the following argument: Let us approximate δ(x - a) through the function δε(x - a), such that:
which approaches δ(x - a) as ε tends to zero.
Clearly, the area covered by δε(x - a) is equal to one;
Furthermore, let F(x) be the primitive of f(x) (that is F'(x) = f(x)), then:
as ε goes to zero, the last expression defines the derivative of F(x) at x = a, which is precisely f(a).
The function δ(x - a) has a number of important applications in mathematical physics, in particular the solution of differential equations. In fact, it belongs to a class of generalized functions called distributions.
REFERENCES
Schwartz, L. (1973) Théorie des Distributions, Hermann, Paris.
参考文献
1. Schwartz, L. (1973) Théorie des Distributions, Hermann, Paris.
| 329
| 1,311
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.875
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.884185
|
http://mathhelpforum.com/calculus/185800-basic-but-frustrating-maximisation-problem.html
| 1,480,950,964,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698541697.15/warc/CC-MAIN-20161202170901-00224-ip-10-31-129-80.ec2.internal.warc.gz
| 175,799,702
| 13,141
|
# Thread: Basic but frustrating maximisation problem
1. ## Basic but frustrating maximisation problem
Consider the following equation
U=(H-h) * (I^(1-e))/1-e
and the following identity
I=h*W
and the following condition
H>h
I want to solve for e when U is at its maximum. By guess is to transform to
U=(H-h) * ((hw)^(1-e))/1-e
and differentiate wrt to h , then set dU/dh to zero and solve. This method should work as it is an inverted U shaped function.
or alternatively express as I
U=(H-I/w) * (I^(1-e))/1-e
and differentiate wrt to I.
The expected result is e=H/(H-h), but I cannot do the proof !
Thanks in advance, my maths is very rusty, and i cannot get any further, unfortunately.
2. ## Re: Basic but frustrating maximisation problem
Originally Posted by frustrated
Consider the following equation
U=(H-h) * (I^(1-e))/1-e
and the following identity
I=h*W
and the following condition
H>h
I want to solve for e when U is at its maximum. By guess is to transform to
U=(H-h) * ((hw)^(1-e))/1-e
and differentiate wrt to h , then set dU/dh to zero and solve. This method should work as it is an inverted U shaped function.
or alternatively express as I
U=(H-I/w) * (I^(1-e))/1-e
and differentiate wrt to I.
The expected result is e=H/(H-h), but I cannot do the proof !
Thanks in advance, my maths is very rusty, and i cannot get any further, unfortunately.
First change your notation, what you have makes it difficult to read and will result in some confusing notation.
Let the free variable be $x=1-e$ rather than $e$, and put $k=H-h>0$, also assume $I>0$. Then your problem is to find the $x$ that maximises:
$U(x)= k \frac{e^{\ln(I)(x)}}{x}$
so:
$U'(x)=k\left[ \frac{\ln(I) e^{\ln(I)x}}{x}-\frac{e^{\ln(I)x}}{x^2}\right]$
so for an extremum you need to solve $U'(x)=0$, which if $x\ne 0$ is equivalent to:
$x=\frac{1}{\ln(I)}$
Note: The maximising $e$ in your notation cannot depend on $H$ as this only appears in the multipler which is by definition a constant greater than zero and so effects the maximum but not the value of $e$ which gives this.
CB
3. ## Re: Basic but frustrating maximisation problem
Thanks, Cptn Black, but i think it is wrong! You cannot treat K as a constant here becuase h=I/W, hence your K=H-I/W. The K contains the I term, so it is not a constant.
4. ## Re: Basic but frustrating maximisation problem
I think i have it solved;
firtsly expand the first section to
U= H* (I^1-e)/(1-e) - I/W * (I^1-e)/(1-e)
bring the I in the second term up;
U= H *1^1-e/(1-e) - I^2-e/(w*(1-e)
differentiate wrt to I
U' = H I ^ e - (2-e)/(w*(1-e))*I^1-e
set U'to zero;
H I ^ e = (2-e)/(W*(1-e)) * I^1-e
divide both terms by I^e
H=(2-e)(W*(1-e)) *I
substitute h/W for I
H= (2-e)(1-e)*h
(2-e)(1-e) = H/h
I think that is right !
5. ## Re: Basic but frustrating maximisation problem
Originally Posted by frustrated
Thanks, Cptn Black, but i think it is wrong! You cannot treat K as a constant here becuase h=I/W, hence your K=H-I/W. The K contains the I term, so it is not a constant.
Unless you are saying that W is an unspecified function of x then that is irrelevant (and if you are then we must assume that h and H are also potentially functions of x).
CB
6. ## Re: Basic but frustrating maximisation problem
Originally Posted by frustrated
I think i have it solved;
firtsly expand the first section to
U= H* (I^1-e)/(1-e) - I/W * (I^1-e)/(1-e)
bring the I in the second term up;
U= H *1^1-e/(1-e) - I^2-e/(w*(1-e)
differentiate wrt to I
U' = H I ^ e - (2-e)/(w*(1-e))*I^1-e
set U'to zero;
H I ^ e = (2-e)/(W*(1-e)) * I^1-e
divide both terms by I^e
H=(2-e)(W*(1-e)) *I
substitute h/W for I
H= (2-e)(1-e)*h
(2-e)(1-e) = H/h
I think that is right !
Please in future identify which symbols represent the variables and which you think are constants or functions. Otherwise we are wasting our time looking at your questions.
CB
7. ## Re: Basic but frustrating maximisation problem
Sorry, my bad. I should have made it clear that W and H are constants, hence, to specify I is to specify h. In this equation h is hours worked and w is wages per hour, and I is income, hence I=hw. I will try to be clearer in the future.
The problem is that I have not done any proper maths for ten years, so I am really out of my depth.
On a differnt note, is it possible to to further simplify the answer given above, assuming it is correct ?
8. ## Re: Basic but frustrating maximisation problem
Originally Posted by frustrated
Sorry, my bad. I should have made it clear that W and H are constants, hence, to specify I is to specify h. In this equation h is hours worked and w is wages per hour, and I is income, hence I=hw. I will try to be clearer in the future.
The problem is that I have not done any proper maths for ten years, so I am really out of my depth.
On a differnt note, is it possible to to further simplify the answer given above, assuming it is correct ?
And what is e? The base of natural logarithms?
CB
9. ## Re: Basic but frustrating maximisation problem
e in this equation is an inequality/risk aversion paramater for the isoelastic utility from income function U=I^1-e/1-e. H is hours available to work or engage in leisure. The H-h term is thus lesure time. There is a straight multiplication of power-discounted income and leisure to give utility ,U
It thus a special form of a Cobb-Douglas function
U = L^a * I^b
where in this instance a=1 and b=1-e so there is 'economies of scale' as a+b>1 when b is positive
I just realised an easier way to perform this type of operation is to take the natural log of the equation, and maximise that instead.
According to a reference book this leads to the generalised result
h/H =b/(b+a)
which from my specific equation form should give
h/H = (1-e)/(2-e)
I have checked the working, and actual numerical results calculated, and
h/H =(1-e)/(2-e) is correct.
I should have used another letter for epsilon !
Working is as above, except that I made a minor typing error leaving out a divisor, proper result is below;
U=(H-I/w) * (I^1-e)/1-e
expand to
U= H(I^1-e)/1-e - I/W(I^1-e)/1-e
Bring the I on the right up and w down
U= H(I^1-e)/1-e - (I^2-e)/w(1-e)
differentiate wrt I gives
Du/DI = HI^-e - (2-e)/w(1-e) * I^(1-e)
Du/Di = 0
HI^-e = (2-e)/w(1-e) * I^(1-e)
divide by I^-e
H= (2-e)/w(1-e) * I
substitute for I =wh
H=(2-e)/(1-e) *h
H/h =(2-e)/(1-e)
b= 1-e
h/H = b/(1+b)
If someone can explaint the log method under the budget constraint, that would be cool.
apparetly we should have
Log U(I,L) =ylnI +(1-y) ln L
where y=b/(b+a)
which under the budget constraints
I = hw
L=H-h
L=H-(I/w)
and the maximum for Log U(I,L) gives;
h/H =y
Cheers.
| 2,012
| 6,713
|
{"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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50
|
longest
|
en
| 0.925626
|
http://www.lawnsite.com/showthread.php?s=bf7349c09982912296505dc9e3ca38d3&p=4864928
| 1,419,229,748,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-52/segments/1418802774718.30/warc/CC-MAIN-20141217075254-00018-ip-10-231-17-201.ec2.internal.warc.gz
| 626,493,367
| 13,808
|
LawnSite.com™ - Lawn Care & Landscaping Business Forum, Discuss News & Reviews Thread: MAG-1000 Balancer VERSUS the Nail in the Wall
Register FAQ Calendar Mark Forums Read
Magna-Matic Lawn Mower Blade Balancers, Sharpeners, and more
#1
09-18-2013, 05:16 PM
Magna-Matic Sponsor Join Date: Jul 2001 Location: Waldo, Wisconsin Posts: 801
MAG-1000 Balancer VERSUS the Nail in the Wall
Hello All,
We have put together a video showing a demonstration to help people understand that using a NAIL in the Wall is not a useful method to balance a lawnmower blade.
Here is the video:
http://youtu.be/x3RMN1QfutM
Here are the calculations and bulletin of the test:
Here is a link to a long lawnsite thread discussing this in length
Thank you,
__________________
Gerd Ferdinand Bauer II
Magna-Matic Corporation
Waldo WI USA
800-328-1110 (toll free USA & Canda)
920-564-2366 (PHONE)
gbauer2@magna-matic.com
www.magna-matic.com
www.magna-matic-direct.com
#2
09-18-2013, 05:20 PM
Magna-Matic Sponsor Join Date: Jul 2001 Location: Waldo, Wisconsin Posts: 801
Hello All,
Here is a part of the bulletin which defines UNBALANCE.
______________________________
Unbalance, what is it?
Only rotors in motion (bodies that are connected to a shaft) have the unique physical property called unbalance.
There are three types of unbalance known today they are as follows:
Static Unbalance is when the center of mass and the center of rotation do not coincide. (not in the same location)
Couple Unbalance is when there are two unbalance conditions that may have the same magnitude, but their direction may be offset by 180º relative to each other. This can create a wobble in a rotor.
Dynamic Unbalance is the same as couple unbalance except for the rotor may have infinite unbalances and may be distributed at random along the axis of its rotation.
How is Unbalance detected or measured?
Static unbalance detection is a measure of gravitational force.
This detection method is commonly referred to as static balancing.
Couple and Dynamic unbalance detection is a measure of centrifugal force and requires the rotor to be power rotated (spun).
This detection method is commonly referred to as dynamic balancing.
What are some examples of rotors that would use static balancing?
The method of static balancing is typically used for single plane rotors that are thin, such as a lawnmower blades, circular saw blades, abrasive cut-off wheels, narrow pulleys, flywheels, impellers, clutches, fan blades, etc.
What are some examples of rotors that would use dynamic balancing?
The method of dynamic balancing is typically used for multiple plane rotors or rotors that are very deep, such as motor armatures, drums, turbines, multiple rotors on the same shaft, complex asymmetrical parts, wide car wheels etc.
Why use static balancing for lawnmower blades?
Static unbalance detection is the most economical for servicing and manufacturing lawnmower blades and discs.
Dynamic balancing is not commonly used for servicing or manufacturing lawnmower blades because dynamic balancing
instruments are very expensive and not required for single plane rotors such as the lawnmower blade.
____________________________
The excerpt above comes from the Magna-Matic EDUCATIONAL BULLETIN SERIES "Understanding Balance, and why a nail in the wall is not a useful measure of balance"
__________________
Gerd Ferdinand Bauer II
Magna-Matic Corporation
Waldo WI USA
800-328-1110 (toll free USA & Canda)
920-564-2366 (PHONE)
gbauer2@magna-matic.com
www.magna-matic.com
www.magna-matic-direct.com
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are Off [IMG] code is On HTML code is Off Forum Rules
Forums
### Corporate
Moose River Media
Dealer Directory
| 906
| 3,871
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52
|
latest
|
en
| 0.861956
|
https://www.experts-exchange.com/questions/28535478/How-to-replace-Jan-Mar-2014-to-Q1-2014-A-Better-way.html
| 1,534,386,634,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-34/segments/1534221210408.15/warc/CC-MAIN-20180816015316-20180816035316-00414.warc.gz
| 883,097,261
| 25,757
|
# How to replace Jan-Mar 2014 to Q1 - 2014 A Better way?
I needed to convert quarter dates in cell Jan-June 2014 to Q1 - 2014
i have came up with this long formula ="Q"&INT((MONTH(MID(A\$1,FIND("-",A\$1)+1,3)&RIGHT(A\$1,4))+2)/3)&" - "&YEAR(MID(A\$1,FIND("-",A\$1)+1,3)&RIGHT(A\$1,4))
i was wondering if there is any shortcut to achieve the same result without having this long formula.
LVL 27
Asked:
###### Who is Participating?
I wear a lot of hats...
"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.
Excel VBA DeveloperCommented:
How about this:
="Q"&INT((MONTH(LEFT(A\$1,FIND("-",A\$1)-1)&RIGHT(A\$1,4))+2)/3)&" - "&RIGHT(A\$1,4)
-Glenn
0
Commented:
Not much you can do to shorten text parsing but...
MID(A\$1,FIND("-",A\$1)+1,3)&RIGHT(A\$1,4))
could probably be replaced with
MID(A\$1,FIND("-",A\$1)+1,999)
which will give 'June 2014', which is not quite the same as what you originally had but I think it will work for you.
Also I'm not sure you need the year when working with the MONTH function so you might be able to take that out except when you concatentate it at the end.
0
Author Commented:
Yes, i made a mistake in the beginning, the result should have been from Jan-Mar 2014 to Q2 - 2014
Not Q1 rspahitz formula gives me the result, but since i made a mistake Glenn formula is much shorter, however it gives me Q1
Glenn,
is there any way that you change your formula to result Q2?
0
Excel VBA DeveloperCommented:
So, is your fiscal year starting in October?
Oct - Dec ==== Q1
Jan - Mar ==== Q2
Apr - Jun ==== Q3
Jul - Sep ==== Q4
0
Author Commented:
not at all
the fiscal year starts in Jan
so my quarters are like this
Jan-Mar 2014 First Q April-June Second Q , July to Sep Third Q and Oct to Dec Fourth Q
0
Excel VBA DeveloperCommented:
If you're on a calendar year, my formula produces that result. The quarter is based on the first month shown. So, "Jan" is the first month in Q1. Why do you want it to display Q2?
The only thing I did to simplify your formula was change from MID to LEFT functions and eliminate the YEAR function at the end as it was not needed.
0
Author Commented:
again another mistake by me. today we had staff bar, so dont blame me for it :)
what i meant was
Jan-Mar 2014 to Q1 - 2014
Jan-June 2014 to Q2 - 2014
your formula results Q1 - 2014 even for Jan-June 2014
0
Excel VBA DeveloperCommented:
Well, that's because I'm looking at the first month and your original question had "Jan-Jun" returning Q1. It wasn't until this post that you referenced another quarter - and then you didn't mention June; it still showed "Jan-Mar". Hence, the confusion on your fiscal year and my follow-up question.
So, to be clear, you want the quarter for the second month listed to be returned by the formula, correct?
-Glenn
0
Author Commented:
yes yes. second month listed to be returned by formula
sorry for causing confusion Glenn
0
Excel VBA DeveloperCommented:
Okay, try this:
="Q"&INT((MONTH(MID(A1,5,3)&RIGHT(A1,4))+2)/3)&" - "&RIGHT(A1,4)
Produces the quarter that the second-listed month falls in on a calendar year.
-Glenn
0
Experts Exchange Solution brought to you by
Your issues matter to us.
Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.
Commented:
I think my original post would have helped. All you need to do is find the "-" and take everything after it:
MID(A\$1,FIND("-",A\$1)+1,999)
Take the month of that and +2 /3 like you're doing:
="Q"&INT((MONTH(MID(A\$1,FIND("-",A\$1)+1,999))+2)/3)&" - "&RIGHT(A\$1,4)
0
Author Commented:
Thank you very much both.
first one arrived raspahits
second one is also raspahits
third line is most shortest version Glenn
here are the outcome
LEN 92 ="Q"&INT((MONTH(MID(A\$1,FIND("-",A\$1)+1,999))+2)/3)&" - "&YEAR(MID(A\$1,FIND("-",A\$1)+1,999))
LEN 70 ="Q"&INT((MONTH(MID(A\$1,FIND("-",A\$1)+1,999))+2)/3)&" - "&RIGHT(A\$1,4)
LEN 64 ="Q"&INT((MONTH(MID(A1,5,3)&RIGHT(A1,4))+2)/3)&" - "&RIGHT(A1,4)
thank you veyr much.
0
Author Commented:
I've requested that this question be closed as follows:
Accepted answer: 300 points for Glenn Ray's comment #a40373879
Assisted answer: 0 points for ProfessorJimJam's comment #a40373856
Assisted answer: 200 points for rspahitz's comment #a40373962
for the following reason:
Thank you. i have now the shorter version. your quick and professional responses are much appreciated.
thank you both and have a great weekend.
0
###### It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Microsoft Excel
From novice to tech pro — start learning today.
| 1,492
| 5,324
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.78125
| 3
|
CC-MAIN-2018-34
|
latest
|
en
| 0.918295
|
http://www.gurufocus.com/term/Book%20Value%20Per%20Share/JPM/Book%252BValue%252BPer%252BShare/JPMorgan%2BChase%2B%2526%2BCo
| 1,455,259,651,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-07/segments/1454701163438.83/warc/CC-MAIN-20160205193923-00057-ip-10-236-182-209.ec2.internal.warc.gz
| 458,510,275
| 27,568
|
Switch to:
JPMorgan Chase & Co (NYSE:JPM)
Book Value Per Share
\$59.67 (As of Sep. 2015)
JPMorgan Chase & Co's book value per share for the quarter that ended in Sep. 2015 was \$59.67.
During the past 12 months, JPMorgan Chase & Co's average Book Value Per Share Growth Rate was 5.60% per year. During the past 3 years, the average Book Value Per Share Growth Rate was 7.00% per year. During the past 5 years, the average Book Value Per Share Growth Rate was 7.50% per year. During the past 10 years, the average Book Value Per Share Growth Rate was 7.00% per year. Please click Growth Rate Calculation Example (GuruFocus) to see how GuruFocus calculates Wal-Mart Stores Inc (WMT)'s revenue growth rate. You can apply the same method to get the book value growth rate using book value per share data.
During the past 13 years, the highest 3-Year average Book Value Per Share Growth Rate of JPMorgan Chase & Co was 14.80% per year. The lowest was 1.40% per year. And the median was 7.35% per year.
JPMorgan Chase & Co's current price is \$53.07. Its book value per share for the quarter that ended in Sep. 2015 was \$59.67. Hence, today's P/B Ratio of JPMorgan Chase & Co is 0.89.
During the past 13 years, the highest P/B Ratio of JPMorgan Chase & Co was 1.54. The lowest was 0.43. And the median was 1.06.
Good Sign:
JPMorgan Chase & Co stock P/B Ratio (=0.92) is close to 3-year low of 0.91
Definition
JPMorgan Chase & Co's Book Value Per Share for the fiscal year that ended in Dec. 2014 is calculated as:
Book Value Per Share = (Total Equity - Preferred Stock) / Shares Outstanding (EOP) = (232,065 - 20,063) / 3,715 = 57.07
JPMorgan Chase & Co's Book Value Per Share for the quarter that ended in Sep. 2015 is calculated as:
Book Value Per Share = (Total Equity - Preferred Stock) / Shares Outstanding (EOP) = (245,728 - 26,068) / 3,681 = 59.67
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Theoretically it is what the shareholders will receive if the company is liquidated. Total equity is a balance sheet item and equal to total assets less total liabilities of the company.
Book value may include intangible items which may come from the companys past acquisitions. Book value less intangibles is called Tangible Book.
Explanation
Usually a companys book value and Tangible Book Value per Share may not reflect its true value. The assets may be carried on the balance sheets at the original cost minus depreciation. This may underestimate the true economic values of the assets. It also may over-estimate their true economic value because the assets can become obsolete.
For financial companies such as banks and insurance companies, their assets may be reported in current market value of the assets owned. Book values of financial companies are more accurate indicator of the economic value of the company.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
JPMorgan Chase & Co Annual Data
Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Book Value Per Share 33.45 36.59 36.15 39.88 43.04 46.59 51.27 53.25 57.07 60.17
JPMorgan Chase & Co Quarterly Data
Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Book Value Per Share 52.01 53.25 54.05 55.53 56.50 57.07 57.76 58.49 59.67 60.17
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More)
| 954
| 3,605
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375
| 3
|
CC-MAIN-2016-07
|
longest
|
en
| 0.930478
|
https://www.webmasterworld.com/forum3/26417.htm
| 1,455,150,960,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-07/segments/1454701160918.28/warc/CC-MAIN-20160205193920-00215-ip-10-236-182-209.ec2.internal.warc.gz
| 893,726,192
| 17,395
|
Welcome to WebmasterWorld Guest from 54.159.50.111
Forum Moderators: open
Message Too Old, No Replies
# Toolbar PageRank Scale
#### tonyww
2:22 pm on Oct 27, 2004 (gmt 0)
#### Junior Member
joined:Jan 10, 2004
posts:56
Is it true that the gaps between Google PRs are logarithmic and it's base 8?
If that is true, does this mean that:
it takes 8 PR 1s to = a PR2
it takes 64 PR 2s to = a PR3
it takes 512 PR 3s to = a PR4
it takes 4096 PR 4s to = a PR5
it takes 32,768 PR 5s to = a PR6
Is that correct?
#### ogletree
2:54 pm on Oct 27, 2004 (gmt 0)
#### Senior Member from US
joined:Apr 14, 2003
posts:4249
It is much more complicated than that. If that were true it would only work if your were the only link on one of those pages. Number of links on a page is a factor. I'm sure there are even more factors than that.
#### diamondgrl
6:12 pm on Oct 27, 2004 (gmt 0)
#### Senior Member
joined:Jan 12, 2004
posts:961
many others claim that the formula is logarithmic and i think that's very misleading. your formulation is not correct. it is not nearly that hard to become a PR6.
the originally published formula relies is a series, with each element in the series representing the PR of the site linking to you divided by the number of outbound links on that page. then there is an iterative damping factor that is applied so that the top value will be 10 (you can't have PR 11).
i would not be surprised if the PR formula has not changed at all - although there are many things other than PR that determine your search engine ranking.
without knowing what every other link on the web is, you can't do a simple calculation as to what your PR will be based on the links to your site.
7:18 pm on Oct 27, 2004 (gmt 0)
#### New User
joined:Sept 29, 2004
posts:10
Yeah, I find it very hard to determine how PR is calc'd. My site is a PR5 site (used to be PR6), and it only has ~50 backlinks, and not all of them from high ranking sites. (Many of the links are PR0 sites.) And I'm ranking on the 2nd page of a very competitive keyword, as well.
DS
#### tonyww
8:05 pm on Oct 27, 2004 (gmt 0)
#### Junior Member
joined:Jan 10, 2004
posts:56
I probably mis-phrased the question. I wasn't asking "how does G calculate PR" - I heard there were more than 100 variables in the algorithm and I did a little bit of analysis of PR and backlinks and it is far from consistent.
What I am trying to get is a feel for how difficult it is to move from one PR to the next PR above.
If we kept the other variables more or less stable, e.g. link reputation, the average PR of the sites linking to us etc, is it harder to go from a PR4 to a PR5 than from a PR2 to a PR3. If 'yes' then why?
Also, can we say how many (for example) PR4s have the same value as a PR5?
#### tonyww
8:10 pm on Oct 27, 2004 (gmt 0)
#### Junior Member
joined:Jan 10, 2004
posts:56
darqShadow: I'm not expert at this and I don't want to teach granny to suck eggs, but maybe one of the sites linking to you has a high PR, or the links to you have very strong reputation with the right balance of keyword density etc. There is a software tool that shows you all this - I don't know if I'm allowed to mention the name of it (I'm not an affiliate btw!).
#### wellzy
8:14 pm on Oct 27, 2004 (gmt 0)
#### Preferred Member
joined:Dec 20, 2002
posts:352
That can't be a correct formula. I have a PR5 site with only 125 baclinks (400 page site)Each individual page has around 10 incoming links and points back to the index page. Icluding my internal pages I have about 500 PR4 pages pointing to my index page. I think you would have to know the exact PR of the page pointing to you. Toolbar PR is rounded off.
wellzy
#### diamondgrl
9:33 pm on Oct 27, 2004 (gmt 0)
#### Senior Member
joined:Jan 12, 2004
posts:961
I heard there were more than 100 variables in the algorithm and I did a little bit of analysis of PR and backlinks and it is far from consistent.
I doubt this is true. My hunch is the basic algorithm is the originally published one. I described it in words above, but here is the original formula.
PR(A) = (1-d) + d (PR(T1)/C(T1) + ... + PR(Tn)/C(Tn))
PR(x) is the PageRank of x, C(x) is the number of outbound links on a page x, d is a damping factor set between 0 and 1 and is controlled by Google.
Note that it is an iterative process because in order to determine the PR of the other pages that link to you, you have to have know the PR of all pages that link to it. It's iterative because it's like a chicken and egg problem as to which to calculate first.
With that said, PR and backlinks are very inconsistent because of numerous factors, including that your toolbar value is not necessarily the actual value Google uses (both because it might be out of date or an approximation) and backlinks shown are notoriously incomplete.
The 100 different factors is probably more accurately attributed not to PR but how well you rank in the SERPs, with PR being one of those 100 factors.
If we kept the other variables more or less stable, e.g. link reputation, the average PR of the sites linking to us etc, is it harder to go from a PR4 to a PR5 than from a PR2 to a PR3.
Yes. If it was just as easy to get from one level to another, don't you think all PR 5s would simply spend twice as much time and money to get to twice as high, or PR10? I sure would.
Also, can we say how many (for example) PR4s have the same value as a PR5?
Good question. Wish I knew something more definitive. But it changes over time. As we all build more and more links, it takes more and more links to get to a certain PR value since the competition has gotten better. You can simply slip in PR if you sit on your duff and don't constantly strive for more links (or naturally get people to link to you).
And note that assuming the above formula is correct, suggests that you will be a lot better of being linked from a PR5 page with 2 links on it than a PR7 page with 10 links, since PR5/2 = 2.5 whereas PR7/10 is only .7.
#### ciml
12:42 pm on Oct 29, 2004 (gmt 0)
#### Senior Member
joined:June 22, 2001
posts:3805
diamondgrl, your last paragraph shows why a log scale would be used.
tonyww:
it takes 8 PR 1s to = a PR2
it takes 64 PR 2s to = a PR3
it takes 512 PR 3s to = a PR4
it takes 4096 PR 4s to = a PR5
it takes 32,768 PR 5s to = a PR6
Not quite, it would be more like these (assuming log 8):
it takes 8 PR 1s to = a PR2
it takes 64 PR 1s to = a PR3
it takes 512 PR 1s to = a PR4
it takes 4096 PR 1s to = a PR5
it takes 32,768 PR 1s to = a PR6
or
it takes 8 PR 1s to = a PR2
it takes 8 PR 2s to = a PR3
it takes 8 PR 3s to = a PR4
it takes 8 PR 4s to = a PR5
it takes 8 PR 5s to = a PR6
Of course this is answering 'how many PRx pages have the same PR as a PRy page', not 'how many links from PRx pages give PRy'.
For the latter you need to change the numbers by some amount, and take into account that the number of links on each page affects the answer.
#### tonyww
1:43 pm on Oct 29, 2004 (gmt 0)
#### Junior Member
joined:Jan 10, 2004
posts:56
Anyone any comment on this:
[seocompany.ca...]
#### asomani
5:45 am on Oct 30, 2004 (gmt 0)
#### Junior Member
joined:Sept 22, 2002
posts:80
My clients' websites homepage got a PR0, but tthe inside pages got PR3, how is it possible.
Initially the root url was holding a PR0, but the index.php under it was a pr4, but oday the root url along with the index.php are all PR0.
If so, that can i do to get the penalty lifted?
#### anallawalla
10:52 am on Nov 1, 2004 (gmt 0)
#### Moderator from AU
joined:Mar 3, 2003
posts:3701
My clients' websites homepage got a PR0, but tthe inside pages got PR3, how is it possible.
Initially the root url was holding a PR0, but the index.php under it was a pr4, but oday the root url along with the index.php are all PR0.
Search for earlier discussions on this topic - I didn't read them because I have long had deep pages with higher PR than the home page. In most cases the home page has no link to them, but the deep pages are good enough to attract numerous direct links.
I have also had the home page drop to PR <1, i.e. white bar, but it was explained away as a technical glitch. Sure enough, the page showed normal PR some months later.
#### brixton
12:29 pm on Nov 1, 2004 (gmt 0)
#### Junior Member
joined:Oct 10, 2004
posts:50
My index PR5 page droped to a few BW links BUT my internal pages got over hundr...s of BWL's ,strange, but i love that :)
#### the_nerd
1:02 pm on Nov 1, 2004 (gmt 0)
#### Senior Member
joined:June 29, 2003
posts:790
diamondgrl
And note that assuming the above formula is correct, suggests that you will be a lot better of being linked from a PR5 page with 2 links on it than a PR7 page with 10 links, since PR5/2 = 2.5 whereas PR7/10 is only .7.
don't forget the log-scale. If we take base 8 as given then a pr 7 link is about 64 times more valuable than a PR 5 , so the link from the PR7 page a waaaayyyyy better. (more than 10 times as good)
#### diamondgrl
4:29 pm on Nov 1, 2004 (gmt 0)
#### Senior Member
joined:Jan 12, 2004
posts:961
thanx, the_nerd. i think you may be incorrect about the logarithmic base assuption, but if so, my post was probably even more inaccurate so i don't blame you. i have since i have since read a little more that helps me clarify a few points in my own mind.
actual pr is different from toolbar pr in that toolbar pr is limited from 0 to 10. actual pr has no such limit, and that is exactly what you would predict with the pr formula. so a "high" 7 as measured in toolbar terms would probably pass a heck of a lot more pr than a "low" 7 since the actual pr might be dramatically different.
however, it would also follow from this that it is not logarithmic in that the range of pr 7 values scales logarithmically over pr 6 value. otherwise, there should not be a limit of 10 since some site might, in theory anyway, qualify for 11 or 12. instead, my hunch is that the actual pr ranges that are associated with each toolbar ranking are picked so there is a logarithmically diminishing number of sites that qualify in each toolbar ranking.
does this make sense to anyone, or strike anyone as off-based?
#### the_nerd
6:55 pm on Nov 1, 2004 (gmt 0)
#### Senior Member
joined:June 29, 2003
posts:790
diamondgrl,
I guess scaling is one of the minor problems Google has, so they probably do just this:
show PR in 10 steps:
0.5 - 1.5 -> TB 1
1.5 - 2.5 -> TB 1
...
9.5 - 10.5 -> TB 10
All you have to do now is take the page with the highest actual pagerank and choose a log base so this highest PR maps to 10.5 (note that in the PR alg every single page has actual PR = 1 to spare, but every link-hop is subject to a dampening factor, otherwise everybody could create his own PR 10 site just with internal links)
Let's assume these links have an average of PR 5 and the log scale is 8. Then a PR 5 Page is worth 8^4 = 4096.
4096 * 2.6000.000 = around 10^10 (10 billion)
If we scale this with our log base of 8 we should get around 10.5 so it can be shown as a TB 10 and leave some space for "lesser" PR 10 pages.
log (8) 10^10 = 17,8. That's far too high. So either the base is too low or the assumption about those links averaging PR 5 is to high.
Let's calculate with a log base of 9:
A PR5 page then is worth 9^4 = 6561
6561 * 2.6000.000 = 17 billion
log (9) 1.7 billion = 13
still too high, but closer to 10.5
let's try 10:
A PR5 page then is worth 10^4 = 10000
10000 * 2.6000.000 = 26 billion
log (10) 2.1 billion = 10.4 (BINGO!)
That means:
- if the average incoming link to Google.com has PR of around 5
- if the number of 2.600.000 incoming links is accurate
- if TB PR is indeed scaled linear on a log basis (no bumps, or emphasising special areas)
then right now the log base used to calculate Toolbar PR could be around 10.
Does that help me a lot? It tells me that on average a PR 5 link is about 10 times as "heavy" as a PR 4 and 100 times a PR 3.
A TB Pagerank of e.g. 5 reflects anything from 4.5 to 5.5 so a link with the same TB PR and the same number of outgoing links may still be 10 times more valuable than another one that "looks" identical.
The only thing that might be remarkable about this calculation: the higher the actual pagerank of the big guys becomes, the less accurate toolbar PR HAS to become. And: it's getting more difficult to increase or even keep your rank because the steps are getting steeper.
Hmmm. Do I help anybody by posting this junk? Took too long to write it, so here you go ;)
#### neuron
11:01 pm on Nov 1, 2004 (gmt 0)
#### Full Member
joined:Jan 28, 2004
posts:316
That's an interesting way to look at it, reverse engineer the exponential scale by roughly calculating the total of the what we assume to be the highest PR page.
- if the average incoming link to Google.com has PR of around 5
- if the number of 2.600.000 incoming links is accurate
- if TB PR is indeed scaled linear on a log basis (no bumps, or emphasising special areas)
alltheweb: 27.5M
altavista: 46.8M
msn: 9.67M
yahoo: 30.1M
and I'm sure a lot more than the 2.61M are included in the calculation, like 10 times that many.
there are 11 integer values between and including the end points of 0 (zero) and 10 (ten), and only 10 units of value. Thus the rescale might be 1/11 increments in the 10 units of value:
PR0= 0.00000 to 0.90909 (0 to 1/11)
PR1= 0.90909 to 1.81818 (1/11 to 2/11)
PR2= 1.81818 to 2.72727 (2/11 to 3/11)
...
PR8= 7.27272 to 8.18181 (8/11 to 9/11)
PR9= 8.18181 to 9.09090 (9/11 to 10/11)
PR10=9.09090 to 10.0000 (10/11 to 1)
And finally, the_nerd, why would you think that the average page linking to google would be a PR5? Wouldn't a PR of 1 (one) as an average PR for all pages be more reasonable? I have quite a few pages and I bet the average PR of my pages is well below PR5.
#### steve128
12:49 am on Nov 2, 2004 (gmt 0)
#### Full Member
joined:Jan 10, 2003
posts:300
>>Anyone any comment on this:<<
Yes, if the guy owning the webite had 1 pr8 link to the page you show he would be pr7, rather than pr2 -;
g-lag accepted
#### steve128
12:51 am on Nov 2, 2004 (gmt 0)
#### Full Member
joined:Jan 10, 2003
posts:300
It seems your post was removed -;
#### cabowabo
12:58 am on Nov 2, 2004 (gmt 0)
#### Senior Member
joined:June 12, 2003
posts:723
In short, PageRank is computing using an unknown numerial value for a page and then allocating a piece of that value based on the number of outgoing links. This is what allows your site to gain more reputation ... a link from a PR6 site with 4 outgoing links would be worth more, in essence, than a link from a PR8 site with 150 outgoing links.
That is a very basic concept, but describes how the process works in a relative sense.
CaboWabo
#### the_nerd
8:20 am on Nov 2, 2004 (gmt 0)
#### Senior Member
joined:June 29, 2003
posts:790
hey neuron,
And finally, the_nerd, why would you think that the average page linking to google would be a PR5? Wouldn't a PR of 1 (one) as an average PR for all pages be more reasonable? I have quite a few pages and I bet the average PR of my pages is well below PR5.
I'm sure that (if they bother reading this) they'll laugh at us for wasting our time calculating the length of the emperor's nose ....
You're right, the PR5 just is a wild guess. It's somewhere in the middle so I took it. But we KNOW they have lots of 9s coming in. And since a 9 is worth 10^8 = 100.000.000 1s I'm sure all your pages (and most of mine) are being accounted for ;)
Ah, yes: Google normally only shows the higher PR-Links with "link:......" So the other 30 million or so links can be offset with a couple of 8s.
I left out the "0" because I don't think it is a real value - just "not yet" or "no more".
Have a great week,
nerd.
#### phantombookman
9:04 am on Nov 2, 2004 (gmt 0)
#### Preferred Member
joined:Dec 30, 2003
posts:625
There is, in my experience, an important and basic factor.
The PR on a page is one less than the page that gave it the PR.
I got a new site of mine PR4 with one link from one of my own sites which was PR5.
I wonder if there is a level at which the algo kicks in and becomes more important?
The reason I ask is because the numbers of links some people are talking about per site massively exceed the total number of links I have over 15 sites. Granted I cannot get further than PR5 but all sites are #1 or top 5 for the KW searches
In my area backlinks are difficult and there is little SEO so basic trends are easy to identify.
I did see a PR6 site with only 2 incoming links a DMOZ (pr4) backlink but one link from a PR7 site.
#### Macro
12:10 pm on Nov 2, 2004 (gmt 0)
#### Senior Member
joined:July 31, 2003
posts:2280
the_nerd, excellent post! (msg 15)
>> The only thing that might be remarkable about this calculation: the higher the actual pagerank of the big guys becomes, the less accurate toolbar PR HAS to become. And: it's getting more difficult to increase or even keep your rank because the steps are getting steeper.
There have been convincing arguments here before for a log of 5 (with accompanying calculations)... but I can't seem to find them.
one small flaw:
I believe it is no longer the case that it's only the higher PR links. In fact, I suspect, that it shows randomly picked out lower PR links when using "link:".
#### the_nerd
5:26 pm on Nov 2, 2004 (gmt 0)
#### Senior Member
joined:June 29, 2003
posts:790
Thanks, Macro!
I we want to learn more about the scale, we could try this: get a random sample of maybe 1000 links pointing to google.com and look at the Toolbar PR to get an idea about the distribution. Maybe somebody did this before and wants to share her findings?
I wouldn't bet the farm on a base of 10, but I don't think it will be much lower than 9. Think of all the macromedia links, for example. They push up the average a lot.
| 5,118
| 17,908
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.828125
| 3
|
CC-MAIN-2016-07
|
latest
|
en
| 0.963456
|
https://www.lesswrong.com/posts/3PXBK2an9dcRoNoid/on-having-no-clue
| 1,708,859,544,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00343.warc.gz
| 871,814,309
| 44,892
|
# 20
Let's suppose you're trying to figure out if something is going to be A or B, but you have no clue.
What probability should you assign to each option?
Some people might say that you should assign 50/50. After all, the argument goes, if you assigned 40/60 or 60/40, well it sure sounds like you're favoring one option over another.
That's a very persuasive argument, but unfortunately, it's more complicated than that.
It may be the case that A splits into two options A1 and A2 where again we have no clue whether A1 is more likely than A2 or how these compare to B. In which case, the same argument would suggest that we should go with 33/33/33 (rounding down).
This seems to be a contradiction. What should we make of this?
First of all, I think we should accept that this exact process of reasoning leads to a contradiction. There's nothing fancy going on here. No dubious steps that could give us an out.
We tried to say that if we had no clue that the logical thing to do is to assign equal probability to each option, we forgot that we were implicitly assuming that we should favor our particular way of carving up probability space.
In other words, we did need a clue after all. So what kind of clue is this exactly?
Well, surely in order to be justified in carving up the space a particular way, we'd need to have a reason to believe that each possibility is equally likely a priori.
Sadly, this is circular. We wanted to defend equal probabilities by asserting that our way of carving up the space was reasonable, but then we tried to assert that it was reasonable by claiming that each possibility had equal probability.
The problem is that we are making an assumption, but rather than owning it, we're trying to deny that we're making any assumption at all, ie. "I'm not assuming a priori A and B have equal probability based on my subjective judgement, I'm using the principle of indifference". Roll to disbelieve.
# 20
New Comment
Complexity heuristics possibly work to some extent here. If A is a hypothesis of comparable complexity to B, then you at least have some basis for assigning them comparable prior probabilities.
If A can be conveniently split into A1 and A2, then the child hypotheses are often more complex than A in that they each require some additional property on top of A. This isn't always the case, but in the exceptional cases it's likely more useful to consider one or both of the child hypotheses to be "primary" instead of A.
This is all very rough, since complexity is usually only defined up to some constant and is relative to some sort of specification model, but it's something that isn't necessarily contingent on observations of the world. If you're an entity that is considering probabilities and hypotheses at all, the concept of measuring complexity is likely already accessible to you.
This is a good point. I neglected to address this possibility.
If you have two options, A and B, 50% odds is maximal ignorance; you aren't saying they have equivalent odds of being true, you're saying you have no information by which to make an inference which is true.
If you then say we can split A into A1 and A2, you have added information to the problem. Like the Monty Hall problem, information can change the odds in unexpected ways!
There's no contradiction here - you have more information than when you originally assigned odds of 50/50. And the information you have added should, in real situations, inform how to distribute the odds. If A1 and A2 are sufficiently distinct (independent), it is possible that a 33/33/33 split is appropriate; if they aren't, it is possible that a 25/25/50 split is appropriate. In order to make a judgment, we'd have to know more about what exactly A1 and A2 are, and why they can be considered a "split" of A.
Consider, for example, the case of a coin being flipped - we don't know if the coin is fair or not. Let us say A is that the coin comes up "heads" and B is that the coin comes up "tails". The split, then, could reflect a second flip, after the first flip is decided, if and only if it is heads; A1 might be "heads-heads", A2 might be "heads-tails". Then a 25/25/50 split makes sense; A1 and A2 are not independent.
If, on the other hand, we have discovered that it isn't a coin at all, but a three-sided die, two faces of which have the same symbol, and one face of which has another symbol; we label the faces with the similar symbol A1 and A2, and the face with the same edit: other symbol B. We still don't know whether or not the die is fair - maybe it is weighed - but the position of maximal ignorance is 33/33/33, because even if it -is- weighted, we don't know which face it is weighted in favor of; A1 and A2 are independent.
So - what are A1 and A2, and how independent are they? We have equations that can work this out with sample data, and your prior probability should reflect your expectation of their independence. If you insist on maximal ignorance about independence - then you can assume independence. Most things are independent; it is only the way the problem is constructed that leads us to confusion here, because it seems to suggest that they are not independent (consider that we can simply rename the set of conclusions to "A, B, and C" - all the names you have utilized are merely labels, after all, and in effect, what you have actually done is to introduce C, with an implication that A and C should maybe be considered partially dependent variables). If you insist on maximal ignorance about that, as well, then you can, I suppose, assume 50% independence, which would be something like splitting the difference between the die and the coin. And there's an argument to be made there, in that you have, in fact, implied that they should maybe be considered partially dependent variables - but this comes down to trying to interpret what you have said, rather than trying to understand the nature of probability itself.
“If you then say we can split A into A1 and A2, you have added information to the problem. Like the Monty Hall problem, information can change the odds in unexpected ways!” - It’s not clear which is the baseline.
The point there is that there is no contradiction because the informational content is different. "Which is the baseline" is up to the person writing the problem to answer. You've asserted that the baseline is A vs B; then you've added information that A is actually A1 and A2.
The issue here is entirely semantic ambiguity.
Observe what happens when we remove the semantic ambiguity:
You've been observing a looping computer program for a while, and have determined that it shows three videos. The first video portrays a coin showing tails. The second video portrays two coins; the left coin shows heads, the right coin shows tails. The third video also portrays two coins; the left coin shows heads, the right coin shows heads.
You haven't been paying attention to the frequency, but now, having determined there are three videos you can see, you want to figure out how frequently each video shows up. What are your prior odds for each video?
33/33/33 seems reasonable. I've specified that you're watching videos; the event is which video you are watching, not the events that unfold within the video.
Now, consider an alternative framing: You are watching somebody as they repeat a series of events. You have determined the events unfold in three distinct ways; all three begin the same way, with a coin being flipped. If the coin shows heads, it is flipped again. If the coin shows tails, it is not. What are your prior odds for each sequence of events?
25/25/50 seems reasonable.
Now, consider yet another framing: You are shown something on a looping computer screen. You have determined the visuals unfold in three distinct ways; all three begin the same way, with a coin being flipped. If the coin shows heads, it is flipped again. If the coin shows tails, it is not. What are your prior odds here?
Both 25/25/50 and 33/33/33 are reasonable. Why? Because it is unclear whether or not you are watching a simulation of coin flips, or something like prerecorded videos; it is unclear whether or not you should treat the events within what you are watching as events, or whether you should treat the visuals themselves you are watching as the event.
Because it is unclear, I'd lean towards treating the visuals you are watching as the event - that is, assume independence. However, it would be perfectly fair to treat the coin tosses as events also. Or you could split the difference. Prior probabilities are just your best guess given the information you have available - and given that I don't have access to all the information you have available, both options are fair.
Now, the semantic ambiguity you have introduced, in the context of this, is like this:
You're told you are going to watch a computer program run, and what you see will begin with a coin being flipped, showing heads or tails. What are your probabilities that it will show heads or tails?
Okay, 50/50. Now, if you see the coin shows heads, you will see that it is flipped again; we now have three possibilities, HT, HH, and TT. What are your probabilities for each event?
Notice: You didn't specify enough to know what the relevant events we're assigning probabilities to even are! We're in the third scenario; we don't know if it's a video, in which case the relevant event is "Which video we are watching", or if it is a simulation, in which case the relevant event is "The outcome of each coin toss." Either answer works, or you can split the difference, because at this point a large part of the probability-space is devoted, not to the events unfolding, but towards the ambiguity in what events we're even evaluating.
Agreed, great post. But I think you are trying to push Bayesian Statistics past what it SHOULD be used for.
Bayesian Statistics are only useful because we approach the correct answer as we gain all the information possible. Only in this limit (of infinite information) is Bayesian useful. Priors based off no information are, well, useless.
Scenario 1: You flip a fair coin and have a 50/50 chance of it landing heads
Scenario 2: (to steal xepo's example) are bloxors greeblic? You have NO IDEA, so your priors are 50/50
Even though in both scenarios the chances are 50/50, I would feel much more confident betting money on scenario 1 than scenario 2. Therefore my model of choices contains something MORE than probabilities. As far as I know Bayesian statistics just doesn't convey this NEEDED information. You cant use Bayesian probabilities here in a useful way. It's the wrong tool for the job.
Even frequentest statistics is useless here.
A lot of day-to-day decisions are based off very limited information. I am not able to lay out a TRUE model of how we intuitively make those decisions but "how much information I have to work with" is definitely an aspect in my mental model that is not entirely captured by Bayes Theorem.
So the one we know the most about gets the heavier cumulative weight, because it has more sub-classes in our reasoning, because we know the most about it.
I suspect that without a solid base case, this process of over-weighting familiar options could be weaponized to form an argument against pursuing novel ideas about familiar topics. If I have "no clue" about, say, how a new model of the universe would compare to the old models, I could split "possibilities from old models" into so many fragments that the probability of the new model, weighted equally to all the tiny slivers of the familiar old model, approaches 0.
It seems like preventing this may require that the things being compared seem of similar size, but if you can guess the size of something, you're no longer entirely clueless about it.
It seems like the problem might be in the assumption of having "no clue". I think someone truly clueless about a question would be unable to divide it into parts in order to compare the parts' probabilities. I imagine being asked an advanced question about the grammar or meaning of a passage in a language that I can neither speak nor read, and I would be unable to even formulate a question to assign probabilities to.
I claim the problem is that our model is insufficient to capture our true beliefs.
There’s a difference in how we act between a coin flip (true 50/50) and “are bloxors greeblic?” (a question we have no info about).
For example, if our friend came and said “Yes, i know this one, the answer is (heads|yes)”. For coin flip you’d say “are you out of your mind?” and for bloxors you’d say “Ok, sure, you know better than me”
I’ve been idly pondering over this since Scott Alexander’s post. What is a better model?
One option would be to have another percentage — a meta-percentage. e.g. “What credence do i give to “this is an accurate model of the world””? For coin flips, you’re 99.999% that 50% is a good model. For bloxors, you’re ~0% that 50% is a good model.
I don’t love it, but it’s better than presuming anything on the base level, i think.
One option would be to have another percentage — a meta-percentage. e.g. “What credence do i give to “this is an accurate model of the world””? For coin flips, you’re 99.999% that 50% is a good model. For bloxors, you’re ~0% that 50% is a good model.
This is a model that I always tend to fall back on but I can never find a name for it so find it hard to look into. I have always figured I am misunderstanding Bayesian statistics and somehow credence is all factored in somehow. That doesn't really seem like the case though.
Does the Scott Alexander post lay this out? I am having difficulty finding it.
The closest term I have been able to find is Kelly constants, which is a measure of how much "wealth" you should rationally put into a probabilistic outcome. Replace "wealth" with credence and maybe it could be useful for decisions but even this misses the point!
Does the Scott Alexander post lay this out? I am having difficulty finding it.
He doesn’t really. Here’s the original article:
https://www.astralcodexten.com/p/mr-tries-the-safe-uncertainty-fallacy
Also there was a long follow-up where he insists 50% is the right answer, but it’s subscriber-only:
https://www.astralcodexten.com/p/but-seriously-are-bloxors-greeblic
What is a better model?
It's possible to do such modelling with beta-distributions (actually similar to meta-probabilities).
Combination of (something like non-informative prior) and (information obtained from friend) will be - moved from equal probabilities far more than combination .
| 3,338
| 14,694
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.977802
|
https://electronics2electrical.com/15496/
| 1,563,625,830,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-30/segments/1563195526508.29/warc/CC-MAIN-20190720111631-20190720133631-00171.warc.gz
| 378,411,052
| 24,856
|
# Compare star and Delta Connection.
272 views
Compare star and Delta Connection.
by
Sr.No. Star connection Delta connection 1. Star connection have a neutral or star point. Delta connection have no star or delta point. 2. Star connection do not form a close loop. Delta connection forms a close loop. 3. One end of each three loads connected together and forms a neutral and remaining three ends are taken out connection. One terminal of load is connected to another terminal of load and forms a closed loop and common ends are taken out for connection. 4. Line voltage (VL) = √3 Phase voltage (Vph) Line voltage (VL) = Phase voltage (Vph) 5. Line current (IL) = Phase current (IPH) Line current (IL) = √3 Phase current (IPH) 6. Star connection can be use as 3 phase 3 wire system or 3 phase 4 wire system. Delta connection can only be used as 3 phase 3 wire system. 7. Star connection is generally used in power transmission. Delta connection is generally used in power distribution. 8. Low insulation requires as phase voltage is low. Heavy insulation is required as line voltage is equal to phase voltage. 9.
by
Compare star and Delta Connection.
Parameter Star Connection Delta Connection Basic definition One terminal of each of the three branches are connected together to form a common point. Such a connection is known as Star Connection The three branches of the network are connected in such a way that it forms a closed loop. Such a connection is known as Delta Connection Connection of terminals The similar ends of the three coils are connected together to form a common point. The end of each coil is connected to the starting point of the other coil that means the opposite terminals of the coils are connected together to form a closed loop. Neutral point Neutral or the star point exists in the star connection. Neutral point does not exist in the delta connection. Relation between line and phase current Line current is equal to the Phase current. Line current is equal to √3 times the Phase Current. Relation between line and phase voltage Line voltage is equal to √3 times the Phase Voltage Line voltage is equal to the Phase voltage. Diagram
## Related questions
What is difference between Star and Delta Connection?
Write the relation between phase and line current, phase and line voltage for Star connection.
If an unsymmetrical line to ground fault occurs at the secondary terminals of a delta/star, ungrounded transformer, then (A) zero sequence currents are present on both sides of the transformer (B) zero sequence currents are absent on both sides of the transformer (C) zero sequence currents are present on the secondary side but not on the primary side (D) zero sequence currents are present on the primary side but not on the secondary side
A balanced Delta connected load supplied with 440 V 50 Hz three phase AC supply has R=10 Ohm and L=0.6 mH in is each arm. Calculate line and phase current, active power.
Draw the Delta connected three phase system showing all the voltages and currents. Write the power equation.
Draw the phasor diagram for Star connected balanced system.
Compare CC and PC of wattmeter on the basis of:- (i) Connection (ii) Status (iii) Number of turns (iv) Gauge of wire.
A star circuit has element of resistance R/2. The equivalent delta elements will be?
What is line voltage and phase voltage? Draw balanced three phase star connected system. Show all the voltages and currents.
Wye and delta conversion formula
Define service connection and state types connection and state types of overhead service connection
State the principle of solar cell. What is necessity of series and parallel connection of solar cells.
State necessity of series connection and parallel connection of batteries.
Compare industrial electrical installation and residential electrical installation.
Classify the solar collectors and compare them.
Compare primary and secondary instruments.
Compare intrinsic and extrinsic semiconductor
Compare RC and LC filter.
Compare copper and aluminium.
Compare electric and magnetic circuit.
Compare BJT, MOSFET and IGBT .
Three identical resistors connected in star consume 5000 W. If the resistors are connected in delta across the same supply, the power consumed will be
A 3-phase voltage source inverter operates in 180° conduction mode with a star connected resistive load. If input dc voltage is 100 V, then the peak to peak output line voltage is :
Wye Delta Transformation
The line current in a balanced delta system is 30 A. The load current is (A) 30√3 A (B) 30/√3 A (C) 10 A (D) 30/√2 A
Explain Three Phase Bidirectional Delta Connected Controller.
State functions of ; i) Cable box ii) Guard wire iii) Shackle insulator iv) Stay wire used in service connection
Why underground service connections are service connections are costly than overhead service connection
number of windings of a coil in a stator of ac motor for ey slot ac motor
Which of the following connection of transformers will give highest value of secondary voltage ? (A) Delta primary delta secondary (B) Delta primary star secondary (C) Star primary star secondary (D) Star primary delta secondary
Which of the following 3-phase connection of a transformer causes interference with the neighbouring communication lines: (A) Delta-Star (B) Star-Delta (C) Delta-Delta (D) Star-Star
What is a three phase connection?
How does the resistance of a conductor compare to that of an insulator?
Compare base load plant with peak load plant on any two points.
Compare conventional energy sources with renewable energy sources on any four points.
Compare PMMC & MI instrument.
Compare JFET with MOSFET.
Compare analog & digital multimeter.
Compare BJT with FET
Compare between CB , CE & CC transistor configuration.
Compare statically induced emf with dynamically induced emf on any four points.
for Delta-Star connection of 3-phase transformer: i) Draw the connection diagram ii) List any two advantages of this connection iii) State its area of application.
What is the equivalent resistance of one limb A when delta connection is transformed in to star? (A) R1R3/R1+R2+R3 (B) R2R3/R1+R2+R3 (C) R1R2R3/R1+R2+R3 (D) R1+R2+R3
For a balanced three phase star connected load for which line voltage is 230V and per phase resistance and reactance is 6 Ohm and 8 Ohm respectively. Calculate: 1) phase voltage 2) line current 3) power factor 4) total power absorbed
Draw and explain star delta starter for 3 phase induction motor
Describe the power and control circuit for automatic Star-Delta starter for 3-phase induction motor.
| 1,438
| 6,635
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.359375
| 3
|
CC-MAIN-2019-30
|
longest
|
en
| 0.934175
|
http://mathhelpforum.com/differential-geometry/174878-riesz-s-lemma.html
| 1,529,467,286,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00133.warc.gz
| 194,012,560
| 10,613
|
1. ## Riesz's lemma
With Riesz's lemma, I know that if i let X be a normed space. and Y and Z be subspaces of a normed space X then if Y is closed proper subset of Z then for every real number theta on interval (0,1) there exist a ||z||=1 s.t.
||z-y||>= theta
My question:
When Y is finite dimensional then theta can be (0,1]
i am having trouble proofing this.
I believe that i want to show that since Y is bounded and closed it is compact. Therefore
their exists value v in Z-Y s.t.
a=inf||v-y||
since Y is compact there exists y_o in Y such that
a=||v-y_o||
and if this is true i could easily finish the proof from there.
Basically my two questions are:
How do i Show that Y is bounded. Is it automatically implied because it is finite dimensional and closed?
If Y is compact can i assume that there exists a y_0 in Y s.t. a=||v-y_o||?
2. Originally Posted by macrone
With Riesz's lemma, I know that if i let X be a normed space. and Y and Z be subspaces of a normed space X then if Y is closed proper subset of Z then for every real number theta on interval (0,1) there exist a ||z||=1 s.t.
||z-y||>= theta
My question:
When Y is finite dimensional then theta can be (0,1]
i am having trouble proofing this.
I believe that i want to show that since Y is bounded and closed it is compact. Therefore
their exists value v in Z-Y s.t.
a=inf||v-y||
since Y is compact there exists y_o in Y such that
a=||v-y_o||
and if this is true i could easily finish the proof from there.
Basically my two questions are:
How do i Show that Y is bounded. Is it automatically implied because it is finite dimensional and closed?
If Y is compact can i assume that there exists a y_0 in Y s.t. a=||v-y_o||?
A subspace of a vector space can never be compact because it is not bounded. What is true is that if the subspace is finite-dimensional then its closed unit ball is compact. More generally, any closed ball in a finite-dimensional space is compact.
The proof of Riesz's lemma shows that you can take $\displaystyle \theta = 1$ provided that the distance $\displaystyle \inf_{y\in Y}\|v-y\|$ is attained. It easily follows from the triangle inequality that the smallest values of $\displaystyle \|v-y\|$ must occur inside the closed ball $\displaystyle \|y\|\leqslant2$. The function $\displaystyle y\mapsto\|v-y\|$ is continuous, and will therefore attain its infimum on any compact set.
Put those facts together and you have your proof.
,
,
,
,
,
,
,
,
# prove Riesz lemma
Click on a term to search for related topics.
| 661
| 2,528
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625
| 4
|
CC-MAIN-2018-26
|
latest
|
en
| 0.932461
|
https://physics.stackexchange.com/questions/460441/different-schwinger-dyson-equations
| 1,713,094,705,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00156.warc.gz
| 427,640,634
| 40,782
|
# Different Schwinger-Dyson Equations
In the literature on QFT there are a lot of different equations that are all called "Schwinger-Dyson equation" so I wanted to know how are they related and if they have proper names.
1. The first equation can be obtained by making a change of variables $$$$\phi \rightarrow \phi + \epsilon$$$$ where $$\epsilon$$ is an arbitrary variation of the fields. Then you write down the generating functional $$Z[J]$$ and use the fact that it should be invariant under changes of variables to get $$$$Z[J]\rightarrow\int\mathcal{D}\phi e^{iS[\phi+\epsilon]+\int J(\phi+\epsilon)}=Z[J].$$$$ Expanding in powers of $$\epsilon$$ you get the following equation $$$$\frac{\delta S}{\delta \phi(x) }[\frac{1}{i}\frac{\delta }{\delta J(x)}]Z[J]+J(x)Z[J]=0.\tag{1}$$$$
2. The second one is pretty similar but instead of seeing how the generating functional changes, you see how does the $$n$$-point correlation function change. After expanding in powers of $$\epsilon$$ you get $$$$\big(\square_x +m^2 \big)\langle \phi(x)\phi(y_1)\dots \phi(y_n)\rangle=\langle\mathcal{L}'_{int}[\phi(x)]\phi(y_1)\dots\phi(y_n)\rangle-i\sum_i\delta(x-y_i)\langle \phi(y_1)\dots\phi(y_{i-1})\phi(y_{i+1})\dots \phi(y_n)\rangle.\tag{2}$$$$ It is my understanding that the first equation is the "generating" equation for the $$n$$ equations on this second case. If you take $$n$$ derivatives $$\frac{\delta}{\delta J(y_i)}$$ on the first one you get the second one. However, I'm not sure about this so I'd like some insight.
