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http://stackoverflow.com/questions/19134443/query-to-convert-millions-of-rows-horizontally | 1,395,004,280,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678703748/warc/CC-MAIN-20140313024503-00055-ip-10-183-142-35.ec2.internal.warc.gz | 129,288,002 | 14,650 | # Query to convert millions of rows horizontally [duplicate]
This question already has an answer here:
i have the following table:
``````RECID OBJECTIF
1001 1001
1001 1002
1001 1003
2001 2001
2001 2002
``````
and i need a result like this:
``````RECID MyFIELD
1001 1001-1002-1003
2001 2001-2002
``````
just if you can give me a hint on what to use, no need to write the code
please take into consideration that i have millions of records in my table
thanks
-
You already found the answer:pivot. You can try some code and ask about it if you get stuck. – Kuzgun Oct 2 '13 at 10:04
I am using SQL Server. So will the "pivot" be the ideal solution over millions of records in 1 table? – Abd H. Oct 2 '13 at 12:52
add comment
## marked as duplicate by Luv, bluefeet♦, Nija, Kuba Ober, TalhaOct 3 '13 at 16:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 1 Answer
Here you go....I am not sure about the performance with millions of records...But still I feel this is better than going for a cursor based approach....
``````SELECT * FROM [PivotSample]
RECID OBJECTIFSEQUENCE
1001 1002 1
1001 1003 3
1001 1001 2
2001 2001 3
2001 2002 4
2001 2003 2
2001 2004 1
``````
Answer query is given below..
``````SELECT [RECID]
,STUFF((SELECT '- ' + CAST([OBJECTIF] AS varchar), REPLACE(MAX(SEQUENCE),MAX(SEQUENCE),'') FROM [PivotSample] B WHERE B.[RECID]=A.[RECID]
GROUP BY [OBJECTIF],SEQUENCE ORDER BY SEQUENCE FOR XML PATH('')), 1, 1, '' ) AS OBJECTIVE
FROM [PivotSample] A
GROUP BY [RECID]
RECID OBJECTIVE
1001 1002- 1001- 1003
2001 2004- 2003- 2001- 2002
``````
-
your answer id pretty good .. +1 to encourage you .. but please do use code formatting ({}) part while you answer .. for more information on code block formatting pls see this .. meta.stackoverflow.com/a/22189/232122 – Dhaval Oct 2 '13 at 11:38
Thanks Dhaval for formatting the results.. i am new to Stackoverflow.. :) – S Thomas Oct 2 '13 at 11:40
that is why i put a link for your information you can have a look at the link if you wish .. and again thanks for your answer – Dhaval Oct 2 '13 at 11:41
thank you, very nice. i would like to know, will this code give the best performance when applied on millions of records in a table? and i will not be able to create any other table/query on the database, i replace the [PivotSample] by the select statement, but it didnt work, as i am having an order by inside, is there any other option to consider? – Abd H. Oct 2 '13 at 12:20
Could you please share your original query here? – S Thomas Oct 2 '13 at 13:26
show 6 more comments | 836 | 2,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.780504 |
http://us.metamath.org/mpeuni/dvcof.html | 1,638,140,432,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00453.warc.gz | 82,938,100 | 10,825 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dvcof Structured version Visualization version GIF version
Theorem dvcof 23612
Description: The chain rule for everywhere-differentiable functions. (Contributed by Mario Carneiro, 10-Aug-2014.) (Revised by Mario Carneiro, 10-Feb-2015.)
Hypotheses
Ref Expression
dvcof.s (𝜑𝑆 ∈ {ℝ, ℂ})
dvcof.t (𝜑𝑇 ∈ {ℝ, ℂ})
dvcof.f (𝜑𝐹:𝑋⟶ℂ)
dvcof.g (𝜑𝐺:𝑌𝑋)
dvcof.df (𝜑 → dom (𝑆 D 𝐹) = 𝑋)
dvcof.dg (𝜑 → dom (𝑇 D 𝐺) = 𝑌)
Assertion
Ref Expression
dvcof (𝜑 → (𝑇 D (𝐹𝐺)) = (((𝑆 D 𝐹) ∘ 𝐺) ∘𝑓 · (𝑇 D 𝐺)))
Proof of Theorem dvcof
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 dvcof.f . . . . 5 (𝜑𝐹:𝑋⟶ℂ)
21adantr 481 . . . 4 ((𝜑𝑥𝑌) → 𝐹:𝑋⟶ℂ)
3 dvcof.df . . . . . 6 (𝜑 → dom (𝑆 D 𝐹) = 𝑋)
4 dvbsss 23567 . . . . . 6 dom (𝑆 D 𝐹) ⊆ 𝑆
53, 4syl6eqssr 3640 . . . . 5 (𝜑𝑋𝑆)
65adantr 481 . . . 4 ((𝜑𝑥𝑌) → 𝑋𝑆)
7 dvcof.g . . . . 5 (𝜑𝐺:𝑌𝑋)
87adantr 481 . . . 4 ((𝜑𝑥𝑌) → 𝐺:𝑌𝑋)
9 dvcof.dg . . . . . 6 (𝜑 → dom (𝑇 D 𝐺) = 𝑌)
10 dvbsss 23567 . . . . . 6 dom (𝑇 D 𝐺) ⊆ 𝑇
119, 10syl6eqssr 3640 . . . . 5 (𝜑𝑌𝑇)
1211adantr 481 . . . 4 ((𝜑𝑥𝑌) → 𝑌𝑇)
13 dvcof.s . . . . 5 (𝜑𝑆 ∈ {ℝ, ℂ})
1413adantr 481 . . . 4 ((𝜑𝑥𝑌) → 𝑆 ∈ {ℝ, ℂ})
15 dvcof.t . . . . 5 (𝜑𝑇 ∈ {ℝ, ℂ})
1615adantr 481 . . . 4 ((𝜑𝑥𝑌) → 𝑇 ∈ {ℝ, ℂ})
177ffvelrnda 6316 . . . . 5 ((𝜑𝑥𝑌) → (𝐺𝑥) ∈ 𝑋)
183adantr 481 . . . . 5 ((𝜑𝑥𝑌) → dom (𝑆 D 𝐹) = 𝑋)
1917, 18eleqtrrd 2707 . . . 4 ((𝜑𝑥𝑌) → (𝐺𝑥) ∈ dom (𝑆 D 𝐹))
209eleq2d 2689 . . . . 5 (𝜑 → (𝑥 ∈ dom (𝑇 D 𝐺) ↔ 𝑥𝑌))
2120biimpar 502 . . . 4 ((𝜑𝑥𝑌) → 𝑥 ∈ dom (𝑇 D 𝐺))
222, 6, 8, 12, 14, 16, 19, 21dvco 23611 . . 3 ((𝜑𝑥𝑌) → ((𝑇 D (𝐹𝐺))‘𝑥) = (((𝑆 D 𝐹)‘(𝐺𝑥)) · ((𝑇 D 𝐺)‘𝑥)))
2322mpteq2dva 4709 . 2 (𝜑 → (𝑥𝑌 ↦ ((𝑇 D (𝐹𝐺))‘𝑥)) = (𝑥𝑌 ↦ (((𝑆 D 𝐹)‘(𝐺𝑥)) · ((𝑇 D 𝐺)‘𝑥))))
24 dvfg 23571 . . . . 5 (𝑇 ∈ {ℝ, ℂ} → (𝑇 D (𝐹𝐺)):dom (𝑇 D (𝐹𝐺))⟶ℂ)
2515, 24syl 17 . . . 4 (𝜑 → (𝑇 D (𝐹𝐺)):dom (𝑇 D (𝐹𝐺))⟶ℂ)
26 recnprss 23569 . . . . . . . 8 (𝑇 ∈ {ℝ, ℂ} → 𝑇 ⊆ ℂ)
2715, 26syl 17 . . . . . . 7 (𝜑𝑇 ⊆ ℂ)
28 fco 6017 . . . . . . . 8 ((𝐹:𝑋⟶ℂ ∧ 𝐺:𝑌𝑋) → (𝐹𝐺):𝑌⟶ℂ)
291, 7, 28syl2anc 692 . . . . . . 7 (𝜑 → (𝐹𝐺):𝑌⟶ℂ)
3027, 29, 11dvbss 23566 . . . . . 6 (𝜑 → dom (𝑇 D (𝐹𝐺)) ⊆ 𝑌)
31 recnprss 23569 . . . . . . . . . . 11 (𝑆 ∈ {ℝ, ℂ} → 𝑆 ⊆ ℂ)
3214, 31syl 17 . . . . . . . . . 10 ((𝜑𝑥𝑌) → 𝑆 ⊆ ℂ)
3316, 26syl 17 . . . . . . . . . 10 ((𝜑𝑥𝑌) → 𝑇 ⊆ ℂ)
34 fvex 6160 . . . . . . . . . . 11 ((𝑆 D 𝐹)‘(𝐺𝑥)) ∈ V
3534a1i 11 . . . . . . . . . 10 ((𝜑𝑥𝑌) → ((𝑆 D 𝐹)‘(𝐺𝑥)) ∈ V)
36 fvex 6160 . . . . . . . . . . 11 ((𝑇 D 𝐺)‘𝑥) ∈ V
3736a1i 11 . . . . . . . . . 10 ((𝜑𝑥𝑌) → ((𝑇 D 𝐺)‘𝑥) ∈ V)
38 dvfg 23571 . . . . . . . . . . . 12 (𝑆 ∈ {ℝ, ℂ} → (𝑆 D 𝐹):dom (𝑆 D 𝐹)⟶ℂ)
39 ffun 6007 . . . . . . . . . . . 12 ((𝑆 D 𝐹):dom (𝑆 D 𝐹)⟶ℂ → Fun (𝑆 D 𝐹))
40 funfvbrb 6287 . . . . . . . . . . . 12 (Fun (𝑆 D 𝐹) → ((𝐺𝑥) ∈ dom (𝑆 D 𝐹) ↔ (𝐺𝑥)(𝑆 D 𝐹)((𝑆 D 𝐹)‘(𝐺𝑥))))
4114, 38, 39, 404syl 19 . . . . . . . . . . 11 ((𝜑𝑥𝑌) → ((𝐺𝑥) ∈ dom (𝑆 D 𝐹) ↔ (𝐺𝑥)(𝑆 D 𝐹)((𝑆 D 𝐹)‘(𝐺𝑥))))
4219, 41mpbid 222 . . . . . . . . . 10 ((𝜑𝑥𝑌) → (𝐺𝑥)(𝑆 D 𝐹)((𝑆 D 𝐹)‘(𝐺𝑥)))
43 dvfg 23571 . . . . . . . . . . . 12 (𝑇 ∈ {ℝ, ℂ} → (𝑇 D 𝐺):dom (𝑇 D 𝐺)⟶ℂ)
44 ffun 6007 . . . . . . . . . . . 12 ((𝑇 D 𝐺):dom (𝑇 D 𝐺)⟶ℂ → Fun (𝑇 D 𝐺))
45 funfvbrb 6287 . . . . . . . . . . . 12 (Fun (𝑇 D 𝐺) → (𝑥 ∈ dom (𝑇 D 𝐺) ↔ 𝑥(𝑇 D 𝐺)((𝑇 D 𝐺)‘𝑥)))
4616, 43, 44, 454syl 19 . . . . . . . . . . 11 ((𝜑𝑥𝑌) → (𝑥 ∈ dom (𝑇 D 𝐺) ↔ 𝑥(𝑇 D 𝐺)((𝑇 D 𝐺)‘𝑥)))
4721, 46mpbid 222 . . . . . . . . . 10 ((𝜑𝑥𝑌) → 𝑥(𝑇 D 𝐺)((𝑇 D 𝐺)‘𝑥))
48 eqid 2626 . . . . . . . . . 10 (TopOpen‘ℂfld) = (TopOpen‘ℂfld)
492, 6, 8, 12, 32, 33, 35, 37, 42, 47, 48dvcobr 23610 . . . . . . . . 9 ((𝜑𝑥𝑌) → 𝑥(𝑇 D (𝐹𝐺))(((𝑆 D 𝐹)‘(𝐺𝑥)) · ((𝑇 D 𝐺)‘𝑥)))
50 reldv 23535 . . . . . . . . . 10 Rel (𝑇 D (𝐹𝐺))
5150releldmi 5326 . . . . . . . . 9 (𝑥(𝑇 D (𝐹𝐺))(((𝑆 D 𝐹)‘(𝐺𝑥)) · ((𝑇 D 𝐺)‘𝑥)) → 𝑥 ∈ dom (𝑇 D (𝐹𝐺)))
5249, 51syl 17 . . . . . . . 8 ((𝜑𝑥𝑌) → 𝑥 ∈ dom (𝑇 D (𝐹𝐺)))
5352ex 450 . . . . . . 7 (𝜑 → (𝑥𝑌𝑥 ∈ dom (𝑇 D (𝐹𝐺))))
5453ssrdv 3594 . . . . . 6 (𝜑𝑌 ⊆ dom (𝑇 D (𝐹𝐺)))
5530, 54eqssd 3605 . . . . 5 (𝜑 → dom (𝑇 D (𝐹𝐺)) = 𝑌)
5655feq2d 5990 . . . 4 (𝜑 → ((𝑇 D (𝐹𝐺)):dom (𝑇 D (𝐹𝐺))⟶ℂ ↔ (𝑇 D (𝐹𝐺)):𝑌⟶ℂ))
5725, 56mpbid 222 . . 3 (𝜑 → (𝑇 D (𝐹𝐺)):𝑌⟶ℂ)
5857feqmptd 6207 . 2 (𝜑 → (𝑇 D (𝐹𝐺)) = (𝑥𝑌 ↦ ((𝑇 D (𝐹𝐺))‘𝑥)))
5915, 11ssexd 4770 . . 3 (𝜑𝑌 ∈ V)
607feqmptd 6207 . . . 4 (𝜑𝐺 = (𝑥𝑌 ↦ (𝐺𝑥)))
6113, 38syl 17 . . . . . 6 (𝜑 → (𝑆 D 𝐹):dom (𝑆 D 𝐹)⟶ℂ)
623feq2d 5990 . . . . . 6 (𝜑 → ((𝑆 D 𝐹):dom (𝑆 D 𝐹)⟶ℂ ↔ (𝑆 D 𝐹):𝑋⟶ℂ))
6361, 62mpbid 222 . . . . 5 (𝜑 → (𝑆 D 𝐹):𝑋⟶ℂ)
6463feqmptd 6207 . . . 4 (𝜑 → (𝑆 D 𝐹) = (𝑦𝑋 ↦ ((𝑆 D 𝐹)‘𝑦)))
65 fveq2 6150 . . . 4 (𝑦 = (𝐺𝑥) → ((𝑆 D 𝐹)‘𝑦) = ((𝑆 D 𝐹)‘(𝐺𝑥)))
6617, 60, 64, 65fmptco 6352 . . 3 (𝜑 → ((𝑆 D 𝐹) ∘ 𝐺) = (𝑥𝑌 ↦ ((𝑆 D 𝐹)‘(𝐺𝑥))))
6715, 43syl 17 . . . . 5 (𝜑 → (𝑇 D 𝐺):dom (𝑇 D 𝐺)⟶ℂ)
689feq2d 5990 . . . . 5 (𝜑 → ((𝑇 D 𝐺):dom (𝑇 D 𝐺)⟶ℂ ↔ (𝑇 D 𝐺):𝑌⟶ℂ))
6967, 68mpbid 222 . . . 4 (𝜑 → (𝑇 D 𝐺):𝑌⟶ℂ)
7069feqmptd 6207 . . 3 (𝜑 → (𝑇 D 𝐺) = (𝑥𝑌 ↦ ((𝑇 D 𝐺)‘𝑥)))
7159, 35, 37, 66, 70offval2 6868 . 2 (𝜑 → (((𝑆 D 𝐹) ∘ 𝐺) ∘𝑓 · (𝑇 D 𝐺)) = (𝑥𝑌 ↦ (((𝑆 D 𝐹)‘(𝐺𝑥)) · ((𝑇 D 𝐺)‘𝑥))))
7223, 58, 713eqtr4d 2670 1 (𝜑 → (𝑇 D (𝐹𝐺)) = (((𝑆 D 𝐹) ∘ 𝐺) ∘𝑓 · (𝑇 D 𝐺)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 384 = wceq 1480 ∈ wcel 1992 Vcvv 3191 ⊆ wss 3560 {cpr 4155 class class class wbr 4618 ↦ cmpt 4678 dom cdm 5079 ∘ ccom 5083 Fun wfun 5844 ⟶wf 5846 ‘cfv 5850 (class class class)co 6605 ∘𝑓 cof 6849 ℂcc 9879 ℝcr 9880 · cmul 9886 TopOpenctopn 15998 ℂfldccnfld 19660 D cdv 23528 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-inf2 8483 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 ax-pre-sup 9959 ax-addf 9960 ax-mulf 9961 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-rmo 2920 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-uni 4408 df-int 4446 df-iun 4492 df-iin 4493 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-se 5039 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-isom 5859 df-riota 6566 df-ov 6608 df-oprab 6609 df-mpt2 6610 df-of 6851 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-2o 7507 df-oadd 7510 df-er 7688 df-map 7805 df-pm 7806 df-ixp 7854 df-en 7901 df-dom 7902 df-sdom 7903 df-fin 7904 df-fsupp 8221 df-fi 8262 df-sup 8293 df-inf 8294 df-oi 8360 df-card 8710 df-cda 8935 df-pnf 10021 df-mnf 10022 df-xr 10023 df-ltxr 10024 df-le 10025 df-sub 10213 df-neg 10214 df-div 10630 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-q 11733 df-rp 11777 df-xneg 11890 df-xadd 11891 df-xmul 11892 df-icc 12121 df-fz 12266 df-fzo 12404 df-seq 12739 df-exp 12798 df-hash 13055 df-cj 13768 df-re 13769 df-im 13770 df-sqrt 13904 df-abs 13905 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-starv 15872 df-sca 15873 df-vsca 15874 df-ip 15875 df-tset 15876 df-ple 15877 df-ds 15880 df-unif 15881 df-hom 15882 df-cco 15883 df-rest 15999 df-topn 16000 df-0g 16018 df-gsum 16019 df-topgen 16020 df-pt 16021 df-prds 16024 df-xrs 16078 df-qtop 16083 df-imas 16084 df-xps 16086 df-mre 16162 df-mrc 16163 df-acs 16165 df-mgm 17158 df-sgrp 17200 df-mnd 17211 df-submnd 17252 df-mulg 17457 df-cntz 17666 df-cmn 18111 df-psmet 19652 df-xmet 19653 df-met 19654 df-bl 19655 df-mopn 19656 df-fbas 19657 df-fg 19658 df-cnfld 19661 df-top 20616 df-bases 20617 df-topon 20618 df-topsp 20619 df-cld 20728 df-ntr 20729 df-cls 20730 df-nei 20807 df-lp 20845 df-perf 20846 df-cn 20936 df-cnp 20937 df-haus 21024 df-tx 21270 df-hmeo 21463 df-fil 21555 df-fm 21647 df-flim 21648 df-flf 21649 df-xms 22030 df-ms 22031 df-tms 22032 df-cncf 22584 df-limc 23531 df-dv 23532 This theorem is referenced by: dvmptco 23636 dvsinax 39419 dvcosax 39434
Copyright terms: Public domain W3C validator | 6,178 | 9,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-49 | longest | en | 0.191556 |
https://metanumbers.com/218675 | 1,638,386,901,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00034.warc.gz | 461,895,396 | 7,341 | # 218675 (number)
218,675 (two hundred eighteen thousand six hundred seventy-five) is an odd six-digits composite number following 218674 and preceding 218676. In scientific notation, it is written as 2.18675 × 105. The sum of its digits is 29. It has a total of 3 prime factors and 6 positive divisors. There are 174,920 positive integers (up to 218675) that are relatively prime to 218675.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 29
• Digital Root 2
## Name
Short name 218 thousand 675 two hundred eighteen thousand six hundred seventy-five
## Notation
Scientific notation 2.18675 × 105 218.675 × 103
## Prime Factorization of 218675
Prime Factorization 52 × 8747
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 43735 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 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 218,675 is 52 × 8747. Since it has a total of 3 prime factors, 218,675 is a composite number.
## Divisors of 218675
6 divisors
Even divisors 0 6 3 3
Total Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 271188 Sum of all the positive divisors of n s(n) 52513 Sum of the proper positive divisors of n A(n) 45198 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 467.627 Returns the nth root of the product of n divisors H(n) 4.83816 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 218,675 can be divided by 6 positive divisors (out of which 0 are even, and 6 are odd). The sum of these divisors (counting 218,675) is 271,188, the average is 45,198.
## Other Arithmetic Functions (n = 218675)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 174920 Total number of positive integers not greater than n that are coprime to n λ(n) 87460 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 19454 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 174,920 positive integers (less than 218,675) that are coprime with 218,675. And there are approximately 19,454 prime numbers less than or equal to 218,675.
## Divisibility of 218675
m n mod m 2 3 4 5 6 7 8 9 1 2 3 0 5 2 3 2
The number 218,675 is divisible by 5.
• Arithmetic
• Deficient
• Polite
• Frugal
## Base conversion (218675)
Base System Value
2 Binary 110101011000110011
3 Ternary 102002222002
4 Quaternary 311120303
5 Quinary 23444200
6 Senary 4404215
8 Octal 653063
10 Decimal 218675
12 Duodecimal a666b
20 Vigesimal 176df
36 Base36 4oqb
## Basic calculations (n = 218675)
### Multiplication
n×y
n×2 437350 656025 874700 1093375
### Division
n÷y
n÷2 109338 72891.7 54668.8 43735
### Exponentiation
ny
n2 47818755625 10456766386296875 2286633389523469140625 500029556454044614326171875
### Nth Root
y√n
2√n 467.627 60.2467 21.6247 11.6939
## 218675 as geometric shapes
### Circle
Diameter 437350 1.37398e+06 1.50227e+11
### Sphere
Volume 4.38012e+16 6.00908e+11 1.37398e+06
### Square
Length = n
Perimeter 874700 4.78188e+10 309253
### Cube
Length = n
Surface area 2.86913e+11 1.04568e+16 378756
### Equilateral Triangle
Length = n
Perimeter 656025 2.07061e+10 189378
### Triangular Pyramid
Length = n
Surface area 8.28245e+10 1.23234e+15 178547
## Cryptographic Hash Functions
md5 8a291ab01ed72bde29cd71fdce6b8e93 d7d64c81a619174d1d1338f6f754e2a1de161208 71a1374035912db3369c8e94af8c99b2f9be59b48934a775720ed7c1d038bcb5 2821e77821a03b43a36a3364a4aa5e9439bbb7d9b7e4b60081992ae9eb35921091977c3d625e7377450b8d24467afe560a0a3bd4e2beb115394f09b0e332f79e 078232f0f49c60337b2808ac83eb156eec5884e8 | 1,429 | 4,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-49 | latest | en | 0.81095 |
https://www.mathworks.com/matlabcentral/answers/557869-how-to-draw-a-graph-for-function-with-constraints | 1,721,920,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00190.warc.gz | 736,872,386 | 24,993 | # How to draw a graph for function with constraints?
23 views (last 30 days)
Trang To on 1 Jul 2020
Commented: Trang To on 2 Jul 2020
How can I draw the contour plot of the function z = 3x + 4y and the constraint curve x^2 + 4xy +5y^2 = 10 in the same figure ?
Matt J on 1 Jul 2020
fimplicit(@(x,y) x.^2 + 4*x.*y +5*y.^2 - 10,'r'); hold on; fcontour(@(x,y) 3*x+4*y) ; hold off
Trang To on 2 Jul 2020
thank you so much
Gurpreet Singh on 1 Jul 2020 | 172 | 449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.788201 |
https://performingineducation.com/product/5th-grade-math-centers-divide-fractions-mixed-numbers/ | 1,696,346,467,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511106.1/warc/CC-MAIN-20231003124522-20231003154522-00587.warc.gz | 489,377,932 | 56,373 | # 5th Grade Math Centers | Divide Fractions & Mixed Numbers
\$4.99
CCSS Standards: , , ,
# 5th Grade Math Centers | Divide Fractions & Mixed Numbers
\$4.99
CCSS Standards: , , ,
## Description
Everything you need to practice and reteach dividing fractions and mixed numbers can be found inside of this set of 5th grade math centers. There are 10 different divide fractions and mixed numbers centers included in this set.
Along with 10 engaging math centers, this set also includes a teacher guide that walks you through prepping each center, storing your centers, and implementing the centers.
Activities Included:
There are 5 different types of activities included in this set of centers. Each activity includes 2 different centers. Check out the included preview to see an example of each of the math activities below.
• Task Cards: divide a whole number by a fraction using models and/or number lines
• Board Games: solve division problems involving fraction division and fractional quotients
• Error Analysis: compare division fraction expressions, estimate quotients of fraction division
• Secret Picture: divide whole numbers, fractions, and/or mixed numbers
• Square Puzzles: divide whole numbers, fractions, and/or mixed numbers
How to Use These Math Centers:
• Each math center has an answer sheet that can be turned in and graded for student accountability.
• The centers can be reused over and over again. Just prep them once and use them for years to come!
• If you prefer, you can also create single-use centers with the included black and white, printable options.
• The teacher guide includes a printable packet cover page that you can use if you prefer to create weekly math center work packets for your students.
Skills Covered:
• CCSS.MATH.CONTENT.5.NF.B.7: Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
• CCSS.MATH.CONTENT.5.NF.B.7.A: Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) × 4 = 1/3.
• CCSS.MATH.CONTENT.5.NF.B.7.B: Interpret division of a whole number by a unit fraction, and compute such quotients. For example, create a story context for 4 ÷ (1/5), and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that 4 ÷ (1/5) = 20 because 20 × (1/5) = 4.
• CCSS.MATH.CONTENT.5.NF.B.7.C: Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem. For example, how much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How many 1/3-cup servings are in 2 cups of raisins?
All of the centers are included in one ZIP download. After unzipping this resource, you will see 6 PDFs: the teacher guide, the task card centers, the board game centers, the error analysis centers, the secret picture centers, and the square puzzle centers. All of the teacher instructions can be found inside of the teacher guide. You will find everything you need for each center inside of the corresponding PDFs.
## Standards & Level
CCSS Standards: , , ,
Select
Select | 793 | 3,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-40 | longest | en | 0.883511 |
https://graphicmaths.com/pure/integration/integration-by-parts/ | 1,725,845,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00171.warc.gz | 260,119,240 | 8,383 | # Integration by parts
By Martin McBride, 2024-07-10
Tags: integration by parts product rule
Categories: integration calculus
Finding integrals can be challenging because, unlike differentiation, we can't usually integrate a function from first principles. Instead, we usually find integrals by identifying a related differential equation and reversing it. For example, if we happen to know that the first derivative of the function f(x) is g(x), then (subject to a few conditions) we know that the integral of g(x) will be f(x).
This gives us a repertoire of standard integrals that we can use to solve problems. However, it is always useful to expand our repertoire. One way to do that is to investigate some of the rules we use for differentiating more complex expressions. These can sometimes be "reversed" to provide a new weapon in our integration armoury.
One well-known rule we use in differentiation is the product rule. The equivalent rule for integration is integration by parts. We will introduce that method in this article, with examples and a proof using calculus.
## Integration by parts
Let's start by stating the rule as an indefinite integral (we will prove it later):
How is this useful? Well, suppose we have an expression to integrate but there is no known way to do it. But we notice that the expression is a factor of two other functions, like this:
Now suppose we happen to know that the integral of h is some other function g. In other words, h is equal to g'. So now we have this:
According to the formula above, we can rewrite this as:
This gives us a second chance at solving the integral. It might be that f' g is easier to integrate than f g'. This sometimes happens if f' is simpler than f. And since finding f(x) times g(x) is easy (it isn't under the integral sign) we have found a solution. This doesn't always work, but it sometimes does, so it is worth a try.
Let's look at a simple example:
We can't integrate this directly, so we will try integration by parts. We need to split the function into two factors, one that will represent f, and the other that will represent the derivative of some other function g:
Our function can be split into two factors fairly easily, it is x times sin x. Let's try defining f and g' like this:
We can now find f' by differentiating f:
We can find g by integrating g':
Looking back to the original rule:
We can substitute f, g and their derivatives:
Simplifying gives:
Evaluating the integral gives us the final result:
Here is a graph showing the original function on the left, and the integral on the right. As you would expect, the places where the left-hand graph is zero correspond to the turning points in the right-hand graph. Also, the regions where the left-hand graph is positive correspond to the regions where the right-hand graph has a positive slope, and vice versa. This gives us a small measure of confidence that the integral is correct:
## Integration by parts doesn't always work
This technique is great when it works, but in many cases (in fact most cases) it won't work. As an example, consider what might happen if we had allocated f and g' differently. In this equation:
We could have chosen to swap the definitions of f and g':
We are allowed to do this, of course, because the order of multiplication makes no difference. This would have created different f' and g functions:
This in turn would have created a different equation for the integration by parts:
This equation is perfectly valid, but it isn't very useful because the integral on the RHS is still not solvable. If anything, it is more difficult than the original.
## Standard form of integration by parts
You will often see the rule for integration by parts written in a standard, almost idiomatic, form. We will derive that form here. Taking the formula above:
By definition, we have:
This can be rearranged to give:
We can derive a similar result for g:
This allows us to replace dx in the original to obtain the standard formulation of the integration by parts rule:
This is often written using u and v rather than f and g.
## Proof
The proof of the method can be derived from the product rule of differentiation. We will assume that our functions f and g are continuously differentiable. When we say that a function is continuously differentiable, we mean that the function is differentiable everywhere and that the first derivative itself is a continuous function.
This proof assumes the product rule is already proven, see the proof from first principles here. In summary, the product rule states that, if we have a function p that is the product of two other functions:
Then, using the product rule, the derivative of p will be:
We can now integrate both sides with respect to x:
The integral of p' is just p:
And p is the product of f and g:
Rearranging this formula gives us the integration by parts rule:
## Finding the definite integral
To find the definite integral, we need to evaluate the integral term over the required range (let's call it a to b). But we also need to evaluate the product term between a and b. So:
This is equivalent to:
## Example
As a second example, we will evaluate this definite integral:
We will choose f and g' as follows:
This means that f' will be 1, as in the previous example. To find g, we rely on the special property of the exponential function, that it is its own derivative. So g is the same as g':
So, using the equation for the definite integral by parts we saw in the previous section, we have:
Evaluating the second integral between 0 and 2 gives:
We can gather both terms on the RHS and evaluate them both between 0 and 2:
Substituting the values gives:
These graphs show the original function of the left, with the calculated integral on the right. The integral graph has been adjusted by adding a constant 1 to ensure that the integral starts at zero when x is zero (because the definite integral range starts at zero):
The value of the graph on the right, when x is 2, is about 8.3891 as calculated above. For the curve on the left, we would therefore expect the area under the curve between 0 and 2 to also be 0.83891.
As a quick check, the yellow triangle also has an area of 0.83891. Visually, it is certainly plausible that the area of the triangle is equal to the area under the curve, which gives an informal validation of the result. | 1,388 | 6,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.946335 |
https://dsoftschools.com/order-of-operations-worksheets/ | 1,702,248,493,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00073.warc.gz | 246,761,033 | 17,818 | # Order Of Operations Worksheets
Some of the worksheets below are Order Of Operations Worksheets, learn how simplify numerical expressions involving positive exponent using order of operations and properties of operations with several exam style questions with answer keys.
## Basic Instructions
### Order of Operations : Evaluate each expression below.
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### Order of Operations with Integers Worksheet : Using integers, write an expression that shows the meaning of these words.
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### Order of Operations : PEMDAS Practice Worksheets.
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### Using Order of Operations and Exploring Properties : Evaluate whole number numerical expressions using the order of operations.
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### Solving Problems Using Order of Operations with Answer Key.
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### Order of Operations Notes and Activities : State which operation (addition, subtraction, multiplication or division) you would perform first.
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### Order of Operations Numerical Problems : Perform the operations in the correct order.
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### Order of Operations Word Problems : Write the expression and solve each given problem below.
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### Order of Operations with Exponents : Use the correct order of operations to determine each answer. Circle your answer for each problem.
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### Reviewing Order of Operations : Solve the following expression using the correct order of operations. Answers are available at the end of the page.
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### Evaluate Expressions : Solve the problems below using your of knowledge of the order of operations.
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| Open in new tab
Bonus Worksheet. | 371 | 1,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-50 | latest | en | 0.851083 |
http://mathhelpforum.com/algebra/81563-problem-based-ratio-print.html | 1,526,839,340,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863662.15/warc/CC-MAIN-20180520170536-20180520190536-00006.warc.gz | 188,352,880 | 3,317 | # Problem based on ratio
• Mar 30th 2009, 07:17 PM
siddscool19
Problem based on ratio
Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..
The answer is 4:11. But I am not getting how to get it.
• Mar 30th 2009, 07:30 PM
e^(i*pi)
Quote:
Originally Posted by siddscool19
Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..
The answer is 4:11. But I am not getting how to get it.
If we assume the first ship has a volume of VL then:
In ship one there will be $\displaystyle V\frac{2}{2+3}$ of coke and so $\displaystyle V(1-\frac{2}{2+3})$ of soda
In ship two there will be $\displaystyle 2V\frac{3}{10}$ of coke and $\displaystyle 2V(1-\frac{3}{10})$ of soda
In ship three there will be $\displaystyle 3V\frac{1}{5}$ of coke and $\displaystyle 3V(1-\frac{1}{5})$
Add the sum of the coke from the three ships and factor out V.