3. The third one is a bit different. Instead of making an arbitrary change of variables we do a symmetry transformation $$$$\phi \rightarrow \phi + A\phi \epsilon$$$$ that keeps the action unchanged. If we treat the transformation parameter $$\epsilon$$ as a function of spacetime we get $$$$\partial_{\mu}\langle j^{\mu}(x)\phi(y_1)\dots \phi(y_n)\rangle-i\sum_i\delta(x-y_i)\langle A \phi(y_1)\dots \phi(y_n)\rangle.\tag{3}$$$$ where $$j^{\mu}$$ is the conserved Noether current associated with the symmetry we are using. This equation is usually known as the Ward-Takahashi identity but many books call it "the Schwinger-Dyson equation for global symmetries" which is a big point of confusion.
So the question is: What is the name for each equation? How are they related? Are there more equations related to this approach? What is the intuition behind these different equations?
• Comment to the post (v4): The term $\frac{\delta S}{\delta \phi(x)}$ in eq. (1) should be inside the path integral $Z[J]$ in order to make sense. May 15, 2019 at 21:38
All OP's 3 versions of the Schwinger-Dyson (SD) equations are consequences of the following SD equation
$$\langle \delta_{\epsilon}F[\phi]\rangle + \frac{i}{\hbar} \langle F[\phi]\delta_{\epsilon}S[\phi]\rangle~=~0.\tag{A}$$
Here it is implicitly assumed that the path integral measure is invariant under the infinitesimal transformation $$\delta_{\epsilon}\phi^{\alpha}(x).\tag{B}$$
In OP's eqs. (1) & (2) the transformation (B) is a just shift. Eq. (3) is explained on the Wikipedia for the Ward-Takahashi identity.
• Notes for later: In a 0+0D matrix model with action $S(M)={\rm tr}V(M)$ then (i) an integration by parts $\quad\int\!dM\frac{\partial}{\partial M^i{}_j}\left[G(M)^i{}_j{\rm tr}F(M)\exp\left(-\frac{1}{\hbar}{\rm tr}V(M)\right)\right]=0$ ignoring boundary contributions, or (ii) an infinitesimal change $\delta M^i{}_j=G(M)^i{}_j$ of the integration variable $M^i{}_j$, both lead to the SD loop equation $\quad\langle\frac{\partial G(M)^i{}_j}{\partial M^i{}_j}{\rm tr}F(M)\rangle+\langle {\rm tr}[G(M)F^{\prime}(M)]\rangle=\frac{1}{\hbar}\langle{\rm tr}F(M){\rm tr}[G(M)V^{\prime}(M)]\rangle$. Feb 8 at 8:54
• If $G(M)=\frac{1}{z-M}$ with $|z|\gg 1$ is the resolvent then $\quad\frac{\partial G(M)^i{}_j}{\partial M^i{}_j}=\left({\rm tr}\frac{1}{z-M}\right)^2$. If $G(M)=e^{zM}$ then $\quad\frac{\partial G(M)^i{}_j}{\partial M^i{}_j}=z\int_0^1\!d\alpha~{\rm tr}e^{\alpha zM}{\rm tr}e^{(1-\alpha) zM}$. Feb 8 at 9:42
• Next define the $\beta$-matrix model with integration measure $\quad dM=\Delta(\lambda)^{\beta}\left[\prod_kd\lambda_k\right]$, where $\Delta(\lambda)=\prod_{i<j}(\lambda_j-\lambda_i)$ is the Vandermonde matrix, cf. e.g. arXiv:1510.04430. Feb 8 at 12:53
• The SD loop equation for the $\beta$-matrix model reads $\quad\left(1-\frac{\beta}{2}\right)\langle{\rm tr}\frac{1}{(z-M)^2}{\rm tr}F(M)\rangle+\frac{\beta}{2}\langle\left({\rm tr}\frac{1}{z-M}\right)^2{\rm tr}F(M)\rangle+\langle {\rm tr}[\frac{1}{z-M}F^{\prime}(M)]\rangle$ $=\frac{1}{\hbar}\langle{\rm tr}F(M){\rm tr}[\frac{1}{z-M}V^{\prime}(M)]\rangle=\frac{1}{\hbar}\langle{\rm tr}F(M){\rm tr}\frac{1}{z-M}\rangle V^{\prime}(z)-\frac{1}{\hbar}\langle{\rm tr}F(M){\rm tr}[\frac{V^{\prime}(z)-V^{\prime}(M)}{z-M}]\rangle.\tag{4.1.20}$ Feb 8 at 17:51
• Sketched proof: $\quad\sum_kG(\lambda_k)\frac{\partial\ln\Delta(\lambda)}{\partial\lambda_k}=\sum_{i<j}\frac{G(\lambda_j)-G(\lambda_i)}{\lambda_j-\lambda_i}$. If $G(\lambda_k)=\frac{1}{z-\lambda_k}$ then $\quad 2\sum_kG(\lambda_k)\frac{\partial\ln\Delta(\lambda)}{\partial\lambda_k}=\sum_{i\neq j}\frac{1}{z-\lambda_j}\frac{1}{z-\lambda_i}$ $=\left({\rm tr}\frac{1}{z-M}\right)^2-{\rm tr}\frac{1}{(z-M)^2}$ $=\left({\rm tr}\frac{1}{z-M}\right)^2+\frac{\partial}{\partial z}{\rm tr}\frac{1}{z-M}$. $\Box$ Feb 8 at 18:35
| 1,789
| 5,356
|
{"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": 18, "wp-katex-eq": 0, "align": 0, "equation": 6, "x-ck12": 0, "texerror": 0}
| 3.328125
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.813825
|
https://math.answers.com/Q/What_is_12x2-14x_plus_4_equals_0
| 1,701,787,762,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100551.2/warc/CC-MAIN-20231205140836-20231205170836-00189.warc.gz
| 441,334,289
| 44,614
|
0
# What is 12x2-14x plus 4 equals 0?
Updated: 10/24/2022
Wiki User
11y ago
That is a quadratic equation. You can solve it easily if you factor it to 2(2x - 1)(3x - 2) and find what makes each factor 0. The solutions are 1/2 and 2/3; you can see the steps taken to get here using the link under "related links".
Wiki User
11y ago
| 113
| 336
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.4375
| 3
|
CC-MAIN-2023-50
|
latest
|
en
| 0.885338
|
https://slideplayer.com/slide/1559355/
| 1,532,098,845,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676591683.78/warc/CC-MAIN-20180720135213-20180720155213-00412.warc.gz
| 749,706,368
| 32,807
|
# Chapter 12 SUPERVISED LEARNING Rule Algorithms and their Hybrids Part 2 Cios / Pedrycz / Swiniarski / Kurgan.
## Presentation on theme: "Chapter 12 SUPERVISED LEARNING Rule Algorithms and their Hybrids Part 2 Cios / Pedrycz / Swiniarski / Kurgan."— Presentation transcript:
Chapter 12 SUPERVISED LEARNING Rule Algorithms and their Hybrids Part 2 Cios / Pedrycz / Swiniarski / Kurgan
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Rule Algorithms Rule algorithms are also referred to as rule learners. Rule induction/generation is distinct from generation of decision trees. In general, it is more complex to generate rules directly from data than to write a set of rules from a decision tree.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Rule Algorithms AlgorithmComplexity ID3O(n) C4.5 rulesO(n 3 ) C5.0O(n log n) DataSqeezerO(n log n) CN2O(n 2 ) CLIP4O(n 2 )
© 2007 Cios / Pedrycz / Swiniarski / Kurgan DataSqueezer Algorithm Let us denote training dataset by D, consisting of s examples and k attributes. The subsets of positive examples, DP, and negative examples, DN, satisfy these properties: DP DN = D, DP DN =,DN, and DP
© 2007 Cios / Pedrycz / Swiniarski / Kurgan DataSqueezer Algorithm The matrix of positive examples is denoted as POS and their number as NPOS; similarly NEG denotes matrix of negative examples and their number is NNEG. The POS and NEG matrices are formed by using all positive and negative examples, where examples are represented by rows, and features/attributes by columns.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan DataSqueezer Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan DataSqueezer Algorithm Given:POS, NEG, k (number of attributes), s (number of examples) Step1. 1.1G POS = DataReduction(POS, k); 1.2G NEG = DataReduction(NEG k); Step2. 2.1Initialize RULES = []; i=1;// where rules i denotes i th rule stored in RULES 2.2create LIST = list of all columns in G POS 2.3within every G POS column that is on LIST, for every non missing value a from selected column j compute sum, s aj, of values of gpos i [k+1] for every row i, in which a appears and multiply s aj, by the number of values the attribute j has 2.4select maximal s aj, remove j from LIST, add j = a selector to rules i 2.5.1if rules i does not describe any rows in G NEG 2.5.2 then remove all rows described by rules i from G POS, i=i+1; 2.5.3 if G POS is not empty go to 2.2, else terminate 2.5.4 else go to 2.3 Output:RULES describing POS DataReduction (D, k)// data reduction procedure for D=POS or D=NEG DR.1Initialize G = []; i=1; tmp = d 1 ; g 1 = d 1 ; g 1 [k+1]=1; DR.2.1for j=1 to N D // for positive/negative data; N D is N POS or N NEG DR.2.2 for kk = 1 to k// for all attributes DR.2.3 if (d j [kk] tmp[kk] or d j [kk] = ) DR.2.4 then tmp[kk] = ;// denotes missing do not care value DR.2.5 if (number of non missing values in tmp 2) DR.2.6 then g i = tmp; g i [k+1]++; DR.2.7 else i++; g i = d j ; g i [k+1]=1; tmp = d j ; DR.2.8return G;
Summed-up values F1F2F3F4Class adio aeip afjp afko bgmq Feature Total number of values Summed-up values F12 values {a, b}v 11 =4x2, v 41 =1x2 F24 values {d, e, f, g}v 12 =1x4, v 22 =1x4,v 42 =2x4,v 52 =1x4 F34 values {i, j, k, m}v 13 =2x4, v 23 =1x4,v 43 =1x4,v 53 =1x4 F43 values {o, p, q}v 14 =2x3, v 24 =2x3, v 44 =1x3 F1, F2, and F3 have the same maximal summed-up values for the following values of features: a for F1, f for F2, and i for F3: v 11 = v 42 = v 13 = 8 Threshold (pruning) on the summed-up values is used to control selection of feature selectors, which are used in the process of rule-generation.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan
DataSqueezer Algorithm As result of the above operations the following two rules are generated that cover all 5 POS training examples: IF TypeofCall = Local AND LangFluency = Fluent THEN Buy IF Age = Very oldTHEN Buy or IF F1=1 AND F2=1THEN F5=1(covers 3 examples) IF F4=5 THEN F5=1 (covers 2 examples) Or, in fact: R1: F1=1, F2=1 R2: F4=5
© 2007 Cios / Pedrycz / Swiniarski / Kurgan DataSqueezer Algorithm Pruning Threshold is used to prune very specific rules. The rule generation process is terminated if the first selector added to rulei has summed-up value, saj, equal to or smaller than the thresholds value. Generalization Threshold is used to allow for rules that cover a small number of negative data. It allows for accepting rules that cover some negative examples: number <= than this threshold.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan DataSqueezer Algorithm DataSqueezer generates a set of rules for each class. Only two outcomes are possible: a test example is assigned to a particular class, or it is left unclassified. To resolve possible conflicts: all rules that cover a given example are found. If no rules cover it then it is left unclassified for every class, the goodness of rules, describing this class, and covering the example is summed; the example is assigned to the class with the highest value. In case of a tie the example is left unclassified. The goodness value for each rule is equal to the percentage (or number) of the POS examples that it covers.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan DataSqueezer Algorithm All unclassified examples are treated as incorrect classifications. Because of this the algorithms classification accuracy is lower. This is in contrast to C5.0 and many other algorithms that use default hypothesis, which states that if an example is not covered by any rule it is assigned to the class with the highest frequency (the default class) in the training data. This means that each example is always classified; this mechanism may lead to significant but artificial improvement in terms of accuracy of the model. For highly skewed / unbalanced data (where one of the classes has significantly larger number of training examples) it leads to generation of the default hypothesis as the only rule.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan DataSqueezer Algorithm # abbr. setsize #clas s #attrib. test data #abbr.setsize #clas s #attrib. test data 1adultAdult488422141628112ledLED display60001074000 2bcw Wisconsin breast cancer 6992910CV13pidPIMA indian diabetes7682810CV 3bldBUPA liver disorder3452610CV14satStatLog satellite image64356372000 4bosBoston housing50631310CV15segimage segmentation231071910CV 5cidcensus-income2992852409976216smoattitude smoking restr.28553131000 6cmccontraceptive method14733910CV17spectSPECT heart imaging267222187 7dnaStatLog DNA3190361119018taeTA evaluation1513510CV 8forcForest cover58101275456589219thythyroid disease72003213428 9heaStatLog heart disease27021310CV20veh StatLog vehicle silhouette 84641810CV 10ipumIPUMS census2335843617007621vot congressional voting rec 43521610CV 11kddIntrusion (kdd cup 99)805050404231102922wavwaveform36003213000
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Data setC5.0CLIP4 DataSqueezer accuracysensitivityspecificity bcw94 (±2.6)95 (±2.5)94 (±2.8)92 (±3.5)98 (±3.3) bld68 (±7.2)63 (±5.4)68 (±7.1)86 (±18.5)44 (±21.5) bos75 (±6.1)71 (±2.7)70 (±6.4)70 (±6.1)88 (±4.3) cmc53 (±3.4)47 (±5.1)44 (±4.3)40 (±4.2)73 (±2.0) dna949192 97 hea78 (±7.6)72 (±10.2)79 (±6.0)89 (±8.3)66 (±13.5) led747168 97 pid75 (±5.0)71 (±4.5)76 (±5.6)83 (±8.5)61 (±10.3) sat8680 7896 seg93 (±1.2)86 (±1.9)84 (±2.5)83 (±2.1)98 (±0.4) smo68 3367 tae52 (±12.5)60 (±11.8)55 (±7.3)53 (±8.4)79 (±3.8) thy99 969599 veh75 (±4.4)56 (±4.5)61 (±4.2)61 (±3.2)88 (±1.6) vot96 (±3.9)94 (±2.2)95 (±2.8)93 (±3.3)96 (±5.2) wav767577 89 MEAN (stdev)78.5 (±14.4)74.9 (±15.0)75.4 (±14.9)74.6 (±19.1)83.5 (±16.7) adult8583829441 cid9589919445 forc6554555690 ipums100-848297 kdd92-961291 spect7686794781 MEAN all (stdev)80.4 (±14.1)75.6 (±14.8)77.0 (±14.6)71.7 (±23.0)80.9 (±19.0)
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Data set C5.0CLIP4 DataSqueezer mean # rules mean # select # select / rule mean # rules mean # select # select / rule mean # rules mean # select # select / rule bcw16 1.0 4122 30.54 13 3.3 bld1442 3.0 10 272 27.23 14 4.7 bos1868 3.8 10 133 13.320 107 5.4 cmc48184 3.8 861 7.620 70 3.5 dna40107 2.7 890 11.339 97 2.5 hea1021 2.1 12 192 16.05 17 3.4 led2079 4.0 41189 4.651 194 3.8 pid1022 2.2 464 16.02 8 4.0 sat96498 5.2 61 3199 52.457 257 4.5 seg42181 4.3 391170 30.057 219 3.8 smo00 0 18242 13.46 12 2.0 tae1233 2.8 9273 30.321 57 2.7 thy715 2.1 4119 29.87 28 4.0 veh37142 3.8 21381 18.124 80 3.3 vot46 1.5 1052 5.21 2 2.0 wav30119 4.0 985 9.422 65 3.0 MEAN Stdev 25.3 (±23.9) 95.8 (±123.5) 2.9 (±1.4) 16.8 (±16.3) 415.3 (±789.1) 18.9 (±12.7) 21.2 (±19.8) 77.5 (±80.3) 3.4 (±0.9) Adult54181 3.3 727561105.0613956.5 cid146412 2.8 19189599.715956.3 forc4321731 4.0 63243838.759210535.7 Ipums75197 2.6 ---108149213.8 kdd108354 3.3 ---2640915.7 spect46 1.5 199.019 MEAN all stdev 55.6 (±92.3) 200.6 (±368.6) 2.9 (±1.2) 21.2 (±21.8) 927.4 (±1800.6) 28.4 (±28.2) 27.7 (±27.6) 261.1 (±520.2) 6.5 (±7.4)
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Hybrid Algorithms A hybrid algorithm combines methods from two or more types of algorithms The goal of a hybrid algorithm design is to combine the most useful mechanisms of two or more algorithms to achieve better robustness, speed, accuracy, etc.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Hybrid Algorithms Hybrid algorithms that combined decision trees and rule algorithms: - CN2 algorithm (Clark and Niblett, 1989) - CLIP algorithms CLILP2(Cios and Liu, 1995) CLIP3(Cios, Wedding and Liu, 1997) CLIP4(Cios and Kurgan, 2004)
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm An important characteristic distinguishing CLIP4 from majority of ML algorithms is that it generates production rules that involve inequalities. This results in generating small number of compact rules in from data with attributes having large number of values, and when they are correlated with the target class. Key characteristic of CLIP4 is dividing the task of rule generation into subtasks, posing each subtask as a set covering (SC) problem and its efficient (by a special alg. within CLIP4) solution. Specifically, the SC alg. is used to: - select the most discriminating features - grow new branches of the tree -select data subsets from which to generate the least overlapping rules, and - generate final rules from the (virtual) tree leafs (that store subsets of the data).
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4s Set Covering Algorithm CLIP4s set covering algorithm is a simplified version of integer programming (IP). Four simplifications are made to the IP model to transform it into the SC problem: - function that is subject of optimization has all coefficients set to one, - all variables are binary, x i ={0,1} - constraint function coefficients are also binary - all constraint functions are >= 1 The SC problem is NP-hard.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4s Set Covering Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4s Set Covering Algorithm Given: BINary matrix, Initialize: Remove all empty (non-active) rows from the BINary matrix; if the matrix has no 1s then return error. 1. Select active rows that have the minimum number of 1s in rows – min-rows 2. Select columns that have the maximum number of 1s within the min-rows – max-columns 3. Within max-columns find columns that have the maximum number of 1s in all active rows – max-max-columns, if there is more than one max-max-column go to 4., otherwise go to 5. 4. Within max-max-columns find the first column that has the lowest number of 1s in the inactive rows 5. Add the selected column to the solution 6. Mark the inactive rows, if all the rows are inactive then terminate; otherwise go to 1. Active row is one that is not covered by the partial solution, and the inactive row is the row that is already covered by the partial solution.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4s Set Covering Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm The set of all training examples is denoted by S. A subset of positive examples is denoted by SP and the subset of negative examples by SN. SP and SN are represented by matrices whose rows represent examples and columns represent attributes. Matrix of the positive examples is denoted as POS and their number by NPOS. Similarly, for the negative examples we have matrices NEG and NNEG. The following properties are satisfied for the subsets: SP SN=S, SP SN=, SN, and SP
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm Examples are described by a set of K attribute-value pairs: where a j denotes j th attribute with value v j d j, # is a relation (, =, <,,, etc.), where K is the number of attributes. An example e consists of set of selectors The CLIP4 algorithm generates rules in the form: IF (s 1 … s m ) THEN class = class i where all selectors are only in the form s i = [a j v j ], namely, we use only inequalities.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm Phase 1: Use the first negative example [1,3,2,1] and matrix POS to create the BINARY matrix
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm Phase 2: After repeating the process illustrated above, at the end of Phase 1 we end up with just two matrices - the leaf nodes of the virtual decision tree (matrix numbers (8 & 9) are not important)
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm From this solution and from the backproj NEG matrix we generate the first rule: IF (F1 3) AND (F2 3) AND (F2 4) THEN F5=Buy (covers examples e1,e2 and e5) By the same process, using POS8, we generate one more rule: IF (F4 1) AND (F4 3) AND (F4 2) AND (F4 4) THEN F5=Buy (covers examples e3 and e4)
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm Phase 3: Using the CLIP4s heuristic, however, we choose only the first rule and remove from matrix POS all examples covered by the first rule. Next, we repeat the entire process on the reduced matrix POS: After going again through all the phases of the algorithm we generate just one rule: IF (F4 1) AND (F4 3) AND (F4 2) AND (F4 4) THEN F5=Buy
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm As the final outcome, in two iterations, the algorithm generated a set of rules that covers all positive examples and none of the negative: IF (F1 3) AND (F2 3) AND (F2 4)THEN F5=Buy IF (F4=5) THEN F5=Buy Notice that by knowing feature values for attribute F4 it is possible to convert the second rule into the simple equality rule shown above. Verbally the two rules say: IF Call International AND Language Fluency Bad AND Language Fluency Foreign THEN Buy IF Customer is 80 years or olderTHEN Buy
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan CLIP4 Algorithm
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Handling of Missing Values ex. #F1F2F3F4class 1123*1 213121 3*3251 433221 511131 631252 712242 821*32 IF F1 3 AND F1 2 AND F3 2 THEN class 1(covers 1,2,5) IF F2 2 AND F2 1 THEN class 1(covers 2,3,4) They cover all positive examples, including those with missing values and none of the negative examples. Notice that both rules cover the second example.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Thresholds Noise Threshold determines which nodes are pruned from the tree grown in Phase 1. The threshold prunes every node that contains less number of examples than its value. Pruning Threshold is used to prune nodes from the generated tree. It uses a goodness value to perform selection of the nodes. The threshold selects the first few nodes with the highest value and removes the remaining nodes from the tree. Stop Threshold stops the algorithm when smaller than the threshold number of positive examples remains uncovered. CLIP4 generates rules by partitioning the data into subsets containing similar examples, and removes examples that are covered by the already generated rules. The noise and stop thresholds are specified as percentage of the size of positive data and thus are easily scalable.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Evolutionary Computing Genetic / evolutionary computing ideas Fundamental components Genetic computing
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Evolutionary computing is concerned with population-oriented, evolution-like optimization It exploits the entire population of potential solutions, and evolves (converges) according to genetics-driven principles Genetic algorithms (GA) are search algorithms based on mechanisms of natural selection and genetics Evolutionary Computing
© 2007 Cios / Pedrycz / Swiniarski / Kurgan GA: Algorithmic Aspects GA exploits the mechanism of natural selection – survival of the fittest - via: collecting an initial population of N individuals determining suitability for survival of the individuals evolving the population to retain the individuals with the highest values of the fitness function eliminating the weakest individuals Result: Individuals with the highest ability to survive
© 2007 Cios / Pedrycz / Swiniarski / Kurgan GA uses the concept of recombination and mutation of individual elements/chromosomes to: generate new offspring, and increase diversity, respectively GA: Algorithmic Aspects
© 2007 Cios / Pedrycz / Swiniarski / Kurgan To perform genetic operations the original space has to be transformed into a GA search space (encoding). GA: Algorithmic Aspects
© 2007 Cios / Pedrycz / Swiniarski / Kurgan GA Pseudocode
© 2007 Cios / Pedrycz / Swiniarski / Kurgan GA pseudocode: start with an initial population and evaluate each of its elements by a fitness function: elements with high fitness have high chance of survival while those with low fitness are gradually eliminated GA Pseudocode
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Fundamental Components of GAs The main functional components of genetic computing are: encoding and decoding selection crossover mutation
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Encoding Encoding transforms a real number into its binary equivalent. It transforms the original problem into a format suitable for genetic computations. Decoding Decoding transforms elements from the GA search space to the original search space Fundamental Components of GAs
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Selection Mechanism When a population of chromosomes is established, we must define a way in which the chromosomes are selected for further optimization steps. Selection methods include: roulette wheel elitist strategy
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Roulette Wheel Fitness values of the elements are normalized to 1 The normalized values are viewed as probabilities The sum of fitness in the denominator describes total fitness of the population P.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Construct a roulette wheel with sectors reflecting probabilities of the strings and spin it N times. Roulette Wheel
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Elitist Strategy Select the best individuals in the population and carry them over, without any alteration, to the next population of strings.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Once the selection is completed, the resulting new population is subject to two GA mechanisms: crossover mutation Fundamental Components of GAs
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Crossover A one-point crossover mechanism chooses two strings and randomly selects a position in the strings at which they interchange their content, thus producing two new offsprings / strings.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Crossover leads to an increased diversity of the population of strings, as the new individuals emerge The intensity of crossover is characterized in terms of the probability at which the elements of strings are affected. The higher the probability, the more individuals are affected by the crossover. Crossover
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Mutation adds additional diversity of a stochastic nature. It is implemented by flipping the values of some randomly selected bits. Mutation
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Mutation rate is related to the probability at which individual bits are affected Example 5% mutation: If applied to a population of 50 strings, each 20 bits long Then 5% of the 1000 bits will be changed = 50 bits Mutation
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Task: derive rules that describe classes Rule Encoding Example
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Structure of the rule: where i=1,2,3,4 and j=1,2 and k=1,2,3 More generally: Rule Encoding Example
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Assuming a single bit per value for encoding each attribute, we have: 4 bit string: 1100for the 1 st attribute 2 bit string: 01 3 bit string: 001for the 3 rd Therefore, each rule encodes as a string of 9 bits: 110001001 This string decodes as: Rule Encoding / Decoding Example
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Fitness function describes how well the rule describes the data: e + is a fraction of positive instances covered by the rule e - is a fraction of the instances identified by the rule that does not belong to the class Rule Encoding Example
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Crossover Mechanism Example Start with two strings (examples): 100010101 101101001 Swapping after the fifth bit results in: 100011001 101100101
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Mutation Mechanism Example Applied to the rule/string 100010101 changes it into its mutated version 100000101
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Use of GA Operators to Improve Accuracy CLIP4 uses the GA in Phase I to enhance the partitioning of the data and obtain more general leaf node subsets. The components of the genetic module are: population and individual Individual/chromosome is defined as a node in the tree and consists of: POS i,j matrix (jth matrix at the ith tree level) and SOL i,j (the solution to the SC problem obtained from POS i,j matrix) Population is defined as a set of nodes at the same level of the tree. encoding and decoding scheme There is no need for encoding using the individuals defined above since GA operators are used on the SOL i,j vector
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Use of GA Operators to Improve Accuracy selection of the new population Initial population is the first tree level that consists of at least two nodes. CLIP4 uses the following fitness function to select the most suitable individuals for the next generation: The fitness value is calculated as the number of rows of the POS i,j matrix divided by the number of 1s from the SOL i,j vector. The fitness function has high values for the tree nodes that consist of large number of examples with low branching factor.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Use of GA Operators to Improve Accuracy The mechanism for selecting individuals for the next population: all individuals are ranked using their fitness function half of the individuals with the highest fitness are automatically selected for the next population (they will branch to create nodes for the next tree level) the second half of the next population is generated by matching the best with the worst individuals (the best with the worst, the second best with the second worst, etc.) and applying GA operators to obtain new individuals (new nodes in the tree).
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Use of GA Operators to Improve Accuracy
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Use of GA Operators to Improve Accuracy
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Use of GA Operators to Improve Accuracy
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Pruning CLIP4 prunes the tree grown in Phase 1 as follows: first, it selects a number (via the pruning threshold) of best (highest fitness) nodes on the ith tree level. Only the selected nodes are used to branch into new nodes, and are passed to the (i+1)th tree level. second, all redundant nodes that resulted from the branching process are removed. Two nodes are redundant if one mode contains positive examples that are identical, or form a subset of positive examples of the other node. third, after the redundant nodes are removed, each new node is evaluated using the noise threshold. If it contains less examples than the one specified by the noise threshold then it is pruned.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Pruning
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Feature and Selector Ranking Goodness of each attribute and selector is computed from the generated rules. Attributes with goodness value greater than zero are relevant and cannot be removed without decreasing accuracy. The attribute and selector goodness values are computed in these steps: Each rule has a goodness value equal to the percentage of the training positive examples it covers Each selector has a goodness value equal to the goodness of the rule it comes from Each attribute has a goodness value equal to the sum of scaled goodness values of all its selectors divided by the total number of attribute values
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Feature and Selector Ranking Suppose we have a two-category data, described by five attributes: a1 = {1, 2, 3}, a2 = {1, 2, 3}, a3 = {1, 2}, a4 = {1, 2, 3}, a5 = {1, 2, 3, 4}, and a6 = {1, 2} a decision attribute. Suppose CLIP4 generated these rules with their % goodness: IF a5 2 and a5 3 and a5 4 THEN class = 1 (covers 46% (29/62) positive examples) IF a1 1 and a1 2 and a2 2 and a2 1 THEN class = 1 (covers 27% (17/62) positive examples) IF a1 1 and a1 3 and a2 3 and a2 1 THEN class = 1 (covers 24% (15/62) positive examples) IF a1 2 a1 3 and a2 2 and a2 3 THEN class = 1 (covers 14% (9/62) positive examples)
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Feature and Selector Ranking Using the information about attribute values we can write the equality rules: IF a5=1THEN class = 1 (covers 46% (29/62) positive examples) IF a1=3 and a2=3THEN class = 1 (covers 27% (17/62) positive examples) IF a1=2 and a2=2THEN class = 1 (covers 24% (15/62) positive examples) IF a1=1 and a2=1THEN class = 1 (covers 14% (9/62) positive examples)
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Feature and Selector Ranking We calculate goodness values for the selectors first and then we can calculate the goodness of attributes: (a5, 1); goodness 46 (a1, 3) and (a2, 3); goodness 27 (a1, 2) and (a2, 2); goodness 24 (a1, 1) and (a2, 1); goodness 14 In order to show their relative goodness they are scaled to the 0- 100 range: (a5, 1); goodness 100 (a1, 3) and (a2, 3); goodness 58.7 (a1, 2) and (a2, 2); goodness 52.2 (a1, 1) and (a2, 1); goodness 30.4
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Feature and Selector Ranking For attribute a1 we have these selectors and their goodness values: (a1,3) with goodness 58.7, (a1,2) with goodness 52.2, and (a1,1) with goodness 30.4. Thus we calculate goodness of the first attribute a1 as: (58.7+52.2+30.4)/3 = 47.1 Similarly we calculate goodness of a2. For attribute a5, we have the following selectors and their goodness values: (a5,1) with goodness 100, AND (a5,2) through (a5,4) each with goodness of 0, thus a5 goodness is: (100+0+0+0)/4 = 25.0 Attributes, a3, a4 and a6, have all goodness value of 0 because they were not used in the generated rules.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan Feature and Selector Ranking The feature and selector ranking performed by CLIP4 algorithm can be used to: Select only relevant attributes/features and discard the irrelevant ones The user can discard all attributes with goodness of 0 and still have correct (with the same accuracy) model of the data. Provide additional insight into data properties The selector ranking can help in analyzing the data in terms of relevance of the selectors to the classification task.
© 2007 Cios / Pedrycz / Swiniarski / Kurgan References Cios K.J. and Liu N. 1992. Machine learning in generation of a neural network architecture: a Continuous ID3 approach. IEEE Trans. on Neural Networks, 3(2):280 291 Cios, K.J., Pedrycz, W. and Swiniarski, R. 1998. Data Mining Methods for Knowledge Discovery. Kluwer Cios, K.J. and Kurgan, L. 2004. CLIP4: Hybrid Inductive Machine Learning Algorithm that Generates Inequality Rules. Information Sciences, 163 (1-3): 37-83 Kurgan L., Cios K.J. and Dick S. 2006. Highly Scalable and Robust Rule Learner: Performance Evaluation and Comparison, IEEE Trans. on Systems Man and Cybernetics, Part B, 36(1):32-53 Kurgan, L. and Cios, K.J. 2002. CAIM Discretization Algorithm, IEEE Trans. on Knowledge and Data Engineering, 16(2): 145-153
Download ppt "Chapter 12 SUPERVISED LEARNING Rule Algorithms and their Hybrids Part 2 Cios / Pedrycz / Swiniarski / Kurgan."
Similar presentations
| 8,494
| 28,977
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.140625
| 3
|
CC-MAIN-2018-30
|
latest
|
en
| 0.803455
|
https://metanumbers.com/53994
| 1,621,125,365,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243991488.53/warc/CC-MAIN-20210515223209-20210516013209-00207.warc.gz
| 416,962,116
| 7,519
|
## 53994
53,994 (fifty-three thousand nine hundred ninety-four) is an even five-digits composite number following 53993 and preceding 53995. In scientific notation, it is written as 5.3994 × 104. The sum of its digits is 30. It has a total of 3 prime factors and 8 positive divisors. There are 17,996 positive integers (up to 53994) that are relatively prime to 53994.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 30
• Digital Root 3
## Name
Short name 53 thousand 994 fifty-three thousand nine hundred ninety-four
## Notation
Scientific notation 5.3994 × 104 53.994 × 103
## Prime Factorization of 53994
Prime Factorization 2 × 3 × 8999
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 53994 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 53,994 is 2 × 3 × 8999. Since it has a total of 3 prime factors, 53,994 is a composite number.
## Divisors of 53994
1, 2, 3, 6, 8999, 17998, 26997, 53994
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 108000 Sum of all the positive divisors of n s(n) 54006 Sum of the proper positive divisors of n A(n) 13500 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 232.366 Returns the nth root of the product of n divisors H(n) 3.99956 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 53,994 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 53,994) is 108,000, the average is 13,500.
## Other Arithmetic Functions (n = 53994)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 17996 Total number of positive integers not greater than n that are coprime to n λ(n) 8998 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5497 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 17,996 positive integers (less than 53,994) that are coprime with 53,994. And there are approximately 5,497 prime numbers less than or equal to 53,994.
## Divisibility of 53994
m n mod m 2 3 4 5 6 7 8 9 0 0 2 4 0 3 2 3
The number 53,994 is divisible by 2, 3 and 6.
## Classification of 53994
• Arithmetic
• Abundant
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
• Sphenic
## Base conversion (53994)
Base System Value
2 Binary 1101001011101010
3 Ternary 2202001210
4 Quaternary 31023222
5 Quinary 3211434
6 Senary 1053550
8 Octal 151352
10 Decimal 53994
12 Duodecimal 272b6
20 Vigesimal 6eje
36 Base36 15nu
## Basic calculations (n = 53994)
### Multiplication
n×i
n×2 107988 161982 215976 269970
### Division
ni
n⁄2 26997 17998 13498.5 10798.8
### Exponentiation
ni
n2 2915352036 157411517831784 8499277493809345296 458909989000741789912224
### Nth Root
i√n
2√n 232.366 37.7962 15.2436 8.84034
## 53994 as geometric shapes
### Circle
Diameter 107988 339254 9.15885e+09
### Sphere
Volume 6.59364e+14 3.66354e+10 339254
### Square
Length = n
Perimeter 215976 2.91535e+09 76359
### Cube
Length = n
Surface area 1.74921e+10 1.57412e+14 93520.4
### Equilateral Triangle
Length = n
Perimeter 161982 1.26238e+09 46760.2
### Triangular Pyramid
Length = n
Surface area 5.04954e+09 1.85511e+13 44085.9
## Cryptographic Hash Functions
md5 f23b0156c2fee3a7f4c921ccfdcecabd e87700822e3f522f7f057710c65105dcfba81350 115ebd03f111e6ca4696dc1ec92b2e2fc766f89424d046f4bb33dea1fb74539b 6ed12717d255824121e7dbb178c0ee69b4c9f13421f09230264959190f9157e2cd27da83ad998ee9904dd23fa5bd6f8ede3232d6327f81decede168da5d6b0e3 6d1f815a436944560420bed7a4571d1a26c4dbbb
| 1,476
| 4,193
|
{"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}
| 3.671875
| 4
|
CC-MAIN-2021-21
|
latest
|
en
| 0.816572
|
http://www.physicsforums.com/showthread.php?t=371935
| 1,409,684,837,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-35/segments/1409535922763.3/warc/CC-MAIN-20140909045618-00210-ip-10-180-136-8.ec2.internal.warc.gz
| 1,264,087,273
| 8,122
|
# Arithmetic mean of two unknowns in a system of equations
by travism123
Tags: arithmetic mean, easy, gre
P: 9 1. The problem statement, all variables and given/known data When finding the arithmetic mean in a system of equations is there any reason why the method that I am using is wrong? Find the arithmetic mean of x and y in the following set of equations 2. Relevant equations 3x + 5y = 65 and 7x + 14y = 175 3. The attempt at a solution I treated it as a statistics problem where x's and y's are individual units, so... 10x + 19y = 240 mean = 240/29 since mean = total sum/N My answer is 8 and 8/29 I understand that this method might not be the correct way of doing it, but if the number of units is known (10x, and not 10xy) is there any reason that this method of finding the arithmetic mean will not work. Thank you to anyone who takes the time to look at this. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
Mentor
P: 21,313
Quote by travism123 1. The problem statement, all variables and given/known data When finding the arithmetic mean in a system of equations is there any reason why the method that I am using is wrong?
Yes. See below.
Quote by travism123 Find the arithmetic mean of x and y in the following set of equations 2. Relevant equations 3x + 5y = 65 and 7x + 14y = 175 3. The attempt at a solution I treated it as a statistics problem where x's and y's are individual units, so... 10x + 19y = 240
You can do this, but it is of no help.
Quote by travism123 mean = 240/29 since mean = total sum/N My answer is 8 and 8/29
This part is not valid. If the left side were 10x + 19x, you could combine these terms to 29x, and it would make sense to divide by 29.
Quote by travism123 I understand that this method might not be the correct way of doing it, but if the number of units is known (10x, and not 10xy) is there any reason that this method of finding the arithmetic mean will not work. Thank you to anyone who takes the time to look at this.
The more obvious approach to this problem is to find the solutions x and y to the system of equations, and then find the arithmetic mean of these numbers. This system can be solved pretty easily, giving integer solutions.
P: 9 Thank you Mark. I guess that I still don't understand why this is wrong. If I solve the equations I get: x=5 and y=10 So the way that I am looking at it, I have a group of ten 5's and nineteen 10's. If I want to find the mean value for this group then wouldn't it make sense to add them to get my total sum, or 240, and then divide by N, or 29. x and y are equal between equations, so I am not looking for the mean of x and the mean of y, only the mean of every unknown in the set of equations, or x and y together. Obviously I would need to solve the equations first if there were an equation with an unknown being multiplied by an unknown like: 10xy + xy + 14y = 392. Then I completely understand why this would not work, but if I know the exact number of x's and y's why can I not simply add the n of each group together to get my N, and then divide the total sum by N? In the end, isn't it accomplishing the same thing as solving the equations to integers, and then finding the mean. Just to clarify...I am only looking for the correct answer, and not the correct method. The way that I did it originally seems to be a big shortcut. Again, thanks very much.
Mentor
P: 21,313
Arithmetic mean of two unknowns in a system of equations
Quote by travism123 In the end, isn't it accomplishing the same thing as solving the equations to integers, and then finding the mean.
Well, no, because you get different answers. The arith. mean of 5 and 10 is 7.5, which is different from the value you got, which is about 8.28.
What you have said makes some sense, but I have never seen a problem where you are expected to take a system of equations and do anything but find its solution. I guess some context for this problem would be helpful--can you give us the exact wording of the problem?
P: 9 I am positive that you are right Mark. The actual question was easier: If a + 2b = 14 and 5a + 4b = 16 what is the average (arithmetic mean) of a and b? (a) 1.5 (b) 2 (c) 2.5 (d) 3 (e) 3.5 The answer is 2.5. (6a + 6b = 30)/6 = a + b = 5 5/2 = 2.5 I think that I made it more complicated than necessary. Thanks, and sorry for taking up your time with something so obvious.
Mentor P: 21,313 No problem. I don't mind helping at all. A favorite saying of mine is, "Keep things as simple as possible, but no simpler."
Related Discussions Engineering, Comp Sci, & Technology Homework 9 Calculus & Beyond Homework 5 Precalculus Mathematics Homework 13 Calculus & Beyond Homework 6 General Math 5
| 1,256
| 4,752
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35
|
latest
|
en
| 0.955853
|
https://nl.mathworks.com/matlabcentral/answers/803101-how-to-error-check-a-number-for-strings-and-blank-inputs
| 1,713,393,640,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00255.warc.gz
| 397,221,119
| 28,582
|
# How to error check a number for strings and blank inputs?
29 views (last 30 days)
Sourav Anilkumar on 15 Apr 2021
Commented: Daniel Pollard on 16 Apr 2021
I have some issues with a unit converter program I'm writing. Currently I have error checking code for negative values but I also want it to check for string and blank inputs. what if statments should I add to achieve this? Any other improvements I can make?
I've added a condensed version of my code below, the full version has 6 different conversions.
clear, clc
n = 1;
m = 1;
o = 1;
fprintf('1. Celsius ↔ Fahrenheit \n2. Millilitre ↔ Fluid Ounce')
menu = input('Enter a number to choose a conversion: ')
while n == 1
% string and blank error checking here
disp('Millilitre ↔ Fluid Ounce Converter')
disp('1. Millilitre to Fluid Ounce')
disp('2. Fluid Ounce to Millilitre')
unit = input('What conversion should be performed?: ')
while m == 1
if unit == 1
mil = input('Input a volume in millilitres: ');
while o == 1
% string and blank error checking here
if (mil < 0)
disp('A negative number has been entered')
mil = input('Input a volume in millilitres: ');
else
floz = (mil/29.5735);
fprintf('The volume in fluid ounces is %1.2f\n', floz)
n = 0;
m = 0;
o = 0;
end
end
elseif unit == 2
floz = input('Input a volume in fluid ounces: ');
while o == 1
% string and blank error checking here
if (floz < 0)
disp('A negative number has been entered')
floz = input('Input a volume in fluid ounces: ');
else
mil = (floz*29.5735);
fprintf('The volume in millilitres is %1.2f\n', mil)
n = 0;
m = 0;
o = 0;
end
end
else % string and blank error checking here
disp('An invalid input has been entered')
unit = input('Please enter a valid number: ')
end
end
end
end
Daniel Pollard on 15 Apr 2021
Are you looking for validateattributes or assert? For example,
will throw an error if menu is not a number equal to 1 or 2.
Sourav Anilkumar on 16 Apr 2021
Yeah sure:
Error using input
Unrecognized function or variable 'w'.
Error in s3784513_Milestone_2 (line 6)
menu = input('Enter a number to choose a conversion: ')
Pretend thats all red haha
Daniel Pollard on 16 Apr 2021
Ok, so according to the documentation, by default, MATLAB thinks that an input is numerical. Give it something like w and it says "hey! w isn't a number!"
You can tell it to read the input in as a string by default by writing something like
Then, you can test whether the input can be converted into a number by using code from the top answer to this question.
If it can, convert it to a number using str2num. From there you can check it has other properties you want using commands like isempty.
### Categories
Find more on Characters and Strings in Help Center and File Exchange
R2020b
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
| 774
| 2,838
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.618297
|
https://www.traditionaloven.com/tutorials/power/convert-mechanical-horsepower-hp-i-to-atm-cfm-atmosphere.html
| 1,524,208,492,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125937161.15/warc/CC-MAIN-20180420061851-20180420081851-00359.warc.gz
| 905,671,372
| 11,726
|
Convert hp I to atm cfm | Mechanical horsepower to atmosphere cubic feet / minute
# power conversion
## Amount: 1 Mechanical horsepower (hp I) of power Equals: 15.59 atmosphere cubic feet / minute (atm cfm) in power
Converting Mechanical horsepower to atmosphere cubic feet / minute value in the power units scale.
TOGGLE : from atmosphere cubic feet / minute into Mechanical horsepower in the other way around.
## power from Mechanical horsepower to atmosphere cubic foot per minute Conversion Results:
### Enter a New Mechanical horsepower Amount of power to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other power measuring units - complete list.
Conversion calculator for webmasters.
## Power units
Power units represent power physics, which is the rate at which energy is used-up, either transformed or transferred from its source to elsewhere, by various ways within the nature of physics. Conversion tool with multiple power units.
Convert power measuring units between Mechanical horsepower (hp I) and atmosphere cubic feet / minute (atm cfm) but in the other reverse direction from atmosphere cubic feet / minute into Mechanical horsepower.
conversion result for power: From Symbol Equals Result To Symbol 1 Mechanical horsepower hp I = 15.59 atmosphere cubic feet / minute atm cfm
# Converter type: power units
This online power from hp I into atm cfm converter is a handy tool not just for certified or experienced professionals.
First unit: Mechanical horsepower (hp I) is used for measuring power.
Second: atmosphere cubic foot per minute (atm cfm) is unit of power.
## 15.59 atm cfm is converted to 1 of what?
The atmosphere cubic feet / minute unit number 15.59 atm cfm converts to 1 hp I, one Mechanical horsepower. It is the EQUAL power value of 1 Mechanical horsepower but in the atmosphere cubic feet / minute power unit alternative.
How to convert 2 Mechanical horsepower (hp I) into atmosphere cubic feet / minute (atm cfm)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
15.5938667381 * 2 (or divide it by / 0.5)
QUESTION:
1 hp I = ? atm cfm
1 hp I = 15.59 atm cfm
## Other applications for this power calculator ...
With the above mentioned two-units calculating service it provides, this power converter proved to be useful also as a teaching tool:
1. in practicing Mechanical horsepower and atmosphere cubic feet / minute ( hp I vs. atm cfm ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with power's values and properties.
International unit symbols for these two power measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for Mechanical horsepower is:
hp I
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for atmosphere cubic foot per minute is:
atm cfm
### One Mechanical horsepower of power converted to atmosphere cubic foot per minute equals to 15.59 atm cfm
How many atmosphere cubic feet / minute of power are in 1 Mechanical horsepower? The answer is: The change of 1 hp I ( Mechanical horsepower ) unit of power measure equals = to 15.59 atm cfm ( atmosphere cubic foot per minute ) as the equivalent measure for the same power type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in hp I - Mechanical horsepower for power amount, the rule is that the Mechanical horsepower number gets converted into atm cfm - atmosphere cubic feet / minute or any other power unit absolutely exactly.
Conversion for how many atmosphere cubic feet / minute ( atm cfm ) of power are contained in a Mechanical horsepower ( 1 hp I ). Or, how much in atmosphere cubic feet / minute of power is in 1 Mechanical horsepower? To link to this power Mechanical horsepower to atmosphere cubic feet / minute online converter simply cut and paste the following.
The link to this tool will appear as: power from Mechanical horsepower (hp I) to atmosphere cubic feet / minute (atm cfm) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
| 976
| 4,555
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375
| 3
|
CC-MAIN-2018-17
|
latest
|
en
| 0.809619
|
https://www.mathworks.com/examples/wavelet/mw/wavelet-ex94687989-lifting-a-filter-bank
| 1,519,502,749,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891815934.81/warc/CC-MAIN-20180224191934-20180224211934-00443.warc.gz
| 912,530,628
| 14,860
|
MATLAB Examples
# Lifting a Filter Bank
This example shows how to use lifting to progressively change the properties of a perfect reconstruction filter bank. The following figure shows the three canonical steps in lifting: split, predict, and update.
The first step in lifting is simply to split the signal into its even- and odd-indexed samples. These are called polyphase components and that step in the lifting process is often referred to as the "lazy" lifting step because you really are not doing that much work. You can do this in MATLAB by creating a "lazy" lifting scheme.
LS = liftwave('lazy');
Apply the lifting scheme to some data.
x = randn(8,1); [ALazy,DLazy] = lwt(x,LS);
MATLAB™ indexes from 1 so ALazy contains the odd-indexed samples of x and DLazy contains the even-indexed samples. Most explanations of lifting assume that the signal starts with sample 0, so ALazy would be the even-indexed samples and DLazy the odd-indexed samples. This examples follows that latter convention. The "lazy" wavelet transform treats one half of the signal as wavelet coefficients, DLazy, and the other half as scaling coefficients, ALazy. This is perfectly consistent within the context of lifting, but a simple split of the data does really sparsify or capture any relevant detail.
The next step in the lifting scheme is to predict the odd samples based on the even samples. The theoretical basis for this is that most natural signals and images exhibit correlation among neighboring samples. Accordingly, you can "predict" the odd-indexed samples using the even-indexed samples. The difference between your prediction and the actual value is the "detail" in the data missed by the predictor. That missing detail comprises the wavelet coefficients.
The prediction step is also referred to as a "dual lifting step". In equation form, you can write the prediction step as where are the wavelet coefficients at the finer scale and is some number of finer-scale scaling coefficients. is the prediction operator.
Add a simple (Haar) dual lifting step that subtracts the even (approximation) coefficient from the odd (detail) coefficient. In this case the prediction operator is simply . In other words, it predicts the odd samples based on the immediately preceding even sample.
ElemLiftStep = {'d',-1,0};
The above code says "create an elementary dual (predict) lifting step using a polynomial in with the highest power . The coefficient is -1. Update the lazy lifting scheme.
LSN = addlift(LS,ElemLiftStep,'end');
Apply the new lifting scheme to the signal.
[A,D] = lwt(x,LSN);
Note that the elements of A are identical to those in ALazy. This is expected because you did not modify the approximation coefficients. If you look at the elements of D, you see that they are equal to
Dnew = DLazy-ALazy;
Compare Dnew to D. Imagine an example where the signal was piecewise constant over every two samples.
v = [1 -1 1 -1 1 -1]; u = repelem(v,2);
Apply the new lifting scheme to u.
[Au,Du] = lwt(u,LSN);
You see that all the Du are zero. This signal has been compressed because all the information is now contained in 6 samples instead of 12 samples. You can easily reconstruct the original signal
urecon = ilwt(Au,Du,LSN);
In your prediction (dual lifting) step, you predicted that the adjacent odd sample in your signal had the same value as the immediately preceding even sample. Obviously, this is true only for trivial signals. The wavelet coefficients capture the difference between the prediction and the actual values (at the odd samples). Finally, use the update step to update the even samples based on differences obtained in the prediction step. In this case, update using the following . This replaces each even-indexed coefficient by the arithmetic average of the even and odd coefficients. An update step is also referred to as a primal lifting step.
elsprimal = {'p',1/2,0}; LSupdated = addlift(LSN,elsprimal,'end');
Obtain the wavelet transform of the signal with the updated lifting scheme.
[A,D] = lwt(x,LSupdated);
If you compare A to the original signal, x, you see that the signal mean is captured in the approximation coefficients.
mean(A) mean(x)
ans = -0.0131 ans = -0.0131
In fact, the elements of A are easily obtainable from x by the following.
n = 1; for ii = 1:2:numel(x) meanz(n) = mean([x(ii) x(ii+1)]); n = n+1; end
Compare meanz and A. As always, you can invert the lifting scheme to obtain a perfect reconstruction of the data.
xrec = ilwt(A,D,LSupdated); max(abs(x-xrec))
ans = 2.2204e-16
It is common to add a normalization step at the end so that the energy in the signal ( norm) is preserved as the sum of the energies in the scaling and wavelet coefficients. Without this normalization step, the energy is not preserved.
norm(x,2)^2 norm(A,2)^2+norm(D,2)^2
ans = 11.6150 ans = 16.8091
LSscaled = LSupdated; LSscaled(end,1:2) = {sqrt(2), sqrt(2)/2}; [A,D] = lwt(x,LSscaled); norm(A,2)^2+norm(D,2)^2
ans = 11.6150
Now the norm of the signal is equal to the sum of the energies in the scaling and wavelet coefficients. The lifting scheme you developed in this example is the Haar lifting scheme.
The Wavelet Toolbox™ supports many commonly used lifting schemes through liftwave with pre-defined dual, primal, and normalization steps. For example, you can obtain the Haar lifting scheme with the following.
lshaar = liftwave('haar');
If you compare lshaar to LSUpdated, you see that our step-by-step lifting scheme matches the Haar lifting scheme. To see that not all lifting schemes consist of single dual and primal lifting steps, examine the lifting scheme that corresponds to the 'bior3.1' wavelet.
lsbior3_1 = liftwave('bior3.1')
lsbior3_1 = 4x3 cell array {'d' } {[ 0.3333]} {[ 1]} {'p' } {1x2 double} {[ 0]} {'d' } {[ -0.4444]} {[ 0]} {[0.4714]} {[ 2.1213]} {0x0 double}
You can also use liftfilt if you want to start with a set of biorthogonal (or orthogonal) scaling and wavelet filters and "lift" them to another set. For example, start with the Haar scaling and lifting filters.
[LoD,HiD,LoR,HiR] = wfilters('haar');
Lift the Haar filters with two primal lifting steps.
twoels(1) = struct('type','p','value',... laurpoly([0.125 -0.125],0)); twoels(2) = struct('type','p','value',... laurpoly([0.125 -0.125],1)); [LoDN,HiDN,LoRN,HiRN] = liftfilt(LoD,HiD,LoR,HiR,twoels);
Plot the resulting scaling and wavelet functions.
[phia,psia,phis,psis,xval] = bswfun(LoDN,HiDN,LoRN,HiRN); subplot(2,2,1) plot(xval,phia,'r','linewidth',2); title('Analysis Scaling Function'); axis tight; grid on; subplot(2,2,2) plot(xval,phis,'linewidth',2); axis tight; grid on; title('Synthesis Scaling Function'); subplot(2,2,3); plot(xval,psia,'r','linewidth',2); axis tight; grid on; title('Analysis Wavelet'); subplot(2,2,4); plot(xval,psis,'linewidth',2); axis tight; grid on; title('Synthesis Wavelet');
If you plot the analysis and synthesis scaling functions and wavelets for the 'bior1.3' wavelet, you see that lifting the Haar wavelet as in the previous example has essentially provided the 'bior1.3' wavelet to within a change of sign on the synthesis wavelet.
[LoD,HiD,LoR,HiR] = wfilters('bior1.3'); [phia,psia,phis,psis,xval] = bswfun(LoD,HiD,LoR,HiR);
| 1,911
| 7,258
|
{"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-2018-09
|
longest
|
en
| 0.934184
|
https://marilynburnsmath.com/category/interview-one-on-one/
| 1,726,098,949,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00433.warc.gz
| 355,283,635
| 23,676
|
Category
# Interviews One-on-One
This is a question that I’ve thought about for a while. It seems that I embrace the same pedagogical approach with both, but I know there are differences. I think I have some clarity now. Read on.
June 5, 2023
During a Listening to Learn interview, when asked to figure out mentally the missing number in the problem 90 – ___ = 75, Meeyah and Rocco explained their reasoning in different ways. Listen to how they thought and how I used these two videos for a class lesson.
March 12, 2023
When asked in a Listening to Learn interview which fraction was greater, it was clear to Adrian that 5/6 was greater than 1/4. His explanation, however, was anything but clear. Listen to Adrian reason and learn how I used this video in a class lesson.
February 19, 2023
Rebecca solved 7 – 3 by adding, demonstrating the important numerical reasoning strategy of applying the inverse relationship between addition and subtraction. Read about what happened when I used the video clip of Rebecca in a class lesson.
April 10, 2022
In a Listening to Learn interview, Nathan knew from memory that 6 x 5 equals 30. He explained to Rusty Bresser, "So, 6 divided by 2 equals 3, and you just add a zero behind it." Here's a suggestion for using that video clip in a lesson and what happened in two different classes.
February 22, 2022
I love incorporating children’s books into math lessons. Since most of my teaching focuses on math, it’s a treat for me to read a book aloud to a class. After the students have a chance to enjoy the story and respond to the illustrations, then I use the book as a springboard for a math lesson.
August 31, 2021
Over the years, I’ve collaborated with Lynne Zolli and Patty Clark on a variety of math education projects. For this blog, we worked together to share our thinking about how Listening to Learn math interviews can serve teachers and students.
August 16, 2021
David Brooks wrote an opinion column in The New York Times on November 19, 2020, “Nine Nonobvious Ways to Have Deeper Conversations.” K–5 math wasn’t his focus or even hinted at in his message, but his suggestions jumped out at me as useful and important for connecting with students.
June 8, 2021
Yes, that’s a photo of me, taken about 30 years ago when I was conducting my first ever math interview. That was an extraordinary experience. It dramatically shifted my professional focus and, after all these years, has finally resulted in Listening to Learn, a digital interview tool to help K–5 teachers learn about how their students reason.
May 25, 2021
On Wednesday, May 5, 2021, I posted the sixth in my Wednesday Twitter series of video clips from Listening to Learn math interviews. The response to this Tweet amazed me―it received over 100,000 impressions! I was appreciative of the many supportive and insightful replies. Read more.
May 11, 2021
| 671
| 2,870
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.53125
| 4
|
CC-MAIN-2024-38
|
latest
|
en
| 0.964428
|
http://slideplayer.com/slide/3306762/
| 1,550,382,781,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-09/segments/1550247481624.10/warc/CC-MAIN-20190217051250-20190217073250-00635.warc.gz
| 226,848,551
| 29,961
|
Primitive Recursive Functions (Chapter 3)
Presentation on theme: "Primitive Recursive Functions (Chapter 3)"— Presentation transcript:
Primitive Recursive Functions (Chapter 3)
Preliminaries: partial and total functions
The domain of a partial function on set A contains the subset of A. The domain of a total function on set A contains the entire set A. A partial function f is called partially computable if there is some program that computes it. Another term for such functions partial recursive. Similarly, a function f is called computable if it is both total and partially computable. Another term for such function is recursive.
Composition Let f : A → B and g : B → C g ͦ f : A → C
Composition of f and g can then be expressed as: g ͦ f : A → C (g ͦ f)(x) = g(f(x)) h(x) = g(f(x)) NB: In general composition is not commutative: ( g ͦ f )(x) ≠ ( f ͦ g )(x)
Composition Definition: Let g be a function containing k variables and f1 ... fk be functions of n variables, so the composition of g and f is defined as: h(0) = k Base step h( x1, ... , xn) = g( f1(x1 ,..., xn), … , fk(x1 ,..., xn) ) Inductive step Example: h(x , y) = g( f1(x , y), f2(x , y), f3(x , y) ) h is obtained from g and f1... fk by composition. If g and f1...fk are (partially) computable, then h is (partially) computable. (Proof by construction)
Recursion From programming experience we know that recursion refers to a function calling upon itself within its own definition. Definition: Let g be a function containing k variables then h is obtained through recursion as follows: h(x1 , … , xn) = g( … , h(x1 , … , xn) ) Example: x + y f( x , 0 ) = x (1) f(x , y+1 ) = f( x , y ) + 1 (2) Input: f ( 3, 2 ) => f ( 3 , 1 ) + 1 => ( f ( 3 , 0 ) + 1 ) + 1 => ( ) + 1 => 5
PRC: Initial functions
Primitive Recursively Closed (PRC) class of functions. Initial functions: Example of a projection function: u2 ( x1 , x2 , x3 , x4 , x5 ) = x2 Definition: A class of total functions C is called PRC² class if: The initial functions belong to C. Function obtained from functions belonging to C by either composition or recursion belongs to C. s(x) = x + 1 n(x) = 0 ui (x1 , … , xn) = xi
PRC: primitive recursive functions
There exists a class of computable functions that is a PRC class. Definition: Function is considered primitive recursive if it can be obtained from initial functions and through finite number of composition and recursion steps. Theorem: A function is primitive recursive iff it belongs to the PRC class. (see proof in chapter 3) Corollary: Every primitive recursive function is computable.