$\displaystyle V(\frac{2}{5} + 2\frac{3}{10} + \frac{1}{5}) = \frac{8V}{5}$
Add the sum of the soda from the three ships and factor out V
$\displaystyle V(1-\frac{2}{5} + 2(1-\frac{3}{10}) + 3(1-\frac{1}{5})) = \frac{22V}{5}$
To find the ratio we divide the cola by the soda:
$\displaystyle \frac{8V}{5} \div \frac{22V}{5} = \frac{8V}{5} \times \frac{5}{22V} = \frac{8}{22} = \frac{4}{11}$
Now we have a ratio of $\displaystyle \frac{4}{11} : 1$ so we multiply by 11 to get an integer ratio of 4:11
(if you're having trouble visualising it try putting V=5)
• Mar 31st 2009, 08:25 AM
stapel
Quote:
Originally Posted by siddscool19
Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..
The first vessel is coke:soda = 2:3, or five parts.
The second vessel is coke:soda = 3:7, or ten parts.
The third vessel is coke:soda = 1:4, or five parts.
Overall, their volumes are 1st:2nd:3rd = 1:2:3, but "5:10:5" doesn't match this. The first bit, the "5:10", matches the "1:2", but the second "5" doesn't match the "3". So make it match:
. . . . .third vessel: 3:12, or fifteen parts
Now you have:
The first vessel is coke:soda = 2:3, or five parts.
The second vessel is coke:soda = 3:7, or ten parts.
The third vessel is coke:soda = 3:12, or fifteen parts.
This gives you the right ratio, assuming all the "parts" are of the same volume, so the mixture will then be:
. . . . .fourth vessel: (2+3+3) : (3+7+12)
...or 8:22, which simplifies as 4:11.
(Wink) | 995 | 3,033 | {"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} | 4.4375 | 4 | CC-MAIN-2018-22 | latest | en | 0.883508 |
http://www.thestudentroom.co.uk/showthread.php?t=1310974&page=154&p=37601860 | 1,386,906,061,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164844212/warc/CC-MAIN-20131204134724-00048-ip-10-33-133-15.ec2.internal.warc.gz | 564,867,613 | 39,222 | You are Here: Home
# The big fat STEP megathread (NOT for getting help with maths questions)
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
Announcements Posted on
1. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by number23)
Does anyone know how to do the last part of 5i in the STEP I June 2003 paper:
Show that there is no term indpendent of x in the expansion of f(x) in powers of x, for lxl > 3.
f(x)=(x^6 + 3x^5)^(0.5) which is also x^3(1+3/x)^(0.5).
Thanks
Binomially expand the second bracket and show no term has power of x -3.
2. Re: The big fat STEP megathread (NOT for getting help with maths questions)
Hi,
I can vaguely remember a problem from STEP I did for preparation and I would like to know from which exam it was. Maybe someone knows the problem and can help me find it!
The question was about the power series of and integrals of the form . You eventually had to prove .
Thanks
3. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by kitkat19)
when youve done AS maths and further maths....so u have module c1 c2 m1 fp1 d1 s1 ...which STEP questions can you do...i wannna start practising asap
Start with STEP I, but expect to run into problems on some questions until you know C3-4. There are questions that don't require them but many do. If I were you I'd focus more on learning C3-4 - you can do it by yourself!
4. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by DamoclesAustria)
Hi,
I can vaguely remember a problem from STEP I did for preparation and I would like to know from which exam it was. Maybe someone knows the problem and can help me find it!
The question was about the power series of and integrals of the form . You eventually had to prove .
Thanks
I've just found
- it turned out it wasn't even a problem from STEP...
5. Re: The big fat STEP megathread (NOT for getting help with maths questions)
Does anyone know any London test centres which take private candidates for STEP? I've been able to find couple but they are charging an extortionate amount! I mean, £140 per paper seriously? Get out of here. It's weird none of those listed in the Cambridge Advisement website host the exam anymore.
6. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by DamoclesAustria)
I've just found
- it turned out it wasn't even a problem from STEP...
I can't blame you. That sheet is very 'stepy'.
7. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by DamoclesAustria)
I've just found
- it turned out it wasn't even a problem from STEP...
Great find, thanks!
8. Re: The big fat STEP megathread (NOT for getting help with maths questions)
People who have done STEP.
What did you do to revise. Apart from past papers
did you have a tutor?
Did you do it all by yourself? If so HOW!!!
Did you go to the cambridge easter STEP school?
did your school give you tuition?
anything else i havent thought of. feel free to add^
and...if you don't mind then tell me what you got too
9. Re: The big fat STEP megathread (NOT for getting help with maths questions)
Can anyone give me the past step paper form 1987 t0 1999 pleasee
10. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by mathsymathsy)
Can anyone give me the past step paper form 1987 t0 1999 pleasee
11. Re: The big fat STEP megathread (NOT for getting help with maths questions)
Thanks
12. Re: The big fat STEP megathread (NOT for getting help with maths questions)
has anyone here got 110+ in a STEP paper? just curious
Last edited by oh_1993; 10-04-2012 at 17:23.
13. Re: The big fat STEP megathread (NOT for getting help with maths questions)
Do people just read through the STEP booklet Advanced Problems in Core Mathematics and start doing the past papers directly?
14. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by johnconnor92)
Do people just read through the STEP booklet Advanced Problems in Core Mathematics and start doing the past papers directly?
I dont know but I would like to know this too....i posted here(few posts above this one) about revision techniques that people may have used but noone got back to me about this...ill just have to wait..
15. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by johnconnor92)
Do people just read through the STEP booklet Advanced Problems in Core Mathematics and start doing the past papers directly?
(Original post by kitkat19)
I dont know but I would like to know this too....i posted here(few posts above this one) about revision techniques that people may have used but noone got back to me about this...ill just have to wait..
Pretty much yeah. Although with the Siklos booklet, it's best if you have a go at the questions before reading his solution and even before reading his comments on the question as you progress.
Plowing through questions is the best preparation, then doing a few papers in exam conditions as you get closer to the exam to gauge speed/timings etc.
16. Re: The big fat STEP megathread (NOT for getting help with maths questions)
i read already that no calculators are allowed, so does that mean that unsimplified surds are acceptable.
regarding the data sheet, can you only bring one? do you have to bring them with you on the day? i am NOT sitting the exam at my college as they don't run it, its at the local adult education center.
17. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by zack-w)
i read already that no calculators are allowed, so does that mean that unsimplified surds are acceptable.
regarding the data sheet, can you only bring one? do you have to bring them with you on the day? i am NOT sitting the exam at my college as they don't run it, its at the local adult education center.
Surds are of course acceptable, but you should always simplify them when possible. (i.e. don't put when you could put )
The "data sheet" is a formula book and they should come delivered with the exam paper AFAIK. It's the same as the OCR A-level one.
18. Re: The big fat STEP megathread (NOT for getting help with maths questions)
Hi guys! How do you think about the difficulty of STEP I papers before 1994? They are the only core maths papers at that time, so are they approximately equivalent to the current STEP II papers or they are easier than the current STEP II papers? I am thinking about if I need to go through them to prepare for my STEP II. Thx!
19. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by klausw)
Hi guys! How do you think about the difficulty of STEP I papers before 1994? They are the only core maths papers at that time, so are they approximately equivalent to the current STEP II papers or they are easier than the current STEP II papers? I am thinking about if I need to go through them to prepare for my STEP II. Thx!
They were the "paper I" so that would imply they're closer to STEP I. However, older papers are a bit less relevant... you might find some nice questions, but only use them if you've dried up all the recent ones IMO.
20. Re: The big fat STEP megathread (NOT for getting help with maths questions)
(Original post by Xero Xenith)
They were the "paper I" so that would imply they're closer to STEP I. However, older papers are a bit less relevant... you might find some nice questions, but only use them if you've dried up all the recent ones IMO.
Thank you!!!
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https://ansanswers.com/mathematics/question12832603 | 1,628,152,318,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155458.35/warc/CC-MAIN-20210805063730-20210805093730-00719.warc.gz | 116,566,775 | 16,248 | , 22.06.2019 20:20 Officaljazz18
# Gary is lining up his three butterflies for biology class. his pan has a ruler (in inches) along the bottom like a number line. he puts his first butterfly in the middle. he leaves a 1 inch space and then lays out the other two butterflies to the left and to the right. what is the total horizontal distance (in inches) between the left edge of the left butterfly and the right edge of the right butterfly? explain how you found the distance.
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Ahiker stands at an unknown elevation before he ascends 2 miles up a mountain to an elevation of 9 miles. find the elevation of the hiker before his ascent. | 264 | 995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-31 | latest | en | 0.9041 |
https://www.teacherspayteachers.com/Product/Multiply-by-Multiples-of-10-Number-Puzzles-3143090 | 1,513,173,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948523222.39/warc/CC-MAIN-20171213123757-20171213143757-00632.warc.gz | 815,026,304 | 27,057 | Total:
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Help students practice multiplying by multiples of 10. There are a five different number puzzles in this product that reinforce multiplication concepts and patterns such as base-10, properties of operations, and place value strategies. This covers NBT-3.
Included are puzzles for:
✔ Properties of Operations Puzzles
✔ Word Problems
✔ Base-10 / Place Value Strategies
★ Save 20% by purchasing the BUNDLE of Third Grade Number Puzzles. ★
Simple print each puzzle piece on colored paper, laminate for durability, and cut apart. I encourage students to use whiteboard markers on some pieces. If laminated, the whiteboard marker ink wipes off easily.
OA1 - Interpret Multiplication
OA2 - Interpret Division
OA3 - Multiplication & Division Word Problems
OA4 - Determine Unknown Numbers for Multiplication and Division Equations
OA5 - Apply Properties of Operations - Multiplication & Division
OA7 - Relationship Between Multiplication & Division
OA8 - Two-Step Word Problems
NBT1 - Rounding to the Nearest 10 & 100
NBT2 - Three-Digit Subtraction (same for second grade)
NBT3 - Multiply by Multiples of 10
NF 1-3 - Fraction Number Lines, Visual Models & Compare Fractions Number Puzzles
MD1 - Time to the One Minute & Interval Word Problems Number Puzzles
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\$4.79 | 496 | 2,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-51 | longest | en | 0.783023 |
http://mathoverflow.net/questions/8912/analogue-of-the-chebyshev-polynomials-over-c/8915 | 1,469,633,950,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826908.63/warc/CC-MAIN-20160723071026-00075-ip-10-185-27-174.ec2.internal.warc.gz | 169,795,820 | 15,801 | # Analogue of the Chebyshev polynomials over C?
I've been driven up a wall by the following question: let p be a complex polynomial of degree d. Suppose that |p(z)|≤1 for all z such that |z|=1 and |z-1|≥δ (for some small δ>0). Then what's the best upper bound one can prove on |p(1)|? (I only care about the asymptotic dependence on d and δ, not the constants.)
For the analogous question where p is a degree-d real polynomial such that |p(x)|≤1 for all 0≤x≤1-δ, I know that the right upper bound on |p(1)| is |p(1)|≤exp(d√δ). The extremal example here is p(x)=Td((1+δ)x), where Td is the dth Chebyshev polynomial.
Indeed, by using the Chebyshev polynomial, it's not hard to construct a polynomial p in z as well as its complex conjugate z*, such that
(i) |p(z)|≤1 for all z such that |z|=1 and |z-1|≥δ, and
(ii) p(1) ~ exp(dδ).
One can also show that this is optimal, for polynomials in both z and its complex conjugate.
The question is whether one can get a better upper bound on |p(1)| by exploiting the fact that p is really a polynomial in z. The fastest-growing example I could find has the form p(z)=Cd,δ(1+z)d. Here, if we choose the constant Cd,δ so that |p(z)|≤1 whenever |z|=1 and |z-1|≥δ, we find that
p(1) ~ exp(dδ2)
For my application, the difference between exp(dδ) and exp(dδ2) is all the difference in the world!
I searched about 6 approximation theory books---and as often the case, they answer every conceivable question except the one I want. If anyone versed in approximation theory can give me a pointer, I'd be incredibly grateful.
Thanks so much! --Scott Aaronson
PS. The question is answered below by David Speyer. For anyone who wants to see explicitly the polynomial implied by David's argument, here it is:
pd,δ(z) = zd Td((z+z-1)(1+δ)/2+δ),
where Td is the dth Chebyshev polynomial.
-
Scott: Good question, but please register with MO. Join the club! – Greg Kuperberg Dec 14 '09 at 21:49
@Scott: I am curious: what was the intended application of this result? – John Jiang Jul 12 at 15:52
I may be missing something obvious here. Let $f(z, z^{*})$ be the polynomial in $z$ and $z^{*}$ of degree $d$ which achieves $\exp(d \delta)$. Let $g(z)$ be the Laurent polynomial obtained from $f$ by replacing $z^{*}$ by $z^{-1}$. On the unit circle, we have $f=g$.
Now, let $h$ be the polynomial $z^d g$. This is an honest polynomial, because we multiplied by a high enough power of $z$ to clear out all the denominators and, for $z$ on the unit circle, we have $|h|=|f|$.
Doesn't this mean that $h$ is a polynomial of degree $2d$, obeying your conditions, with $|h(1)| \sim \exp(d \delta)$? | 774 | 2,627 | {"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-2016-30 | latest | en | 0.922862 |
https://convertoctopus.com/233-7-pounds-to-kilograms | 1,679,526,729,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.97/warc/CC-MAIN-20230322211955-20230323001955-00030.warc.gz | 221,096,679 | 7,426 | ## Conversion formula
The conversion factor from pounds to kilograms is 0.45359237, which means that 1 pound is equal to 0.45359237 kilograms:
1 lb = 0.45359237 kg
To convert 233.7 pounds into kilograms we have to multiply 233.7 by the conversion factor in order to get the mass amount from pounds to kilograms. We can also form a simple proportion to calculate the result:
1 lb → 0.45359237 kg
233.7 lb → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 233.7 lb × 0.45359237 kg
M(kg) = 106.004536869 kg
The final result is:
233.7 lb → 106.004536869 kg
We conclude that 233.7 pounds is equivalent to 106.004536869 kilograms:
233.7 pounds = 106.004536869 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 0.0094335585017064 × 233.7 pounds.
Another way is saying that 233.7 pounds is equal to 1 ÷ 0.0094335585017064 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred thirty-three point seven pounds is approximately one hundred six point zero zero five kilograms:
233.7 lb ≅ 106.005 kg
An alternative is also that one kilogram is approximately zero point zero zero nine times two hundred thirty-three point seven pounds.
## Conversion table
### pounds to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from pounds to kilograms
pounds (lb) kilograms (kg)
234.7 pounds 106.458 kilograms
235.7 pounds 106.912 kilograms
236.7 pounds 107.365 kilograms
237.7 pounds 107.819 kilograms
238.7 pounds 108.272 kilograms
239.7 pounds 108.726 kilograms
240.7 pounds 109.18 kilograms
241.7 pounds 109.633 kilograms
242.7 pounds 110.087 kilograms
243.7 pounds 110.54 kilograms | 496 | 1,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-14 | latest | en | 0.778755 |
http://docs.scipy.org/doc/scipy-dev/reference/generated/scipy.sparse.coo_matrix.html | 1,444,367,405,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737913815.59/warc/CC-MAIN-20151001221833-00187-ip-10-137-6-227.ec2.internal.warc.gz | 83,704,610 | 5,278 | # scipy.sparse.coo_matrix¶
class scipy.sparse.coo_matrix(arg1, shape=None, dtype=None, copy=False)[source]
A sparse matrix in COOrdinate format.
Also known as the ‘ijv’ or ‘triplet’ format.
This can be instantiated in several ways:
coo_matrix(D)
with a dense matrix D
coo_matrix(S)
with another sparse matrix S (equivalent to S.tocoo())
coo_matrix((M, N), [dtype])
to construct an empty matrix with shape (M, N) dtype is optional, defaulting to dtype=’d’.
coo_matrix((data, (i, j)), [shape=(M, N)])
to construct from three arrays:
1. data[:] the entries of the matrix, in any order
2. i[:] the row indices of the matrix entries
3. j[:] the column indices of the matrix entries
Where A[i[k], j[k]] = data[k]. When shape is not specified, it is inferred from the index arrays
Notes
Sparse matrices can be used in arithmetic operations: they support addition, subtraction, multiplication, division, and matrix power.
• facilitates fast conversion among sparse formats
• permits duplicate entries (see example)
• very fast conversion to and from CSR/CSC formats
• does not directly support:
• arithmetic operations
• slicing
Intended Usage
• COO is a fast format for constructing sparse matrices
• Once a matrix has been constructed, convert to CSR or CSC format for fast arithmetic and matrix vector operations
• By default when converting to CSR or CSC format, duplicate (i,j) entries will be summed together. This facilitates efficient construction of finite element matrices and the like. (see example)
Examples
>>> from scipy.sparse import coo_matrix
>>> coo_matrix((3, 4), dtype=np.int8).toarray()
array([[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]], dtype=int8)
>>> row = np.array([0, 3, 1, 0])
>>> col = np.array([0, 3, 1, 2])
>>> data = np.array([4, 5, 7, 9])
>>> coo_matrix((data, (row, col)), shape=(4, 4)).toarray()
array([[4, 0, 9, 0],
[0, 7, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 5]])
>>> # example with duplicates
>>> row = np.array([0, 0, 1, 3, 1, 0, 0])
>>> col = np.array([0, 2, 1, 3, 1, 0, 0])
>>> data = np.array([1, 1, 1, 1, 1, 1, 1])
>>> coo_matrix((data, (row, col)), shape=(4, 4)).toarray()
array([[3, 0, 1, 0],
[0, 2, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 1]])
Attributes
nnz Get the count of explicitly-stored values (nonzeros)
dtype (dtype) Data type of the matrix shape (2-tuple) Shape of the matrix ndim (int) Number of dimensions (this is always 2) data COO format data array of the matrix row COO format row index array of the matrix col COO format column index array of the matrix
Methods
arcsin() Element-wise arcsin. arcsinh() Element-wise arcsinh. arctan() Element-wise arctan. arctanh() Element-wise arctanh. asformat(format) Return this matrix in a given sparse format asfptype() Upcast matrix to a floating point format (if necessary) astype(t) ceil() Element-wise ceil. conj() conjugate() copy() deg2rad() Element-wise deg2rad. diagonal() Returns the main diagonal of the matrix dot(other) Ordinary dot product expm1() Element-wise expm1. floor() Element-wise floor. getH() get_shape() getcol(j) Returns a copy of column j of the matrix, as an (m x 1) sparse matrix (column vector). getformat() getmaxprint() getnnz([axis]) Get the count of explicitly-stored values (nonzeros) getrow(i) Returns a copy of row i of the matrix, as a (1 x n) sparse matrix (row vector). log1p() Element-wise log1p. max([axis]) Maximum of the elements of this matrix. maximum(other) mean([axis]) Average the matrix over the given axis. min([axis]) Minimum of the elements of this matrix. minimum(other) multiply(other) Point-wise multiplication by another matrix nonzero() nonzero indices power(n[, dtype]) This function performs element-wise power. rad2deg() Element-wise rad2deg. reshape(shape) rint() Element-wise rint. set_shape(shape) setdiag(values[, k]) Set diagonal or off-diagonal elements of the array. sign() Element-wise sign. sin() Element-wise sin. sinh() Element-wise sinh. sqrt() Element-wise sqrt. sum([axis]) Sum the matrix over the given axis. sum_duplicates() Eliminate duplicate matrix entries by adding them together tan() Element-wise tan. tanh() Element-wise tanh. toarray([order, out]) See the docstring for spmatrix.toarray. tobsr([blocksize]) tocoo([copy]) tocsc() Return a copy of this matrix in Compressed Sparse Column format tocsr() Return a copy of this matrix in Compressed Sparse Row format todense([order, out]) Return a dense matrix representation of this matrix. todia() todok() tolil() transpose([copy]) trunc() Element-wise trunc.
#### Previous topic
scipy.sparse.bsr_matrix.trunc
#### Next topic
scipy.sparse.coo_matrix.nnz | 1,303 | 4,588 | {"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} | 2.578125 | 3 | CC-MAIN-2015-40 | latest | en | 0.655791 |
https://www.solid-earth.net/7/817/2016/ | 1,580,007,204,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00148.warc.gz | 1,090,752,360 | 16,225 | Journal topic
Solid Earth, 7, 817–825, 2016
https://doi.org/10.5194/se-7-817-2016
Solid Earth, 7, 817–825, 2016
https://doi.org/10.5194/se-7-817-2016
Research article 24 May 2016
Research article | 24 May 2016
# Multi-quadric collocation model of horizontal crustal movement
Gang Chen1,2, Anmin Zeng3,4, Feng Ming3,4, and Yifan Jing3,4 Gang Chen et al.
• 1Faculty of Information Engineering, China University of Geosciences, Wuhan 430074, China
• 2National Engineering Research Center for Geographic Information System, Wuhan 430074, China
• 3Faculty of Geospatial Information, Information Engineering University, Zhengzhou 410052, China
• 4State Key Laboratory of Geographic Information Engineering, Xi'an 710054, China
Abstract. To establish the horizontal crustal movement velocity field of the Chinese mainland, a Hardy multi-quadric fitting model and collocation are usually used. However, the kernel function, nodes, and smoothing factor are difficult to determine in the Hardy function interpolation. Furthermore, the covariance function of the stochastic signal must be carefully constructed in the collocation model, which is not trivial. In this paper, a new combined estimation method for establishing the velocity field, based on collocation and multi-quadric equation interpolation, is presented. The crustal movement estimation simultaneously takes into consideration an Euler vector as the crustal movement trend and the local distortions as the stochastic signals, and a kernel function of the multi-quadric fitting model substitutes for the covariance function of collocation. The velocities of a set of 1070 reference stations were obtained from the Crustal Movement Observation Network of China, and the corresponding velocity field was established using the new combined estimation method. A total of 85 reference stations were used as checkpoints, and the precision in the north and east component was 1.25 and 0.80 mm yr−1, respectively. The result obtained by the new method corresponds with the collocation method and multi-quadric interpolation without requiring the covariance equation for the signals. | 482 | 2,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.866089 |
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# MS Finance at Univ of Houston
Author Message
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Joined: 13 Sep 2008
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MS Finance at Univ of Houston [#permalink]
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13 Sep 2008, 08:43
Can any one tell me how reputed is the MS finance course at Univ of Houston. Do they help with placements and how good are these.
I am also wondering if it is better to get an MBA with the Univ of Houston or an MS in Finance.
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06 Oct 2012, 00:13
I'm seeking for this information
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06 Oct 2012, 00:17
thank for posting !!!
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06 Oct 2012, 00:18
thank for posting !!!
Re: MS Finance at Univ of Houston [#permalink] 06 Oct 2012, 00:18
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Display posts from previous: Sort by | 631 | 2,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.838294 |
http://blog.oddhead.com/category/people/page/2/ | 1,620,765,288,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989856.11/warc/CC-MAIN-20210511184216-20210511214216-00038.warc.gz | 8,368,411 | 13,787 | # Where is the betting market for P=NP when you need it?
HP research scientist Vinay Deolalikar has constructed the most credible proof yet of the most important open question in computer science. If his proof is validated (and there are extremely confident skeptics as you’ll see) he proved that P≠NP, or loosely speaking that some of the most widespread computational problems — everything from finding a good layout of circuits on a chip to solving Sudoku puzzles to computing LMSR prices in a combinatorial market — cannot be solved efficiently. Most computer scientists believe that P≠NP, but after decades of some of the smartest people in the world trying, and despite the promise of worldwide accolades and a cool \$1 million from the Clay Mathematics Institute, no one has been able to prove it, until possibly now.
Scott Aaronson is a skeptic, to say the least. He made an amazing public bet to demonstrate his confidence. He pledged that if Deolalikar wins the \$1 million prize, Aaronson will top it off with \$200,000 of own money. Even more amazing: Aaronson made the bet without even reading the proof. [Update: I should have said “without reading the proof in detail”: see comments] (Perhaps more amazing still: a PC World journalist characterized Aaronson’s stance as “noncommittal” without a drip of sarcasm.) [Hat tip to Dan Reeves.]
As Aaronson explains:
The point is this: I really, really doubt that Deolalikar’s proof will stand. And while I haven’t studied his long, interesting paper and pinpointed the irreparable flaw… I have a way of stating my prediction that no reasonable person could hold against me: I’ve literally bet my house on it.
Aaronson is effectively offering infinite odds [Update: actually more like 2000/1 odds: see comments] that the question “P=NP?” will not be resolved in the near future. Kevin McCurley and Ron Fagin made a different (conditional) bet: Fagin offered 5/1 odds (at much lower stakes) that if the question is resolved in 2010, the answer will be P≠NP. Bill Gasarch says that he, like Aaronson, would bet that the proof is wrong… if only he were a betting man. Richard Lipton recounts a discussion about the odds of P=NP with Ken Steiglitz.
But beyond a few one-off bets and declarations, where is the central market where I can bet on P=NP? I don’t even necessarily want in on the action, I just want the odds. (Really!)
My first thought was the Foresight Exchange. It does list one related contract — Good 3SAT Algorithm by 2020 — which should presumably go to zero if Deolalikar’s proof is correct. It hasn’t budged much, consistent with skepticism (or with apathy). My second thought was the PopSci Predictions Exchange (PPX), though sadly it has retired. InklingMarkets has a poll about whether P=NP will be resolved before the other Clay Institute prize questions, but not about how it will be resolved or the odds of it happening. (The poll is one of several markets sponsored by the Woodrow Wilson Center’s Science and Technology Innovation Program — hat tip to Vince Conitzer.) I don’t see anything at longbets, and anyway longbets doesn’t provide odds despite it’s name.
In 1990 Robin Hanson provocatively asked: Could gambling save science?. That question and his thoughtful answers inspired a number of people, including me, to study prediction markets. Indeed, the Foresight Exchange was built largely in his image. P=NP seems one of the most natural claims for any scitech prediction market.
All these years later, when I really need my fix, I can’t seem to get it!
2010/08/14 Update: Smarkets comes the closest: they have real-money betting on whether P=NP will be resolved before the other Clay Institute prize questions. They report a 53% chance as of 2010/08/14 (for the record, I would bet against that). What’s missing is when the award might happen and how the question might be resolved, P=NP or P≠NP. I also don’t see a graph to check whether Deolalikar’s proof had any effect.
If it wasn’t clear in my original post, I found Aaronson’s bet incredibly useful and I am thrilled he did it. I believe he should be commended: his bet was exactly what more scientists should do. Scientists should express their opinion, and betting is a clear, credible, and quantitative way to express it. It would be as shame if some of the negative reactions caused him or others not to make similar bets in the future.
I just wish there were a central place to make bets on scientific claims and follow the odds in the vision of Robin Hanson, rather than every scientist having to declare their bet on their own individual blogs.
# The most prescient footnote ever
Back in 2004, who could have imagined Apple’s astonishing rise to overtake Microsoft as the most valuable tech company in the world?
At least one person.
Paul Graham wins the award for the most prescient parenthetical statement inside a footnote ever.
In footnote 14 of Chapter 5 (p. 228) of Graham’s classic Hackers and Painters, published in 2004, Graham asks “If the the Mac was so great why did it lose?”. His explanation ends with this caveat, in parentheses:
And it hasn’t lost yet. If Apple were to grow the iPod into a cell phone with a web browser, Microsoft would be in big trouble.
Wow. Count me impressed.
To find this quote, search inside the book for “ipod cell”.
To get into a 2004 midset, look here and here.
# 2 weeks, 2 geeks: My two new fearless leaders
Well, geeks are certainly inheriting my earth.
On January 13, my company named Carol Bartz, a self-avowed math nerd and former punch-card carrying member of her college computer club, as its CEO. In her own words:
I was a real nerd. I love, love, love, love math. Back in the late ’60s, math meant being a teacher if you were a woman. I wasn’t interested in teaching. Then I took my first computer course. It was crazy. It was like math, only more fun. I switched to computer science.
Exactly one week later, on January 20, my country turned over executive control to Barack Obama, a CrackBerry addicted comic book geek. In his inauguration speech, Obama vowed to “restore science to its rightful place”, “wield technology’s wonders”, and even addressed “non-believers” — wording that in any sane universe should be entirely unremarkable, yet in ours appears to represent an unprecedented milestone.
I can’t recall a two-week span filled with so much geek pride and cautious optimism.
Back to the Carol Bartz quote. Reading it brings a smile to my face. It also reminds me of my mom, who, convinced it was her only option, taught middle school for a few years before returning to medical school to pursue her passion, enjoying a successful career as one of the first women radiologists.
I highly recommend Bartz’s essay, which mixes biography with prescience and insight. Bartz describes how technology and the Internet are transforming collaboration and improving productivity, at the same time ushering in an era of information overload, email bankruptcy, and misuse of the extra time technology affords. Remarkably, she wrote about these things in 1997!
It’s amazing to think how things have changed since 1997. My own first web experience, courtesy Mosaic, came in 1994, the same year Yahoo! was founded. In 1996, PayPal predecessor and public company First Virtual wrote their own keystroke-sniffing malware as a stunt to bolster their urgent call to “NEVER TYPE YOUR CREDIT CARD NUMBER INTO A COMPUTER”. Ebay was founded in 1995, PayPal in 1998. In 1997, Friendster had neither come nor gone, and Facebook CEO Mark Zuckerberg was 13.
Yet Bartz’s words seem more relevant than ever today.
# Death in artificial intelligence
Until just reading about it in Wired, I knew little1 of the apparent suicide of Push Singh, a rising star in the field of artificial intelligence.
Singh seemed to have everything going for him: brilliant and driven, he became the protégé of his childhood hero Marvin Minsky, eventually earning a faculty position alongside him at MIT. Professionally, Singh earned praise from everyone from IEEE Intelligent Systems, who named Singh one of AI’s Ten to Watch (apparently revised), to Bill Gates, who asked Singh to keep him appraised of his latest publications. Singh’s social life seemed healthy and happy. The article struggles to uncover a hint of why Singh would take his own life, mentioning his excruciating chronic back pain (and linking it to a passage on the evolutionary explanation of pain as “programming bug” in Minsky’s new book, a book partly inspired by Singh).
The article weaves Push’s story with the remarkable parallel life and death of Chris McKinstry, a man with similar lofty goals of solving general AI, and even a similar approach of eliciting common sense facts from the public. (McKinstry’s Mindpixel predated Singh’s OpenMind initiative.) McKinstry’s path was less socially revered, and he seemed on a never ending and aching quest for credibility. The article muses whether there might be some direct or indirect correlation between the eerily similar suicides of the two men, even down to their methods.
For me, the story felt especially poignant, as growing up I was nourished on nearly the same computer geek diet as Singh: Vic 20, Apple II, Star Trek, D&D, HAL 9000, etc. In Singh I saw a smarter and more determined version of myself. Like many, I dreamt of solved AI, and of solving AI, even at one point wondering if a neural network trained on yes/no questions might suffice, the framework proposed by McKinstry. My Ph.D. is in artificial intelligence, though like most AI researchers my work is far removed from the quest for general AI. Over the years, I’ve become at once disillusioned with the dream2 and, hypocritically, upset that so many in the field have abandoned the dream in pursuit of a fractured set of niche problems with questionable relevance to whole.
Increasingly, researchers are calling for a return to the grand challenge of general AI. It’s sad that Singh, one of the few people with a legitimate shot at leading the way, is now gone.
Push Singh Memorial Fund
1Apparently details about Singh’s death have been slow to emerge, with MIT staying mostly quiet, for example not discussing the cause of death and taking down a memorial wiki built for Singh. 1 My colleague Fei Sha, a new father, put it nicely, saying he is “constantly amazed by the abilities of children to learn and adapt and is losing bit by bit his confidence in the romantic notion of artificial intelligence”. | 2,325 | 10,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-21 | latest | en | 0.955938 |
https://www.nagwa.com/en/worksheets/734141960145/ | 1,550,928,035,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249501174.94/warc/CC-MAIN-20190223122420-20190223144420-00177.warc.gz | 913,130,134 | 11,627 | # Worksheet: Derivatives of Polar Functions
Q1:
Consider the polar equation .
Calculate for .
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Find the slope of the tangent to when . Give your answer accurate to three significant figures.
Q2:
Consider the polar equation . We can calculate the derivative by dividing the derivative by the derivative .
To calculate the derivative , we first need to introduce the variable by multiplying both sides of the equation by and then substituting. Write this equation in terms of .
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Calculate the derivative .
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Similarly, to calculate the derivative , we first need to introduce the variable by multiplying both sides of the original equation by and then substituting. Write this equation in terms of .
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• D
• E
Calculate the derivative .
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The derivative is equal to . Calculate .
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Use the derivative function to calculate the slope of the tangent to at .
• A
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• D
• E | 262 | 994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-09 | latest | en | 0.843569 |
https://en.neurochispas.com/trigonometry/cylindrical-coordinates-formulas-and-diagrams/ | 1,685,813,521,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00059.warc.gz | 265,539,064 | 28,764 | # Cylindrical coordinates – Formulas and diagrams
Coordinate systems can be defined as ways of locating points in space. In three-dimensional space, the Cartesian coordinate system has the form (x, y, z). However, this system is not always the most convenient, so we have alternative coordinate systems. One of these systems is the cylindrical coordinate system. The cylindrical coordinates are considered as an extension of the polar coordinates towards the third dimension.
The general form of the cylindrical coordinates is (r, θ, z), where, r is the distance from the origin to the point in the xy plane, θ is the angle formed with respect to the x-axis, and z is the same z component as in Cartesian coordinates.
##### TRIGONOMETRY
Relevant for
Learning about cylindrical coordinates with examples.
See definitions
##### TRIGONOMETRY
Relevant for
Learning about cylindrical coordinates with examples.
See definitions
## What are cylindrical coordinates?
Cylindrical coordinates are defined as an alternate three-dimensional coordinate system to the Cartesian system. Cylindrical coordinates are written in the form (r, θ, z), where, r represents the distance from the origin to the point in the xy plane, θ represents the angle formed with respect to the x-axis and z is the z component, which is the same as in Cartesian coordinates.
This coordinate system is used mainly to graph cylindrical-shaped figures such as tubes or tanks. This is mainly convenient in calculus since depending on the given equations, finding their derivatives or integrals can be easier.
## Examples of cylindrical coordinates
To locate a point in cylindrical coordinates, we start by locating it in the xy plane by measuring the distance from the origin and measuring the angle from the x-axis. Then, we add the z component. In the following graph, we can look at the point $latex (3, \frac{\pi}{3}, 4)$.
A characteristic of cylindrical coordinates is that we can describe a point using several coordinates. That is, there is an infinite number of coordinates for each point. This is because the angle θ can be written in different ways.
If we add or subtract 2π, we get an equivalent angle. For example, the angles $latex \frac{\pi}{2}, \frac{5 \pi}{2}$, and $latex – \frac{3 \pi}{2}$ are the same.
## Cylindrical to Cartesian conversion formulas
We use the following diagram to derive the conversion formulas from cylindrical coordinates to Cartesian coordinates:
We can see that the z coordinate is the same in both systems. Therefore, we just have to find the values of x and y in terms of r and θ. We can use the cosine function to find the x component and the sine function to find the y component. Therefore, we have:
#### EXAMPLE 1
If we have the cylindrical coordinates $latex (3, \frac{\pi}{4}, 4)$, what is their equivalent in Cartesian coordinates?
We can recognize the values $latex r=3, ~\theta=\frac{\pi}{4}$. We use the formulas seen above with these values:
$latex x=3 \cos(\frac{\pi}{4})$
$latex x=2.1$
$latex y=3 \sin(\frac{\pi}{4})$
$latex y=2.1$
The Cartesian coordinates are (2.1, 2.1, 4).
#### EXAMPLE 2
We have the cylindrical coordinates $latex (10, \frac{7 \pi}{4}, 5)$. What is its equivalent in Cartesian coordinates?
We have the values $latex r=10, ~\theta=\frac{7\pi}{4}$. Using these values, we find the values of x and y:
$latex x=10 \cos(\frac{7\pi}{4})$
$latex x=7.07$
$latex y=10 \sin(\frac{7\pi}{4})$
$latex y=-7.07$
The Cartesian coordinates are (7.07, -7.07, 5).
## Cylindrical to Polar conversion Formulas
To derive the conversion formulas from Cartesian to cylindrical coordinates, we are going to use the same diagram:
The value of r is found using the Pythagorean theorem with the x and y components. So, we have:
$latex {{r}^2}={{x}^2}+{{y}^2}$
The angle θ can be found using the inverse tangent. We know that the tangent is equal to the opposite side divided by the adjacent side. In the diagram, we see that the opposite side is y and the adjacent side is x. Therefore, we have:
Something to keep in mind with this angle is that sometimes the value given by the calculator is wrong. This is because the inverse tangent outputs with a range from $latex – \frac {\pi}{2}$ to $latex \frac{\pi}{2}$ and this does not cover all four quadrants.
We can correct this by adding 180° or π when the point is in the second and third quadrants and adding 360° or 2π when the point is in the fourth quadrant. When the point is in the first quadrant, the value given by the calculator is correct.
#### EXAMPLE 1
The point (4, 6, 7) is in Cartesian coordinates. What is its equivalent in cylindrical coordinates?
We have the values $latex x=4,~y=6$. We can find r and θ using the formulas given above. Therefore, the value of r is:
$latex r=\sqrt{{{x}^2}+{{y}^2}}$
$latex r=\sqrt{{{4}^2}+{{6}^2}}$
$latex r=\sqrt{16+36}$
$latex r=\sqrt{52}$
$latex r=7.2$
The angle θ is:
$latex \theta={{\tan}^{-1}}(\frac{y}{x})$
$latex \theta={{\tan}^{-1}}(\frac{6}{4})$
$latex \theta=0.98$ rad
Both the x component and the y component are positive, so the point is in the first quadrant and the angle obtained is correct.
The coordinates of the point are (7.2, 0.98 rad, 7).
#### EXAMPLE 2
We have the point (-3, -6, 5) in Cartesian coordinates. What is its equivalence in cylindrical coordinates?
We have $latex x=-3,~y=-6$. We use the formulas above to find r and θ:
$latex r=\sqrt{{{x}^2}+{{y}^2}}$
$latex r=\sqrt{{{(-3)}^2}+{{(-6)}^2}}$
$latex r=\sqrt{9+36}$
$latex r=\sqrt{47}$
$latex r=6.9$
The angle θ is:
$latex \theta={{\tan}^{-1}}(\frac{y}{x})$
$latex \theta={{\tan}^{-1}}(\frac{-6}{-3})$
$latex \theta=1.11$ rad
Both the x component and the y component are negative, so the point is in the third quadrant and we have to add π to get the correct angle. Therefore, the correct angle is $latex \theta = 1.11 + \pi = 4.25$ rad.
The coordinates of the point are (6.9, 4.25 rad, 5).
## See also
Interested in learning more about cylindrical coordinates? Take a look at these pages:
### Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
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3年以上 前 | 2
Plot the trapeoidal waveform using given equations
syms theta y=piecewise(... 0<=theta<2*pi/3,1,... 2*pi/3<=theta<pi,(1-6/pi*(theta-2*pi/3)),... pi<=theta<5*pi/3,-...
4年弱 前 | 0
How to draw an ellipse | 1,729 | 5,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.69705 |
https://sportskeptic.wordpress.com/2011/06/30/retrodiction-contest-part-3-the-conclusions/ | 1,539,950,415,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512395.23/warc/CC-MAIN-20181019103957-20181019125457-00553.warc.gz | 799,852,840 | 29,294 | ## Retrodiction Contest Part 3: The Conclusions
If you haven’t already, make sure you check out part 1 for the methods and part 2 for the results. In this post I’ll talk about what the results for the retrodiction contest might mean and some caveats.
One clear result is that APM is not especially accurate in evaluating players. It could be argued, as I suggested in part 2, that part of the poor performance is my assumption of replacement level productivity. But, as I said in part 2, given that these players are not on the court much, it doesn’t seem like that should lead to the big drop in predictive power.
Following that idea, it is clear that the regularization procedure improves on APM. That isn’t surprising, given that Joe Sill demonstrated with another analysis that RAPM is clearly better. But it appears to help enough to vault the non-boxscore measure past WP. With only one season of ezPM to compare to, it’s unclear if RAPM is clearly better than that boxscore metric.
Why might RAPM do so well? The main feature of RAPM is to move all players closer to average. In 2010, RAPM thought that about 55% of players were between -1 and 1 points per 100 possessions. On the other hand, ezPM100 has a much larger range of points per 100 possession values and only placed about 26% of players in the -1 to 1 range in 2010, and it did well. Assuming I converted correctly, WP places 20% of players in an equivalent productivity range (again using 2010 so as to compare the metrics on the same data). So simply assuming that many players are average is not sufficient since ezPM did pretty well and WP isn’t too far off. Given that RAPM is a black box, it’s hard to say what exactly it’s picking up on.
In contrast, ezPM has a boxscore formula, so we could try in theory to look at why it did well in its one season. I’m sure that many APBRers will want to suggest that rebounds are the main issue. However, ezPM differs from WP virtually across the board in how it weighs different boxscore stats. It also uses more stats than WP since it’s based on play-by-play; it knows if a shot has been assisted or not, if a free throw was an and-one, etc. Additionally, it incorporates counterpart defense from play-by-play data, so it arguably has a richer picture of player defense than WP uses. It would be interesting to see how Arturo or Ty’s version of WP with counterpart defense would fare in this test. In short, ezPM uses much more information than WP does, and it is hard to say which part of that lead to its better performance in retrodicting 2011.
A different kind of explanation was suggested by Guy in the comments to one of my previous retrodiction posts. Wins Produced, as the name suggests, attempts to predict team wins. But as you probably know, point differential is more predictive of team quality and future performance. Predicting wins adds a certain amount of random noise. RAPM, APM, and ezPM, on the other hand, are inherently predicting point differential. Thus they may get a certain advantage by predicting point differential instead of going through the conversion to wins.
One last thought has to do with the change in accuracy when metrics were allowed to use actual rookie performance as opposed to giving each rookie an average rookie productivity. RAPM tended to do worse with the actual values (as did ezPM to a small degree in the one season available) while APM and WP got better. This is an intriguing result. One might conclude that APM and WP tend to overfit the data; they do better at predicting outcomes when known data (e.g. actual performance from that season) is included; that idea has at least been proposed for APM’s failures in other tests. Another possible conclusion is the inverse for RAPM; it has good predictive power at the cost of explaining current results. Within a season, rookies are valued alongside veterans that are assumed to be close to average. This may lead to players being slightly misvalued if most players are not in fact close to average, and thus RAPM does worse when using its own rookie values. But across seasons, the assumption of being average serves as adding a regression to the mean component which benefits prediction. Given the results of the contest it would appear that the regression to the mean outweighs the potentially mistaken values, but it is something to think about.
In terms of my own feelings, I’m surprised that RAPM did so well. I’ll certainly give more weight to the method than I did in the past, although I’ll continue to wish that it weren’t a black box. I’m not surprised that APM did so poorly, although it was nice to see it spelled out in the numbers. Hopefully we can never speak of it again, although since it seems to keep coming up on ESPN I doubt that will be the case. And more broadly, I hope that this kind of analysis catches on as a fair method of comparing metrics. When ten people use ten different sets of assumptions as inputs into their predictions, it’s hard to tell why one person ‘wins’ and another person ‘loses’. But when the predictions are made as similarly as possible, then it becomes easier to tell which metrics are honing in on true player value. I wouldn’t consider this the final word, but a somewhat thorough first good attempt.
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### 17 Responses to Retrodiction Contest Part 3: The Conclusions
1. Guy says:
Excellent work, Alex. And a fair-minded writeup. A few miscellaneous thoughts:
It’s not surprising that ezPM outperforms WP. The two are similar in many ways, but ezPM fixes some of the most obvious errors in WP. I don’t think the valuations of the boxscore stats in the two metrics differ as much as you suggest — in most cases they seem pretty similar.
What’s most interesting IMO is the strong performance of RAPM. One advantage of RAPM (and ezPM to a lesser extent) is capturing defensive contributions more than WP can, as you suggest. Another potentially important difference is that RAPM works hard to isolate a players’ impact from the performance of teammates. Ironically, WP fans have long criticized PM metrics for being biased by other players’ performance — which can indeed be a problem — but have downplayed the same danger in WP because of the unspoken assumption that players completely “own” their boxscore stats. But Berri’s own estimate of diminishing returns means that each extra win a player delivers is partially offset by a loss of about .3 wins from other players. We can argue about whether that’s really a function of reducing teammates’ productivity vs. a flaw in the metric that misallocates value, but the result is the same: WP will overestimate a good player’s actual win contribution and exaggerates the damage done by weak players (another reason to think WP’s implied spread of talent is too wide). It appears that RAPM is doing a better job of isolating an individual player’s true impact.
That would also explain why using actual rookie performance improves WP so much. Because a significant part of WP is other players’ work, using the rookie’s actual WP is giving the metric an artificial boost. “Actual WP” and “actual rookie performance” are not necessarily the same thing — if they were, you wouldn’t need to to this test! The fact that RAPM actually loses accuracy when using actual rookie ratings is an interesting puzzle. I’ll have to give that some more thought.
I’ll repeat two old suggestions, in case they interest you:
1) run this analysis on the top 50% of teams in term of personnel change from prior year, to see if the differences in predictions are more pronounced or any different.
2) give us a “stupid model” benchmark to compare these all to. How well can you predict using only players’ points per minute? Or just prior season MP? How much do these metrics improve on a very simple model like that?
2. Guy says:
I also think your findings on the implied spread of talent are interesting. My own guess is that the reality is somewhere between the very narrow RAPM spread and the much wider spread in ezPM and (especially) WP. After all, we know that the spread among entire teams isn’t that large — I think the SD is around 4 or 4.5 points. That would seem to suggest a SD at the player level of maybe about 2.0 points, or about 38% of players between -1 and +1. (And this is complicated by the fact that player talent is presumably more of a pyramid than a normal distribution.)
In any case, how much variance there is probably matters less than whether a metric actually measures useful contributions to winning. But I do think that the very large spread in the other two metrics is another sign that they are including outcomes not solely due to the contribution of the player being measured. And this seems to be more true of WP than ezPM (and you can probably guess why I think that’s the case).
• Alex says:
My other reply has some range info. But from 2006-2011, the SD in team point differentials is 4.55. Why does that suggest a player SD of 2?
• Guy says:
The team variance will be the sum of the player variances, if you assume teams are constructed such that players are reasonably independent. So if the player SD were 2, then at the team level the SD would be sqrt (5 * 2^2) = 4.5. This is roughly what we see in baseball: team variance in runs scored is about 9X the player variance. Now, it may not be strictly true in basketball. I suppose the salary cap, for example, might artificially limit the disparity between teams. But I think it gives you a decent ballpark estimate of what a plausible spread of talent is. It seems implausible, for example, that the player SD could be as high as 3 (which would produce a team sd of about 6.7 points).
BTW, this same logic is one reason we can be sure that players with high rebound totals “take” rebounds from teammates. The player SD for rebounds is as large or larger than the team SD. If there wasn’t huge negative correlation among teammates, then the variation among teams in rebounding would be vastly greater than it is.
• Alex says:
That would seem to make a lot of assumptions besides independent players. Doesn’t it also assume that the players all get equal minutes? Also, doesn’t adding up the players on a team give you the expected standard deviation for that team across games, not across teams?
And not to whip a dead horse, but the same logic would also imply we should severely curb assists as well.
• Guy says:
No, it gives you the SD for teams that you would get if you randomly assigned all NBA players to teams. If you think it’s obvious that existing teams have more variance (or less) than you’d get with random assignment, then you’d need to adjust for that (but it seems like a decent rough approximation to me).
But yes, I was using a very simple model, in which teams have 5 full time players. So really what it says is the SD for productivity teams get from each position — not each player — is around 2 points. You could obviously develop a more complex model. If you thought that the talent spread for bench players was rather narrow, for example, then that could mean the spread for starters is a bit larger. So maybe the SD for starters could be 2.5? Anyway, I was just suggesting this as a rough sanity check for the metrics. Would it be easy for you to report the SD for each metric for players with non-trivial numpber of MP?
And yes, the same logic could apply to assists. Clearly, some players (mostly PGs) are given vastly more opportunities to record assists than other players. That probably does mean you need to evaluate PGs (especially) in comparison to what a replacement level PG would give you, rather than just assigning X value to all assists. A parallel situation is the pitcher in baseball. He runs the defense like a PG runs the offense, but with even more control. Some defensive accomplishments can be done ONLY by the pitcher (e.g. a strikeout). And even the worst pitcher will get some strikeouts (as I would guess the worst PG will get some assists). So we don’t compare a pitcher to other fielders in those categories — we compare them only to a replacement level pitcher. I think Evan would agree that valuing assists properly is one of the bigger challenges in constructing these kind of metrics.
• Alex says:
2010 is the only year I have overlapping data for each method. If I use a 1000-minute cutoff, that leaves 243 players. The SD in points per 100 possessions for WP (converted from wins), ezPM, RAPM, and APM are 3.32, 2.59, 1.83, and 4.67. The ranges are [-8.69, 11.9], [-6.78, 11.49], [-4, 6.7], and [-13.47, 18.64]. The standard deviations don’t change much if I move the cutoff down to 800 minutes, which adds 38 players.
• Guy says:
Thanks for the data. To me, the SD for APM is clearly too large, and I think WP is too large as well. It’s hard to see how the player SD can be nearly as large as the team SD, unless you believe in very large diminishing returns. And even if you do, then it seems to me you need to incorporate that into your estimate of player values. What’ s the point of saying “player X added 12 wins,” if you know that he also took 4 wins away from other players? In fact, that player is going to add 8 wins to your team. RAPM looks like it may regress players too much, but hard to say for sure. All of that said, measuring production correctly is more important than getting the right SD — figuring out the right regression for forecasting is relatively easy once you know how to properly measure productivity.
Have you looked at the distribution of ratings within any of the metrics? What I think it should show is a skewed distribution in which a majority of players have negative values. I’d expect a lot of guys to be between, say, -2 and zero. And I’d expect to see a few +4 and +5 SD players, but less extreme values on the negative side.
• Alex says:
I’ve looked at the plots for 2010, since it’s the only year where they all overlap (presumably talent distribution is fairly stable over years, but who knows). RAPM only places 54% of players below average; ezPM and WP are more like 63%. APM places 69%, but part of that is due to the replacement player level. If you take out my assumed level of performance, 60% of the rest of the group is below average. RAPM places 45% of players between -2 and 0 while the number is 25% for ezPM and 21% for WP and APM. RAPM and APM have longer positive tails while WP has a longer negative tail and ezPM is fairly symmetric and very wide. Neither RAPM nor WP place anyone over 4.5 SD; ezPM goes as high as 11.9. APM is closer to RAPM and WP, particularly if you take out replacement players. RAPM and APM put the worst player at about -2.7 SD while it’s -8 for WP and -12.5 for ezPM. So I guess RAPM sounds closest to what you describe, but I’m still pretty curious about its failure to describe the current season as well as any of the other metrics.
3. EvanZ says:
Great stuff! Sounds like you’re starting to come around on the RAPM, which I appreciate.
I can’t remember now if I suggested you take a look at LambdaPM which is essentially like a blended boxscore-APM metric that tries to optimize both. There’s a thread on APBR, but unfortunately, only 2011 data are available so far.
oh, of course, also happy that ezPM beat WP 😛
• Alex says:
I did see your comment about lambda. I can’t use any metric with only one season, especially since it would have to predict a season that may not be played 😦
4. EvanZ says:
“I’ll certainly give more weight to the method than I did in the past, although I’ll continue to wish that it weren’t a black box.”
One thing that still isn’t clear to me is why you keep insisting that RAPM is a “black box”. I know you understand the basics of the method. It’s been published and made publicly available. It’s no more a black box than WP, which also uses regression to determine the weights of various box score categories. Why is that any less of a black box?
• Alex says:
It’s a black box in that I can’t tell why someone gets the rating they do. As I’ve said in the past, even if you don’t like WP (or ezPM, or Win Shares, or PER…) you could adjust it if you wanted to. If someone thinks that Kevin Love is overvalued because of his rebounds, you can discount it. But if RAPM likes Kevin Love, I have no idea why. Does he set good picks? Does he spread the floor with his inside-outside game? Is it because he rebounds a lot? I have no idea. Someone could create a ‘statistical RAPM’, where they predict RAPM from boxscore or other measures, but I’m guessing that it wouldn’t have great predictive power if APM is any indication. So even if you assume that RAPM is “correct”, you would still be limited in using it for team construction because you don’t know what a player does to get his rating or how two players would overlap or complement each other in their abilities. I guess you could argue that WP is a black box in terms of determining why a point/possession is worth .033, but beyond that it’s transparent and player ratings are obvious as well. I know exactly why Kevin Love gets the value he does for any boxscore-based metric.
• EvanZ says:
“Someone could create a ‘statistical RAPM’, where they predict RAPM from boxscore or other measures, but I’m guessing that it wouldn’t have great predictive power if APM is any indication.”
Daniel’s ASPM does exactly this, although with older APM data, I believe. In fact, his ratings are currently leading on the APBR retrodiction contest, beating RAPM and ezPM.
• Alex says:
I know of Daniel’s model, although I don’t think I’ve ever seen him explain how he picked the values for his ratings. It’s a very complicated model, to say the least. I also haven’t seen retrodiction or current season ‘prediction’ for anything beyond this season, although I expect his current season predictions are fine since he uses a team adjustment. | 4,110 | 18,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-43 | latest | en | 0.954745 |
https://learnmech.com/projection-method-first-and-third-angle/ | 1,725,706,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00698.warc.gz | 334,608,713 | 21,406 | # Projection Method | First And Third Angle Projection | Difference
## Projection Methods:
Projection Methods are used to show drawing on a paper . Projection Methods has mainly two types used in Engineering Drawing :
1) First Angle Projection Method
2) Third Angle Projection Method
The principal projection planes and quadrants used to create drawings are shown in figure. The object can be considered to be in any of the four quadrant.
## First Angle Projection
In this the object in assumed to be positioned in the first quadrant and is shown in figure.
The object is assumed to be positioned in between the projection planes and the observer. The views are obtained by projecting the images on the respective planes. Note that the right hand side view is projected on the plane placed at the left of the object. After projecting on to the respective planes, the bottom plane and left plane is unfolded on to the front view plane. i.e. the left plane is unfolded towards the left side to obtain the Right hand side view on the left side of the Front view and aligned with the Front view. The bottom plane is unfolded towards the bottom to obtain the Top view below the Front view and aligned with the Front View.
## Third Angle Projection
In the third angle projection method, the object is assumed to be in the third quadrant. i.e. the object behind vertical plane and below the horizontal plane. In this projection technique, Placing the object in the third quadrant puts the projection planes between the viewer and the object and is shown in figure
## Difference between first- and third-angle projections
Either first angle projection or third angle projection are used for engineering drawing. Second angle projection and fourth angle projections are not used since the drawing becomes complicated.
## Symbol of projection
As per BIS standards, Drawing symbols for 1st angle and 3rd angle are shown in below figure. as previously explained that in case of first angle method the left hand side view is shown on right side of front view and vice verse.The symbol recommended by BIS is to draw the two sides of a frustum of a cone placed with its axis horizontal The left view is drawn.
###### Mechanical Subjectwise Basic Concept Notes ,Articles
Sachin Thorat
Sachin is a B-TECH graduate in Mechanical Engineering from a reputed Engineering college. Currently, he is working in the sheet metal industry as a designer. Additionally, he has interested in Product Design, Animation, and Project design. He also likes to write articles related to the mechanical engineering field and tries to motivate other mechanical engineering students by his innovative project ideas, design, models and videos.
## 3 thoughts on “Projection Method | First And Third Angle Projection | Difference”
1. Chandru77 says:
thank you
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 574 | 2,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-38 | latest | en | 0.912423 |
https://gomathanswerkey.com/mcgraw-hill-math-grade-1-chapter-11-test-answer-key/ | 1,716,987,614,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00191.warc.gz | 233,450,590 | 53,199 | All the solutions provided in McGraw Hill Math Grade 1 Answer Key PDF Chapter 11 Test as per the latest syllabus guidelines.
Question 1.
Circle the name of the shape.
circle
rectangle
triangle
Rectangle
Explanation:
A rectangle has 4 sides and 4 corners
Opposite sides are equal.
Question 2.
Circle the name of the solid figure.
cube
cone
cylinder
Cube
Explanation:
A cube has 6 faces and 8 vertices
A faces are in same size.
Question 3.
Color 3 of the 4 equal parts.
Explanation:
The rectangle is divided into 4 equal parts
3 out of 4 is colored.
Question 4.
How many equal parts does the shape have?
Is the shape divided into halves or fourths?
Explanation:
The circle is divided in to 2 equal parts.
Line of symmetry a line divides the shapes into 2 equal halves.
Question 5.
Find the shape that matches. Circle it.
Explanation:
The given shape is a triangle so, circled the triangle.
Question 6.
Color the squares.
Explanation:
In the given shapes there are 2 squares
so, colored the squares.
Question 7.
What shape do you see? Circle the name.
rectangle
square
circle
triangle
Square
Explanation:
The face of the cube is a square, a cube has 6 faces.
All the faces are small sizes.
Question 8.
Look at the solid figure. Circle the object that has the same shape.
Explanation:
The given shape is a cylinder.
so, circled the cylinder shape.
Question 9.
Look at the two red shapes. What new shape can you make? Circle the name of the new shape.
square
triangle
cone
circle
Square
Explanation:
The given shapes are rectangles if we join the rectangles it forms a square.
Question 10.
Color 3 of the 4 equal parts.
Color 1 of the 2 equal parts.
Circle the rectangle with the smaller parts. Tell why the parts are smaller. | 422 | 1,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-22 | latest | en | 0.882869 |
http://www.talisman.org/~erlkonig/misc/cs345-prog-languages/schintro-v13/schintro_95.html | 1,508,465,717,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823605.33/warc/CC-MAIN-20171020010834-20171020030834-00279.warc.gz | 564,765,618 | 2,258 | ### Using Type Predicates
We can easily write a procedure `pair-tree-sum` to give us the sum of a binary tree of integers, whose interior nodes are pairs and whose leaves are the integers.
[ blah blah blah... rewrite some of the following, simplifying to only handle trees, not proper lists. ]
Our notion of a "pair tree" is a binary tree of pairs. Here we're doing something a little strange, because in general we're improper lists. We'll regard the `car` and `cdr` fields of a pair as the "left child" and "right child" fields of a tree node. A proper list wouldn't be a pair tree, because it the last pair in the list would point to the empty list object, not a number.
(Later, I'll show a record facility that allows us to build "tree node" records that are not pairs. That's nicer, because it doesn't confuse pairs' roles in improper lists with their roles in trees. For now, we'll stick with pairs, because the point of this example is recursion, not the details of records.)
Just as we did for proper lists, we start by characterizing this data structure recursively. We'll consider any subtree of a pair-tree to be a pair-tree. This includes the leaves, e.g., the numbers in a tree of numbers. (This is analogous to the way we considered the empty list to be a kind of list in the recursive characterization of lists.)
A pair tree is either
• a leaf (not a pair), or
• a pair, whose car and cdr values are pair-trees.
Our recursive summing procedure will have to deal with these two cases:
• a numbers, i.e., leaves of a tree of numbers, and
• pairs, in which case it should sum the left and right subtrees, and add those sums together.
The first case is the base case for the recursion. The sum of a leaf is the numeric of that leaf.
The second case is the recursive case, where we have a subtree to sum.
```Scheme>(define (pair-tree-sum pair-tree)
(cond ((number? pair-tree)
pair-tree)
(else
(+ (pair-tree-sum (car pair-tree))
(pair-tree-sum (cdr pair-tree))))))
```
Try this out, and make sure you understand why it works.
```Scheme>(pair-tree-sum 1)
1
Scheme>(pair-tree-sum '(1 . 2))
3
Scheme>(pair-tree-sum '((40 . 30) . (20 . 10)))
100
```
Notice how simple `pair-tree-sum` is, and how it depends on getting the base case for the recursion right. If we hadn't considered the leaves to be pair-trees in their own right, it would have gotten much uglier. For example, if we'd "bottomed out" at pairs whose left and right children weren't both pairs, we'd have had more cases to deal with--cases where one child is a leaf but the other's not.
Add `display` and `newline` expressions at the beginning of `pair-tree-sum`, as we did for `list-sum`, and try it out again. Be sure you understand the output in terms of the recursive call pattern. | 668 | 2,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-43 | latest | en | 0.921833 |
http://www.beyondtheboxscore.com/2010/11/22/1829554/can-someone-explain-the-bill-james-projection-of-scott-baker | 1,409,173,313,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829839.93/warc/CC-MAIN-20140820021349-00404-ip-10-180-136-8.ec2.internal.warc.gz | 275,994,146 | 22,954 | ## Can Someone Explain the Bill James Projection of Scott Baker?
I am doing some fantasy previews and got to Scott Baker and looked at his stats at Fangraphs which currently includes Bill James's projections. Bill's projections have always been optimistic compared to other projection systems, but Scott's was low and I just can't follow it. Take a look:
Season Team W L SV G GS IP K/9 BB/9 HR/9 BABIP ERA FIP 2005 Twins 3 3 0 10 9 53.2 5.37 2.35 0.84 0.265 3.35 3.82 2006 Twins 5 8 0 16 16 83.1 6.7 1.73 1.84 0.359 6.37 4.99 2007 Twins 9 9 0 24 23 143.2 6.39 1.82 0.94 0.327 4.26 3.89 2008 Twins 11 4 0 28 28 172.1 7.36 2.19 1.04 0.290 3.45 3.79 2009 Twins 15 9 0 33 33 200 7.29 2.16 1.26 0.287 4.37 4.08 2010 Twins 12 9 0 29 29 170.1 7.82 2.27 1.22 0.329 4.49 3.96 2011 Bill James 9 8 0 26 26 152 7.16 2.19 1.18 0.311 4.03 4.01
Problem 1: Why does he have him starting 29 games in 2011, at least 2 less than any of the past 3 years. This tough does match with the drop in IP of 18 from 2010 to 2011. Maybe he is looking at other Twin pitchers getting some starts. Definitely no kind of standard weighting used in other systems is being used here.
Problem 2. The K/9 drop is quite rough. It is lower than any of the previous 3 seasons. He must be factor in a pretty heavy aging factor.
Problem 3. ERA of 4.03. Some his ERA is going to be lower than anytime in the past 2 years, but his K/9 is going down. Also with a 4.03 ERA, you would think the Twins will play him more.
I just can't make any sense of this at all. He is is supposed to allow a half a run less than 2010, by having his K/9 drop by 0.76. With his ERA dropping, the Twins also decide to play him less.
I just don't get it.
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We have our own Community Guidelines at Beyond the Box Score. You should read them. | 816 | 2,584 | {"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} | 2.625 | 3 | CC-MAIN-2014-35 | latest | en | 0.933921 |
https://brilliant.org/discussions/thread/chemistry-ideal-gases/ | 1,532,097,877,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591683.78/warc/CC-MAIN-20180720135213-20180720155213-00632.warc.gz | 617,930,779 | 21,795 | # Chemistry - Ideal Gases
For Ideal Gases, the important assumptions are as below:
1. The intermolecular forces (Attractive Force or Repulsive Force) between the gas molecules are entirely negligible.
2. The volume occupied by the gas molecules themselves is entirely negligible relative to the volume of the container.
By the assumptions above, we can sure that there are no "Ideal Gas" in our real world.
However, scientists formulate the concept of Ideal Gases to make sure some gas law will always true. Or else, more errors will present in our calculations or data.
On the other hand, errors do exist when we dealt with Real Gases. We shall discuss this later.
If you are well with Ideal Gases, then here is the Ideal Gas Law:
$$PV=nRT$$
$$P$$ is for Pressure, with the unit $$Pa$$.
$$V$$ is for Volume, with the unit $$m^3$$.
$$n$$ is the Number of moles, with the unit $$mol$$.
$$R$$ is a Constant, which is $$8.314 m^3mol^{-1}K^{-1}$$.
$$T$$ is for Temperature, with the unit $$K$$.
Now, there is an example:
At a standard status (we mean by $$P=1 \text{atm}$$ and $$T=0^\circ C$$), what is the volume of a $$25g$$ $$O_2$$?
First, we convert the units.
$$P=1 \text{atm}=101325 Pa$$
$$T=0^\circ C=(0+273)K=273K$$.
We now have all the details we wanted, by the formula we have
$$PV=nRT$$
$$(101325)V=\frac{25}{32}\times8.314\times273$$
$$V=0.0175m^3$$
Thus, the volume is $$0.0175m^3$$.
Note by Christopher Boo
4 years, 5 months ago
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## Comments
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It has been a long time since my last post. This is because I went to my hometown to celebrate Chinese New Year. However, Anish and Vishnuram had post some chemistry notes on Electrochemistry and Organic Chemistry. Check it out here!
- 4 years, 5 months ago
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This is good stuff Christopher. Keep it up!
Staff - 4 years, 5 months ago
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Superb. Waiting for gas law problems.
- 4 years, 5 months ago
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Happy Chinese New Year!! ^^
- 4 years, 5 months ago
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Can I know what is this topic?
- 4 years, 5 months ago
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Hi Welthan, this topic is about about the laws governing ideal gases as defined by Christopher. It helps us to link up the temperature, volume and pressure of the gases along with the mass i.e. the number of moles. It is derived from the various gas laws given by Boyle, Charles, Gay-Lussac, etc. It has its applications in various fields.
- 4 years, 5 months ago
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I've noticed I missed our many thing too. I will try my best to edit it. :)
- 4 years, 5 months ago
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This is quite of a common sense once you get the intuition for it. ^^ But this formula, is very useful in physics and chemistry. (y) A chemistry toolkit that is used a lot in thermodynamics. ^^ I guess, that's a bit on the application side. Hahhaa. =)
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Set Loading... | 1,140 | 3,814 | {"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} | 3.25 | 3 | CC-MAIN-2018-30 | longest | en | 0.866556 |
https://oeis.org/A121152 | 1,624,157,721,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00534.warc.gz | 379,947,715 | 3,822 | 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!)
A121152 Dimension of the space spanned by the symmetric functions L_lambda of Gessel and Reutenauer, where lambda ranges over all partitions of n. 0
1, 1, 2, 3, 4, 6, 10, 13, 19, 26, 38, 52, 70, 91, 123, 161 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 REFERENCES I. M. Gessel and C. Reutenauer, J. Combinatorial Theory (A) 64 (1993), 189-215. R. P. Stanley, Enumerative Combinatorics, vol. 2 (Exercise 7.89). LINKS EXAMPLE In terms of Schur functions we have L[4] = s[3,1] + s[2,1,1], L[3,1] = s[3,1] + s[2,2] + s[1,1,1,1], L[2,2] = s[2,2] + s[1,1,1,1], L[2,1,1] = s[3,1] + s[2,1,1], L[4] = s[4]. There is one linear dependence relation, viz., L[4] = L[2,1,1], so for n=4 we get the value 5-1=4. CROSSREFS Sequence in context: A061018 A130126 A288338 * A229863 A215255 A200928 Adjacent sequences: A121149 A121150 A121151 * A121153 A121154 A121155 KEYWORD nonn AUTHOR Richard Stanley, Aug 12 2006 STATUS approved
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Last modified June 19 22:51 EDT 2021. Contains 345152 sequences. (Running on oeis4.) | 514 | 1,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-25 | latest | en | 0.65753 |
https://lawofgamedesign.com/2016/08/12/seeking-a-new-algorithm/?replytocom=843 | 1,653,171,616,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00361.warc.gz | 420,960,365 | 26,801 | # Seeking a New Algorithm
I’ve run into an efficiency issue while coding the new version of Phalanx. While it’s not game-breaking, I very much want a game that’s meant to run 24/7 on arcade cabinets to be as easy on the hardware as possible. Thus, I thought I would raise it here, in the hope that someone might have insight.