Primitive recursive functions: sum
We have already seen the addition function, which can be rewritten in LRR as follows: sum( x, succ(y) ) => succ( sum( x , y)) ; sum( x , 0 ) => x ; Example: sum(succ(0),succ(succ(succ(0)))) => succ(sum(succ(0),succ(succ(0)))) => succ(succ(sum(succ(0),succ(0)))) => succ(succ(succ(sum(succ(0),0) => succ(succ(succ(succ(0))) => succ(succ(succ(1))) => succ(succ(2)) => succ(3) => 4 NB: To prove that a function is primitive recursive you need show that it can be obtained from the initial functions using only concatenation and recursion.
Primitive recursive functions: multiplication
h( x , 0 ) = 0 h( x , y + 1) = h( x , y ) + x In LRR this can be written as: mult(x,0) => 0 ; mult(x,succ(y)) => sum(mult(x,y),x) ; What would happen on the following input? mult(succ(succ(0)),succ(succ(0)))
Primitive recursive functions: factorial
0! = 1 ( x + 1 ) ! = x ! * s( x ) LRR implementation would be as follows: fact(0) => succ(null(0)) ; fact(succ(x)) => mult(fact(x),succ(x)) ; Output for the following? fact(succ(succ(null(0))))
Primitive recursive functions: power and predecessor
Power function In LRR the power function can be expressed as follows: pow(x,0) => succ(null(0)) ; pow(x,succ(y)) => mult(pow(x,y),x) ; Predecessor function In LRR the predecessor is as follows: pred(1) => 0 ; pred(succ(x)) => x ; p (0) = 0 p ( t + 1 ) = t
Primitive recursive functions: ∸, | x – y | and α
dotsub(x,x) => 0 ; dotsub(x,succ(y)) => pred(dotsub(x,y)) ; x ∸ 0 = x x ∸ ( t + 1) = p( x ∸ t ) What would be the output? dotsub(succ(succ(succ(0))),succ(0)) | x – y | = ( x ∸ y ) + ( y ∸ x ) abs(x,y) => sum(dotsub(x,y),dotsub(y,x)) ; α(x) = 1 ∸ x α(x) => dotsub(1,x) ; Output for the following? a(succ(succ(0))) a(null(0))
Primitive recursive functions
x + y f( x , 0 ) = x f( x , y + 1 ) = f( x , y ) + 1 x * y h( x , 0 ) = 0 h( x , y + 1 ) = h( x , y ) + x x! 0! = 1 ( x + 1 )! = x! * s(x) x^y x^0 = 1 x^( y + 1 ) = x^y * x p(x) p( 0 ) = 0 p( x + 1 ) = x x ∸ y if x ≥ y then x ∸ y = x – y; else x ∸ y = 0 x ∸ 0 = x x ∸ ( t + 1) = p( x ∸ t ) | x – y | | x – y | = ( x ∸ y ) + ( y ∸ x ) α(x) α(x) = 1 ∸ x
Bounded quantifiers Theorem: Let C be a PRC class. If f( t , x1 , … , xn) belongs to C then so do the functions g( y , x1 , ... , xn ) = f( t , x1 , …, xn ) g( y , x1 , ... , xn ) = f( t , x1 , …, xn )
Bounded quantifiers Theorem: Let C be a PRC class. If f( t , x1 , … , xn) belongs to C then so do the functions g( y , x1 , ... , xn ) = f( t , x1 , …, xn ) g( y , x1 , ... , xn ) = f( t , x1 , …, xn ) Theorem: If the predicate P( t, x1 , … , xn ) belongs to some PRC class C, then so do the predicates: ( t)≤y P(t, x1, … , xn ) (∃t)≤y P(t, x1, … , xn )
Primitive recursive predicates
x = y d( x , y ) = α( | x – y | ) x ≤ y α ( x ∸ y ) ~P α( P ) P & Q P * Q P v Q ~ ( ~P & ~Q ) y | x y | x = (∃t)≤x { y * t = x } Prime(x) Prime(x) = x > 1 & ( t)≤x { t = 1 v t = x v ~( t | x ) } Exercises for Chapter 3: page 62 Questions 3,4 and 5. Fibonacci function
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows:
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows: ,where
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows: ,where Function g also belongs to C as it is attained from composition of primitive recursive functions.
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows: ,where Function g also belongs to C as it is attained from composition of primitive recursive functions. t0 is the least value for for which the predicate P is true (1).
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows: ,where Function g also belongs to C as it is attained from composition of primitive recursive functions. t0 is the least value for for which the predicate P is true (1).
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows: ,where Function g also belongs to C as it is attained from composition of primitive recursive functions. t0 is the least value for for which the predicate P is true (1).
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows: ,where Function g also belongs to C as it is attained from composition of primitive recursive functions. t0 is the least value for for which the predicate P is true (1).
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows: ,where Function g also belongs to C as it is attained from composition of primitive recursive functions. t0 is the least value for for which the predicate P is true (1). g( y , x1 , ... , xn ) produces the least value for which P is true. Finally the definition for bounded minimalization can be given as:
Bounded minimalization
Let P(t, x1, … ,xn) be in some PRC class C and we can define a function g as follows: ,where Function g also belongs to C as it is attained from composition of primitive recursive functions. t0 is the least value for for which the predicate P is true (1). g( y , x1 , ... , xn ) produces the least value for which P is true. Finally the definition for bounded minimalization can be given as: Theorem: If P(t,x1, … ,xn) belongs to some PRC class C and there is function g that does the bounded minimalization for P, then f belongs to C.
Unbounded minimalization
Definition: y is the least value for which predicate P is true if it exists. If there is no value of y for which P is true, the unbounded minimalization is undefined.
Unbounded minimalization
Definition: y is the least value for which predicate P is true if it exists. If there is no value of y for which P is true, the unbounded minimalization is undefined. We can then define this as a non-total function in the following way:
Unbounded minimalization
Definition: y is the least value for which predicate P is true if it exists. If there is no value of y for which P is true, the unbounded minimalization is undefined. We can then define this as a non-total function in the following way: Theorem: If P(x1, … , xn, y) is a computable predicate and if then g is a partially computable function. (Proof by construction)
[ x / y ] , the whole part of the division i.e. [10/4]=2 R(x,y) , remainder of the division of x by y. pn , nth prime number i.e p1=2 , p2=3 etc.
Pairing functions Let us consider the following primitive recursive function that provides a coding for two numbers x and y.
Pairing functions Let us consider the following primitive recursive function that provides a coding for two numbers x and y. Example: <1,1> = 2 * ( ) ∸ 1 = 6 ∸ 1 = 5 <1,2> = 2 * ( 2*2 + 1) ∸ 1 = 10 ∸ 1 = 9
Pairing functions Let us consider the following primitive recursive function that provides a coding for two numbers x and y. Example: <1,1> = 2 * ( ) ∸ 1 = 6 ∸ 1 = 5 <1,2> = 2 * ( 2*2 + 1) ∸ 1 = 10 ∸ 1 = 9
Pairing functions Let us consider the following primitive recursive function that provides a coding for two numbers x and y. Example: <1,1> = 2 * ( ) ∸ 1 = 6 ∸ 1 = 5 <1,2> = 2 * ( 2*2 + 1) ∸ 1 = 10 ∸ 1 = 9 Define z to be as: < x , y > = z Then for any z there is always a unique solution x and y.
Pairing functions Let us consider the following primitive recursive function that provides a coding for two numbers x and y. Example: <1,1> = 2 * ( ) ∸ 1 = 6 ∸ 1 = 5 <1,2> = 2 * ( 2*2 + 1) ∸ 1 = 10 ∸ 1 = 9 Define z to be as: < x , y > = z Then for any z there is always a unique solution x and y.
Pairing functions Let us consider the following primitive recursive function that provides a coding for two numbers x and y. Example: <1,1> = 2 * ( ) ∸ 1 = 6 ∸ 1 = 5 <1,2> = 2 * ( 2*2 + 1) ∸ 1 = 10 ∸ 1 = 9 Define z to be as: < x , y > = z Then for any z there is always a unique solution x and y. Thus we have the solutions for x and y which can then be defined using the following functions:
Pairing functions More formally this can written as:
Pairing Function Theorem: functions <x,y>, l(z), r(z) have the following properties: are primitive recursive l(<x,y>) = x and r(<x,y>) = y < l(z) , r(z) > = z l(z) , r(z)≤ z
Gödel numbers Let (a1, … , an) be any sequence, then the Gödel number is computed as follows:
Gödel numbers Let (a1, … , an) be any sequence, then the Gödel number is computed as follows: Example: Take a sequence (1,2,3,4), the Gödel number will be computed as follows:
Gödel numbers Let (a1, … , an) be any sequence, then the Gödel number is computed as follows: Example: Take a sequence (1,2,3,4), the Gödel number will be computed as follows: Gödel numbering has a special uniqueness property: If [a1, … , an ] = [ b1, … , bn ] then ai = bi , where i = 1, … , n
Gödel numbers Let (a1, … , an) be any sequence, then the Gödel number is computed as follows: Example: Take a sequence (1,2,3,4), the Gödel number will be computed as follows: Gödel numbering has a special uniqueness property: If [a1, … , an ] = [ b1, … , bn ] then ai = bi , where i = 1, … , n Also notice: [ a1, … , an ] = [ a1, … , an, 0 ]
Gödel numbers Given that x = [a1, … , an ], we can now define two important functions:
Gödel numbers Given that x = [a1, … , an ], we can now define two important functions: Example: Let x = [ 4 , 3 , 2 , 1 ], then (x)2 = 3 and (x)4=1 and (x)0 = 0
Gödel numbers Given that x = [a1, … , an ], we can now define two important functions: Example: Let x = [ 4 , 3 , 2 , 1 ], then (x)2 = 3 and (x)4=1 and (x)0 = 0 Lt(10) = will be the length of the sequence derived using Gödel numbering
Gödel numbers Given that x = [a1, … , an ], we can now define two important functions: Example: Let x = [ 4 , 3 , 2 , 1 ], then (x)2 = 3 and (x)4=1 and (x)0 = 0 Lt(10) = will be the length of the sequence derived using Gödel numbering So: 10 = 2^1 * 3^0 * 5^1 = [ 1, 0, 1 ] => Lt(10) = 3
Gödel numbers Given that x = [a1, … , an ], we can now define two important functions: Example: Let x = [ 4 , 3 , 2 , 1 ], then (x)2 = 3 and (x)4=1 and (x)0 = 0 Lt(10) = will be the length of the sequence derived using Gödel numbering So: 10 = 2^1 * 3^0 * 5^1 = [ 1, 0, 1 ] => Lt(10) = 3 Sequence Number Theorem: (1)
Gödel numbers Given that x = [a1, … , an ], we can now define two important functions: Example: Let x = [ 4 , 3 , 2 , 1 ], then (x)2 = 3 and (x)4=1 and (x)0 = 0 Lt(10) = will be the length of the sequence derived using Gödel numbering So: 10 = 2^1 * 3^0 * 5^1 = [ 1, 0, 1 ] => Lt(10) = 3 Sequence Number Theorem: (1) (2)
Similar presentations
| 4,203
| 13,216
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.21875
| 4
|
CC-MAIN-2019-09
|
latest
|
en
| 0.823694
|
https://www.physicsforums.com/threads/volume-of-gas-by-burning-gasoline.422259/
| 1,521,488,097,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257647044.86/warc/CC-MAIN-20180319175337-20180319195337-00562.warc.gz
| 883,535,031
| 14,226
|
Volume of Gas by Burning Gasoline
1. Aug 14, 2010
korneld
What is the volume of gas produced for a unit of gasoline burnt?
2. Aug 14, 2010
Bob S
It is straight forward to calculate it. Just insert molecular weights in this and use 1 molar volume of a gas is 22.41 liters at STP.
2·C8H18 + 25·O2 => 16·CO2 + 18·H2O
Bob S
3. Aug 14, 2010
korneld
Thanks for that. Isn't that 9 H2O, though?
4. Aug 14, 2010
xxChrisxx
Can't be, 2 moles of octane were combust in that formula.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
| 192
| 584
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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
|
longest
|
en
| 0.913498
|
http://howtoprogram.eu/tag/gis,436/page/1
| 1,505,834,085,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-39/segments/1505818685850.32/warc/CC-MAIN-20170919145852-20170919165852-00420.warc.gz
| 148,883,298
| 8,393
|
## Create a unique legend based on a contingency (2x2) table in geom_map or ggplot2?
How can I do this based on this contingency table? I'm not entirely sure how to create a custom legend in R based on the indicator table I made (crimes). Reproducible code in R: require(maps) set.seed(123) # randomly assign 2 variable...
more »
## Create choropleth map from coordinate points
I have a data frame consisting of multiple data points with specific geocoordinates (latitude and longitude). I'm looking to create a choropleth-style world map where geographical regions are shaded according to how many data points fall within the b...
more »
## How to calculate 3D distance (including altitude) between two points in GeoDjango
Prologue: This is a question arising often in SO: 3d distance calculations with GeoDjango Calculating distance between two points using latitude longitude and altitude (elevation) Distance between two 3D point in geodjango (postgis) I wanted to ...
more »
## Great circles Map within a country in R
I am working on a commuter travel Pattern (Origin - Destination) Flow Map in R. The Data i have is daily transaction of commuters (Date,Card,Entry_lat,Entry_Long,Exit_Lat,Exit_Long). The Travel Path might be similar (as they commute to work). I nee...
more »
## r raster cover function - error with landsat stacks
I'm running into an unusual issue with the cover function in R. We are trying to fill cloudy pixels with values from another layer. I can make it work just fine with stacks as follows - library(raster) r1 <- raster(ncols=36, nrows=18) r1[] <- ...
more »
## Flow map(Travel Path) Using Lat and Long in R
I am trying to plot flow map (for singapore) . I have Entry(Lat,Long) and Exit (Lat,long). I am trying to map the flow from entry to exit in singapore map. structure(list(token_id = c(1.12374e+19, 1.12374e+19, 1.81313e+19, 1.85075e+19, 1.30752e+19,...
more »
## How to add a temporary field for calculation in django model?
I am developing a GIS based django app in which i am allocating the nearest exam center to a student based on his current location. I have my centre model like this class Center(models.Model): name = models.CharField(max_length=500) city = ...
more »
## Does SQL Server support measuring distance via routes?
I've been experimenting with SQL Server spatial functions and seen that the STDistance method allows you to determine the distance between two geometries. Is there any functionality that allows you to measure distance between two points via roads, li...
more »
## Convert GeoJSON polygon to point before drawing in Leaflet Map
I have a GeoJSON dataset which has points and polygons. I have a simple Leaflet code which reads them into a map, like this: var MyLayer = new L.GeoJSON.AJAX("/UrbanSyntax/Desarrollo/twitter /data/boxtest.json", { pointToLayer: function(feature,...
more »
## How can calculate the real distance between two points with GeoDjango?
from django.contrib.gis.geos import Point p1 = Point(36.74851779201058, -6.429006806692149, srid=4326) p2 = Point(37.03254161520977, -8.98366068931684, srid=4326) p1.distance(p2) Out: 2.5703941316759376 But what is the unit of this float number? ...
more »
## Querying json in postgres
I have to extract data from a json file who contains spatial information. The content of this file is {"vertices":[{"lat":46.744628268759314,"lon":6.569952920654968}, {"lat":46.74441692818192,"lon":6.570487107359068}, {"lat":46.744...
more »
## Avoiding hoizontal lines and crazy shapes when plotting maps in ggplot2
I want a plot of an area, such as Latin America, using the world shape file from IPUMSI... https://international.ipums.org/international/resources/gis/IPUMSI_world.zip ... I will add some more IPUMS districts later so I really want to use this as m...
more »
## Unable to get the location bound of Sketchup model using Sketchup Ruby API
I have a Sketchup 3d model that is geo-located. I can get the geo-location of the model as follows :- latitude = Sketchup.active_model.attribute_dictionaries["GeoReference"]["Latitude"] longitude = Sketchup.active_model.attribute_dictionaries["GeoR...
more »
| 1,024
| 4,153
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39
|
latest
|
en
| 0.864964
|
https://www.myessaygram.com/fin303-exam-type-questions-final-exam/
| 1,627,766,008,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00429.warc.gz
| 932,544,272
| 13,446
|
# FIN303 Exam-type questions For Final exam
Question
FIN303
Exam-type questions
Don't use plagiarized sources. Get Your Custom Essay on
FIN303 Exam-type questions For Final exam
Just from \$13/Page
For Final exam
Chapter 9
1. Stock A has a required return of 10 percent. Its dividend is expected to grow at a constant rate of 7 percent per year. Stock B has a required return of 12 percent. Its dividend is expected to grow at a constant rate of 9 percent per year. Stock A has a price of \$25 per share, while Stock B has a price of \$40 per share. Which of the following statements is most correct?
a. The two stocks have the same dividend yield.
b. If the stock market were efficient, these two stocks should have the same price.
c. If the stock market were efficient, these two stocks should have the same expected return.
d. Statements a and c are correct.
2. If D1 = \$2.00, g (which is constant) = 6%, and P0 = \$40, what is the stock’s expected capital gains yield for the coming year?
a. 5.2%
b. 5.4%
c. 5.6%
d. 6.0%
3. The Lashgari Company is expected to paya dividend of \$1 per share at the end of the year, and that dividend is expected to grow at a constant rate of 5% per year in the future. The company’s beta is 1.2, the market risk premium is 5%, and the risk-free rate is 3%. What is the company’s current stock price?
a. \$15.00
b. \$20.00
c. \$25.00
d. \$30.00
4. McKenna Motors is expected to pay a \$1.00 per-share dividend at the end of the year (D1 = \$1.00). The stock sells for \$20 per share and its required rate of return is 11 percent. The dividend is expected to grow at a constant rate, g, forever. What is the growth rate, g, for this stock?
a. 5%
b. 6%
c. 7%
d. 8%
5. The last dividend paid by Klein Company was \$1.00. Klein’s growth rate is expected to be a constant 5 percent for 2 years, after which dividends are expected to grow at a rate of 10 percent forever. Klein’s required rate of return on equity (ks) is 12 percent. What is the current price of Klein’s common stock?
a. \$21.00
b. \$33.33
c. \$42.25
d. \$50.16
6. You must estimate the intrinsic value of Gallovits Technologies’ stock. Gallovits’s end-of-year free cash flow (FCF) is expected to be \$25 million, and it is expected to grow at a constant rate of 8.5% a year thereafter. The company’s WACC is 11%. Gallovits has \$200 million of long-term debt plus preferred stock, and there are 30 million shares of common stock outstanding. What is Gallovits’ estimated intrinsic value per share of common stock?
a. \$22.67
b. \$24.00
c. \$25.33
d. \$26.67
Chapter 10
7. Campbell Co. is trying to estimate its weighted average cost of capital (WACC). Which of the following statements is most correct?
a. The after-tax cost of debt is generally cheaper than the after-tax cost of equity.
b. Since retained earnings are readily available, the cost of retained earnings is generally lower than the cost of debt.
c. The after-tax cost of debt is generally more expensive than the before-tax cost of debt.
d. Statements a and c are correct.
8. Wyden Brothers has no retained earnings. The company uses the CAPM to calculate the cost of equity capital. The company’s capital structure consists of common stock, preferred stock, and debt. Which of the following events will reduce the company’s WACC?
a. A reduction in the market risk premium.
b. An increase in the flotation costs associated with issuing new common stock.
c. An increase in the company’s beta.
d. An increase in expected inflation.
9. Dick Boe Enterprises, an all-equity firm, has a corporate beta coefficient of 1.5. The financial manager is evaluating a project with an expected return of 21 percent, before any risk adjustment. The risk-free rate is 10 percent, and the required rate of return on the market is 16 percent. The project being evaluated is riskier than Boe’s average project, in terms of both beta risk and total risk. Which of the following statements is most correct?
a. The project should be accepted since its expected return (before risk adjustment) is greater than its required return.
b. The project should be rejected since its expected return (before risk adjustment) is less than its required return.
c. The accept/reject decision depends on the risk-adjustment policy of the firm. If the firm’s policy were to reduce a riskier-than-average project’s expected return by 1 percentage point, then the project should be accepted.
d. Riskier-than-average projects should have their expected returns increased to reflect their added riskiness. Clearly, this would make the project acceptable regardless of the amount of the adjustment.
10. Conglomerate Inc. consists of 2 divisions of equal size, and Conglomerate is 100 percent equity financed. Division A’s cost of equity capital is 9.8 percent, while Division B’s cost of equity capital is 14 percent. Conglomerate’s composite WACC is 11.9 percent. Assume that all Division A projects have the same risk and that all Division B projects have the same risk. However, the projects in Division A are not the same risk as those in Division B. Which of the following projects should Conglomerate accept?
a.Division A project with an 11 percent return.
b. Division B project with a 12 percent return.
c. Division B project with a 13 percent return.
d. Statements a and c are correct.
11. Billick Brothers is estimating its WACC. The company has collected the following information:
· Its capital structure consists of 40 percent debt and 60 percent common equity.
· The company has 20-year bonds outstanding with a 9 percent annual coupon that are trading at par.
· The company’s tax rate is 40 percent.
· The risk-free rate is 5.5 percent.
· The market risk premium is 5 percent.
· The stock’s beta is 1.4.
What is the company’s WACC?
a. 9.71%
b. 9.66%
c. 8.31%
d. 11.18%
12. Flaherty Electric has a capital structure that consists of 70 percent equity and 30 percent debt. The company’s long-term bonds have a before-tax yield to maturity of 8.4 percent. The company uses the DCF approach to determine the cost of equity. Flaherty’s common stock currently trades at \$40.5 per share. The year-end dividend (D1) is expected to be \$2.50 per share, and the dividend is expected to grow forever at a constant rate of 7 percent a year. The company estimates that it will have to issue new common stock to help fund this year’s projects. The company’s tax rate is 40 percent. What is the company’s weighted average cost of capital, WACC?
13. Hamilton Company’s 8 percent coupon rate, quarterly payment, \$1,000 par value bond, which matures in 20 years, currently sells at a price of \$686.86. The company’s tax rate is 40 percent. What is the firm’s component cost of debt for purposes of calculating the WACC?
a. 3.05%
b. 7.32%
c. 7.36%
d. 12.20%
14. For a typical firm, which of the following is correct? All rates are after taxes, and assume the firm operates at its target capital structure. Note. d is for debt; e is for equity
a. rd > re > WACC.
b. re > rd > WACC.
c. WACC > re > rd.
d. re > WACC > rd.
Chapter 11
15. Which of the following statements is most correct?
a. The NPV method assumes that cash flows will be reinvested at the cost of capital, while the IRR method assumes reinvestment at the IRR.
b. The NPV method assumes that cash flows will be reinvested at the risk-free rate, while the IRR method assumes reinvestment at the IRR.
c. The NPV method assumes that cash flows will be reinvested at the cost of capital, while the IRR method assumes reinvestment at the risk-free rate.
d. The NPV method does not consider the inflation premium.
16. A major disadvantage of the payback period is that it
a. Is useless as a risk indicator.
b. Ignores cash flows beyond the payback period.
c. Does not directly account for the time value of money.
d. Statements b and c are correct.
17. Which of the following statements is most correct?
a. If a project’s internal rate of return (IRR) exceeds the cost of capital, then the project’s net present value (NPV) must be positive.
b. If Project A has a higher IRR than Project B, then Project A must also have a higher NPV.
c. The IRR calculation implicitly assumes that all cash flows are reinvested at a rate of return equal to the cost of capital.
d. Statements a and c are correct.
18. The Seattle Corporation has been presented with an investment opportunity that will yield cash flows of \$30,000 per year in Years 1 through 4, \$35,000 per year in Years 5 through 9, and \$40,000 in Year 10. This investment will cost the firm \$150,000 today, and the firm’s cost of capital is 10 percent. Assume cash flows occur evenly during the year, 1/365th each day. What is the payback period for this investment?
a. 5.23 years
b. 4.86 years
c. 4.00 years
d. 6.12 years
19. Coughlin Motors is considering a project with the following expected cash flows:
Project
Year Cash Flow
0 -\$700 million
1 200 million
2 370 million
3 225 million
4 700 million
The project’s WACC is 10 percent. What is the project’s discounted payback?
a. 3.15 years
b. 4.09 years
c. 1.62 years
d. 3.09 years
20. As the director of capital budgeting for Denver Corporation, you are evaluating two mutually exclusive projects with the following net cash flows:
Project X Project Z
Year Cash Flow Cash Flow
0 -\$100,000 -\$100,000
1 50,000 10,000
2 40,000 30,000
3 30,000 40,000
4 10,000 60,000
If Denver’s cost of capital is 15 percent, which project would you choose?
a. Neither project.
b. Project X, since it has the higher IRR.
c. Project Z, since it has the higher NPV.
d. Project X, since it has the higher NPV.
21. Your company is choosing between the following non-repeatable, equally risky, mutually exclusive projects with the cash flows shown below. Your cost of capital is 10 percent. How much value will your firm sacrifice if it selects the project with the higher IRR?
22. Assume a project has normal cash flows. All else equal, which of the following statements is CORRECT?
a. The project’s IRR increases as the WACC declines.
b. The project’s NPV increases as the WACC declines.
c. The project’s MIRR is unaffected by changes in the WACC.
d. The project’s regular payback increases as the WACC declines.
23. Which of the following statements is CORRECT?
a. One defect of the IRR method is that it does not take account of cash flows over a project’s full life.
b. One defect of the IRR method is that it does not take account of the time value of money.
c. One defect of the IRR method is that it does consider the time value of money.
d. One defect of the IRR method is that it assumes that the cash flows to be received from a project can be reinvested at the IRR itself, and that assumption is often not valid.
Pages (550 words)
Approximate price: -
Why Work with Us
Top Quality and Well-Researched Papers
We always make sure that writers follow all your instructions precisely. You can choose your academic level: high school, college/university or professional, and we will assign a writer who has a respective degree.
We have a team of professional writers with experience in academic and business writing. Many are native speakers and able to perform any task for which you need help.
Free Unlimited Revisions
If you think we missed something, send your order for a free revision. You have 10 days to submit the order for review after you have received the final document. You can do this yourself after logging into your personal account or by contacting our support.
Prompt Delivery and 100% Money-Back-Guarantee
All papers are always delivered on time. In case we need more time to master your paper, we may contact you regarding the deadline extension. In case you cannot provide us with more time, a 100% refund is guaranteed.
Original & Confidential
We use several writing tools checks to ensure that all documents you receive are free from plagiarism. Our editors carefully review all quotations in the text. We also promise maximum confidentiality in all of our services.
Our support agents are available 24 hours a day 7 days a week and committed to providing you with the best customer experience. Get in touch whenever you need any assistance.
Try it now!
## Calculate the price of your order
Total price:
\$0.00
How it works?
Fill in the order form and provide all details of your assignment.
Proceed with the payment
Choose the payment system that suits you most.
Our Services
No need to work on your paper at night. Sleep tight, we will cover your back. We offer all kinds of writing services.
## Essay Writing Service
No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system.
| 3,167
| 12,932
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.953125
| 3
|
CC-MAIN-2021-31
|
latest
|
en
| 0.931284
|
https://tuftsenigma.org/urbrainy-year-4.html
| 1,603,623,043,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107888931.67/warc/CC-MAIN-20201025100059-20201025130059-00258.warc.gz
| 553,106,345
| 8,024
|
# Urbrainy Year 4
### Statistics this two week block covers interpreting pictograms bar charts and line graphs with plenty of fun graphs to find information from and takes us into the second half of the summer term.
Urbrainy year 4. Go to maths mastery. All this will be based on sound mental methods of calculating. 3 45 as comments off on year 4 money continue reading. It is a time for children to really master understanding pounds and pence and the use of decimals.
One hundredth of 2 000 is equivalent to 2 000 divided by 100. I have recommended your site to so many people that some of them suspect i am working for your company. Number and place value in year 4 including maths mastery resources a fantastic selection of worksheets has just been published for year 4 covering. By the end of year 4 children will be using standard written methods of addition and subtraction and will be developing their written methods of multiplication and division.
They will have a sound knowledge of all times tables up to 10x10 and use this to work out corresponding division facts. Posted on august 25 2019. By the end of the year they should be confident with writing money using decimal notation e g. Year 3 age 7 8 year 4 age 8 9 year 5 age 9 10 year 6 age 10 11 ks1 maths sats ks2 maths sats booster times tables further resources.
They should also be shown the connection between hundredths and tenths and that hundredths arise by dividing tenths by ten e g 0 5 divided by ten. All this will be based on sound mental methods of calculating. Year 4 money in year 4 i cannot think of a more important topic in maths than money. In year 4 children will begin to work with hundredths understanding that finding one hundredth is equivalent to dividing by 100 e g.
I consider your site a complete godsend no exaggeration intended here. Year 4 maths mastery. Subscribe to our newsletter. I have been a great fan of urbrainy for about two years now.
It will be an on line on screen assessment which will take less than 5 minutes to complete. Finding the value of digits in numbers up to 4 digits rounding to the nearest 10 including using roman numerals rounding to the nearest 100. Year 3 age 7 8 year 4 age 8 9 year 5 age 9 10 year 6 age 10 11 ks1 maths sats ks2 maths. The year 4 programme of study states.
It will be automatically scored and results will be available to schools. Year 4 maths mastery for summer term. They will have a sound knowledge of all times tables up to 12x12 and use this to work out corresponding division facts. In what s new year 4.
### Urbrainy Com Shapes Quarter
Source : pinterest.com
| 608
| 2,622
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.15625
| 3
|
CC-MAIN-2020-45
|
latest
|
en
| 0.935784
|
https://codereview.stackexchange.com/questions/190197/implemnting-a-trie-data-structures-problem-from-hackerrank-using-python3
| 1,653,690,784,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00052.warc.gz
| 228,610,231
| 69,365
|
# Implemnting a Trie Data structures problem from HackerRank using Python3
I'm trying to write a better code in python. I don't know where to start checking. I want to make sure that I do write a best practice code.
This is the second problem in Hackerrank trie data structure:
Given N strings. Each string contains only lowercase letters from (both inclusive). The set of strings is said to be GOOD SET if no string is prefix of another string else, it is BAD SET. (If two strings are identical, they are considered prefixes of each other.)
For example, aab, abcde, aabcd is BAD SET because aab is prefix of aabcd.
Print GOOD SET if it satisfies the problem requirement. Else, print BAD SET and the first string for which the condition fails.
Input Format
First line contains , the number of strings in the set. Then next lines follow, where line contains string.
Output Format
Output GOOD SET if the set is valid. Else, output BAD SET followed by the first string for which the condition fails.
Sample Input00
7
aab
defgab
abcde
aabcde
cedaaa
bbbbbbbbbb
Sample Output00
BAD SET
aabcde
Sample Input01
4
aab
aac
aacghgh
aabghgh
Sample Output01
BAD SET
aacghgh
from sys import stdin
class Node:
def __init__(self,char):
self.character = char
self.children = {}
self.counter = 0
self.end_word = False
class Trie:
def __init__(self):
self.root = Node('*')
current = self.root
fail = False
for char in word:
if char not in current.children:
new_node = Node(char)
current.children[char] = new_node
current = new_node
new_node.counter += 1
else:
current = current.children[char]
current.counter += 1
if current.end_word:
fail = True
current.end_word = True
# first word > second word : second word is prefix of first word
if current.counter >=2:
fail = True
return fail
if __name__ == "__main__":
tree = Trie()
for i in n[1:]:
i = i.strip()
print(i)
break
print("GOOD SET")
• You could run autopep8 for minor style corrections (e.g. whitespaces).
• Naming variables correctly is always extremely important, especially in a dynamic language, without indications about object type:
• Your trie is a Trie, not just a tree.
• n sounds like an integer. It could be called lines.
• i also sounds like an integer. It could be line or word.
• Using for/else syntax, you can refactor your main method without any flag.
if __name__ == "__main__":
trie = Trie()
for line in lines:
word = line.strip()
print(word)
break
else:
print("GOOD SET")
• If you wish, you could replace the 3 last lines of add by return fail or current.counter >= 2
I think the only change I would make is removing node.counter, and instead detecting if a previous word is a prefix of the current word by checking if current.fail==True.
• Thank you for your raply. May I know Why did you mention taht there are Performance Issues? Mar 22, 2018 at 20:44
• oops, I combined 2 separate (and unrelated) questions in my head, and pulled out that you thought this was a bottleneck somehow. Edited. Mar 22, 2018 at 21:01
There is a lot of repetition in your add method. This can be simplified, ditching the fail flag to
def add_unless_prefix(self,word):
current = self.root
for char in word:
if char not in current.children:
current.children[char] = Node(char)
current = current.children[char]
current.counter += 1
if current.end_word:
return True
current.end_word = True
return current.counter >=2
This can be simplified a bit using dict.setdefault.
if char not in current.children:
current.children[char] = Node(char)
current = current.children[char]
can become current = current.children.setdefault(char, Node(char))
• I also tried to remove the fail flag, but .... I failed. By returning early, you prevent word from being added to the trie. return fail was just some extra piece of information, not the main purpose of the add method. At the very least, your new method should be called differently, e.g. add_unless_prefix. Mar 25, 2018 at 19:52
• Oops, small bug. That was meant to be return True, signalling that it matches a prefix Mar 26, 2018 at 6:51
• It still doesn't add the word to the trie, though. Given the name of the class and method, it still sounds like a bug. Mar 26, 2018 at 6:57
• this isn't mean to be a complete Trie. This is meant to detect whether there is a prefix, so in the context of this HackerRank issue, this suffices. Mar 26, 2018 at 7:08
• Fine, then you should change the method name to make it clear. Mar 26, 2018 at 7:20
| 1,113
| 4,457
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.546875
| 3
|
CC-MAIN-2022-21
|
longest
|
en
| 0.779295
|
https://www.mankier.com/3/ssyr2.f
| 1,726,535,338,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00527.warc.gz
| 814,876,636
| 3,427
|
# ssyr2.f - Man Page
BLAS/SRC/ssyr2.f
## Synopsis
### Functions/Subroutines
subroutine ssyr2 (uplo, n, alpha, x, incx, y, incy, a, lda)
SSYR2
## Function/Subroutine Documentation
### subroutine ssyr2 (character uplo, integer n, real alpha, real, dimension(*) x, integer incx, real, dimension(*) y, integer incy, real, dimension(lda,*) a, integer lda)
SSYR2
Purpose:
``` SSYR2 performs the symmetric rank 2 operation
A := alpha*x*y**T + alpha*y*x**T + A,
where alpha is a scalar, x and y are n element vectors and A is an n
by n symmetric matrix.```
Parameters
UPLO
``` UPLO is CHARACTER*1
On entry, UPLO specifies whether the upper or lower
triangular part of the array A is to be referenced as
follows:
UPLO = 'U' or 'u' Only the upper triangular part of A
is to be referenced.
UPLO = 'L' or 'l' Only the lower triangular part of A
is to be referenced.```
N
``` N is INTEGER
On entry, N specifies the order of the matrix A.
N must be at least zero.```
ALPHA
``` ALPHA is REAL
On entry, ALPHA specifies the scalar alpha.```
X
``` X is REAL array, dimension at least
( 1 + ( n - 1 )*abs( INCX ) ).
Before entry, the incremented array X must contain the n
element vector x.```
INCX
``` INCX is INTEGER
On entry, INCX specifies the increment for the elements of
X. INCX must not be zero.```
Y
``` Y is REAL array, dimension at least
( 1 + ( n - 1 )*abs( INCY ) ).
Before entry, the incremented array Y must contain the n
element vector y.```
INCY
``` INCY is INTEGER
On entry, INCY specifies the increment for the elements of
Y. INCY must not be zero.```
A
``` A is REAL array, dimension ( LDA, N )
Before entry with UPLO = 'U' or 'u', the leading n by n
upper triangular part of the array A must contain the upper
triangular part of the symmetric matrix and the strictly
lower triangular part of A is not referenced. On exit, the
upper triangular part of the array A is overwritten by the
upper triangular part of the updated matrix.
Before entry with UPLO = 'L' or 'l', the leading n by n
lower triangular part of the array A must contain the lower
triangular part of the symmetric matrix and the strictly
upper triangular part of A is not referenced. On exit, the
lower triangular part of the array A is overwritten by the
lower triangular part of the updated matrix.```
LDA
``` LDA is INTEGER
On entry, LDA specifies the first dimension of A as declared
in the calling (sub) program. LDA must be at least
max( 1, n ).```
Author
Univ. of Tennessee
Univ. of California Berkeley
NAG Ltd.
Further Details:
``` Level 2 Blas routine.
-- Written on 22-October-1986.
Jack Dongarra, Argonne National Lab.
Jeremy Du Croz, Nag Central Office.
Sven Hammarling, Nag Central Office.
Richard Hanson, Sandia National Labs.```
Definition at line 146 of file ssyr2.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code.
## Referenced By
The man page ssyr2(3) is an alias of ssyr2.f(3).
Tue Nov 28 2023 12:08:41 Version 3.12.0 LAPACK
| 854
| 3,072
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.65625
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.748075
|
http://mathoverflow.net/feeds/question/27008
| 1,371,704,503,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368710299158/warc/CC-MAIN-20130516131819-00041-ip-10-60-113-184.ec2.internal.warc.gz
| 157,646,438
| 2,392
|
Polygon Chain - Conversion to non-crossing while preserving shape? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:01:41Z http://mathoverflow.net/feeds/question/27008 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27008/polygon-chain-conversion-to-non-crossing-while-preserving-shape Polygon Chain - Conversion to non-crossing while preserving shape? Monte 2010-06-04T04:20:58Z 2010-06-04T07:15:59Z <p>I have polygon chains similar to the following...</p> <p><a href="http://upload.wikimedia.org/wikipedia/commons/thumb/6/62/Self_crossed_polygonal_chain.svg/220px-Self_crossed_polygonal_chain.svg.png" rel="nofollow">http://upload.wikimedia.org/wikipedia/commons/thumb/6/62/Self_crossed_polygonal_chain.svg/220px-Self_crossed_polygonal_chain.svg.png</a></p> <p>...given the chain in the image, how would I go about calculating a chain that defines <em>the same shape</em> but <em>without</em> crossing paths?</p> <p>Specifically, in the case of the image, the result I want looks like this:</p> <p>A1,<br> A2,<br> Intersect between A2 and A3,<br> Intersect between A3 and A4,<br> A4,<br> A5,<br> Intersect between A3 and A4,<br> A3,<br> Intersect between A3 and A2,<br> A6</p> <p>I'm looking for an algorithm to accomplish this for any chain, but I'm not sure what I'm trying to do is even called, which makes searching for a solution tricky. </p> <p>If there's a name for what I'm trying to do it would be of great help to know it. </p> <p>Thank you for any assistance!</p> http://mathoverflow.net/questions/27008/polygon-chain-conversion-to-non-crossing-while-preserving-shape/27013#27013 Answer by Jason Dyer for Polygon Chain - Conversion to non-crossing while preserving shape? Jason Dyer 2010-06-04T05:22:04Z 2010-06-04T05:22:04Z <p>While I've never heard of your exact algorithm being done before, the sweep line algorithm can be used to detect all the intersections quickly. (As mentioned <a href="http://compgeom.cs.uiuc.edu/~jeffe/teaching/algorithms/notes/xo-sweepline.pdf" rel="nofollow">here</a>, for instance.)</p> http://mathoverflow.net/questions/27008/polygon-chain-conversion-to-non-crossing-while-preserving-shape/27024#27024 Answer by David Eppstein for Polygon Chain - Conversion to non-crossing while preserving shape? David Eppstein 2010-06-04T07:15:59Z 2010-06-04T07:15:59Z <p>There is a fair amount of research on algorithms for this problem in the computational geometry, under the name "arrangement of line segments" (the line segments are the ones in the polygonal chain, and the arrangement is the planar graph you get when you place vertices at all the endpoints and crossing points).</p> <p>In practice, Jason Dyer's suggestion of using a sweep line algorithm is a good one. In theory, there are better algorithms: for \$n\$ line segments with \$k\$ crossing points, the sweep line gives you time \$O((n + k)\log n)\$, and some other methods lead to \$O(n\log n+k)\$ running times; see e.g. <a href="http://dx.doi.org/10.1007/BF02187740" rel="nofollow">Clarkson and Shor, Discrete Comput. Geom. 1989</a>.</p> <p>When the line segments themselves form a connected nonplanar graph (as in your case, where they form a path graph) even faster algorithms are possible (at most an iterated log factor away from linear time, and linear whenever \$k\$ is not too close to \$n\$): see my paper <a href="http://arxiv.org/abs/0812.0893" rel="nofollow">"Linear-time algorithms for geometric graphs with sublinearly many crossings"</a> and its references. But I think these methods are too complicated to be practically useful.</p>
| 1,003
| 3,615
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.109375
| 3
|
CC-MAIN-2013-20
|
latest
|
en
| 0.745749
|
https://www.annytab.com/neural-network-classification-in-python/
| 1,726,452,473,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00291.warc.gz
| 591,671,734
| 22,962
|
Skip to content
Neural Network Classification in Python
I am going to perform neural network classification in this tutorial. I am using a generated data set with spirals, the code to generate the data set is included in the tutorial. I am going to train and evaluate two neural network models in Python, an MLP Classifier from scikit-learn and a custom model created with keras functional API.
A neural network tries to depict an animal brain, it has connected nodes in three or more layers. A neural network includes weights, a score function and a loss function. A neural network learns in a feedback loop, it adjusts its weights based on the results from the score function and the loss function. A simple neural network includes three layers, an input layer, a hidden layer and an output layer. More than 3 layers is often referred to as deep learning.
Keras functional API can be used to build very complex deep learning models with multiple layers, the image above is a plot of the model used in this tutorial. I takes a 2 dimensional array as input (x,y), the input layer is connected to a hidden layer with 64 nodes (you can test with more and less) and the output layer make predictions for 3 classes. Each class is assigned a probability, and I select the class with the highest probability as my prediction.
Data set and libraries
I generate a data set with three spirals by using the code below, it is a non-linear data set that a linear classifier has difficulties to learn. I save the data set to a `.csv` file. You need to install Graphviz if you want to plot a model like the one above, you also need to add a path to its binaries as environment variables. I am using the following libraries: pandas, joblib, numpy, matplotlib, keras, tensorflow (tensorflow-gpu) and scikit-learn.
``````import numpy
import pandas
import matplotlib.pyplot as plt
# Generate a data set with spirals
# http://cs231n.github.io/neural-networks-case-study/
def generate_spirals():
N = 400 # number of points per class
D = 2 # dimensionality
K = 3 # number of classes
data = numpy.zeros((N*K,D)) # data matrix (each row = single example)
labels = numpy.zeros(N*K, dtype='uint8') # class labels
for j in range(K):
ix = range(N*j,N*(j+1))
r = numpy.linspace(0.0,1,N) # radius
t = numpy.linspace(j*4,(j+1)*4,N) + numpy.random.randn(N)*0.2 # theta
data[ix] = numpy.c_[r*numpy.sin(t), r*numpy.cos(t)]
labels[ix] = j
# Save to a csv file
f = open('files\\spirals.csv', 'w')
f.write('x,y,label\n')
for i in range(len(labels)):
f.write(str(data[i][0]) + ',' + str(data[i][1]) + ',' + str(labels[i]) + '\n')
f.close()
# Visualize data set
def visualize_data_set():
# Load data set
ds = pandas.read_csv('files\\spirals.csv')
# Print first 5 rows in data set
print('--- First 5 rows ---')
print(ds.head())
# Print the shape
print('\n--- Shape of data set ---')
print(ds.shape)
# Print class distribution
print('\n--- Class distribution ---')
print(ds.groupby('label').size())
# Visualize data set
figure = plt.figure(figsize = (12, 8))
figure.suptitle('Spirals', fontsize=16)
grouped_dataset = ds.groupby('label')
labels = ['0', '1', '2']
for i, group in grouped_dataset:
plt.scatter(group['x'], group['y'], label=labels[int(i)])
plt.ylabel('y')
plt.xlabel('x')
plt.legend()
#plt.show()
plt.savefig('plots\\spirals.png')
# Generate spirals
generate_spirals()
# Visualize data set
visualize_data_set()``````
Visualize data set
The data set is well balanced by design, it has 1 200 data points and 3 classes (400 per class). A linear classifier would be very bad in classifying this data set as it is impossible to divide data points by lines as you can see in the image below. It is a 33.33 % probabilty (400/1200) to classify a data point correctly and this is our baseline performance, our models need to perform better than this.
``````--- First 5 rows ---
x y label
0 0.000000 0.000000 0
1 -0.000156 0.002501 0
2 -0.001956 0.004615 0
3 0.000877 0.007468 0
4 0.004620 0.008897 0
--- Shape of data set ---
(1200, 3)
--- Class distribution ---
label
0 400
1 400
2 400
dtype: int64``````
MLP Classifier
MLP Classifier is a neural network classifier in scikit-learn and it has a lot of parameters to fine-tune. I am using default parameters when I train my model. I load the data set, slice it into data and labels and split the set in a training set and a test set. I am making sure that the split will be the same each time by using a random state and I am making sure that sets is balanced after the split. The code and the output from the evaluation process is shown below.
``````# Import libraries
import pandas
import joblib
import numpy as np
import matplotlib.pyplot as plt
import sklearn.model_selection
import sklearn.metrics
import sklearn.neural_network
# Train and evaluate
def train_and_evaluate(X_train, Y_train, X_test, Y_test):
# Create a model
model = sklearn.neural_network.MLPClassifier(hidden_layer_sizes=(100, ), activation='relu', solver='adam',
alpha=0.0001, batch_size='auto', learning_rate='constant', learning_rate_init=0.001, power_t=0.5,
max_iter=1000, shuffle=True, random_state=None, tol=0.0001, verbose=False, warm_start=False, momentum=0.9,
nesterovs_momentum=True, early_stopping=False, validation_fraction=0.1, beta_1=0.9, beta_2=0.999, epsilon=1e-08,
n_iter_no_change=10)
# Train the model on the whole data set
model.fit(X_train, Y_train)
# Save the model (Make sure that the folder exists)
joblib.dump(model, 'models\\mlp_classifier.jbl')
# Evaluate on training data
print('\n-- Training data --')
predictions = model.predict(X_train)
accuracy = sklearn.metrics.accuracy_score(Y_train, predictions)
print('Accuracy: {0:.2f}'.format(accuracy * 100.0))
print('Classification Report:')
print(sklearn.metrics.classification_report(Y_train, predictions))
print('Confusion Matrix:')
print(sklearn.metrics.confusion_matrix(Y_train, predictions))
print('')
# Evaluate on test data
print('\n---- Test data ----')
predictions = model.predict(X_test)
accuracy = sklearn.metrics.accuracy_score(Y_test, predictions)
print('Accuracy: {0:.2f}'.format(accuracy * 100.0))
print('Classification Report:')
print(sklearn.metrics.classification_report(Y_test, predictions))
print('Confusion Matrix:')
print(sklearn.metrics.confusion_matrix(Y_test, predictions))
# Plot the classifier
def plot_classifier(X, Y):
# Load the model
model = joblib.load('models\\mlp_classifier.jbl')
# Calculate
h = 0.02
x_min, x_max = X[:, 0].min() - 1, X[:, 0].max() + 1
y_min, y_max = X[:, 1].min() - 1, X[:, 1].max() + 1
xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h))
# Make predictions
Z = model.predict(np.c_[xx.ravel(), yy.ravel()])
Z = Z.reshape(xx.shape)
# Plot diagram
fig = plt.figure(figsize = (12, 8))
plt.contourf(xx, yy, Z, cmap='ocean', alpha=0.25)
plt.contour(xx, yy, Z, colors='w', linewidths=0.4)
plt.scatter(X[:, 0], X[:, 1], c=Y, s=40, cmap='Spectral')
plt.xlim(xx.min(), xx.max())
plt.ylim(yy.min(), yy.max())
plt.savefig('plots\\mlp_classifier.png')
# The main entry point for this module
def main():
# Load data set (includes header values)
dataset = pandas.read_csv('files\\spirals.csv')
# Slice data set in data and labels (2D-array)
X = dataset.values[:,0:2] # Data
Y = dataset.values[:,2].astype(int) # Labels
# Split data set in train and test (use random state to get the same split every time, and stratify to keep balance)
X_train, X_test, Y_train, Y_test = sklearn.model_selection.train_test_split(X, Y, test_size=0.2, random_state=5, stratify=Y)
# Make sure that data still is balanced
print('\n--- Class balance ---')
print(np.unique(Y_train, return_counts=True))
print(np.unique(Y_test, return_counts=True))
# Train and evaluate
train_and_evaluate(X_train, Y_train, X_test, Y_test)
# Plot classifier
plot_classifier(X, Y)
# Tell python to run main method
if __name__ == "__main__": main()``````
``````--- Class balance ---
(array([0, 1, 2]), array([320, 320, 320], dtype=int64))
(array([0, 1, 2]), array([80, 80, 80], dtype=int64))
-- Training data --
Accuracy: 99.38
Classification Report:
precision recall f1-score support
0 1.00 0.98 0.99 320
1 0.99 1.00 0.99 320
2 0.99 1.00 1.00 320
accuracy 0.99 960
macro avg 0.99 0.99 0.99 960
weighted avg 0.99 0.99 0.99 960
Confusion Matrix:
[[315 4 1]
[ 0 319 1]
[ 0 0 320]]
---- Test data ----
Accuracy: 99.17
Classification Report:
precision recall f1-score support
0 1.00 0.99 0.99 80
1 0.98 1.00 0.99 80
2 1.00 0.99 0.99 80
accuracy 0.99 240
macro avg 0.99 0.99 0.99 240
weighted avg 0.99 0.99 0.99 240
Confusion Matrix:
[[79 1 0]
[ 0 80 0]
[ 0 1 79]]``````
Keras Neural Network Classifier
I load and prepare the data set in the same way as before by splitting it into a training set and a test set, sets is still balanced after the split. A neural network model is built with keras functional API, it has one input layer, a hidden layer and an output layer. Keras functional API can be used to build very complex deep learning models with many layers. Training is evaluated on accuracy and the loss function is categorical crossentropy. The code and the evaluation output is shown below.
``````# Import libraries
import pandas
import numpy as np
import matplotlib.pyplot as plt
import sklearn.model_selection
import sklearn.metrics
import sklearn.preprocessing
import keras
# Train and evaluate
def train_and_evaluate(X_train, Y_train, X_test, Y_test):
# Create layers (Functional API)
inputs = keras.layers.Input(shape=(2,), dtype='float32', name='input_layer') # Input (2 dimensions)
outputs = keras.layers.Dense(64, activation='relu', name='hidden_layer')(inputs) # Hidden layer
outputs = keras.layers.Dense(3, activation='softmax', name='output_layer')(outputs) # Output layer (3 labels)
# Create a model from input layer and output layers
model = keras.models.Model(inputs=inputs, outputs=outputs, name='neural_network')
# Compile the model (binary_crossentropy if 2 classes)
model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
# Convert labels to categorical: categorical_crossentropy expects targets
# to be binary matrices (1s and 0s) of shape (samples, classes)
Y_binary = keras.utils.to_categorical(Y_train, num_classes=3, dtype='int')
# Train the model on the train set (output debug information)
model.fit(X_train, Y_binary, batch_size=1, epochs=100, verbose=1)
# Save the model (Make sure that the folder exists)
model.save('models\\keras_nn.h5')
# Evaluate on training data
print('\n-- Training data --')
predictions = model.predict(X_train)
accuracy = sklearn.metrics.accuracy_score(Y_train, np.argmax(predictions, axis=1))
print('Accuracy: {0:.2f}'.format(accuracy * 100.0))
print('Classification Report:')
print(sklearn.metrics.classification_report(Y_train, np.argmax(predictions, axis=1)))
print('Confusion Matrix:')
print(sklearn.metrics.confusion_matrix(Y_train, np.argmax(predictions, axis=1)))
print('')
# Evaluate on test data
print('\n---- Test data ----')
predictions = model.predict(X_test)
accuracy = sklearn.metrics.accuracy_score(Y_test, np.argmax(predictions, axis=1))
print('Accuracy: {0:.2f}'.format(accuracy * 100.0))
print('Classification Report:')
print(sklearn.metrics.classification_report(Y_test, np.argmax(predictions, axis=1)))
print('Confusion Matrix:')
print(sklearn.metrics.confusion_matrix(Y_test, np.argmax(predictions, axis=1)))
# Plot the classifier
def plot_classifier(X, Y):
# Load the model
model = keras.models.load_model('models\\keras_nn.h5')
# Plot model (Requires Graphviz)
#keras.utils.plot_model(model, show_shapes=True, rankdir='LR', expand_nested=True, to_file='plots\\keras_nn_model.png')
# Calculate
h = 0.02
x_min, x_max = X[:, 0].min() - 1, X[:, 0].max() + 1
y_min, y_max = X[:, 1].min() - 1, X[:, 1].max() + 1
xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h))
# Make predictions
Z = model.predict(np.c_[xx.ravel(), yy.ravel()])
Z = np.argmax(Z, axis=1)
Z = Z.reshape(xx.shape)
# Plot diagram
fig = plt.figure(figsize = (12, 8))
plt.contourf(xx, yy, Z, cmap='ocean', alpha=0.25)
plt.contour(xx, yy, Z, colors='w', linewidths=0.4)
plt.scatter(X[:, 0], X[:, 1], c=Y, s=40, cmap='Spectral')
plt.xlim(xx.min(), xx.max())
plt.ylim(yy.min(), yy.max())
plt.savefig('plots\\keras_nn_classifier.png')
# The main entry point for this module
def main():
# Load data set (includes header values)
dataset = pandas.read_csv('files\\spirals.csv')
# Slice data set in data and labels (2D-array)
X = dataset.values[:,0:2].astype(float) # Data
Y = dataset.values[:,2].astype(int) # Labels
# Split data set in train and test (use random state to get the same split every time, and stratify to keep balance)
X_train, X_test, Y_train, Y_test = sklearn.model_selection.train_test_split(X, Y, test_size=0.2, random_state=5, stratify=Y)
# Make sure that data still is balanced
print('\n--- Class balance ---')
print(np.unique(Y_train, return_counts=True))
print(np.unique(Y_test, return_counts=True))
# Train and evaluate
train_and_evaluate(X_train, Y_train, X_test, Y_test)
# Plot classifier
plot_classifier(X, Y)
# Tell python to run main method
if __name__ == "__main__": main()``````
``````--- Class balance ---
(array([0, 1, 2]), array([320, 320, 320], dtype=int64))
(array([0, 1, 2]), array([80, 80, 80], dtype=int64))
-- Training data --
Accuracy: 99.69
Classification Report:
precision recall f1-score support
0 1.00 0.99 1.00 320
1 1.00 1.00 1.00 320
2 0.99 1.00 1.00 320
accuracy 1.00 960
macro avg 1.00 1.00 1.00 960
weighted avg 1.00 1.00 1.00 960
Confusion Matrix:
[[318 0 2]
[ 0 319 1]
[ 0 0 320]]
---- Test data ----
Accuracy: 99.58
Classification Report:
precision recall f1-score support
0 1.00 1.00 1.00 80
1 0.99 1.00 0.99 80
2 1.00 0.99 0.99 80
accuracy 1.00 240
macro avg 1.00 1.00 1.00 240
weighted avg 1.00 1.00 1.00 240
Confusion Matrix:
[[80 0 0]
[ 0 80 0]
[ 0 1 79]]``````
Tags:
8 thoughts on “Neural Network Classification in Python”
1. Hi! I’m looking for a classification program for medical diagnoses. (eg diagnosis and classification of eye diseases). I need your advice I don’t know where to start.
2. thank you, it is very helpful. but i got error “ValueError: Error when checking input: expected input_layer to have shape (28,) but got array with shape (2,)” could you please help?
I have 28 input features and 6 output classes.
1. Hi!
Change the shape on the input layer to (28,):
```# Create layers (Functional API)
inputs = keras.layers.Input(shape=(28,), dtype='float32', name='input_layer') # Input (2 dimensions)
```
1. thank you for your reply. I have six output classes. but I got an error message
“categorical[np.arange(n), y] = 1
IndexError: index 6 is out of bounds for axis 1 with size 6”
could you please help me how to debug it?
1. Hi!
Have you set 6 on the output layer and converted y-labels to a matrix? Number of classes should be 6 in your case.
``` # Convert labels to categorical: categorical_crossentropy expects targets
# to be binary matrices (1s and 0s) of shape (samples, classes)
Y_binary = keras.utils.to_categorical(Y_train, num_classes=6, dtype='int')
```
3. very clear and helpful. awesome.
| 4,378
| 15,825
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.897937
|
http://lambdapage.org/30459/2nd-grade-daily-math-review-worksheets/lone-star-daily-rigor-math-staar-review-mathwarm-ups-free-with-regard-to-2nd-grade-daily-math-review-worksheets/
| 1,619,122,222,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618039604430.92/warc/CC-MAIN-20210422191215-20210422221215-00155.warc.gz
| 61,067,907
| 7,669
|
# lone star daily rigor math staar review mathwarm ups free with regard to 2nd grade daily math review worksheets
##### lone star daily rigor math staar review mathwarm ups free with regard to 2nd grade daily math review worksheets.
If you think that spelling words and recognizing the number is too complex, then you can try a simple worksheet that has main purpose of tracing skill. It’s a line tracing worksheet that can be used to achieve this skill.
If you think that your kids coloring process is not really tidy, then you can repeat the same exercise using the different pictures. It will help them to improve their coloring skill.
Playing music as the background for this activity is also can be fun. It helps the situation to be more cheerful. Your kid will also excite for the activity. Thing that you also need to prepare are coloring pencil or crayon.
This can be a creative solution that will make your kid enhance their imagination and creativity. This free activity is also good for redeem the boredom. During this session, you can play a cheerful music on the background. Ask your kid to dance together and they will enjoy for this activity.
March 10, 2021
March 8, 2021
March 10, 2021
March 6, 2021
### Color Blue Worksheets For Preschool
March 7, 2021
This can be very fun and make your kid enjoy to get more exercise. This learning progress, of course is recommended for upgrading kids’ soft motoric skills.
This is the very first step before your kid can write on a blank paper. Never push them to do so until they can manage to trace perfectly. You also need to make a summative evaluation by start counting the number together with kid.
It will also teach your kid to recognize the color. By using this way, there are more skills that can be achieved using a single tool. You also can be more creative by making a jingle during the tracing process. Your kid will see that this activity is fun and they will be encouraged to learn more about shapes, colors and traces.
For the advanced exercise, if you think that your kids have already got their line traced tidily, then you also can make the other line examples with more complex shapes.
March 5, 2021
March 4, 2021
March 11, 2021
March 9, 2021
### Decimal Computation Worksheets
March 3, 2021
Although this activity is free, you can still accompany your kid to finish the project. Give an argument about the color that they should use. For example, if the picture is about a scenery, then you can go by saying that the color of the sky is usually blue. Your kids will understand and start making a progress by coloring the sheet using their creative ideas.
Encourage them once they’ve done their first letter even they need an assistance. Repeat this process after they complete with the letter. Do not move to the next letter if they still need a helping-hand for the first letter.
Sometimes, your kids will get bored to trace the line or shape. If this happens, remember that you also can ask them to draw freely. For example, after finished tracing the objects, you can give your kids a blank paper and ask them to draw freely.
You also can make a creative process by also singing the alphabet. It helps your kids to memorize about the letters of the alphabet. Of course, it also can be a fun activity to do with your kid. After finished singing together, instruct your kid to follow your writing.