Phalanx is played in real time on a board.
Although the game is played in real time, it is effectively divided into turns–one turn per frame. In each turn, the movement rules are as follows:
1. A piece can move if the space it is moving into:
1. is currently empty, or
2. will be empty this turn (e.g., because the piece in it is leaving).
2. A piece cannot move off the board, or into a space containing water (the blue area in the middle), either of the goals, or another piece that is not going to leave this turn.
3. Pieces move in number order, starting with player 1’s pieces and then proceeding through player 2’s pieces. (This is terrible, and messing with the algorithm is a good time to change it.) In the event of a conflict during movement (e.g., two pieces want to enter the same space), the earlier-moving piece takes priority.
4. If a piece is part of a formation, the entire formation cannot move if any individual member of the formation cannot move.
To implement those rules, the game proceeds as follows:
1. The game loops through all the pieces that want to move.
1. All those that are moving into an empty space do so.
2. All those that definitely will not move (e.g., pieces blocked by water) are marked as not moving.
3. If a piece wants to move and is blocked by something that might get out of the way (e.g., another piece that hasn’t had its turn in the loop yet), that piece is skipped for this loop.
2. The game repeats step (1) until all pieces have moved or been marked as not moving.
This algorithm consistently produces the correct results, but the performance hit is rough. Is there a better way?
## One thought on “Seeking a New Algorithm”
1. Can you iterate through the loop only once per frame instead of repeating until it’s resolved every frame? | 487 | 2,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.964279 |
https://www.prepbharat.com/EntranceExams/JEEMain/AtomicStructure/atomicstructure-1.html | 1,713,038,049,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00147.warc.gz | 912,561,005 | 5,753 | ## Atomic Structure Questions and Answers Part-1
1. The electronic configuration of an element is 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1. This represents its:
a) excited state
b) ground state
c) cationic form
d) anionic form
Explanation: The given electronic configuration is ground state for chromium.
2. Select the pairs of ions which have same electronic configuration?
a) Cr3+, Fe3+
b) Fe3+, Mn2+
c) Fe3+, Co3+
d) Se3+, Cr3+
Explanation: Fe3+ and Mn2+ have same electronic configuration
3. Many elements have non-integral atomic masses because:
a) they have isotopes
b) their isotopes have non-integral masses
c) their isotopes have different masses
d) Both a and c
Explanation: Non-integral atomic masses are due to isotopes which have different masses.
4. Light of wavelength l shines on a metal surface with intensity x and the metal emits electrons per second of average energy, Z. What will happen to Y and Z if x is doubled?
a) Y will be double and Z will become half
b) Y will remain same and Z will be doubled
c) Both Y and Z will be doubled
d) Y will be doubled but Z will remain same
Explanation: When intensity is doubled, number of electrons emitted per second is also doubled but average energy of photoelectrons emitted remains the same.
5. Difference between nth and (n + 1)th Bohr's radius of H-atom is equal to its (n – 1)th Bohr's radius. The value of n is:
a) 1
b) 2
c) 3
d) 4
Explanation: rn n2
But rn + 1 – rn = rn – 1
(n + 1)2 – n2 = (n – 1)2
n = 4
6. Atoms consists of protons, neutrons and electrons. If the mass of neutrons and electrons were made half and two times respectively to their actual masses, then the atomic mass of $_{6}{{C}^{12}}$
a) Will remain approximately the same
b) Will become approximately two times
c) Will remain approximately half
d) Will be reduced by 25%
Explanation: No change by doubling mass of electrons however by reducing mass of neutron to half total atomic mass becomes $6+3$ instead of $6+6$ . Thus, reduced by 25%.
7. Principal, azimuthal and magnetic quantum numbers are respectively related to:
a) size, orientation and shape
b) size, shape and orientation
c) shape, size and orientation
d) none of these
Explanation: Principal gives size, azimuthal gives shape and magnetic quantum number gives the orientation.
8. The number of d-electrons retained in Fe2+ (At.no. of Fe = 26) ion is
a) 4
b) 5
c) 6
d) 3
Explanation: Configuration of Fe+2 = 3d6.4s°
9. Consider the ground state of Cr atom (Z=24). The number of electrons with the azimuthal quantum numbers 1 = 1 and 2 are respectively
a) 12 and 4
b) 16 and 5
c) 16 and 4
d) 12 and 5
Explanation: Configuration of Cr (Z = 24) is 1s22s22p63s23p63d54s*
1 = 1, denotes p-electrons (2p6, 3p6)
1 = 2, denotes d-electrons (3d5)
10. The electrons identified by quantum numbers n and l
1) n = 4, l = 1 2) n = 4, l = 0
3) n = 3, l = 2 4) n = 3, l = 1
can be placed in order of increasing energy as
a) c < d < b < a
b) d < b < c < a
c) b < d < a < c
d) a < c < b < d | 920 | 2,989 | {"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 | 0.909955 |
https://homepage.univie.ac.at/franz.vesely/sp_english/sp/node12.html | 1,701,284,890,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100135.11/warc/CC-MAIN-20231129173017-20231129203017-00469.warc.gz | 355,083,242 | 3,254 | Next: 3. Phase space Up: 2. Elements of Kinetic Previous: 2.5 Just in case
# 2.6 Problems for Chapter 2
Tutors' page
EXERCISES:
2.1 Variational method of Lagrange: Determine the minimum of the function under the condition (i.e. along the curve) , by two different methods:
a) by inserting in f(x,y) and differentiating by ;
b) using a Lagrange multiplier and evaluating Lagrange's equations and . (Shorthand notation: ; ... Variation of '').
For your better understanding sketch the functions and .
2.2 Method of the most probable distribution: Having understood the principle of Lagrange variation, reproduce the derivation of the Boltzmann distribution of energies (see text). Compare your result with the bottom right histogram in Applet LBRoulette.
2.3 Moments of the Maxwell-Boltzmann distribution: Verify the expressions for the most probable velocity, the average and the r.m.s. velocity.
2.4 Pressure in a dilute gas: Verify the formula .
2.5 Transport properties: For nitrogen under standard conditions, estimate the mean free path and the transport coefficients , and . ( ).
TEST YOUR UNDERSTANDING OF CHAPTER 2:
2.1 Maxwell-Boltzmann distribution: What is the formula for the distribution density of the molecular velocities in equilibrium; what is the respective formula for the speeds (absolute values of the velocity), ?
2.2 Pressure in an ideal gas: Derive the pressure equation of state for the ideal gas from simple kinetic theory.
2.3 Transport coefficients:
- Write down the defining equation for one of the three transport coefficients.
- What is the mean free path in a gas of spheres with diameter ?
- How does the viscosity of a dilute gas depend on the density?
Next: 3. Phase space Up: 2. Elements of Kinetic Previous: 2.5 Just in case
Franz Vesely
2005-01-25 | 415 | 1,797 | {"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} | 3.234375 | 3 | CC-MAIN-2023-50 | latest | en | 0.806783 |
https://research.stlouisfed.org/fred2/series/FARG038TRAD | 1,448,652,279,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450559.94/warc/CC-MAIN-20151124205410-00025-ip-10-71-132-137.ec2.internal.warc.gz | 859,919,361 | 19,295 | # All Employees: Trade, Transportation, and Utilities in Fargo, ND-MN (MSA)
2015-10: 30.3 Thousands of Persons
1yr | 5yr | 10yr | Max
The data services of the Federal Reserve Bank of St. Louis include series that are seasonally adjusted. To make these adjustments, we use the X-12 Procedure of SAS to remove the seasonal component of the series so that non-seasonal trends can be analyzed. This procedure is based on the U.S. Bureau of the Census X-12-ARIMA Seasonal Adjustment Program. More information on this program can be found at http://www.census.gov/srd/www/x12a/.
The seasonal moving average function used is that of the Census Bureau’s X-11-ARIMA program. This includes a 3x3 moving average for the initial seasonal factors and a 3x5 moving average to calculate the final seasonal factors. The D11 function is also used to output the entire seasonally adjusted series that is displayed. For specific information on the SAS X-12 procedure, please visit their website: http://support.sas.com/documentation/cdl/en/etsug/60372/HTML/default/viewer.htm#etsug_x12_sect001.htm.
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
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``` Federal Reserve Bank of St. Louis, All Employees: Trade, Transportation, and Utilities in Fargo, ND-MN (MSA) [FARG038TRAD], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/FARG038TRAD/, November 27, 2015. ```
Retrieving data.
Graph updated. | 611 | 2,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2015-48 | latest | en | 0.865854 |
https://www.jiqizhixin.com/articles/2017-07-04-3 | 1,582,826,667,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146744.74/warc/CC-MAIN-20200227160355-20200227190355-00349.warc.gz | 768,826,913 | 13,721 | # 如何用Python和机器学习炒股赚钱？
「星期天《纽约时报》上发表的一篇关于癌症治疗新药开发潜力的文章导致 EntreMed 的股价从周五收盘时的 12.063 飙升至 85，在周一收盘时接近 52。在接下来的三周，它的收盘价都在 30 以上。这股投资热情也让其它生物科技股得到了溢价。但是，这个癌症研究方面的可能突破在至少五个月前就已经被 Nature 期刊和各种流行的报纸报道过了，其中甚至包括《泰晤士报》！因此，仅仅是热情的公众关注就能引发股价的持续上涨，即便实际上并没有出现真正的新信息。」
「（股价）运动可能会集中于有一些共同之处的股票上，但这些共同之处不一定要是经济基础。」
``````import numpy as npimport pandas as pdfrom sklearn.decomposition import PCAfrom sklearn.cluster import KMeansimport matplotlib.pyplot as pltimport seaborn as sb
np.seterr(divide='ignore', invalid='ignore')# Quick way to test just a few column features# stocks = pd.read_csv('supercolumns-elements-nasdaq-nyse-otcbb-general-UPDATE-2017-03-01.csv', usecols=range(1,16))stocks = pd.read_csv('supercolumns-elements-nasdaq-nyse-otcbb-general-UPDATE-2017-03-01.csv')
str_list = []for colname, colvalue in stocks.iteritems(): if type(colvalue[1]) == str:
str_list.append(colname)# Get to the numeric columns by inversionnum_list = stocks.columns.difference(str_list)
stocks_num = stocks[num_list]
``````stocks_num = stocks_num.fillna(value=0, axis=1)
X = stocks_num.valuesfrom sklearn.preprocessing import StandardScaler
X_std = StandardScaler().fit_transform(X)
f, ax = plt.subplots(figsize=(12, 10))
plt.title('Pearson Correlation of Concept Features (Elements & Minerals)')# Draw the heatmap using seabornsb.heatmap(stocks_num.astype(float).corr(),linewidths=0.25,vmax=1.0, square=True, cmap="YlGnBu", linecolor='black', annot=True)
sb.plt.show()``````
## 测量「已解释方差（Explained Variance）」和主成分分析（PCA）
``````# Calculating Eigenvectors and eigenvalues of Cov matirxmean_vec = np.mean(X_std, axis=0)
cov_mat = np.cov(X_std.T)
eig_vals, eig_vecs = np.linalg.eig(cov_mat)# Create a list of (eigenvalue, eigenvector) tupleseig_pairs = [ (np.abs(eig_vals[i]),eig_vecs[:,i]) for i in range(len(eig_vals))]# Sort from high to loweig_pairs.sort(key = lambda x: x[0], reverse= True)# Calculation of Explained Variance from the eigenvaluestot = sum(eig_vals)
var_exp = [(i/tot)*100 for i in sorted(eig_vals, reverse=True)]
cum_var_exp = np.cumsum(var_exp) # Cumulative explained variance# Variances plotmax_cols = len(stocks.columns) - 1plt.figure(figsize=(10, 5))
plt.bar(range(max_cols), var_exp, alpha=0.3333, align='center', label='individual explained variance', color = 'g')
plt.step(range(max_cols), cum_var_exp, where='mid',label='cumulative explained variance')
plt.ylabel('Explained variance ratio')
plt.xlabel('Principal components')
plt.legend(loc='best')
plt.show()``````
``````pca = PCA(n_components=9)
x_9d = pca.fit_transform(X_std)
plt.figure(figsize = (9,7))
plt.scatter(x_9d[:,0],x_9d[:,1], c='goldenrod',alpha=0.5)
plt.ylim(-10,30)
plt.show()``````
## K-均值聚类（K-Means Clustering）
``````# Set a 3 KMeans clusteringkmeans = KMeans(n_clusters=3)# Compute cluster centers and predict cluster indicesX_clustered = kmeans.fit_predict(x_9d)# Define our own color mapLABEL_COLOR_MAP = {0 : 'r',1 : 'g',2 : 'b'}
label_color = [LABEL_COLOR_MAP[l] for l in X_clustered]# Plot the scatter digramplt.figure(figsize = (7,7))
plt.scatter(x_9d[:,0],x_9d[:,2], c= label_color, alpha=0.5)
plt.show()``````
``````# Create a temp dataframe from our PCA projection data "x_9d"df = pd.DataFrame(x_9d)
df = df[[0,1,2]]
df['X_cluster'] = X_clustered# Call Seaborn's pairplot to visualize our KMeans clustering on the PCA projected datasb.pairplot(df, hue='X_cluster', palette='Dark2', diag_kind='kde', size=1.85)
sb.plt.show()`````` | 1,156 | 3,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-10 | longest | en | 0.223876 |
https://datascience.stackexchange.com/questions/37069/elastic-net-regression-in-orange/37354 | 1,563,545,824,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526254.26/warc/CC-MAIN-20190719140355-20190719162355-00008.warc.gz | 373,104,617 | 36,108 | # Elastic net regression in orange
The penalty term for Elastic regression is written as
How are the values of lambda's calculated if the slider is moved from the right to the left. I have read most the material available on the internet on Elastic Net Regression. Here they define, $\alpha=\lambda_2/(\lambda_1 + \lambda_2)$. If the value of $\lambda_2$ and $\alpha$ is given, then $\lambda_1$ is calculated. In Orange the slider is provided.
$$\lambda_1 = 1 - \lambda_2$$
Finally I found the answer. In elastic net the cost function is written as $J(\theta)=MSE + r\alpha\sum_{i=1}^n \vert\theta_i \vert + \frac {1-r}{2}\alpha\sum_{1=1}^n \theta_i ^2$. Here r is the ratio, $\alpha$ is the hyperparameter and n is the number of features. The ratio slider controls r and $\alpha$ slider controls the value of hyperparameter. If the ratio r=0, then Elastic net regression is equal to ridge regression. For r=1, the Elastic Net Regression is equal to Lasso regression. I hope this perfectly answers the question I posed. I would like to have comments from followers of this post. | 282 | 1,081 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.796875 | 3 | CC-MAIN-2019-30 | latest | en | 0.876539 |
http://math.stackexchange.com/questions/43311/choosing-points-in-fractions-of-the-unit-interval | 1,469,795,413,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00091-ip-10-185-27-174.ec2.internal.warc.gz | 165,482,370 | 17,876 | # Choosing points in fractions of the unit interval
How long a series of points in (0,1) can be chosen such that the first two are in different halves, the first three are in different thirds, ... the first $n$ are in different $n^{\text{th}}$s? My first try of $(0+,1-,\frac{1}{2}-,\frac{3}{4}-, \frac{1}{5}+,\frac{5}{8}-,\frac{1}{3}-,\frac{7}{8}-,\frac{1}{3}+)$ works through $9$, but there are two points in $(0.3,0.4)$. The plus and minus signs indicate a shift of some distance away from the given point small enough not to move over any fraction of interest.
-
Mathworld calls this the "18-point problem": mathworld.wolfram.com/18-PointProblem.html – joriki Jun 5 '11 at 2:06
Fascinating. Interestingly, the solution roughly matches Ross' up to $5/8-$. – joriki Jun 5 '11 at 1:59 | 248 | 788 | {"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} | 3.4375 | 3 | CC-MAIN-2016-30 | latest | en | 0.777015 |
https://training.incf.org/search?f%5B0%5D=topics%3A22&f%5B1%5D=topics%3A34&f%5B2%5D=topics%3A47&f%5B3%5D=topics%3A52&f%5B4%5D=topics%3A58&f%5B5%5D=topics%3A77&%3Bf%5B1%5D=topics%3A61 | 1,571,826,527,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00433.warc.gz | 740,290,361 | 14,458 | ## Difficulty level
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https://codeforces.com/blog/entry/69725 | 1,591,502,528,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00156.warc.gz | 281,923,677 | 18,951 | ### SupaHotFire's blog
By SupaHotFire, history, 9 months ago, ,
let's say we have an array={9,1,7,4,5} and I want to query what is the nearest bigger element to the left and to the right of the index I'm querying for example the nearest bigger element for 5 is 7 (it doesn't have bigger element to the right because it's the last element)and for 1 is 7 to the right and 9 to the left .
I think it needs some data structure(to make it nlog(n) instead of n^2) so if someone can provide some code it would be better and thanks in advance
• -11
» 9 months ago, # | ← Rev. 4 → 0 For one direction, you can do it with a stack:At each element a[i], if the top of the stack is bigger than a[i], then the answer of a[i] is this top of stack, else if a[i] is bigger or equal to the top of stack, then the top of stack is no longer an answer for any element from now, so you will pop all the element from the top of the stack untill the stack becomes empty, or the top of the stack becomes bigger than a[i], so you reached the answer of a[i].And in both cases, you have to push a[i] in the stack. So, each element will be pushed once and poped at most once, so the time complexity is O(n).The solution with data structure can be as follows:You can maintain a maximum segment tree on the range [minVal, maxVal]. Initially, all the position of the segTree will be -1, and at each element a[i], you will update the segTree at position a[i] with i, and for the answer of a[i], you will do a maximum query on the range [a[i]+1, maxVal].And in both solutions, you will do the approach two times, one forward, and one backward, (for both directions, left and right, with some small differences in the segTree solution).
» 9 months ago, # | ← Rev. 2 → 0 you don't need any data structure, you can precompute suffix array such that suf[i]=min(a[i],suf[i+1]). After that all you need to do is binary search the smallest value in suffix array which is >a[i] on range i+1 to end of array. You can do for left of array similarly by precomputing prefix array. A problem using similar idea:Queue.Code: My solution | 554 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-24 | latest | en | 0.919116 |
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Finance
# The following information is available for the first month
Customer Question
of operations of Zahorik Company...
The following information is available for the first month of operations of Zahorik Company, a manufacturer of mechanical pencils:
Salaries360,000
Gross profit210,000
Cost of goods manufactured180,000
Indirect labor78,000
Factory depreciation12,000
Material purchased111,000
Total manufacturing costs for the period207,000
Materials inventory15,000
Using the above information, determine the following missing amounts:
a)Cost of goods sold 150,000
b)Finished goods inventory
c)Direct materials cost
d)Direct labor cost
e)Work in process inventory
Submitted: 6 years ago.Category: Finance
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7/2/2011
Financial Professional: DanielleCPA, Certified Public Accountant (CPA) replied 6 years ago
DanielleCPA, Certified Public Accountant (CPA)
Category: Finance
Satisfied Customers: 794
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DMC :
Hi and welcome to Just Answer! I'm happy to help answer your Finance questions. Feel free to interject at any time if you need clarification.
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Please give me a few moments to do the necessary calculations to get you an answer.
Customer reply replied 6 years ago
please let me know step by step
Financial Professional: DanielleCPA, Certified Public Accountant (CPA) replied 6 years ago
Sure. I will let you know step by step. I am figuring everything out for you right now.
Customer reply replied 6 years ago
Thanks
Financial Professional: DanielleCPA, Certified Public Accountant (CPA) replied 6 years ago
Just to confirm - is the first item salaries of 360,000 or sales of 360,000?
Customer reply replied 6 years ago
Sales
Financial Professional: DanielleCPA, Certified Public Accountant (CPA) replied 6 years ago
Cost of Goods Sold is Calculated as Sales less Gross Profit.
360,000 - 210,000 = 150,000
Finished Goods Inventory = Cost of Goods Manufactured Less Cost of Goods Sold
180,000 - 150,000 = 30,000
Direct Materials Cost = Direct Materials Purchased Less Materials Inventory
111,000 - 15,000 = 96,000
Work in Process Inventory = Manufacturing Costs less Cost of Goods Manufactured
207,000 - 180,000 = 27,000
Direct Labor = Manufacturing Costs Less Direct Materials Less Overhead.
Overhead in this problem is Indirect Labor Plus Depreciation
207,000 - 96,000 - 78,000 -12,000= 21,000
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Satisfied Customers: 794
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Show MoreShow Less | 2,873 | 12,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.826521 |
https://mymathangels.com/problem-507/ | 1,675,171,570,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00247.warc.gz | 437,452,230 | 13,400 | # Problem 507
Triangle ABC is congruent to triangle DFE. Find x.
Solution:-
In congruent triangle angle A = angle D, angle B = angle F, and angle C = angle E.
100 = 2x + 1
2x = 100 – 1
x = 99/2
x=49.5 | 78 | 209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-06 | longest | en | 0.727286 |
https://artofproblemsolving.com/wiki/index.php/2006_Canadian_MO_Problems/Problem_3 | 1,701,268,001,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00087.warc.gz | 150,749,576 | 10,310 | # 2006 Canadian MO Problems/Problem 3
## Problem
In a rectangular array of nonnegative real numbers with m rows and n columns, each row and each column contains at least one positive element. Moreover, if a row and a column intersect in a positive element, then the sums of their elements are the same. Prove that m = n.
## Solution
Assume that $m$ does not equal $n$. WLOG, m < n. Now we proceed by strong induction. We will assume that the statement is
true if a < m and b < n. The base case, where either m or n is 1, is clearly true. Consider the bipartite graph between the set
of rows and the set of column, with an edge between a row and a column if they intersect in a positive element.
Now, we invoke the Marriage Lemma. Consider any subset S of the rows, and the set C of columns that have an edge between the
column and a member of S. If |C| $\ge$ |S| for all such sets S, then there exists $V_1, V_2,... V_m$ such that they are all
positive and no two are in the same row or column, and adding them all up, we get that the sum of the numbers of all the rows is
the same as the sum of part of the columns, which imply that there are columns with only 0s, a contradiction. So there exist S
such that |S| > |C|. We apply the inductive hypothesis to get that this is impossible. | 332 | 1,297 | {"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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-50 | latest | en | 0.90635 |
https://oeis.org/A337925 | 1,701,941,750,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00881.warc.gz | 485,790,164 | 4,825 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 60th year, we have over 367,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”). Other ways to Give
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A337925 Digits of n rearranged to be the smallest number with the fewest possible prime factors, counted with multiplicity. Terms retain the same number of digits as n, i.e. leading digits may not be zero. 1
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 21, 13, 41, 15, 61, 17, 18, 19, 20, 21, 22, 23, 42, 25, 26, 27, 82, 29, 30, 13, 23, 33, 43, 53, 63, 37, 83, 39, 40, 41, 42, 43, 44, 45, 46, 47, 84, 49, 50, 15, 25, 53, 45, 55, 65, 57, 58, 59, 60, 61, 26, 63, 46, 65, 66, 67, 86, 69, 70, 17, 27, 37, 47, 57, 67, 77 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Rémy Sigrist, Table of n, a(n) for n = 1..10000 FORMULA a(a(n)) = a(n). - Rémy Sigrist, Oct 22 2020 MATHEMATICA a[n_] := Module[{p = FromDigits /@ Select[Permutations @ IntegerDigits[n], First[#] > 0 &]}, o = PrimeOmega[p]; Min[p[[Position[o, Min[o]] // Flatten]]]]; Array[a, 100] (* Amiram Eldar, Oct 19 2020 *) PROG (PARI) a(n) = {my(d = digits(n), v = select(x->#(digits(x))==#d, vector((#d)!, i, fromdigits(vector(#d, k, d[numtoperm(#d, i-1)[k]])))), b = vecmin(vector(#v, k, bigomega(v[k])))); vecmin(select(x->(bigomega(x)==b), v)); } \\ Michel Marcus, Oct 19 2020 CROSSREFS Cf. A001222, A004186, A328447. Sequence in context: A188650 A132578 A101318 * A130575 A330760 A064222 Adjacent sequences: A337922 A337923 A337924 * A337926 A337927 A337928 KEYWORD nonn,look,base AUTHOR Roderick Kimball, Sep 30 2020 STATUS approved
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Last modified December 7 04:15 EST 2023. Contains 367629 sequences. (Running on oeis4.) | 796 | 2,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-50 | latest | en | 0.691383 |
http://www.jiskha.com/display.cgi?id=1311907098 | 1,498,189,767,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319992.22/warc/CC-MAIN-20170623031127-20170623051127-00457.warc.gz | 553,287,160 | 3,727 | # Algebra
posted by on .
4x squared+9x+5=0?
• Algebra - ,
4x^2 + 9x + 5 = 0
(x+1)(4x+5) = 0
x = -1 or x = -5/4
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Post a New Question | 88 | 246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-26 | latest | en | 0.815176 |
https://www.thestudentroom.co.uk/showthread.php?t=3912459&page=33 | 1,529,669,883,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00443.warc.gz | 913,618,169 | 46,114 | You are Here: Home >< Maths
# Edexcel AS/A2 Mathematics M1 - 8th June 2016 - Official Thread watch
1. (Original post by KloppOClock)
vectors questions are pretty much always the same i think. but id still rather have 7 questions so i can have more time to check answers
Let's just hope it's easy. How have the C1 and C2 papers been?
2. (Original post by particlestudent)
Let's just hope it's easy. How have the C1 and C2 papers been?
c2 was pretty sound altho i messed up the discrimination question on c1 :/
you?
3. is there a proof we need to learn for m1? I doubt it but I need to check
4. For this question, why is it not possible to just say x + 9 = 41 --> X = 41 - 9 = 32N? I thought resultant forces are the sum of the other forces :/?
5. (Original post by KloppOClock)
c2 was pretty sound altho i messed up the discrimination question on c1 :/
you?
I'm in A2 so I've got C3 and 4 this year, along with M1. I just have to do well in M1 tomorrow to make life easier in C4 .
I'm just working through Gold papers today, I don't know what else I can do...
6. (Original post by Don Pedro K.)
For this question, why is it not possible to just say x + 9 = 41 --> X = 41 - 9 = 32N? I thought resultant forces are the sum of the other forces :/?
the forces are not acting in the same direction. You need to draw a force diagram (triangle in this case), showing the forces in series (one after the other). The resultant is the diagonal joining the start point to the end point
7. Good luck for tomorrow guys I'm ******** it as always!!
8. (Original post by candol)
the forces are not acting in the same direction. You need to draw a force diagram (triangle in this case), showing the forces in series (one after the other). The resultant is the diagonal joining the start point to the end point
Ah alright, thanks
Also, is it just me being stupid (again) or is the equation for solving (-->) in this solution wrong? Surely it should be
F + Rcos(60) = 30cos(30)?
9. (Original post by Don Pedro K.)
Ah alright, thanks
Also, is it just me being stupid (again) or is the equation for solving (--> in this solution wrong? Surely it should be
F + Rcos(60) = 30cos(30)?
that's the same thing
10. (Original post by candol)
that's the same thing
But in their equation, F - 30cos(30) = Rcos(60) so F = Rcos(60) + 30cos(30)
but if it was F + Rcos(60) = 30cos(30) to begin with, F would = 30cos(30) - Rcos(60), which isn't the same as above?
11. (Original post by Don Pedro K.)
But in their equation, F - 30cos(30) = Rcos(60) so F = Rcos(60) + 30cos(30)
but if it was F + Rcos(60) = 30cos(30) to begin with, F would = 30cos(30) - Rcos(60), which isn't the same as above?
quite right its not - do apologise. wasn't paying full attention. There equation looks right, but why would you draw 2 diagrams for a question?
12. (Original post by candol)
quite right its not - do apologise. wasn't paying full attention. There equation looks right, but why would you draw 2 diagrams for a question?
Hmm I'm not sure :/ I have a feeling that it's wrong but I don't know to be honest! Zacken could you have a look?
13. (Original post by Don Pedro K.)
Hmm I'm not sure :/ I have a feeling that it's wrong but I don't know to be honest! Zacken could you have a look?
Have you got the original question - the diagram is fine, (it is BASIC forces) but you really should have everything on the one diagram. It is not obvious to me what the original question was (cant be statics, it would be moving down??)
14. (Original post by Don Pedro K.)
Hmm I'm not sure :/ I have a feeling that it's wrong but I don't know to be honest! Zacken could you have a look?
It's correct.
If you resolve to the right, you get F - 30 cos 30.
This is the resultant force to the right. The resultant force to the right is also given by R cos 60
So F - 30 \cos 30= R \cos 60.
They've been drawn on different diagrams for a reason, you're treating them as though they are the same diagram.
15. sorry for the crap quality but does anyone know how to do this question? I can't find a mark scheme for it anywhere
16. (Original post by Zacken)
It's correct.
If you resolve to the right, you get F - 30 cos 30.
This is the resultant force to the right. The resultant force to the right is also given by R cos 60
So F - 30 \cos 30= R \cos 60.
They've been drawn on different diagrams for a reason, you're treating them as though they are the same diagram.
I see, I see. Why do we use two separate diagrams then?
17. (Original post by Strom)
Good luck for tomorrow guys I'm ******** it as always!!
Ayy, optimism ftw
(Original post by candol)
the forces are not acting in the same direction. You need to draw a force diagram (triangle in this case), showing the forces in series (one after the other). The resultant is the diagonal joining the start point to the end point
Post 666 and you mention triangle
18. (Original post by bertolli1)
sorry for the crap quality but does anyone know how to do this question? I can't find a mark scheme for it anywhere
part a) suvat
s=h
A=-9.8
t=5
u=20
partb)
Find V using suvat for the first ball
You can now work out 3/4 v for the second ball
Do suvat with v=3/4V, u=w, a=9.8, s= height found from part a
19. (Original post by bertolli1)
sorry for the crap quality but does anyone know how to do this question? I can't find a mark scheme for it anywhere
For (a)
s = -h, u = 20, v = v, a = -9.8, t = 5
Shove into suvat and solve for h.
(Original post by Don Pedro K.)
I see, I see. Why do we use two separate diagrams then?
I wouldn't use two diagrams. I'd just say resultant to the right is F - 30 cos 30 and to the bottom is 30 sin 30 or w/e and work from there without using R.
20. (Original post by candol)
part a) suvat
s=h
A=-9.8
t=5
u=20
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Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 1,813 | 6,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-26 | latest | en | 0.959829 |
https://rdrr.io/cran/FactoMineR/man/AovSum.html | 1,624,024,841,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487637721.34/warc/CC-MAIN-20210618134943-20210618164943-00324.warc.gz | 446,308,776 | 10,794 | # AovSum: Analysis of variance with the contrasts sum (the sum of the... In FactoMineR: Multivariate Exploratory Data Analysis and Data Mining
## Description
Analysis of variance with the contrasts sum (the sum of the coefficients is 0)
Test for all the coefficients
Handle missing values
## Usage
`1` ```AovSum(formula, data, na.action = na.omit, ...) ```
## Arguments
`formula` the formula for the model 'y~x1+x2+x1:x2' `data` a data-frame `na.action` (where relevant) information returned by model.frame on the special handling of NAs. `...` other arguments, cf the function `lm`
## Value
Retourne des objets
`Ftest` a table with the F-tests `Ttest` a table with the t-tests
## Author(s)
Francois Husson husson@agrocampus-ouest.fr
`aov`, `lm`
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14``` ```## Example two-way anova data(senso) res <- AovSum(Score~ Product + Day , data=senso) res ## Example two-way anova with interaction data(senso) res2 <- AovSum(Score~ Product + Day + Product : Day, data=senso) res2 ## Example ancova data(footsize) res3 <- AovSum(footsize ~ size + sex + size : sex, data=footsize) res3 ```
### Example output
```Ftest
SS df MS F value Pr(>F)
Product 2.0278 2 1.0139 3.4428 0.04148 *
Day 25.4725 1 25.4725 86.4959 1.18e-11 ***
Residuals 12.0742 41 0.2945
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Ttest
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.590736 0.081464 56.3528 < 2e-16 ***
Product - 1 0.237057 0.120675 1.9644 0.05628 .
Product - 2 -0.284745 0.114663 -2.4833 0.01720 *
Product - 3 0.047688 0.118456 0.4026 0.68935
Day - 1 0.803378 0.086382 9.3003 < 2e-16 ***
Day - 2 -0.803378 0.086382 -9.3003 < 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Ftest
SS df MS F value Pr(>F)
Product 1.7773 2 0.8886 2.9876 0.06206 .