Labeled:
March 9, 2021
March 3, 2021
March 6, 2021
March 3, 2021
March 8, 2021
March 10, 2021
March 9, 2021
March 9, 2021
March 10, 2021
March 11, 2021
March 5, 2021
March 4, 2021
March 4, 2021
March 11, 2021
March 3, 2021
| 859
| 3,638
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17
|
latest
|
en
| 0.951637
|
https://library.fiveable.me/ap-psych/unit-1/stats-psych/study-guide/St9jOZWjJV5GZX6HA9Uo
| 1,713,676,824,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00050.warc.gz
| 335,147,444
| 144,976
|
or
Find what you need to study
Light
# 1.5 Statistical Analysis in Psychology
###### Attend a live cram event
Review all units live with expert teachers & students
## Types of Statistics
involves the use of numerical data to measure and describe the characteristics of groups, and this includes and variation. We'll be focusing on in this study guide! It does not involve making inferences about a population based on sample data.
, on the other hand, involves using statistical methods to make inferences about a population based on data. It allows you to draw conclusions about a population based on the characteristics of a sample. Specifically, it provides a way to see validity drawn from the results of the experiment🧪🔬.
Therefore, describe the data, while tell us what the data means.
## Summarizing Data
When one has a ton of data, how do they begin to go through it? Typically, a researcher would construct and interpret a graph with their data, and they use to do so. 📈
### Measures of Central Tendency
are statistical values that represent the center or typical value of a dataset. The three most commonly used are the , , and .
• The is the average of a set of scores. You can calculate the by summing all of the values in a dataset and dividing by the total number of values. The is sensitive to , or unusually large or small values, and can be affected by them.
• The is the middle score of distribution, separating the higher half of the data from the lower half. The is not affected by and can be a better measure of central tendency when the dataset contains .
• The is the most frequently recurring score in a dataset. A dataset can have one , more than one , or no . If two scores appear the most frequently, the distribution is bimodal. If three or more scores appear most frequently, the distribution is multimodal.
Let's practice calculating the three , , , and , using the following data set: 5, 10, 5, 7, 12, 15, 18
The easiest to spot is the : which value, if any, appears more often than others? Here, we can see 5 twice, so the of this dataset is 5.
Then, you may want to calculate the by adding all of these data values and dividing by the total. Since we have seven values, we have to divide by seven: (5 + 10 + 5 + 7 + 12 + 15 + 18)/7 = 10.286
The is the middle of the data set when the numbers are in order. Make sure you always put them in order!! If you do so here, you will find that the is 10.
### Measures of Variation
describe how spread out or dispersed the values in a dataset are. The most commonly used measure of variation is the , which is a measure of how much the values in a dataset deviate from the . It is basically used to assess how far the values are spread below and above the . A dataset with a low has values that are relatively close to the , while a dataset with a high has values that are more spread out.
Another, less complex, measure of variation you should be familiar with for this course is the of a dataset. is just the difference between the highest and lowest values in the dataset.
## Correlation
The is a statistical measure that describes the strength and direction of the relationship between two variables. It can from -1 to 1. A value of -1 indicates a strong negative relationship, a value of 1 indicates a strong positive relationship, and a value of 0 indicates no relationship.
You can simply think of it as a measure of how well two variables are correlated, and the closer it is to -1 or +1, the stronger the correlation.
### Positive Correlation
shows that as one variable increases ⬆️, the other variable increases ⬆️. For example, a positively correlated group may show that as height increases, weight increases as well.
Image courtesy of Expii
### Negative Correlation
shows that as one variable increases ⬆️, the other decreases ⬇️. An example of a could be how as the number of hours of sleep increases, tiredness decreases.
Image courtesy of Expii
### No Correlation
shows that there is no connection between the two variables. An example of could be IQ and how many pairs of pants an individual owns.
Image courtesy of Expii.
Remember, correlation does not imply causation, even if the is -1 or +1. You must run an experiment to prove there is causation.
## Skews
A is a breakdown of how the scores fall into different categories or ranges. There are several types of frequency distributions:
• 🔔A is a bell-shaped that is symmetrical about the . We'll delve into this deeper, but graph (b) on the image below shows a !
• 2️⃣A is a with two peaks. This occurs when the dataset has two distinct groups of values that occur with different frequencies.
• 👍A positively skewed distribution has a tail extending to the right (towards larger values). This occurs when the dataset has a few unusually large values that pull the to the right. Graph (a) in the image below is a , where the is greater than the .
• 👎A negatively skewed distribution has a tail extending to the left (towards smaller values). This occurs when the dataset has a few unusually small values that pull the to the left. Graph (c) in the image below is a , where the is greater than the .
It might be hard to remember which way the skew is. If the tail on the right is longer like it is in (a), then it's a skew to the right. If the tail on the left is longer like it is in (c), then it's a skew to the left.
Image courtesy of ResearchGate
## Normal Distributions
The normal curve, or (b) in the above image, is the only one you have to really be familiar with for this course. There are two important values that you should memorize: 68% and 95%.
This is a normal curve that includes data about intelligence📖. Basically, 68% of the data falls within one of the . Here, one is equivalent to 15, so the data falls between 85 and 115, or +- 15 points of 100.
95% of the data falls within two standard deviations of the . Since 2 standard deviations are equal to 30, the data falls between 70 and 130, or +-30 points of 100.
Another term that you should be somewhat familiar with is , or the likelihood that something occurs by chance😲. If something is statistically significance, it did not occur by chance (some outside factor influenced the data). If something isn't statistically significant, it occurred completely by chance. To determine this, you would compare the of the control group and the of the experimental group.
## Practice AP FRQ
The following question is taken from the College Board website (2017 AP Exam - Part B of #1).
A study was conducted to investigate the role of framing on concern for healthy eating🍏. Each participant (N = 100) was randomly assigned to one of the two conditions. In the first condition, the participants read an article indicating that obesity is a disease🦠. Participants in the second condition read an article indicating that obesity is the result of personal behaviors and decisions.
Participants were asked to indicate how important it would be for them to eat a healthy diet. Scores ranged from 1 (not very important) to 9 (very important). The results are presented in the table below.
Group Score - Concern for Healthy Eating Disease 3.4 1.4 Behavior 6.1 1.2
Table Courtesy of College Board
• Operationally define the dependent variable.
• What makes the study experimental rather than correlational?
• What is the most appropriate conclusion the researchers can draw about the relationship between the variables in the study?
The scoring guidelines provide the rubric for this question. You should be able to answer all three parts. If not, just go through this unit’s guides one more time and you’ll nail this FRQ.
🎥Watch AP Psychology teacher John Mohl review major statistical themes that may be part of the AP Psychology exam, including central tendency, variation, percentile, and statistical significance.
# Key Terms to Review (22)
Bimodal Distribution
: A bimodal distribution occurs when two different values appear most frequently (modes) in the data set.
Correlation Coefficient
: The correlation coefficient measures the strength and direction of a linear relationship between two variables on a scatterplot. It ranges from -1 to 1 where -1 indicates perfect negative correlation, 0 indicates no correlation, and 1 indicates perfect positive correlation.
Descriptive Statistics
: Descriptive statistics are numerical data used to measure and describe characteristics of groups. They do not allow us to make conclusions beyond the data we have analysed or reach conclusions regarding any hypotheses we might have made.
Frequency Distribution
: Frequency distribution refers to how often something happens within certain ranges or intervals for a set of data points.
Inferential Statistics
: Inferential statistics are procedures used that allow researchers to infer or generalize observations made with samples to the larger population from which they were drawn.
Mean
: The mean is simply the sum of all values in a dataset divided by the total number of values. It's often referred to as the "average."
Measures of Central Tendency
: Measures of central tendency are statistical indicators that identify the center, or average, of a data set. These measures include mean, median, and mode.
Measures of Variation
: Measures of variation describe how spread out or scattered the values in a data set are. They include range, variance, and standard deviation among others.
Median
: The median is defined as the middle value when all values within a dataset are arranged from smallest to largest. If there is an even number of observations, then there is no single middle value; so we take an average (mean) between two middle numbers instead.
Mode
: The mode is the most frequently occurring score in a set of given numbers.
Multimodal Distribution
: A multimodal distribution is a probability distribution with more than one peak, or "mode." This means that there are multiple values that appear most frequently in the data set.
Negative Correlation
: A negative correlation is a relationship between two variables in which one variable increases as the other decreases.
Negatively Skewed Distribution
: A negatively skewed distribution is a type of distribution in which more data values fall to the right side (higher end) of the distribution graph, with the tail on the left side (lower end).
No Correlation
: No correlation exists when there is no relationship between two variables; changes in one do not affect changes in another.
Normal Distribution
: A normal distribution, also known as a bell curve, is a statistical concept that refers to a type of continuous probability distribution for a real-valued random variable. In this distribution, most of the data falls near the mean (average), with frequencies decreasing away from the mean.
Outliers
: An outlier is an observation that lies an abnormal distance from other values in a random sample from a population.
Positive Correlation
: Positive Correlation occurs when both variables increase together or decrease together; as one variable increases, so does the other.
Positively Skewed Distribution
: A positively skewed distribution is a type of distribution where the values are more spread out on the right side (tail) of the distribution graph. This means that there are some unusually high values in your data.
Range
: In statistics, range refers to the difference between the highest and lowest scores in a data set.
Standard Deviation
: Standard deviation is a measure used to quantify the amount of variation or dispersion in a set of values. It tells us how much on average scores deviate from their mean value.
Statistical Significance
: Statistical significance refers to whether any differences observed between groups being studied are "real" or if they’re likely due just to chance. It's often determined by p-values less than 0.05.
Summarizing Data
: Summarizing data involves simplifying collected information into smaller, understandable parts. This can involve using measures such as averages, percentages, or graphs.
# 1.5 Statistical Analysis in Psychology
###### Attend a live cram event
Review all units live with expert teachers & students
## Types of Statistics
involves the use of numerical data to measure and describe the characteristics of groups, and this includes and variation. We'll be focusing on in this study guide! It does not involve making inferences about a population based on sample data.
, on the other hand, involves using statistical methods to make inferences about a population based on data. It allows you to draw conclusions about a population based on the characteristics of a sample. Specifically, it provides a way to see validity drawn from the results of the experiment🧪🔬.
Therefore, describe the data, while tell us what the data means.
## Summarizing Data
When one has a ton of data, how do they begin to go through it? Typically, a researcher would construct and interpret a graph with their data, and they use to do so. 📈
### Measures of Central Tendency
are statistical values that represent the center or typical value of a dataset. The three most commonly used are the , , and .
• The is the average of a set of scores. You can calculate the by summing all of the values in a dataset and dividing by the total number of values. The is sensitive to , or unusually large or small values, and can be affected by them.
• The is the middle score of distribution, separating the higher half of the data from the lower half. The is not affected by and can be a better measure of central tendency when the dataset contains .
• The is the most frequently recurring score in a dataset. A dataset can have one , more than one , or no . If two scores appear the most frequently, the distribution is bimodal. If three or more scores appear most frequently, the distribution is multimodal.
Let's practice calculating the three , , , and , using the following data set: 5, 10, 5, 7, 12, 15, 18
The easiest to spot is the : which value, if any, appears more often than others? Here, we can see 5 twice, so the of this dataset is 5.
Then, you may want to calculate the by adding all of these data values and dividing by the total. Since we have seven values, we have to divide by seven: (5 + 10 + 5 + 7 + 12 + 15 + 18)/7 = 10.286
The is the middle of the data set when the numbers are in order. Make sure you always put them in order!! If you do so here, you will find that the is 10.
### Measures of Variation
describe how spread out or dispersed the values in a dataset are. The most commonly used measure of variation is the , which is a measure of how much the values in a dataset deviate from the . It is basically used to assess how far the values are spread below and above the . A dataset with a low has values that are relatively close to the , while a dataset with a high has values that are more spread out.
Another, less complex, measure of variation you should be familiar with for this course is the of a dataset. is just the difference between the highest and lowest values in the dataset.
## Correlation
The is a statistical measure that describes the strength and direction of the relationship between two variables. It can from -1 to 1. A value of -1 indicates a strong negative relationship, a value of 1 indicates a strong positive relationship, and a value of 0 indicates no relationship.
You can simply think of it as a measure of how well two variables are correlated, and the closer it is to -1 or +1, the stronger the correlation.
### Positive Correlation
shows that as one variable increases ⬆️, the other variable increases ⬆️. For example, a positively correlated group may show that as height increases, weight increases as well.
Image courtesy of Expii
### Negative Correlation
shows that as one variable increases ⬆️, the other decreases ⬇️. An example of a could be how as the number of hours of sleep increases, tiredness decreases.
Image courtesy of Expii
### No Correlation
shows that there is no connection between the two variables. An example of could be IQ and how many pairs of pants an individual owns.
Image courtesy of Expii.
Remember, correlation does not imply causation, even if the is -1 or +1. You must run an experiment to prove there is causation.
## Skews
A is a breakdown of how the scores fall into different categories or ranges. There are several types of frequency distributions:
• 🔔A is a bell-shaped that is symmetrical about the . We'll delve into this deeper, but graph (b) on the image below shows a !
• 2️⃣A is a with two peaks. This occurs when the dataset has two distinct groups of values that occur with different frequencies.
• 👍A positively skewed distribution has a tail extending to the right (towards larger values). This occurs when the dataset has a few unusually large values that pull the to the right. Graph (a) in the image below is a , where the is greater than the .
• 👎A negatively skewed distribution has a tail extending to the left (towards smaller values). This occurs when the dataset has a few unusually small values that pull the to the left. Graph (c) in the image below is a , where the is greater than the .
It might be hard to remember which way the skew is. If the tail on the right is longer like it is in (a), then it's a skew to the right. If the tail on the left is longer like it is in (c), then it's a skew to the left.
Image courtesy of ResearchGate
## Normal Distributions
The normal curve, or (b) in the above image, is the only one you have to really be familiar with for this course. There are two important values that you should memorize: 68% and 95%.
This is a normal curve that includes data about intelligence📖. Basically, 68% of the data falls within one of the . Here, one is equivalent to 15, so the data falls between 85 and 115, or +- 15 points of 100.
95% of the data falls within two standard deviations of the . Since 2 standard deviations are equal to 30, the data falls between 70 and 130, or +-30 points of 100.
Another term that you should be somewhat familiar with is , or the likelihood that something occurs by chance😲. If something is statistically significance, it did not occur by chance (some outside factor influenced the data). If something isn't statistically significant, it occurred completely by chance. To determine this, you would compare the of the control group and the of the experimental group.
## Practice AP FRQ
The following question is taken from the College Board website (2017 AP Exam - Part B of #1).
A study was conducted to investigate the role of framing on concern for healthy eating🍏. Each participant (N = 100) was randomly assigned to one of the two conditions. In the first condition, the participants read an article indicating that obesity is a disease🦠. Participants in the second condition read an article indicating that obesity is the result of personal behaviors and decisions.
Participants were asked to indicate how important it would be for them to eat a healthy diet. Scores ranged from 1 (not very important) to 9 (very important). The results are presented in the table below.
Group Score - Concern for Healthy Eating Disease 3.4 1.4 Behavior 6.1 1.2
Table Courtesy of College Board
• Operationally define the dependent variable.
• What makes the study experimental rather than correlational?
• What is the most appropriate conclusion the researchers can draw about the relationship between the variables in the study?
The scoring guidelines provide the rubric for this question. You should be able to answer all three parts. If not, just go through this unit’s guides one more time and you’ll nail this FRQ.
🎥Watch AP Psychology teacher John Mohl review major statistical themes that may be part of the AP Psychology exam, including central tendency, variation, percentile, and statistical significance.
# Key Terms to Review (22)
Bimodal Distribution
: A bimodal distribution occurs when two different values appear most frequently (modes) in the data set.
Correlation Coefficient
: The correlation coefficient measures the strength and direction of a linear relationship between two variables on a scatterplot. It ranges from -1 to 1 where -1 indicates perfect negative correlation, 0 indicates no correlation, and 1 indicates perfect positive correlation.
Descriptive Statistics
: Descriptive statistics are numerical data used to measure and describe characteristics of groups. They do not allow us to make conclusions beyond the data we have analysed or reach conclusions regarding any hypotheses we might have made.
Frequency Distribution
: Frequency distribution refers to how often something happens within certain ranges or intervals for a set of data points.
Inferential Statistics
: Inferential statistics are procedures used that allow researchers to infer or generalize observations made with samples to the larger population from which they were drawn.
Mean
: The mean is simply the sum of all values in a dataset divided by the total number of values. It's often referred to as the "average."
Measures of Central Tendency
: Measures of central tendency are statistical indicators that identify the center, or average, of a data set. These measures include mean, median, and mode.
Measures of Variation
: Measures of variation describe how spread out or scattered the values in a data set are. They include range, variance, and standard deviation among others.
Median
: The median is defined as the middle value when all values within a dataset are arranged from smallest to largest. If there is an even number of observations, then there is no single middle value; so we take an average (mean) between two middle numbers instead.
Mode
: The mode is the most frequently occurring score in a set of given numbers.
Multimodal Distribution
: A multimodal distribution is a probability distribution with more than one peak, or "mode." This means that there are multiple values that appear most frequently in the data set.
Negative Correlation
: A negative correlation is a relationship between two variables in which one variable increases as the other decreases.
Negatively Skewed Distribution
: A negatively skewed distribution is a type of distribution in which more data values fall to the right side (higher end) of the distribution graph, with the tail on the left side (lower end).
No Correlation
: No correlation exists when there is no relationship between two variables; changes in one do not affect changes in another.
Normal Distribution
: A normal distribution, also known as a bell curve, is a statistical concept that refers to a type of continuous probability distribution for a real-valued random variable. In this distribution, most of the data falls near the mean (average), with frequencies decreasing away from the mean.
Outliers
: An outlier is an observation that lies an abnormal distance from other values in a random sample from a population.
Positive Correlation
: Positive Correlation occurs when both variables increase together or decrease together; as one variable increases, so does the other.
Positively Skewed Distribution
: A positively skewed distribution is a type of distribution where the values are more spread out on the right side (tail) of the distribution graph. This means that there are some unusually high values in your data.
Range
: In statistics, range refers to the difference between the highest and lowest scores in a data set.
Standard Deviation
: Standard deviation is a measure used to quantify the amount of variation or dispersion in a set of values. It tells us how much on average scores deviate from their mean value.
Statistical Significance
: Statistical significance refers to whether any differences observed between groups being studied are "real" or if they’re likely due just to chance. It's often determined by p-values less than 0.05.
Summarizing Data
: Summarizing data involves simplifying collected information into smaller, understandable parts. This can involve using measures such as averages, percentages, or graphs.
| 5,126
| 24,204
|
{"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}
| 4.0625
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.925715
|
http://nixdoc.net/man-pages/NetBSD/man3/atanh.3.html
| 1,550,677,511,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-09/segments/1550247495147.61/warc/CC-MAIN-20190220150139-20190220172139-00443.warc.gz
| 198,641,159
| 4,590
|
· Home
+ man pages
-> Linux -> FreeBSD -> OpenBSD -> NetBSD -> Tru64 Unix -> HP-UX 11i -> IRIX
· Linux HOWTOs
· FreeBSD Tips
· *niX Forums
man pages->NetBSD man pages -> atanh (3)
Title
Content
Arch
Section All Sections 1 - General Commands 2 - System Calls 3 - Subroutines 4 - Special Files 5 - File Formats 6 - Games 7 - Macros and Conventions 8 - Maintenance Commands 9 - Kernel Interface n - New Commands
## ATANH(3)
```
```
### NAME[Toc][Back]
``` atanh, atanhf - inverse hyperbolic tangent function
```
### LIBRARY[Toc][Back]
``` Math Library (libm, -lm)
```
### SYNOPSIS[Toc][Back]
``` #include <math.h>
double
atanh(double x);
float
atanhf(float x);
```
### DESCRIPTION[Toc][Back]
``` The atanh() and atanhf() functions compute the inverse hyperbolic tangent
of the real argument x.
```
### RETURN VALUES[Toc][Back]
``` If |x|>=1, atanh(x) and atanhf(x) return +inf, -inf or NaN, and sets the
global variable errno to EDOM.
```
``` acosh(3), asinh(3), exp(3), math(3)
```
### HISTORY[Toc][Back]
``` The atanh() function appeared in 4.3BSD.
BSD May 6, 1991 BSD
```
[ Back ]
Similar pages
Name OS Title atanh OpenBSD inverse hyperbolic tangent functions atanh FreeBSD inverse hyperbolic tangent functions atanhf FreeBSD inverse hyperbolic tangent functions atanhf OpenBSD inverse hyperbolic tangent functions tanh NetBSD hyperbolic tangent function tanh Linux hyperbolic tangent function tanhf NetBSD hyperbolic tangent function asinh NetBSD inverse hyperbolic sine function acosh NetBSD inverse hyperbolic cosine function acosh Linux inverse hyperbolic cosine function
| 472
| 1,693
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.671875
| 3
|
CC-MAIN-2019-09
|
latest
|
en
| 0.340729
|
https://zh.wikipedia.org/wiki/%E9%9C%8D%E6%99%AE%E5%85%8B%E6%B4%9B%E5%A4%AB%E7%89%B9%EF%BC%8D%E5%8D%A1%E6%99%AE%E7%AE%97%E6%B3%95
| 1,550,959,900,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-09/segments/1550249550830.96/warc/CC-MAIN-20190223203317-20190223225317-00449.warc.gz
| 1,007,298,990
| 17,147
|
# 霍普克洛夫特－卡普算法
program Project1;
const maxn=1000;
var dx,dy,mx,my,q:array[1..maxn]of longint;
n,m,e,i,j,ans,ff,rr:longint;
function bfs:boolean;
var i,u,j:longint;
begin
bfs:=false;
fillchar(q,sizeof(q),0);
rr:=1;
ff:=1;
for i:=1 to n do
if mx[i]=-1
then begin
q[ff]:=i;
inc(ff);
end;
for i:=1 to n do dx[i]:=0;
for i:=1 to m do dy[i]:=0;
while rr<ff do
begin
u:=q[rr];
inc(rr);
begin
if dy[i]=0
then begin
dy[i]:=dx[u]+1;
if my[i]=-1
then bfs:=true
else begin
dx[my[i]]:=dy[i]+1;
q[ff]:=my[i];
inc(ff);
end;
end;
end;
end;
end;
function dfs(x:longint):boolean;
var i,j:longint;
begin
begin
if dy[i]=dx[x]+1
then begin
dy[i]:=0;
if(my[i]=-1)or dfs(my[i])
then begin
mx[x]:=i;
my[i]:=x;
exit(true);
end;
end;
end;
exit(false);
end;
begin
for i:=1 to e do
begin
end;
for i:=1 to n do mx[i]:=-1;
for i:=1 to m do my[i]:=-1;
ans:=0;
while bfs do
for i:=1 to n do
if(mx[i]=-1)and(dfs(i))
then inc(ans);
writeln(ans);
end.
| 381
| 916
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.953125
| 3
|
CC-MAIN-2019-09
|
latest
|
en
| 0.222333
|
http://math.stackexchange.com/questions/73544/isolate-a-variable-in-a-polynomial-function
| 1,469,654,150,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257827079.61/warc/CC-MAIN-20160723071027-00164-ip-10-185-27-174.ec2.internal.warc.gz
| 162,847,860
| 19,275
|
# Isolate a variable in a polynomial function
How would I go about isolating $y$ in this function? I'm going crazy right now because I can't figure this out.
The purpose of this is to allow me to derive $f(x)$ afterwards.
$$x = \frac{y^2}{4} + 2y .$$
-
Try completing the square. Note that you can simply apply implicit differentiation also, to find the derivative. – JavaMan Oct 18 '11 at 3:32
Is your function $x = \frac{y^2}{4} + 2y$, or $x = \frac{y^2}{4+2y}$? The problem with the slash notation is that the absence of parentheses makes it ambiguous. – Arturo Magidin Oct 18 '11 at 3:33
What you have is a quadratic equation for $y$; $(y^2/4)+2y-x=0$. What methods do you know for solving quadratic equations? – Gerry Myerson Oct 18 '11 at 3:49
When you have a nice relationship, differentiate immediately. That usually works out better. Presumably the problem in the post arose when you were trying to solve some other problem. If one knew what that problem was, an efficient approach could be suggested. – André Nicolas Oct 18 '11 at 3:52
The whole point of implicit differentiation is to find $y'$ without having to first "solve for $y$". Don't try: use implicit differentiation! – Arturo Magidin Oct 18 '11 at 4:01
Point the Zeroth: ignore points the first and points the second (in the sense that they aren't really the 'right' way of proceeding; they are presented so you can see that they are not the right path to take).
Point the First: $y$ is not a function of $x$; if you plot this equation on the plane, you'll have a parabola, $$x = \frac{y^2}{4} + 2y = \left(\frac{y}{2}\right)^2 + 2\left(\frac{y}{2}\right)(2) + 2^2 - 2^2 = \left(\frac{y}{2} + 2\right)^2 - 4.$$ This parabola opens right, so it is not the graph of a function of $x$.
Point the Second: You can break up the graph into two functions by using the quadratic formula: $$\frac{y^2}{4} + 2y - x = 0$$ gives $$y^2 + 8y - 4x = 0,$$ so $$y = \frac{-8+\sqrt{64+16x}}{2},\quad\text{or}\quad y = \frac{-8-\sqrt{64+16x}}{2}.$$ We would then need to find the derivatives of each of these two separately, and for any given value of $x$ and $y$, determine which of the two formulas to use. They are not hard, but they are somewhat annoying.
If $y = -4 + \frac{1}{2}\sqrt{64+16x}$, then $$\frac{dy}{dx} = \frac{1}{4}(64+16x)^{-1/2}(16) = \frac{4}{\sqrt{64+16x}}.$$ Similarly, if $y=-4-\frac{1}{2}\sqrt{64+16x}$, then $$\frac{dy}{dx} = -\frac{4}{\sqrt{64+16x}}.$$
Point the Third: What you really want to do here is implicit differentiation, which is a way of handling all of these difficulties without having to solve for $y$ first, and without having to worry about "which formula" to use later. Explicitly, from $$x = \frac{y^2}{4} + 2y,$$ take derivatives on both sides, using the Chain Rule and remembering that $y$ is an (implicit) function of $x$, so that $y'$ needs to be left indicated (we don't know what it is right now): \begin{align*} x & = \frac{y^2}{4} + 2y\\ \frac{d}{dx}x &= \frac{d}{dx}\left( \frac{y^2}{4} + 2y\right)\\ 1 &= \frac{2y}{4}y' + 2y'\\ 1&= \frac{y}{2}y' + 2y'\\ 1 &= y'\left(\frac{y}{2} + 2\right).\end{align*} Solving for $y'$ gives an implicit definition for $\frac{dy}{dx}$ in terms of $y$ and $x$ (though in this case, $x$ plays no role): $$y' = \frac{1}{\frac{y}{2}+2} = \frac{2}{y+4}.$$
Point the Fourth: Alternatively, since you have $x$ explicitly as a function of $y$, use the Inverse Function Theorem: taking derivatives with respect to $y$, we have: $$\frac{dx}{dy} = \frac{1}{2}y + 2,$$ so $$\frac{dy}{dx} = \frac{1}{\quad\frac{dx}{dy}\quad} = \frac{1}{\frac{1}{2}y + 2} = \frac{2}{y+4}.$$
Point the Fifth: So, do these "implicit formulas" give the same answer as the "explicit ones" we got in Point the Second? Yes!
If $y = \frac{-8+\sqrt{64+16x}}{2}$, then $y+4 = \frac{\sqrt{64+16x}}{2}$, so $$\frac{2}{y+4} = \frac{2}{\frac{1}{2}\sqrt{64+16x}} = \frac{4}{\sqrt{64+16x}},$$ same answer as in Point the Second; and if $y=\frac{-8-\sqrt{64+16x}}{2}$ then $y+4 = -\frac{1}{2}\sqrt{64+16x}$, so $$\frac{2}{y+4} = \frac{2}{-\frac{1}{2}\sqrt{64+16x}} = -\frac{4}{\sqrt{64+16x}},$$ again, same answer as in Point the Second.
But using implicit differentiation (or in cases like this, when $x$ is an explicit function of $y$, the inverse function theorem) is much easier than first solving for $y$, possibly requiring breaking up the original implicit function into several different explicit functions, and then differentiating. If you were trying to work with the Folium of Descartes ($x^3+y^3=3xy$), you would have to consider three different formulas, each involving a sum of cubic roots that has square roots inside the radicals; if you were trying to work with a function like $y = \sin(x+y)$, you would have a hard time solving for $y$, but using implicit differentiation is pretty easy.
-
| 1,583
| 4,816
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.21875
| 4
|
CC-MAIN-2016-30
|
latest
|
en
| 0.885891
|
https://numberworld.info/root-of-268
| 1,723,375,213,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00653.warc.gz
| 352,286,331
| 3,014
|
# Root of 268
#### [Root of two hundred sixty-eight]
square root
16.3707
cube root
6.4473
fourth root
4.0461
fifth root
3.0593
In mathematics extracting a root is declared as the determination of the unknown "x" in the equation $y=x^n$ The outcome of the extraction of the root is known as a so-called root. In the case of "n equals 2", one talks about a square root or sometimes a second root also, another possibility could be that n equals 3 by this time one would consider it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 268 is represented as this: $$\sqrt[]{268}=16.370705543745$$
Moreover it is legit to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 268 is 16.370705543745. The cube root of 268 is 6.4473057272669. The fourth root of 268 is 4.0460728544781 and the fifth root is 3.0593344616085.
Look Up
| 281
| 967
|
{"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}
| 3.671875
| 4
|
CC-MAIN-2024-33
|
latest
|
en
| 0.921614
|
https://www.distancesto.com/fuel-cost/la/phonsavan-to-thakhek/history/232023.html
| 1,719,070,684,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00589.warc.gz
| 643,093,224
| 14,654
|
# INR14.78 Total Cost of Fuel from Phonsavan to Thakhèk
Your trip to Thakhèk will consume a total of 5.91 gallons of fuel.
Trip start from Phonsavan, LA and ends at Thakhèk, LA.
Trip (236.4 mi) Phonsavan » Thakhèk
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Phonsavan, Laos and end at Thakhèk, Laos.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Phonsavan to Thakhèk plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Phonsavan to Thakhèk.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Phonsavan to Thakhèk.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Phonsavan to Thakhèk.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to Thakhèk are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Phonsavan to Thakhèk.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Phonsavan to Thakhèk.
Speaking of travel time, a flight to Thakhèk takes up a lot less. How much less? Flight time from Phonsavan to Thakhèk.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Phonsavan to Thakhèk.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
Recent Fuel Calculations for Phonsavan LA:
Fuel Cost from Phonsavan to Muang Hiam
Fuel Cost from Phonsavan to 337/32 Nittayo Rd.
Fuel Cost from Phonsavan to Luang Prabang
Fuel Cost from Phonsavan to Phonsavan
Fuel Cost from Phonsavan to Oudomxay Province
| 672
| 2,753
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26
|
latest
|
en
| 0.958703
|
http://openstudy.com/updates/50d1e938e4b069abbb70fda2
| 1,448,748,683,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-48/segments/1448398454160.51/warc/CC-MAIN-20151124205414-00094-ip-10-71-132-137.ec2.internal.warc.gz
| 167,505,216
| 13,763
|
## A community for students. Sign up today
Here's the question you clicked on:
## chevygirl 2 years ago Please help me. 6+x over 3=1
• This Question is Closed
1. soty2013
0k
2. surdawi
$\frac{6+x}{3}=1$ multiply by LCD (3) $3*\frac{6+x}{3}=3*1$ $6+x=3$ subtract 6 from both sides and you should get the answer
3. soty2013
-3...
4. chevygirl
|dw:1355934285826:dw|
5. surdawi
|dw:1355934391619:dw|
6. chevygirl
how is it negative?
7. surdawi
when u subtract 6 1-6=-5
8. hager
6+x/3=1 multiply both sides by 3 18+x=3 18+x-3
9. chevygirl
but how does that make it a negative??? if ur just subtracting and there is no negatives in this problem??
10. surdawi
who is greater the 6 or the 1?
11. soty2013
its too easy, its general arithmetics..
12. chevygirl
6 duh sorry i thought i posted it
13. chevygirl
That is the answer that is my book
14. surdawi
thats what i did hehe
15. soty2013
its correct @chevygirl Well Done
16. chevygirl
why did you say -3? @soty2013
17. surdawi
we thought its was $\frac{6+x}{3}$
18. chevygirl
oh sorry my bad
19. soty2013
yes so we said that , now we all are correct lol :)
20. chevygirl
ok well thanks
#### Ask your own question
Sign Up
Find more explanations on OpenStudy
Privacy Policy
| 424
| 1,258
|
{"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}
| 3.328125
| 3
|
CC-MAIN-2015-48
|
longest
|
en
| 0.841287
|
https://socratic.org/questions/how-do-you-graph-y-sin-2x
| 1,579,440,275,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00158.warc.gz
| 669,934,961
| 6,083
|
# How do you graph y = -sin(2x)?
##### 1 Answer
Jun 15, 2015
These are three steps (transformations) that you can apply in every order you want,and you'll obtain $y = - \sin \left(2 x\right)$.
#### Explanation:
1. Start graphing the trig function you know, $y = \sin \left(x\right)$.
graph{sinx [-10, 10, -5, 5]}
1. Doubling the argument you have a compression in the period, that means that the function double the frequency: if we had the solutions in $0 , \pi , 2 \pi , 3 \pi , \ldots$ now we have solutions in $0 , \frac{\pi}{2} , \pi , \frac{3}{2} \pi , 2 \pi , \ldots$
graph{sin(2x) [-10, 10, -5, 5]}
1. The minus sign in front of the function is a reflection on the x-axis, so for each point of the function $\left(x , y\right) \to \left(x , - y\right)$:
graph{-sin(2x) [-10, 10, -5, 5]}
| 277
| 803
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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}
| 4.5
| 4
|
CC-MAIN-2020-05
|
longest
|
en
| 0.728928
|
https://www.kylesconverter.com/speed-or-velocity/microns-per-second-to-meters-per-year
| 1,718,676,181,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00304.warc.gz
| 731,263,119
| 5,650
|
# Convert Microns Per Second to Meters Per Year
### Kyle's Converter > Speed Or Velocity > Microns Per Second > Microns Per Second to Meters Per Year
Microns Per Second (µ) Meters Per Year (m/yr) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Meters Per Year to Microns Per Second
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Micron per Second:
1 Micron per second is equal to 0.000 001 meters per second (SI unit).
1 Meter per Year:
1 Meter per year is approximately 3.168 874 x 10-8 meters per second (SI unit). Using Gregorian year of 365.2425 days.
Conversions Table
1 Microns Per Second to Meters Per Year = 31.55770 Microns Per Second to Meters Per Year = 2208.9866
2 Microns Per Second to Meters Per Year = 63.113980 Microns Per Second to Meters Per Year = 2524.5562
3 Microns Per Second to Meters Per Year = 94.670990 Microns Per Second to Meters Per Year = 2840.1257
4 Microns Per Second to Meters Per Year = 126.2278100 Microns Per Second to Meters Per Year = 3155.6952
5 Microns Per Second to Meters Per Year = 157.7848200 Microns Per Second to Meters Per Year = 6311.3904
6 Microns Per Second to Meters Per Year = 189.3417300 Microns Per Second to Meters Per Year = 9467.0856
7 Microns Per Second to Meters Per Year = 220.8987400 Microns Per Second to Meters Per Year = 12622.7808
8 Microns Per Second to Meters Per Year = 252.4556500 Microns Per Second to Meters Per Year = 15778.476
9 Microns Per Second to Meters Per Year = 284.0126600 Microns Per Second to Meters Per Year = 18934.1712
10 Microns Per Second to Meters Per Year = 315.5695800 Microns Per Second to Meters Per Year = 25245.5616
20 Microns Per Second to Meters Per Year = 631.139900 Microns Per Second to Meters Per Year = 28401.2568
30 Microns Per Second to Meters Per Year = 946.70861,000 Microns Per Second to Meters Per Year = 31556.952
40 Microns Per Second to Meters Per Year = 1262.278110,000 Microns Per Second to Meters Per Year = 315569.52
50 Microns Per Second to Meters Per Year = 1577.8476100,000 Microns Per Second to Meters Per Year = 3155695.2
60 Microns Per Second to Meters Per Year = 1893.41711,000,000 Microns Per Second to Meters Per Year = 31556952
| 703
| 2,234
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26
|
latest
|
en
| 0.564404
|
https://mailman.ucar.edu/pipermail/ncl-talk/2017-November/010860.html
| 1,725,850,915,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00810.warc.gz
| 347,119,032
| 3,205
|
# [ncl-talk] "sum" and "avg" functions without ignoring missing values
Rashed Mahmood rashidcomsis at gmail.com
Fri Nov 24 12:52:11 MST 2017
```Thanks Dave, that is exactly what I was looking for...
On Fri, Nov 24, 2017 at 11:46 AM, Dave Allured - NOAA Affiliate <
dave.allured at noaa.gov> wrote:
> Rasheed,
>
> I think dim_cumsum and dim_cumsum_n with opt=0 are intended for this
> purpose. There is no corresponding function for "avg". However, simply
> dividing the result array from dim_cumsum by N should give the correct
> array result, with no extra "if" or "where" statement.
>
> --Dave
>
>
> On Fri, Nov 24, 2017 at 12:29 PM, Rashed Mahmood <rashidcomsis at gmail.com>
> wrote:
>
>> Hi Arne,
>>
>> Thanks for trying to answer the question, however, the things you
>> mentioned would add confusion and possibly wrong results.
>> Actually I know how to achieve what I want using a do loop. For example,
>> let us say we have variable a = (ntimes,nlevel,nlat,nlon) and we want to do
>> sum over time domain which can easily be done using: (dim_sum_n(a,0)), Now
>> the results would be sum over all non-missing values of
>> time1,time2,time2...timeN, which is correct.
>>
>> But if I want to do sum ONLY when all of the values of
>> time1,time2,time3...timeN are non-missing then we can do this:
>>
>> dimz = dimsizes(a)
>> b = new(/dimz(1),dimz(2),dimz(3),float)
>> b at _FillValue = a at _FillValue
>>
>> b(:,:,:) = 0.
>> do n=0,dimz(0)-1
>> b = b+a(n,:,:,:)
>> end do
>>
>> I just wanted to avoid the loop and use the "sum" function for
>> efficiency. I hope this clarifies the question a bit more.
>>
>> Cheers,
>> Rashed
>>
>>
>> On Fri, Nov 24, 2017 at 3:01 AM, Arne Melsom <arne.melsom at met.no> wrote:
>>
>>> Hi!
>>> Since 'music piano' replied, I thought I might chime in ;)
>>> I'm pretty sure that no option(s) to functions allow you to do what you
>>> want.
>>> So you'll need to write alternative functions yourself (perhaps not what
>>> you consider 'an easy way'...)
>>> You may use something like
>>>
>>> orgFill = a at _FillValue
>>> newFill = -min(abs(a)) - 1 ; make sure newFill is out of range and <0
>>> if (newFill .eq. orgFill) then ; newFill must be different from orgFill
>>> newFill = newFill - 1
>>> end if
>>> newVar = ndtooned(a)
>>> newVar at _FillValue = newFill
>>> indx = ind(ismissing(newVar))
>>> newVar(indx) = orgFill
>>>
>>> ...and, if you later want to restore:
>>> a = newa
>>> a at _FillValue = orgFill
>>>
>>> Hope this helps/clarifies!
>>> Best regards,
>>> Arne
>>>
>>>
>>> 2017-11-24 1:49 GMT+01:00 Rashed Mahmood <rashidcomsis at gmail.com>:
>>>
>>>> Hello all,
>>>> I am just wondering if there is an easy way to tell the "sum" and "avg"
>>>> functions NOT ignore the missing values. Below lines clarify my question
>>>> where I would like to see same values for c and d:
>>>>
>>>> ncl 0> a = (/(/-9.,2,-9./),(/1,-9.,2./)/)
>>>> ncl 1> a at _FillValue = -9.
>>>> ncl 2> c = (a(0,:)+a(1,:))
>>>> ncl 3> d = dim_sum_n(a,0)
>>>> ncl 4> print(" "+c+ " ... "+d)
>>>> (0) -9 ... 1
>>>> (1) -9 ... 2
>>>> (2) -9 ... 2
>>>>
>>>> Thanks,
>>>> Rashed
>>>>
>>>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mailman.ucar.edu/pipermail/ncl-talk/attachments/20171124/a6b6f7b9/attachment.html>
```
| 1,054
| 3,330
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38
|
latest
|
en
| 0.891343
|
http://gmatclub.com/forum/unique-151397.html
| 1,484,940,878,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560280850.30/warc/CC-MAIN-20170116095120-00031-ip-10-171-10-70.ec2.internal.warc.gz
| 116,228,380
| 52,212
|
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 20 Jan 2017, 11:34
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Unique
Author Message
Manager
Joined: 24 Feb 2013
Posts: 105
GMAT 1: 660 Q47 V35
GMAT 2: 690 Q46 V38
GMAT 3: 680 Q46 V37
GMAT 4: 680 Q45 V39
GMAT 5: 760 Q48 V47
GPA: 3.97
Followers: 0
Kudos [?]: -8 [0], given: 45
### Show Tags
22 Apr 2013, 07:37
I apologize for the length, but I felt it appropriate. My target school is Tuck (Early Action).
I grew up in a single parent household in Philadelphia and attended three high schools. The first was closed after a riot during my freshman year. The second had metal detectors and x-ray machines for our bags. The third, and final, was a small private school- I worked as a janitor after hours in lieu of tuition.
My peers had no aspiration. Of my mother's twelve brothers and sisters, three graduated high school. Given my environment, college was never on my radar. One day, I had a revelation, and decided to take my SAT's.
I did well enough without study to get into the common Philadelphia schools, but ultimately decided to chose a lesser ranked University that I had visited last-minute on the recommendation of a mentor. This was a pivotal moment in my life.
I chose Eastern University for what would seem an odd reason to most. When I visited, I felt extremely out of place. Think Beverly Hillbillies, except Suburban City Boy. Having been a boy scout, the "camp" feeling was familiar to me. That is how I felt every single day. I knew, though, that removing myself from the urban environment would be key.
Uprooting myself was a painful experience. I did not fit in. I was extremely bored, and it was difficult to relate to my new surroundings. I had set my sights on a goal however, and my emotional reactions were not going to stop me; I was going to better position myself to take care of my mother and sister, and it did not matter how painful of an experience that meant enduring.
I graduated a semester early with a B.S. in Accounting & Finance, a minor in Psychology, a 3.9 GPA, and I did so while employed (I did not have outside financial help). I landed a position with Edward Jones a few months after graduation, passed my Series 7 and 66, and knew that things were going to change for my family and I.
But that didn't last. The economy upended, and I did not agree with the way that I was taught to advice clients. I backed out of brokerage and ended up at a small company as a Junior Staff Accountant. Two years later, I leveraged this experience into my current role: I work for American Infrastructure, a large heavy civil construction company (bridges, highways, excavation). My position is unorthodox- I am a "Financial Career Partner". My program consists of 4 to 8 month inter-company rotations in Job Cost, Finance, Audit, Equipment, Estimating, Purchasing, and Aggregates. The goal is to graduated into management at the end, as an employee that they have trained from top-to-bottom. My program ends in December 2013 (I extended Job Cost from 8 months to 16).
Most people would be content. I came from nothing and am doing well for myself now, with a solid career and guaranteed advancement. I believe that I am capable of more however, and that's where the idea of an MBA began dancing around my mind. I sat for my GMAT this prior Thursday and 690'ed (Q46 V38). I was a bit disappointed after 740'ing both GMATPreps, but I believe I've made myself competitive.
After months of research, I am absolutely in love with Tuck. It fits the construct of small-school, strong-relationship, amazing-people, style learning that I have thrived in. I have asked to be put into contact with a current Tuckie to ask a few questions:
I am afraid that regardless of a compelling story, that my background just isn't competitive enough. Is there any insight/advice that you could offer? I know that given a shot, I will succeed.. I just don't know how much faith others will have in that.
SVP
Status: Trust Experience, Trust Success
Affiliations: U. Chicago, Johns Hopkins, AIGAC
Joined: 12 Dec 2005
Posts: 1775
Followers: 15
Kudos [?]: 184 [1] , given: 0
### Show Tags
22 Apr 2013, 09:52
1
KUDOS
Shanek,
You definitely have a distinctive story, one that admissions officers can warm to, if executed well. Tuck is the kind of school that responds to applicants who understand its uniqueness and show that. Early Action would help, but I would need to get a better idea of your leadership experience (including extracurricular) to give you my take on your odds of admission. If you send me your resume/CV, I can give you more focused feedback: paulbodine@yahoo.com.
--Paul Bodine, Great Applications for Business School, http://www.paulsbodine.com/testimonials
Manager
Joined: 24 Feb 2013
Posts: 105
GMAT 1: 660 Q47 V35
GMAT 2: 690 Q46 V38
GMAT 3: 680 Q46 V37
GMAT 4: 680 Q45 V39
GMAT 5: 760 Q48 V47
GPA: 3.97
Followers: 0
Kudos [?]: -8 [0], given: 45
### Show Tags
22 Apr 2013, 10:16
Paul,
Thank you for your response. I have sent you an email, which I have posted below in case others are following. I have also uploaded a copy of my resume for other consultants.
Quote:
I have attached the resume that I used to secure my most recent position. As I have just finished taking my GMAT, it is not updated to reflect the last 2 years of my career. Bullet points are as follows:
Currently employed in what amounts to an accelerated management program for a large construction company. It consists of 4-8 month rotations in a variety of business functions. The goal at the end is a large promotion (think "graduation") into a senior/managerial role.
I have completed the following rotations:
Materials (4 months)- Took over the role of a senior accountant that had left recently.
Job Cost (16 months)- Originally an 8 month rotation, I recognized it as the core of the business and voluntarily extended it by another 8 months.
Purchasing/Estimating (4 months)- Self-explanatory. Was asked to manage a co-op from Drexel University. Trained her to take over some of my responsibilities.
Equipment (2 months)- Began this rotation last week. I will be involved with the business decisions regarding our heavy machinery (which are, in their own way, our revenue stream).
As for extra curriculars, I worked close to full time in college to pay for my degree and did not have the space to invest myself in clubs. The majority of my free time now is spent on one of my many hobbies. Of these, it is dietetics and exercise that I am most passionate about. I am a bit confused as to how to present this as leadership experience though; I can deadlift more than double my body weight, which is commendable, but what truly excites me is the impact that my efforts make on those around me. I am directly attributable for more then 100+ pounds of weight loss, and field a constant stream of questions as to which foods to eat, how to exercise, ways to maintain motivation, etc. That being said, I have no paper evidence- how do you jump this hurdle?
Attachments
Konowal, Shane_Resume AI.doc [42 KiB]
Re: Unique [#permalink] 22 Apr 2013, 10:16
Similar topics Replies Last post
Similar
Topics:
Unique Profile Eval 1 01 Apr 2010, 20:21
Profile / Unique Timing Questions 3 07 Mar 2010, 10:54
Profile / Unique Timing Questions 3 05 Mar 2010, 21:06
Profile evaluation - sort of unique! 1 30 Sep 2009, 04:39
Unique Profile Evaluation 5 29 Aug 2009, 07:10
Display posts from previous: Sort by
| 2,001
| 8,068
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04
|
latest
|
en
| 0.956739
|
https://oeis.org/A014990
| 1,582,493,073,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875145839.51/warc/CC-MAIN-20200223185153-20200223215153-00165.warc.gz
| 478,460,350
| 4,237
|
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A014990 a(n) = (1 - (-8)^n)/9. 10
1, -7, 57, -455, 3641, -29127, 233017, -1864135, 14913081, -119304647, 954437177, -7635497415, 61083979321, -488671834567, 3909374676537, -31274997412295, 250199979298361, -2001599834386887, 16012798675095097 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS q-integers for q=-8. LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (-7,8). FORMULA a(n) = a(n-1) + q^{(n-1)} = {(q^n - 1) / (q - 1)} From Philippe Deléham, Feb 13 2007: (Start) a(1)=1, a(2)=-7, a(n) = -7*a(n-1) + 8*a(n-2) for n > 2. a(n) = (-1)^(n+1)*A015565(n). G.f.: x/(1 + 7*x - 8*x^2). (End) a(n) = (1/9)*(1 + 8*(-8)^n), with n >= 0. - Paolo P. Lava, Nov 21 2008 E.g.f.: (exp(x) - exp(-8*x))/9. - G. C. Greubel, May 26 2018 MAPLE a:=n->sum ((-8)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008 MATHEMATICA QBinomial[Range[20], 1, -8] (* or *) LinearRecurrence[{-7, 8}, {1, -7}, 20] (* Harvey P. Dale, Dec 19 2011 *) PROG (Sage) [gaussian_binomial(n, 1, -8) for n in range(1, 20)] # Zerinvary Lajos, May 28 2009 (MAGMA) I:=[1, -7]; [n le 2 select I[n] else -7*Self(n-1) +8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012 (PARI) a(n)=(1-(-8)^n)/9 \\ Charles R Greathouse IV, Oct 07 2015 CROSSREFS Cf. A015565, A077925, A014983, A014985, A014986, A014987, A014989, A014991, A014992, A014993, A014994. Sequence in context: A218587 A218838 A082310 * A015565 A268316 A291537 Adjacent sequences: A014987 A014988 A014989 * A014991 A014992 A014993 KEYWORD sign,easy AUTHOR EXTENSIONS Better name from Ralf Stephan, Jul 14 2013 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified February 23 16:23 EST 2020. Contains 332177 sequences. (Running on oeis4.)
| 860
| 2,196
|
{"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}
| 3.28125
| 3
|
CC-MAIN-2020-10
|
latest
|
en
| 0.457704
|
http://mathhelpforum.com/calculus/170716-integrate-partial-fractions.html
| 1,527,243,310,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00251.warc.gz
| 191,601,228
| 10,219
|
# Thread: Integrate by partial fractions
1. ## Integrate by partial fractions
$\displaystyle \displaystyle \int \frac{6 x^3+4 x^2 - 3 x - 5}{x^2-1}\, dx$
What am I doing wrong here?
First, I long divide & get:
$\displaystyle \int 6x+4+\frac{3x-1}{x^2-1}dx$
$\displaystyle \int 6x+4+\frac{3x-1}{(x+1)(x-1)}dx= \int 6x+4 +\frac{A}{x+1}+\frac{B}{x-1} dx$
$\displaystyle A(x-1)+B(x+1)=3x-1$
I figure that A = 1, B = 2
$\displaystyle \int 6x+4 + \frac{2}{x+1} + \frac{1}{x-1} dx$
This should integrate to...
$\displaystyle 3x^2+4x+2ln(x+1) + ln(x-1) + C$
this answer appears to be wrong...
Where did I go wrong?
Thank you!
Check it here.
3. Originally Posted by Vamz
$\displaystyle \displaystyle \int \frac{6 x^3+4 x^2 - 3 x - 5}{x^2-1}\, dx$
What am I doing wrong here?
First, I long divide & get:
$\displaystyle \int 6x+4+\frac{3x-1}{x^2-1}dx$
$\displaystyle \int 6x+4+\frac{3x-1}{(x+1)(x-1)}dx= \int 6x+4 +\frac{A}{x+1}+\frac{B}{x-1} dx$
$\displaystyle A(x-1)+B(x+1)=3x-1$
I figure that A = 1, B = 2
$\displaystyle \int 6x+4 + \frac{2}{x+1} + \frac{1}{x-1} dx$
This should integrate to...
$\displaystyle 3x^2+4x+2ln(x+1) + ln(x-1) + C$
this answer appears to be wrong...
Where did I go wrong?
Thank you!
I think you'll find that A = 2 and B = 1.
4. Originally Posted by Vamz
$\displaystyle 3x^2+4x+2ln(x+1) + ln(x-1) + C$
I must be getting picky in my old age....
The solution is
$\displaystyle 3x^2+4x+2ln|x+1| + ln|x-1| + C$
-Dan
| 620
| 1,458
|
{"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}
| 4.34375
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.667106
|
https://www.scribd.com/document/371668321/2017-2018-observation-2
| 1,571,644,081,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00274.warc.gz
| 1,072,343,435
| 67,248
|
You are on page 1of 4
# ANNUAL PROFESSIONAL PERFORMANCE REVIEW (APPR)
## TEACHER OBSERVATION REPORT
Teacher Name: CAITLIN FREDRICKS Teacher ID: 1565925
29Q109-Jean Nuzzi Intermediate
School Year: 2017-2018 School Name/DBN: School
## CLASSROOM OBSERVATION (OBS):
In each observation, all components for which there is observed evidence must be rated. Each form must
contain lesson-specific evidence for each of the components observed during a classroom observation.
## This observation was: (check one)
Formal Observation (full period) Informal Observation (15 minute minimum)
## Date of Observation: 12/04/2017 Time/Period: 12:50 Period 7
Component Ratings
## 1a (obs): Demonstrating knowledge of content and pedagogy 4- Highly Effective
Ms. Fredricks displays extensive knowledge of the important concepts in the
discipline and how one-step/two-step equations relate to the inverse operation of
multiplication and division.
## 1e (obs): Designing coherent instruction 3- Effective
The learning activities are aligned with the instructional outcomes and follow an
organized progression suitable to groups of students. The learning activities have
reasonable time allocations, with some differentiation for different groups of
students. During my observation the lesson opened up with a warm up from the
Go Math textbooks. The the direction instruction you reviewed the difference of
one-step and two step equations. The student activity was for the students to write
and solve the equation for the geometric shape.
## 2a: Creating an environment of respect and rapport 4- Highly Effective
Classroom interactions between Ms. Fredricks and her students are highly
respectful. This is evident by the Ms. Fredricks praising her students conceptional
understanding as they performed the warm up.
## 2d: Managing student behavior 4- Highly Effective
Student behavior is entirely appropriate. Ms. Fredricks circulated the classroom to
ensure that the students were on task.
## 3b: Using questioning and discussion techniques 3- Effective
Throughout the lesson probing questions were asked to assess students
understanding of the process of writing and solving equations.
## Last Revised: 12/19/17 11:20:07 AM By dmarsh5
1. What do we need to have in order to have an algebraic expression? Sabrina,"A
variable"
## 3c: Engaging students in learning 4- Highly Effective
All students are engaged in challenging content through well-designed learning
tasks and activities that require complex thinking by students. As I circulated the
classroom the students were writing equation for the distance between the three
walls. In addition students had to decide which unit if measure to convert in as it
relates to inches and feet. Most of the students converted into inches.
## 3d: Using assessment in instruction 3- Effective
Questions and assessments are regularly used to diagnose evidence of learning.
Students were assessed by their explanation of knowing which operation to use in
solving an equation. Ms. Fredricks provided feedback on the purpose of using the
inverse operation to balance equations as they isolate the variable.
## 4e (obs): Growing and developing professionally 3- Effective
Ms. Fredrick continues to implement technology in her instruction to engage her
students in the lesson.
## Last Revised: 12/19/17 11:20:07 AM By dmarsh5
Teacher ID 1565925 Teacher Name CAITLIN FREDRICKS
## ASSESSMENT OF PREPARATION AND PROFESSIONALISM (P&P):
In this section of the form, evaluators should rate evidence for components 1a, 1e, and 4e that was
observed within fifteen (15) school days prior to the classroom observation as part of an assessment
of a teacher’s preparation and professionalism. Each form must contain teacher-specific evidence
for each of the components observed.
Component Ratings
## 4e (p&p): Growing and developing professionally N/A
On 12/19/17 an email was sent to provide feedback on the informal observation performed on 12/4/17
## LEQ: How do we write algebraic expression?
Ms. Fredricks as I walked into your classroom the students were working on the warm up from the Go Math
books. After the warm up you did a short review of one and two-step equations on the smartboard. During
your direct instruction you wanted the students to understand the process of using the inverse operation to
solve equations. In your presentation you displayed key points using the inverse operation to balance
equations, reminding students what they perform on side they should perform on the other side.
Commendations: Your students were actively engage in the student activity. You provided a rigorous task for
your students in writing expressions and equations using geometric shapes. During your direct instruction
you asked probing questions to gauge students prior knowledge of the process of solving two-step equations
and understanding the difference in using the inverse operation to isolate the variable.
Recommendations: Ms. Fredricks as you provide instruction in your direct instruction I need to see more
students come to the board to show their work and explain the process. Midway throught he students activity
have some of the students share their reasoning for converting from feet to inches. I believe this part of the
activity is essential to students learning which init of measure may be the better option to convert into.
## Last Revised: 12/19/17 11:20:07 AM By dmarsh5
Teacher ID 1565925 Teacher Name CAITLIN FREDRICKS
## Teacher's signature: Date
(I have read and received a copy of the above and understand that a copy will be placed in my file.)
| 1,224
| 5,621
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2019-43
|
latest
|
en
| 0.911817
|
http://blog.csdn.net/Miracle_ma/article/details/51557673
| 1,503,049,477,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-34/segments/1502886104631.25/warc/CC-MAIN-20170818082911-20170818102911-00463.warc.gz
| 53,520,164
| 22,612
|
# 离线专题学习
149人阅读 评论(0)
hdu 4288
1ikai[i%5=3]
sum[rt][i]=sum[lrt][i]+sum[rrt][5((num[lrt]i1)%5+5)%51]
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 100005
#define MAXN 500005
#define maxnode 105
#define sigma_size 2
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000
//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const ull mx = 1333333331;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
char op[MAX][10];
int a[MAX];
int b[MAX];
LL sum[MAX<<2][5];
int num[MAX<<2];
void build(int l,int r,int rt){
num[rt]=0;
for(int i=0;i<5;i++) sum[rt][i]=0;
if(l==r) return;
middle;
build(lson);
build(rson);
}
void pushup(int rt){
for(int i=0;i<5;i++)
sum[rt][i]=sum[lrt][i]+sum[rrt][5-((num[lrt]-i-1)%5+5)%5-1];
}
void update(int l,int r,int rt,int pos,int d,int v){
num[rt]+=v;
if(l==r){
sum[rt][0]+=d;
return;
}
middle;
if(pos<=m) update(lson,pos,d,v);
else update(rson,pos,d,v);
pushup(rt);
}
int main(){
//freopen("test.txt","r",stdin);
int n;
while(~scanf("%d",&n)){
int tmp=0;
for(int i=0;i<n;i++){
scanf("%s",op[i]);
if(op[i][0]!='s'){
scanf("%d",&a[i]);
b[tmp++]=a[i];
}
}
build(1,n,1);
sort(b,b+tmp);
int tot=unique(b,b+tmp)-b;
for(int i=0;i<n;i++){
int pos=lower_bound(b,b+tot,a[i])-b+1;
if(op[i][0]=='a') update(1,n,1,pos,a[i],1);
else if(op[i][0]=='d') update(1,n,1,pos,-a[i],-1);
else printf("%I64d\n",sum[1][2]);
}
}
return 0;
}
hdu 4417
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 100005
#define MAXN 500005
#define maxnode 105
#define sigma_size 2
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000
//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const ull mx = 1333333331;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
struct num{
int a,pos;
bool operator < (const num &b) const{
return a<b.a;
}
}pp[MAX];
struct Node{
int l,r,h,id,ans;
}p[MAX];
bool cmp1(Node a,Node b){
return a.h<b.h;
}
bool cmp2(Node a,Node b){
return a.id<b.id;
}
int sum[MAX<<2];
void build(int l,int r,int rt){
sum[rt]=0;
if(l==r) return ;
middle;
build(lson);
build(rson);
}
void pushup(int rt){
sum[rt]=sum[lrt]+sum[rrt];
}
void update(int l,int r,int rt,int pos){
if(l==r){
sum[rt]++;
return;
}
middle;
if(pos<=m) update(lson,pos);
else update(rson,pos);
pushup(rt);
}
int query(int l,int r,int rt,int L,int R){
if(L<=l&&r<=R) return sum[rt];
middle;
int ans=0;
if(L<=m) ans+=query(lson, L,R);
if(R>m) ans+=query(rson,L,R);
return ans;
}
int main(){
//freopen("test.txt","r",stdin);
int t,kase=0;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
for(int i=0;i<n;i++){
int a;
scanf("%d",&a);
pp[i]=(num){a,i+1};
}
sort(pp,pp+n);
for(int i=0;i<m;i++){
int l,r,h;
scanf("%d%d%d",&l,&r,&h);
p[i]=(Node){l+1,r+1,h,i,0};
}
sort(p,p+m,cmp1);
int ret=0;
build(1,n,1);
for(int i=0;i<m;i++){
while(pp[ret].a<=p[i].h&&ret<n){
update(1,n,1,pp[ret].pos);
ret++;
}
p[i].ans=query(1,n,1,p[i].l,p[i].r);
}
sort(p,p+m,cmp2);
kase++;
printf("Case %d:\n",kase);
for(int i=0;i<m;i++) printf("%d\n",p[i].ans);
}
return 0;
}
hdu 3874
http://acm.hdu.edu.cn/showproblem.php?pid=3874
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 50005
#define MAXN 6005
#define maxnode 15
#define sigma_size 30
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000
//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
//const double inf = 1e18;
const double eps = 1e-8;
const LL mod = 1e9+7;
const ull mx = 133333331;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
struct Node{
int l,r,id;
bool operator < (const Node&e)const{
return r>e.r;
}
}p[MAX*4];
LL ans[MAX*4];
int a[MAX];
int pos[1000005];
int pre[MAX];
LL sum[MAX<<2];
void pushup(int rt){
sum[rt]=sum[lrt]+sum[rrt];
}
void build(int l,int r,int rt){
sum[rt]=0;
if(l==r) return;
middle;
build(lson);
build(rson);
pushup(rt);
}
void update(int l,int r,int rt,int pos,int d){
if(l==r){
sum[rt]=d;
return;
}
middle;
if(pos<=m) update(lson,pos,d);
else update(rson,pos,d);
pushup(rt);
}
LL query(int l,int r,int rt,int L,int R){
if(L<=l&&r<=R) return sum[rt];
middle;
LL ans=0;
if(L<=m) ans+=query(lson,L,R);
if(R>m) ans+=query(rson,L,R);
return ans;
}
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
mem(pre,-1);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
pos[a[i]]=-1;
}
for(int i=1;i<=n;i++){
if(pos[a[i]]!=-1){
pre[i]=pos[a[i]];
pos[a[i]]=i;
}
else pos[a[i]]=i;
}
build(1,n,1);
int m;
cin>>m;
for(int i=0;i<m;i++){
int l,r;
scanf("%d%d",&l,&r);
p[i]=(Node){l,r,i};
}
sort(p,p+m);
for(int i=1;i<=n;i++){
if(pre[i]!=-1) update(1,n,1,pre[i],0);
update(1,n,1,i,a[i]);
}
int r=n;
for(int i=0;i<m;i++){
while(r>p[i].r){
if(pre[r]!=-1) update(1,n,1,pre[r],a[r]);
r--;
}
ans[p[i].id]=query(1,n,1,p[i].l,p[i].r);
}
for(int i=0;i<m;i++){
printf("%I64d\n",ans[i]);
}
}
return 0;
}
hdu 4638
http://acm.hdu.edu.cn/showproblem.php?pid=4638
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 100005
#define MAXN 6005
#define maxnode 15
#define sigma_size 30
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000
//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
//const double inf = 1e18;
const double eps = 1e-8;
const LL mod = 1e9+7;
const ull mx = 133333331;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
int pos[MAX];
int a[MAX];
struct que{
int l,r,id;
bool operator < (const que&e)const{
return r<e.r;
}
}p[MAX];
int ans[MAX];
int sum[MAX<<2];
void pushup(int rt){
sum[rt]=sum[lrt]+sum[rrt];
}
void build(int l,int r,int rt){
if(l==r){
sum[rt]=0;
return;
}
middle;
build(lson);
build(rson);
pushup(rt);
}
void update(int l,int r,int rt,int pos,int d){
if(l==r){
sum[rt]+=d;
return;
}
middle;
if(pos<=m) update(lson,pos,d);
else update(rson,pos,d);
pushup(rt);
}
int query(int l,int r,int rt,int L,int R){
if(L<=l&&r<=R) return sum[rt];
middle;
int ans=0;
if(L<=m) ans+=query(lson,L,R);
if(R>m) ans+=query(rson,L,R);
return ans;
}
int main(){
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
pos[a[i]]=i;
}
for(int i=0;i<m;i++){
int l,r;
scanf("%d%d",&l,&r);
p[i]=(que){l,r,i};
}
sort(p,p+m);
build(1,n,1);
int tot=0;
for(int i=1;i<=n;i++){
update(1,n,1,i,1);
if(pos[a[i]-1]&&pos[a[i]-1]<=i) update(1,n,1,pos[a[i]-1],-1);
if(pos[a[i]+1]&&pos[a[i]+1]<=i) update(1,n,1,pos[a[i]+1],-1);
while(tot<m&&p[tot].r==i){
ans[p[tot].id]=query(1,n,1,p[tot].l,p[tot].r);
tot++;
}
}
for(int i=0;i<m;i++) printf("%d\n",ans[i]);
}
return 0;
}
0
0
* 以上用户言论只代表其个人观点，不代表CSDN网站的观点或立场
个人资料
• 访问：119049次
• 积分：4227
• 等级：
• 排名：第7182名
• 原创：314篇
• 转载：5篇
• 译文：0篇
• 评论：45条
阅读排行
最新评论
| 3,629
| 10,431
|
{"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.828125
| 3
|
CC-MAIN-2017-34
|
longest
|
en
| 0.131794
|
https://pomegranate.readthedocs.io/en/v0.8.1/BayesianNetwork.html
| 1,680,381,074,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00401.warc.gz
| 508,845,930
| 15,300
|
# Bayesian Networks¶
Bayesian networks are a probabilistic model that are especially good at inference given incomplete data. Much like a hidden Markov model, they consist of a directed graphical model (though Bayesian networks must also be acyclic) and a set of probability distributions. The edges encode dependency statements between the variables, where the lack of an edge between any pair of variables indicates a conditional independence. Each node encodes a probability distribution, where root nodes encode univariate probability distributions and inner/leaf nodes encode conditional probability distributions. Bayesian networks are exceptionally flexible when doing inference, as any subset of variables can be observed, and inference done over all other variables, without needing to define these groups in advance. In fact, the set of observed variables can change from one sample to the next without needing to modify the underlying algorithm at all.
Currently, pomegranate only supports discrete Bayesian networks, meaning that the values must be categories, i.e. ‘apples’ and ‘oranges’, or 1 and 2, where 1 and 2 refer to categories, not numbers, and so 2 is not explicitly ‘bigger’ than 1.
## Initialization¶
Bayesian networks can be initialized in two ways, depending on whether the underlying graphical structure is known or not: (1) the graphical structure can be built one node at a time with pre-initialized distributions set for each node, or (2) both the graphical structure and distributions can be learned directly from data. This mirrors the other models that are implemented in pomegranate. However, typically expectation maximization is used to fit the parameters of the distribution, and so initialization (such as through k-means) is typically fast whereas fitting is slow. For Bayesian networks, the opposite is the case. Fitting can be done quickly by just summing counts through the data, while initialization is hard as it requires an exponential time search through all possible DAGs to identify the optimal graph. More is discussed in the tutorials above and in the fitting section below.
Let’s take a look at initializing a Bayesian network in the first manner by quickly implementing the Monty Hall problem. The Monty Hall problem arose from the gameshow Let’s Make a Deal, where a guest had to choose which one of three doors had a prize behind it. The twist was that after the guest chose, the host, originally Monty Hall, would then open one of the doors the guest did not pick and ask if the guest wanted to switch which door they had picked. Initial inspection may lead you to believe that if there are only two doors left, there is a 50-50 chance of you picking the right one, and so there is no advantage one way or the other. However, it has been proven both through simulations and analytically that there is in fact a 66% chance of getting the prize if the guest switches their door, regardless of the door they initially went with.
Our network will have three nodes, one for the guest, one for the prize, and one for the door Monty chooses to open. The door the guest initially chooses and the door the prize is behind are uniform random processes across the three doors, but the door which Monty opens is dependent on both the door the guest chooses (it cannot be the door the guest chooses), and the door the prize is behind (it cannot be the door with the prize behind it).
from pomegranate import *
guest = DiscreteDistribution({'A': 1./3, 'B': 1./3, 'C': 1./3})
prize = DiscreteDistribution({'A': 1./3, 'B': 1./3, 'C': 1./3})
monty = ConditionalProbabilityTable(
[['A', 'A', 'A', 0.0],
['A', 'A', 'B', 0.5],
['A', 'A', 'C', 0.5],
['A', 'B', 'A', 0.0],
['A', 'B', 'B', 0.0],
['A', 'B', 'C', 1.0],
['A', 'C', 'A', 0.0],
['A', 'C', 'B', 1.0],
['A', 'C', 'C', 0.0],
['B', 'A', 'A', 0.0],
['B', 'A', 'B', 0.0],
['B', 'A', 'C', 1.0],
['B', 'B', 'A', 0.5],
['B', 'B', 'B', 0.0],
['B', 'B', 'C', 0.5],
['B', 'C', 'A', 1.0],
['B', 'C', 'B', 0.0],
['B', 'C', 'C', 0.0],
['C', 'A', 'A', 0.0],
['C', 'A', 'B', 1.0],
['C', 'A', 'C', 0.0],
['C', 'B', 'A', 1.0],
['C', 'B', 'B', 0.0],
['C', 'B', 'C', 0.0],
['C', 'C', 'A', 0.5],
['C', 'C', 'B', 0.5],
['C', 'C', 'C', 0.0]], [guest, prize])
s1 = Node(guest, name="guest")
s2 = Node(prize, name="prize")
s3 = Node(monty, name="monty")
model = BayesianNetwork("Monty Hall Problem")
model.bake()
Note
The objects ‘state’ and ‘node’ are really the same thing and can be used interchangable. The only difference is the name, as hidden Markov models use ‘state’ in the literature frequently whereas Bayesian networks use ‘node’ frequently.
The conditional distribution must be explicitly spelled out in this example, followed by a list of the parents in the same order as the columns take in the tabble that is provided (e.g. the columns in the table correspond to guest, prize, monty, probability.)
However, one can also initialize a Bayesian network based completely on data. As mentioned before, the exact version of this algorithm takes exponential time with the number of variables and typically can’t be done on more than ~25 variables. This is because there are a super-exponential number of directed acyclic graphs that one could define over a set of variables, but fortunately one can use dynamic programming in order to reduce this complexity down to “simply exponential.” The implementation of the exact algorithm actually goes further than the original dynamic programing algorithm by implementing an A* search to somewhat reduce computational time but drastically reduce required memory, sometimes by an order of magnitude.
from pomegranate import *
import numpy
model = BayesianNetwork.from_samples(X, algorithm='exact')
The exact algorithm is not the default, though. The default is a novel greedy algorithm that greedily chooses a topological ordering of the variables, but optimally identifies the best parents for each variable given this ordering. It is significantly faster and more memory efficient than the exact algorithm and produces far better estimates than using a Chow-Liu tree. This is set to the default to avoid locking up the computers of users that unintentionally tell their computers to do a near-impossible task.
## Probability¶
You can calculate the probabiity of a sample under a Bayesian network as the product of the probability of each variable given its parents, if it has any. This can be expressed as $$P = \prod\limits_{i=1}^{d} P(D_{i}|Pa_{i})$$ for a sample with $d$ dimensions. For example, in the Monty Hal problem, the probability of a show is the probability of the guest choosing the respective door, times the probability of the prize being behind a given door, times the probability of Monty opening a given door given the previous two values. For example, using the manually initialized network above:
>>> print model.probability([['A', 'A', 'A'],
['A', 'A', 'B'],
['C', 'C', 'B']])
[ 0. 0.05555556 0.05555556]
## Prediction¶
Bayesian networks are frequently used to infer/impute the value of missing variables given the observed values. In other models, typically there is either a single or fixed set of missing variables, such as latent factors, that need to be imputed, and so returning a fixed vector or matrix as the predictions makes sense. However, in the case of Bayesian networks, we can make no such assumptions, and so when data is passed in for prediction it should be in the format as a matrix with None in the missing variables that need to be inferred. The return is thus a filled in matrix where the Nones have been replaced with the imputed values. For example:
>>> print model.predict([['A', 'B', None],
['A', 'C', None],
['C', 'B', None]])
[['A' 'B' 'C']
['A' 'C' 'B']
['C' 'B' 'A']]
In this example, the final column is the one that is always missing, but a more complex example is as follows:
>>> print model.predict([['A', 'B', None],
['A', None, 'C'],
[None, 'B', 'A']])
[['A' 'B' 'C']
['A' 'B' 'C']
['C' 'B' 'A']]
## Fitting¶
Fitting a Bayesian network to data is a fairly simple process. Essentially, for each variable, you need consider only that column of data and the columns corresponding to that variables parents. If it is a univariate distribution, then the maximum likelihood estimate is just the count of each symbol divided by the number of samples in the data. If it is a multivariate distribution, it ends up being the probability of each symbol in the variable of interest given the combination of symbols in the parents. For example, consider a binary dataset with two variables, X and Y, where X is a parent of Y. First, we would go through the dataset and calculate P(X=0) and P(X=1). Then, we would calculate P(Y=0|X=0), P(Y=1|X=0), P(Y=0|X=1), and P(Y=1|X=1). Those values encode all of the parameters of the Bayesian network.
## API Reference¶
class pomegranate.BayesianNetwork.BayesianNetwork
A Bayesian Network Model.
A Bayesian network is a directed graph where nodes represent variables, edges represent conditional dependencies of the children on their parents, and the lack of an edge represents a conditional independence.
Parameters: name : str, optional The name of the model. Default is None states : list, shape (n_states,) A list of all the state objects in the model graph : networkx.DiGraph The underlying graph object.
add_edge()
Add a transition from state a to state b which indicates that B is dependent on A in ways specified by the distribution.
add_node()
Add a node to the graph.
add_nodes()
Add multiple states to the graph.
add_state()
Another name for a node.
add_states()
Another name for a node.
add_transition()
Transitions and edges are the same.
bake()
Finalize the topology of the model.
Assign a numerical index to every state and create the underlying arrays corresponding to the states and edges between the states. This method must be called before any of the probability-calculating methods. This includes converting conditional probability tables into joint probability tables and creating a list of both marginal and table nodes.
Parameters: None None
clear_summaries()
Clear the summary statistics stored in the object.
copy()
Return a deep copy of this distribution object.
This object will not be tied to any other distribution or connected in any form.
Parameters: None distribution : Distribution A copy of the distribution with the same parameters.
dense_transition_matrix()
Returns the dense transition matrix. Useful if the transitions of somewhat small models need to be analyzed.
edge_count()
Returns the number of edges present in the model.
fit()
Fit the model to data using MLE estimates.
Fit the model to the data by updating each of the components of the model, which are univariate or multivariate distributions. This uses a simple MLE estimate to update the distributions according to their summarize or fit methods.
This is a wrapper for the summarize and from_summaries methods.
Parameters: X : array-like, shape (n_samples, n_nodes) The data to train on, where each row is a sample and each column corresponds to the associated variable. weights : array-like, shape (n_nodes), optional The weight of each sample as a positive double. Default is None. inertia : double, optional The inertia for updating the distributions, passed along to the distribution method. Default is 0.0. pseudocount : double, optional A pseudocount to add to the emission of each distribution. This effectively smoothes the states to prevent 0. probability symbols if they don’t happen to occur in the data. Only effects hidden Markov models defined over discrete distributions. Default is 0. verbose : bool, optional Whether or not to print out improvement information over iterations. Only required if doing semisupervised learning. Default is False. n_jobs : int The number of jobs to use to parallelize, either the number of threads or the number of processes to use. -1 means use all available resources. Default is 1. self : BayesianNetwork The fit Bayesian network object with updated model parameters.
freeze()
Freeze the distribution, preventing updates from occuring.
from_json()
Read in a serialized Bayesian Network and return the appropriate object.
Parameters: s : str A JSON formatted string containing the file. model : object A properly initialized and baked model.
from_samples()
Learn the structure of the network from data.
Find the structure of the network from data using a Bayesian structure learning score. This currently enumerates all the exponential number of structures and finds the best according to the score. This allows weights on the different samples as well. The score that is optimized is the minimum description length (MDL).
Parameters: X : array-like, shape (n_samples, n_nodes) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : array-like, shape (n_nodes), optional The weight of each sample as a positive double. Default is None. algorithm : str, one of ‘chow-liu’, ‘greedy’, ‘exact’, ‘exact-dp’ optional The algorithm to use for learning the Bayesian network. Default is ‘greedy’ that greedily attempts to find the best structure, and frequently can identify the optimal structure. ‘exact’ uses DP/A* to find the optimal Bayesian network, and ‘exact-dp’ tries to find the shortest path on the entire order lattice, which is more memory and computationally expensive. ‘exact’ and ‘exact-dp’ should give identical results, with ‘exact-dp’ remaining an option mostly for debugging reasons. ‘chow-liu’ will return the optimal tree-like structure for the Bayesian network, which is a very fast approximation but not always the best network. max_parents : int, optional The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. root : int, optional For algorithms which require a single root (‘chow-liu’), this is the root for which all edges point away from. User may specify which column to use as the root. Default is the first column. constraint_graph : networkx.DiGraph or None, optional A directed graph showing valid parent sets for each variable. Each node is a set of variables, and edges represent which variables can be valid parents of those variables. The naive structure learning task is just all variables in a single node with a self edge, meaning that you know nothing about pseudocount : double, optional A pseudocount to add to the emission of each distribution. This effectively smoothes the states to prevent 0. probability symbols if they don’t happen to occur in the data. Default is 0. state_names : array-like, shape (n_nodes), optional A list of meaningful names to be applied to nodes name : str, optional The name of the model. Default is None. reduce_dataset : bool, optional Given the discrete nature of these datasets, frequently a user will pass in a dataset that has many identical samples. It is time consuming to go through these redundant samples and a far more efficient use of time to simply calculate a new dataset comprised of the subset of unique observed samples weighted by the number of times they occur in the dataset. This typically will speed up all algorithms, including when using a constraint graph. Default is True. n_jobs : int, optional The number of threads to use when learning the structure of the network. If a constraint graph is provided, this will parallelize the tasks as directed by the constraint graph. If one is not provided it will parallelize the building of the parent graphs. Both cases will provide large speed gains. model : BayesianNetwork The learned BayesianNetwork.
from_structure()
Return a Bayesian network from a predefined structure.
Pass in the structure of the network as a tuple of tuples and get a fit network in return. The tuple should contain n tuples, with one for each node in the graph. Each inner tuple should be of the parents for that node. For example, a three node graph where both node 0 and 1 have node 2 as a parent would be specified as ((2,), (2,), ()).
Parameters: X : array-like, shape (n_samples, n_nodes) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. structure : tuple of tuples The parents for each node in the graph. If a node has no parents, then do not specify any parents. weights : array-like, shape (n_nodes), optional The weight of each sample as a positive double. Default is None. pseudocount : double, optional A pseudocount to add to the emission of each distribution. This effectively smoothes the states to prevent 0. probability symbols if they don’t happen to occur in the data. Default is 0. name : str, optional The name of the model. Default is None. state_names : array-like, shape (n_nodes), optional A list of meaningful names to be applied to nodes n_jobs : int The number of jobs to use to parallelize, either the number of threads or the number of processes to use. -1 means use all available resources. Default is 1. model : BayesianNetwoork A Bayesian network with the specified structure.
from_summaries()
Use MLE on the stored sufficient statistics to train the model.
This uses MLE estimates on the stored sufficient statistics to train the model.
Parameters: inertia : double, optional The inertia for updating the distributions, passed along to the distribution method. Default is 0.0. pseudocount : double, optional A pseudocount to add to the emission of each distribution. This effectively smoothes the states to prevent 0. probability symbols if they don’t happen to occur in the data. Default is 0. None
log_probability()
Return the log probability of samples under the Bayesian network.
The log probability is just the sum of the log probabilities under each of the components. The log probability of a sample under the graph A -> B is just P(A)*P(B|A). This will return a vector of log probabilities, one for each sample.
Parameters: X : array-like, shape (n_samples, n_dim) The sample is a vector of points where each dimension represents the same variable as added to the graph originally. It doesn’t matter what the connections between these variables are, just that they are all ordered the same. n_jobs : int The number of jobs to use to parallelize, either the number of threads or the number of processes to use. -1 means use all available resources. Default is 1. logp : numpy.ndarray or double The log probability of the samples if many, or the single log probability.
marginal()
Return the marginal probabilities of each variable in the graph.
This is equivalent to a pass of belief propogation on a graph where no data has been given. This will calculate the probability of each variable being in each possible emission when nothing is known.
Parameters: None marginals : array-like, shape (n_nodes) An array of univariate distribution objects showing the marginal probabilities of that variable.
node_count()
Returns the number of nodes/states in the model
plot()
Draw this model’s graph using NetworkX and matplotlib.
Note that this relies on networkx’s built-in graphing capabilities (and not Graphviz) and thus can’t draw self-loops.
See networkx.draw_networkx() for the keywords you can pass in.
Parameters: **kwargs : any The arguments to pass into networkx.draw_networkx() None
predict()
Predict missing values of a data matrix using MLE.
Impute the missing values of a data matrix using the maximally likely predictions according to the forward-backward algorithm. Run each sample through the algorithm (predict_proba) and replace missing values with the maximally likely predicted emission.
Parameters: X : array-like, shape (n_samples, n_nodes) Data matrix to impute. Missing values must be either None (if lists) or np.nan (if numpy.ndarray). Will fill in these values with the maximally likely ones. max_iterations : int, optional Number of iterations to run loopy belief propogation for. Default is 100. n_jobs : int The number of jobs to use to parallelize, either the number of threads or the number of processes to use. -1 means use all available resources. Default is 1. X : numpy.ndarray, shape (n_samples, n_nodes) This is the data matrix with the missing values imputed.
predict_proba()
Returns the probabilities of each variable in the graph given evidence.
This calculates the marginal probability distributions for each state given the evidence provided through loopy belief propogation. Loopy belief propogation is an approximate algorithm which is exact for certain graph structures.
Parameters: data : dict or array-like, shape <= n_nodes, optional The evidence supplied to the graph. This can either be a dictionary with keys being state names and values being the observed values (either the emissions or a distribution over the emissions) or an array with the values being ordered according to the nodes incorporation in the graph (the order fed into .add_states/add_nodes) and None for variables which are unknown. If nothing is fed in then calculate the marginal of the graph. Default is {}. max_iterations : int, optional The number of iterations with which to do loopy belief propogation. Usually requires only 1. Default is 100. check_input : bool, optional Check to make sure that the observed symbol is a valid symbol for that distribution to produce. Default is True. probabilities : array-like, shape (n_nodes) An array of univariate distribution objects showing the probabilities of each variable.
probability()
Return the probability of the given symbol under this distribution.
Parameters: symbol : object The symbol to calculate the probability of probability : double The probability of that point under the distribution.
sample()
Return a random item sampled from this distribution.
Parameters: n : int or None, optional The number of samples to return. Default is None, which is to generate a single sample. sample : double or object Returns a sample from the distribution of a type in the support of the distribution.
state_count()
Returns the number of states present in the model.
summarize()
Summarize a batch of data and store the sufficient statistics.
This will partition the dataset into columns which belong to their appropriate distribution. If the distribution has parents, then multiple columns are sent to the distribution. This relies mostly on the summarize function of the underlying distribution.
Parameters: X : array-like, shape (n_samples, n_nodes) The data to train on, where each row is a sample and each column corresponds to the associated variable. weights : array-like, shape (n_nodes), optional The weight of each sample as a positive double. Default is None. None
thaw()
Thaw the distribution, re-allowing updates to occur.
to_json()
Serialize the model to a JSON.
Parameters: separators : tuple, optional The two separaters to pass to the json.dumps function for formatting. indent : int, optional The indentation to use at each level. Passed to json.dumps for formatting. json : str A properly formatted JSON object.
class pomegranate.BayesianNetwork.ParentGraph
Generate a parent graph for a single variable over its parents.
This will generate the parent graph for a single parents given the data. A parent graph is the dynamically generated best parent set and respective score for each combination of parent variables. For example, if we are generating a parent graph for x1 over x2, x3, and x4, we may calculate that having x2 as a parent is better than x2,x3 and so store the value of x2 in the node for x2,x3.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. parent_set : tuple, default () The variables which are possible parents for this variable. If nothing is passed in then it defaults to all other variables, as one would expect in the naive case. This allows for cases where we want to build a parent graph over only a subset of the variables. structure : tuple, shape=(d,) The parents for each variable in this SCC
pomegranate.BayesianNetwork.discrete_exact_a_star()
Find the optimal graph over a set of variables with no other knowledge.
This is the naive dynamic programming structure learning task where the optimal graph is identified from a set of variables using an order graph and parent graphs. This can be used either when no constraint graph is provided or for a SCC which is made up of a node containing a self-loop. It uses DP/A* in order to find the optimal graph without considering all possible topological sorts. A greedy version of the algorithm can be used that massively reduces both the computational and memory cost while frequently producing the optimal graph.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. n_jobs : int The number of threads to use when learning the structure of the network. This parallelizes the creation of the parent graphs. structure : tuple, shape=(d,) The parents for each variable in this SCC
pomegranate.BayesianNetwork.discrete_exact_component()
Find the optimal graph over a multi-node component of the constaint graph.
The general algorithm in this case is to begin with each variable and add all possible single children for that entry recursively until completion. This will result in a far sparser order graph than before. In addition, one can eliminate entries from the parent graphs that contain invalid parents as they are a fast of computational time.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. n_jobs : int The number of threads to use when learning the structure of the network. This parallelizes the creation of the parent graphs. structure : tuple, shape=(d,) The parents for each variable in this SCC
pomegranate.BayesianNetwork.discrete_exact_dp()
Find the optimal graph over a set of variables with no other knowledge.
This is the naive dynamic programming structure learning task where the optimal graph is identified from a set of variables using an order graph and parent graphs. This can be used either when no constraint graph is provided or for a SCC which is made up of a node containing a self-loop. This is a reference implementation that uses the naive shortest path algorithm over the entire order graph. The ‘exact’ option uses the A* path in order to avoid considering the full order graph.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. n_jobs : int The number of threads to use when learning the structure of the network. This parallelizes the creation of the parent graphs. structure : tuple, shape=(d,) The parents for each variable in this SCC
pomegranate.BayesianNetwork.discrete_exact_slap()
Find the optimal graph in a node with a Self Loop And Parents (SLAP).
Instead of just performing exact BNSL over the set of all parents and removing the offending edges there are efficiencies that can be gained by considering the structure. In particular, parents not coming from the main node do not need to be considered in the order graph but simply added to each entry after creation of the order graph. This is because those variables occur earlier in the topological ordering but it doesn’t matter how they occur otherwise. Parent graphs must be defined over all variables however.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. n_jobs : int The number of threads to use when learning the structure of the network. This parallelizes the creation of the parent graphs. structure : tuple, shape=(d,) The parents for each variable in this SCC
pomegranate.BayesianNetwork.discrete_exact_with_constraints()
This returns the optimal Bayesian network given a set of constraints.
This function controls the process of learning the Bayesian network by taking in a constraint graph, identifying the strongly connected components (SCCs) and solving each one using the appropriate algorithm. This is mostly an internal function.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. constraint_graph : networkx.DiGraph A directed graph showing valid parent sets for each variable. Each node is a set of variables, and edges represent which variables can be valid parents of those variables. The naive structure learning task is just all variables in a single node with a self edge, meaning that you know nothing about n_jobs : int The number of threads to use when learning the structure of the network. This parallelized both the creation of the parent graphs for each variable and the solving of the SCCs. -1 means use all available resources. Default is 1, meaning no parallelism. structure : tuple, shape=(d,) The parents for each variable in the network.
pomegranate.BayesianNetwork.discrete_exact_with_constraints_task()
This is a wrapper for the function to be parallelzied by joblib.
This function takes in a single task as an id and a set of parents and children and calls the appropriate function. This is mostly a wrapper for joblib to parallelize.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. task : tuple A 3-tuple containing the id, the set of parents and the set of children to learn a component of the Bayesian network over. The cases represent a SCC of the following: 0 - Self loop and no parents 1 - Self loop and parents 2 - Parents and no self loop 3 - Multiple nodes n_jobs : int The number of threads to use when learning the structure of the network. This parallelizes the creation of the parent graphs for each task or the finding of best parents in case 2. structure : tuple, shape=(d,) The parents for each variable in this SCC
pomegranate.BayesianNetwork.discrete_greedy()
Find the optimal graph over a set of variables with no other knowledge.
This is the naive dynamic programming structure learning task where the optimal graph is identified from a set of variables using an order graph and parent graphs. This can be used either when no constraint graph is provided or for a SCC which is made up of a node containing a self-loop. It uses DP/A* in order to find the optimal graph without considering all possible topological sorts. A greedy version of the algorithm can be used that massively reduces both the computational and memory cost while frequently producing the optimal graph.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. greedy : bool, default is True Whether the use a heuristic in order to massive reduce computation and memory time, but without the guarantee of finding the best network. n_jobs : int The number of threads to use when learning the structure of the network. This parallelizes the creation of the parent graphs. structure : tuple, shape=(d,) The parents for each variable in this SCC
pomegranate.BayesianNetwork.generate_parent_graph()
Generate a parent graph for a single variable over its parents.
This will generate the parent graph for a single parents given the data. A parent graph is the dynamically generated best parent set and respective score for each combination of parent variables. For example, if we are generating a parent graph for x1 over x2, x3, and x4, we may calculate that having x2 as a parent is better than x2,x3 and so store the value of x2 in the node for x2,x3.
Parameters: X : numpy.ndarray, shape=(n, d) The data to fit the structure too, where each row is a sample and each column corresponds to the associated variable. weights : numpy.ndarray, shape=(n,) The weight of each sample as a positive double. Default is None. key_count : numpy.ndarray, shape=(d,) The number of unique keys in each column. pseudocount : double A pseudocount to add to each possibility. max_parents : int The maximum number of parents a node can have. If used, this means using the k-learn procedure. Can drastically speed up algorithms. If -1, no max on parents. Default is -1. parent_set : tuple, default () The variables which are possible parents for this variable. If nothing is passed in then it defaults to all other variables, as one would expect in the naive case. This allows for cases where we want to build a parent graph over only a subset of the variables. structure : tuple, shape=(d,) The parents for each variable in this SCC
| 8,045
| 36,495
|
{"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": 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}
| 2.515625
| 3
|
CC-MAIN-2023-14
|
latest
|
en
| 0.914578
|
https://www.spreadsheetweb.com/excel-yield-function/
| 1,695,769,192,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00740.warc.gz
| 1,099,901,335
| 13,307
|
Select Page
The Excel YIELD function is a Financial formula that calculates and returns the yield on a security that pays a periodic interest. A common use case for the YIELD function is calculating bond yields. In this guide, we’re going to show you how to use the Excel YIELD function, and also go over some tips and error handling methods.
Supported versions
• All Excel versions
Excel YIELD Function Syntax
YIELD(settlement, maturity, rate, pr, redemption, frequency, [basis])
Arguments
settlement The security's settlement date. The security settlement date is the date after the issue date when the security is traded to the buyer. maturity The security's maturity date. The maturity date is the date when the security expires. rate The security's interest rate at date of issue. pr The security's price per \$100 face value. redemption The security's redemption value per \$100 face value. frequency The number of coupon payments per year. For annual payments, frequency = 1; for semiannual, frequency = 2; for quarterly, frequency = 4. [basis] Optional. The type of day count basis to be used. 0 or omitted: US (NASD) 30/360 1: Actual/actual 2: Actual/360 3: Actual/365 4: European 30/360
Example
The following example shows how to calculate the yield on a bond purchased on August 8, 2019, with maturity date of February 2nd 2024. The annual rate of interest is 5.0% on the price per \$100 face value is \$101, and the redemption value is \$100. The payments will be made based on the US (NASD) 30/360 day count.
Tips
• Microsoft recommends using the DATE or other similar functions that can return a date serial number.
• Excel keeps date and time values as numbers. Excel assumes that Jan 1st, 1900 is 1, and every subsequent date value is based on this. While whole numbers represent days, decimals represent time values. For example; 1/1/2018 is equal to 43101, and 12:00 is equal to 5.
• Settlement, maturity, and basis are truncated to integers.
• Other price-related functions:
• PRICE returns the price per \$100 face value of a security that pays periodic interest.
• PRICEDISC returns the price per \$100 face value of a discounted security.
• PRICEMAT returns the returns the price per \$100 face value of a security that pays interest at maturity.
• DURATION returns the Macauley duration for an assumed par value of \$100.
Common Issues
• If settlement or maturity are not valid dates, the Excel YIELD function returns the #VALUE! error.
• If rate < 0, YIELD returns the #NUM! error value.
• The YIELD returns the #NUM! error value if pr ≤ 0 or if redemption ≤ 0.
• If [basis] < 0 or if [basis] > 4, YIELD returns the #NUM! error value.
• If settlementmaturity, YIELD returns the #NUM! error value.
| 664
| 2,735
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.46875
| 3
|
CC-MAIN-2023-40
|
latest
|
en
| 0.863181
|
https://palm.muk.uni-hannover.de/trac/changeset/991/palm/trunk/TUTORIAL/SOURCE
| 1,714,043,321,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00877.warc.gz
| 393,370,238
| 5,467
|
# Changeset 991 for palm/trunk/TUTORIAL/SOURCE
Ignore:
Timestamp:
Sep 5, 2012 1:09:35 PM (12 years ago)
Message:
small modifications to sgs_models.tex
File:
1 edited
### Legend:
Unmodified
r987 \end{itemize} $\frac{\partial \bar{e}}{\partial t} = -\bar{u_k} \frac{\partial \bar{e}}{\partial x_k} - \tau_{ki} \frac{\partial \bar{u_i}}{\partial x_k} + \frac{g}{\Theta_0} \overline{u_3' \Theta'} - \frac{\partial}{\partial x_k} \left\{ \overline{u_k' \left( e' + \frac{\pi'}{\rho_0} \right)} \right\} - \epsilon$ \par\bigskip \par\bigskip $\tau_{ki} = -K_{m} \left(\frac{\partial \bar{u_{i}}}{\partial x_{k}} + \frac{\partial \bar{u_{k}}}{\partial x_{i}}\right) + \frac{2}{3}\delta_{ik}\bar{e} \qquad \textnormal{with} \qquad K_{m}=0.1\cdot \Lambda \sqrt{\bar{e}}$ \par\bigskip $H_{k}=\overline{u_k'\Theta'} = -K_{h}\frac{\partial\bar{\Theta}}{\partial x_{k}} \qquad \textnormal{with} \qquad K_{h}= \left(1+2\frac{\Lambda}{\Delta}\right)$ \par\bigskip $W_{k}=\overline{u_k'q'} = -K_{h}\frac{\partial\bar{q}}{\partial x_{k}}$ \par\bigskip $\frac{\partial}{\partial x_k} \left[ \overline{u_k' \left( e' + \frac{\pi'}{\rho_0} \right)} \right] = - \frac{\partial}{\partial x_k} \nu_e \frac{\partial \bar{e}}{\partial x_k}$ \par\bigskip
| 496
| 1,234
|
{"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}
| 3.109375
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.38903
|
https://id.scribd.com/document/106237131/Simulated-Annealing
| 1,571,651,528,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00307.warc.gz
| 531,263,361
| 68,722
|
Anda di halaman 1dari 2
# SIMULATED ANNEALING It is a computation method to find the global minimum of a function having many local minima.
The concept of Simulated Annealing was independently described by Scott Kirkpatrick, C. Daniel Gelatt and Mario P. Vecchi(Optimization by Simulated Annealing) in 1983 and by Vlado ern(Thermodynamical Approach to the Traveling Salesman Problem) in 1985. Simulated Annealing is derived from the physical process of ANNEALING in metallurgy in which a material is heated and then cooled in a controlled manner as a result of which atoms place themselves in a pattern that corresponds to the global energy minimum of a perfect crystal. In this process each point on the function represents a state of a physical system and the value of the function at that point represent the internal energy of the system at that state. Using simulated annealing we try to bring the system to from an arbitrary point to the minimum energy state. We start by comparing the energy of the initial state with a neighbouring state and depending upon the acceptance probability function we either move to neighbouring point or stay at the original point. Let the current state be s and the neighbouring state be s. Now the probability to make the transition from s to s depends upon acceptance probability function P(e,e,T) where e and e represent the energy function as e=E(s) and T is a global varying parameter called temperature. We perform the iterations till we get satisfactory point where the state has minimum internal energy. Here I will try to show how SA is better than traditional iterative methods.
Consider a cost function dependent on a single parameter. The ball represents the current configuration and we want to reach the global minima. Now traditional methods try to go downhill and while reaching a local minima they get stuck at that point .
The SA differs from the traditional method as it allows the ball to move in upward direction in a controlled fashion so that we dont get stuck in local minima. As a heuristic algorithm Simulated Annealing is always useful in problems where the objective function is non-linear with lots of local minimums. It has been used in structural optimization where the problem of designing a sustaining structure with a minimum of materials is studied. SA has also been used in water distribution systems and also in designing diesel engines.
Bibliography:
## 1. Kirkpatrick, S.; Gelatt, C. D.; Vecchi, M. P. (1983). "Optimization by Simulated
Annealing" 2. Rutenbar, R. A.(1989). Simulated Annealing Algorithms: An Overview
| 545
| 2,601
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.109375
| 3
|
CC-MAIN-2019-43
|
latest
|
en
| 0.900694
|
https://ero-blog.com/date/2022/11
| 1,675,938,169,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764501555.34/warc/CC-MAIN-20230209081052-20230209111052-00495.warc.gz
| 247,204,837
| 15,701
|
Unit 10 Circles Homework 5 Inscribed Angles Answer Key
Unit 10 Circles Homework 5 Inscribed Angles Answer Key. Israel67371 israel67371 04/26/2022 mathematics high school. Unit 10 circles homework 5 inscribed angles answer key.
Israel67371 israel67371 04/26/2022 mathematics high school. Find the measures of the labeled angles. Unit 10 circles homework 5 inscribed angles answer key.
Which Factors Affect The Rate Of Osmotic Movement Of Water
Which Factors Affect The Rate Of Osmotic Movement Of Water. What are the factors that affect the rate of osmosis? The higher the temperature, the faster the rate of.
There are many different factors that affect the movement of water in the soil. Osmotic pressure is defined as. Osmosis is the diffusion of water.
You Can Often Tell What A Word Means By
You Can Often Tell What A Word Means By. Spelling 2answers anonymous 1 month ago a. You can infer the meaning of a word.
Web you can often tell what a word means by its _____ in a sentence. The correct answer is b. You can often tell what a word means by its_____in a sentence.
Picking Wildflowers While Walking Through The Woods.
Picking Wildflowers While Walking Through The Woods.. “picking wild flowers while walking in the. ~ ☽ 🌲🦉🔮hello magical beings!🔮🦉🌲 ☾ ~we are the wild mystics and today we are walking through the woods and connecting with nature spirits.
Baby, whatcha say we go pickin’ wildflowers got a spot way back in the woods sneak away for a couple of hours you and me, baby, pickin’ wildflowers take a trail ride if you know what i. Gonna get a little peace on earth. English high school answered identify this group of words.
160 Is What Percent Of 40
160 Is What Percent Of 40. % percent, ÷ division, × multiplication, = the equal sign, / the fraction bar, ≈ approximately the same. 40 is 25 percent of 160 :
% = (number1 ÷ number2) × 100. Steps to solve what percent is 40 of 160? 40 of 160 can be written as: So, 40 is out of 160 = 40 / 160 x 100 =.
A Chromosome Has An Inversion Which Describes A Pericentric Inversion
A Chromosome Has An Inversion Which Describes A Pericentric Inversion. Later it was detected in a wide variety of plant and animal species. Web a basic type of chromosome rearrangement in which a segment that includes the centromere (and so is pericentric) has been snipped out of a chromosome, turned.
Web inversions are rearrangements in chromosomes where small segments break, rotate about 180° and reattach themselves. Pericentric inversion includes centromere, so genes are switched around it. Web mutation a chromosome has an inversion.
A Corporate Vms Has The Advantage Of Controlling
A Corporate Vms Has The Advantage Of Controlling. The vms includes travel agents selling. Web 51) a corporate vms has the advantage of controlling the entire distribution chain under _____.
A corporate vms has the advantage of controlling the entire distribution chain under _____. Web a corporate vms has the advantage of controlling the entire distribution chain under _____ single ownership; Which of the following is a major disadvantage of a.
4/5 Divided By 3/4 In Fraction
4/5 Divided By 3/4 In Fraction. You can also add, subtract, multiply, and. Here is the answer to questions like:
Using a calculator, if you typed in 13 divided by 3, you’d get 4.3333. Divide 4/5 with 3/10 4 5 ÷ 3 10 is 8 3. The fraction calculator will reduce a fraction to its simplest form.
Silver Has A Density Of 10.5 G Cm3
Silver Has A Density Of 10.5 G Cm3. Web the density of silver is 10.5 g/cm 3. Knowing r we can find the diameter.
What is its density in kg/m3 ? Mass =density (ρ) × volume mass of silver =. M g = (19.3 g/cm 3 ) (5 cm 3) = 96.5 g.
| 950
| 3,694
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06
|
latest
|
en
| 0.881095
|
https://forum.bebac.at/forum_entry.php?id=21552
| 1,601,253,272,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00612.warc.gz
| 382,099,203
| 6,890
|
## Reference ranges, prediction intervals [GxP / QC / QA]
Hi all,
pardon me if this is the wrong category. Not sure where else to put it?!?
I am reading up on prediction intervals.
Wikipedia is a good place to start for someone like me. Mainly because I know nothing and am uncritical of most new things.
Wikipedia says: "Prediction intervals are commonly used as definitions of reference ranges, such as reference ranges for blood tests to give an idea of whether a blood test is normal or not. For this purpose, the most commonly used prediction interval is the 95% prediction interval, and a reference range based on it can be called a standard reference range."
I find that very interesting and the idea is appealing and intuitive: We have n observations from a group of people we think are normal. We define the reference range form them. Then we sample the next subjects and checks if they are within the interval of observations defining normal ("standard"), too. And so forth. That interval defining normal builds the sample mean and a sampling variance from the normal ones into the prediction. Sounds right. At least to me when I read it. To you, too?
I have been an auditor and inspector doing work at various path labs for BE on four continents. When I ask to see how they define the reference ranges, the story is often that CROs or path labs sample some subjects and take the 2.5th percentile and the 97.5th percentile to define reference ranges*. I have generally accepted that. I have generally accepted anything as long as the CRO or path lab has given the ref. range some consideration.
Thinking very hard about it I have not ever seen any CRO use the prediction interval approach to define ref. ranges. Should they? What is your opinion here? Have you seen the approach with prediction intervals in use in any operations relating to BE?
*: the exception is hemoglobin and hematocrit. After WHO published their rule about aberrant hemoglobin a few years back, some CROs revised their reference ranges in funny ways so that anyone on the brink of keeling over due to anemia could still be enrolled. Or they revised their SOPs on ways PI's or delegates could clear such subjects for participation. But I digress, this latter aspect is more a question of low ethical standards than a question of reference ranges. OK, I will take a deep inhalation from the Schütozomycin bong now, I promise to be calm the rest of the day.
I could be wrong, but...
Best regards,
ElMaestro
R's base package has 274 reserved words and operators, along with 1761 functions. I can use 18 of them (about 14 of them properly). I believe this makes me the Donald Trump of programming.
| 580
| 2,686
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40
|
longest
|
en
| 0.947194
|
https://www.rcgroups.com/forums/showthread.php?62038-How-To-Convert-dm2-to-oz-sq-ft
| 1,511,121,779,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934805761.46/warc/CC-MAIN-20171119191646-20171119211646-00143.warc.gz
| 869,744,687
| 15,692
|
How To Convert dm2 to oz/sq.ft - RC Groups
Sep 17, 2002, 04:35 PM
Registered User
# How To Convert dm2 to oz/sq.ft
Hi
I was looking to get a plane but they have 27dm2 for there wing loading. How do I convert to oz/sq.ft. The plane is Fieseler Fi 156 Storch and 56" span.
Thks
Wayne
Sep 17, 2002, 04:49 PM Registered User Look on the internet for a program called "Convert" - do a google search with the keywords "convert Measurements program" and you should find it Its a great tool that among other things will do this conversion Alan W Last edited by Alan W; Sep 17, 2002 at 04:56 PM.
Sep 17, 2002, 05:10 PM Simian Member You need more information. You said dm2. Decimeters squared, right? That's area. You need the weight too to convert to oz/ft2.
Sep 17, 2002, 05:18 PM Registered User Presumably g/dm2. So 27 g/dm2 = 8.848 oz/ft2 (approximately)
Sep 17, 2002, 05:18 PM Registered User Hi This is the link to the page of the plane. Like to know wing loading in oz/ft2 something i can relate to. http://mujweb.atlas.cz/www/bohemia/storch.htm Thks Wayne
Sep 17, 2002, 05:26 PM Registered User Dave At bottom of page they say G-55,5g/dm2 and E-66,7g/dm2 I assume one is gas the the other is electric can you give me a approx on these numbers. Thks Wayne
Sep 17, 2002, 06:42 PM Registered User It's a 3 to 1 ratio approx. So your 27g/sq.dm is 9 oz./sq.ft Cheers, Martin (who prefers to do it in his head)
Sep 17, 2002, 08:01 PM Registered User The E-66,7 g/dm2 means 66.7 g/dm2 for the electric version which is about 21.9 oz/ft2.
Sep 18, 2002, 10:02 AM Registered User Thanks Dave seems like its not a slow flyer which I thought of that plane design. thks Wayne
Sep 18, 2002, 11:05 AM Registered User A very nice kit that you can fly electric with a 400/480 geared. I haven't done all the weights yet, but I doubt that the wing loading is as high as 21 oz. I suspect its closer to 16-18. It is not a slow flyer, but you probably could fly it around a baseball field.
Sep 18, 2002, 11:41 AM
Registered User
# Re: How To Convert dm2 to oz/sq.ft
Quote:
Originally posted by wayne Hi I was looking to get a plane but they have 27dm2 for there wing loading. How do I convert to oz/sq.ft. The plane is Fieseler Fi 156 Storch and 56" span. Thks Wayne
A great little program that I found, will convert just about anything to anything.
To convert gm/dm^2 to oz/ft^2, for example you would type :
j <enter>
27 gm/dm^2 <enter>
oz/ft^2 <enter>
Mike
| 765
| 2,457
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2017-47
|
latest
|
en
| 0.919282
|
https://sigmapublications.co.nz/index.php/math-workbooks/ncea-level-1/14-linear-algebra/
| 1,604,104,579,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107912593.62/warc/CC-MAIN-20201031002758-20201031032758-00259.warc.gz
| 467,846,060
| 7,771
|
# NCEA Mathematics
ISBN 978-1-877567-55-1 Curriculum Level NCEA Achievement Standard 1.4 Recommended For Year 11 Number of Pages 52 Edition First
## MWB AS 1.4 Linear Algebra Workbook
#### Author : Wiesje Geldof
Total of 3 Credits - Internally Assessed
The Sigma Workbook 'AS 1.4 - Linear Algebra' is a write-on student workbook covering the mathematics students must understand to gain AchievedMerit, or Excellence in the internally assessed Achievement Standard 1.4 – ‘Apply linear algebra in solving problems’ (AS 91029).
In this workbook students will find clear explanations and detailed worked examples showing the step-by-step processes that will be needed to solve each type of problem, these are followed by a large number of practice exercises and problems of increasing difficulty. At the end of the workbook there is an Achievement Standard Test. Candidates should use this to evaluate their strengths and weaknesses before the internal assessment.
A full set of answers is included, which acts as a guide to structuring quality answers. An abbreviated version of the Achievement Criteria and Explanatory Notes of the Achievement Standard itself complete the resource.
Topics Covered in this Workbook Include :
• Solving Linear Equations
• Solving Inequations
• Forming an Equation
• Equations for Lines
• Linear Problems
• Using Formulas
• Plotting a Graph
• Plotting Lines
• Interpreting Graphs
• Features of Line Graphs
• Writing an Equation for a Line
• Parallel and Crossing Lines
• Solving Simultaneous Equations
• Linear Patterns
• Solving Linear Problems
• Inequations with Two Variables
• Linear programming
The Sigma Workbook 'AS 1.4 Linear Algebra' is best bought before the standard is taught at school and used throughout this time as a classroom and homework resource. Alternatively, the workbook would make an excellent independent study resource in the lead-up to the internal assessment of Achievement Standard 1.4.
Once completed, 'AS 1.4 Linear Algebra' provides a useful set of revision notes at assessment time. It can be kept as a reference book if students intend taking a Year 12 mathematics course.
• 1-2: \$5.95
• 3-19: \$5.00
• 20+: \$4.50
*
| 511
| 2,193
|
{"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}
| 2.921875
| 3
|
CC-MAIN-2020-45
|
longest
|
en
| 0.825785
|
https://www.physicsforums.com/threads/linear-transformation-question.531914/
| 1,511,234,682,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934806310.85/warc/CC-MAIN-20171121021058-20171121041058-00650.warc.gz
| 844,119,817
| 15,121
|
# Linear Transformation Question
1. Sep 19, 2011
### seansrk
Question about linear transformations if you have a matrix such as
| 5 6 9 |
| 5 0 3 |
| 9 -3 -7 |
Can it be a matrix transformation? Or does it have to follow the identity matrix?
Can be a transformation and the "y" transformation being just makes the it flat on the y axis? or does it have to be a form of the identity matrix?
Or am I totally misunderstanding this?
2. Sep 19, 2011
### Alchemista
A matrix is a linear transformation expressed with respect to a basis for the source space and the target space.
Given a linear transformation $T:\mathbb{F}^n \to \mathbb{F}^m$, the corresponding matrix written with respect to a basis $\alpha$ for the source space and a basis $\beta$ for the target space is as follows:
$\left[ \begin{array}{cccc} [T(\alpha_1)]_\beta & [T(\alpha_2)]_\beta & ... & [T(\alpha_n)]_\beta \end{array} \right]$
The theory behind this is as follows. Since any vector in a given vector space can be expressed as a linear combination of a set of basis vectors for that vector space, we need only transform an arbitrary basis to capture the transformation.
Given some vector $v \in \mathbb{F}^n$ and a basis $\alpha$ we can write $v = a_1\alpha_1 + a_2\alpha_2 + ... + a_n\alpha_n$. Then $v$ transformed is as follows
$\begin{eqnarray*} T(v) &=& T(a_1\alpha_1 + a_2\alpha_2 + ... + a_n\alpha_n) \\ &=& T(a_1\alpha_1) + T(a_2\alpha_2) + ... + T(a_n\alpha_n) \\ &=& a_1T(\alpha_1) + a_2T(\alpha_2) + ... + a_nT(\alpha_n) \end{eqnarray*}$
3. Sep 19, 2011
### Fredrik
Staff Emeritus
I don't understand your questions. I don't know what you mean by "follow the identity matrix" or "a form of the identity matrix". Also, how do you define "matrix transformation" if you don't mean "a function defined by a matrix"?
4. Sep 20, 2011
### HallsofIvy
Staff Emeritus
Any m by n matrix is a linear transformation from $R^m$ to $R^n$.
What you have given is a perfectly good linear transformation- although the way you have written it, with the "straight" vertical sides, makes it look more like a determinant than a matrix!
The matrix you give represents the linear transformation that maps a vector, $a\vec{i}+ b\vec{j}+ c\vec{k}[itex] into [itex]a(5\vec{i}+ 5\vec{j}+ 9\vec{k})+ b(6\vec{i}- 3\vec{k})+ c(9\vec{i}+ 3\vec{j}- 7\vec{kl})$$= (5a+ 6b+ 9c)\vec{i}+ (5a+ 3b)\vec{i}+ (9a- 3b- 7c)\vec{k}$.
I wonder if you aren't confusing "matrix", in general, with "invertible matrix".
Last edited: Sep 20, 2011
| 795
| 2,502
|
{"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-2017-47
|
longest
|
en
| 0.830032
|
https://pressbooks.online.ucf.edu/osuniversityphysics/chapter/7-3-work-energy-theorem/
| 1,702,307,784,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00075.warc.gz
| 510,019,863
| 31,205
|
7 Work and Kinetic Energy
# 7.3 Work-Energy Theorem
### Learning Objectives
By the end of this section, you will be able to:
• Apply the work-energy theorem to find information about the motion of a particle, given the forces acting on it
• Use the work-energy theorem to find information about the forces acting on a particle, given information about its motion
We have discussed how to find the work done on a particle by the forces that act on it, but how is that work manifested in the motion of the particle? According to Newton’s second law of motion, the sum of all the forces acting on a particle, or the net force, determines the rate of change in the momentum of the particle, or its motion. Therefore, we should consider the work done by all the forces acting on a particle, or the net work, to see what effect it has on the particle’s motion.
Let’s start by looking at the net work done on a particle as it moves over an infinitesimal displacement, which is the dot product of the net force and the displacement: $d{W}_{\text{net}}={\mathbf{\overset{\to }{F}}}_{\text{net}}\cdot d\mathbf{\overset{\to }{r}}.$ Newton’s second law tells us that ${\mathbf{\overset{\to }{F}}}_{\text{net}}=m(d\mathbf{\overset{\to }{v}}\text{/}dt),$, so $d{W}_{\text{net}}=m(d\mathbf{\overset{\to }{v}}\text{/}dt)\cdot d\mathbf{\overset{\to }{r}}.$ For the mathematical functions describing the motion of a physical particle, we can rearrange the differentials dt, etc., as algebraic quantities in this expression, that is,
$d{W}_{\text{net}}=m(\frac{d\mathbf{\overset{\to }{v}}}{dt})\cdot d\mathbf{\overset{\to }{r}}=md\mathbf{\overset{\to }{v}}\cdot (\frac{d\mathbf{\overset{\to }{r}}}{dt})=m\mathbf{\overset{\to }{v}}\cdot d\mathbf{\overset{\to }{v}},$
where we substituted the velocity for the time derivative of the displacement and used the commutative property of the dot product [(Equation 2.30)]. Since derivatives and integrals of scalars are probably more familiar to you at this point, we express the dot product in terms of Cartesian coordinates before we integrate between any two points A and B on the particle’s trajectory. This gives us the net work done on the particle:
$\begin{array}{cc}\hfill {W}_{\text{net},AB}& ={\int }_{A}^{B}(m{v}_{x}d{v}_{x}+m{v}_{y}d{v}_{y}+m{v}_{z}d{v}_{z})\hfill \\ & =\frac{1}{2}m{|{v}_{x}^{2}+{v}_{y}^{2}+{v}_{z}^{2}|}_{A}^{B}={|\frac{1}{2}m{v}^{2}|}_{A}^{B}={K}_{B}-{K}_{A}.\hfill \end{array}$
In the middle step, we used the fact that the square of the velocity is the sum of the squares of its Cartesian components, and in the last step, we used the definition of the particle’s kinetic energy. This important result is called the work-energy theorem (Figure).
#### Work-Energy Theorem
The net work done on a particle equals the change in the particle’s kinetic energy:
${W}_{\text{net}}={K}_{B}-{K}_{A}.$
According to this theorem, when an object slows down, its final kinetic energy is less than its initial kinetic energy, the change in its kinetic energy is negative, and so is the net work done on it. If an object speeds up, the net work done on it is positive. When calculating the net work, you must include all the forces that act on an object. If you leave out any forces that act on an object, or if you include any forces that don’t act on it, you will get a wrong result.
The importance of the work-energy theorem, and the further generalizations to which it leads, is that it makes some types of calculations much simpler to accomplish than they would be by trying to solve Newton’s second law. For example, in Newton’s Laws of Motion, we found the speed of an object sliding down a frictionless plane by solving Newton’s second law for the acceleration and using kinematic equations for constant acceleration, obtaining
${v}_{\text{f}}^{2}={v}_{\text{i}}^{2}+2g({s}_{\text{f}}-{s}_{\text{i}})\text{sin}\,\theta ,$
where s is the displacement down the plane.
We can also get this result from the work-energy theorem. Since only two forces are acting on the object—gravity and the normal force—and the normal force doesn’t do any work, the net work is just the work done by gravity. This only depends on the object’s weight and the difference in height, so
${W}_{\text{net}}={W}_{\text{grav}}=\text{−}mg({y}_{\text{f}}-{y}_{\text{i}}),$
where y is positive up. The work-energy theorem says that this equals the change in kinetic energy:
$\text{−}mg({y}_{\text{f}}-{y}_{\text{i}})=\frac{1}{2}m({v}_{\text{f}}^{2}-{v}_{\text{i}}^{2}).$
Using a right triangle, we can see that $({y}_{\text{f}}-{y}_{\text{i}})=({s}_{\text{f}}-{s}_{\text{i}})\text{sin}\,\theta ,$ so the result for the final speed is the same.
What is gained by using the work-energy theorem? The answer is that for a frictionless plane surface, not much. However, Newton’s second law is easy to solve only for this particular case, whereas the work-energy theorem gives the final speed for any shaped frictionless surface. For an arbitrary curved surface, the normal force is not constant, and Newton’s second law may be difficult or impossible to solve analytically. Constant or not, for motion along a surface, the normal force never does any work, because it’s perpendicular to the displacement. A calculation using the work-energy theorem avoids this difficulty and applies to more general situations.
#### Problem-Solving Strategy: Work-Energy Theorem
1. Draw a free-body diagram for each force on the object.
2. Determine whether or not each force does work over the displacement in the diagram. Be sure to keep any positive or negative signs in the work done.
3. Add up the total amount of work done by each force.
4. Set this total work equal to the change in kinetic energy and solve for any unknown parameter.
5. Check your answers. If the object is traveling at a constant speed or zero acceleration, the total work done should be zero and match the change in kinetic energy. If the total work is positive, the object must have sped up or increased kinetic energy. If the total work is negative, the object must have slowed down or decreased kinetic energy.
### Example
#### Loop-the-Loop
The frictionless track for a toy car includes a loop-the-loop of radius R. How high, measured from the bottom of the loop, must the car be placed to start from rest on the approaching section of track and go all the way around the loop?
#### Strategy
The free-body diagram at the final position of the object is drawn in Figure. The gravitational work is the only work done over the displacement that is not zero. Since the weight points in the same direction as the net vertical displacement, the total work done by the gravitational force is positive. From the work-energy theorem, the starting height determines the speed of the car at the top of the loop,
$mg({y}_{2}-{y}_{1})=\frac{1}{2}m{v}_{2}{}^{2},$
where the notation is shown in the accompanying figure. At the top of the loop, the normal force and gravity are both down and the acceleration is centripetal, so
${a}_{\text{top}}=\frac{F}{m}=\frac{N+mg}{m}=\frac{{v}_{2}^{2}}{R}.$
The condition for maintaining contact with the track is that there must be some normal force, however slight; that is, $N \gt 0$. Substituting for ${v}_{2}^{2}$ and N, we can find the condition for ${y}_{1}$.