Day 25.8418 1 25.8418 86.8787 1.804e-11 ***
Product:Day 0.4738 2 0.2369 0.7964 0.45813
Residuals 11.6004 39 0.2974
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Ttest
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.5842929 0.0871239 52.6181 < 2e-16 ***
Product - 1 0.2557071 0.1225836 2.0860 0.04357 *
Product - 2 -0.2509596 0.1203159 -2.0858 0.04358 *
Product - 3 -0.0047475 0.1266521 -0.0375 0.97029
Day - 1 0.8120707 0.0871239 9.3209 < 2e-16 ***
Day - 2 -0.8120707 0.0871239 -9.3209 < 2e-16 ***
Product - 1 : Day - 1 0.0279293 0.1225836 0.2278 0.82096
Product - 2 : Day - 1 -0.1454040 0.1203159 -1.2085 0.23412
Product - 3 : Day - 1 0.1174747 0.1266521 0.9275 0.35935
Product - 1 : Day - 2 -0.0279293 0.1225836 -0.2278 0.82096
Product - 2 : Day - 2 0.1454040 0.1203159 1.2085 0.23412
Product - 3 : Day - 2 -0.1174747 0.1266521 -0.9275 0.35935
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Ftest
SS df MS F value Pr(>F)
size 72.322 1 72.322 70.4901 1.353e-12 ***
sex 3.114 1 3.114 3.0353 0.08531 .
size:sex 3.789 1 3.789 3.6930 0.05820 .
Residuals 82.079 80 1.026
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Ttest
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.973278 4.214939 1.1799 0.24153
size 0.204172 0.024318 8.3958 < 2e-16 ***
sex - F 7.343305 4.214939 1.7422 0.08531 .
sex - M -7.343305 4.214939 -1.7422 0.08531 .
sex - F : size -0.046733 0.024318 -1.9217 0.05820 .
sex - M : size 0.046733 0.024318 1.9217 0.05820 .
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
```
FactoMineR documentation built on Jan. 8, 2021, 2:18 a.m. | 1,702 | 3,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | longest | en | 0.368841 |
https://dailydevsblog.com/python/learn-python-heap-sort-in-python-basic-and-advance-2672/ | 1,675,924,044,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00321.warc.gz | 207,532,190 | 18,327 | # Learn Python – Heap Sort in Python- Basic and advance
The heap kind is quite the identical as the determination sort, where we locate the most component and put it at the end. It is based totally on a contrast sorting algorithm which works on Binary heap statistics structure. It is the nice instance of an efficient sorting algorithm.
## What is Heap Sort?
Heap type is an efficient and famous sorting algorithm. The heap kind concept is to “eliminate” the aspect from the heap section of the listing one-by-one and insert them to the sorted phase of the list. Before mastering greater about the heap sorting algorithm, let’s talk about the heap statistics structure.
It is an in-place algorithm, which potential a fixed quantity of reminiscence is used to shop the sorted list, or the reminiscence dimension does not remember on the dimension of the preliminary list.
For example – We don’t want the extra reminiscence stack to store the sorted array and neither recursive call stack. The heapsort algorithm commonly implements the use of the 2d array to kind the fixed values. This system is quick, simple, herbal and effortless to implement.
On the other hand, heap sort is unstable, which capacity it does not hold the comparative order of elements with equal values. It can rapidly kind primitive sorts such as integers and characters, but it has a trouble with the complex kinds and objects.
Let’s understand it by the following example –
We have a custom class Student with houses age and name, and quite a few objects of that category in an array, which include a pupil known as “Thomas” a long time “20” and additionally “Peter,” have aged 20 show up in the identical order.
If we sort the array of humans by age, then there is no warranty that “Thomas” would appear earlier than the “Peter” in the sorted array. It can be defined order, however there is no guarantee.
## Heap Data Structure
The heap records structure is a entire binary tree that fulfills the heap property. It is additionally known as the binary heap.
A whole binary tree satisfies the following properties.
Every level should be filled.
All the nodes are as far left as possible.
As we can see in the above image of the heap, but it is now not sorted. We will no longer dig in-depth this article due to the fact our center of attention is to explain the Heap type algorithm, now not a heap. In the heap sort, the subsequent smallest thing is constantly the first element.
The heap tree can be the two types – min-heap and max tree. A min-heap is saved a record of the maximum element. A max heap maintains song of the biggest element. Heap generally helps the following operations – delete_minimum(), get_minimum() and add().
The first thing of the heap can delete after restoring it. It takes O(log N) time, and that is fairly effective.
## Implementation
Python presents the in-built functions for sorting factors the usage of heap sort. The features are given below.
heappush(list, item) -It is used to add the heap element and re-sort it.
heappop(list) – It is used to remove the element and return the element.
heapfy() – It is used to turn the given list into a heap.
Consider the following example for heap sort.
Example –
``````from heapq import heappop, heappush
def heapsort(list1):
heap = []
for ele in list1:
heappush(heap, ele)
sort = []
# the elements are lift in the heap
while heap:
sort.append(heappop(heap))
return sort
list1 = [27, 21, 55, 15, 60, 4, 11, 17, 2, 87]
print(heapsort(list1))
``````
Output:
``[2, 4, 11, 15, 17, 21, 27, 55, 60, 87]``
Explanation
In the above code, we have imported the heapq module which consist heappop() and heappush() method. We created the Heapsort Heapsort () method, which takes list1 as an argument. A for loop iterated the list1 and pushed the elements to the empty heap. We used the while loop and sorted component delivered to the empty sort.
We called the Heapsort Heapsort () function and exceeded a list. It back the sorted list.
## Sorting Custom Objects
Heap type is beneficial for predefined information types, but it is greater intricate to manage the user-define records types, such as type objects. We will type the customized objects in this section.
As we can see, our implementation depends upon the built-in methods. Python affords the following methods.
heapq.nlargest(*n*, *iterable*, *key = None) – This method is used to get a list with the n largest element from the dataset, defined by the iterable.
heapq.nsmallest(*n*, *iterable*, *key = None) – This method is used to get a list with the n smallest elements from the dataset, which is defined by the iterable.
Let’s understand the following implementation of customized objects.
Example –
``````from heapq import heappop, heappush
class Car:
def __init__(self, model, year):
self.model = model
self.year = year
def __str__(self):
return str.format("Model Name: {}, Year: {}", self.model, self.year)
def __lt__(self, other):
return self.year < other.year
def __gt__(self, other):
return other.__lt__(self)
def __eq__(self, other):
return self.year == other.year
def __ne__(self, other):
return not self.__eq__(other)
def heapsort(list1):
heap = []
for element in list1:
heappush(heap, element)
ordered = []
while heap:
ordered.append(heappop(heap))
return ordered
car1 = Car("Renault", 2001)
car2 = Car("Bentley", 2005)
car3 = Car("Kia", 2014)
car4 = Car("Maruti Suzuki", 1999);
car5 = Car("Nano", 2012)
list1 = [car1, car2, car3, car4, car5]
for c in Heapsort Heapsort (list1):
print(c)
``````
Output:
``````Model Name: Maruti Suzuki, Year: 1999
Model Name: Renault, Year: 2001
Model Name: Bentley, Year: 2005
Model Name: Nano, Year: 2012
Model Name: Kia, Year: 2014``````
We have sorted the objects on the year base.
## Comparison between Heap sort and Other Algorithm
One of the famous quick type algorithms is additionally very efficient, but heap type is legally used because of its reliability. The heap sort’s key advantage is O(nlogn) upper certain as far as the time complexity is fretful.
The heap kind algorithm takes O(nlogn) time in each common and worst-case eventualities whilst the speedy sort is 20% quicker in the average case.
The quick kind algorithm becomes slow in predictable situations. There is a hazard of the protection breach in quick sort on the grounds that the foul O(n2) can be without problems triggered.
Now we compare to the Merge sort, which takes the equal time as the heap sort.
Merge kind is a lot steady and intuitively parallelizable, the place heap type does not have such advantages.
Furthermore, Merge type is faster than the Heap Sort in most cases considering they have the equal time complexity.
In contrast, HeapsortHeapsort can be applied lots rapidly in-place than Marge sort can.
## Conclusion
Heapsort is no longer so famous and faster, but it is greater predictable than any different sorting algorithm. This algorithm is preferred the place memory and safety are a priority.
It can be shortly implemented using Python. We require to insert the elements in a heap and take them out. | 1,682 | 7,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.934281 |
https://rosettacode.org/wiki/User_talk:Heuberg?oldid=148154 | 1,709,611,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00396.warc.gz | 489,432,289 | 64,469 | # Brownian tree
This page uses content from Wikipedia. The original article was at Brownian_tree. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Brownian tree
You are encouraged to solve this task according to the task description, using any language you may know.
Generate and draw a Brownian Tree.
A Brownian Tree is generated as a result of an initial seed, followed by the interaction of two processes.
1. The initial "seed" is placed somewhere within the field. Where is not particularly important; it could be randomized, or it could be a fixed point.
2. Particles are injected into the field, and are individually given a (typically random) motion pattern.
3. When a particle collides with the seed or tree, its position is fixed, and it's considered to be part of the tree.
Because of the lax rules governing the random nature of the particle's placement and motion, no two resulting trees are really expected to be the same, or even necessarily have the same general shape.
## AutoHotkey
Works with: AutoHotkey_L
Takes a little while to run, be patient. Requires the GDI+ Standard Library by Tic <lang AHK>SetBatchLines -1 Process, Priority,, high size := 400 D := .08 num := size * size * d field:= Object() field[size//2, size//2] := true ; set the seed lost := 0
Loop % num { x := Rnd(1, size), y := Rnd(1, size) Loop { oldX := X, oldY := Y x += Rnd(-1, 1), y += Rnd(1, -1) If ( field[x, y] ) { field[oldX, oldY] := true break } If ( X > Size ) or ( Y > Size) or ( X < 1 ) or ( Y < 1 ) { lost++ break } } }
pToken := Gdip_startup() pBitmap := Gdip_CreateBitmap(size, size) loop %size% { x := A_index Loop %size% { If ( field[x, A_Index] ) { Gdip_SetPixel(pBitmap, x, A_Index, 0xFF0000FF) } } } Gdip_SaveBitmapToFile(pBitmap, "brownian.png") Gdip_DisposeImage(pBitmap) Gdip_Shutdown(pToken) Run brownian.png
MsgBox lost %lost%
Rnd(min, max){ Random, r, min, max return r }</lang>Sample output file here
## BBC BASIC
<lang bbcbasic> SYS "SetWindowText", @hwnd%, "Brownian Tree"
``` SIZE = 400
VDU 23,22,SIZE;SIZE;8,16,16,0
GCOL 10
REM set the seed:
PLOT SIZE, SIZE
OFF
REPEAT
REM set particle's initial position:
REPEAT
X% = RND(SIZE)-1
Y% = RND(SIZE)-1
UNTIL POINT(2*X%,2*Y%) = 0
REPEAT
oldX% = X%
oldY% = Y%
X% += RND(3) - 2
Y% += RND(3) - 2
UNTIL POINT(2*X%,2*Y%)
IF X%>=0 IF X%<SIZE IF Y%>=0 IF Y%<SIZE PLOT 2*oldX%,2*oldY%
UNTIL FALSE
```
</lang>
## C
Library: FreeImage
<lang c>#include <string.h>
1. include <stdlib.h>
2. include <time.h>
3. include <math.h>
4. include <FreeImage.h>
1. define NUM_PARTICLES 1000
2. define SIZE 800
void draw_brownian_tree(int world[SIZE][SIZE]){
``` int px, py; // particle values
int dx, dy; // offsets
int i;
// set the seed
world[rand() % SIZE][rand() % SIZE] = 1;
```
``` for (i = 0; i < NUM_PARTICLES; i++){
// set particle's initial position
px = rand() % SIZE;
py = rand() % SIZE;
```
``` while (1){
// randomly choose a direction
dx = rand() % 3 - 1;
dy = rand() % 3 - 1;
```
``` if (dx + px < 0 || dx + px >= SIZE || dy + py < 0 || dy + py >= SIZE){
// plop the particle into some other random location
px = rand() % SIZE;
py = rand() % SIZE;
}else if (world[py + dy][px + dx] != 0){
// bumped into something
world[py][px] = 1;
break;
}else{
py += dy;
px += dx;
}
}
}
```
}
int main(){
``` int world[SIZE][SIZE];
FIBITMAP * img;
int x, y;
memset(world, 0, sizeof world);
srand((unsigned)time(NULL));
```
``` draw_brownian_tree(world);
```
``` img = FreeImage_Allocate(SIZE, SIZE, 32, 0, 0, 0);
```
``` for (y = 0; y < SIZE; y++){
for (x = 0; x < SIZE; x++){
rgb.rgbRed = rgb.rgbGreen = rgb.rgbBlue = (world[y][x] ? 255 : 0);
FreeImage_SetPixelColor(img, x, y, &rgb);
}
}
FreeImage_Save(FIF_BMP, img, "brownian_tree.bmp", 0);
```
}</lang>
Bold text
### Alternative Version
Translation of: D
This version writes the image as Portable Bit Map to stdout and doesn't move already set particles. <lang c>#include <stdio.h>
1. include <stdlib.h>
2. include <time.h>
3. include <stdbool.h>
1. define SIDE 600
2. define NUM_PARTICLES 10000
bool W[SIDE][SIDE];
int main() {
``` srand((unsigned)time(NULL));
W[SIDE / 2][SIDE / 2] = true;
```
``` for (int i = 0; i < NUM_PARTICLES; i++) {
unsigned int x, y;
OVER: do {
x = rand() % (SIDE - 2) + 1;
y = rand() % (SIDE - 2) + 1;
} while (W[y][x]);
```
``` while (W[y-1][x-1] + W[y-1][x] + W[y-1][x+1] +
W[y][x-1] + W[y][x+1] +
W[y+1][x-1] + W[y+1][x] + W[y+1][x+1] == 0) {
unsigned int dxy = rand() % 8;
if (dxy > 3) dxy++;
x += (dxy % 3) - 1;
y += (dxy / 3) - 1;
if (x < 1 || x >= SIDE - 1 || y < 1 || y >= SIDE - 1)
goto OVER;
}
```
``` W[y][x] = true;
}
```
``` printf("P1\n%d %d\n", SIDE, SIDE);
for (int r = 0; r < SIDE; r++) {
for (int c = 0; c < SIDE; c++)
printf("%d ", W[r][c]);
putchar('\n');
}
return 0;
```
}</lang> Run-time about 12.4 seconds with SIDE=600, NUM_PARTICLES=10000.
## C++
For an animated version based on this same code see: Brownian tree/C++ animated <lang cpp>#include <windows.h>
1. include <iostream>
2. include <string>
//-------------------------------------------------------------------- using namespace std;
//-------------------------------------------------------------------- enum states { SEED, GROWING, MOVING, REST }; enum treeStates { NONE, MOVER, TREE }; const int MAX_SIDE = 480, MAX_MOVERS = 511, MAX_CELLS = 15137;
//-------------------------------------------------------------------- class point { public:
``` point() { x = y = 0; }
point( int a, int b ) { x = a; y = b; }
void set( int a, int b ) { x = a; y = b; }
```
``` int x, y;
```
}; //-------------------------------------------------------------------- class movers { public:
``` point pos;
bool moving;
movers() : moving( false ){}
```
}; //-------------------------------------------------------------------- class myBitmap { public:
``` myBitmap() : pen( NULL ) {}
~myBitmap()
{
```
DeleteObject( pen ); DeleteDC( hdc ); DeleteObject( bmp );
``` }
```
``` bool create( int w, int h )
{
```
BITMAPINFO bi; ZeroMemory( &bi, sizeof( bi ) );
HDC dc = GetDC( GetConsoleWindow() ); bmp = CreateDIBSection( dc, &bi, DIB_RGB_COLORS, &pBits, NULL, 0 ); if( !bmp ) return false;
hdc = CreateCompatibleDC( dc ); SelectObject( hdc, bmp ); ReleaseDC( GetConsoleWindow(), dc );
width = w; height = h;
return true;
``` }
```
``` void clear()
{
```
ZeroMemory( pBits, width * height * sizeof( DWORD ) );
``` }
```
``` void setPenColor( DWORD clr )
{
```
if( pen ) DeleteObject( pen ); pen = CreatePen( PS_SOLID, 1, clr ); SelectObject( hdc, pen );
``` }
```
``` void saveBitmap( string path )
{
```
GetObject( bmp, sizeof( bitmap ), &bitmap );
dwpBits = new DWORD[bitmap.bmWidth * bitmap.bmHeight]; ZeroMemory( dwpBits, bitmap.bmWidth * bitmap.bmHeight * sizeof( DWORD ) ); ZeroMemory( &infoheader, sizeof( BITMAPINFO ) ); ZeroMemory( &fileheader, sizeof( BITMAPFILEHEADER ) );
GetDIBits( hdc, bmp, 0, height, ( LPVOID )dwpBits, &infoheader, DIB_RGB_COLORS );
file = CreateFile( path.c_str(), GENERIC_WRITE, 0, NULL,
``` CREATE_ALWAYS, FILE_ATTRIBUTE_NORMAL, NULL );
```
delete [] dwpBits;
``` }
```
``` HDC getDC() { return hdc; }
int getWidth() { return width; }
int getHeight() { return height; }
```
private:
``` HBITMAP bmp;
HDC hdc;
HPEN pen;
void *pBits;
int width, height;
```
}; //-------------------------------------------------------------------- class brownianTree { public:
``` brownianTree()
{
```
_bmp.create( MAX_SIDE, MAX_SIDE ); init();
``` }
```
``` void init()
{
```
_cellCount = 0; ZeroMemory( _grid, sizeof( _grid ) ); _bmp.clear(); _state = SEED;
``` }
```
bool mainLoop()
``` {
```
switch( _state ) { case REST: saveTree(); return false; case SEED: doSeed(); break; case GROWING: startMovers(); break; case MOVING: moveMovers(); } return true; }
``` myBitmap* getBmp() { return &_bmp; }
```
private:
``` void saveTree()
{
```
for( int y = 0; y < MAX_SIDE; y++ ) for( int x = 0; x < MAX_SIDE; x++ ) if( _grid[x][y] == TREE ) SetPixel( _bmp.getDC(), x, y, RGB( 255, 120, 0 ) );
``` _bmp.saveBitmap( "f:\\rc\\tree.bmp" );
}
```
``` void doSeed()
{
```
int x = MAX_SIDE - MAX_SIDE / 2, y = MAX_SIDE / 4; _grid[rand() % x + y][rand() % x + y] = TREE; _cellCount++; _state = GROWING;
``` }
```
``` void addMover( movers* m )
{
```
m->moving = true; int x = MAX_SIDE - MAX_SIDE / 2, y = MAX_SIDE / 4, a, b; while( true ) { a = rand() % x + y; b = rand() % x + y; if( _grid[a][b] == NONE ) break; }
m->pos.set( a, b ); _grid[a][b] = MOVER;
``` }
```
``` void startMovers()
{
```
movers* m; for( int c = 0; c < MAX_MOVERS; c++ ) { m = &_movers[c]; addMover( m ); } _state = MOVING;
``` }
```
``` void addToTree( movers* m )
{
```
m->moving = false; _grid[m->pos.x][m->pos.y] = TREE; if( ++_cellCount >= MAX_CELLS ) _state = REST;
COORD c = { 0, 1 }; SetConsoleCursorPosition( GetStdHandle( STD_OUTPUT_HANDLE ), c ); cout << "Cells added: " << _cellCount
``` << " from " << MAX_CELLS << " => "
<< static_cast<float>( 100 * _cellCount ) /
static_cast<float>( MAX_CELLS )
<< "% ";
}
```
``` bool moveIt( movers* m )
{
```
point f[8]; int ff = 0; for( int y = -1; y < 2; y++ ) { for( int x = -1; x < 2; x++ ) { if( !x && !y ) continue; int a = m->pos.x + x, b = m->pos.y + y; if( a < 0 || b < 0 || a >= MAX_SIDE || b >= MAX_SIDE ) { addToTree( m ); return true; } switch( _grid[a][b] ) { case TREE: addToTree( m ); return true; case NONE: f[ff++].set( a, b ); } }
``` }
```
if( ff < 1 ) return false;
_grid[m->pos.x][m->pos.y] = NONE; m->pos = f[rand() % ff]; _grid[m->pos.x][m->pos.y] = MOVER;
return false;
``` }
```
``` void moveMovers()
{
```
movers* mm; for( int m = 0; m < MAX_MOVERS; m++ ) { mm = &_movers[m]; if( !mm->moving ) continue; if( moveIt( mm ) && _cellCount < MAX_CELLS ) addMover( mm ); }
``` }
```
``` states _state;
BYTE _grid[MAX_SIDE][MAX_SIDE];
myBitmap _bmp;
int _cellCount;
movers _movers[MAX_MOVERS];
```
}; //-------------------------------------------------------------------- int main( int argc, char* argv[] ) {
``` ShowWindow( GetConsoleWindow(), SW_MAXIMIZE );
srand( GetTickCount() );
```
``` brownianTree tree;
```
``` DWORD now = GetTickCount();
while( tree.mainLoop() );
```
``` now = GetTickCount() - now;
cout << endl << endl << "It took "
<< now / 1000
<< " seconds to complete the task!" << endl << endl;
```
``` BitBlt( GetDC( GetConsoleWindow() ), 20, 90, MAX_SIDE, MAX_SIDE,
tree.getBmp()->getDC(), 0, 0, SRCCOPY );
```
``` system( "pause" );
return 0;
```
} //--------------------------------------------------------------------</lang>
## C#
Works with: C# version 3.0
<lang csharp>using System; using System.Drawing;
namespace BrownianTree {
``` class Program
{
static Bitmap BrownianTree(int size, int numparticles)
{
Bitmap bmp = new Bitmap(size, size);
Rectangle bounds = new Rectangle { X = 0, Y = 0, Size = bmp.Size };
using (Graphics g = Graphics.FromImage(bmp))
{
g.Clear(Color.Black);
}
Random rnd = new Random();
bmp.SetPixel(rnd.Next(size), rnd.Next(size), Color.White);
Point pt = new Point(), newpt = new Point();
for (int i = 0; i < numparticles; i++)
{
pt.X = rnd.Next(size);
pt.Y = rnd.Next(size);
do
{
newpt.X = pt.X + rnd.Next(-1, 2);
newpt.Y = pt.Y + rnd.Next(-1, 2);
if (!bounds.Contains(newpt))
{
pt.X = rnd.Next(size);
pt.Y = rnd.Next(size);
}
else if (bmp.GetPixel(newpt.X, newpt.Y).R > 0)
{
bmp.SetPixel(pt.X, pt.Y, Color.White);
break;
}
else
{
pt = newpt;
}
} while (true);
}
return bmp;
}
```
``` static void Main(string[] args)
{
BrownianTree(300, 3000).Save("browniantree.png");
}
}
```
}</lang>
## D
Uses the module of the Grayscale Image task. Partial
Translation of: PureBasic
<lang d>void main() {
``` import core.stdc.stdio, std.random, grayscale_image;
```
``` enum uint side = 600; // Square world side.
enum uint num_particles = 10_000;
static assert(side > 2 && num_particles < (side ^^ 2 * 0.7));
```
``` auto rng = unpredictableSeed.Xorshift;
ubyte[side][side] W; // World.
W[side / 2][side / 2] = 1; // Set tree root.
```
``` foreach (immutable _; 0 .. num_particles) {
// Random initial particle position.
OVER: uint x, y;
do {
x = uniform(1, side - 1, rng);
y = uniform(1, side - 1, rng);
} while (W[y][x]); // Assure the chosen cell is empty.
```
``` while (W[y-1][x-1] + W[y-1][x] + W[y-1][x+1] +
W[y][x-1] + W[y][x+1] +
W[y+1][x-1] + W[y+1][x] + W[y+1][x+1] == 0) {
// Randomly choose a direction (Moore neighborhood).
uint dxy = uniform(0, 8, rng);
if (dxy > 3) dxy++; // To avoid the center.
x += (dxy % 3) - 1;
y += (dxy / 3) - 1;
if (x < 1 || x >= side - 1 || y < 1 || y >= side - 1)
goto OVER;
}
```
``` W[y][x] = 1; // Touched, set the cell.
}
```
``` ubyte[] data = (&W[0][0])[0 .. side ^^ 2]; // Flat view.
data[] += 255;
Image!ubyte.fromData(data, side, side).savePGM("brownian_tree.pgm");
```
}</lang> World side = 600, num_particles = 10_000, cropped (about 20 seconds run-time with dmd, about 4.3 seconds with ldc2):
## Delphi
<lang delphi>const
``` SIZE = 256;
NUM_PARTICLES = 1000;
```
procedure TForm1.Button1Click(Sender: TObject); type
``` TByteArray = array[0..0] of Byte;
PByteArray = ^TByteArray;
```
var
``` B: TBitmap;
I: Integer;
P, D: TPoint;
```
begin
``` Randomize;
B := TBitmap.Create;
try
B.Width := SIZE;
B.Height := SIZE;
B.PixelFormat := pf8bit;
```
``` B.Canvas.Brush.Color := clBlack;
B.Canvas.FillRect(B.Canvas.ClipRect);
B.Canvas.Pixels[Random(SIZE), Random(SIZE)] := clWhite;
```
``` For I := 0 to NUM_PARTICLES - 1 do
Begin
P.X := Random(SIZE);
P.Y := Random(SIZE);
```
``` While true do
Begin
D.X := Random(3) - 1;
D.Y := Random(3) - 1;
Inc(P.X, D.X);
Inc(P.Y, D.Y);
```
``` If ((P.X or P.Y) < 0) or (P.X >= SIZE) or (P.Y >= SIZE) Then
Begin
P.X := Random(SIZE);
P.Y := Random(SIZE);
end
else if PByteArray(B.ScanLine[P.Y])^[P.X] <> 0 then
begin
PByteArray(B.ScanLine[P.Y-D.Y])^[P.X-D.X] := \$FF;
Break;
end;
end;
end;
```
``` Canvas.Draw(0, 0, B);
finally
FreeAndNil(B);
end;
```
end;</lang>
## Fantom
<lang fantom> using fwt using gfx
class Main {
``` public static Void main ()
{
particles := Particles (300, 200)
Window // open up a display for the final tree
{
title = "Brownian Tree"
EdgePane
{
center = ScrollPane { content = ParticleCanvas(particles) }
},
}.open
}
```
}
class Particles {
``` Bool[][] image
Int height
Int width
```
``` new make (Int height, Int width)
{
this.height = height
this.width = width
// set up initial image as an array of booleans with one set cell
image = [,]
width.times |w|
{
row := [,]
}
image[Int.random(0..<width)][Int.random(0..<height)] = true
}
```
``` Bool get (Int w, Int h) { return image[w][h] }
```
``` Void addParticle ()
{
x := Int.random(0..<width)
y := Int.random(0..<height)
```
``` Int dx := 0
Int dy := 0
while (!image[x][y]) // loop until hit existing part of the tree
{
dx = [-1,0,1].random
dy = [-1,0,1].random
```
``` if ((0..<width).contains(x + dx))
x += dx
else // did not change x, so set dx = 0
dx = 0
if ((0..<height).contains(y + dy))
y += dy
else
dy = 0
}
```
``` // put x,y back to just before move onto existing part of tree
x -= dx
y -= dy
```
``` image[x][y] = true
}
```
}
class ParticleCanvas : Canvas {
``` Particles particles
new make (Particles particles) { this.particles = particles }
```
``` // provides canvas size for parent scrollpane
override Size prefSize(Hints hints := Hints.defVal)
{
Size(particles.width, particles.height)
}
```
``` // repaint the display
override Void onPaint (Graphics g)
{
g.brush = Color.black
g.fillRect(0, 0, size.w, size.h)
g.brush = Color.green
particles.width.times |w|
{
particles.height.times |h|
{
if (particles.get(w, h)) // draw a 1x1 square for each set particle
g.fillRect (w, h, 1, 1)
}
}
}
```
} </lang>
## Fortran
Works with: Fortran version 95 and later
Translation of: C
For RCImageBasic and RCImageIO, see Basic bitmap storage/Fortran and Write ppm file#Fortran
<lang fortran>program BrownianTree
``` use RCImageBasic
use RCImageIO
```
``` implicit none
```
``` integer, parameter :: num_particles = 1000
integer, parameter :: wsize = 800
```
``` integer, dimension(wsize, wsize) :: world
type(rgbimage) :: gworld
integer :: x, y
```
``` ! init seed
call init_random_seed
world = 0
call draw_brownian_tree(world)
```
``` call alloc_img(gworld, wsize, wsize)
call fill_img(gworld, rgb(0,0,0))
do y = 1, wsize
do x = 1, wsize
if ( world(x, y) /= 0 ) then
call put_pixel(gworld, x, y, rgb(255, 255, 255))
end if
end do
end do
```
``` open(unit=10, file='browniantree.ppm', action='write')
call output_ppm(10, gworld)
close(10)
```
``` call free_img(gworld)
```
contains
``` ! this code is taken from the GNU gfortran online doc
subroutine init_random_seed
integer :: i, n, clock
integer, dimension(:), allocatable :: seed
```
``` call random_seed(size = n)
allocate(seed(n))
call system_clock(count = clock)
seed = clock + 37 * (/ ( i - 1, i = 1, n) /)
call random_seed(put = seed)
deallocate(seed)
end subroutine init_random_seed
```
``` function randbetween(a, b) result(res) ! suppose a < b
integer, intent(in) :: a, b
integer :: res
```
``` real :: r
```
``` call random_number(r)
```
``` res = a + int((b-a)*r + 0.5)
```
``` end function randbetween
```
``` function bounded(v, ll, ul) result(res)
integer, intent(in) :: v, ll, ul
logical res
```
``` res = ( v >= ll ) .and. ( v <= ul )
end function bounded
```
``` subroutine draw_brownian_tree(w)
integer, dimension(:,:), intent(inout) :: w
```
``` integer :: px, py, dx, dy, i
integer :: xsize, ysize
```
``` xsize = size(w, 1)
ysize = size(w, 2)
```
``` w(randbetween(1, xsize), randbetween(1, ysize)) = 1
do i = 1, num_particles
px = randbetween(1, xsize)
py = randbetween(1, ysize)
do
dx = randbetween(-1, 1)
dy = randbetween(-1, 1)
if ( .not. bounded(dx+px, 1, xsize) .or. .not. bounded(dy+py, 1, ysize) ) then
px = randbetween(1, xsize)
py = randbetween(1, ysize)
else if ( w(px+dx, py+dy) /= 0 ) then
w(px, py) = 1
exit
else
py = py + dy
px = px + dx
end if
end do
end do
end subroutine draw_brownian_tree
```
end program</lang>
## Go
The interpretation here of "collide" in the case of a new particle generated on top of a pixel of the existing tree is not to ignore the particle, but to find a place for it nearby. This properly increases the brightness of the area, reflecting that a particle was generated in the area. Visually, it appears to strengthen existing spines of the tree.
Using standard image library: <lang go>package main
import (
``` "fmt"
"image"
"image/color"
"image/png"
"math/rand"
"os"
```
)
const w = 400 // image width const h = 300 // image height const n = 15000 // number of particles to add const frost = 255 // white
var g *image.Gray
func main() {
``` g = image.NewGray(image.Rectangle{image.Point{0, 0}, image.Point{w, h}})
// off center seed position makes pleasingly asymetrical tree
g.SetGray(w/3, h/3, color.Gray{frost})
```
generate:
``` for a := 0; a < n; {
// generate random position for new particle
rp := image.Point{rand.Intn(w), rand.Intn(h)}
if g.At(rp.X, rp.Y).(color.Gray).Y == frost {
// position is already set. find a nearby free position.
for {
rp.X += rand.Intn(3) - 1
rp.Y += rand.Intn(3) - 1
// execpt if we run out of bounds, consider the particle lost.
if !rp.In(g.Rect) {
continue generate
}
if g.At(rp.X, rp.Y).(color.Gray).Y != frost {
break
}
}
} else {
// else particle is in free space. let it wander
// until it touches tree
for !hasNeighbor(rp) {
rp.X += rand.Intn(3) - 1
rp.Y += rand.Intn(3) - 1
// but again, if it wanders out of bounds consider it lost.
if !rp.In(g.Rect) {
continue generate
}
}
}
// x, y now specify a free position toucing the tree.
g.SetGray(rp.X, rp.Y, color.Gray{frost})
a++
// progress indicator
if a%100 == 0 {
fmt.Println(a, "of", n)
}
}
f, err := os.Create("tree.png")
if err != nil {
fmt.Println(err)
return
}
err = png.Encode(f, g)
if err != nil {
fmt.Println(err)
}
f.Close()
```
}
var n8 = []image.Point{
``` {-1, -1}, {-1, 0}, {-1, 1},
{0, -1}, {0, 1},
{1, -1}, {1, 0}, {1, 1}}
```
func hasNeighbor(p image.Point) bool {
``` for _, n := range n8 {
if o.In(g.Rect) && g.At(o.X, o.Y).(color.Gray).Y == frost {
return true
}
}
return false
```
}</lang> Nearly the same, version below works with code from the bitmap task: <lang go>package main
// Files required to build supporting package raster are found in: // * Bitmap // * Grayscale image // * Write a PPM file
import (
``` "fmt"
"math/rand"
"raster"
```
)
const w = 400 // image width const h = 300 // image height const n = 15000 // number of particles to add const frost = 65535 // white
var g *raster.Grmap
func main() {
``` g = raster.NewGrmap(w, h)
// off center seed position makes pleasingly asymetrical tree
g.SetPx(w/3, h/3, frost)
var x, y int
```
generate:
``` for a := 0; a < n; {
// generate random position for new particle
x, y = rand.Intn(w), rand.Intn(h)
switch p, ok := g.GetPx(x, y); p {
case frost:
// position is already set. find a nearby free position.
for p == frost {
x += rand.Intn(3) - 1
y += rand.Intn(3) - 1
p, ok = g.GetPx(x, y)
```
``` // execpt if we run out of bounds, consider the particle lost.
if !ok {
continue generate
}
}
default:
// else particle is in free space. let it wander
// until it touches tree
for !hasNeighbor(x, y) {
x += rand.Intn(3) - 1
y += rand.Intn(3) - 1
// but again, if it wanders out of bounds consider it lost.