#### Solution
Implement the steps in the strategy to arrive at the desired result:
$N=\frac{\text{−}mg+m{v}_{2}^{2}}{R}=\frac{\text{−}mg+2mg({y}_{1}-2R)}{R} \gt 0\text{ }\text{or}\text{ }{y}_{1} \gt \frac{5R}{2}.$
#### Significance
On the surface of the loop, the normal component of gravity and the normal contact force must provide the centripetal acceleration of the car going around the loop. The tangential component of gravity slows down or speeds up the car. A child would find out how high to start the car by trial and error, but now that you know the work-energy theorem, you can predict the minimum height (as well as other more useful results) from physical principles. By using the work-energy theorem, you did not have to solve a differential equation to determine the height.
Suppose the radius of the loop-the-loop in Figure is 15 cm and the toy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop?
Show Solution
$\sqrt{3}\,\text{m/s}$
Visit Carleton College’s site to see a video of a looping rollercoaster.
In situations where the motion of an object is known, but the values of one or more of the forces acting on it are not known, you may be able to use the work-energy theorem to get some information about the forces. Work depends on the force and the distance over which it acts, so the information is provided via their product.
### Example
#### Determining a Stopping Force
A bullet from a 0.22LR-caliber cartridge has a mass of 40 grains (2.60 g) and a muzzle velocity of 1100 ft./s (335 m/s). It can penetrate eight 1-inch pine boards, each with thickness 0.75 inches. What is the average stopping force exerted by the wood, as shown in Figure?
#### Strategy
We can assume that under the general conditions stated, the bullet loses all its kinetic energy penetrating the boards, so the work-energy theorem says its initial kinetic energy is equal to the average stopping force times the distance penetrated. The change in the bullet’s kinetic energy and the net work done stopping it are both negative, so when you write out the work-energy theorem, with the net work equal to the average force times the stopping distance, that’s what you get. The total thickness of eight 1-inch pine boards that the bullet penetrates is $8\times \frac{3}{4}\,\text{in}\text{.}=6\,\text{in}\text{.}=15.2\,\text{cm}\text{.}$
#### Solution
Applying the work-energy theorem, we get
${W}_{\text{net}}=\text{−}{F}_{\text{ave}}\Delta {s}_{\text{stop}}=\text{−}{K}_{\text{initial}},$
so
${F}_{\text{ave}}=\frac{\frac{1}{2}m{v}^{2}}{\Delta {s}_{\text{stop}}}=\frac{\frac{1}{2}(2.6\times {10}^{-3}\text{kg}){(335\,\text{m/s})}^{2}}{0.152\,\text{m}}=960\,\text{N}\text{.}$
#### Significance
We could have used Newton’s second law and kinematics in this example, but the work-energy theorem also supplies an answer to less simple situations. The penetration of a bullet, fired vertically upward into a block of wood, is discussed in one section of Asif Shakur’s recent article [“Bullet-Block Science Video Puzzle.” The Physics Teacher (January 2015) 53(1): 15-16]. If the bullet is fired dead center into the block, it loses all its kinetic energy and penetrates slightly farther than if fired off-center. The reason is that if the bullet hits off-center, it has a little kinetic energy after it stops penetrating, because the block rotates. The work-energy theorem implies that a smaller change in kinetic energy results in a smaller penetration. You will understand more of the physics in this interesting article after you finish reading Angular Momentum.
Learn more about work and energy in this PhET simulation called “the ramp.” Try changing the force pushing the box and the frictional force along the incline. The work and energy plots can be examined to note the total work done and change in kinetic energy of the box.
### Summary
• Because the net force on a particle is equal to its mass times the derivative of its velocity, the integral for the net work done on the particle is equal to the change in the particle’s kinetic energy. This is the work-energy theorem.
• You can use the work-energy theorem to find certain properties of a system, without having to solve the differential equation for Newton’s second law.
### Conceptual Questions
The person shown below does work on the lawn mower. Under what conditions would the mower gain energy from the person pushing the mower? Under what conditions would it lose energy?
Show Solution
The mower would gain energy if $-90^\circ\lt\theta \lt 90^\circ.$ It would lose energy if $90^\circ\lt\theta \lt 270^\circ.$ The mower may also lose energy due to friction with the grass while pushing; however, we are not concerned with that energy loss for this problem.
Work done on a system puts energy into it. Work done by a system removes energy from it. Give an example for each statement.
Two marbles of masses m and 2m are dropped from a height h. Compare their kinetic energies when they reach the ground.
Show Solution
The second marble has twice the kinetic energy of the first because kinetic energy is directly proportional to mass, like the work done by gravity.
Compare the work required to accelerate a car of mass 2000 kg from 30.0 to 40.0 km/h with that required for an acceleration from 50.0 to 60.0 km/h.
Suppose you are jogging at constant velocity. Are you doing any work on the environment and vice versa?
Show Solution
Unless the environment is nearly frictionless, you are doing some positive work on the environment to cancel out the frictional work against you, resulting in zero total work producing a constant velocity.
Two forces act to double the speed of a particle, initially moving with kinetic energy of 1 J. One of the forces does 4 J of work. How much work does the other force do?
### Problems
(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).
A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 m while bringing a 900-kg car to rest from an initial speed of 1.1 m/s.
Show Solution
2.72 kN
Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted by a boxing glove on an opponent’s face, if the glove and face compress 7.50 cm during a blow in which the 7.00-kg arm and glove are brought to rest from an initial speed of 10.0 m/s. (b) Calculate the force exerted by an identical blow in the gory old days when no gloves were used, and the knuckles and face would compress only 2.00 cm. Assume the change in mass by removing the glove is negligible. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to cause damage even though it is lower than the force with no glove?
Using energy considerations, calculate the average force a 60.0-kg sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.
Show Solution
102 N
A 5.0-kg box has an acceleration of $2.0{\,\text{m/s}}^{2}$ when it is pulled by a horizontal force across a surface with ${\mu }_{K}=0.50.$ Find the work done over a distance of 10 cm by (a) the horizontal force, (b) the frictional force, and (c) the net force. (d) What is the change in kinetic energy of the box?
A constant 10-N horizontal force is applied to a 20-kg cart at rest on a level floor. If friction is negligible, what is the speed of the cart when it has been pushed 8.0 m?
Show Solution
2.8 m/s
In the preceding problem, the 10-N force is applied at an angle of $45^\circ$ below the horizontal. What is the speed of the cart when it has been pushed 8.0 m?
Compare the work required to stop a 100-kg crate sliding at 1.0 m/s and an 8.0-g bullet traveling at 500 m/s.
Show Solution
$W(\text{bullet})=20\times W(\text{crate})$
A wagon with its passenger sits at the top of a hill. The wagon is given a slight push and rolls 100 m down a $10^\circ$ incline to the bottom of the hill. What is the wagon’s speed when it reaches the end of the incline. Assume that the retarding force of friction is negligible.
An 8.0-g bullet with a speed of 800 m/s is shot into a wooden block and penetrates 20 cm before stopping. What is the average force of the wood on the bullet? Assume the block does not move.
Show Solution
12.8 kN
A 2.0-kg block starts with a speed of 10 m/s at the bottom of a plane inclined at $37^\circ$ to the horizontal. The coefficient of sliding friction between the block and plane is ${\mu }_{k}=0.30.$ (a) Use the work-energy principle to determine how far the block slides along the plane before momentarily coming to rest. (b) After stopping, the block slides back down the plane. What is its speed when it reaches the bottom? (Hint: For the round trip, only the force of friction does work on the block.)
When a 3.0-kg block is pushed against a massless spring of force constant constant $4.5\times {10}^{3}\,\text{N/m,}$ the spring is compressed 8.0 cm. The block is released, and it slides 2.0 m (from the point at which it is released) across a horizontal surface before friction stops it. What is the coefficient of kinetic friction between the block and the surface?
Show Solution
0.25
A small block of mass 200 g starts at rest at A, slides to B where its speed is ${v}_{B}=8.0\,\text{m/s,}$ then slides along the horizontal surface a distance 10 m before coming to rest at C. (See below.) (a) What is the work of friction along the curved surface? (b) What is the coefficient of kinetic friction along the horizontal surface?
A small object is placed at the top of an incline that is essentially frictionless. The object slides down the incline onto a rough horizontal surface, where it stops in 5.0 s after traveling 60 m. (a) What is the speed of the object at the bottom of the incline and its acceleration along the horizontal surface? (b) What is the height of the incline?
Show Solution
a. 24 m/s, −4.8 m/s2; b. 29.4 m
When released, a 100-g block slides down the path shown below, reaching the bottom with a speed of 4.0 m/s. How much work does the force of friction do?
A 0.22LR-caliber bullet like that mentioned in Figure is fired into a door made of a single thickness of 1-inch pine boards. How fast would the bullet be traveling after it penetrated through the door?
Show Solution
310 m/s
A sled starts from rest at the top of a snow-covered incline that makes a $22^\circ$ angle with the horizontal. After sliding 75 m down the slope, its speed is 14 m/s. Use the work-energy theorem to calculate the coefficient of kinetic friction between the runners of the sled and the snowy surface.
### Glossary
net work
work done by all the forces acting on an object
work-energy theorem
net work done on a particle is equal to the change in its kinetic energy
| 4,754
| 18,651
|
{"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}
| 4.75
| 5
|
CC-MAIN-2023-50
|
longest
|
en
| 0.813467
|
https://howkgtolbs.com/convert/19.98-kg-to-lbs
| 1,642,501,291,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00178.warc.gz
| 366,196,010
| 12,135
|
# 19.98 kg to lbs - 19.98 kilograms to pounds
Do you need to know how much is 19.98 kg equal to lbs and how to convert 19.98 kg to lbs? Here it is. You will find in this article everything you need to make kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to emphasize that whole this article is dedicated to one amount of kilograms - this is one kilogram. So if you need to know more about 19.98 kg to pound conversion - keep reading.
Before we go to the more practical part - this is 19.98 kg how much lbs calculation - we are going to tell you few theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 19.98 kg to lbs? 19.98 kilograms it is equal 44.0483599476 pounds, so 19.98 kg is equal 44.0483599476 lbs.
## 19.98 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, known also as International System of Units (in short form SI).
From time to time the kilogram could be written as kilogramme. The symbol of this unit is kg.
Firstly the kilogram was defined in 1795. The kilogram was defined as the mass of one liter of water. First definition was not complicated but hard to use.
Later, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was switched by a new definition.
The new definition of the kilogram is based on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It 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⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It can be also divided to 100 decagrams and 1000 grams.
## 19.98 kilogram to pounds
You learned a little about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to point out that there are more than one kind of pound. What are we talking about? For example, there are also pound-force. In this article we are going to to concentrate only on pound-mass.
The pound is in use in the Imperial and United States customary systems of measurements. To be honest, this unit is used also in other systems. The symbol of the pound is lb or “.
The international avoirdupois pound has no descriptive definition. It is defined as exactly 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 19.98 kg?
19.98 kilogram is equal to 44.0483599476 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 19.98 kg in lbs
Theoretical section is already behind us. In next part we will tell you how much is 19.98 kg to lbs. Now you know that 19.98 kg = x lbs. So it is high time to get the answer. Just see:
19.98 kilogram = 44.0483599476 pounds.
That is an accurate result of how much 19.98 kg to pound. You may also round off the result. After it your outcome is as following: 19.98 kg = 43.956 lbs.
You learned 19.98 kg is how many lbs, so have a look how many kg 19.98 lbs: 19.98 pound = 0.45359237 kilograms.
Obviously, this time you may also round off this result. After rounding off your outcome is exactly: 19.98 lb = 0.45 kgs.
We are also going to show you 19.98 kg to how many pounds and 19.98 pound how many kg outcomes in charts. See:
We are going to start with a chart for how much is 19.98 kg equal to pound.
### 19.98 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
19.98 44.0483599476 43.9560
Now look at a chart for how many kilograms 19.98 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
19.98 0.45359237 0.45
Now you know how many 19.98 kg to lbs and how many kilograms 19.98 pound, so we can move on to the 19.98 kg to lbs formula.
### 19.98 kg to pounds
To convert 19.98 kg to us lbs you need a formula. We will show you two versions of a formula. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 44.0483599476 outcome in pounds
The first version of a formula give you the most exact outcome. In some situations even the smallest difference could be considerable. So if you need an exact result - this version of a formula will be the best solution to convert how many pounds are equivalent to 19.98 kilogram.
So move on to the second version of a formula, which also enables calculations to know how much 19.98 kilogram in pounds.
The second version of a formula is as following, let’s see:
Number of kilograms * 2.2 = the outcome in pounds
As you see, the second version is simpler. It can be better choice if you need to make a conversion of 19.98 kilogram to pounds in quick way, for example, during shopping. You only need to remember that final outcome will be not so accurate.
Now we want to show you these two versions of a formula in practice. But before we will make a conversion of 19.98 kg to lbs we are going to show you another way to know 19.98 kg to how many lbs without any effort.
### 19.98 kg to lbs converter
Another way to learn what is 19.98 kilogram equal to in pounds is to use 19.98 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Converter is based on first version of a formula which we gave you above. Thanks to 19.98 kg pound calculator you can easily convert 19.98 kg to lbs. You only have to enter amount of kilograms which you need to calculate and click ‘calculate’ button. You will get the result in a second.
So try to convert 19.98 kg into lbs using 19.98 kg vs pound converter. We entered 19.98 as a number of kilograms. Here is the outcome: 19.98 kilogram = 44.0483599476 pounds.
As you see, our 19.98 kg vs lbs converter is intuitive.
Now we can go to our primary topic - how to convert 19.98 kilograms to pounds on your own.
#### 19.98 kg to lbs conversion
We will begin 19.98 kilogram equals to how many pounds conversion with the first version of a formula to get the most exact outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 44.0483599476 the outcome in pounds
So what need you do to learn how many pounds equal to 19.98 kilogram? Just multiply amount of kilograms, in this case 19.98, by 2.20462262. It gives 44.0483599476. So 19.98 kilogram is 44.0483599476.
You can also round off this result, for instance, to two decimal places. It is equal 2.20. So 19.98 kilogram = 43.9560 pounds.
It is high time for an example from everyday life. Let’s convert 19.98 kg gold in pounds. So 19.98 kg equal to how many lbs? And again - multiply 19.98 by 2.20462262. It gives 44.0483599476. So equivalent of 19.98 kilograms to pounds, if it comes to gold, is exactly 44.0483599476.
In this example it is also possible to round off the result. It is the result after rounding off, in this case to one decimal place - 19.98 kilogram 43.956 pounds.
Now we can go to examples calculated using a short version of a formula.
#### How many 19.98 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 43.956 the outcome in pounds
So 19.98 kg equal to how much lbs? And again, you have to multiply amount of kilogram, in this case 19.98, by 2.2. Let’s see: 19.98 * 2.2 = 43.956. So 19.98 kilogram is exactly 2.2 pounds.
Make another calculation using shorer version of a formula. Now calculate something from everyday life, for instance, 19.98 kg to lbs weight of strawberries.
So let’s calculate - 19.98 kilogram of strawberries * 2.2 = 43.956 pounds of strawberries. So 19.98 kg to pound mass is exactly 43.956.
If you learned how much is 19.98 kilogram weight in pounds and are able to calculate it using two different versions of a formula, let’s move on. Now we are going to show you these outcomes in charts.
#### Convert 19.98 kilogram to pounds
We are aware that results presented in tables are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in tables for your convenience. Due to this you can easily compare 19.98 kg equivalent to lbs outcomes.
Let’s begin with a 19.98 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
19.98 44.0483599476 43.9560
And now look 19.98 kg equal pound table for the second formula:
Kilograms Pounds
19.98 43.956
As you see, after rounding off, when it comes to how much 19.98 kilogram equals pounds, the outcomes are not different. The bigger number the more considerable difference. Please note it when you want to make bigger amount than 19.98 kilograms pounds conversion.
#### How many kilograms 19.98 pound
Now you learned how to calculate 19.98 kilograms how much pounds but we want to show you something more. Are you curious what it is? What about 19.98 kilogram to pounds and ounces calculation?
We will show you how you can calculate it little by little. Let’s start. How much is 19.98 kg in lbs and oz?
First things first - you need to multiply number of kilograms, this time 19.98, by 2.20462262. So 19.98 * 2.20462262 = 44.0483599476. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To check how much 19.98 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces.
So your result is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then final result will be equal 2 pounds and 33 ounces.
As you can see, calculation 19.98 kilogram in pounds and ounces is not complicated.
The last conversion which we want to show you is calculation of 19.98 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we show you it, look:
• 19.98 kilograms meters = 7.23301385 foot pounds,
• 19.98 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 19.98 foot pounds to kilograms meters you have to multiply 19.98 by 0.13825495. It is 0.13825495. So 19.98 foot pounds is equal 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 19.98 foot pounds will be equal 0.14 kilogram meters.
We hope that this calculation was as easy as 19.98 kilogram into pounds calculations.
We showed you not only how to do a conversion 19.98 kilogram to metric pounds but also two other conversions - to know how many 19.98 kg in pounds and ounces and how many 19.98 foot pounds to kilograms meters.
We showed you also other way to make 19.98 kilogram how many pounds conversions, that is using 19.98 kg en pound calculator. This will be the best solution for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you can make 19.98 kilogram equal to how many pounds conversion - on your own or with use of our 19.98 kgs to pounds calculator.
Don’t wait! Convert 19.98 kilogram mass to pounds in the best way for you.
Do you want to make other than 19.98 kilogram as pounds calculation? For instance, for 15 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 19.98 kilogram equal many pounds.
### How much is 19.98 kg in pounds
We want to sum up this topic, that is how much is 19.98 kg in pounds , we prepared one more section. Here you can find the most important information about how much is 19.98 kg equal to lbs and how to convert 19.98 kg to lbs . It is down below.
What is the kilogram to pound conversion? The conversion kg to lb is just multiplying 2 numbers. How does 19.98 kg to pound conversion formula look? . Check it down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 19.98 kilogram to pounds? The exact answer is 44.0483599476 lb.
You can also calculate how much 19.98 kilogram is equal to pounds with another, easier type of the equation. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So now, 19.98 kg equal to how much lbs ? The result is 44.0483599476 pounds.
How to convert 19.98 kg to lbs in a few seconds? You can also use the 19.98 kg to lbs converter , which will do whole mathematical operation for you and give you an exact answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
| 3,581
| 13,685
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2022-05
|
longest
|
en
| 0.955104
|
http://www.textbooks.com/Geometry-for-Dummies-With-Geometry-Wo-2-Edition/9780470537022/Ryan-mark.php?CSID=AODJWKCJKKS3AQOKUDUKU2CDB
| 1,397,671,714,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00288-ip-10-147-4-33.ec2.internal.warc.gz
| 707,516,959
| 25,446
|
0
# *Geometry for Dummies [With Geometry Wo - 2nd edition
## by Ryan mark
ISBN13: 978-0470537022
ISBN10: 0470537027
Summary: Learning geometry doesn't have to hurt. With a little bit of friendly guidance, it can even be fun! Geometry For Dummies, 2nd Edition, helps you make friends with lines, angles, theorems, and postulates. It eases you into all the principles and formulas you need to analyze two- and three-dimensional shapes, and it gives you the skills and strategies you need to write geometry proofs. Before you know it, you'll be understanding proofs like an expert. You'll find out how a proof's chain of logic works and discover some basic secrets for getting past rough spots. Soon, you'll be proving triangles congruent, calculating circumferences, using formulas, and serving up pi. The non-proof parts of the book contain helpful formulas and tips that you can use anytime you need to shape up your knowledge of shapes. You'll even get a feel for why geometry continues to draw people to careers in art, engineering, carpentry, robotics, physics, and computer animation, among others. You'll discover how to:
• Identify lines, angles, and planes
• Measure segments and angles
• Calculate the area of a triangle
• Use tips and strategies to make proofs easier
• Figure the volume and surface area of a pyramid
• Bisect angles and construct perpendicular lines
• Work with 3-D shapes
• Work with figures in the x-y coordinate system
So quit scratching your head. Geometry For Dummies, 2nd Edition, gets you un-stumped in a hurry. When you need to shape up, open up the included Geometry Workbook For Dummies, which contains over 290 pages with hundreds of practice problems featuring ample workspace to work out the problems. Each problem includes a step-by-step answer set to identify where you went wrong (or right). You'll be proving yourself proof-worthy in no time! AUTHOR BIO: Mark Ryan owns and operates The Math Center in Chicago, a teaching and tutoring service for all math subjects as w ...show less
Cover: Paperback
Publisher: Wiley, John & Sons, Incorporated
Published: 06/22/2009
International: No
List Price: \$29.99
Used Currently Sold Out
New Currently Sold Out
Rental \$14.99
Due back 08/01/2014
Save \$15.00 (50%)
• Free return shipping
• In stock
• 21-day satisfaction guarantee
CDs or access codes may not be included
Marketplace sellers starting at \$16.07
More prices and sellers below.
Additional Sellers for *Geometry for Dummies [With Geometry Wo
Just for you, we curate a growing list of independent booksellers, giving you even MORE choices when shopping for your textbooks.
Filter by: All (14) | New (11) | Like New (1) | Very Good (2)
Price
Condition
Seller
\$24.51
+\$3.99 s/h
New
Barnes & Noble.com
Various Shipping Locations, USA
Brand new, ships from the closest shipping location to your shipping address!
\$16.07
+\$3.99 s/h
New
PaperbackshopUS
Secaucus, NJ
New Book. Shipped from US within 4 to 14 business days. Established seller since 2000
\$16.54
+\$3.99 s/h
New
GreatBookPrices
Westminster, MD
Brand New, Perfect Condition, Please allow 4-14 business days for delivery. 100% Money Back Guarantee, Over 1,000,000 customers served.
\$16.55
+\$3.99 s/h
New
Stage_Coah_Books NJ
Linden, NJ
PAPERBACK New 0470537027 OUTER PLASTIC HAS SOME DAMAGE. Unread copy excellent condition....Satisfaction Guaranteed!
\$17.31
+\$3.99 s/h
New
indoo
Avenel, NJ
BRAND NEW
\$19.97
+\$3.99 s/h
New
BargainBookStores
Grand Rapids, MI
BRAND NEW ITEM! 100% Satisfaction Guarantee. Tracking provided on most orders. Buy with Confidence! Millions of items sold!.
\$21.24
+\$3.99 s/h
New
MovieMars-Books
Indian Trail, NC
"BRAND NEW. 30 Day Satisfaction Guarantee. Quick International Airmail!"
\$21.46
+\$3.99 s/h
LikeNew
GreatBookPrices
Westminster, MD
Brand New, Perfect Condition, Please allow 4-14 business days for delivery. 100% Money Back Guarantee, Over 1,000,000 customers served.
\$21.47
+\$3.99 s/h
VeryGood
Vitalbooks PA
Southampton, PA
PAPERBACK Very Good 0470537027 May have slight shelf wear. ! ! KNOWLEDGE IS POWER! ! ENJOY OUR BEST PRICES! ! ! Ships Fast. All standard orders delivered within 5 to 12 business days.
\$21.47
+\$3.99 s/h
VeryGood
BookHouse COM LLC
| 1,130
| 4,231
|
{"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}
| 2.8125
| 3
|
CC-MAIN-2014-15
|
latest
|
en
| 0.915841
|
https://dougshaw.com/you-cant-predict-a-caucus/
| 1,722,642,663,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00862.warc.gz
| 176,292,669
| 14,152
|
# You Can’t Predict a Caucus
This essay was written in 2016. Change the names and it applies to 2020.
There’s a trite joke in Iowa – “Don’t vote for a candidate you haven’t met.” My city isn’t even in the top ten largest cities in Iowa, but there’s Mike Huckabee going from table to table at the coffee shop where I’m grading papers. There’s Hillary Clinton at the bike shop where we bought our daughter her two-wheeler. Last Saturday, Martin O’ Malley’s campaign office contained, in addition to file drawers and desks, Martin O’ Malley himself. Parking was tricky to find a couple of weeks ago when I went to practice, because Donald Trump was speaking in the building next door. Bernie Sanders was kind to my daughter, Jeb Bush is in town tomorrow, Rand Paul is as well. You get the idea.
So, who is winning? I don’t know. You don’t either. And the press and the pundits have NO CLUE. But they think they do, because they don’t know the difference between a primary and a caucus. Iowa caucuses are not primaries. They are nothing like primaries. Primaries are predictable; caucuses are not. To put it mathematically, caucuses behave like non-linear dynamical systems, or in lay-terms Chaos Theory, with a bit of Catastrophe Theory mixed in. When I teach statistics, we spend a lot of time on interpreting poll results. And I tell my students at University of Northern Iowa that there are two phrases that tell us to discount a poll entirely, without thinking any further:
1. Online poll
2. Iowa caucus
Even if CNN and FOX don’t understand why online polls are meaningless, you probably do. But you probably don’t know why Iowa caucus polls are meaningless, nor do your local reporters, nor do your national reporters, so let’s talk.
Let’s start with the Democratic caucuses, and then we’ll visit a Republican one.
Come caucus with me! But first – do you have three hours?
Before we get in the car, you need to realize that this isn’t as quick as waiting in line to pull a lever. We are going to hear speeches, go through round 1, then the caucus round, then round 2, then the delegate election. If you aren’t able to spend three mostly boring hours with us, you aren’t going be there with me. The sad thing is that even if you are going to stay home, pollsters will probably have asked your opinion more than once. It is easy to participate in a primary. It is a pain in the ass to participate in a caucus. The pollsters and reporters don’t understand this factor, yet this factor is crucial to the outcome.
I’m glad you made it! Look who else came! And look at the walls and the tables!
When you walk in the building, you notice that you are surrounded by people who live near you, what in Iowa we call “neighbors.” And they are feeling chatty about their candidate. As are we, actually. Those of us who braved the cold and the time commitment are interested in politics and usually pretty well-informed. That means conversation. There are also people here from outside Iowa, who are not allowed to participate, but may observe. They often are chatty as well. There is campaign literature everywhere, and posters. In 2004, I was ready to caucus for Edwards, and told pollsters that information. Lots of pollsters. They are everywhere, always calling and talking to you. But then when I got in the building, I saw a giant Why You Should Support Edwards sign, and it was awful, and it cost him my vote. There’s also food. In 2008, Hillary Clinton provided brownies, John Edwards provided cookies, and Barack Obama provided little sandwiches. In a primary, campaign literature and campaign staff are not allowed within (about) 100 yards of the voting location. In a caucus, they are all over the place, and they affect votes. The pollsters and reporters don’t understand this factor, yet this factor is crucial to the outcome.
Put that cookie down, it is time for Round One! Oh – you thought this was anonymous?
Here is where the Republicans and Democrats diverge. Each Democratic candidate is assigned a place in the room, and people literally stand for their candidate. The people who live near you (“neighbors” if you want to speak like an Iowan!) see who you are supporting. Your spouse sees. Your boss may see. Your pastor, church-buddies, childrens’ teachers, clients… they all see. You are the only African American in the room. Or the only Muslim. Or your Muslim best friend makes casual eye-contact with you as you are about to stand for a candidate who talks dirt about them. In a primary, your vote is secret. A caucus is a party meeting, and your vote is public. The pollsters and reporters don’t understand this factor, yet this factor is crucial to the outcome.
We haven’t even caucused yet! That comes next! Come with me; you will love it!
After Round One, there is time for people to… change each other’s mind. And this does happen, because Iowa caucus-goers tend to be informed and interested, and informed, interested people have good discussions. Which sometimes results in minds changing. I remember in 2004, the Howard Dean people had “specialists” for the other candidates. So one Dean person was prepared specifically to talk to Kerry people, another for Edwards people, another for Clark people, etc. This is my favorite part of the process. I’ve had some of the best political conversations of my life during caucus time. And even if it doesn’t affect your first choice, it will probably affect your second choice. “Who cares about your SECOND choice,” you ask? You, you tourist from a state that cannot feed itself like mine can? Well, funny you should ask, we’ll get to that next. But first: The polls take place before the discussion round of a caucus. But the discussion round is the soul of the caucus. The pollsters and reporters don’t understand this factor, yet this factor is crucial to the outcome.
Round Two! Oh, hey, quick, switch circles; I’ll tell you why…
The people running the circus do a head-count, and figure out how many people a candidate needs to have a delegate. Let’s say the number is 11. I see you are standing with Clinton, and she has 24 people. There’s no way you are going to get 9 more people to join you at this stage to make 33. So she gets two delegates. But look! Martin O Malley has 9. Two of your Clinton people can walk to that circle, to give him the 11 he needs to make a delegate! And it won’t hurt Clinton, but it will screw Sanders. Oh, the Sanders people don’t like that, and they are calling to you not to do that. The O’ Malley people disagree, and your best friend is giving you an imploring look. Everyone is arguing and stuff, and something will happen. WHAT will happen? I don’t know. You don’t know either. And I can tell you for sure the polls didn’t predict it.
What happens if your candidate doesn’t get the requisite 11 (in this case) supporters? That candidate is said not to be “viable” and his or her supporters… can go somewhere else! To their second choice! Or their third! I’ve been to caucuses where I wound up standing with my third or fourth choice. The pollsters never ask what your second choice is, because that doesn’t matter in a primary, but it is important in a caucus.
Oh, and the polls don’t tell you about the deals, either. “Hey, gang, Kucinich and Kerry have this deal where if he isn’t viable his people are supposed to go to Kerry, and in exchange, they will get to go to the regional convention where they will be free to stand for Kucinich there.” Yes, after the results are declared and the press leaves Iowa, there are regional and district conventions. And you are not bound to stand for the same person there. The polls don’t reflect these deals either. Most people outside of Iowa don’t know about them.
Once people have declared in round two, the arithmetic allows for strategic vote-switching. There are also deals between candidate organizations. Many peoples’ candidates turn out not to be “viable” at their location and they have to go with a second or third choice. Finally, the delegates can change their minds at the regional convention. The pollsters and reporters don’t understand these factors, yet these factors are crucial to the outcome.
So, get a babysitter for your small children, take the evening off of your second-shift job, reconcile yourself to missing your favorite show, and I’ll pick you up at 6:30… wait, where are you going?
What about the Republican caucus? Surely they won’t stand for that foolishness!
The Republican caucuses don’t have the circles. But they do have speakers first, trying to persuade people at the last minute. Then there is a paper ballot, counted by hand. After that, they talk about viability and have a second binding ballot round. So they are definitely less chaotic than Democratic caucuses, but still unpredictable.
I wrote this essay because reporters, candidates, and regular people keep talking about who is “winning Iowa.” Who is winning Iowa? I don’t know. You don’t know either. It is unknowable. When I moved to Iowa, I thought this was weird and crazy. Now? I kind of like it. It is archaic and unfair and undemocratic, but there is something about getting together with your neighbors and choosing your party’s candidate together that fosters a real investment in the outcome. And you know what? I kind of like that the outcome is unpredictable. Thanks for reading. Now tell your favorite journalist.
| 2,093
| 9,393
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.945799
|
http://www.ionizationx.com/index.php?PHPSESSID=9d7rmcifevbj5pdcru8e46kts3&topic=2946.64
| 1,652,988,929,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00569.warc.gz
| 88,762,619
| 8,768
|
Author Topic: How the VIC Works - IMPORTANT! (Read 37378 times)
0 Members and 1 Guest are viewing this topic.
• Moderator
• Sr. member
• Posts: 421
Re: How the VIC Works - Induced DC Current Voltage
« Reply #64 on: March 09, 2015, 19:23:28 pm »
We also hooked the cell different up. In Stans schematic it is betweeb the 2 chokes.
In our schematic the cell is in parallel with the first choke.......
Put that into yr simulators.
Andrei Puharich talked about ac electrolysis....
Interesting thought about thin core. Not sure it should really matter. I'm thinking it may keep the impedances low on the coils but this can also be accomplished by smaller coils.
Also, Meyer did indicate that there are different ways to implement the VIC for those who were familiar with the art. I have no doubt that there are other configurations that may work. I may give it a shot in simulation but am continuing my real one in its current direction.
« Last Edit: March 09, 2015, 22:35:51 pm by timeshell »
• Global Moderator
• Hero member
• Posts: 3630
Re: How the VIC Works - Induced DC Current Voltage
« Reply #65 on: March 10, 2015, 15:59:59 pm »
So far i understood there are two ways of getting there... one applying high voltage low power and one applying lower voltage probably with higher output...
The point is this is two different systems they don't even look much the same disregarding the cell connections...
I think we need to split into there parts to get it right on the second method
input, transmission , and resonant inductor + cell
Now when the frequency of the cel+inductor is matched to the resonant 1/4 or ((1/4)+(1/2)*n) frequency of the line the voltage is amplified because the pulses are ordered.. frequency is doubled when the resonant inductor + cell collapse generating a pulse that travels back in the line, when it gets to the diode its reflected back to the cell because the diode only allow electric current in one direction, here another pulse comes summing creating the step charging effect ...
the intrinsic series inductance of the line also collapses amplifying voltage
the number of pulses is proportional to the wavelength and resonant frequency.
when impedance of the load = that of the line source there is no reflection... of course at resonance the impedance will not reach exactly zero... it can be a more positive number to allow accommodate to the contaminants on water.. frequency doubly occur any way since the resonant inductor is located between two cells forming a series resonant ckt..
theres still some magic about the cell connections and the tuning procedure..
• Global Moderator
• Hero member
• Posts: 3630
Re: How the VIC Works - Induced DC Current Voltage
« Reply #66 on: March 10, 2015, 23:24:30 pm »
i could be wrong but the next step in the meyer dream was injector system so i think that he was indeed generating gas on demand,....
so theres no to small pressure on the cell or little vaccum indeed... this way the air intake could go straight to the cell (butterfly valve) so the gas is mixed with air than goes to a bubbler and than goes to the engine... i think i would use a small electric turbine to than pressurize the gas before injecting to the engine...
• Moderator
• Sr. member
• Posts: 421
Re: How the VIC Works - Induced DC Current Voltage
« Reply #67 on: March 11, 2015, 01:30:45 am »
i could be wrong but the next step in the meyer dream was injector system so i think that he was indeed generating gas on demand,....
so theres no to small pressure on the cell or little vaccum indeed... this way the air intake could go straight to the cell (butterfly valve) so the gas is mixed with air than goes to a bubbler and than goes to the engine... i think i would use a small electric turbine to than pressurize the gas before injecting to the engine...
Kind of off topic isn't it?
• Global Moderator
• Hero member
• Posts: 3630
Re: How the VIC Works - Induced DC Current Voltage
« Reply #68 on: March 11, 2015, 11:31:30 am »
i wanted to mean that the gas is generated on demand, there is no need to worry about getting any reserve of gas probably even just using the resonant cell with no injectors..
and so we are talking about the vic this is what it should be its power output indication!
did you tried the frequency doubling?
simply put a coil between two cells and pulse it with a transformer thru a diode
« Last Edit: March 11, 2015, 17:09:24 pm by sebosfato »
• Moderator
• Sr. member
• Posts: 421
Re: How the VIC Works - Induced DC Current Voltage
« Reply #69 on: March 14, 2015, 08:42:39 am »
Here's a new tidbit to think about.
If you have the coils in the VIC opposing each other, what is the total inductance going to be? For example, let's say choke L1 is 1.2H and the secondary on the same core is 1.4H and they are wired to oppose each other. Their net inductance will be less than the two of them. For me, although just subtracting the lesser from the greater didn't result in the actual inductance so I am assuming there is another factor involved such as mutual inductance. At any rate, let's say that the net difference leaves us with 200mH.
200mH doesn't match up with a 1nF capacitor to give a frequency of let's say 5kHz. So, the final choke, L2 needs to make up for the difference. We need a total of 1.1H to make resonance at 5kHz. So, L2 in this example would need to be 900mH to achieve proper resonance in this circuit.
« Last Edit: March 14, 2015, 16:51:58 pm by timeshell »
• Moderator
• Sr. member
• Posts: 421
Re: How the VIC Works - Induced DC Current Voltage
« Reply #70 on: March 14, 2015, 16:25:27 pm »
I am presently experimenting with inline resistors in the VIC. While not desirable in the end result, it has amplified the resonant effect I have been getting and actually increased the RMS volts. At resonance, the transformer sounds like it is producing static and you will see higher peaks in the voltage waveform associated with this "static". I believe this is where the step up is occuring.
« Last Edit: March 14, 2015, 16:48:05 pm by timeshell »
| 1,528
| 6,091
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.927463
|
http://mathoverflow.net/questions/17138?sort=newest
| 1,369,245,334,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368702185502/warc/CC-MAIN-20130516110305-00003-ip-10-60-113-184.ec2.internal.warc.gz
| 167,251,409
| 12,280
|
## Dirichlet and the prime number theorem
I browsed Dirichlets Werke today and was kind of surprised by two remarks that he made on p. 354 (Über die Bestimmung ...) and p. 372 (Sur l'usage ...). In the second paper, he claims (my translation)
I have applied these principles to a demonstration of the remarkable formula given by Legendre for expressing in an approximate manner how many prime numbers there are below an arbitrary, but very large, limit.
In a handwritten note on the reprint he sent to Gauss he remarked that $\sum 1/\log n$ (this is Gauss's version of the PNT, at least if you replace the sum by an integral) is a better estimate than Legendre's.
I am a little bit puzzled as to why Dirichlet's claim to have proved the prime number theorem is not discussed anywhere in the literature. Or is it?
-
Good question!! – John Stillwell Mar 4 2010 at 22:37
Dirichlet's remark from the first paper is extracted and translated on page 98 of The Development of Prime Number Theory by Narkiewicz. So this has not passed completely unnoticed. Narkiewicz remarks that Dirichlet believed that his analytic methods would enable him to prove Legendre's conjecture, and that Dirichlet never returned to the problem.
Dirichlet remained interested in the asymptotic growth laws ("Asymptotische Gesetze") of arithmetic functions for the rest of his life, as seen from his 1849 paper with the estimate
$$\sum_{n \leq x}d(n) = x\log(x) + (2\gamma - 1)x + O(x^{1/2}),$$
and a couple of other estimates, and a letter of 1858 to Kronecker reprinted in Dirichlet's Werke, where he mentions having obtained a substantial improvement of the error term $O(x^{1/2})$ by a new method.
Since Dirichlet demonstrably did not lose interest in such questions, and never returned to the PNT in print, it seems reasonable to believe that he discovered that his real-variable method would not yield the PNT.
-
Narkiewicz, the "modern-day Dickson". I accepted this answer because of the reference - thanks! – Franz Lemmermeyer Mar 9 2010 at 18:43
-
When voting down you are asked to add a comment, so I will. What? – Dror Speiser May 6 2010 at 22:34
The second paper is Sur l'usage des s\'eries infinies dan la th\'eorie des nombres Crelle $\mathbf{18}$ (1838), 259--274. The quote in Crelle is near the end, at the top of p. 272. After it he says he has determined some mean-value formulas for arithmetic functions (like sums of divisors) by similar techniques. The technique he is describing is that of encoding a sequence of interest as the coefficients in a Dirichlet series and then looking at its (real) pole. This method is indeed one of Dirichlet's important discoveries, but the prime number theorem is such an order of magnitude harder than the other results he lists here that he must have erroneously convinced himself that he could derive the prime number theorem by his new method just like he had derived other number-theoretic limit laws.
Amusingly, his notation for the Riemann zeta-function (on p. 272) is $\varphi(s)$!
-
So, we have an apparent counterexample to the famous saying of Jacobi: Dirichlet alone, not I, nor Cauchy, nor Gauss knows what a completely rigorous mathematical proof is. Rather we learn it first from him. When Gauss says that he has proved something, it is very clear; when Cauchy says it, one can wager as much pro as con; when Dirichlet says it, it is certain ... Quoted in G Schubring, Zur Modernisierung des Studiums der Mathematik in Berlin, 1820-1840. – John Stillwell Mar 5 2010 at 1:10
@John Stillwell: Certainly he never claimed that he proved the PNT, did he? – Harry Gindi Mar 5 2010 at 6:21
Guess that's exactly what he was implying when he wrote "a demonstration of the remarkable formula given by Legendre". Nice catch, Professor Stillwell! – J. H. S. May 6 2010 at 22:02
| 976
| 3,822
|
{"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}
| 2.578125
| 3
|
CC-MAIN-2013-20
|
latest
|
en
| 0.936539
|
https://aircraftflightmechanics.com/_sources/StaticStability/Nondimensionalrates.ipynb
| 1,701,227,366,000,000,000
|
text/plain
|
crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00632.warc.gz
| 108,353,923
| 1,707
|
{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Non-dimensional rates\n", "\n", "In the following, the stability in the other two axes will be described. Note that this is a simplified analysis to facilitate undergraduate understanding at this point in the course. In the full aircraft equations of motion which will be derived in the next module, there are 108 stability derivatives (and we could easily add more). A subset will be spoken about from first principles pertaining to the aircraft static stability.\n", "\n", "In the formulations used in this section for the roll, pitch, and yaw stability:\n", "\n", "\n", "\\begin{align}\n", " C_{\\ell} & = C_{\\ell_\\beta}\\cdot\\beta + C_{\\ell_P}\\cdot P + C_{\\ell_{\\delta_a}}\\cdot\\delta_a\\\\\n", " C_m & = C_{m_\\alpha}\\cdot\\alpha + C_{m_Q}\\cdot Q + C_{m_{\\delta_e}}\\cdot\\delta_e\\\\\n", " C_n & = C_{n_\\beta}\\cdot\\beta + C_{n_R}\\cdot R + C_{n_{\\delta_r}}\\cdot\\delta_r\n", "\\end{align}\n", "\n", "\n", "the product of the right hand terms are unitless, but the derivatives themselves are dimensional. To compare different aircraft, the angular rates are often expressed in non-dimensional forms:\n", "\n", "\n", "\\begin{align}\n", " \\bar{p}&\\triangleq\\frac{P\\, b}{2\\,V_\\infty}\\\\\n", " \\bar{q}&\\triangleq\\frac{Q\\, \\bar{c}}{2\\,V_\\infty}\\\\\n", " \\bar{r}&\\triangleq\\frac{R\\, b}{2\\,V_\\infty}\n", "\\end{align}\n", "\n", "\n", "this enables simpler expressions to be developed for the corresponding stability derivatives, *e.g.,* $C_{\\ell_{\\bar{p}}}$ and $C_{\\ell_{P}}$ both express *roll damping*, but the dimensional derivative is function of the aircraft geometry and forward speed whereas the nondimensional expression is not." ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.8.5" } }, "nbformat": 4, "nbformat_minor": 4 }
| 642
| 2,126
|
{"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.96875
| 4
|
CC-MAIN-2023-50
|
latest
|
en
| 0.77615
|
https://discourse.mc-stan.org/t/are-transformed-parameters-included-in-the-search-space-for-optimizations/7185
| 1,652,832,180,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00281.warc.gz
| 260,944,929
| 5,361
|
# Are transformed parameters included in the search space for optimizations?
Here is a simplified version of the regression model that I’m working on:
``````data {
int N;
vector[N] x;
vector[N] y;
}
parameters {
real<lower=0> shape;
real<lower=0> a;
real<lower=0> b;
}
transformed parameters {
vector[N] mu = a + b * x;
vector[N] rate = shape ./ mu;
}
model {
y ~ gamma(shape, rate);
}
``````
Below is the rstan code. First, generate fake data:
``````set.seed(1)
a <- 3
b <- 7
shape <- 20
x <- runif(1000, min = 0, max = 10)
mu <- a + b * x
rate <- shape / mu
y <- rgamma(1000, shape = shape, rate = rate)
``````
Use Stan to get maximum-likelihood estimates of parameters:
``````stanmodel <- stan_model('stan_discourse_question.stan')
standata <- list(N = length(x), x = x, y = y)
mle <- optimizing(stanmodel, data = standata)
``````
Print out the results:
``````str(mle)
# List of 3
# \$ par : Named num [1:2003] 18.48 2.92 6.98 21.44 28.88 ...
# ..- attr(*, "names")= chr [1:2003] "shape" "a" "b" "mu[1]" ...
# \$ value : num -3367
# \$ return_code: int 0
mle\$par[1:20]
# shape a b mu[1] mu[2] mu[3] mu[4] mu[5] mu[6] mu[7] mu[8] mu[9] mu[10] mu[11]
# 18.476546 2.915327 6.976342 21.438120 28.875964 42.879538 66.275011 16.985349 65.590066 68.819107 49.014843 46.804476 7.225748 17.284818
# mu[12] mu[13] mu[14] mu[15] mu[16] mu[17]
# 15.232530 50.844392 29.711717 56.622099 37.636530 52.978850
``````
The estimated parameters include the estimates for the vectors `mu` and `rate`. Are the `mu` and `rate` vectors included in the search space for the optimization? I suspect that they’re not, since the transformed parameters are simple functions of the parameters and the data, but I want to double check that the search space for my model is not substantially increased by including transformed parameters.
No
1 Like
| 658
| 1,921
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.875
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.629418
|
http://www.realself.com/question/250-ml-cc
| 1,448,683,851,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-48/segments/1448398450762.4/warc/CC-MAIN-20151124205410-00352-ip-10-71-132-137.ec2.internal.warc.gz
| 646,500,775
| 22,256
|
# 250 Ml Would Be How Much Cc?
### 250 ml=250 cc
Millileters and cubic centimeters are equivalent measures of volume. One ml is equal to one cc.
Thanks and best wishes.
### Ml = cc
Millileters (ml's) and cubic centimeters (cc's) are equivalent measures of volume. Therefore one ml equals one cc.
Larry S. Nichter, MD, MS, FACS
Orange County Plastic Surgeon
5.0 out of 5 stars 76 reviews
### Milliliter or cc with Breast Implants
A milliliter and a cubic centimer (cc) represent the same volume. 250 cc is 250 ml. Kenneth Hughes, MD breast implants Los Angeles, CA
### Breast Implants CC's
1ml is the same volume as 1cc. So 250ml is the same as 250cc. A cc weighs about 1g. It is important to get a size of implant in cc's that fits your body and gives you the look that you want. It helps to try implant sizes in a bra. Also, in my office I use computer 3d images of each patient to visualize what different implant sizes look like.
### 250 Ml Would Be How Much CC's?
250cc = 250ml
250 cc of water or saline = 250 grams = 0.55 pounds = 8.82 ounces
Peter A. Aldea, MD
Memphis Plastic Surgeon
5.0 out of 5 stars 83 reviews
### How to convert ml to cc.
The conversion is 1 ml equals 1 cc. Also if you are interested 1 ml which equals 1 cc also equals 1 gram. So 1000ccs equals 1 kilogram which equals approximately 2.2 pounds
Jay M. Pensler, MD
Chicago Plastic Surgeon
4.5 out of 5 stars 13 reviews
### 1 mL = 1 cc
Dear Rockinggirl,
1 milliliter (mL) is the equivalent of 1 cubic centimeter (cc),
ie. 1 mL = 1 cc.
Hope this helps.
Warmly,
Larry Fan, MD
Larry Fan, MD
San Francisco Plastic Surgeon
4.5 out of 5 stars 14 reviews
### MLs and CCs for breast implants
A milliliter (mL) and cubic centimeter (cc) are equivalent units of measurement of volume. Normally, breast implant volumes are described in cc's, but it is interchangeable with mL's.
Naveen Setty, MD
Dallas Plastic Surgeon
5.0 out of 5 stars 46 reviews
### Milliliter versus Cubic Centimeter Difference in Breast Implant Volume?
The milliliter (ml) and cubic centimeter (cc) are equivalent measurements used to measure the volume of saline or silicone in breast implants.
### CC = ML
CC's are the same thing as ML's as they both measure volume. Most breast implants are meausred in cc's as it is the usual custom.
These answers are for educational purposes and should not be relied upon as a substitute for medical advice you may receive from your physician. If you have a medical emergency, please call 911. These answers do not constitute or initiate a patient/doctor relationship.
| 707
| 2,583
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.125
| 3
|
CC-MAIN-2015-48
|
latest
|
en
| 0.784379
|
https://oeis.org/A342164
| 1,628,048,750,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00182.warc.gz
| 414,461,179
| 4,270
|
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A342164 A self-describing sequence: start with 0, then for each digit in each successive term, starting from the first term, append to the sequence its most recent position in the string formed by the concatenation of all previous terms. 0
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 17, 19, 16, 23, 18, 27, 30, 27, 22, 27, 24, 39, 41, 44, 28, 47, 50, 41, 52, 56, 50, 56, 56, 56, 50, 56, 53, 72, 42, 75, 54, 80, 80, 76, 83, 80, 85, 92, 90, 80, 54, 99, 94, 99, 86, 99, 98, 99, 108, 99, 108, 99, 108, 99, 126, 99 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS After the leading zero taking the a(n)-th digit of the sequence returns the digits of the sequence. LINKS EXAMPLE The second term is 1 as the 0 in the first term appears as the first digit in the sequence. Likewise the third term is 2 as the 1 in the second term is the second digit of the sequence, and so on to the eleventh term. As the eleventh term is 10 and has two digits, the twelfth and thirteenth terms give the most recent position of a 1 and 0 in the sequence, and they appear at the eleventh and twelfth position. As the twelfth term is 11, the fourteenth and fifteenth terms give the most recent position of the two 1's. The last 1 appears at the fifteenth position, and after appending 15, which contains a 1, the most recent 1 now appears at the seventeenth position. CROSSREFS Cf. A125132, A264646, A114308, A263563, A308387 Sequence in context: A135578 A050607 A238368 * A263808 A180477 A271239 Adjacent sequences: A342161 A342162 A342163 * A342165 A342166 A342167 KEYWORD nonn,base AUTHOR Scott R. Shannon, Mar 03 2021 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 3 22:03 EDT 2021. Contains 346441 sequences. (Running on oeis4.)
| 696
| 2,164
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.4375
| 3
|
CC-MAIN-2021-31
|
latest
|
en
| 0.864616
|
https://la.mathworks.com/matlabcentral/cody/solutions/2571678
| 1,610,834,439,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00777.warc.gz
| 433,079,796
| 18,911
|
Cody
# Problem 44309. Pi Digit Probability
Solution 2571678
Submitted on 17 Jun 2020 by Li Kehan Li
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
N = 101; n = 3; y_correct = 0.1200; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match')))) % modified from the comment of Alfonso on https://www.mathworks.com/matlabcentral/cody/problems/44343
a = '3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989' c = '3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067' M = '3' d = 1 11 17 19 26 27 29 45 48 66 88 93 e = 1 f = 12 y = 0.1200 y = 0.1200
2 Pass
N = 201; n = 6; y_correct = 0.0750; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
a = '3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989' c = '3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819' M = '6' d = 9 22 24 43 71 74 77 84 100 110 119 120 129 183 186 e = 1 f = 15 y = 0.0750 y = 0.0750
3 Pass
N = 202; n = 6; y_correct = 0.0796; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
a = '3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989' c = '3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196' M = '6' d = 9 22 24 43 71 74 77 84 100 110 119 120 129 183 186 202 e = 1 f = 16 y = 0.0796 y = 0.0796
4 Pass
N = 203; n = 6; y_correct = 0.0792; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
a = '3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989' c = '3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964' M = '6' d = 9 22 24 43 71 74 77 84 100 110 119 120 129 183 186 202 e = 1 f = 16 y = 0.0792 y = 0.0792
5 Pass
N = 1001; n = 9; y_correct = 0.1050; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
a = '3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989' c = '3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420198' M = '9' d = Columns 1 through 15 7 14 16 32 40 44 46 47 57 60 64 81 82 102 124 Columns 16 through 30 131 146 171 182 189 192 195 201 210 216 249 251 261 286 296 Columns 31 through 45 330 333 338 343 355 358 390 393 401 416 418 420 424 435 442 Columns 46 through 60 461 462 467 484 489 498 500 503 529 531 535 544 551 555 566 Columns 61 through 75 574 596 638 641 660 666 677 688 692 695 699 707 708 716 720 Columns 76 through 90 734 740 749 750 764 765 766 767 768 769 774 779 780 785 796 Columns 91 through 105 802 809 816 846 896 904 910 926 943 948 978 988 990 992 1000 e = 1 f = 105 y = 0.1050 y = 0.1050
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
| 3,392
| 9,402
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04
|
latest
|
en
| 0.3228
|
https://socratic.org/questions/how-do-you-find-the-x-and-y-intercepts-of-2-5x-5y-15
| 1,575,557,049,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540481076.11/warc/CC-MAIN-20191205141605-20191205165605-00503.warc.gz
| 551,042,080
| 5,990
|
# How do you find the x and y intercepts of 2.5x-5y=-15?
Feb 15, 2017
see the entire solution process below:
#### Explanation:
To find the y-intercept, set $x = 0$ and solve for $y$:
$\left(2.5 \times 0\right) - 5 y = - 15$
$0 - 5 y = - 15$
$- 5 y = - 15$
$\frac{- 5 y}{\textcolor{red}{- 5}} = - \frac{15}{\textcolor{red}{- 5}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5}}} y}{\cancel{\textcolor{red}{- 5}}} = 3$
$y = 3$
The y-intercept is $3$ or $\left(0 , 3\right)$
To find the x-intercept, set $y = 0$ and solve for $x$:
$2.5 x - \left(5 \times 0\right) = - 15$
$2.5 x - 0 = - 15$
$2.5 x = - 15$
$\frac{2.5 x}{\textcolor{red}{2.5}} = - \frac{15}{\textcolor{red}{2.5}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2.5}}} x}{\cancel{\textcolor{red}{2.5}}} = - 6$
$x = - 6$
The x-intercept is $- 6$ or $\left(- 6 , 0\right)$
| 382
| 867
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "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}
| 4.65625
| 5
|
CC-MAIN-2019-51
|
longest
|
en
| 0.568377
|
http://serc.carleton.edu/admin/manage/view_vocab.php?vocab_id=199
| 1,418,857,546,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-52/segments/1418802765002.8/warc/CC-MAIN-20141217075245-00084-ip-10-231-17-201.ec2.internal.warc.gz
| 299,165,467
| 23,579
|
Central Administration Page Science Education Resource Center | Carleton College
## SERC Vocabularies
These controlled vocabularies are under development at SERC. They vary widely in terms of refinement and the degree to which they have been tested with users. None represents a finished product. Caveat emptor.
## Energy Literacy Principles
• Energy is a physical quantity Energy is a physical quantity that follows precise natural laws.
199:2 end
• 1.1 Energy is a quantity Energy is a quantity that is transferred from system to system.
Energy is the ability of a system to do work. A system has done work if it has exerted a force on another system over some distance. When this happens energy is transferred from one system to another. At least some of the energy is also transformed from one type into another during this process. One can keep track of how much energy transfers into or out of a system. 199:2 199:3 end
• 1.2 Thermal energy The energy of a system or object that results in its temperature is called thermal energy.
When there is a net transfer of energy from one system to another, due to a difference in temperature, we call the energy transferred heat. Heat transfer happens in three ways: convection, conduction and radiation. Like all energy transfer, heat transfer involves forces exerted over a distance at some level as systems interact. 199:2 199:4 end
• 1.3 Energy is neither created nor destroyed Energy is neither created nor destroyed.
The change in the total amount of energy in a system is always equal to the difference between the amount of energy transferred in and the amount transferred out. The total amount of energy in the universe is finite and constant. 199:2 199:7 end
• 1.4 Energy quality degrades over time Energy available to do useful work decreases as it is transferred from system to system.
Although energy is neither created nor destroyed, its quality degrades over time. During all transfers of energy between two systems, some energy is lost to the surroundings. In a practical sense, this lost energy has been “used up,” even though it is still around somewhere. A more efficient system will lose less energy, up to a theoretical limit. 199:2 199:8 end
• 1.5 Forms of energy Energy comes in different forms and can be divided into categories.
Forms of energy include light energy, elastic energy, chemical energy and more. There are two categories that all energy falls into, kinetic and potential. Kinetic describes types of energy associated with motion and the word potential describes energy possessed by an object or system due to its position relative to another object or system and forces between the two. Some forms of energy are part kinetic and part potential energy. 199:2 199:9 end
• 1.6 Chemical and nuclear reactions Chemical and nuclear reactions involve transfer and transformation of energy.
The energy associated with nuclear reactions is much larger than that associated with chemical reactions for a given amount of mass. Nuclear reactions take place at the centers of stars, in nuclear bombs and in both fission and fusion-based nuclear reactors. Chemical reactions are pervasive in living and non-living Earth systems. 199:2 199:10 end
• 1.7 Units of energy Many different units are used to quantify energy.
As with other physical quantities, many different units are associated with energy. For example, joules, calories, ergs, kilowatt-hours and BTUs are all units of energy. Given a quantity of energy in one set of units, one can always convert it to another (E.g., 1 calorie = 4.186 joules). 199:2 199:11 end
• 1.8 Power Power is a measure of energy transfer rate.
It is useful to talk about the rate at which energy is transferred from one system to another (energy per time). This rate is called power. One joule of energy transferred in one second is called a Watt (I.e., 1 joule/second = 1 Watt). 199:2 199:12 end
• Physical processes on Earth are the result of energy flow Physical processes on Earth are the result of energy flow through the Earth system.
199:5 end
• 2.1 Changes in energy flow over time Earth is constantly changing as energy flows through the system.
Geologic, fossil and ice records provide evidence of significant changes throughout Earth's history. These changes are always associated with changes in the flow of energy through the Earth system. Both living and non-living processes have contributed to this change. 199:5 199:13 end
• 2.2 Sources of energy on Earth Sunlight, gravitational potential, decay of radioactive isotopes, and rotation of the Earth are the major sources of energy driving physical processes on Earth.
Sunlight is a source external to Earth while radioactive isotopes and gravitational potential, with the exception of tides, are internal. Radioactive isotopes and gravity work together to produce geothermal energy beneath Earth's surface. Earth's rotation influences global flow of air and water. 199:5 199:14 end
• 2.3 Earth's climate driven by the Sun Earth's weather and climate is mostly driven by energy from the Sun.
For example, unequal warming of Earth's surface and atmosphere by the Sun drives convection within the atmosphere, producing winds, and influencing ocean currents. 199:5 199:15 end
• 2.4 Water stores and transfers energy Water plays a major role in the storage and transfer of energy in the Earth system.
This is due to water's prevalence, high heat capacity and the fact that phase changes of water occur regularly on Earth. The Sun provides the energy that drives the water cycle on Earth. 199:5 199:16 end
• 2.5 Energy moves between reservoirs Movement of matter between reservoirs is driven by Earth's internal and external sources of energy.
These movements are often accompanied by a change in the physical and chemical properties of the matter. Carbon, for example, occurs in carbonate rocks such as limestone, in the atmosphere as carbon dioxide gas, in water as dissolved carbon dioxide, and in all organisms as complex molecules that control the chemistry of life. Energy drives the flow of carbon between these different reservoirs. 199:5 199:17 end
• 2.6 Greenhouse gases affect energy flow Greenhouse gases affect energy flow through the Earth system.
Greenhouse gases in the atmosphere, such as carbon dioxide and water vapor, are transparent to much of the incoming Sunlight but not to the infrared light from the warmed surface of Earth. These gases play a major role in determining average global surface temperatures. When Earth emits the same amount of energy as it absorbs its average temperature remains stable. 199:5 199:18 end
• 2.7 Effects of changes in Earth's energy system The effects of changes in Earth's energy system are often not immediately apparent.