_, ok = g.GetPx(x, y)
if !ok {
continue generate
}
}
}
// x, y now specify a free position toucing the tree.
g.SetPx(x, y, frost)
a++
// progress indicator
if a%100 == 0 {
fmt.Println(a, "of", n)
}
}
g.Bitmap().WritePpmFile("tree.ppm")
```
}
var n8 = [][]int{
``` {-1, -1}, {-1, 0}, {-1, 1},
{ 0, -1}, { 0, 1},
{ 1, -1}, { 1, 0}, { 1, 1}}
```
func hasNeighbor(x, y int) bool {
``` for _, n := range n8 {
if p, ok := g.GetPx(x+n[0], y+n[1]); ok && p == frost {
return true
}
}
return false
```
}</lang>
The modules `Bitmap`, `Bitmap.Netpbm`, and `Bitmap.BW` are on Rosetta Code. The commented-out type signatures require scoped type variables in order to function.
<lang haskell>import Control.Monad import Control.Monad.ST import Data.STRef import Data.Array.ST import System.Random import Bitmap import Bitmap.BW import Bitmap.Netpbm
main = do
``` g <- getStdGen
(t, _) <- stToIO \$ drawTree (50, 50) (25, 25) 300 g
writeNetpbm "/tmp/tree.pbm" t
```
drawTree :: (Int, Int) -> (Int, Int) -> Int -> StdGen -> ST s (Image s BW, StdGen) drawTree (width, height) start steps stdgen = do
``` img <- image width height off
setPix img (Pixel start) on
gen <- newSTRef stdgen
let -- randomElem :: [a] -> ST s a
randomElem l = do
let (i, stdgen') = randomR (0, length l - 1) stdgen
writeSTRef gen stdgen'
return \$ l !! i
-- newPoint :: ST s (Int, Int)
newPoint = do
p <- randomElem border
c <- getPix img \$ Pixel p
if c == off then return p else newPoint
-- wander :: (Int, Int) -> ST s ()
wander p = do
next <- randomElem \$ filter (inRange pointRange) \$ adjacent p
c <- getPix img \$ Pixel next
if c == on then setPix img (Pixel p) on else wander next
replicateM_ steps \$ newPoint >>= wander
return (img, stdgen)
where pointRange = ((0, 0), (width - 1, height - 1))
adjacent (x, y) = [(x - 1, y - 1), (x, y - 1), (x + 1, y - 1),
(x - 1, y), (x + 1, y),
(x - 1, y + 1), (x, y + 1), (x + 1, y + 1)]
border = liftM2 (,) [0, width - 1] [0 .. height - 1] ++
liftM2 (,) [1 .. width - 2] [0, height - 1]
off = black
on = white</lang>
```
## Icon and Unicon
In this version the seed is randomly set within an inner area and particles are injected in an outer ring.
procedure main() # brownian tree
Density := .08 # % particles to area SeedArea := .5 # central area to confine seed ParticleArea := .7 # central area to exclude particles appearing Height := Width := 400 # canvas
Particles := Height * Width * Density Field := list(Height) every !Field := list(Width)
Size := sprintf("size=%d,%d",Width,Height) Fg := sprintf("fg=%s",?["green","red","blue"]) Label := sprintf("label=Brownian Tree %dx%d PA=%d%% SA=%d%% D=%d%%",
``` Width,Height,ParticleArea*100,SeedArea*100,Density*100)
```
WOpen(Label,Size,Fg,"bg=black") | stop("Unable to open Window")
sx := Height * SetInside(SeedArea) sy := Width * SetInside(SeedArea) Field[sx,sy] := 1 DrawPoint(sx,sy) # Seed the field
Lost := 0
every 1 to Particles do {
``` repeat {
px := Height * SetOutside(ParticleArea)
py := Width * SetOutside(ParticleArea)
if /Field[px,py] then
break # don't materialize in the tree
}
repeat {
dx := delta()
dy := delta()
if not ( xy := Field[px+dx,py+dy] ) then {
Lost +:= 1
next # lost try again
}
if \xy then
break # collision
px +:= dx # move to clear spot
py +:= dy
}
Field[px,py] := 1
DrawPoint(px,py) # Stick the particle
}
```
printf("Brownian Tree Complete: Particles=%d Lost=%d.\n",Particles,Lost) WDone() end
procedure delta() #: return a random 1 pixel perturbation
``` return integer(?0 * 3) - 1
```
end
procedure SetInside(core) #: set coord inside area
``` return core * ?0 + (1-core)/2
```
end
procedure SetOutside(core) #: set coord outside area
``` pt := ?0 * (1 - core)
pt +:= ( pt > (1-core)/2, core)
return pt
```
end</lang>
## J
<lang j>brtr=:4 :0
``` seed=. ?x
clip=. 0 >. (<:x) <."1 ]
near=. [: clip +"1/&(,"0/~i:1)
p=.i.0 2
mask=. 1 (<"1 near seed)} x\$0
field=.1 (<seed)} x\$0
for.i.y do.
p=. clip (p +"1 <:?3\$~\$p),?x
fix=. b#p
if.#fix do. NB. if. works around j602 bug: 0(0#a:)}i.0 0
p=. (-.b)# p
field=. 1 (<"1 fix)} field
end.
end.
field
```
)</lang>
Example use:
<lang j> require'viewmat'
``` viewmat 480 640 brtr 30000</lang>
```
Note that building a brownian tree like this takes a while and would be more interesting if this were animated.
## Java
Library: Swing
Library: AWT
<lang java>import java.awt.Graphics; import java.awt.image.BufferedImage; import java.util.*; import javax.swing.JFrame;
public class BrownianTree extends JFrame implements Runnable {
``` BufferedImage I;
private List<Particle> particles;
static Random rand = new Random();
```
``` public BrownianTree() {
super("Brownian Tree");
setBounds(100, 100, 400, 300);
setDefaultCloseOperation(EXIT_ON_CLOSE);
I = new BufferedImage(getWidth(), getHeight(), BufferedImage.TYPE_INT_RGB);
I.setRGB(I.getWidth() / 2, I.getHeight() / 2, 0xff00);
}
```
``` @Override
public void paint(Graphics g) {
g.drawImage(I, 0, 0, this);
}
```
``` public void run() {
for (int i = 0; i < 20000; i++) {
}
while (!particles.isEmpty()) {
for (Iterator<Particle> it = particles.iterator(); it.hasNext();) {
if (it.next().move()) {
it.remove();
}
}
repaint();
}
}
```
``` public static void main(String[] args) {
BrownianTree b = new BrownianTree();
b.setVisible(true);
}
```
``` private class Particle {
```
``` private int x, y;
```
``` private Particle() {
x = rand.nextInt(I.getWidth());
y = rand.nextInt(I.getHeight());
}
```
``` /* returns true if either out of bounds or collided with tree */
private boolean move() {
int dx = rand.nextInt(3) - 1;
int dy = rand.nextInt(3) - 1;
if ((x + dx < 0) || (y + dy < 0)
|| (y + dy >= I.getHeight()) || (x + dx >= I.getWidth())) {
return true;
}
x += dx;
y += dy;
if ((I.getRGB(x, y) & 0xff00) == 0xff00) {
I.setRGB(x - dx, y - dy, 0xff00);
return true;
}
return false;
}
}
```
}</lang>
This is an alternate version which is a port of most of the code here. This code does not use a GUI and saves the output to image.png. <lang Java>import java.awt.Point; import java.awt.image.BufferedImage; import java.io.File; import java.io.IOException;
import javax.imageio.ImageIO;
public class BasicBrownianTree {
``` private int pixelsLost;
private Point p;
private Point nextP;
private int pixelCount;
private int width;
private int height;
private int color;
private BufferedImage img;
```
``` public BasicBrownianTree( int argb, int size, double density ) {
pixelsLost = 0;
p = new Point();
nextP = new Point();
width = size;
height = size;
color = argb;
pixelCount = (int) ( width * height * density );
img = new BufferedImage( width, height, BufferedImage.TYPE_INT_ARGB );
}
```
``` public void generate() {
// print text to the console
System.out.println( "Drawing " + pixelCount + " pixels" );
int background = img.getRGB( 0, 0 );
img.setRGB( width / 2, height / 2, color );
```
``` for( int i = 0; i < pixelCount; i++ ) {
p.x = (int) ( Math.random() * width );
p.y = (int) ( Math.random() * height );
```
``` while ( true ) {
int dx = (int) ( Math.random() * 3 ) - 1;
int dy = (int) ( Math.random() * 3 ) - 1;
nextP.setLocation( p.x + dx, p.y + dy );
// handle out-of-bounds
if ( nextP.x < 0 || nextP.x >= width || nextP.y < 0
|| nextP.y >= height ) {
// increment the number of pixels lost and escape the loop
pixelsLost++;
break;
}
if ( img.getRGB( nextP.x, nextP.y ) != background ) {
img.setRGB( p.x, p.y, color );
break;
}
p.setLocation( nextP );
}
// Print a message every 2%
if ( i % ( pixelCount / 50 ) == 0 ) {
System.out.println( "Done with " + i + " pixels" );
}
}
// We're done. Let the user know how many pixels were lost
System.out.println( "Finished. Pixels lost = " + pixelsLost );
}
```
``` public BufferedImage getImage() {
return img;
}
```
``` public int getWidth() {
return width;
}
```
``` public int getHeight() {
return height;
}
```
``` public static void main( String[] args ) {
// create the new generator
BasicBrownianTree generator = new BasicBrownianTree( 0x664444ff, 400, 0.4 );
// generate the image
generator.generate();
try {
// save the image to the file "image.png"
ImageIO.write( generator.getImage(), "png", new File( "image.png" ) );
} catch ( IOException e ) {
e.printStackTrace();
}
}
```
}</lang>
## JavaScript + <canvas>
Live version <lang javascript>function brownian(canvasId, messageId) {
``` var canvas = document.getElementById(canvasId);
var ctx = canvas.getContext("2d");
```
``` // Options
var drawPos = true;
var seedResolution = 50;
var clearShade = 0; // 0..255
// Static state
var width = canvas.width;
var height = canvas.height;
var cx = width/2;
var cy = height/2;
```
``` // Utilities
return Math.sqrt((x-cx)*(x-cy)+(y-cx)*(y-cy));
}
function test(x, y) {
if (x < 0 || y < 0 || x >= width || y >= height)
return false;
var data = ctx.getImageData(x, y, 1, 1).data;
}
function setc(x, y, c) {
//var imgd = ctx.createImageData(1, 1);
//var pix = imgd.data;
//pix[0] = pix[1] = pix[2] = c == 255 ? 255 : shade;
//pix[3] = 255;
//ctx.putImageData(imgd, x, y);
//ctx.fillStyle = "rgba("+c+", "+c+", "+c+", 1)";
if (c) {
} else {
ctx.fillStyle = clearStyle;
}
ctx.fillRect (x, y, 1, 1);
}
function set(x,y) {
setc(x,y,true);
}
function clear(x,y) {
setc(x,y,false);
}
```
``` // Initialize canvas to blank opaque
ctx.fillStyle = clearStyle;
ctx.fillRect (0, 0, width, height);
```
``` // Current position
var x;
var y;
```
``` // Farthest distance from center a particle has yet been placed.
```
``` // Place seed
set(cx, cy);
```
``` // Choose a new random position for a particle (not necessarily unoccupied)
function newpos() {
// Wherever particles are injected, the tree will tend to grow faster
// toward it. Ideally, particles wander in from infinity; the best we
// could do is to have them wander in from the edge of the field.
// But in order to have the rendering occur in a reasonable time when
// the seed is small, without too much visible bias, we instead place
// the particles in a coarse grid. The final tree will cover every
// point on the grid.
//
// There's probably a better strategy than this.
x = Math.floor(Math.random()*(width/seedResolution))*seedResolution;
y = Math.floor(Math.random()*(height/seedResolution))*seedResolution;
}
newpos();
```
``` var animation;
animation = window.setInterval(function () {
if (drawPos) clear(x,y);
for (var i = 0; i < 10000; i++) {
var ox = x;
var oy = y;
// Changing this to use only the first four directions will result
// in a denser tree.
switch (Math.floor(Math.random()*8)) {
case 0: x++; break;
case 1: x--; break;
case 2: y++; break;
case 3: y--; break;
case 4: x++; y++; break;
case 5: x--; y++; break;
case 6: x++; y--; break;
case 7: x--; y--; break;
}
if (x < 0 || y < 0 ||
x >= width || y >= height ||
// wandered out of bounds or out of interesting range of the
// tree, so pick a new spot
var progress = 1000;
do {
newpos();
progress--;
} while ((test(x-1,y-1) || test(x,y-1) || test(x+1,y-1) ||
test(x-1,y ) || test(x,y ) || test(x+1,y ) ||
test(x-1,y+1) || test(x,y+1) || test(x+1,y+1)) && progress > 0);
if (progress <= 0) {
document.getElementById(messageId).appendChild(
document.createTextNode("Stopped for lack of room."));
clearInterval(animation);
break;
}
}
if (test(x, y)) {
// hit something, mark where we came from and pick a new spot
set(ox,oy);
newpos();
}
}
if (drawPos) set(x,y);
}, 1);
```
}</lang>
<lang html><html>
```<head>
<script src="brownian.js"></script>
<canvas id="canvas" width="402" height="402" style="border: 2px inset;"></canvas>
```
```</body>
```
</html></lang>
## Liberty BASIC
<lang lb>'[RC]Brownian motion tree
``` nomainwin
dim screen(600,600)
WindowWidth = 600
WindowHeight = 600
open "Brownian" for graphics_nsb_nf as #1
#1 "trapclose [quit]"
#1 "down ; fill blue"
particles=500
```
``` 'draw starting circle and mid point
for n= 1 to 360
#1, "color white ; set ";x;" ";y
screen(x,y)=1
next n
#1, "color white ; set 300 300"
screen(300,300)=1
```
``` 'set up initial particles
dim particle(particles,9)'x y deltax deltay rotx roty
for n = 1 to particles
gosub [randomparticle]
next
```
``` 'start timed drawing loop
timer 17, [draw]
wait
```
``` [draw]
scan
for n = 1 to particles
oldx=particle(n,1)
oldy=particle(n,2)
```
``` 'erase particle
if not(screen(oldx,oldy)) then
#1 "color blue ; set ";oldx;" ";oldy
end if
```
``` 'move particle x
particle(n,5)=particle(n,5)+6 mod 360
if particle(n,1)>599 or particle(n,1)<1 then gosub [randomparticle]
```
``` 'move particle y
particle(n,6)=particle(n,6)+6 mod 360
if particle(n,2)>599 or particle(n,2)<1 then gosub [randomparticle]
```
``` 'checkhit
x=particle(n,1)
y=particle(n,2)
if screen(x-1,y-1) or screen(x-1,y) or screen(x-1,y+1)_
or screen(x,y-1) or screen(x,y) or screen(x,y+1)_
or screen(x+1,y-1) or screen(x+1,y) or screen(x+1,y+1) then
#1 "color white ; set ";particle(n,1);" ";particle(n,2)
screen(particle(n,1),particle(n,2))=1
else
#1 "color red ; set ";particle(n,1);" ";particle(n,2)
end if
next
wait
```
``` [randomparticle]
particle(n,1)=int(rnd(0)*599)+1
particle(n,2)=int(rnd(0)*599)+1
particle(n,3)=int(2-rnd(0)*4)
particle(n,4)=int(2-rnd(0)*4)
particle(n,5)=int(rnd(0)*360)
particle(n,6)=int(rnd(0)*360)
return
```
``` [quit]
timer 0
close #1
end</lang>
```
## Lua
The output is stored in as a ppm-image. The source code of these output-functions is located at Bitmap/Write a PPM file#Lua, Grayscale image#Lua, Basic bitmap storage#Lua. <lang lua>function SetSeed( f )
``` for i = 1, #f[1] do -- the whole boundary of the scene is used as the seed
f[1][i] = 1
f[#f][i] = 1
end
for i = 1, #f do
f[i][1] = 1
f[i][#f[1]] = 1
end
```
end
function SetParticle( f )
``` local pos_x, pos_y
repeat
pos_x = math.random( #f )
pos_y = math.random( #f[1] )
until f[pos_x][pos_y] == 0
return pos_x, pos_y
```
end
function Iterate( f, num_particles )
``` for i = 1, num_particles do
local pos_x, pos_y = SetParticle( f )
while true do
local dx = math.random(5) - 3
local dy = math.random(5) - 3
```
``` if ( pos_x+dx >= 1 and pos_x+dx <= #f and pos_y+dy >= 1 and pos_y+dy <= #f[1] ) then
if f[pos_x+dx][pos_y+dy] ~= 0 then
f[pos_x][pos_y] = 1
break
else
pos_x = pos_x + dx
pos_y = pos_y + dy
end
end
end
end
```
end
size_x, size_y = 400, 400 -- size of the scene num_particles = 16000
math.randomseed( os.time() )
f = {} for i = 1, size_x do
``` f[i] = {}
for j = 1, size_y do
f[i][j] = 0
end
```
end
SetSeed( f ) Iterate( f, num_particles )
-- prepare the data for writing into a ppm-image file for i = 1, size_x do
``` for j = 1, size_y do
if f[i][j] == 1 then f[i][j] = 255 end
end
```
end Write_PPM( "brownian_tree.ppm", ConvertToColorImage(f) )</lang>
## Mathematica
There is a prettier version at the Mathematica demo site. Its source code is also available there but it is not mine.
Loose
Translation of: D
<lang Mathematica>canvasdim = 1000; n = 0.35*canvasdim^2; canvas = ConstantArray[0, {canvasdim, canvasdim}]; init = Floor@(0.5*{canvasdim, canvasdim}); (*RandomInteger[canvasdim,2]*) canvas[[init1, init2]] = 1; (*1st particle initialized to midpoint*)
Monitor[ (*Provides real-time intermediate result monitoring*)
```Do[
particle = RandomInteger[canvasdim, 2];
While[True,
ds = RandomInteger[{-1, 1}, 2];
While[ (*New Particle Domain Limit Section*)
!And @@ (0 < (particle + ds)# <= canvasdim & /@ {1, 2}),
particle = RandomInteger[canvasdim, 2];
];
(* Particle Aggregation Section *)
If[canvas[[(particle + ds)1, (particle + ds)2]] > 0,
canvas[[particle1, particle2]] = i;
Break[],
particle += ds
];
],
{i, n}],
{i, (particle + ds), MatrixPlot@canvas}
]
```
MatrixPlot[canvas,FrameTicks->None,ColorFunction->"DarkRainbow",ColorRules->{0 -> None}]</lang>
Result:
## OCaml
Translation of: D
<lang ocaml>let world_width = 400 let world_height = 400 let num_particles = 20_000
let () =
``` assert(num_particles > 0);
assert(world_width * world_height > num_particles);
```
let dla ~world =
``` (* put the tree seed *)
world.(world_height / 2).(world_width / 2) <- 1;
for i = 1 to num_particles do
(* looping helper function *)
let rec aux px py =
(* randomly choose a direction *)
let dx = (Random.int 3) - 1 (* offsets *)
and dy = (Random.int 3) - 1 in
```
``` if dx + px < 0 || dx + px >= world_width ||
dy + py < 0 || dy + py >= world_height then
(* plop the particle into some other random location *)
aux (Random.int world_width) (Random.int world_height)
else if world.(py + dy).(px + dx) <> 0 then
(* bumped into something, particle set *)
world.(py).(px) <- 1
else
aux (px + dx) (py + dy)
in
(* set particle's initial position *)
aux (Random.int world_width) (Random.int world_height)
done
```
let to_pbm ~world =
``` print_endline "P1"; (* Type=Portable bitmap, Encoding=ASCII *)
Printf.printf "%d %d\n" world_width world_height;
Array.iter (fun line ->
Array.iter print_int line;
print_newline()
) world
```
let () =
``` Random.self_init();
let world = Array.make_matrix world_width world_height 0 in
dla ~world;
to_pbm ~world;
```
</lang>
better to compile to native code to get a faster program:
```\$ ocamlopt -o brownian_tree.opt brownian_tree.ml
\$ ./brownian_tree.opt | display -```
## Octave
Translation of: C
<lang octave>function r = browniantree(xsize, ysize = xsize, numparticle = 1000)
``` r = zeros(xsize, ysize, "uint8");
r(unidrnd(xsize), unidrnd(ysize)) = 1;
for i = 1:numparticle
px = unidrnd(xsize-1)+1;
py = unidrnd(ysize-1)+1;
while(1)
dx = unidrnd(2) - 1;
dy = unidrnd(2) - 1;
if ( (dx+px < 1) || (dx+px > xsize) || (dy+py < 1) || (dy+py > ysize) )
```
px = unidrnd(xsize-1)+1; py = unidrnd(ysize-1)+1;
``` elseif ( r(px+dx, py+dy) != 0 )
```
r(px, py) = 1; break;
``` else
```
px += dx; py += dy;
``` endif
endwhile
endfor
```
endfunction
r = browniantree(200); r( r > 0 ) = 255; jpgwrite("browniantree.jpg", r, 100); % image package</lang>
## Perl
Simulation code. Tremendously slow, partly because it doesn't use a grid-based collision checking. Showing three sample images with different STEP and ATTRACT parameters, to demonstrate how sensitive the result is to them.
Code runs until the tree reached specified radius. Output is written to "test.eps" of wherever the current directory is. The 0-0 sample took maybe 3 hours (I don't really know, I went for dinner.) <lang perl>sub PI() { atan2(1,1) * 4 } # The, er, pi sub STEP() { .5 } # How far does the particle move each step. Affects
``` # both speed and accuracy greatly
```
sub STOP_RADIUS() { 100 } # When the tree reaches this far from center, end
1. At each step, move this much towards center. Bigger numbers help the speed because
2. particles are less likely to wander off, but greatly affects tree shape.
3. Should be between 0 and 1 ish. Set to 0 for pain.
sub ATTRACT() { .2 }
my \$r_start = 3; my \$max_dist = 0;
sub dist2 {
``` my (\$dx, \$dy) = (\$_[0][0] - \$_[1][0], \$_[0][1] - \$_[1][1]);
\$dx * \$dx + \$dy * \$dy
```
}
sub move {
``` my \$p = shift;
# moved too far, kill particle
# return if dist2(\$p, [0, 0]) > 2 * \$r_start * \$r_start;
\$p->[0] += 2 * \$r_start while \$p->[0] < -\$r_start;
\$p->[0] -= 2 * \$r_start while \$p->[0] > \$r_start;
\$p->[1] += 2 * \$r_start while \$p->[1] < -\$r_start;
\$p->[1] -= 2 * \$r_start while \$p->[1] > \$r_start;
```
``` my (\$ix, \$iy) = (int(\$p->[0]), int(\$p->[1]));
my \$dist = 2 * \$r_start * \$r_start;
my \$nearest;
```
``` # see if the particle is close enough to stick to an exist one
for (\$ix - 1 .. \$ix + 1) {
my \$idx = STOP_RADIUS + \$_;
next if \$idx > 2 * STOP_RADIUS || \$idx < 0;
my \$xs = \$particles[ \$idx ];
for (\$iy - 1 .. \$iy + 1) {
my \$idx = STOP_RADIUS + \$_;
next if \$idx > 2 * STOP_RADIUS || \$idx < 0;
for (@{ \$xs->[ \$idx ] }) {
my \$d = dist2(\$p, \$_);
next if \$d > 2;
next if \$d > \$dist;
```
``` \$dist = \$d;
\$nearest = \$_;
}
}
}
```
``` # yes, found one
if (\$nearest) {
my \$displace = [ \$p->[0] - \$nearest->[0],
\$p->[1] - \$nearest->[1] ];
my \$angle = atan2(\$displace->[1], \$displace->[0]);
\$p->[0] = \$nearest->[0] + cos(\$angle);
\$p->[1] = \$nearest->[1] + sin(\$angle);
```
``` push @{\$particles[\$ix + STOP_RADIUS][\$iy + STOP_RADIUS]}, [ @\$p ];
\$dist = sqrt dist2(\$p);
```
``` if (\$dist + 10 > \$r_start && \$r_start < STOP_RADIUS + 10) {
\$r_start = \$dist + 10
}
if (int(\$dist + 1) > \$max_dist) {
\$max_dist = int(\$dist + 1);
# write_eps();
# system('pstopnm -portrait -xborder 0 -yborder 0 test.eps 2> /dev/null');
# system('pnmtopng test.eps001.ppm 2>/dev/null > test.png');
return 3 if \$max_dist >= STOP_RADIUS;
}
return 2;
}
```
``` # random walk
my \$angle = rand(2 * PI);
\$p->[0] += STEP * cos(\$angle);
\$p->[1] += STEP * sin(\$angle);
```
``` # drag particle towards center by some distance
my \$nudge;
if (sqrt(dist2(\$p, [0, 0])) > STOP_RADIUS + 1) {
\$nudge = 1;
} else {
\$nudge = STEP * ATTRACT;
}
```
``` if (\$nudge) {
\$angle = atan2(\$p->[1], \$p->[0]);
\$p->[0] -= \$nudge * cos(\$angle);
\$p->[1] -= \$nudge * sin(\$angle);
}
```
``` return 1;
```
}
my \$count; PARTICLE: while (1) {
``` my \$a = rand(2 * PI);
my \$p = [ \$r_start * cos(\$a), \$r_start * sin(\$a) ];
while (\$_ = move(\$p)) {
given (\$_) {
when (1) { next }
when (2) { \$count++; last; }
when (3) { last PARTICLE }
default { last }
}
}
print STDERR "\$count \$max_dist/@{[int(\$r_start)]}/@{[STOP_RADIUS]}\r" unless \$count% 7;
```
}
sub write_eps {
``` my \$size = 128;
my \$p = \$size / (STOP_RADIUS * 1.05);
my \$b = STOP_RADIUS * \$p;
if (\$p < 1) {
\$p = 1;
}
```
``` my \$hp = \$p / 2;
```
``` open OUT, ">", "test.eps";
```
``` # print EPS to standard out
```
%!PS-Adobe-3.0 EPSF-3.0 %%BoundingBox: 0 0 @{[\$size*2, \$size*2]} \$size \$size translate /l{ rlineto }def /c{ \$hp 0 360 arc fill }def -\$size -\$size moveto \$size 2 mul 0 l 0 \$size 2 mul l -\$size 2 mul 0 l closepath 0 setgray fill 0 setlinewidth .1 setgray 0 0 \$b 0 360 arc stroke .8 setgray /TimesRoman findfont 16 scalefont setfont -\$size 10 add \$size -16 add moveto (Step = @{[STEP]} Attract = @{[ATTRACT]}) show 0 1 0 setrgbcolor newpath HEAD
``` for (@particles) {
for (@\$_) {
printf OUT "%.3g %.3g c ", map { \$_ * \$p } @\$_ for @\$_;
}
}
print OUT "\n%%EOF";
close OUT;
```
}
write_eps;</lang>
## Perl 6
This solution spawns new Particles at a growing square border and displays the Tree every 50 particles and at the end using unicode UPPER/LOWER HALF BLOCK and FULL BLOCK.
With the given size of 100 and particle count of 1000, this takes about 25 seconds with Niecza on my notebook.
<lang Perl6>constant size = 100; constant particlenum = 1_000;
constant mid = size div 2;
my \$spawnradius = 5; my @map;
sub set(\$x, \$y) {
``` @map[\$x][\$y] = True;
```
}
sub get(\$x, \$y) {
``` return @map[\$x][\$y] || False;
```
}
set(mid, mid); my @blocks = " ","\c[UPPER HALF BLOCK]", "\c[LOWER HALF BLOCK]","\c[FULL BLOCK]";
sub infix:<█>(\$a, \$b) {
``` @blocks[\$a + 2 * \$b]
```
}
sub display {
``` my \$start = 0;
my \$end = size;
say (for \$start, \$start + 2 ... \$end -> \$y {
(for \$start..\$end -> \$x {
if abs((\$x&\$y) - mid) < \$spawnradius {
get(\$x, \$y) █ get(\$x, \$y+1);
} else {
" "
}
}).join
}).join("\n")
```
}
for ^particlenum -> \$progress {
``` my Int \$x;
my Int \$y;
my &reset = {
repeat {
(\$x, \$y) = (\$y, \$x) if (True, False).pick();
} while get(\$x,\$y);
}
reset;
```
``` while not get(\$x-1|\$x|\$x+1, \$y-1|\$y|\$y+1) {
\$x = (\$x-1, \$x, \$x+1).pick;
\$y = (\$y-1, \$y, \$y+1).pick;
if (False xx 3, True).pick {
\$x = \$x >= mid ?? \$x - 1 !! \$x + 1;
\$y = \$y >= mid ?? \$y - 1 !! \$y + 1;
}
if abs((\$x | \$y) - mid) > \$spawnradius {
reset;
}
}
set(\$x,\$y);
display if \$progress %% 50;
if \$spawnradius < mid && abs((\$x|\$y) - mid) > \$spawnradius - 5 {
}
```
}
say ""; display; say ""; say "time elapsed: ", (now - BEGIN { now }).Num.fmt("%.2f"), " seconds"; say "";</lang>
## PicoLisp
(de brownianTree (File Size Cnt)
``` (let Img (grid Size Size)
(put Img (/ Size 2) (/ Size 2) 'pix T)
(use (P Q)
(do Cnt
(setq P (get Img (rand 1 Size) (rand 1 Size)))
(loop
(setq Q ((if2 (rand T) (rand T) north east south west) P))
(T (; Q pix) (put P 'pix T))
(setq P (or Q (get Img (rand 1 Size) (rand 1 Size)))) ) ) )
(out "img.pbm"
(prinl "P1")
(prinl Size " " Size)
(for L Img
(for This L
(prin (if (: pix) 1 0)) )
(prinl) ) ) ) )</lang>
```
Use:
```(brownianTree "img.pbm" 300 9000)
(call 'display "img.pbm")```
## PureBasic
<lang PureBasic>#Window1 = 0
1. Image1 = 0
1. NUM_PARTICLES = 3000
2. width = 200
3. height = 200
4. xmax = #width -3
5. ymax = #height -3
Define.i Event ,i ,x,y
If OpenWindow(#Window1, 0, 0, #width, #height, "Brownian Tree PureBasic Example", #PB_Window_SystemMenu )
``` If CreateImage(#Image1, #width, #height)
StartDrawing(ImageOutput(#Image1))
FrontColor(\$FFFFFF)
Plot( Random(#xmax) , Random(#ymax ))
StopDrawing()
For i = 1 To #NUM_PARTICLES
x = Random(#xmax)+1 : y = Random (#ymax)+1
StartDrawing(ImageOutput(#Image1))
While Point(x+1, y+1) + Point(x, y+1)+Point(x+1, y)+Point(x-1, y-1)+Point(x-1, y)+Point(x, y-1) = 0
x = x + (Random(2)-1) : y = y + (Random(2)-1)
If x < 1 Or x > #xmax Or y < 1 Or y > #ymax
x = Random(#xmax)+1 : y = Random (#ymax)+1
EndIf
Wend
Plot(x,y)
StopDrawing()
Next
EndIf
```
``` Repeat
Event = WaitWindowEvent()
Until Event = #PB_Event_CloseWindow
```
EndIf</lang>
## Python
Library: pygame
<lang python>import pygame, sys, os from pygame.locals import * from random import randint pygame.init()
MAXSPEED = 15 SIZE = 3 COLOR = (45, 90, 45) WINDOWSIZE = 400 TIMETICK = 1 MAXPART = 50
freeParticles = pygame.sprite.Group() tree = pygame.sprite.Group()
window = pygame.display.set_mode((WINDOWSIZE, WINDOWSIZE)) pygame.display.set_caption("Brownian Tree")
screen = pygame.display.get_surface()
class Particle(pygame.sprite.Sprite):
``` def __init__(self, vector, location, surface):
pygame.sprite.Sprite.__init__(self)
self.vector = vector
self.surface = surface
self.accelerate(vector)
self.rect = pygame.Rect(location[0], location[1], SIZE, SIZE)
self.surface.fill(COLOR, self.rect)
```
``` def onEdge(self):
if self.rect.left <= 0:
self.vector = (abs(self.vector[0]), self.vector[1])
elif self.rect.top <= 0:
self.vector = (self.vector[0], abs(self.vector[1]))
elif self.rect.right >= WINDOWSIZE:
self.vector = (-abs(self.vector[0]), self.vector[1])
elif self.rect.bottom >= WINDOWSIZE:
self.vector = (self.vector[0], -abs(self.vector[1]))
```
``` def update(self):
if freeParticles in self.groups():
self.surface.fill((0,0,0), self.rect)
self.remove(freeParticles)
if pygame.sprite.spritecollideany(self, freeParticles):
self.accelerate((randint(-MAXSPEED, MAXSPEED),
randint(-MAXSPEED, MAXSPEED)))
elif pygame.sprite.spritecollideany(self, tree):
self.stop()
else:
self.onEdge()
```
``` if (self.vector == (0,0)) and tree not in self.groups():
self.accelerate((randint(-MAXSPEED, MAXSPEED),
randint(-MAXSPEED, MAXSPEED)))
self.rect.move_ip(self.vector[0], self.vector[1])
self.surface.fill(COLOR, self.rect)
```
``` def stop(self):
self.vector = (0,0)
self.remove(freeParticles)
```
``` def accelerate(self, vector):
self.vector = vector
```
NEW = USEREVENT + 1 TICK = USEREVENT + 2
pygame.time.set_timer(NEW, 50) pygame.time.set_timer(TICK, TIMETICK)
def input(events):
``` for event in events:
if event.type == QUIT:
sys.exit(0)
elif event.type == NEW and (len(freeParticles) < MAXPART):
Particle((randint(-MAXSPEED,MAXSPEED),
randint(-MAXSPEED,MAXSPEED)),
(randint(0, WINDOWSIZE), randint(0, WINDOWSIZE)),
screen)
elif event.type == TICK:
freeParticles.update()
```
half = WINDOWSIZE/2 tenth = WINDOWSIZE/10
root = Particle((0,0),
``` (randint(half-tenth, half+tenth),
randint(half-tenth, half+tenth)), screen)
```
root.stop()
while True:
``` input(pygame.event.get())
pygame.display.flip()</lang>
```
## Racket
<lang racket>#lang racket (require 2htdp/image)
The unsafe fixnum ops are faster than the checked ones,
but if you get anything wrong with them, they'll bite.
If you experience any problems reactivate the
(require racket/fixnum) and instead of the unsafe requirement
below...
we have tested this...
1. (require racket/fixnum)
so we can use this...
(require racket/require
``` (only-in racket/fixnum make-fxvector in-fxvector)
(filtered-in
(? (name) (regexp-replace #rx"unsafe-" name ""))
racket/unsafe/ops))
```
This implementation uses a 1d, mutable, fixnum vector
there's a lot of work done making the tree, so this optimisation
at the expense of clarity has been made. Sorry, guys!
(define (brownian-tree w h collisions n-particles seed-tree
``` generate-particle walk-particle)
(define w*h (fx* w h))
(define V (make-fxvector w*h))
(define (collision? x.y) (fx> (fxvector-ref V x.y) 0))
;; The main loop
(define (inner-b-t collisions particles)
(cond
[(fx= 0 collisions) V]
[else
(define-values (new-particles new-collisions)
(for/fold
((prtcls null)
(clsns 0))
((x.y particles)
#:break (fx= collisions clsns))
(define new-particle (walk-particle x.y w h w*h))
(cond
[(not new-particle) ; it died!
(values (cons (generate-particle V w h w*h) prtcls) clsns)]
[(collision? new-particle)
(fxvector-set! V x.y 1)
(values (cons (generate-particle V w h w*h) prtcls) (add1 clsns))]
[else
(values (cons new-particle prtcls) clsns)])))
(when (fx> new-collisions 0)
(define remain (fx- collisions new-collisions))
(unless (fx= (exact-floor (* 10 (log collisions)))
(exact-floor (* 10 (log (fxmax 1 remain)))))
(eprintf "~a (e^~a)~%" remain (log (fxmax 1 remain))))
(log-info "~a new collisions: ~a remain~%" new-collisions remain))
(inner-b-t (fxmax 0 (fx- collisions new-collisions)) new-particles)]))
;; Seed the tree
(seed-tree V w h)
(inner-b-t collisions
(build-list n-particles
(lambda (x) (generate-particle V w h w*h)))))
```
See below for why we do the (fxremainder ...) test
(define (uniform-particle-generator v w h w*h)
``` (define x.y (random w*h))
(define valid-x.y?
(and
(fx= (fxvector-ref v x.y) 0) ; start on empty cell
(fx> (fxremainder x.y w) 0))) ; not on left edge
; if it's valid take it otherwise regenerate
(if valid-x.y? x.y (uniform-particle-generator v w h w*h)))
```
The boundaries to the walker are to remain within the limits of
the vector... however, unless we stop particles going off the
east/west edges, the tree will be formed on a cylinder as the
particles wrap. So we kill particles that reach the left edge
either by decrement from the right or by incrementing and wrapping.
This is is tested with (= 0 (remainder x.y w)).
(define (brownian-particle-walker x.y w h w*h)
``` (define dx (fx- (random 3) 1))
(define dy (fx- (random 3) 1))
(define new-x.y (fx+ x.y (fx+ dx (fx* w dy))))
(and (fx> new-x.y 0) (fx< new-x.y w*h)
(fx> (fxremainder new-x.y w) 0)
new-x.y))
```
These seed functions modify v however you want!
(define (seed-middle v w h)
``` (fxvector-set! v (+ (quotient w 2) (* w (quotient h 2))) 1))
```
(define (seed-circle v w h)
``` (for ((a (in-range 0 360 120)))
(define x (exact-floor (* w 1/8 (+ 4 (sin (* pi 1/180 a))))))
(define y (exact-floor (* h 1/8 (+ 4 (cos (* pi 1/180 a))))))
(fxvector-set! v (+ x (* w y)) 1)))
```
SCALE is a general purpose knob for modifying the size of the problem
complexity increases with the sqaure of SCALE (at least)
(define SCALE 1) (define tree-W (* SCALE 320)) (define tree-H (* SCALE 240)) (define tree-W.H (* tree-W tree-H))
play with tree-PARTICLES -- small values will lead to a smaller tree
as the tree moves towards the edges, more particles might affect its shape
(define tree-PARTICLES (quotient tree-W.H 4))
these are the particles that are bimbling around at any one time. If it's
too low, you might get bored waiting for a collision... if it's too high
you might get inappropriate collisions
(define working-PARTICLES (quotient tree-W.H 300))
(define b-t (time
``` (brownian-tree
tree-W tree-H tree-PARTICLES working-PARTICLES
seed-middle
uniform-particle-generator
brownian-particle-walker)))
```
(define (b-t-value->color c) (case c ((1) "black") (else "white"))) (define img (color-list->bitmap
``` (for*/list ((x (in-fxvector b-t)))
(b-t-value->color x))
tree-W tree-H))
```
img (save-image img "brownian-tree.png")</lang>
## REXX
A large part of the REXX program's prologue was to handle the various options.
With a little more REXX code, a petri dish option could be added, that is, when a particle hits the edge,
it "bounces" back. Also, the field could then be displayed as a round area (a petri dish).
REXX code was added to display snapshots of the field, either after so many cycles, and/or after some
elapsed time has elapsed (whole seconds only). This makes for some fascinating observations.
Program note: to keep things simple, the (system) command to clear the screen was hard-coded as CLS. <lang rexx>/*REXX program shows Brownian motion of dust in a field with one seed.*/ parse arg height width motes randSeed . /*get args from the C.L. */ if height== | height==',' then height=0 /*None? Use the default*/ if width== | width==',' then width=0 /* " " " " */ if motes== | motes==',' then motes='10%' /*% dust motes in field, */
``` /* ··· otherwise just the number.*/
```
tree = '*' /*an affixed dust speck (tree). */ mote = '·' /*char for a loose mote (of dust)*/ hole = ' ' /*char for an empty spot in field*/ clearScr = 'CLS' /*(DOS?) command to clear screen.*/ eons = 1000000 /* # cycles for Brownian movement*/ snapshot = 0 /*every n winks, show snapshot.*/ snaptime = 1 /*every n secs, show snapshot.*/ seedPos = 30 45 /*place seed in this field pos. */ seedPos = 0 /*if =0, use middle of the field.*/
``` /*if -1, use a random placement. */
/*otherwise, place it at seedPos.*/
/*set RANDSEED for repeatability.*/
```
if datatype(randSeed,'W') then call random ,,randSeed /*if #, use it*/
``` /* [↑] set the 1st random number*/
```
if height==0 | width==0 then _=scrsize() /*not all REXXes have SCRSIZE.*/ if height==0 then height=word(_,1)-3 /*adjust for border.*/ if width==0 then width=word(_,2)-1 /* " " " */
``` seedAt=seedPos
```
if seedPos== 0 then seedAt=width%2 height%2 if seedPos==-1 then seedAt=random(1,width) random(1,height) parse var seedAt xs ys . /*obtain X & Y seed coördinates*/
``` /* [↓] if right-most≡'%', use %.*/
```
if right(motes,1)=='%' then motes=height * width * strip(motes,,'%') %100 @.=hole /*create the field, all empty. */
``` do j=1 for motes /*sprinkle # dust motes randomly.*/
rx=random(1, width); ry=random(1,height); @.rx.ry=mote
end /*j*/ /* [↑] place a mote at random. */
/*plant the seed from which the */
/*tree will grow from dust motes */
```
@.xs.ys=tree /*that affixed themselves. */ call show /*show field before we mess it up*/ tim=0 /*the time (in secs) of last show*/ loX=1; hiX= width /*used to optimize mote searching*/ loY=1; hiY=height /* " " " " " */
``` /*═════════════════════════════soooo, this is Brownian motion.*/
do winks=1 for eons until \motion /*EONs is used instead of ∞. */
motion=0 /*turn off Brownian motion flag. */
if snapshot\==0 then if winks//snapshot==0 then call show
if snaptime\==0 then do; t=time('S')
if t\==tim & t//snaptime==0 then do
tim=time('s')
call show
end
end
minX=loX; maxX=hiX /*as the tree grows, the search */
minY=loY; maxY=hiY /* for dust motes gets faster. */
loX= width; hiX=1 /*used to limit mote searching. */
loY=height; hiY=1 /* " " " " " */
```
``` do x =minX to maxX; xm=x-1; xp=x+1
do y=minY to maxY; if @.x.y\==mote then iterate
if x<loX then loX=x; if x>hiX then hiX=x /*is faster than: hiX=max(X hiX) */
if y<loY then loY=y; if y>hiY then hiY=y /*is faster than: hiY=max(y hiY) */
if @.xm.y ==tree then do; @.x.y=tree; iterate; end /*neighbor?*/
if @.xp.y ==tree then do; @.x.y=tree; iterate; end /*neighbor?*/
ym=y-1
if @.x.ym ==tree then do; @.x.y=tree; iterate; end /*neighbor?*/
if @.xm.ym==tree then do; @.x.y=tree; iterate; end /*neighbor?*/
if @.xp.ym==tree then do; @.x.y=tree; iterate; end /*neighbor?*/
yp=y+1
if @.x.yp ==tree then do; @.x.y=tree; iterate; end /*neighbor?*/
if @.xm.yp==tree then do; @.x.y=tree; iterate; end /*neighbor?*/
if @.xp.yp==tree then do; @.x.y=tree; iterate; end /*neighbor?*/
motion=1 /* [↓] Brownian motion is coming.*/
xb=x+random(1,3)-2 /* apply Brownian motion for X.*/
yb=y+random(1,3)-2 /* " " " " Y.*/
if @.xb.yb\==hole then iterate /*can mote actually move there ? */
@.x.y=hole /*empty out the old mote position*/
@.xb.yb=mote /*move the mote (or possibly not)*/
if xb<loX then loX=max(1,xb); if xb>hiX then hiX=min( width, xb)
if yb<loY then loY=max(1,yb); if yb>hiY then hiY=min(height, yb)
end /*y*/ /* [↑] limit the motes movement.*/
end /*x*/
call crop /*crops/truncates the mote field.*/
end /*winks*/
```
call show exit /*stick a fork in it, we're done.*/ /*────────────────────────────────CROP subroutine───────────────────────*/ crop: if loX>1 & hiX<width & loY>1 & hiY<height then return /*cropping?*/
``` do yc=-1 to height+1 by height+2 /*delete motes (moved off field).*/
do xc=-1 to width+1; if @.xc.yc==hole then iterate; @.xc.yc=hole
end /*xc*/
end /*yc*/
```
``` do xc=-1 to width+1 by width+2 /*delete motes (moved off field).*/
do yc=-1 to height+1; if @.xc.yc==hole then iterate; @.xc.yc=hole
end /*yc*/
end /*xc*/
```
return /*────────────────────────────────SHOW subroutine───────────────────────*/ show: clearScr /*not necessary, but speeds it up*/
``` do ys=height for height by -1; aRow=
do xs=1 for width; aRow=aRow || @.xs.ys
end /*xs*/
say aRow
end /*ys*/
```
return</lang> This REXX program makes use of SCRSIZE REXX program (or BIF) which is used to determine the screen size of the terminal (console).
The SCRSIZE.REX REXX program is included here ──► SCRSIZE.REX.
output when using the default input
``` * *
**
* *
* * * ****
*** * * * * *
* * * * * * **
** ** * ** *
* ** * * **
** * ** * *
** * ** * * * *
* * * * **
* * ** ** * * *** *
* * * * *
* * * * * * * * *
** * ** * * ** * **
* * * * * * ** * *
* ** * * ** * * *
* * * ** * ** *
* * * * * * *
* * * * ** * * *
* *** * * * * ** * **
* *** * ** * ** * * ** ** * *
** * * * * ** * * * *
* * * * * ** * * *** *
** ** * * * * * * ** ** * * * * **
* * * * ** * * * * ** ** * * *
* * * * * * * * *
* * ** * ** * * * **
* * *** * **** * * * * *
* **** * * * * * **** * ** *
* * * * * *** * * ** ** **
* * * * * * ** * * * * *** * ***
** * * * ** * * * ** * ** ** *
** * * * * * ** ** * * * **
* * * * * * * * ** ** * ** **
* * * ** * * * * * * * * **** *
* * * * * ** * ** * * *** * * * * **
** ** ** * ** ** * * * * * ** *
* * * ** * * * * * * *
* * ** * * * * ** * * * * * * *
* * * * * ** * * * * * ** * * *
* ** * * * * * *** * * ** **
* * * * * * * * ** * * * * * ** *
* * * * ** * ** * * * * * ** * * *
*** * * * * * * * * * * * * * * * *
* * * ** * * * * * * ** * * * * * * * *** * *** * * ** * * *
** * * * * * ** * ** ** ** * * ** * * *** * ** * * * * ** * *
* * * * * * * * * * * * * * * ** * * * ** * ** *
* * * * * ** ** ****** * * ** * ** * * ** * * * * * * * ** * * * ** *
* * ** * ** * * * ** ** ** * ** ** * * * * * * * * * * ** * ** *
** * * * * * * * * * * * * * * * * ** * * * * * * * *
* *** * * * * * * * * ** ** * * * * * * * *** * * * *** *
* * * * * * * * * * * * ** * * * ** * * * ** * * ** * * *
** * ** * * * ** ** * * * * ** * ** * * * * *
* * * * * * ** ***** * * ** * * * ** * *
* * * ** * * ** * * * * * * ** * * ** * ** * ** *
* * * * * * * * * * * * * * * * * * ** *
* * * ** * * * * ** * * **** ** * * *
** * * ** * ** * * * * * * * * ** ** * * *
* ** * * * * * * ** * * * * * * * * * *
* * * ***** *** ** * * ** * ** * *** * ** * * ** ** *
** * * * * * ** * * ** * * * * *
* * * * * * * * * * * * * * * * *
* ** *** * * * * * * ** * * * ***
* * * * * * ** * * * *
** * * * * * * * * * * *
* ** * ** * ** * ** * * * * * *
* ** * * * * * ** * * * * *
* * * * ** * * * * * * * * *
* * * * * ** ** * * * * *
* * * ** * * * * * * *
** *** * * * ** * *
*** * * * * * ** ** * * **
* * *** * * * * **
* * * * * *
** *
* * *
* *
* ***
* * *
* * **
* * *
*
```
## Ruby
Library: RMagick
<lang ruby>require 'rubygems' require 'RMagick'
NUM_PARTICLES = 1000 SIZE = 800
def draw_brownian_tree world
``` # set the seed
world[rand SIZE][rand SIZE] = 1
```
``` NUM_PARTICLES.times do
# set particle's position
px = rand SIZE
py = rand SIZE
```
``` loop do
# randomly choose a direction
dx = rand(3) - 1
dy = rand(3) - 1
```
``` if dx + px < 0 or dx + px >= SIZE or dy + py < 0 or dy + py >= SIZE
# plop the particle into some other random location
px = rand SIZE
py = rand SIZE
elsif world[py + dy][px + dx] != 0
# bumped into something
world[py][px] = 1
break
else
py += dy
px += dx
end
end
end
```
end
world = Array.new(SIZE) { Array.new(SIZE, 0) } srand Time.now.to_i
draw_brownian_tree world
img = Magick::Image.new(SIZE, SIZE) do
``` self.background_color = "black"
```
end
draw = Magick::Draw.new draw.fill "white"
world.each_with_index do |row, y|
``` row.each_with_index do |colour, x|
draw.point x, y if colour != 0
end
```
end
draw.draw img img.write "brownian_tree.bmp"</lang>
## Run BASIC
<lang runbasic>numParticles = 3000 dim canvas(201,201) canvas(rnd(1) * 100 , rnd(1) * 200) = 1 'start point for i = 1 To numParticles
``` x = (rnd(1) * 199) + 1
y = (rnd(1) * 199) + 1
```
``` while canvas(x+1, y+1) + canvas(x, y+1)+canvas(x+1, y)+canvas(x-1, y-1)+canvas(x-1, y)+canvas(x, y-1) = 0
x = x + (rnd(1)* 2) + 1
y = y + (rnd(1)* 2) + 1
If x < 1 Or x > 200 Or y < 1 Or y > 200 then
x = (rnd(1) * 199) + 1
y = (rnd(1) * 199) + 1
end if
wend
canvas(x,y) = 1
```
next i
graphic #g, 200,200 for x = 1 to 200
``` for y = 1 to 200
if canvas(x,y) = 1 then #g "color green ; set "; x; " "; y else #g "color blue ; set "; x; " "; y
next y
```
next x render #g
1. g "flush"</lang>
## Scheme
Works with: Guile
<lang scheme>; Save bitmap to external file (define (save-pbm bitmap filename) (define f (open-output-file filename)) (simple-format f "P1\n~A ~A\n" (list-ref (array-dimensions bitmap) 0) (list-ref (array-dimensions bitmap) 1)) (do ((c 0 (+ c 1))) ((eqv? c (list-ref (array-dimensions bitmap) 1))) (do ((r 0 (+ r 1))) ((eqv? r (list-ref (array-dimensions bitmap) 0))) (display (array-ref bitmap r c) f)) (newline f)) (close-output-port f) )
Return a random coordinate in the bitmap that isn't filled yet along with a direction
(define (new-particle bitmap) (define x (random (list-ref (array-dimensions bitmap) 0))) (define y (random (list-ref (array-dimensions bitmap) 1))) (define dx (- (random 3) 1)) (define dy (- (random 3) 1)) ;Repeat until we find an unused location (if (> (array-ref bitmap x y) 0) (new-particle bitmap) (list (list x y) (list dx dy))))
Check neighboring coordinates to see if a collision occured
(define (collision-check bitmap p) (define c #f) (define oob #f) (define x (list-ref (car p) 0)) (define y (list-ref (car p) 1)) (define dx (list-ref (cadr p) 0)) (define dy (list-ref (cadr p) 1)) (define w (list-ref (array-dimensions bitmap) 0)) (define h (list-ref (array-dimensions bitmap) 1))
; If the particle hasn't gone out of bounds keep checking for a collision (if (or (> 0 x) (> 0 y) (<= w x) (<= h y)) (set! oob #t) (do ((x (- (list-ref (car p) 0) 1) (+ x 1))) ((eqv? x (+ (list-ref (car p) 0) 2))) (do ((y (- (list-ref (car p) 1) 1) (+ y 1))) ((eqv? y (+ (list-ref (car p) 1) 2))) ; Check existing neighbors for collisions (if (and (<= 0 x) (<= 0 y) (> w x) (> h y)) (if (not (zero? (array-ref bitmap x y))) (set! c #t)))))) (if oob #f ; Return false if out of bounds (if c p ; Return the point of collision if a collision occured (if (and (zero? dx) (zero? dy)) #f ; Return false if particle is motionless with no collision (collision-check bitmap (particle-move p))))))
Plot a particle on the bitmap
(define (particle-plot! bitmap p) (array-set! bitmap 1 (list-ref (car p) 0) (list-ref (car p) 1)))
Move a particle along its slope
(define (particle-move p) (list (list (+ (list-ref (car p) 0) (list-ref (cadr p) 0)) (+ (list-ref (car p) 1) (list-ref (cadr p) 1))) (cadr p)))
Grow a brownian tree
(define (grow-brownian-tree! bitmap collisions) (define w (list-ref (array-dimensions bitmap) 0)) (define h (list-ref (array-dimensions bitmap) 1))
; Generate a new particle at a random location (define p (new-particle bitmap))
; Find a collision or lack of one and plot it on the bitmap (set! p (collision-check bitmap p)) (if p (begin ; Display collision number and the place it happened (display collisions)(display ": ")(display (car p))(newline) (set! collisions (- collisions 1)) ; Plot the point (particle-plot! bitmap p)))
; If we're done say so (if (zero? collisions) (display "Done\n"))
; Keep going until we have enough collisions ; or have filled the bitmap (if (and (< 0 collisions) (memq 0 (array->list (array-contents bitmap)))) (grow-brownian-tree! bitmap collisions)))
Plot a random point to seed the brownian tree
(define (seed-brownian-tree! bitmap) (define p (new-particle bitmap)) (particle-plot! bitmap p))
Example usage ;;;
Seed the random number generator
(let ((time (gettimeofday))) (set! *random-state* (seed->random-state (+ (car time) (cdr time)))))
Generate a tree with 320*240 collisions on a bitmap of the size 640x480
The bitmap is zeroed to start and written with a one where a collision occurs
(define bitmap (make-array 0 640 480)) (seed-brownian-tree! bitmap) (grow-brownian-tree! bitmap (* 320 240))
Save to a portable bitmap file
(save-pbm bitmap "brownian-tree.pbm")</lang>
## Seed7
The program below generates a small brownian tree. You can watch how it grows.
<lang seed7>\$ include "seed7_05.s7i";
``` include "draw.s7i";
include "keybd.s7i";
```
const integer: SIZE is 300; const integer: SCALE is 1;
const proc: genBrownianTree (in integer: fieldSize, in integer: numParticles) is func
``` local
var array array integer: world is 0 times 0 times 0;
var integer: px is 0;
var integer: py is 0;
var integer: dx is 0;
var integer: dy is 0;
var integer: i is 0;
var boolean: bumped is FALSE;
begin
world := fieldSize times fieldSize times 0;
world[rand(1, fieldSize)][rand(1, fieldSize)] := 1; # Set the seed
for i range 1 to numParticles do
# Set particle's initial position
px := rand(1, fieldSize);
py := rand(1, fieldSize);
bumped := FALSE;
repeat
# Randomly choose a direction
dx := rand(-1, 1);
dy := rand(-1, 1);
if dx + px < 1 or dx + px > fieldSize or dy + py < 1 or dy + py > fieldSize then
# Plop the particle into some other random location
px := rand(1, fieldSize);
py := rand(1, fieldSize);
elsif world[py + dy][px + dx] <> 0 then
# Bumped into something
world[py][px] := 1;
rect(SCALE * pred(px), SCALE * pred(py), SCALE, SCALE, white);
DRAW_FLUSH;
bumped := TRUE;
else
py +:= dy;
px +:= dx;
end if;
until bumped;
end for;
end func;
```
const proc: main is func
``` begin
screen(SIZE * SCALE, SIZE * SCALE);
KEYBOARD := GRAPH_KEYBOARD;
genBrownianTree(SIZE, 20000);
end func;</lang>
```
Original source: [1]
## Tcl
Library: Tk
<lang tcl>package require Tcl 8.5 package require Tk
set SIZE 300
image create photo brownianTree -width \$SIZE -height \$SIZE interp alias {} plot {} brownianTree put white -to brownianTree put black -to 0 0 [expr {\$SIZE-1}] [expr {\$SIZE-1}] proc rnd {range} {expr {int(rand() * \$range)}}
proc makeBrownianTree count {
``` global SIZE
# Set the seed
plot [rnd \$SIZE] [rnd \$SIZE]
for {set i 0} {\$i<\$count} {incr i} {
```
# Set a random particle's initial position set px [rnd \$SIZE] set py [rnd \$SIZE]
while 1 { # Randomly choose a direction set dx [expr {[rnd 3] - 1}] set dy [expr {[rnd 3] - 1}]
# If we are going out of bounds... if {\$px+\$dx < 0 || \$px+\$dx >= \$SIZE || \$py+\$dy < 0 || \$py+\$dy>=\$SIZE} { # Out of bounds, so move back in set dx [expr {[rnd 3] - 1}] set dy [expr {[rnd 3] - 1}] continue }
set ox \$px set oy \$py # Move/see if we would hit anything incr px \$dx incr py \$dy if {[lindex [brownianTree get \$px \$py] 0]} { # Hit something, so plot where we were plot \$ox \$oy break } } ## For display while things are processing, uncomment next line #update;puts -nonewline .;flush stdout
``` }
```
}
pack [label .l -image brownianTree] update makeBrownianTree 1000 brownianTree write tree.ppm</lang>
## TI-83 BASIC
<lang ti83b>:StoreGDB 0
ClrDraw
FnOff
AxesOff
Pxl-On(31,47)
For(I,1,50)
randInt(1,93)→X
randInt(1,61)→Y
1→A
While A
randInt(1,4)→D
Pxl-Off(Y,X)
If D=1 and Y≥2
Y-1→Y
If D=2 and X≤92
X+1→X
If D=3 and Y≤60
Y+1→Y
If D=4 and X≥2
X-1→X
Pxl-On(Y,X)
If pxl-Test(Y+1,X) or pxl-Test(Y+1,X+1) or pxl-Test(Y+1,X-1) or pxl-Test(Y,X+1) or pxl-Test(Y,X-1) or pxl-Test(Y-1,X) or pxl-Test(Y-1,X-1) or pxl-Test(Y-1,X+1)
0→A
End
End
Pause
RecallGDB 0</lang>
## Visual Basic .NET
Windows Forms Application.
<lang vbnet> Imports System.Drawing.Imaging
Public Class Form1
``` ReadOnly iCanvasColor As Integer = Color.Black.ToArgb
ReadOnly iSeedColor As Integer = Color.White.ToArgb
```
``` Dim iCanvasWidth As Integer = 0
Dim iCanvasHeight As Integer = 0
```
``` Dim iPixels() As Integer = Nothing
```
``` Private Sub BrownianTree()
```
``` Dim oCanvas As Bitmap = Nothing
Dim oRandom As New Random(Now.Millisecond)
Dim oXY As Point = Nothing
Dim iParticleCount As Integer = 0
```
``` iCanvasWidth = ClientSize.Width
iCanvasHeight = ClientSize.Height
```
``` oCanvas = New Bitmap(iCanvasWidth, iCanvasHeight, Imaging.PixelFormat.Format24bppRgb)
```
``` Graphics.FromImage(oCanvas).Clear(Color.FromArgb(iCanvasColor))
```
``` iPixels = GetData(oCanvas)
```
``` ' We'll use about 10% of the total number of pixels in the canvas for the particle count.
iParticleCount = CInt(iPixels.Length * 0.1)
```
``` ' Set the seed to a random location on the canvas.
iPixels(oRandom.Next(iPixels.Length)) = iSeedColor
```
``` ' Run through the particles.
For i As Integer = 0 To iParticleCount
Do
' Find an open pixel.
oXY = New Point(oRandom.Next(oCanvas.Width), oRandom.Next(oCanvas.Height))
Loop While iPixels(oXY.Y * oCanvas.Width + oXY.X) = iSeedColor
```
``` ' Jitter until the pixel bumps another.
oXY.X += oRandom.Next(-1, 2)
oXY.Y += oRandom.Next(-1, 2)
```
``` ' Make sure we don't jitter ourselves out of bounds.
If oXY.X < 0 Then oXY.X = 0 Else If oXY.X >= oCanvas.Width Then oXY.X = oCanvas.Width - 1
If oXY.Y < 0 Then oXY.Y = 0 Else If oXY.Y >= oCanvas.Height Then oXY.Y = oCanvas.Height - 1
End While
```
``` iPixels(oXY.Y * oCanvas.Width + oXY.X) = iSeedColor
```
``` ' If you'd like to see updates as each particle collides and becomes
' part of the tree, uncomment the next 4 lines (it does slow it down slightly).
' SetData(oCanvas, iPixels)
' BackgroundImage = oCanvas
' Invalidate()
' Application.DoEvents()
Next
```
``` oCanvas.Save("BrownianTree.bmp")
BackgroundImage = oCanvas
```
``` End Sub
```
``` ' Check adjacent pixels for an illuminated pixel.
Private Function CheckAdjacency(ByVal XY As Point) As Boolean
```
``` Dim n As Integer = 0
```
``` For y As Integer = -1 To 1
' Make sure not to drop off the top or bottom of the image.
If (XY.Y + y < 0) OrElse (XY.Y + y >= iCanvasHeight) Then Continue For
```
``` For x As Integer = -1 To 1
' Make sure not to drop off the left or right of the image.
If (XY.X + x < 0) OrElse (XY.X + x >= iCanvasWidth) Then Continue For
```
``` ' Don't run the test on the calling pixel.
If y <> 0 AndAlso x <> 0 Then
n = (XY.Y + y) * iCanvasWidth + (XY.X + x)
If iPixels(n) = iSeedColor Then Return True
End If
Next
Next
```
``` Return False
```
``` End Function
```
``` Private Function GetData(ByVal Map As Bitmap) As Integer()
```
``` Dim oBMPData As BitmapData = Nothing
Dim oData() As Integer = Nothing
```
``` oBMPData = Map.LockBits(New Rectangle(0, 0, Map.Width, Map.Height), ImageLockMode.ReadOnly, PixelFormat.Format32bppArgb)
```
``` Array.Resize(oData, Map.Width * Map.Height)
```
``` Runtime.InteropServices.Marshal.Copy(oBMPData.Scan0, oData, 0, oData.Length)
```
``` Map.UnlockBits(oBMPData)
```
``` Return oData
```
``` End Function
```
``` Private Sub SetData(ByVal Map As Bitmap, ByVal Data As Integer())
```
``` Dim oBMPData As BitmapData = Nothing
```
``` oBMPData = Map.LockBits(New Rectangle(0, 0, Map.Width, Map.Height), ImageLockMode.WriteOnly, PixelFormat.Format32bppArgb)
```
``` Runtime.InteropServices.Marshal.Copy(Data, 0, oBMPData.Scan0, Data.Length)
```
``` Map.UnlockBits(oBMPData)
```
``` End Sub
```
``` Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
DoubleBuffered = True
BackgroundImageLayout = ImageLayout.Center
Show()
Activate()
Application.DoEvents()
BrownianTree()
End Sub
```
End Class </lang>
Final output:
## XPL0
<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations def W=128, H=W; \width and height of field int X, Y; [SetVid(\$13); \set 320x200 graphic video mode Point(W/2, H/2, 6\brown\); \place seed in center of field loop [repeat X:= Ran(W); Y:= Ran(H); \inject particle
``` until ReadPix(X,Y) = 0; \ in an empty location
loop [Point(X, Y, 6\brown\); \show particle
Point(X, Y, 0\black\); \erase particle
X:= X + Ran(3)-1; \(Brownian) move particle
Y:= Y + Ran(3)-1;
if X<0 or X>=W or Y<0 or Y>=H then quit; \out of bounds
];
if KeyHit then [SetVid(3); quit]; \restore text mode
];
```
]</lang>
## zkl
This grows rather slowly, so I've added a circle for barnacles to attach to. It looks like tendrils growing from the center to the circle and vice versa. The tree type is similar to that shown in the XPLO and Visual Basic .NET solutions. Also, the image is written to disk as each particle attaches so EventViewer will auto update to show the progression.
Uses the PPM class from http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm#zkl <lang zkl>w:=h:=400; numParticles:=20000; bitmap:=PPM(w+2,h+2,0); // add borders as clip regions
bitmap[w/2,h/2]=0xff|ff|ff; // plant seed bitmap.circle(w/2,h/2,h/2,0x0f|0f|0f); // plant seeds
fcn touching(x,y,bitmap){ // is (x,y) touching another pixel?
``` // (x,y) isn't on the border/edge of bitmap so no edge conditions
var [const] box=T(T(-1,-1),T(0,-1),T(1,-1),
```
T(-1, 0), T(1, 0), T(-1, 1),T(0, 1),T(1, 1));
``` box.filter1('wrap([(a,b)]){ bitmap[a+x,b+y] }); //-->False: not touching, (a,b) if is
```
}
while(numParticles){
``` c:=(0x1|00|00).random(0x1|00|00|00) + (0x1|00).random(0x1|00|00) + (0x1).random(0x1|00);
reg x,y;
do{ x=(1).random(w); y=(1).random(h); }while(bitmap[x,y]); // find empty spot
while(1){ // stagger around until bump into a particle, then attach barnicle
if(touching(x,y,bitmap)){
bitmap[x,y]=c;
```
bitmap.write(f:=File("foo.ppm","wb")); // tell ImageViewer to update image numParticles-=1; break;
``` }
x+=(-1).random(2); y+=(-1).random(2); // [-1,0,1]
if( not ((0<x<w) and (0<y<h)) ){ // next to border --> color border
bitmap[x,y]=c;
```
break;
``` }
}
```
} bitmap.write(File("foo.ppm","wb")); // the final image</lang> | 26,020 | 83,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-10 | latest | en | 0.681781 |
https://socratic.org/questions/58ea497cb72cff4ff4af3bdb | 1,656,648,203,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00187.warc.gz | 581,860,183 | 6,866 | # Question #f3bdb
May 5, 2017
Alternating current $A C$ is the type of electrical current that is generated by an alternator in a car or power plant. Direct current $D C$ comes from a voltage regulator or battery.
#### Explanation:
$A C$ can be generated and used in many different forms such as sine waves that look like a swimming snake, or square waves that look like building blocks alternating above and below a center line.
Each of these waveforms start at zero potential and they gradually or suddenly jump up in the positive direction to the full potential of the force applied on them. Then the waveform gradually or suddenly returns to zero and immediately repeats the same waveform in the negative direction. This entire transition (one plus and one minus pulse) is the cycle or period of the wave, and it will continue repeating as long as the force is applied.
The sine wave has the gradual change, and the square wave has the sudden jump. Sine waves are produced by alternators in cars and power plants by the force of the car engine or the power plant turbines. Square waves are produced by inverters that can change $D C$ from batteries to $A C$ to run a building's appliances and computers in the event of a mains power failure.