Responses to changes in Earth's energy system, input versus output, are often only noticeable over the course of months, years or even decades. 199:5 199:19 end
• Biological processes depend on energy flow Biological processes depend on energy flow through the Earth system.
199:6 end
• 3.1 The Sun is major source of energy for organisms and ecosystems The Sun is the major source of energy for organisms and the ecosystems of which they are a part.
Producers such as plants, algae and cyanobacteria use the energy from sunlight to make organic matter from carbon dioxide and water. This establishes the beginning of energy flow through almost all food webs. 199:6 199:20 end
• 3.2 Food is a biofuel Food is a biofuel used by organisms to acquire energy for internal living processes.
Food is composed of molecules that serve as fuel and building material for all organisms as energy stored in the molecules is released and used. The breakdown of food molecules enables cells to store energy in new molecules that are used to carry out the many functions of the cell and thus the organism. 199:6 199:21 end
• 3.3 Continual input of energy is needed Energy available to do useful work decreases as it is transferred from organism to organism.
The chemical elements that make up the molecules of living things are passed through food chains and are combined and recombined in different ways. At each level in a food chain, some energy is stored in newly made chemical structures, but most is dissipated into the environment. Continual input of energy, mostly from sunlight, keeps the process going. 199:6 199:22 end
• 3.4 Energy flows through food webs Energy flows through food webs in one direction, from producers to consumers and decomposers.
An organism that eats lower on a food chain is more energy efficient than one eating higher on a food chain. Eating producers is the lowest, and thus most energy efficient, level at which an animal can eat. 199:6 199:23 end
• 3.5 Ecosystems are affected by availability of energy. Ecosystems are affected by changes in the availability of energy and matter.
The amount and kind of energy and matter available constrains the distribution and abundance of organisms in an ecosystem and the ability of the ecosystem to recycle materials. 199:6 199:24 end
• 3.6 Humans live within Earth's ecosystems. Humans live within Earth's ecosystems.
Increasingly, humans modify the energy balance of Earth's ecosystems. The changes happen, for example, as a result of changes in agricultural and food processing technology, consumer habits, and human population size. 199:6 199:25 end
• Various sources of energy are used to power human activities
199:26 end
• 4.1 Humans transfer and transform energy Humans transfer and transform energy from the environment into forms useful for human endeavors.
The primary sources of energy in the environment include renewable sources such as sunlight, wind, moving water, and geothermal energy. Primary sources also include fuels like coal, oil, natural gas, uranium and biomass. All primary source fuels except biomass are non-renewable. 199:26 199:29 end
• 4.2 Human use of energy is subject to limits and constraints Human use of energy is subject to limits and constraints.
Industry, transportation, urban development, agriculture, and most other human activities are closely tied to the amount and kind of energy available.The availability of energy resources is constrained by the distribution of natural resources, availability of affordable technologies and socio-economic policies and status. 199:26 199:30 end
• 4.3 Fossil and bio fuels contain energy captured from sunlight Fossil and bio fuels are organic matter that contain energy captured from sunlight.
The energy in fossil fuels such as oil, natural gas and coal comes from energy that producers such as plants, algae and cyanobacteria, captured from sunlight long ago. The energy in biofuels such as food, wood, and ethanol comes from energy that producers captured from sunlight very recently. Energy stored in these fuels is released during chemical reactions, such as combustion and respiration, which also release carbon dioxide into the atmosphere. 199:26 199:31 end
• 4.4 Humans transport energy Humans transport energy from place to place.
Fuels are often not used at their source but are transported, sometimes over long distances. Fuels are transported primarily by pipelines, trucks, ships, and trains. Electrical energy can be generated from a variety of energy resources and can be transformed into almost any other form of energy. Electric circuits are used to distribute energy to distant locations. Electricity is not a primary energy source, but a means of transmitting energy. 199:26 199:32 end
• 4.5 Electricity generation Electricity is usually generated in one of two ways.
When a magnet moves or magnetic field changes relative to a coil of wire, electrons are induced to flow in the wire. Most human generation of electricity happens in this way. Electrons can also be induced to flow through direct interaction with light particles; this is the basis upon which a solar cell operates. 199:26 199:33 end
• 4.6 Humans store energy Humans intentionally store energy for later use in a number of different ways.
Examples include batteries, water reservoirs, compressed air, hydrogen and thermal storage. Storage of energy involves many technological, environmental and social challenges. 199:26 199:34 end
• 4.7 Different sources of energy have different benefits and drawbacks Different sources of energy and the different ways energy can be transformed, transported and stored each have different benefits and drawbacks.
A given energy system, from source to sink, will have an inherent level of energy efficiency, monetary cost and environmental risk. Each system will also have national security, access and equity implications. 199:26 199:35 end
• Energy decisions are influenced by several factors
199:27 end
• 5.1 Energy decisions are made at many levels Decisions concerning the use of energy resources are made at many levels.
Humans make individual, community, national and international energy decisions. Each of these levels of decision making have some common and some unique aspects. In the developed world, decisions made beyond the individual level usually involve a formally established process. 199:27 199:36 end
• 5.2 Energy infrastructure has inertia Energy infrastructure has inertia.
The decisions that governments, corporations, and individuals made in the past have created today's energy infrastructure. The large amount of money, time, and technology invested in these systems make changing the infrastructure difficult, but not impossible. The decisions of one generation both provide and limit the range of possibilities open to the future generations. 199:27 199:37 end
• 5.3 Systems-based approach Energy decisions can be made using a systems-based approach.
As individuals and societies make energy decisions they can consider the costs and benefits of each decision. Some costs and benefits are more obvious than others. Identifying all costs and benefits requires a careful and informed systems-based approach to decision making. 199:27 199:38 end
• 5.4 Economic factors Energy decisions are influenced by economic factors.
Monetary costs of energy affect energy decision making at all levels. Energy exhibits characteristics of both a commodity and a differentiable product. Energy costs are often subject to market fluctuations and energy choices made by individuals and societies affect these fluctuations. Cost differences also arise as a result of differences in energy source and as a result of tax-based incentives and rebates. 199:27 199:39 end
• 5.5 Political factors Energy decisions are influenced by political factors.
Political factors play a role in energy decision making at all levels. These factors include, but are not limited to, governmental structure and power balances, actions taken by politicians, and partisan-based or self serving actions taken by individuals and groups. 199:27 199:40 end
• 5.6 Environmental factors Energy decisions are influenced by environmental factors.
Environmental costs of energy decisions affect energy decision making at all levels. All energy decisions have environmental consequences. These consequences can be positive or negative. 199:27 199:41 end
• 5.7 Social Factors Energy decisions are influenced by social factors.
Questions of ethics, morality and social norms affect energy decision making at all levels. Social factors often involve economic, political and environmental factors. 199:27 199:42 end
• Human use of energy
199:28 end
• 6.1 Two meanings 'conservation of energy' Conservation of energy has two very different meanings.
There is the physical law of conservation of energy, also known as the First Law of Thermodynamics. This law says that the total amount of energy in the Universe is constant. Conserving energy is also commonly used to mean the decreased use of societal energy resources. When speaking of people conserving energy this second meaning is always intended. 199:28 199:43 end
• 6.2 Conserving energy One way to manage energy resources is through conservation.
Conservation includes reducing wasteful energy use, using energy for a given purpose more efficiently, making strategic choices as to sources of energy and reducing energy use altogether. 199:28 199:44 end
• 6.3 Demand for energy is increasing Human demand for energy is increasing.
Population growth, industrialization and socio-economic development result in increased demand for energy. Societies have choices with regard to how they respond to this increase. Each of these choices has consequences. 199:28 199:45 end
• 6.4 Earth has finite energy resources Earth has finite energy resources.
Increasing human energy consumption places stress on the natural processes that renew some energy resources and it depletes those that cannot be renewed. 199:28 199:46 end
• 6.5 Social and technological innovation Social and technological innovation affects the amount of energy used by human society.
The amount of energy society uses per capita or in total can be decreased. Decreases will happen as a result of technological and social innovation and change. Decreased use of energy does not equate to decreased quality of life. In many cases it will be associated with increased quality of life in the form of increased economic and national security, reduced environmental risks and monetary savings. 199:28 199:47 end
• 6.6 Behavior and design Behavior and design affect the amount of energy used by human society.
There are actions individuals and society can take to conserve energy. These actions might come in the form of changes in behavior or in changes to the design of technology and infrastructure. Some of these actions have more impact than others. 199:28 199:48 end
• 6.7 Embedded energy Products and services carry with them embedded energy.
The energy needed for the entire lifecycle of a product or service is called the “embedded” or “embodied” energy. An accounting of the embedded energy in a product or service along with knowledge of the source(s) of the energy is essential when calculating the amount of energy used and in assessing impacts and consequences. 199:28 199:49 end
• 6.8 Calculating and monitoring energy use Amount of energy used can be calculated and monitored.
An individual, organization or government can monitor, measure and control energy use in many ways. Understanding utility costs, knowing where consumer goods and food come from, and understanding energy efficiency as it relates to home, work and transportation are essential to this process. 199:28 199:50 end
• Energy affects quality of life The quality of life of individuals and societies is affected by energy choices.
199:1 end
• 7.1 Economic security Economic security is impacted by energy choices.
Individuals and society continually make energy choices that have economic consequences. These consequences come in the form of monetary cost in general and in the form of price fluctuation and instability specifically. 199:1 199:51 end
• 7.2 National security National security is impacted by energy choices.
The security of a nation is dependent, in part, on the sources of that nation's energy supplies. For example, a nation that has diverse sources of energy that come mostly from within its borders is more secure than a nation largely dependent on foreign energy supplies. 199:1 199:52 end
• 7.3 Environmental quality Environmental quality is impacted by energy choices.
Energy choices made by humans have environmental consequences. The quality of life of humans and other organisms on Earth can be significantly affected by these consequences. 199:1 199:53 end
• 7.4 Fossil fuel supplies are limited Increasing demand for and limited supplies of fossil fuels affects quality of life.
Fossil fuels provide the vast majority of the world's energy. Fossil fuel supplies are limited. If society has not transitioned to sources of energy that are renewable before depleting Earth's fossil fuel supplies, it will find itself in a situation where energy demand far exceeds energy supply. This will have many social and economic consequences. 199:1 199:54 end
| 4,144
| 20,064
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.203125
| 3
|
CC-MAIN-2014-52
|
longest
|
en
| 0.953576
|
https://slideplayer.com/slide/5701150/
| 1,722,648,875,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00582.warc.gz
| 418,080,894
| 18,062
|
# Apply the Pythagorean Theorem Chapter 7.1. Sides of a Right Triangle Hypotenuse – the side of a right triangle opposite the right angle and the longest.
## Presentation on theme: "Apply the Pythagorean Theorem Chapter 7.1. Sides of a Right Triangle Hypotenuse – the side of a right triangle opposite the right angle and the longest."— Presentation transcript:
Apply the Pythagorean Theorem Chapter 7.1
Sides of a Right Triangle Hypotenuse – the side of a right triangle opposite the right angle and the longest side. Legs – the sides of a right triangle that are not the hypotenuse.
The Pythagorean Theorem In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the legs.
Proving the Pythagorean Theorem There are multiple ways of proving the Pythagorean Theorem. If you can find a legitimate proof of the Pythagorean theorem you may earn 2 bonus points.
Using the Pythagorean Theorem 68x²x² 36 + 64 = x ² 100= x ²
Find the missing leg x3 5² x ² + 9 = 25 x ² = 16
Find the Area of the Triangle What is the formula for the area of a triangle? A = ½bh How will we find the height?
Find the Area of the Triangle
Pythagorean Triples A Pythagorean Triple is a set of 3 positive integers or whole numbers that satisfies the Pythagorean theorem.
Is it a Pythagorean Triple? 3, 4, and 5 21, 28, and 35 30, 72, and 91 14, 48, and 50 yes no yes
If I am given 2 sides of a right triangle make up the sides of Pythagorean triple, how do you find the missing side? There are 2 possible scenarios: 1.You are given both legs of the right triangle and need to solve for the hypotenuse. 2.You are given one leg and one hypotenuse and need to solve for the other leg.
Given 2 sides: 20 and 25 Scenario 1. 20 and 25 are the legs 20 2 + 25 2 = x 2 400 + 625 = x 2 1025 = x 2 32.02 = x This cannot be the answer for a Pythagorean triple because it is not a whole number. Scenario 2. 20 and 25 are the leg and hypotenuse 20 2 + x 2 = 25 2 400 + x 2 = 625 x 2 = 225 X = 25 This can be the answer for a Pythagorean triple because it is not a whole number.
Given 2 sides: 28 and 96
Find the area when given a leg and the hypotenuse 1.Find the other leg by plugging the known values into the Pythagorean Theorem. 2.Use the 2 legs in the formula for area of a triangle A = ½BH
Find the area when given a leg and a hypotenuse. L=8 and h= 16 A ò+8ò=16ò A ò+64=256 A ò=192
Find the area when given a leg and a hypotenuse. L=13 and h= 17 A ò+13ò=17ò A ò+169=289 A ò=120
#3 – 7, 11- 16, 18 - 23
Download ppt "Apply the Pythagorean Theorem Chapter 7.1. Sides of a Right Triangle Hypotenuse – the side of a right triangle opposite the right angle and the longest."
Similar presentations
| 816
| 2,723
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.84375
| 5
|
CC-MAIN-2024-33
|
latest
|
en
| 0.807539
|
https://www.unitconverters.net/magnetic-flux-density/gamma-to-gauss.htm
| 1,675,605,560,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00038.warc.gz
| 1,044,447,052
| 2,916
|
Home / Magnetic Flux Density Conversion / Convert Gamma to Gauss
# Convert Gamma to Gauss
Please provide values below to convert gamma to gauss [Gs, G], or vice versa.
From: gamma To: gauss
### Gamma to Gauss Conversion Table
GammaGauss [Gs, G]
0.01 gamma1.0E-7 Gs, G
0.1 gamma1.0E-6 Gs, G
1 gamma1.0E-5 Gs, G
2 gamma2.0E-5 Gs, G
3 gamma3.0E-5 Gs, G
5 gamma5.0E-5 Gs, G
10 gamma0.0001 Gs, G
20 gamma0.0002 Gs, G
50 gamma0.0005 Gs, G
100 gamma0.001 Gs, G
1000 gamma0.01 Gs, G
### How to Convert Gamma to Gauss
1 gamma = 1.0E-5 Gs, G
1 Gs, G = 100000 gamma
Example: convert 15 gamma to Gs, G:
15 gamma = 15 × 1.0E-5 Gs, G = 0.00015 Gs, G
| 268
| 644
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2023-06
|
longest
|
en
| 0.420419
|
http://us.metamath.org/mpeuni/lmatcl.html
| 1,638,035,632,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00264.warc.gz
| 93,114,791
| 9,388
|
Mathbox for Thierry Arnoux < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > lmatcl Structured version Visualization version GIF version
Theorem lmatcl 29656
Description: Closure of the literal matrix. (Contributed by Thierry Arnoux, 12-Sep-2020.)
Hypotheses
Ref Expression
lmatfval.m 𝑀 = (litMat‘𝑊)
lmatfval.n (𝜑𝑁 ∈ ℕ)
lmatfval.w (𝜑𝑊 ∈ Word Word 𝑉)
lmatfval.1 (𝜑 → (#‘𝑊) = 𝑁)
lmatfval.2 ((𝜑𝑖 ∈ (0..^𝑁)) → (#‘(𝑊𝑖)) = 𝑁)
lmatcl.b 𝑉 = (Base‘𝑅)
lmatcl.1 𝑂 = ((1...𝑁) Mat 𝑅)
lmatcl.2 𝑃 = (Base‘𝑂)
lmatcl.r (𝜑𝑅𝑋)
Assertion
Ref Expression
lmatcl (𝜑𝑀𝑃)
Distinct variable groups: 𝑖,𝑀 𝑖,𝑁 𝑖,𝑊 𝜑,𝑖
Allowed substitution hints: 𝑃(𝑖) 𝑅(𝑖) 𝑂(𝑖) 𝑉(𝑖) 𝑋(𝑖)
Proof of Theorem lmatcl
Dummy variables 𝑗 𝑘 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 lmatfval.m . . . 4 𝑀 = (litMat‘𝑊)
2 lmatfval.w . . . . 5 (𝜑𝑊 ∈ Word Word 𝑉)
3 lmatval 29653 . . . . 5 (𝑊 ∈ Word Word 𝑉 → (litMat‘𝑊) = (𝑘 ∈ (1...(#‘𝑊)), 𝑗 ∈ (1...(#‘(𝑊‘0))) ↦ ((𝑊‘(𝑘 − 1))‘(𝑗 − 1))))
42, 3syl 17 . . . 4 (𝜑 → (litMat‘𝑊) = (𝑘 ∈ (1...(#‘𝑊)), 𝑗 ∈ (1...(#‘(𝑊‘0))) ↦ ((𝑊‘(𝑘 − 1))‘(𝑗 − 1))))
51, 4syl5eq 2672 . . 3 (𝜑𝑀 = (𝑘 ∈ (1...(#‘𝑊)), 𝑗 ∈ (1...(#‘(𝑊‘0))) ↦ ((𝑊‘(𝑘 − 1))‘(𝑗 − 1))))
6 lmatfval.1 . . . . 5 (𝜑 → (#‘𝑊) = 𝑁)
76oveq2d 6621 . . . 4 (𝜑 → (1...(#‘𝑊)) = (1...𝑁))
8 lmatfval.n . . . . . . 7 (𝜑𝑁 ∈ ℕ)
9 lbfzo0 12445 . . . . . . 7 (0 ∈ (0..^𝑁) ↔ 𝑁 ∈ ℕ)
108, 9sylibr 224 . . . . . 6 (𝜑 → 0 ∈ (0..^𝑁))
11 0nn0 11252 . . . . . . . 8 0 ∈ ℕ0
1211a1i 11 . . . . . . 7 (𝜑 → 0 ∈ ℕ0)
13 simpr 477 . . . . . . . . 9 ((𝜑𝑖 = 0) → 𝑖 = 0)
1413eleq1d 2688 . . . . . . . 8 ((𝜑𝑖 = 0) → (𝑖 ∈ (0..^𝑁) ↔ 0 ∈ (0..^𝑁)))
1513fveq2d 6154 . . . . . . . . . 10 ((𝜑𝑖 = 0) → (𝑊𝑖) = (𝑊‘0))
1615fveq2d 6154 . . . . . . . . 9 ((𝜑𝑖 = 0) → (#‘(𝑊𝑖)) = (#‘(𝑊‘0)))
1716eqeq1d 2628 . . . . . . . 8 ((𝜑𝑖 = 0) → ((#‘(𝑊𝑖)) = 𝑁 ↔ (#‘(𝑊‘0)) = 𝑁))
1814, 17imbi12d 334 . . . . . . 7 ((𝜑𝑖 = 0) → ((𝑖 ∈ (0..^𝑁) → (#‘(𝑊𝑖)) = 𝑁) ↔ (0 ∈ (0..^𝑁) → (#‘(𝑊‘0)) = 𝑁)))
19 lmatfval.2 . . . . . . . 8 ((𝜑𝑖 ∈ (0..^𝑁)) → (#‘(𝑊𝑖)) = 𝑁)
2019ex 450 . . . . . . 7 (𝜑 → (𝑖 ∈ (0..^𝑁) → (#‘(𝑊𝑖)) = 𝑁))
2112, 18, 20vtocld 3248 . . . . . 6 (𝜑 → (0 ∈ (0..^𝑁) → (#‘(𝑊‘0)) = 𝑁))
2210, 21mpd 15 . . . . 5 (𝜑 → (#‘(𝑊‘0)) = 𝑁)
2322oveq2d 6621 . . . 4 (𝜑 → (1...(#‘(𝑊‘0))) = (1...𝑁))
24 eqidd 2627 . . . 4 (𝜑 → ((𝑊‘(𝑘 − 1))‘(𝑗 − 1)) = ((𝑊‘(𝑘 − 1))‘(𝑗 − 1)))
257, 23, 24mpt2eq123dv 6671 . . 3 (𝜑 → (𝑘 ∈ (1...(#‘𝑊)), 𝑗 ∈ (1...(#‘(𝑊‘0))) ↦ ((𝑊‘(𝑘 − 1))‘(𝑗 − 1))) = (𝑘 ∈ (1...𝑁), 𝑗 ∈ (1...𝑁) ↦ ((𝑊‘(𝑘 − 1))‘(𝑗 − 1))))
265, 25eqtrd 2660 . 2 (𝜑𝑀 = (𝑘 ∈ (1...𝑁), 𝑗 ∈ (1...𝑁) ↦ ((𝑊‘(𝑘 − 1))‘(𝑗 − 1))))
27 lmatcl.1 . . 3 𝑂 = ((1...𝑁) Mat 𝑅)
28 lmatcl.b . . 3 𝑉 = (Base‘𝑅)
29 lmatcl.2 . . 3 𝑃 = (Base‘𝑂)
30 fzfid 12709 . . 3 (𝜑 → (1...𝑁) ∈ Fin)
31 lmatcl.r . . 3 (𝜑𝑅𝑋)
3223ad2ant1 1080 . . . . 5 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → 𝑊 ∈ Word Word 𝑉)
33 simp2 1060 . . . . . . 7 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → 𝑘 ∈ (1...𝑁))
34 fz1fzo0m1 12453 . . . . . . 7 (𝑘 ∈ (1...𝑁) → (𝑘 − 1) ∈ (0..^𝑁))
3533, 34syl 17 . . . . . 6 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (𝑘 − 1) ∈ (0..^𝑁))
3663ad2ant1 1080 . . . . . . 7 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (#‘𝑊) = 𝑁)
3736oveq2d 6621 . . . . . 6 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (0..^(#‘𝑊)) = (0..^𝑁))
3835, 37eleqtrrd 2707 . . . . 5 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (𝑘 − 1) ∈ (0..^(#‘𝑊)))
39 wrdsymbcl 13252 . . . . 5 ((𝑊 ∈ Word Word 𝑉 ∧ (𝑘 − 1) ∈ (0..^(#‘𝑊))) → (𝑊‘(𝑘 − 1)) ∈ Word 𝑉)
4032, 38, 39syl2anc 692 . . . 4 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (𝑊‘(𝑘 − 1)) ∈ Word 𝑉)
41 simp3 1061 . . . . . 6 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → 𝑗 ∈ (1...𝑁))
42 fz1fzo0m1 12453 . . . . . 6 (𝑗 ∈ (1...𝑁) → (𝑗 − 1) ∈ (0..^𝑁))
4341, 42syl 17 . . . . 5 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (𝑗 − 1) ∈ (0..^𝑁))
44 ovex 6633 . . . . . . . . . . 11 (𝑘 − 1) ∈ V
4544a1i 11 . . . . . . . . . 10 (𝜑 → (𝑘 − 1) ∈ V)
46 simpr 477 . . . . . . . . . . . 12 ((𝜑𝑖 = (𝑘 − 1)) → 𝑖 = (𝑘 − 1))
47 eqidd 2627 . . . . . . . . . . . 12 ((𝜑𝑖 = (𝑘 − 1)) → (0..^𝑁) = (0..^𝑁))
4846, 47eleq12d 2698 . . . . . . . . . . 11 ((𝜑𝑖 = (𝑘 − 1)) → (𝑖 ∈ (0..^𝑁) ↔ (𝑘 − 1) ∈ (0..^𝑁)))
4946fveq2d 6154 . . . . . . . . . . . . 13 ((𝜑𝑖 = (𝑘 − 1)) → (𝑊𝑖) = (𝑊‘(𝑘 − 1)))
5049fveq2d 6154 . . . . . . . . . . . 12 ((𝜑𝑖 = (𝑘 − 1)) → (#‘(𝑊𝑖)) = (#‘(𝑊‘(𝑘 − 1))))
5150eqeq1d 2628 . . . . . . . . . . 11 ((𝜑𝑖 = (𝑘 − 1)) → ((#‘(𝑊𝑖)) = 𝑁 ↔ (#‘(𝑊‘(𝑘 − 1))) = 𝑁))
5248, 51imbi12d 334 . . . . . . . . . 10 ((𝜑𝑖 = (𝑘 − 1)) → ((𝑖 ∈ (0..^𝑁) → (#‘(𝑊𝑖)) = 𝑁) ↔ ((𝑘 − 1) ∈ (0..^𝑁) → (#‘(𝑊‘(𝑘 − 1))) = 𝑁)))
5345, 52, 20vtocld 3248 . . . . . . . . 9 (𝜑 → ((𝑘 − 1) ∈ (0..^𝑁) → (#‘(𝑊‘(𝑘 − 1))) = 𝑁))
5453imp 445 . . . . . . . 8 ((𝜑 ∧ (𝑘 − 1) ∈ (0..^𝑁)) → (#‘(𝑊‘(𝑘 − 1))) = 𝑁)
5534, 54sylan2 491 . . . . . . 7 ((𝜑𝑘 ∈ (1...𝑁)) → (#‘(𝑊‘(𝑘 − 1))) = 𝑁)
56553adant3 1079 . . . . . 6 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (#‘(𝑊‘(𝑘 − 1))) = 𝑁)
5756oveq2d 6621 . . . . 5 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (0..^(#‘(𝑊‘(𝑘 − 1)))) = (0..^𝑁))
5843, 57eleqtrrd 2707 . . . 4 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → (𝑗 − 1) ∈ (0..^(#‘(𝑊‘(𝑘 − 1)))))
59 wrdsymbcl 13252 . . . 4 (((𝑊‘(𝑘 − 1)) ∈ Word 𝑉 ∧ (𝑗 − 1) ∈ (0..^(#‘(𝑊‘(𝑘 − 1))))) → ((𝑊‘(𝑘 − 1))‘(𝑗 − 1)) ∈ 𝑉)
6040, 58, 59syl2anc 692 . . 3 ((𝜑𝑘 ∈ (1...𝑁) ∧ 𝑗 ∈ (1...𝑁)) → ((𝑊‘(𝑘 − 1))‘(𝑗 − 1)) ∈ 𝑉)
6127, 28, 29, 30, 31, 60matbas2d 20143 . 2 (𝜑 → (𝑘 ∈ (1...𝑁), 𝑗 ∈ (1...𝑁) ↦ ((𝑊‘(𝑘 − 1))‘(𝑗 − 1))) ∈ 𝑃)
6226, 61eqeltrd 2704 1 (𝜑𝑀𝑃)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 384 ∧ w3a 1036 = wceq 1480 ∈ wcel 1992 Vcvv 3191 ‘cfv 5850 (class class class)co 6605 ↦ cmpt2 6607 0cc0 9881 1c1 9882 − cmin 10211 ℕcn 10965 ℕ0cn0 11237 ...cfz 12265 ..^cfzo 12403 #chash 13054 Word cword 13225 Basecbs 15776 Mat cmat 20127 litMatclmat 29651 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1841 ax-6 1890 ax-7 1937 ax-8 1994 ax-9 2001 ax-10 2021 ax-11 2036 ax-12 2049 ax-13 2250 ax-ext 2606 ax-rep 4736 ax-sep 4746 ax-nul 4754 ax-pow 4808 ax-pr 4872 ax-un 6903 ax-cnex 9937 ax-resscn 9938 ax-1cn 9939 ax-icn 9940 ax-addcl 9941 ax-addrcl 9942 ax-mulcl 9943 ax-mulrcl 9944 ax-mulcom 9945 ax-addass 9946 ax-mulass 9947 ax-distr 9948 ax-i2m1 9949 ax-1ne0 9950 ax-1rid 9951 ax-rnegex 9952 ax-rrecex 9953 ax-cnre 9954 ax-pre-lttri 9955 ax-pre-lttrn 9956 ax-pre-ltadd 9957 ax-pre-mulgt0 9958 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3or 1037 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1883 df-eu 2478 df-mo 2479 df-clab 2613 df-cleq 2619 df-clel 2622 df-nfc 2756 df-ne 2797 df-nel 2900 df-ral 2917 df-rex 2918 df-reu 2919 df-rab 2921 df-v 3193 df-sbc 3423 df-csb 3520 df-dif 3563 df-un 3565 df-in 3567 df-ss 3574 df-pss 3576 df-nul 3897 df-if 4064 df-pw 4137 df-sn 4154 df-pr 4156 df-tp 4158 df-op 4160 df-ot 4162 df-uni 4408 df-int 4446 df-iun 4492 df-br 4619 df-opab 4679 df-mpt 4680 df-tr 4718 df-eprel 4990 df-id 4994 df-po 5000 df-so 5001 df-fr 5038 df-we 5040 df-xp 5085 df-rel 5086 df-cnv 5087 df-co 5088 df-dm 5089 df-rn 5090 df-res 5091 df-ima 5092 df-pred 5642 df-ord 5688 df-on 5689 df-lim 5690 df-suc 5691 df-iota 5813 df-fun 5852 df-fn 5853 df-f 5854 df-f1 5855 df-fo 5856 df-f1o 5857 df-fv 5858 df-riota 6566 df-ov 6608 df-oprab 6609 df-mpt2 6610 df-om 7014 df-1st 7116 df-2nd 7117 df-supp 7242 df-wrecs 7353 df-recs 7414 df-rdg 7452 df-1o 7506 df-oadd 7510 df-er 7688 df-map 7805 df-ixp 7854 df-en 7901 df-dom 7902 df-sdom 7903 df-fin 7904 df-fsupp 8221 df-sup 8293 df-card 8710 df-pnf 10021 df-mnf 10022 df-xr 10023 df-ltxr 10024 df-le 10025 df-sub 10213 df-neg 10214 df-nn 10966 df-2 11024 df-3 11025 df-4 11026 df-5 11027 df-6 11028 df-7 11029 df-8 11030 df-9 11031 df-n0 11238 df-z 11323 df-dec 11438 df-uz 11632 df-fz 12266 df-fzo 12404 df-hash 13055 df-word 13233 df-struct 15778 df-ndx 15779 df-slot 15780 df-base 15781 df-sets 15782 df-ress 15783 df-plusg 15870 df-mulr 15871 df-sca 15873 df-vsca 15874 df-ip 15875 df-tset 15876 df-ple 15877 df-ds 15880 df-hom 15882 df-cco 15883 df-0g 16018 df-prds 16024 df-pws 16026 df-sra 19086 df-rgmod 19087 df-dsmm 19990 df-frlm 20005 df-mat 20128 df-lmat 29652 This theorem is referenced by: lmat22det 29662
Copyright terms: Public domain W3C validator
| 5,462
| 8,247
|
{"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}
| 3.203125
| 3
|
CC-MAIN-2021-49
|
latest
|
en
| 0.28604
|
https://www.bizcalcs.com/tag/health-calculators/page/2/
| 1,723,634,770,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00197.warc.gz
| 545,081,635
| 16,516
|
## Full Body Analysis Calculator
Calculate and show me many of the common health indicators. This calculator combines the output from several individual calculators to give you a full body analysis. Quickly find out your BMI, waist to hip ratio, body frame size, ideal weight, body fat, RMR, calories burned, target heart rate, and maximum heart rate. Field Help Input … Read more
## Smoking Costs Calculator
How much is it costing me to smoke? Smoking cigarettes can be an expensive habit. Use this calculator to see how much cigarettes are costing you each day, week, month, year, or over your lifetime. Of course, if you want to quit smoking these costs represent the money you could save. Field Help Input Fields … Read more
## Ideal Weight Calculator
How much should I weigh? Enter your gender and height to calculate your ideal weight using industry accepted formulas. You can see the ideal weights and corresponding BMIs for the 4 most used formulas (Devine, Hamwi, Miller, and Robinson) by clicking the ‘Show Additional Fields’ icon button. Field Help Input Fields Title: A title for … Read more
## Waist to Hip Ratio Calculator
What do my waist and hip size indicate about my risk of heart disease? Recent research shows that a person’s waist-to-hip ratio may be a better indicator of heart disease risk than body mass index. Enter your waist and hip measurement and this calculator will help you determine if you may be at higher risk. … Read more
## Frame Size (Elbow) Calculator
Use my elbow width to determine my frame size. Since bone structure varies in size from person to person, researchers have added frame size as a factor in helping to determine someone’s ideal weight. Knowing your frame size can help you set reasonable weight goals. Enter your height and elbow width and this calculator will … Read more
## Frame Size (Wrist) Calculator
Use my wrist circumference to determine my frame size. Since bone structure varies in size from person to person, researchers have added frame size as a factor in helping to determine someone’s ideal weight. Knowing your frame size can help you set reasonable weight goals. Enter your height and wrist circumference and this calculator will … Read more
## Medication Costs Calculator
How much is medication costing me? Use this calculator to see how much an over-the-counter or prescription medication is costing you each day, week, month, year, or over your lifetime. Field Help Input Fields Title: A title for these calculator results that will help you identify it if you have printed out several versions of … Read more
## Days Until Next Birthday Calculator
So your last birthday went great and you can’t wait for your next birthday? Don’t worry, we have got you covered with our calculator. This tool lets you know the exact number of days till your next birthday and how old you will be then. Exciting right? With our tool, you can do so much … Read more
## Find Date Given Age Calculator
What date will I turn a specific age? What date will you turn 59 and a half, the age you can begin drawing from your retirement funds? Or what date will you turn 15 and 9 months so you can take Driver’s Ed? Enter your birthdate and the age you want to know about and … Read more
## Body Mass Index (BMI) Calculator
The balance of health and body weight is a pressing concern in today’s world. According to the research of 2022, 43% of adults globally were overweight, and 16% were living with obesity. This alarming trend underscores the importance of monitoring one’s health. To take a step towards better health, we encourage you to use our … Read more
| 755
| 3,630
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33
|
latest
|
en
| 0.868219
|
https://www.topperlearning.com/answer/show-that-a-b-a-b-a-b-are-in-arithmetic-progression/ozzi0zyy
| 1,685,751,663,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00009.warc.gz
| 1,114,969,170
| 59,300
|
Request a call back
Show that (a-b)², (a²+b²), (a+b)² are in arithmetic progression.
Asked by apal63216 | 02 Jan, 2021, 07:40: PM
Hint
Check if the common difference is same. If so, the given progression will be AP.
(a - b)², (a² + b²), (a + b)²
1st term = (a - b)²
2nd term = (a² + b²)
3rd term = (a + b)²
Check 2nd term - 1st term
= (a² + b²) - (a - b)²
= (a² + b²) - (a2 - 2b + b2)
= 2b
Check 3rd term - 2nd term
= (a + b)² - (a² + b²)
= a2 + 2b + b2 - a² - b²
= 2b
So the common difference is same. Its is AP.
Answered by | 03 Jan, 2021, 11:27: AM
## Concept Videos
CBSE 10 - Maths
Asked by dixitayush414 | 04 Jan, 2023, 09:03: PM
CBSE 10 - Maths
Asked by apal63216 | 02 Jan, 2021, 07:40: PM
CBSE 10 - Maths
Asked by arindeep.singh | 10 Jul, 2020, 06:33: PM
CBSE 10 - Maths
Asked by pushpa.Kudle09 | 12 Mar, 2020, 12:37: AM
| 383
| 836
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23
|
longest
|
en
| 0.875209
|
https://www.physicsforums.com/threads/functional-relationship-between-pressure-and-position-1d.773847/
| 1,521,729,684,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257647885.78/warc/CC-MAIN-20180322131741-20180322151741-00402.warc.gz
| 845,591,256
| 14,578
|
# Functional relationship between pressure and position(1d).
1. Oct 1, 2014
Hello there, so today I started doing my research on oscillations in a course on advanced mechanics. The experiment was to mathematically model the speed of sound in air and experimentally prove the usability of the model. To keep it simple and pose my question as directly as possible, my professor told me to use the driving sound wave from the speaker in a open cylindrical tube. Further, the maxima of the voltage as read by the microphone(displayed by the oscilloscope) was to be considered as the peak(amplitude) of the pressure.
What I am trying to establish is the functional relationship between pressure and position. I have a graph of voltage against position(attached), but I do not know what to fit the data points with.
All your suggestions and help is hereby gratefully appreciated. Thank you in advance.
File size:
13.2 KB
Views:
64
2. Oct 6, 2014
### Greg Bernhardt
Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
| 252
| 1,140
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13
|
longest
|
en
| 0.955239
|
https://www.jobduniya.com/Private-Companies/3i-Infotech/Fully-Solved-Papers/Topic-Languages-2/Part-161.html
| 1,638,551,485,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00211.warc.gz
| 913,338,049
| 7,215
|
Languages [3i Infotech Placement]: Sample Questions 391 - 393 of 546
Glide to success with Doorsteptutor material for competitive exams : get questions, notes, tests, video lectures and more- for all subjects of your exam.
Question 391
Edit
Describe in Detail
Essay▾
List out some of the OODBMS available.
Explanation
• GEMSTONE/OPAL of Gemstone systems.
• ONTOS of Ontos.
• Objectivity of Objectivity Inc.
• Versant of Versant objects technology.
• Object store of Object Design.
• ARDENT of ARDENT software.
• POET of POET software.
Question 392
Languages
Edit
Write in Short
What is update method called?
Explanation
• The update () method is defined by the AWT and is called when our applet has requested that a portion of its window be redrawn.
• The problem is that the default version of update () first fills an applet with the default background colour and then calls paint () .
• We can override the update () method.
Question 393
Edit
Describe in Detail
Essay▾
What is the output of the following program?
1. `int swap (int ⚹a, int ⚹b)`
2. `{`
3. ` ⚹a =⚹ a +⚹ b;`
4. ` ⚹b =⚹ a -⚹ b;`
5. ` ⚹a =⚹ a -⚹ b;`
6. `}`
7. `main ()`
8. `{`
9. ` int x =10, y =20;`
10. ` swap (&x, &y);`
11. ` printf ( “x =%d y =%d”, x, y);`
12. `}`
Explanation
In this program
int x = 10, y = 20; Given the integer variable x = 10 and y = 20 swap (&x, &y) ; This is use for swapping two values function int swap (int ⚹ a, int ⚹ b){⚹ a =⚹ a +⚹ b;⚹ b =⚹ a -⚹ b;⚹ a =⚹ a -⚹ b;} In a swap function a = x and b = ySo a = 10 and b = 20a =⚹ a +⚹ b;⚹ a = 10 + 20 = 30⚹ b =⚹ a -⚹ b;⚹ b = 30 - 20 = 10⚹ a =⚹ a -⚹ b⚹ a = 30 - 10 = 20 printf ( “x =% d y =% d” , x, y) ; Now the printf print the value of x = 20and y = 10
Developed by:
| 690
| 1,750
|
{"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}
| 2.5625
| 3
|
CC-MAIN-2021-49
|
latest
|
en
| 0.7301
|
https://menwithoutwork.com/let-r-be-the-region-in-the-xy-plane-that-is/
| 1,620,781,284,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00513.warc.gz
| 422,600,651
| 10,411
|
# Let R Be The Region In The Xy Plane That Is Bounded Between The Xaxis And Graph Y=3x+7 For 1&Lt
=xlt=3.1 = 32+7 -152-3
( a)
4= 34+7
17(3, 16 )
(oj7 )
12
(- 1, 4 ))
14
174
Area of the region R= Area of A+ Area of quadrilateral
= 2x 4x 12 + 4 x 4
24 + 16
40 units square.
( b) fly) = 3u+7
[ -1, 37 n =…Math
| 146
| 307
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.421875
| 3
|
CC-MAIN-2021-21
|
latest
|
en
| 0.539864
|
https://www.authorea.com/users/75574/articles/102301-phys6562-f4-daily-quaetion/_show_article
| 1,495,501,377,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463607245.69/warc/CC-MAIN-20170523005639-20170523025639-00489.warc.gz
| 836,106,001
| 276,142
|
# PHYS6562 F4 Daily Quaetion
## Ergodicity
The system of classical particles bouncing back and forth is not ergodic since each particles can only occupy certain close volume in phase space, with fixed $$x,y$$ position, $$P_x, P_y$$ momenta and $$\pm p_z$$. Therefore, the given volume in phase space of the ensemble will not evolve with time.
## Heat engine
(a) True.
(b) True, since the work in done on the gas.
(c) False, it’s a refrigerator.
(d) True.
$W = (P_0)(3V_0) + \int_{4V_0}^{V_0} \frac{4P_0V_0}{V} dV = 3P_0V_0 - 4P_0V_0\log4$ (e) True, a part does no work so $$\Delta U = Q$$. $\Delta U = \frac{3}{2} nRT_c - \frac{3}{2} nRT_h = -\frac{3}{2} (3P_0V_0) = -\frac{9}{2}P_0V_0$ where n is the number of particles in mole and $$R = 8.317$$ is the ideal gas constant. (f) True. The whole cycle $$\Delta U = 0$$, the heat input and the net work done onto the gas must equal the heat output. $Q_h + W = Q_c$
| 329
| 916
|
{"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}
| 3.171875
| 3
|
CC-MAIN-2017-22
|
longest
|
en
| 0.730547
|
http://mathhelpforum.com/calculus/5500-question-about-tangent-lines-slopes-print.html
| 1,503,252,090,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-34/segments/1502886106865.74/warc/CC-MAIN-20170820170023-20170820190023-00065.warc.gz
| 258,566,457
| 3,029
|
# a question about tangent lines and slopes
• Sep 13th 2006, 11:46 PM
a question about tangent lines and slopes
the question is:
Find the points on the graph of y = x ^ (3/2) - x ^ (1/2) at which the tangent line is parallel to the line y - x = 3.
i have derived the first function to find the tangent line then i have derived it again to find it's slope, then derived the second function to get it's slope, then i have set the first slope = the second slope anyway this is what i got lastly:
(3/4) x ^ (-1/2) + (1/4) x ^ (-3/2) = 1
i could not get the value of x from this equation.is there any formula that will get me the value of x like the quadratic formula?
• Sep 14th 2006, 01:17 AM
CaptainBlack
Quote:
the question is:
Find the points on the graph of y = x ^ (3/2) - x ^ (1/2) at which the tangent line is parallel to the line y - x = 3.
i have derived the first function to find the tangent line then i have derived it again to find it's slope, then derived the second function to get it's slope, then i have set the first slope = the second slope anyway this is what i got lastly:
(3/4) x ^ (-1/2) + (1/4) x ^ (-3/2) = 1
i could not get the value of x from this equation.is there any formula that will get me the value of x like the quadratic formula?
The tangent line to y=x^(3/2)-x^(1/2) is parallel to y-x=3, when the
gradient is equal to that of the line.
The gradient of the line is 1, so you want the solutions of:
dy/dx=1.
RonL
• Sep 14th 2006, 12:29 PM
| 441
| 1,480
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.71875
| 4
|
CC-MAIN-2017-34
|
longest
|
en
| 0.946982
|
https://community.jmp.com/t5/Discussions/how-to-replace-two-row-values-by-one/td-p/235535
| 1,601,286,093,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00535.warc.gz
| 314,575,242
| 75,528
|
Choose Language Hide Translation Bar
Highlighted
Level III
## how to replace two row values by one?
I am working with a data set where in I need to find average of two rows and replace the two rows by average.
the following is the data set in which I want to find average value for each day for a given week and sensor.Once i figure this out I can use transpose/stack functions to obtain desired output.Any help is appreciated.
Sensor Week DoW Value A 1 0.25 A 2 0.33 A 3 0.76 A 4 Sun(D) 0.21 A Sun(N) 0.75 A Mon(D) 0.33 A Mon(N) 0.34 B 1 0.6 B 2 0.82 B 3 0.55 B 4 Sun(D) 0.4 B Sun(N) 0.22 B Mon(D) 0.91 B Mon(N) 0.98 C 1 0.25 C 2 0.35 C 3 0.76 C 4 Sun(D) 0.21 C Sun(N) 0.75 C Mon(D) 0.33 C Mon(N) 0.34
The expected output is -
Sensor Week1 Week2 Week3 Week4.Sun Week4.Mon A 0.25 0.33 0.76 0.48 0.34 B 0.6 0.82 0.55 0.31 0.95 C 0.25 0.35 0.76 0.48 0.34
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Super User
## Re: how to replace two row values by one?
See if this bit of JSL will give you what you want, or at least a starting point. I had to make a couple of data assumptions, but it may work for you
``````Names Default To Here( 1 );
dt = New Table( "Example",
New Script(
"Source",
Data Table( "Untitled 29" ) << Transpose(
columns( :Column 1 ),
By( :Column 2 ),
Output Table( "Transpose of Untitled 29" )
)
),
New Column( "Sensor",
Character,
"Nominal",
Set Values(
{"A", "A", "A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", "B",
"C", "C", "C", "C", "C", "C", "C"}
)
),
New Column( "Week",
Numeric,
"Ordinal",
Format( "Best", 12 ),
Set Values(
[1, 2, 3, 4, ., ., ., 1, 2, 3, 4, ., ., ., 1, 2, 3, 4, ., ., .]
)
),
New Column( "DoW",
Character,
"Nominal",
Set Values(
{"", "", "", "Sun(D)", "Sun(N)", "Mon(D)", "Mon(N)", "", "", "",
"Sun(D)", "Sun(N)", "Mon(D)", "Mon(N)", "", "", "", "Sun(D)", "Sun(N)",
"Mon(D)", "Mon(N)"}
)
),
New Column( "Value",
Numeric,
"Continuous",
Format( "Best", 12 ),
Set Values(
[0.25, 0.33, 0.76, 0.21, 0.75, 0.33, 0.34, 0.6, 0.82, 0.55, 0.4, 0.22,
0.91, 0.98, 0.25, 0.35, 0.76, 0.21, 0.75, 0.33, 0.34]
)
)
);
wait(5); // wait so you can see the original data table
dt << New Column( "The Date",
Character,
"Nominal",
Formula(
dowInclude = "";
If( Is Missing( :Week ) == 0,
theInclude = Char( :Week )
);
If( :DoW != "",
dowInclude = Word( 1, :DoW, "(" )
);
Trim( "Week" || theInclude || " " || dowInclude );
),
Set Selected,
Set Display Width( 83 )
);
wait(5); // wait so you can see the new formula generated column
// Create the means
dtSum = dt << Summary(
invisible,
Group( :Sensor, :The Date ),
Mean( :Value ),
Freq( "None" ),
Weight( "None" ),
statistics column name format( "column" ),
Link to original data table( 0 )
);
// Split the table into the Final Display Format
dtFinal = dtSum << Split(
Split By( :The Date ),
Split( :Value ),
Group( :Sensor ),
Remaining Columns( Drop All ),
Sort by Column Property
);
// Clean up
close( dtSum, nosave );``````
Jim
3 REPLIES 3
Highlighted
Staff
## Re: how to replace two row values by one?
You would use the summary table platform to do this with two Subgroup variables.
First you would need to strip out the extra characters from the DoW values and then perform the Summary so for example:
``````New Column( "Day", Character, "Nominal", Formula( Substr( :DoW, 1, 3 )));
Summary(
Group( :Sensor ),
Mean( :Value ),
Subgroup( :Week, :Day ),
Freq( "None" ),
Weight( "None" )
);``````
Highlighted
Level III
## Re: how to replace two row values by one?
Thanks for your response.I have already tried this approach.But the problem is the column names are in the format Value,week,day and renaming them is tricky because the column names change with change in week and day.
Highlighted
Super User
## Re: how to replace two row values by one?
See if this bit of JSL will give you what you want, or at least a starting point. I had to make a couple of data assumptions, but it may work for you
``````Names Default To Here( 1 );
dt = New Table( "Example",
New Script(
"Source",
Data Table( "Untitled 29" ) << Transpose(
columns( :Column 1 ),
By( :Column 2 ),
Output Table( "Transpose of Untitled 29" )
)
),
New Column( "Sensor",
Character,
"Nominal",
Set Values(
{"A", "A", "A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", "B",
"C", "C", "C", "C", "C", "C", "C"}
)
),
New Column( "Week",
Numeric,
"Ordinal",
Format( "Best", 12 ),
Set Values(
[1, 2, 3, 4, ., ., ., 1, 2, 3, 4, ., ., ., 1, 2, 3, 4, ., ., .]
)
),
New Column( "DoW",
Character,
"Nominal",
Set Values(
{"", "", "", "Sun(D)", "Sun(N)", "Mon(D)", "Mon(N)", "", "", "",
"Sun(D)", "Sun(N)", "Mon(D)", "Mon(N)", "", "", "", "Sun(D)", "Sun(N)",
"Mon(D)", "Mon(N)"}
)
),
New Column( "Value",
Numeric,
"Continuous",
Format( "Best", 12 ),
Set Values(
[0.25, 0.33, 0.76, 0.21, 0.75, 0.33, 0.34, 0.6, 0.82, 0.55, 0.4, 0.22,
0.91, 0.98, 0.25, 0.35, 0.76, 0.21, 0.75, 0.33, 0.34]
)
)
);
wait(5); // wait so you can see the original data table
dt << New Column( "The Date",
Character,
"Nominal",
Formula(
dowInclude = "";
If( Is Missing( :Week ) == 0,
theInclude = Char( :Week )
);
If( :DoW != "",
dowInclude = Word( 1, :DoW, "(" )
);
Trim( "Week" || theInclude || " " || dowInclude );
),
Set Selected,
Set Display Width( 83 )
);
wait(5); // wait so you can see the new formula generated column
// Create the means
dtSum = dt << Summary(
invisible,
Group( :Sensor, :The Date ),
Mean( :Value ),
Freq( "None" ),
Weight( "None" ),
statistics column name format( "column" ),
Link to original data table( 0 )
);
// Split the table into the Final Display Format
dtFinal = dtSum << Split(
Split By( :The Date ),
Split( :Value ),
Group( :Sensor ),
Remaining Columns( Drop All ),
Sort by Column Property
);
// Clean up
close( dtSum, nosave );``````
Jim
Article Labels
There are no labels assigned to this post.
| 2,004
| 5,781
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.65625
| 3
|
CC-MAIN-2020-40
|
latest
|
en
| 0.705128
|
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/7942/2/a/x/1/1/
| 1,656,152,832,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00024.warc.gz
| 931,197,808
| 81,881
|
# Properties
Label 7942.2.a.x.1.1 Level $7942$ Weight $2$ Character 7942.1 Self dual yes Analytic conductor $63.417$ Analytic rank $0$ Dimension $2$ CM no Inner twists $1$
# Learn more
## Newspace parameters
Level: $$N$$ $$=$$ $$7942 = 2 \cdot 11 \cdot 19^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 7942.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$63.4171892853$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{17})$$ Defining polynomial: $$x^{2} - x - 4$$ x^2 - x - 4 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 418) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## Embedding invariants
Embedding label 1.1 Root $$2.56155$$ of defining polynomial Character $$\chi$$ $$=$$ 7942.1
## $q$-expansion
$$f(q)$$ $$=$$ $$q+1.00000 q^{2} -2.56155 q^{3} +1.00000 q^{4} +2.00000 q^{5} -2.56155 q^{6} -0.561553 q^{7} +1.00000 q^{8} +3.56155 q^{9} +O(q^{10})$$ $$q+1.00000 q^{2} -2.56155 q^{3} +1.00000 q^{4} +2.00000 q^{5} -2.56155 q^{6} -0.561553 q^{7} +1.00000 q^{8} +3.56155 q^{9} +2.00000 q^{10} +1.00000 q^{11} -2.56155 q^{12} -0.561553 q^{13} -0.561553 q^{14} -5.12311 q^{15} +1.00000 q^{16} -0.561553 q^{17} +3.56155 q^{18} +2.00000 q^{20} +1.43845 q^{21} +1.00000 q^{22} +1.43845 q^{23} -2.56155 q^{24} -1.00000 q^{25} -0.561553 q^{26} -1.43845 q^{27} -0.561553 q^{28} +5.68466 q^{29} -5.12311 q^{30} -2.00000 q^{31} +1.00000 q^{32} -2.56155 q^{33} -0.561553 q^{34} -1.12311 q^{35} +3.56155 q^{36} +5.12311 q^{37} +1.43845 q^{39} +2.00000 q^{40} -2.00000 q^{41} +1.43845 q^{42} +1.00000 q^{44} +7.12311 q^{45} +1.43845 q^{46} +8.00000 q^{47} -2.56155 q^{48} -6.68466 q^{49} -1.00000 q^{50} +1.43845 q^{51} -0.561553 q^{52} -12.8078 q^{53} -1.43845 q^{54} +2.00000 q^{55} -0.561553 q^{56} +5.68466 q^{58} +7.68466 q^{59} -5.12311 q^{60} +6.24621 q^{61} -2.00000 q^{62} -2.00000 q^{63} +1.00000 q^{64} -1.12311 q^{65} -2.56155 q^{66} +7.68466 q^{67} -0.561553 q^{68} -3.68466 q^{69} -1.12311 q^{70} -6.00000 q^{71} +3.56155 q^{72} -9.68466 q^{73} +5.12311 q^{74} +2.56155 q^{75} -0.561553 q^{77} +1.43845 q^{78} +4.00000 q^{79} +2.00000 q^{80} -7.00000 q^{81} -2.00000 q^{82} +14.2462 q^{83} +1.43845 q^{84} -1.12311 q^{85} -14.5616 q^{87} +1.00000 q^{88} +0.876894 q^{89} +7.12311 q^{90} +0.315342 q^{91} +1.43845 q^{92} +5.12311 q^{93} +8.00000 q^{94} -2.56155 q^{96} +7.12311 q^{97} -6.68466 q^{98} +3.56155 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + 2 q^{2} - q^{3} + 2 q^{4} + 4 q^{5} - q^{6} + 3 q^{7} + 2 q^{8} + 3 q^{9}+O(q^{10})$$ 2 * q + 2 * q^2 - q^3 + 2 * q^4 + 4 * q^5 - q^6 + 3 * q^7 + 2 * q^8 + 3 * q^9 $$2 q + 2 q^{2} - q^{3} + 2 q^{4} + 4 q^{5} - q^{6} + 3 q^{7} + 2 q^{8} + 3 q^{9} + 4 q^{10} + 2 q^{11} - q^{12} + 3 q^{13} + 3 q^{14} - 2 q^{15} + 2 q^{16} + 3 q^{17} + 3 q^{18} + 4 q^{20} + 7 q^{21} + 2 q^{22} + 7 q^{23} - q^{24} - 2 q^{25} + 3 q^{26} - 7 q^{27} + 3 q^{28} - q^{29} - 2 q^{30} - 4 q^{31} + 2 q^{32} - q^{33} + 3 q^{34} + 6 q^{35} + 3 q^{36} + 2 q^{37} + 7 q^{39} + 4 q^{40} - 4 q^{41} + 7 q^{42} + 2 q^{44} + 6 q^{45} + 7 q^{46} + 16 q^{47} - q^{48} - q^{49} - 2 q^{50} + 7 q^{51} + 3 q^{52} - 5 q^{53} - 7 q^{54} + 4 q^{55} + 3 q^{56} - q^{58} + 3 q^{59} - 2 q^{60} - 4 q^{61} - 4 q^{62} - 4 q^{63} + 2 q^{64} + 6 q^{65} - q^{66} + 3 q^{67} + 3 q^{68} + 5 q^{69} + 6 q^{70} - 12 q^{71} + 3 q^{72} - 7 q^{73} + 2 q^{74} + q^{75} + 3 q^{77} + 7 q^{78} + 8 q^{79} + 4 q^{80} - 14 q^{81} - 4 q^{82} + 12 q^{83} + 7 q^{84} + 6 q^{85} - 25 q^{87} + 2 q^{88} + 10 q^{89} + 6 q^{90} + 13 q^{91} + 7 q^{92} + 2 q^{93} + 16 q^{94} - q^{96} + 6 q^{97} - q^{98} + 3 q^{99}+O(q^{100})$$ 2 * q + 2 * q^2 - q^3 + 2 * q^4 + 4 * q^5 - q^6 + 3 * q^7 + 2 * q^8 + 3 * q^9 + 4 * q^10 + 2 * q^11 - q^12 + 3 * q^13 + 3 * q^14 - 2 * q^15 + 2 * q^16 + 3 * q^17 + 3 * q^18 + 4 * q^20 + 7 * q^21 + 2 * q^22 + 7 * q^23 - q^24 - 2 * q^25 + 3 * q^26 - 7 * q^27 + 3 * q^28 - q^29 - 2 * q^30 - 4 * q^31 + 2 * q^32 - q^33 + 3 * q^34 + 6 * q^35 + 3 * q^36 + 2 * q^37 + 7 * q^39 + 4 * q^40 - 4 * q^41 + 7 * q^42 + 2 * q^44 + 6 * q^45 + 7 * q^46 + 16 * q^47 - q^48 - q^49 - 2 * q^50 + 7 * q^51 + 3 * q^52 - 5 * q^53 - 7 * q^54 + 4 * q^55 + 3 * q^56 - q^58 + 3 * q^59 - 2 * q^60 - 4 * q^61 - 4 * q^62 - 4 * q^63 + 2 * q^64 + 6 * q^65 - q^66 + 3 * q^67 + 3 * q^68 + 5 * q^69 + 6 * q^70 - 12 * q^71 + 3 * q^72 - 7 * q^73 + 2 * q^74 + q^75 + 3 * q^77 + 7 * q^78 + 8 * q^79 + 4 * q^80 - 14 * q^81 - 4 * q^82 + 12 * q^83 + 7 * q^84 + 6 * q^85 - 25 * q^87 + 2 * q^88 + 10 * q^89 + 6 * q^90 + 13 * q^91 + 7 * q^92 + 2 * q^93 + 16 * q^94 - q^96 + 6 * q^97 - q^98 + 3 * q^99
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 1.00000 0.707107
$$3$$ −2.56155 −1.47891 −0.739457 0.673204i $$-0.764917\pi$$
−0.739457 + 0.673204i $$0.764917\pi$$
$$4$$ 1.00000 0.500000
$$5$$ 2.00000 0.894427 0.447214 0.894427i $$-0.352416\pi$$
0.447214 + 0.894427i $$0.352416\pi$$
$$6$$ −2.56155 −1.04575
$$7$$ −0.561553 −0.212247 −0.106124 0.994353i $$-0.533844\pi$$
−0.106124 + 0.994353i $$0.533844\pi$$
$$8$$ 1.00000 0.353553
$$9$$ 3.56155 1.18718
$$10$$ 2.00000 0.632456
$$11$$ 1.00000 0.301511
$$12$$ −2.56155 −0.739457
$$13$$ −0.561553 −0.155747 −0.0778734 0.996963i $$-0.524813\pi$$
−0.0778734 + 0.996963i $$0.524813\pi$$
$$14$$ −0.561553 −0.150081
$$15$$ −5.12311 −1.32278
$$16$$ 1.00000 0.250000
$$17$$ −0.561553 −0.136197 −0.0680983 0.997679i $$-0.521693\pi$$
−0.0680983 + 0.997679i $$0.521693\pi$$
$$18$$ 3.56155 0.839466
$$19$$ 0 0
$$20$$ 2.00000 0.447214
$$21$$ 1.43845 0.313895
$$22$$ 1.00000 0.213201
$$23$$ 1.43845 0.299937 0.149968 0.988691i $$-0.452083\pi$$
0.149968 + 0.988691i $$0.452083\pi$$
$$24$$ −2.56155 −0.522875
$$25$$ −1.00000 −0.200000
$$26$$ −0.561553 −0.110130
$$27$$ −1.43845 −0.276829
$$28$$ −0.561553 −0.106124
$$29$$ 5.68466 1.05561 0.527807 0.849364i $$-0.323014\pi$$
0.527807 + 0.849364i $$0.323014\pi$$
$$30$$ −5.12311 −0.935347
$$31$$ −2.00000 −0.359211 −0.179605 0.983739i $$-0.557482\pi$$
−0.179605 + 0.983739i $$0.557482\pi$$
$$32$$ 1.00000 0.176777
$$33$$ −2.56155 −0.445909
$$34$$ −0.561553 −0.0963055
$$35$$ −1.12311 −0.189839
$$36$$ 3.56155 0.593592
$$37$$ 5.12311 0.842233 0.421117 0.907006i $$-0.361638\pi$$
0.421117 + 0.907006i $$0.361638\pi$$
$$38$$ 0 0
$$39$$ 1.43845 0.230336
$$40$$ 2.00000 0.316228
$$41$$ −2.00000 −0.312348 −0.156174 0.987730i $$-0.549916\pi$$
−0.156174 + 0.987730i $$0.549916\pi$$
$$42$$ 1.43845 0.221957
$$43$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$44$$ 1.00000 0.150756
$$45$$ 7.12311 1.06185
$$46$$ 1.43845 0.212087
$$47$$ 8.00000 1.16692 0.583460 0.812142i $$-0.301699\pi$$
0.583460 + 0.812142i $$0.301699\pi$$
$$48$$ −2.56155 −0.369728
$$49$$ −6.68466 −0.954951
$$50$$ −1.00000 −0.141421
$$51$$ 1.43845 0.201423
$$52$$ −0.561553 −0.0778734
$$53$$ −12.8078 −1.75928 −0.879641 0.475638i $$-0.842217\pi$$
−0.879641 + 0.475638i $$0.842217\pi$$
$$54$$ −1.43845 −0.195748
$$55$$ 2.00000 0.269680
$$56$$ −0.561553 −0.0750407
$$57$$ 0 0
$$58$$ 5.68466 0.746432
$$59$$ 7.68466 1.00046 0.500229 0.865893i $$-0.333249\pi$$
0.500229 + 0.865893i $$0.333249\pi$$
$$60$$ −5.12311 −0.661390
$$61$$ 6.24621 0.799745 0.399873 0.916571i $$-0.369054\pi$$
0.399873 + 0.916571i $$0.369054\pi$$
$$62$$ −2.00000 −0.254000
$$63$$ −2.00000 −0.251976
$$64$$ 1.00000 0.125000
$$65$$ −1.12311 −0.139304
$$66$$ −2.56155 −0.315305
$$67$$ 7.68466 0.938830 0.469415 0.882978i $$-0.344465\pi$$
0.469415 + 0.882978i $$0.344465\pi$$
$$68$$ −0.561553 −0.0680983
$$69$$ −3.68466 −0.443581
$$70$$ −1.12311 −0.134237
$$71$$ −6.00000 −0.712069 −0.356034 0.934473i $$-0.615871\pi$$
−0.356034 + 0.934473i $$0.615871\pi$$
$$72$$ 3.56155 0.419733
$$73$$ −9.68466 −1.13350 −0.566752 0.823889i $$-0.691800\pi$$
−0.566752 + 0.823889i $$0.691800\pi$$
$$74$$ 5.12311 0.595549
$$75$$ 2.56155 0.295783
$$76$$ 0 0
$$77$$ −0.561553 −0.0639949
$$78$$ 1.43845 0.162872
$$79$$ 4.00000 0.450035 0.225018 0.974355i $$-0.427756\pi$$
0.225018 + 0.974355i $$0.427756\pi$$
$$80$$ 2.00000 0.223607
$$81$$ −7.00000 −0.777778
$$82$$ −2.00000 −0.220863
$$83$$ 14.2462 1.56372 0.781862 0.623451i $$-0.214270\pi$$
0.781862 + 0.623451i $$0.214270\pi$$
$$84$$ 1.43845 0.156947
$$85$$ −1.12311 −0.121818
$$86$$ 0 0
$$87$$ −14.5616 −1.56116
$$88$$ 1.00000 0.106600
$$89$$ 0.876894 0.0929506 0.0464753 0.998919i $$-0.485201\pi$$
0.0464753 + 0.998919i $$0.485201\pi$$
$$90$$ 7.12311 0.750841
$$91$$ 0.315342 0.0330568
$$92$$ 1.43845 0.149968
$$93$$ 5.12311 0.531241
$$94$$ 8.00000 0.825137
$$95$$ 0 0
$$96$$ −2.56155 −0.261437
$$97$$ 7.12311 0.723242 0.361621 0.932325i $$-0.382223\pi$$
0.361621 + 0.932325i $$0.382223\pi$$
$$98$$ −6.68466 −0.675252
$$99$$ 3.56155 0.357950
$$100$$ −1.00000 −0.100000
$$101$$ −6.87689 −0.684277 −0.342138 0.939650i $$-0.611151\pi$$
−0.342138 + 0.939650i $$0.611151\pi$$
$$102$$ 1.43845 0.142427
$$103$$ 13.3693 1.31732 0.658659 0.752442i $$-0.271124\pi$$
0.658659 + 0.752442i $$0.271124\pi$$
$$104$$ −0.561553 −0.0550648
$$105$$ 2.87689 0.280756
$$106$$ −12.8078 −1.24400
$$107$$ 9.93087 0.960053 0.480027 0.877254i $$-0.340627\pi$$
0.480027 + 0.877254i $$0.340627\pi$$
$$108$$ −1.43845 −0.138415
$$109$$ 6.31534 0.604900 0.302450 0.953165i $$-0.402195\pi$$
0.302450 + 0.953165i $$0.402195\pi$$
$$110$$ 2.00000 0.190693
$$111$$ −13.1231 −1.24559
$$112$$ −0.561553 −0.0530618
$$113$$ 3.12311 0.293797 0.146899 0.989152i $$-0.453071\pi$$
0.146899 + 0.989152i $$0.453071\pi$$
$$114$$ 0 0
$$115$$ 2.87689 0.268272
$$116$$ 5.68466 0.527807
$$117$$ −2.00000 −0.184900
$$118$$ 7.68466 0.707430
$$119$$ 0.315342 0.0289073
$$120$$ −5.12311 −0.467673
$$121$$ 1.00000 0.0909091
$$122$$ 6.24621 0.565505
$$123$$ 5.12311 0.461935
$$124$$ −2.00000 −0.179605
$$125$$ −12.0000 −1.07331
$$126$$ −2.00000 −0.178174
$$127$$ −2.24621 −0.199319 −0.0996595 0.995022i $$-0.531775\pi$$
−0.0996595 + 0.995022i $$0.531775\pi$$
$$128$$ 1.00000 0.0883883
$$129$$ 0 0
$$130$$ −1.12311 −0.0985029
$$131$$ 1.12311 0.0981262 0.0490631 0.998796i $$-0.484376\pi$$
0.0490631 + 0.998796i $$0.484376\pi$$
$$132$$ −2.56155 −0.222955
$$133$$ 0 0
$$134$$ 7.68466 0.663853
$$135$$ −2.87689 −0.247604
$$136$$ −0.561553 −0.0481528
$$137$$ 12.5616 1.07321 0.536603 0.843835i $$-0.319707\pi$$
0.536603 + 0.843835i $$0.319707\pi$$
$$138$$ −3.68466 −0.313659
$$139$$ 7.36932 0.625057 0.312529 0.949908i $$-0.398824\pi$$
0.312529 + 0.949908i $$0.398824\pi$$
$$140$$ −1.12311 −0.0949197
$$141$$ −20.4924 −1.72577
$$142$$ −6.00000 −0.503509
$$143$$ −0.561553 −0.0469594
$$144$$ 3.56155 0.296796
$$145$$ 11.3693 0.944170
$$146$$ −9.68466 −0.801508
$$147$$ 17.1231 1.41229
$$148$$ 5.12311 0.421117
$$149$$ −6.87689 −0.563377 −0.281689 0.959506i $$-0.590894\pi$$
−0.281689 + 0.959506i $$0.590894\pi$$
$$150$$ 2.56155 0.209150
$$151$$ −4.00000 −0.325515 −0.162758 0.986666i $$-0.552039\pi$$
−0.162758 + 0.986666i $$0.552039\pi$$
$$152$$ 0 0
$$153$$ −2.00000 −0.161690
$$154$$ −0.561553 −0.0452512
$$155$$ −4.00000 −0.321288
$$156$$ 1.43845 0.115168
$$157$$ −14.4924 −1.15662 −0.578311 0.815817i $$-0.696288\pi$$
−0.578311 + 0.815817i $$0.696288\pi$$
$$158$$ 4.00000 0.318223
$$159$$ 32.8078 2.60182
$$160$$ 2.00000 0.158114
$$161$$ −0.807764 −0.0636607
$$162$$ −7.00000 −0.549972
$$163$$ 11.3693 0.890514 0.445257 0.895403i $$-0.353112\pi$$
0.445257 + 0.895403i $$0.353112\pi$$
$$164$$ −2.00000 −0.156174
$$165$$ −5.12311 −0.398833
$$166$$ 14.2462 1.10572
$$167$$ 0.630683 0.0488037 0.0244019 0.999702i $$-0.492232\pi$$
0.0244019 + 0.999702i $$0.492232\pi$$
$$168$$ 1.43845 0.110979
$$169$$ −12.6847 −0.975743
$$170$$ −1.12311 −0.0861383
$$171$$ 0 0
$$172$$ 0 0
$$173$$ −18.0000 −1.36851 −0.684257 0.729241i $$-0.739873\pi$$
−0.684257 + 0.729241i $$0.739873\pi$$
$$174$$ −14.5616 −1.10391
$$175$$ 0.561553 0.0424494
$$176$$ 1.00000 0.0753778
$$177$$ −19.6847 −1.47959
$$178$$ 0.876894 0.0657260
$$179$$ −5.75379 −0.430058 −0.215029 0.976608i $$-0.568985\pi$$
−0.215029 + 0.976608i $$0.568985\pi$$
$$180$$ 7.12311 0.530925
$$181$$ 10.8769 0.808473 0.404237 0.914654i $$-0.367537\pi$$
0.404237 + 0.914654i $$0.367537\pi$$
$$182$$ 0.315342 0.0233747
$$183$$ −16.0000 −1.18275
$$184$$ 1.43845 0.106044
$$185$$ 10.2462 0.753316
$$186$$ 5.12311 0.375644
$$187$$ −0.561553 −0.0410648
$$188$$ 8.00000 0.583460
$$189$$ 0.807764 0.0587562
$$190$$ 0 0
$$191$$ −8.31534 −0.601677 −0.300838 0.953675i $$-0.597267\pi$$
−0.300838 + 0.953675i $$0.597267\pi$$
$$192$$ −2.56155 −0.184864
$$193$$ −7.12311 −0.512732 −0.256366 0.966580i $$-0.582525\pi$$
−0.256366 + 0.966580i $$0.582525\pi$$
$$194$$ 7.12311 0.511409
$$195$$ 2.87689 0.206019
$$196$$ −6.68466 −0.477476
$$197$$ 18.2462 1.29999 0.649994 0.759939i $$-0.274771\pi$$
0.649994 + 0.759939i $$0.274771\pi$$
$$198$$ 3.56155 0.253109
$$199$$ 11.6847 0.828303 0.414152 0.910208i $$-0.364078\pi$$
0.414152 + 0.910208i $$0.364078\pi$$
$$200$$ −1.00000 −0.0707107
$$201$$ −19.6847 −1.38845
$$202$$ −6.87689 −0.483857
$$203$$ −3.19224 −0.224051
$$204$$ 1.43845 0.100711
$$205$$ −4.00000 −0.279372
$$206$$ 13.3693 0.931484
$$207$$ 5.12311 0.356080
$$208$$ −0.561553 −0.0389367
$$209$$ 0 0
$$210$$ 2.87689 0.198525
$$211$$ 0.315342 0.0217090 0.0108545 0.999941i $$-0.496545\pi$$
0.0108545 + 0.999941i $$0.496545\pi$$
$$212$$ −12.8078 −0.879641
$$213$$ 15.3693 1.05309
$$214$$ 9.93087 0.678860
$$215$$ 0 0
$$216$$ −1.43845 −0.0978739
$$217$$ 1.12311 0.0762414
$$218$$ 6.31534 0.427729
$$219$$ 24.8078 1.67635
$$220$$ 2.00000 0.134840
$$221$$ 0.315342 0.0212122
$$222$$ −13.1231 −0.880765
$$223$$ −24.7386 −1.65662 −0.828311 0.560269i $$-0.810698\pi$$
−0.828311 + 0.560269i $$0.810698\pi$$
$$224$$ −0.561553 −0.0375203
$$225$$ −3.56155 −0.237437
$$226$$ 3.12311 0.207746
$$227$$ −28.1771 −1.87018 −0.935089 0.354412i $$-0.884681\pi$$
−0.935089 + 0.354412i $$0.884681\pi$$
$$228$$ 0 0
$$229$$ 12.8769 0.850929 0.425465 0.904975i $$-0.360111\pi$$
0.425465 + 0.904975i $$0.360111\pi$$
$$230$$ 2.87689 0.189697
$$231$$ 1.43845 0.0946429
$$232$$ 5.68466 0.373216
$$233$$ −28.7386 −1.88273 −0.941365 0.337389i $$-0.890456\pi$$
−0.941365 + 0.337389i $$0.890456\pi$$
$$234$$ −2.00000 −0.130744
$$235$$ 16.0000 1.04372
$$236$$ 7.68466 0.500229
$$237$$ −10.2462 −0.665563
$$238$$ 0.315342 0.0204406
$$239$$ 23.9309 1.54796 0.773980 0.633210i $$-0.218263\pi$$
0.773980 + 0.633210i $$0.218263\pi$$
$$240$$ −5.12311 −0.330695
$$241$$ 19.6155 1.26355 0.631774 0.775153i $$-0.282327\pi$$
0.631774 + 0.775153i $$0.282327\pi$$
$$242$$ 1.00000 0.0642824
$$243$$ 22.2462 1.42710
$$244$$ 6.24621 0.399873
$$245$$ −13.3693 −0.854134
$$246$$ 5.12311 0.326637
$$247$$ 0 0
$$248$$ −2.00000 −0.127000
$$249$$ −36.4924 −2.31261
$$250$$ −12.0000 −0.758947
$$251$$ 17.1231 1.08080 0.540400 0.841408i $$-0.318273\pi$$
0.540400 + 0.841408i $$0.318273\pi$$
$$252$$ −2.00000 −0.125988
$$253$$ 1.43845 0.0904344
$$254$$ −2.24621 −0.140940
$$255$$ 2.87689 0.180158
$$256$$ 1.00000 0.0625000
$$257$$ 22.0000 1.37232 0.686161 0.727450i $$-0.259294\pi$$
0.686161 + 0.727450i $$0.259294\pi$$
$$258$$ 0 0
$$259$$ −2.87689 −0.178762
$$260$$ −1.12311 −0.0696521
$$261$$ 20.2462 1.25321
$$262$$ 1.12311 0.0693857
$$263$$ 8.87689 0.547373 0.273686 0.961819i $$-0.411757\pi$$
0.273686 + 0.961819i $$0.411757\pi$$
$$264$$ −2.56155 −0.157653
$$265$$ −25.6155 −1.57355
$$266$$ 0 0
$$267$$ −2.24621 −0.137466
$$268$$ 7.68466 0.469415
$$269$$ −2.87689 −0.175407 −0.0877037 0.996147i $$-0.527953\pi$$
−0.0877037 + 0.996147i $$0.527953\pi$$
$$270$$ −2.87689 −0.175082
$$271$$ 25.6847 1.56023 0.780116 0.625635i $$-0.215160\pi$$
0.780116 + 0.625635i $$0.215160\pi$$
$$272$$ −0.561553 −0.0340491
$$273$$ −0.807764 −0.0488881
$$274$$ 12.5616 0.758871
$$275$$ −1.00000 −0.0603023
$$276$$ −3.68466 −0.221790
$$277$$ 8.49242 0.510260 0.255130 0.966907i $$-0.417882\pi$$
0.255130 + 0.966907i $$0.417882\pi$$
$$278$$ 7.36932 0.441982
$$279$$ −7.12311 −0.426449
$$280$$ −1.12311 −0.0671184
$$281$$ 25.3693 1.51341 0.756703 0.653758i $$-0.226809\pi$$
0.756703 + 0.653758i $$0.226809\pi$$
$$282$$ −20.4924 −1.22031
$$283$$ 11.3693 0.675836 0.337918 0.941176i $$-0.390277\pi$$
0.337918 + 0.941176i $$0.390277\pi$$
$$284$$ −6.00000 −0.356034
$$285$$ 0 0
$$286$$ −0.561553 −0.0332053
$$287$$ 1.12311 0.0662948
$$288$$ 3.56155 0.209867
$$289$$ −16.6847 −0.981450
$$290$$ 11.3693 0.667629
$$291$$ −18.2462 −1.06961
$$292$$ −9.68466 −0.566752
$$293$$ −3.93087 −0.229644 −0.114822 0.993386i $$-0.536630\pi$$
−0.114822 + 0.993386i $$0.536630\pi$$
$$294$$ 17.1231 0.998640
$$295$$ 15.3693 0.894836
$$296$$ 5.12311 0.297774
$$297$$ −1.43845 −0.0834672
$$298$$ −6.87689 −0.398368
$$299$$ −0.807764 −0.0467142
$$300$$ 2.56155 0.147891
$$301$$ 0 0
$$302$$ −4.00000 −0.230174
$$303$$ 17.6155 1.01199
$$304$$ 0 0
$$305$$ 12.4924 0.715314
$$306$$ −2.00000 −0.114332
$$307$$ 22.2462 1.26966 0.634829 0.772653i $$-0.281070\pi$$
0.634829 + 0.772653i $$0.281070\pi$$
$$308$$ −0.561553 −0.0319974
$$309$$ −34.2462 −1.94820
$$310$$ −4.00000 −0.227185
$$311$$ −4.80776 −0.272623 −0.136312 0.990666i $$-0.543525\pi$$
−0.136312 + 0.990666i $$0.543525\pi$$
$$312$$ 1.43845 0.0814360
$$313$$ −0.561553 −0.0317408 −0.0158704 0.999874i $$-0.505052\pi$$
−0.0158704 + 0.999874i $$0.505052\pi$$
$$314$$ −14.4924 −0.817855
$$315$$ −4.00000 −0.225374
$$316$$ 4.00000 0.225018
$$317$$ 7.05398 0.396191 0.198095 0.980183i $$-0.436524\pi$$
0.198095 + 0.980183i $$0.436524\pi$$
$$318$$ 32.8078 1.83977
$$319$$ 5.68466 0.318280
$$320$$ 2.00000 0.111803
$$321$$ −25.4384 −1.41984
$$322$$ −0.807764 −0.0450149
$$323$$ 0 0
$$324$$ −7.00000 −0.388889
$$325$$ 0.561553 0.0311493
$$326$$ 11.3693 0.629688
$$327$$ −16.1771 −0.894595
$$328$$ −2.00000 −0.110432
$$329$$ −4.49242 −0.247675
$$330$$ −5.12311 −0.282018
$$331$$ 14.5616 0.800375 0.400188 0.916433i $$-0.368945\pi$$
0.400188 + 0.916433i $$0.368945\pi$$
$$332$$ 14.2462 0.781862
$$333$$ 18.2462 0.999886
$$334$$ 0.630683 0.0345094
$$335$$ 15.3693 0.839715
$$336$$ 1.43845 0.0784737
$$337$$ −11.1231 −0.605914 −0.302957 0.953004i $$-0.597974\pi$$
−0.302957 + 0.953004i $$0.597974\pi$$
$$338$$ −12.6847 −0.689954
$$339$$ −8.00000 −0.434500
$$340$$ −1.12311 −0.0609090
$$341$$ −2.00000 −0.108306
$$342$$ 0 0
$$343$$ 7.68466 0.414933
$$344$$ 0 0
$$345$$ −7.36932 −0.396751
$$346$$ −18.0000 −0.967686
$$347$$ 9.61553 0.516189 0.258094 0.966120i $$-0.416905\pi$$
0.258094 + 0.966120i $$0.416905\pi$$
$$348$$ −14.5616 −0.780581
$$349$$ 21.6155 1.15705 0.578526 0.815664i $$-0.303628\pi$$
0.578526 + 0.815664i $$0.303628\pi$$
$$350$$ 0.561553 0.0300163
$$351$$ 0.807764 0.0431153
$$352$$ 1.00000 0.0533002
$$353$$ 22.1771 1.18037 0.590183 0.807269i $$-0.299055\pi$$
0.590183 + 0.807269i $$0.299055\pi$$
$$354$$ −19.6847 −1.04623
$$355$$ −12.0000 −0.636894
$$356$$ 0.876894 0.0464753
$$357$$ −0.807764 −0.0427514
$$358$$ −5.75379 −0.304097
$$359$$ −15.9309 −0.840799 −0.420400 0.907339i $$-0.638110\pi$$
−0.420400 + 0.907339i $$0.638110\pi$$
$$360$$ 7.12311 0.375421
$$361$$ 0 0
$$362$$ 10.8769 0.571677
$$363$$ −2.56155 −0.134447
$$364$$ 0.315342 0.0165284
$$365$$ −19.3693 −1.01384
$$366$$ −16.0000 −0.836333
$$367$$ 30.2462 1.57884 0.789420 0.613854i $$-0.210382\pi$$
0.789420 + 0.613854i $$0.210382\pi$$
$$368$$ 1.43845 0.0749842
$$369$$ −7.12311 −0.370814
$$370$$ 10.2462 0.532675
$$371$$ 7.19224 0.373402
$$372$$ 5.12311 0.265621
$$373$$ −3.43845 −0.178036 −0.0890180 0.996030i $$-0.528373\pi$$
−0.0890180 + 0.996030i $$0.528373\pi$$
$$374$$ −0.561553 −0.0290372
$$375$$ 30.7386 1.58734
$$376$$ 8.00000 0.412568
$$377$$ −3.19224 −0.164409
$$378$$ 0.807764 0.0415469
$$379$$ −6.56155 −0.337044 −0.168522 0.985698i $$-0.553899\pi$$
−0.168522 + 0.985698i $$0.553899\pi$$
$$380$$ 0 0
$$381$$ 5.75379 0.294776
$$382$$ −8.31534 −0.425450
$$383$$ 2.00000 0.102195 0.0510976 0.998694i $$-0.483728\pi$$
0.0510976 + 0.998694i $$0.483728\pi$$
$$384$$ −2.56155 −0.130719
$$385$$ −1.12311 −0.0572388
$$386$$ −7.12311 −0.362557
$$387$$ 0 0
$$388$$ 7.12311 0.361621
$$389$$ 18.0000 0.912636 0.456318 0.889817i $$-0.349168\pi$$
0.456318 + 0.889817i $$0.349168\pi$$
$$390$$ 2.87689 0.145677
$$391$$ −0.807764 −0.0408504
$$392$$ −6.68466 −0.337626
$$393$$ −2.87689 −0.145120
$$394$$ 18.2462 0.919231
$$395$$ 8.00000 0.402524
$$396$$ 3.56155 0.178975
$$397$$ 38.9848 1.95659 0.978297 0.207209i $$-0.0664381\pi$$
0.978297 + 0.207209i $$0.0664381\pi$$
$$398$$ 11.6847 0.585699
$$399$$ 0 0
$$400$$ −1.00000 −0.0500000
$$401$$ 7.75379 0.387206 0.193603 0.981080i $$-0.437983\pi$$
0.193603 + 0.981080i $$0.437983\pi$$
$$402$$ −19.6847 −0.981782
$$403$$ 1.12311 0.0559459
$$404$$ −6.87689 −0.342138
$$405$$ −14.0000 −0.695666
$$406$$ −3.19224 −0.158428
$$407$$ 5.12311 0.253943
$$408$$ 1.43845 0.0712137
$$409$$ −16.2462 −0.803323 −0.401662 0.915788i $$-0.631567\pi$$
−0.401662 + 0.915788i $$0.631567\pi$$
$$410$$ −4.00000 −0.197546
$$411$$ −32.1771 −1.58718
$$412$$ 13.3693 0.658659
$$413$$ −4.31534 −0.212344
$$414$$ 5.12311 0.251787
$$415$$ 28.4924 1.39864
$$416$$ −0.561553 −0.0275324
$$417$$ −18.8769 −0.924405
$$418$$ 0 0
$$419$$ −14.8769 −0.726784 −0.363392 0.931636i $$-0.618381\pi$$
−0.363392 + 0.931636i $$0.618381\pi$$
$$420$$ 2.87689 0.140378
$$421$$ 29.9309 1.45874 0.729371 0.684119i $$-0.239813\pi$$
0.729371 + 0.684119i $$0.239813\pi$$
$$422$$ 0.315342 0.0153506
$$423$$ 28.4924 1.38535
$$424$$ −12.8078 −0.622000
$$425$$ 0.561553 0.0272393
$$426$$ 15.3693 0.744646
$$427$$ −3.50758 −0.169744
$$428$$ 9.93087 0.480027
$$429$$ 1.43845 0.0694489
$$430$$ 0 0
$$431$$ −31.8617 −1.53473 −0.767363 0.641213i $$-0.778432\pi$$
−0.767363 + 0.641213i $$0.778432\pi$$
$$432$$ −1.43845 −0.0692073
$$433$$ −6.63068 −0.318650 −0.159325 0.987226i $$-0.550932\pi$$
−0.159325 + 0.987226i $$0.550932\pi$$
$$434$$ 1.12311 0.0539108
$$435$$ −29.1231 −1.39635
$$436$$ 6.31534 0.302450
$$437$$ 0 0
$$438$$ 24.8078 1.18536
$$439$$ 9.61553 0.458924 0.229462 0.973318i $$-0.426303\pi$$
0.229462 + 0.973318i $$0.426303\pi$$
$$440$$ 2.00000 0.0953463
$$441$$ −23.8078 −1.13370
$$442$$ 0.315342 0.0149993
$$443$$ 39.8617 1.89389 0.946944 0.321398i $$-0.104153\pi$$
0.946944 + 0.321398i $$0.104153\pi$$
$$444$$ −13.1231 −0.622795
$$445$$ 1.75379 0.0831376
$$446$$ −24.7386 −1.17141
$$447$$ 17.6155 0.833186
$$448$$ −0.561553 −0.0265309
$$449$$ −23.6155 −1.11449 −0.557243 0.830350i $$-0.688141\pi$$
−0.557243 + 0.830350i $$0.688141\pi$$
$$450$$ −3.56155 −0.167893
$$451$$ −2.00000 −0.0941763
$$452$$ 3.12311 0.146899
$$453$$ 10.2462 0.481409
$$454$$ −28.1771 −1.32242
$$455$$ 0.630683 0.0295669
$$456$$ 0 0
$$457$$ 36.5616 1.71028 0.855139 0.518399i $$-0.173472\pi$$
0.855139 + 0.518399i $$0.173472\pi$$
$$458$$ 12.8769 0.601698
$$459$$ 0.807764 0.0377032
$$460$$ 2.87689 0.134136
$$461$$ 26.7386 1.24534 0.622671 0.782484i $$-0.286047\pi$$
0.622671 + 0.782484i $$0.286047\pi$$
$$462$$ 1.43845 0.0669226
$$463$$ 3.50758 0.163011 0.0815055 0.996673i $$-0.474027\pi$$
0.0815055 + 0.996673i $$0.474027\pi$$
$$464$$ 5.68466 0.263904
$$465$$ 10.2462 0.475157
$$466$$ −28.7386 −1.33129
$$467$$ 9.75379 0.451352 0.225676 0.974202i $$-0.427541\pi$$
0.225676 + 0.974202i $$0.427541\pi$$
$$468$$ −2.00000 −0.0924500
$$469$$ −4.31534 −0.199264
$$470$$ 16.0000 0.738025
$$471$$ 37.1231 1.71054
$$472$$ 7.68466 0.353715
$$473$$ 0 0
$$474$$ −10.2462 −0.470624
$$475$$ 0 0
$$476$$ 0.315342 0.0144537
$$477$$ −45.6155 −2.08859
$$478$$ 23.9309 1.09457
$$479$$ −13.3693 −0.610860 −0.305430 0.952215i $$-0.598800\pi$$
−0.305430 + 0.952215i $$0.598800\pi$$
$$480$$ −5.12311 −0.233837
$$481$$ −2.87689 −0.131175
$$482$$ 19.6155 0.893463
$$483$$ 2.06913 0.0941487
$$484$$ 1.00000 0.0454545
$$485$$ 14.2462 0.646887
$$486$$ 22.2462 1.00911
$$487$$ −31.1231 −1.41032 −0.705161 0.709047i $$-0.749125\pi$$
−0.705161 + 0.709047i $$0.749125\pi$$
$$488$$ 6.24621 0.282753
$$489$$ −29.1231 −1.31699
$$490$$ −13.3693 −0.603964
$$491$$ −39.3693 −1.77671 −0.888356 0.459155i $$-0.848152\pi$$
−0.888356 + 0.459155i $$0.848152\pi$$
$$492$$ 5.12311 0.230967
$$493$$ −3.19224 −0.143771
$$494$$ 0 0
$$495$$ 7.12311 0.320160
$$496$$ −2.00000 −0.0898027
$$497$$ 3.36932 0.151135
$$498$$ −36.4924 −1.63526
$$499$$ 9.75379 0.436640 0.218320 0.975877i $$-0.429942\pi$$
0.218320 + 0.975877i $$0.429942\pi$$
$$500$$ −12.0000 −0.536656
$$501$$ −1.61553 −0.0721765
$$502$$ 17.1231 0.764242
$$503$$ 21.6847 0.966871 0.483436 0.875380i $$-0.339389\pi$$
0.483436 + 0.875380i $$0.339389\pi$$
$$504$$ −2.00000 −0.0890871
$$505$$ −13.7538 −0.612036
$$506$$ 1.43845 0.0639468
$$507$$ 32.4924 1.44304
$$508$$ −2.24621 −0.0996595
$$509$$ 31.3693 1.39042 0.695210 0.718806i $$-0.255311\pi$$
0.695210 + 0.718806i $$0.255311\pi$$
$$510$$ 2.87689 0.127391
$$511$$ 5.43845 0.240583
$$512$$ 1.00000 0.0441942
$$513$$ 0 0
$$514$$ 22.0000 0.970378
$$515$$ 26.7386 1.17824
$$516$$ 0 0
$$517$$ 8.00000 0.351840
$$518$$ −2.87689 −0.126403
$$519$$ 46.1080 2.02391
$$520$$ −1.12311 −0.0492514
$$521$$ 7.12311 0.312069 0.156034 0.987752i $$-0.450129\pi$$
0.156034 + 0.987752i $$0.450129\pi$$
$$522$$ 20.2462 0.886153
$$523$$ −31.0540 −1.35790 −0.678948 0.734187i $$-0.737564\pi$$
−0.678948 + 0.734187i $$0.737564\pi$$
$$524$$ 1.12311 0.0490631
$$525$$ −1.43845 −0.0627790
$$526$$ 8.87689 0.387051
$$527$$ 1.12311 0.0489232
$$528$$ −2.56155 −0.111477
$$529$$ −20.9309 −0.910038
$$530$$ −25.6155 −1.11267
$$531$$ 27.3693 1.18773
$$532$$ 0 0
$$533$$ 1.12311 0.0486471
$$534$$ −2.24621 −0.0972031
$$535$$ 19.8617 0.858698
$$536$$ 7.68466 0.331927
$$537$$ 14.7386 0.636019
$$538$$ −2.87689 −0.124032
$$539$$ −6.68466 −0.287929
$$540$$ −2.87689 −0.123802
$$541$$ −39.2311 −1.68667 −0.843337 0.537384i $$-0.819412\pi$$
−0.843337 + 0.537384i $$0.819412\pi$$
$$542$$ 25.6847 1.10325
$$543$$ −27.8617 −1.19566
$$544$$ −0.561553 −0.0240764
$$545$$ 12.6307 0.541039
$$546$$ −0.807764 −0.0345691
$$547$$ 42.7386 1.82737 0.913686 0.406421i $$-0.133223\pi$$
0.913686 + 0.406421i $$0.133223\pi$$
$$548$$ 12.5616 0.536603
$$549$$ 22.2462 0.949445
$$550$$ −1.00000 −0.0426401
$$551$$ 0 0
$$552$$ −3.68466 −0.156829
$$553$$ −2.24621 −0.0955186
$$554$$ 8.49242 0.360808
$$555$$ −26.2462 −1.11409
$$556$$ 7.36932 0.312529
$$557$$ 17.1231 0.725529 0.362765 0.931881i $$-0.381833\pi$$
0.362765 + 0.931881i $$0.381833\pi$$
$$558$$ −7.12311 −0.301545
$$559$$ 0 0
$$560$$ −1.12311 −0.0474599
$$561$$ 1.43845 0.0607313
$$562$$ 25.3693 1.07014
$$563$$ 42.7386 1.80122 0.900609 0.434630i $$-0.143121\pi$$
0.900609 + 0.434630i $$0.143121\pi$$
$$564$$ −20.4924 −0.862887
$$565$$ 6.24621 0.262780
$$566$$ 11.3693 0.477888
$$567$$ 3.93087 0.165081
$$568$$ −6.00000 −0.251754
$$569$$ 21.3693 0.895848 0.447924 0.894072i $$-0.352163\pi$$
0.447924 + 0.894072i $$0.352163\pi$$
$$570$$ 0 0
$$571$$ −5.61553 −0.235003 −0.117501 0.993073i $$-0.537488\pi$$
−0.117501 + 0.993073i $$0.537488\pi$$
$$572$$ −0.561553 −0.0234797
$$573$$ 21.3002 0.889828
$$574$$ 1.12311 0.0468775
$$575$$ −1.43845 −0.0599874
$$576$$ 3.56155 0.148398
$$577$$ 41.0540 1.70910 0.854550 0.519370i $$-0.173833\pi$$
0.854550 + 0.519370i $$0.173833\pi$$
$$578$$ −16.6847 −0.693990
$$579$$ 18.2462 0.758287
$$580$$ 11.3693 0.472085
$$581$$ −8.00000 −0.331896
$$582$$ −18.2462 −0.756330
$$583$$ −12.8078 −0.530443
$$584$$ −9.68466 −0.400754
$$585$$ −4.00000 −0.165380
$$586$$ −3.93087 −0.162383
$$587$$ −40.4924 −1.67130 −0.835651 0.549261i $$-0.814909\pi$$
−0.835651 + 0.549261i $$0.814909\pi$$
$$588$$ 17.1231 0.706145
$$589$$ 0 0
$$590$$ 15.3693 0.632745
$$591$$ −46.7386 −1.92257
$$592$$ 5.12311 0.210558
$$593$$ 34.0000 1.39621 0.698106 0.715994i $$-0.254026\pi$$
0.698106 + 0.715994i $$0.254026\pi$$
$$594$$ −1.43845 −0.0590202
$$595$$ 0.630683 0.0258555
$$596$$ −6.87689 −0.281689
$$597$$ −29.9309 −1.22499
$$598$$ −0.807764 −0.0330319
$$599$$ 17.8617 0.729811 0.364905 0.931045i $$-0.381101\pi$$
0.364905 + 0.931045i $$0.381101\pi$$
$$600$$ 2.56155 0.104575
$$601$$ −21.3693 −0.871673 −0.435836 0.900026i $$-0.643547\pi$$
−0.435836 + 0.900026i $$0.643547\pi$$
$$602$$ 0 0
$$603$$ 27.3693 1.11456
$$604$$ −4.00000 −0.162758
$$605$$ 2.00000 0.0813116
$$606$$ 17.6155 0.715582
$$607$$ 13.6155 0.552637 0.276319 0.961066i $$-0.410885\pi$$
0.276319 + 0.961066i $$0.410885\pi$$
$$608$$ 0 0
$$609$$ 8.17708 0.331352
$$610$$ 12.4924 0.505803
$$611$$ −4.49242 −0.181744
$$612$$ −2.00000 −0.0808452
$$613$$ 18.2462 0.736958 0.368479 0.929636i $$-0.379879\pi$$
0.368479 + 0.929636i $$0.379879\pi$$
$$614$$ 22.2462 0.897784
$$615$$ 10.2462 0.413167
$$616$$ −0.561553 −0.0226256
$$617$$ 44.7386 1.80111 0.900555 0.434743i $$-0.143161\pi$$
0.900555 + 0.434743i $$0.143161\pi$$
$$618$$ −34.2462 −1.37758
$$619$$ −9.75379 −0.392038 −0.196019 0.980600i $$-0.562801\pi$$
−0.196019 + 0.980600i $$0.562801\pi$$
$$620$$ −4.00000 −0.160644
$$621$$ −2.06913 −0.0830313
$$622$$ −4.80776 −0.192774
$$623$$ −0.492423 −0.0197285
$$624$$ 1.43845 0.0575840
$$625$$ −19.0000 −0.760000
$$626$$ −0.561553 −0.0224442
$$627$$ 0 0
$$628$$ −14.4924 −0.578311
$$629$$ −2.87689 −0.114709
$$630$$ −4.00000 −0.159364
$$631$$ 3.50758 0.139634 0.0698172 0.997560i $$-0.477758\pi$$
0.0698172 + 0.997560i $$0.477758\pi$$
$$632$$ 4.00000 0.159111
$$633$$ −0.807764 −0.0321057
$$634$$ 7.05398 0.280149
$$635$$ −4.49242 −0.178276
$$636$$ 32.8078 1.30091
$$637$$ 3.75379 0.148731
$$638$$ 5.68466 0.225058
$$639$$ −21.3693 −0.845357
$$640$$ 2.00000 0.0790569
$$641$$ −31.6155 −1.24874 −0.624369 0.781129i $$-0.714644\pi$$
−0.624369 + 0.781129i $$0.714644\pi$$
$$642$$ −25.4384 −1.00398
$$643$$ −12.6307 −0.498106 −0.249053 0.968490i $$-0.580119\pi$$
−0.249053 + 0.968490i $$0.580119\pi$$
$$644$$ −0.807764 −0.0318304
$$645$$ 0 0
$$646$$ 0 0
$$647$$ −47.5464 −1.86924 −0.934621 0.355646i $$-0.884261\pi$$
−0.934621 + 0.355646i $$0.884261\pi$$
$$648$$ −7.00000 −0.274986
$$649$$ 7.68466 0.301649
$$650$$ 0.561553 0.0220259
$$651$$ −2.87689 −0.112754
$$652$$ 11.3693 0.445257
$$653$$ −15.6155 −0.611083 −0.305541 0.952179i $$-0.598837\pi$$
−0.305541 + 0.952179i $$0.598837\pi$$
$$654$$ −16.1771 −0.632574
$$655$$ 2.24621 0.0877667
$$656$$ −2.00000 −0.0780869
$$657$$ −34.4924 −1.34568
$$658$$ −4.49242 −0.175133
$$659$$ 14.4233 0.561852 0.280926 0.959729i $$-0.409359\pi$$
0.280926 + 0.959729i $$0.409359\pi$$
$$660$$ −5.12311 −0.199417
$$661$$ 26.5616 1.03312 0.516562 0.856250i $$-0.327211\pi$$
0.516562 + 0.856250i $$0.327211\pi$$
$$662$$ 14.5616 0.565951
$$663$$ −0.807764 −0.0313710
$$664$$ 14.2462 0.552860
$$665$$ 0 0
$$666$$ 18.2462 0.707026
$$667$$ 8.17708 0.316618
$$668$$ 0.630683 0.0244019
$$669$$ 63.3693 2.45000
$$670$$ 15.3693 0.593769
$$671$$ 6.24621 0.241132
$$672$$ 1.43845 0.0554893
$$673$$ 27.1231 1.04552 0.522759 0.852480i $$-0.324903\pi$$
0.522759 + 0.852480i $$0.324903\pi$$
$$674$$ −11.1231 −0.428446
$$675$$ 1.43845 0.0553659
$$676$$ −12.6847 −0.487871
$$677$$ −6.17708 −0.237405 −0.118702 0.992930i $$-0.537873\pi$$
−0.118702 + 0.992930i $$0.537873\pi$$
$$678$$ −8.00000 −0.307238
$$679$$ −4.00000 −0.153506
$$680$$ −1.12311 −0.0430691
$$681$$ 72.1771 2.76583
$$682$$ −2.00000 −0.0765840
$$683$$ −28.4924 −1.09023 −0.545116 0.838361i $$-0.683514\pi$$
−0.545116 + 0.838361i $$0.683514\pi$$
$$684$$ 0 0
$$685$$ 25.1231 0.959905
$$686$$ 7.68466 0.293402
$$687$$ −32.9848 −1.25845
$$688$$ 0 0
$$689$$ 7.19224 0.274002
$$690$$ −7.36932 −0.280545
$$691$$ −2.38447 −0.0907096 −0.0453548 0.998971i $$-0.514442\pi$$
−0.0453548 + 0.998971i $$0.514442\pi$$
$$692$$ −18.0000 −0.684257
$$693$$ −2.00000 −0.0759737
$$694$$ 9.61553 0.365000
$$695$$ 14.7386 0.559068
$$696$$ −14.5616 −0.551954
$$697$$ 1.12311 0.0425407
$$698$$ 21.6155 0.818160
$$699$$ 73.6155 2.78439
$$700$$ 0.561553 0.0212247
$$701$$ −40.9848 −1.54798 −0.773988 0.633200i $$-0.781741\pi$$
−0.773988 + 0.633200i $$0.781741\pi$$
$$702$$ 0.807764 0.0304871
$$703$$ 0 0
$$704$$ 1.00000 0.0376889
$$705$$ −40.9848 −1.54358
$$706$$ 22.1771 0.834645
$$707$$ 3.86174 0.145236
$$708$$ −19.6847 −0.739795
$$709$$ −42.4924 −1.59584 −0.797918 0.602766i $$-0.794065\pi$$
−0.797918 + 0.602766i $$0.794065\pi$$
$$710$$ −12.0000 −0.450352
$$711$$ 14.2462 0.534275
$$712$$ 0.876894 0.0328630
$$713$$ −2.87689 −0.107741
$$714$$ −0.807764 −0.0302298
$$715$$ −1.12311 −0.0420018
$$716$$ −5.75379 −0.215029
$$717$$ −61.3002 −2.28930
$$718$$ −15.9309 −0.594535
$$719$$ −10.5616 −0.393879 −0.196940 0.980416i $$-0.563100\pi$$
−0.196940 + 0.980416i $$0.563100\pi$$
$$720$$ 7.12311 0.265462
$$721$$ −7.50758 −0.279597
$$722$$ 0 0
$$723$$ −50.2462 −1.86868
$$724$$ 10.8769 0.404237
$$725$$ −5.68466 −0.211123
$$726$$ −2.56155 −0.0950681
$$727$$ 22.4233 0.831634 0.415817 0.909448i $$-0.363496\pi$$
0.415817 + 0.909448i $$0.363496\pi$$
$$728$$ 0.315342 0.0116873
$$729$$ −35.9848 −1.33277
$$730$$ −19.3693 −0.716891
$$731$$ 0 0
$$732$$ −16.0000 −0.591377
$$733$$ 26.1080 0.964319 0.482160 0.876083i $$-0.339853\pi$$
0.482160 + 0.876083i $$0.339853\pi$$
$$734$$ 30.2462 1.11641
$$735$$ 34.2462 1.26319
$$736$$ 1.43845 0.0530219
$$737$$ 7.68466 0.283068
$$738$$ −7.12311 −0.262205
$$739$$ −44.9848 −1.65479 −0.827397 0.561617i $$-0.810179\pi$$
−0.827397 + 0.561617i $$0.810179\pi$$
$$740$$ 10.2462 0.376658
$$741$$ 0 0
$$742$$ 7.19224 0.264035
$$743$$ 42.2462 1.54986 0.774932 0.632045i $$-0.217784\pi$$
0.774932 + 0.632045i $$0.217784\pi$$
$$744$$ 5.12311 0.187822
$$745$$ −13.7538 −0.503900
$$746$$ −3.43845 −0.125890
$$747$$ 50.7386 1.85643
$$748$$ −0.561553 −0.0205324
$$749$$ −5.57671 −0.203768
$$750$$ 30.7386 1.12242
$$751$$ −41.2311 −1.50454 −0.752271 0.658853i $$-0.771042\pi$$
−0.752271 + 0.658853i $$0.771042\pi$$
$$752$$ 8.00000 0.291730
$$753$$ −43.8617 −1.59841
$$754$$ −3.19224 −0.116254
$$755$$ −8.00000 −0.291150
$$756$$ 0.807764 0.0293781
$$757$$ −42.4924 −1.54441 −0.772207 0.635371i $$-0.780847\pi$$
−0.772207 + 0.635371i $$0.780847\pi$$
$$758$$ −6.56155 −0.238326
$$759$$ −3.68466 −0.133745
$$760$$ 0 0
$$761$$ −42.8078 −1.55178 −0.775890 0.630868i $$-0.782699\pi$$
−0.775890 + 0.630868i $$0.782699\pi$$
$$762$$ 5.75379 0.208438
$$763$$ −3.54640 −0.128388
$$764$$ −8.31534 −0.300838
$$765$$ −4.00000 −0.144620
$$766$$ 2.00000 0.0722629
$$767$$ −4.31534 −0.155818
$$768$$ −2.56155 −0.0924321
$$769$$ −13.6847 −0.493481 −0.246741 0.969082i $$-0.579360\pi$$
−0.246741 + 0.969082i $$0.579360\pi$$
$$770$$ −1.12311 −0.0404739
$$771$$ −56.3542 −2.02955
$$772$$ −7.12311 −0.256366
$$773$$ −37.9309 −1.36428 −0.682139 0.731222i $$-0.738950\pi$$
−0.682139 + 0.731222i $$0.738950\pi$$
$$774$$ 0 0
$$775$$ 2.00000 0.0718421
$$776$$ 7.12311 0.255705
$$777$$ 7.36932 0.264373
$$778$$ 18.0000 0.645331
$$779$$ 0 0
$$780$$ 2.87689 0.103009
$$781$$ −6.00000 −0.214697
$$782$$ −0.807764 −0.0288856
$$783$$ −8.17708 −0.292225
$$784$$ −6.68466 −0.238738
$$785$$ −28.9848 −1.03451
$$786$$ −2.87689 −0.102615
$$787$$ −15.6847 −0.559098 −0.279549 0.960131i $$-0.590185\pi$$
−0.279549 + 0.960131i $$0.590185\pi$$
$$788$$ 18.2462 0.649994
$$789$$ −22.7386 −0.809517
$$790$$ 8.00000 0.284627
$$791$$ −1.75379 −0.0623575
$$792$$ 3.56155 0.126554
$$793$$ −3.50758 −0.124558
$$794$$ 38.9848 1.38352
$$795$$ 65.6155 2.32714
$$796$$ 11.6847 0.414152
$$797$$ 15.1922 0.538137 0.269068 0.963121i $$-0.413284\pi$$
0.269068 + 0.963121i $$0.413284\pi$$
$$798$$ 0 0
$$799$$ −4.49242 −0.158930
$$800$$ −1.00000 −0.0353553
$$801$$ 3.12311 0.110350
$$802$$ 7.75379 0.273796
$$803$$ −9.68466 −0.341764
$$804$$ −19.6847 −0.694224
$$805$$ −1.61553 −0.0569399
$$806$$ 1.12311 0.0395597
$$807$$ 7.36932 0.259412
$$808$$ −6.87689 −0.241928
$$809$$ 8.06913 0.283696 0.141848 0.989888i $$-0.454696\pi$$
0.141848 + 0.989888i $$0.454696\pi$$
$$810$$ −14.0000 −0.491910
$$811$$ −8.31534 −0.291991 −0.145996 0.989285i $$-0.546639\pi$$
−0.145996 + 0.989285i $$0.546639\pi$$
$$812$$ −3.19224 −0.112026
$$813$$ −65.7926 −2.30745
$$814$$ 5.12311 0.179565
$$815$$ 22.7386 0.796500
$$816$$ 1.43845 0.0503557
$$817$$ 0 0
$$818$$ −16.2462 −0.568035
$$819$$ 1.12311 0.0392445
$$820$$ −4.00000 −0.139686
$$821$$ −20.9848 −0.732376 −0.366188 0.930541i $$-0.619337\pi$$
−0.366188 + 0.930541i $$0.619337\pi$$
$$822$$ −32.1771 −1.12230
$$823$$ −13.9309 −0.485600 −0.242800 0.970076i $$-0.578066\pi$$
−0.242800 + 0.970076i $$0.578066\pi$$
$$824$$ 13.3693 0.465742
$$825$$ 2.56155 0.0891818
$$826$$ −4.31534 −0.150150
$$827$$ −49.9309 −1.73627 −0.868133 0.496331i $$-0.834680\pi$$
−0.868133 + 0.496331i $$0.834680\pi$$
$$828$$ 5.12311 0.178040
$$829$$ −12.1771 −0.422928 −0.211464 0.977386i $$-0.567823\pi$$
−0.211464 + 0.977386i $$0.567823\pi$$
$$830$$ 28.4924 0.988986
$$831$$ −21.7538 −0.754631
$$832$$ −0.561553 −0.0194683
$$833$$ 3.75379 0.130061
$$834$$ −18.8769 −0.653653
$$835$$ 1.26137 0.0436514
$$836$$ 0 0
$$837$$ 2.87689 0.0994400
$$838$$ −14.8769 −0.513914
$$839$$ 7.12311 0.245917 0.122958 0.992412i $$-0.460762\pi$$
0.122958 + 0.992412i $$0.460762\pi$$
$$840$$ 2.87689 0.0992623
$$841$$ 3.31534 0.114322
$$842$$ 29.9309 1.03149
$$843$$ −64.9848 −2.23820
$$844$$ 0.315342 0.0108545
$$845$$ −25.3693 −0.872731
$$846$$ 28.4924 0.979590
$$847$$ −0.561553 −0.0192952
$$848$$ −12.8078 −0.439820
$$849$$ −29.1231 −0.999502
$$850$$ 0.561553 0.0192611
$$851$$ 7.36932 0.252617
$$852$$ 15.3693 0.526544
$$853$$ −52.3542 −1.79257 −0.896286 0.443476i $$-0.853745\pi$$
−0.896286 + 0.443476i $$0.853745\pi$$
$$854$$ −3.50758 −0.120027
$$855$$ 0 0
$$856$$ 9.93087 0.339430
$$857$$ 10.0000 0.341593 0.170797 0.985306i $$-0.445366\pi$$
0.170797 + 0.985306i $$0.445366\pi$$
$$858$$ 1.43845 0.0491078
$$859$$ 15.8617 0.541196 0.270598 0.962692i $$-0.412779\pi$$
0.270598 + 0.962692i $$0.412779\pi$$
$$860$$ 0 0
$$861$$ −2.87689 −0.0980443
$$862$$ −31.8617 −1.08522
$$863$$ −0.246211 −0.00838113 −0.00419056 0.999991i $$-0.501334\pi$$
−0.00419056 + 0.999991i $$0.501334\pi$$
$$864$$ −1.43845 −0.0489370
$$865$$ −36.0000 −1.22404
$$866$$ −6.63068 −0.225320
$$867$$ 42.7386 1.45148
$$868$$ 1.12311 0.0381207
$$869$$ 4.00000 0.135691
$$870$$ −29.1231 −0.987366
$$871$$ −4.31534 −0.146220
$$872$$ 6.31534 0.213864
$$873$$ 25.3693 0.858621
$$874$$ 0 0
$$875$$ 6.73863 0.227807
$$876$$ 24.8078 0.838177
$$877$$ 44.5616 1.50474 0.752368 0.658743i $$-0.228911\pi$$
0.752368 + 0.658743i $$0.228911\pi$$
$$878$$ 9.61553 0.324508
$$879$$ 10.0691 0.339623
$$880$$ 2.00000 0.0674200
$$881$$ 20.7386 0.698702 0.349351 0.936992i $$-0.386402\pi$$
0.349351 + 0.936992i $$0.386402\pi$$
$$882$$ −23.8078 −0.801649
$$883$$ −5.26137 −0.177059 −0.0885295 0.996074i $$-0.528217\pi$$
−0.0885295 + 0.996074i $$0.528217\pi$$
$$884$$ 0.315342 0.0106061
$$885$$ −39.3693 −1.32339
$$886$$ 39.8617 1.33918
$$887$$ 32.0000 1.07445 0.537227 0.843437i $$-0.319472\pi$$
0.537227 + 0.843437i $$0.319472\pi$$
$$888$$ −13.1231 −0.440383
$$889$$ 1.26137 0.0423049
$$890$$ 1.75379 0.0587871
$$891$$ −7.00000 −0.234509
$$892$$ −24.7386 −0.828311
$$893$$ 0 0
$$894$$ 17.6155 0.589151
$$895$$ −11.5076 −0.384656
$$896$$ −0.561553 −0.0187602
$$897$$ 2.06913 0.0690863
$$898$$ −23.6155 −0.788060
$$899$$ −11.3693 −0.379188
$$900$$ −3.56155 −0.118718
$$901$$ 7.19224 0.239608
$$902$$ −2.00000 −0.0665927
$$903$$ 0 0
$$904$$ 3.12311 0.103873
$$905$$ 21.7538 0.723120
$$906$$ 10.2462 0.340408
$$907$$ 29.7926 0.989247 0.494624 0.869107i $$-0.335306\pi$$
0.494624 + 0.869107i $$0.335306\pi$$
$$908$$ −28.1771 −0.935089
$$909$$ −24.4924 −0.812362
$$910$$ 0.630683 0.0209069
$$911$$ −18.9848 −0.628996 −0.314498 0.949258i $$-0.601836\pi$$
−0.314498 + 0.949258i $$0.601836\pi$$
$$912$$ 0 0
$$913$$ 14.2462 0.471481
$$914$$ 36.5616 1.20935
$$915$$ −32.0000 −1.05789
$$916$$ 12.8769 0.425465
$$917$$ −0.630683 −0.0208270
$$918$$ 0.807764 0.0266602
$$919$$ 32.5616 1.07411 0.537053 0.843548i $$-0.319537\pi$$
0.537053 + 0.843548i $$0.319537\pi$$
$$920$$ 2.87689 0.0948484
$$921$$ −56.9848 −1.87771
$$922$$ 26.7386 0.880590
$$923$$ 3.36932 0.110902
$$924$$ 1.43845 0.0473214
$$925$$ −5.12311 −0.168447
$$926$$ 3.50758 0.115266
$$927$$ 47.6155 1.56390
$$928$$ 5.68466 0.186608
$$929$$ 11.9309 0.391439 0.195720 0.980660i $$-0.437296\pi$$
0.195720 + 0.980660i $$0.437296\pi$$
$$930$$ 10.2462 0.335987
$$931$$ 0 0
$$932$$ −28.7386 −0.941365
$$933$$ 12.3153 0.403186
$$934$$ 9.75379 0.319154
$$935$$ −1.12311 −0.0367295
$$936$$ −2.00000 −0.0653720
$$937$$ −0.699813 −0.0228619 −0.0114310 0.999935i $$-0.503639\pi$$
−0.0114310 + 0.999935i $$0.503639\pi$$
$$938$$ −4.31534 −0.140901
$$939$$ 1.43845 0.0469419
$$940$$ 16.0000 0.521862
$$941$$ −24.5616 −0.800684 −0.400342 0.916366i $$-0.631109\pi$$
−0.400342 + 0.916366i $$0.631109\pi$$
$$942$$ 37.1231 1.20954
$$943$$ −2.87689 −0.0936846
$$944$$ 7.68466 0.250114
$$945$$ 1.61553 0.0525531
$$946$$ 0 0
$$947$$ −36.9848 −1.20185 −0.600923 0.799307i $$-0.705200\pi$$
−0.600923 + 0.799307i $$0.705200\pi$$
$$948$$ −10.2462 −0.332781
$$949$$ 5.43845 0.176539
$$950$$ 0 0
$$951$$ −18.0691 −0.585932
$$952$$ 0.315342 0.0102203
$$953$$ 0.876894 0.0284054 0.0142027 0.999899i $$-0.495479\pi$$
0.0142027 + 0.999899i $$0.495479\pi$$
$$954$$ −45.6155 −1.47686
$$955$$ −16.6307 −0.538156
$$956$$ 23.9309 0.773980
$$957$$ −14.5616 −0.470708
$$958$$ −13.3693 −0.431943
$$959$$ −7.05398 −0.227785
$$960$$ −5.12311 −0.165348
$$961$$ −27.0000 −0.870968
$$962$$ −2.87689 −0.0927548
$$963$$ 35.3693 1.13976
$$964$$ 19.6155 0.631774
$$965$$ −14.2462 −0.458602
$$966$$ 2.06913 0.0665732
$$967$$ 53.8617 1.73208 0.866038 0.499978i $$-0.166658\pi$$
0.866038 + 0.499978i $$0.166658\pi$$
$$968$$ 1.00000 0.0321412
$$969$$ 0 0
$$970$$ 14.2462 0.457418
$$971$$ −2.24621 −0.0720843 −0.0360422 0.999350i $$-0.511475\pi$$
−0.0360422 + 0.999350i $$0.511475\pi$$
$$972$$ 22.2462 0.713548
$$973$$ −4.13826 −0.132667
$$974$$ −31.1231 −0.997249
$$975$$ −1.43845 −0.0460672
$$976$$ 6.24621 0.199936
$$977$$ −30.6307 −0.979962 −0.489981 0.871733i $$-0.662996\pi$$
−0.489981 + 0.871733i $$0.662996\pi$$
$$978$$ −29.1231 −0.931254
$$979$$ 0.876894 0.0280257
$$980$$ −13.3693 −0.427067
$$981$$ 22.4924 0.718128
$$982$$ −39.3693 −1.25633
$$983$$ 31.7538 1.01279 0.506394 0.862302i $$-0.330978\pi$$
0.506394 + 0.862302i $$0.330978\pi$$
$$984$$ 5.12311 0.163319
$$985$$ 36.4924 1.16275
$$986$$ −3.19224 −0.101662
$$987$$ 11.5076 0.366290
$$988$$ 0 0
$$989$$ 0 0
$$990$$ 7.12311 0.226387
$$991$$ −12.8769 −0.409048 −0.204524 0.978862i $$-0.565565\pi$$
−0.204524 + 0.978862i $$0.565565\pi$$
$$992$$ −2.00000 −0.0635001
$$993$$ −37.3002 −1.18369
$$994$$ 3.36932 0.106868
$$995$$ 23.3693 0.740857
$$996$$ −36.4924 −1.15631
$$997$$ −28.6307 −0.906743 −0.453371 0.891322i $$-0.649779\pi$$
−0.453371 + 0.891322i $$0.649779\pi$$
$$998$$ 9.75379 0.308751
$$999$$ −7.36932 −0.233155
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 7942.2.a.x.1.1 2
19.18 odd 2 418.2.a.e.1.2 2
57.56 even 2 3762.2.a.y.1.1 2
76.75 even 2 3344.2.a.k.1.1 2
209.208 even 2 4598.2.a.bj.1.2 2
By twisted newform
Twist Min Dim Char Parity Ord Type
418.2.a.e.1.2 2 19.18 odd 2
3344.2.a.k.1.1 2 76.75 even 2
3762.2.a.y.1.1 2 57.56 even 2
4598.2.a.bj.1.2 2 209.208 even 2
7942.2.a.x.1.1 2 1.1 even 1 trivial
| 23,439
| 42,705
|
{"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}
| 2.578125
| 3
|
CC-MAIN-2022-27
|
longest
|
en
| 0.281487
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.