$D C$ is much less exciting, with no ups and downs other than $o n \mathmr{and} o f f$. $D C$ looks like a straight line of electromotive force above a ground line or zero potential. The force behind $D C$ may be the chemical reaction occurring in a battery or the $A C$ we already generated in the car alternator. The $A C$ is then fed through diodes to strip away the waves and convert it into a nice straight line. Then we come to the voltage regulator that keeps the $D C$ voltage at a respectable level for use in the car.
You may have noticed that we changed currents from $A C \to D C$ and $D C \to A C$ in our discussion. Get used to it, because our modern machines and computers do this many times right in front of you.
You can find $A C \mathmr{and} D C$ images here: | 465 | 2,033 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-27 | latest | en | 0.937723 |
http://www.docstoc.com/docs/72093671/rpt-maths-form-3 | 1,429,628,293,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246642037.57/warc/CC-MAIN-20150417045722-00280-ip-10-235-10-82.ec2.internal.warc.gz | 457,227,629 | 49,612 | # rpt maths form 3 by umimadihah
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``` MATHEMATICS
FORM 3
SCHEME OF WORK
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
1. LINES AND Students will be taught Students will be able to :
1-2 ANGLES II to :
1.1 Properties of 1.1 Understand and i. Identify : Contextual Interpreting Understanding Parallel lines
Angles use properties of a) transversals learning
Associated angles associated b) corresponding angles Identifying Equity Transversal
with with transversal c) alternate angles Thinking Skill relations
Transversal and parallel lines. d) interior angles Self-confident Alternate angle
and Parallel Constructivism Making
Lines ii. Determine that for parallel lines: inferences Cooperation Interior angle
a) corresponding angles are Enquiry-
equal discovery Translating Corresponding angle
b) alternate angles are equal
c) sum of interior angles is Cooperative Intersecting lines
1800 learning
Acetate overlay
iii. Find the values of : Mastery
a) corresponding angles learning
b) alternate angles
c) interior angles associated Multiple
with parallel lines. Intelligence
iv. Determine if two given Integrating ICT
lines are parallel based on into learning
the properties of angles and teaching
associated with
transversals. Future studies
v. Solve problems involving
properties of angles
associated with
transversals.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
3-4 2. POLYGONS II taught to :
2.1 Regular 2.1 Understand i. Determine if a given Thinking skill. Classifying Responsible Polygon
Polygons the concept of polygon is a regular
regular polygon. Constructivism Drawing Fairness Regular polygon
polygons diagrams
ii. Find: Enquiry- Equity Convex
a) the axes of discovery Comparing
symmetry and The ability to Axes of symmetry
b) the number of axes Cooperative contrasting reason
of symmetry learning Straightedges
of a polygon. Identifying Angle
Mastery relations
iii. Sketch regular learning Equilateral triangle
polygons. Translating
Square
iv. Draw regular polygons
by dividing equally the Regular hexagon
angle at the centre.
v. Construct equilateral
triangles, squares and
regular hexagons.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
Students will be taught Students will be able to :
3-4 2. POLYGONS II to :
2.2 Exterior and 2.2 Understand i. Identify the interior angles and Thinking skill. Classifying Determination Interior angle
Interior Angles and use the exterior angles of a polygon.
of Polygons knowledge of Constructivism Drawing Courtesy Exterior angle
exterior and ii. Find the size of an exterior angle diagrams
interior when the interior angle of a Enquiry- Responsible Complementary angle
angles of polygon is given and vice versa. discovery Comparing
polygons and Understanding
iii. Determine the sum of the Cooperative contrasting
interior angles of polygons learning
Identifying
iv. Determine the sum of the Mastery relations
exterior angles of polygons learning
Translating
v. Find:
a) The size of an interior angle
of a regular polygon given
the number of sides.
b) The size of an exterior angle
of a regular polygon given
the number of sides.
c) The number of sides of a
regular polygon given the
size of the interior or
exterior angle.
vi. Solve problems involving
angles and sides of polygons
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
6-7 3. CIRLCES II taught to :
3.1 Properties 3.1 Understand i. Identify a diameter of a Contextual Identifying Unity Diameter
of circles and use circle as an axis of learning relations
involving properties of symmetry Truthfulness axis of symmetry
symmetry, circles Learning how Translating
chords and involving ii. Determine that: to learn Responsible chord
arcs symmetry, a) A radius that is Classifying
chords and perpendicular to a Self-access Fairness perpendicular
arcs chord divides the chord learning Comparing bisector
into two equal parts and Equity
and vice versa Cooperative contrasting intersect
learning The ability to
b) Perpendicular bisectors reason equidistant
of two chords intersect Thinking skill
at the center arc
Constructivism
symmetry
c) Two chords that are Problem
equal in length are solving center
equidistant from the
d) Chords of the same perpendicular
length cut arcs of the
same length
iii. Solve problems
involving symmetry,
chords and arcs of circles
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
6-7 3. CIRLCES II taught to :
3.2 Properties 3.2 Understand i. Identify angles subtended by Cooperative Identifying Unity angle
of angles in and use and arc the center and at the learning relations
circles properties of circumference of a circle Truthfulness subtended
angles in Constructivism Translating
circles ii. Determine that angles Responsible semicircle
subtended at the Thinking skills Classifying
circumference by the same Fairness circumference
arc are equal Integrating ICT Comparing
into teaching and Equity arc
iii. Determine that angles and learning contrasting
subtended: The ability to chord
a) At the circumference Problem reason
b) At the center by arcs of solving reflex angle
the same length are equal
center
iv. Determine the relationship
between angle at the center
and angle at the
circumference subtended by
an arc
v. Determine the size of an
angle subtended at the
circumference in a
semicircle
vi. Solve problems involving
angles subtended at the
center and angles at the
circumference of circles
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
6-7 3. CIRLCES II taught to :
3.3 Cyclic 3.3 Understand i. Identify cyclic quadrilaterals Cooperative Identifying Unity Cyclic
concept of ii. Identify the interior opposite Truthfulness
cyclic angles of cyclic Integrating ICT Translating Interior
and learning Classifying angle
iii. Determine the relationship Fairness
between interior opposite Constructivism Comparing Exterior angle
angles and cyclic and Equity
The ability to
iv. Identify exterior angles and reason
the corresponding interior
opposite angle of cyclic
v. Determine the relationship
between exterior angles and
the corresponding interior
opposite angel of cyclic
vi. Solve problem involving
angles of cyclic
vii. Solve the problems
involving circles.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
4. STATISTICS Students will be Students will be able to : Drawing Systematic Sector
8-9 II taught to :
4.1 Pie charts 4.1 Represent and i. Obtained interpret Constructivism diagrams Pie chart
interpret data information from pie Cooperation Angle
in pie charts to charts. Problem
solve problems. solving Interpreting Suitable
ii. Construct pie charts to Considerate representation
represent data. Contextual
learning Translating Construct
iii. Solve problems involving Act wisely Size of sector
pie charts. Future studies
Comparing Quantity
iv. Determine suitable ICT and Steadfastness Data
representation of data.
Self-access contrasting Size of angle
learning. Label
Problem Title
Solving Pictograms
Bar chart
Pie chart
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
4. STATISTICS Students will be Students will be able to :
8-9 II taught to :
4.2 Mode, 4.2 Understand i. Determine the mode of: Contextual Interpreting Systematic Data
median and and use the a) sets of data. learning
Translating
mean concept of b) Data given in frequency Mode
mode, tables. Mastery Comparing Cooperation
median and learning Discrete
and
mean to ii. Determine the mode
solve and the respective Integrating contrasting Considerate Frequency
problems. frequency from ICT into
Problem
pictographs, bar charts, teaching and Median
line graphs and pie learning. Solving Act wisely
charts. Arrange
Self-access
iii. Determine the median of learning Steadfastness Odd
sets of data.
Problem Even
iv. Determine the median of solving
data in frequency tables. Middle
v. Calculate the mean of : Frequency table
a) sets of data. mean
b) Data frequency tables.
vi. Solve problems involving
mode, median and mean.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
11-12 5. INDICES taught to :
5.1 Indices 5.1Understand the i. Express repeated ICT Working out Systematic indices
concept of multiplication as an and Contextual
indices vice versa. Mastery mentally Thoughtful base
Learning
ii. Find the value of an. The ability to reason index
iii. Express numbers in index Self-motivated power of
notation.
Act wisely index notation
5.2 5.2 Perform i. Verify am × an = am+n index form
Multiplicati computations
on of involving ii. Simplify multiplication express
numbers in multiplication of :.
index of numbers in a) numbers value
notation index b) algebraic terms
notation. expressed in index real numbers
notation with the same
base. repeated
iii. Simplify multiplication of multiplication
a) numbers
b) algebraic terms factor
expressed in index
notation with different
bases.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBSTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
11-12 5. INDICES taught to :
5.3 Division of 5.3 Perform i. Verify am ÷ an = am-n ICT Working out Systematic multiplication
numbers in computation
index notation involving ii. Simplify division of :. Self accessed mentally Thoughtful simplify
division of a) numbers Learning
numbers in b) algebraic terms The ability to base
index notation expressed in index notation Mastery
with the same base. Learning reason algebraic term
5.4 Raising 5.4 Perform mn mn Constructivism Self-motivated verify
numbers and computations i. Derive a a
algebraic involving ii. Simplify: Act wisely index notation
terms in index raising a) numbers
notation to a b) algebraic terms
indices
power. algebraic expressed in index
terms in index notation raised to a power.
notation to a law of indices
power. iii. Simplify
multiplication and unknown
division of:
a) numbers Raised to a power
b) algebraic terms base
expressed in index
notation with different
bases raised to a power.
iv. Perform combined
operations involving
multiplication,
division, and raised to
a power on:
a) numbers
b) algebraic terms
TOPICS/ LEARNING CCTS MORAL VOCABULARY
WEEK LEARNING OUTCOMES GENERICS
SUBTOPICS OBJECTIVES VALUES
Student will be Student will be able to:
11-12 5. INDICES taught to: n 1
5.5 Negative 5.5 Perform i. Verify a Mastery Working out Systematic verify
indices. computations an Learning
involving n 1 mentally Thoughtful
ii. State a as n and vice
negative a ICT
indices. versa The ability to
iii. Perform combined Self accessed
operations involving Learning reason
multiplication, division, and
raised to a power on: Self-motivated
a) numbers
b) algebraic terms
1
Act wisely
i. Verify n n
5.6 Fractional 5.6 Perform a a
1
indices. computations ii. State as n a and vice
involving n
fractional a
versa. 1
indices. iii. Find the value of a n
iv. State nm
a as:
1 1
a) a m n or a n m
n m
n m
b) a or a
v. Perform combined operations
involving multiplication,
division, and raised to a power
on:
a) numbers
b) algebraic terms
v. Find the value of nm
a
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
Student will be Student will be able to:
11-12 5. INDICES taught to:
5.7 Computations 5.7 Perform i. Perform multiplication, Mastery Working out Systematic
involving laws computations division, raised to a power Learning
of indices. involving or operations on several mentally Thoughtful
laws of numbers expressed in ICT
indices. index notation. The ability to
Self
ii. Perform combined accessed reason
operations multiplication, Learning
division, and raised to a Self-motivated
power involving positive,
negative and fractional Act wisely
indices.
TOPICS/ LEARNING CCTS MORAL VOCABUL
WEEK LEARNING OUTCOMES GENERICS
SUBTOPICS OBJECTIVES VALUES ARY
6. ALGEBRAIC Student will be Students will be able to:
13-15 EXPRESSION taught to:
III 6.1 Understand i. Expand single brackets. Contextual Identifying Act wisely Linear algebraic
6.1 Expansions and use the learning relations Terms
concept of ii. Expand two brackets Systematic Like terms
expending Constructivism Unlike terms
brackets Sincerity Expansion
Expand
Dedication Single brackets
Two brackets
6.2 Factorization of 6.2 Understand i. State factors of Effort Multiply
algebraic and use the an Factorization
expression concept of algebraic term Square
factorization Common factor
of algebraic ii. State common factors and Term
expression to the HCF for Highest
solve several common factor
problems. algebraic terms. (HCF)
difference of
iii. Factorize algebraic two square
expression: numerator
a) using common factor denominator
b) the difference of two algebraic
squares. fraction
factorization
iv Factorize and simplify
algebraic fractions.
TOPICS/ LEARNING CCTS MORAL VOCABU LARY
WEEK LEARNING OUTCOMES GENERICS
SUBTOPICS OBJECTIVES VALUES
13-15 6. ALGEBRAIC Student will be
Students will be able to:
EXPRESSION taught to: Contextual Identifying Act wisely common factor
III 6.3.Perform learning relations
algebraic fractions with
subtraction on subtraction on Constructivism multiple (LCM)
the same denominator.
algebraic algebraic Sincerity
fractions. fractions. multiple
Dedication
denominator
algebraic fractions
Effort
with one denominator as a
simplification
multiple of the other
6.4 Multiplication 6.4.Perform
denominator.
and division on multiplication
algebraic and division
fractions. on algebraic
i. Multiply two algebraic
fractions.
fractions involving
denominator with:
a) one term.
b) two terms
ii. Divide two algebraic
fractions involving
denominator with:
a) one term.
b) two terms
iii. Perform multiplication
and division of two
algebraic fractions using
factorization and the
different of two square.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS MORAL VOCABULARY
SUBTOPICS OBJECTIVES CCTS
16-17 7. 7.1 Variable i) Determine if a quantity in a given situation is a variable or Contextual Identify Thoughtful Linear algebraic
ALGEBRAIC And Constant a constant. learning relation
relations Terms
FORMULAE Constructivism
ii)Determine the variable in a given situation and represent
it with a letter symbol.
iiii) Determine the possible values of a variable in a given
situation.
Students will be able to :
7.2 Formulae
i. Write a formula based on a given:
a. Statement
b. situation.
ii. Identify the subject of a given formula.
iii. Express a specified variable as the subject of a
formula involving:
a. one of the basic operations: +, ,, x, ÷
b. powers or roots
TOPICS/ LEARNING MORAL VOCABULARY
WEEK LEARNING OUTCOMES GENERICS CCTS VALUES
SUBTOPICS OBJECTIVES
8. SOLID Students will be Students will be able to:
18-19 GEOMETRY taught to : i. Derive the formula for Appreciative Derive
III 8.1 Understand volume of : Enquiry - Identifying Prism
8.1 Volume of and use a) prisms discovery. relations Thoughtful Cylinder
the b) cylinders. Right circular
right prisms concept of ii. Calculate the volume of a right prism in Learning how Translating Understanding cylinder
and right volume of cubic units given the height and : to learn. Base
circular right prisms a) the area of the base
b) dimensions of the base. Self-confident Radius
cylinders. and right Contextual Volume
circular iii. Calculate the height of a prism given the learning.
volume and the area of the base. Dedication Area
cylinders to Cubic unit
iv. Calculate the area of the base of a prism
solve Constructivism Square
given the volume and the height.
problems. Rectangle
v. Calculate the volume of a cylinder in
cubic units given: Triangle
a) area of the base and the height. b) Dimension
radius of the base and the height of Height
the cylinder. Cubic metre
vi. Calculate the height of a cylinder, given Cubic
the volume and the radius of the base. centimetre
vii. Calculate the radius of he base of a Cubic
cylinder given the volume and the height. millimeter
viii. Convert volume in one metric unit to Millimeter
another: Litre
a) mm3 , cm3 and m3
b) cm3 , mℓ and ℓ.
ix. Calculate volume of liquid in a container.
x. Solve problems involving volume of
prisms and cylinders.
WEEK TOPICS/ LEARNING
SUBTOPICS OBJECTIVES LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
8. SOLID Students will be Students will be able to :
18-19 GEOMETRY taught to :
III 8.2 Understand i. Derive the formula Self access Identifying Appreciative Convert
8.2 Volume of and use the for the volume of : learning. relations
right concept of a) pyramids Thoughtful Metric unit
pyramids and volume of b) cones Constructivism Translating Liquid
right circular right ii. Calculate the volume of pyramids Understanding
cones. pyramids in mm3 , cm3 and m3 , given Thinking Container
and right the height and : skills. Self-confident
circular a) area of the base Pyramid
cones to b) dimensions of base. Dedication
solve iii. Calculate the height of a pyramid Enquiry - Cone
problems. given the volume and the discovery.
dimension of the base.
iv. Calculate the area of the base of a Learning how Volume
Pyramid given the volume and the to learn. Base
height.
v. Calculate the volume of a cone in Contextual Height
mm3 , cm3 and m3 , given the learning.
height and radius of the base. Dimension
vi. Calculate the height of a cone,
given the volume and the radius of
the base.
vii. Calculate the radius of the base of a
cone given the volume and the
height.
viii. Solve problems involving volume
of pyramids and cones.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS MORAL VOCABULARY
SUBTOPICS OBJECTIVES CCTS
18-19 8. SOLID 8.3 Understand Students will be able to : Contextual Identify Thoughtful Height
GEOMETRY and use the i) i. Calculate the volume of spere given the radius of the learning relation
relations
III concept of ii) sphere. Constructivism Dimensions
volume of ii. Calculate the radius of the sphere given the volume of
8.3 Volume of sphere to solve the sphere. Sphere
sphere problems. iii.Solve problems involving volume of sphere.
Hemisphere
Solid Composite
8.4 Composite 8.4 Apply the i. Calculate the volunme of composite solids.
solids concept of ii. Solve problems involving volumes of composite solids.
volume to
solve problems
involving
composite
solids.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES
22 9. SCALE Students will be Student will be able to : Contextual Identify Thoughtful Height
DRAWINGS taught to : i. Sketch shapes learning Relation
relations
9.1 Understand a) Of the same size as the objects using grid papers Constructivism Dimensions
and the concept b) Smaller than the object using gris papers. Translating
of scale c) Larger than the object using grid papers Sphere
drawings.
ii. Draw geometric shapes according the scale 1: n, Hemisphere
where n = 1, 2, 3 , ½, ......
Solid Composite
iii. Draw the shapes according to a given scale using
a) Grid papaers
b) Blank papers
iv. Redraw the shapes on grids papers of different
sizes.
v.Solve probles involving scale drawings.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
10. TRANSFORMATION Students will be Students should be able to:
23 II taught to : i. Identify if given shapes are
10.1 Similarity 10.1 Understand similar. Contextual Classifying Fairness shape
and use the ii. Calculate the lengths of learning similar
concept of unknown sides of two Identifying Equity side
10.2 Enlargement similarity . similar shapes Multiple relations angle
10.2 Understand intelligence . Effort proportion
and use the i. Identify an enlargement . Interpreting centre of
concept of ii. Find the scale factor, given Thinking Considerate enlargement
the object and its image of an
enlargement. enlargement when : skill. Drawing transformation
a) scale factor > 0 diagrams Thoughtful enlargement
b) scale factor < 0 Integrating scale factor
c) scale factor = negative.
ICT into Translating object
iii. Determine the centre of teaching & image
enlargement , given the learning invariant reduction
object and its image. size
iv. Determine the image of an Self access orientation
object given the centre of learning . similarity
enlargement and the scale
factor.
v. Determine the properties of
enlargement
vi. Calculate :
a) the scale factor
b) length of the side of the
image
c) length of the side of the
object of an enlargement.
vii. Solve problems involving
enlargement.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
11. LINEAR Students will be Students will be able to:
24 EQUATIONS II taught to :
11.1 Linear 11.1 Understand i) Determine if an equation is Enquiry Classifying Determination Equation
equations in and use the a linear equation in two discovery
two variables concept of variables. Identifying Tolerance Variable
two linear Thinking relations
equations in ii) Write linear equations in skills Understanding Linear equation
two variables two variables from given Translating
information. Learning how Courtesy Value
to learn
iii) Determine the value of a Sincerity Possible solution
variable given the other Multiple
variables. Intelligence Simultaneous
Linear equation
iv) Determine the possible Integrating
solutions for a linear ICT Solution
equation in two variables.
Self-access Substitution
learning
11.2 Simultaneous 11.2 Understand i) Determine if two given Elimination
linear equations and use the equations are simultaneous Trial and
in two variables concept of linear equations. improvement
two
simultaneous ii) Solve two simultaneous Problem
linear linear equations in two solving
equations in variables by
two variables a. substitution
to solve b. elimination
problems
iii) Solve problems involving
two simultaneous linear
equations in two variables.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
25 12. INEQUALITIES taught to:
12.1 Inequalities 12.1 Understand i. identify the Contextual Comparing and Determination Inequality
and use the relationship: learning contrasting
concept of a) greater than Tolerance Greater
inequalities. b) less than Constructivis Identifying Less
based on given situations m relations Understanding
Greater than
ii. write the relationship Enquiry- Interpreting Courtesy
between two given discovery Less than
numbers using the Drawing Sincerity
symbol “>” or “<” . Integrating diagrams Equal to
ICT into
iii. identify the teaching and Translating Include
relationship: learning
a) greater than or equal Finding all Equivalent
to possible Solution
b) less than or equal to solutions
based on given situations
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULA
SUBTOPICS OBJECTIVES VALUES RY
Students will be Students will be able to :
25 12. INEQUALITIES taught to:
12.2 Linear 12.2 Understand i. Determine if a given Cooperating Comparing Determination Relationship
inequalities in and use relationship is a linear learning and
one unknown concept of inequality. contrasting Tolerance Linear
linear Contextual
inequalities ii. Determine the possible learning Identifying Understanding Unknown
in one solutions for a given relations
unknown linear inequality in on Multiple Courtesy Number line
unknown intelligence Interpreting
a) x > h; Sincerity
b) x < h; Thinking Drawing
c) x ≥ h; skills diagrams
d) x ≤ h
e) Self access Translating
iii. Represent a linear learning
inequality: Finding all
a) x > h; Integrating possible
b) x < h; ICT into solutions
c) x ≥ h; teaching and
d) x ≤ h learning
on a number line and vice
versa
iv. Construct linear
inequalities using
symbols:
a) “ > ” or “ < ”
b) “ ≥ ” or “ ≤ ”
from given information.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
25 12. INEQUALITIES taught to:
12.3 Computations 12.3 Perform i. State a new inequality Mastery Comparing and Determination Addition
involving computations for a given inequality learning contrasting
subtraction, subtraction, a) added to Constructivism Identifying
multiplication, multiplication and b) subtract from relations Understandin Multiplication
and division on division on both sides of the Multiple g
inequalities inequalities inequalities. intelligence Interpreting Relation
Courtesy
ii. State a new inequality Thinking skills Drawing Equivalent
for a given inequality diagrams Sincerity
both sides of the Integrating ICT Simplest
inequalities are : into teaching Translating
a) multiplied by a and learning Collect
number Finding all
b) divided by a possible Isolate
number solutions
solve
iii. Construct inequalities
a) x k m k
b) x k m k
c) kxkm
x m
d)
k k
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
Students will be Students will be able to : Integrating
25 12. INEQUALITIES taught to: ICT into Comparing and Determination add
12.4 Solving Linear 12.4 Perform i) Solve a linear inequality by: teaching & contrasting
Inequalities In computations a) adding a number learning. Tolerance subtract
one Unknown to solve b) subtracting a number on Identifying
inequalities in both sides of the Contextual relations Understanding multiply
one variable inequality. learning.
Interpreting Courtesy divide
ii) Solve a linear inequality by: Enquiry-
a) multiplying a number discovery Drawing Sincerity
b) dividing a number on diagrams determine
both sides of the
inequality. Translating common value
iii) Solve linear inequalities in Finding all simultaneous
one variable using a possible
combination of operations. solutions Combining
linear inequality
number line
12.5 Simultaneous 12.5 Understand i) Represent the common values
Linear the concept of of two simultaneous linear equivalent
Inequalities In simultaneous inequalities on a number line.
One Variables linear
inequalities in ii) Determine the equivalent
one variable. inequalities for two given
linear inequalities.
iii) Solve two simultaneous
linear inequalities.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
13. GRAPHS OF Students will be Students will be able to:
26 FUNCTIONS taught to :
13.1 Functions 13.1 Understand i. State the relationship Enquiry - Classifying Determination Function
and use the between two variables discovery
concept of based on the given Identifying Tolerance Relationship
functions. information. Thinking relations
ii. Identify the dependent skills Understanding Dependent
and independent Interpreting variable
variables in a given Self access Self-confident
relationship involving learning Drawing Independent
two variables. diagrams Sincerity variable
iii. Calculate the value of the Integrating
dependent variable, given ICT into Translating Self-motivated Ordered pairs
the value of the teaching and
independent variable. learning Coordinate
13.2 Graphs of 13.2 Draw and plane
functions use graphs of
functions i. Construct tables of Table of values
values for given Origin
functions.
ii. Draw graphs of functions Graphs
using given scale.
iii. Determine from graphs x-coordinate
the value of y, given y-coordinate
value of x and vice versa.
iv. Solve problems involving x-axis
graphs of functions. y-axis
Scale
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
14. RATIOS, Students will be Students will be able to :
27 RATES AND taught to :
PROPORTIONS Comparing Systematic Rate
II i. Determine the rates involved in Enquiry- and
14.1 Rate 14.1 Understand given situations and identify the discovery contrasting Determination Quantity
the concept two quantities involved.
of ii. Calculate the rate given two Future Identifying Tolerance Unit of
rate and different quantities. studies relations measurement
perform iii. Calculate a certain quantity given Understanding
computations the rate and the other quantity. Interpreting Speed
involving iv. Convert rates from one unit of Courtesy
rates. measurement to another. Translating Distance
v. Solve problems involving rates. Sincerity
Time
14.2 Speed i. Identify the two quantities Uniform
14.2 Understand involved in speed.
and use the ii. Calculate and interpret speed. Non-uniform
concept of iii. Calculate :
speed. a) the distance, given the speed differentiate
and the time.
b) The time, given the speed
and the distance.
iv. Convert speed from one unit of
measurement to another.
v. Differentiate between uniform
speed and non-uniform speed.
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES
14.3 Average Speed i. Calculate the average speed in various
situations.
ii. Calculate
a) the distance, given the average speed and the
time
b) the time, given the average speed and the
distance
14.4 Acceleration i. Identify the two quantities involved in
acceleration.
ii. Calculate and interpret acceleration
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
Students will be Students will be able to:
28 15. TRIGONOMETRY taught to : Multiple Identifying Thankful Tangent
15.1 Tangent of an 15.1 Understand i. Identify the: Intelligence relations
acute angle and use a) hypotenuse Responsible Acute angle
tangent of b) the opposite side and the Integrating Translating Average speed
an acute the adjacent side with respect ICT into and Self- Right-angled
angle in a to one of the acute angles. teaching and motivated triangle
right-angled ii. Determine the tangent of an learning
triangle. angle. Identifying Act wisely Hypotenuse
iii. Calculate the tangent of an Thinking relations
angle given the lengths of sides skills Opposite sides
of the triangle Interpreting
iv. Calculate the lengths of sides of Cooperative Adjacent side
a triangle given the value of learning Translating
15.2 Understand tangent and the length of Ratio
15.2 Sine of an acute and use sine another side.
angle of an acute . Value
angle in a i. Determine the sine of an angle.
right-angled ii. Calculate the sine of an angle Length
triangle. given the lengths of sides of the
triangle. Size
concept of iii. Calculate the lengths of sides of a
15.3 Cosine of an acute 15.3 Understand triangle given the value of sine and
angle and use the length of another side.
cosine of an
acute angle i) Determine the cosine of an
in a right- angle.
angled ii) Calculate the cosine of an angle
triangle. given the lengths of sides of the
triangle.
iii) Calculate the lengths of sides of a
triangle given the value of cosine
WEEK TOPICS/ LEARNING LEARNING OUTCOMES GENERICS CCTS MORAL VOCABULARY
SUBTOPICS OBJECTIVES VALUES
Students will be Students will be able to :
28 15. TRIGONOMETRY taught to :
15.4 Problem solving 15.4 Use the i) Calculate the value of other Multiple Identifying Thankful Angle
involving the values values of trigonometric ratios given the Intelligence relations
of tangent, sine tangent, value of a trigonometric ratio. Responsible Degree
and cosine sine and ii) Convert the measurement of Integrating Translating
cosine to angles from: ICT into Self- Minute
solve a) degrees to degrees and minutes. teaching and motivated
problems. b) Degrees and minutes to learning Tangent
degrees. Act wisely
iii) Find the value of: Thinking Sine
a) tangent skills
b) sine cosine
c) cosine Cooperative
of 30°, 45° and 60° without using learning
scientific calculator.
iv) Find the value of:
a) tangent
b) sine
c) cosine
using scientific calculator.
v) Find the angles given the
values of:
a) tangent
b) sine
c) cosine
using scientific calculators.
vi) Solve problems involving
trigonometric ratios.
```
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# Counting Mat
Alignments to Content Standards: K.CC.B.4
#### Materials
• Counting Mats
• Small objects to count, such as bears or tiles.
#### Action
The teacher gives students the counting mat and many small objects to count with. Some students will automatically read the numbers and assemble the correct number of object then match them to the dots on the counting mat to verify they counted correctly. Other students who need more scaffolding will match each object to a dot. Students who do it this way should be guided to count the objects once they have assembled them on the dots. Once a student is done with each number they can move on to the next number. The teacher should do a quick check of a student's work before the student begins working on the larger numbers.
## IM Commentary
The teacher can print the counting mats, copy them as double-sided copies with 1-6 on the front and 7-12 on the back, and then laminate them so they can be reused. The mats can also be copied onto larger paper (such as 11x17 paper). This task will probably be used at the start of the kindergarten year and some children may benefit from the larger size mat as their fine-motor skills are still developing.
This task gives students another way to practice counting and gain fluency with connecting a written number with the act of counting. This task should be introduced by the teacher and would then be a good independent center. The number mat could be made with a different counting sequence if the teacher desires.
• Number Mat
• ## Solution
Students must correctly match up the number of objects to the dots and the number.
over 5 years
removed
#### Stacie Kaichi-Imamura says:
over 5 years
Question: With this particular activity, how would I know if my students had a firm understanding of K.CC.B.4c (Understanding that each successive number refers to a quantity that is one larger). Perhaps after the students do this activity at least once, I should make sure that I ask them questions such as, "What do you notice about the number that comes right after the previous number (e.g., what do you notice about the number of counters that you put under the number 5 and the number of counters you put under the number 6?" I think this would be helpful for me to know that the students will be paying attention to the pattern in the numbers and that there is one more than the previous number.
#### Georgia Wood, Teacher says:
over 5 years
Hi Stacie, thanks for your question! I think your suggestion is great. When I do this activity in my classroom I try and observe my students doing the activity and ask questions as they are working. I usually let students do this activity a few times during the fall and I think that as their understanding of numbers deepens they notice different things. For example in September a student might just be able to match up the counters, count out loud and then connect that to the numeral. However six weeks later when a student does this activity for the second or third time they would notice how each number has one more counter then the last. Questioning them, and pointing out patterns, help students get to that deeper level. | 670 | 3,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-43 | latest | en | 0.944321 |
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# 5 Formulas to Calculate the Area of a Triangle
A triangle is a special case of polygon with three sides. It can be formed by three non-parallel lines intersecting each other. The area of a triangle refers to the area enclosed by its three sides. Depending on given conditions, you may choose different formula to calculate the area of a triangle.
## 1. Area of a triangle by base and height
The formula for the area of a triangle by its base and height is
A = \dfrac{1}{2}bh
where b stands for the base and h stands for the altitude (height) drawn to that base.
How to derive the formula? Well, consider a diagonal divides a parallelogram into two congruent triangles. The area of the triangle will be one half of the area of the parallelogram.
Based on the formula of the area of a parallelogram, that is,
A_p = bh
The area of a parallelogram A_p is the product of the base b and the height h of the parallelogram, we get the formula of the area for the triangle.
## 2. Trigonometry Formula for the Area of a Triangle
Given two sides and the Included angle of the two sides, the area of a triangle can be expressed as,
A = \dfrac{1}{2}ab\sin C
That is, the area of a triangle is the half of the product of two sides and the sines of the included angle.
If angle C is equal to 90 \degree , we get the formula for the area of right triangle,
A = \dfrac{1}{2}ab
## 3. Area of a triangle by the length of three sides - Heron’s formula
If three sides of a triangle are known values, the area of the triangle can be calculated with Heron’s formula.
A = \sqrt{s(s-a)(s-b)(s-c)} ,
in which s =\dfrac{a+b+c}{2} ,
Heron's formula can also be written as
A = \dfrac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}
## 4. Area of a triangle with an inscribed circle
Given three sides of a triangle and radius of its inscribed circle, the triangle can be divided into three sub-triangles with the same height. So the area of the triangle is the sum of the three triangles.
A = \dfrac{1}{2}(a+b+c)r
(1)
By Heron’s formula , we get,
r = \dfrac{A}{s}
= \dfrac{ \sqrt{s(s-a)(s-b)(s-c)}}{s}
=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s} }
(2)
## 5. Area of a triangle with circumcircle circle
Given three sides and the radius of the circumcircle circle of a triangle, the area of a triangle can be expressed in simple form .
According g to The Law of Sines,
\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R
in which R is the radius of the circumcircle circle of the triangle.
Then
\sin C = \dfrac{c}{2R}
The equation related to a side and the sines of its opposite angle of a triangle can also be derived geometrically.
Substitute to the trigonometry area formula, we get
A = \dfrac{abc}{4R}
Steven Zheng posted 3 years ago
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A particle is thrown vertically upwards from the ground. If the ratio of time of ascent to time of descent is √2/√3, then retardation produced due to resistive force acting on the particle is (a) 4 m/s^2 (b) 2 m/s^2 (c) 3 m/s^2 (d) 6 m/s^2
A particle is thrown vertically upwards from the ground. If the ratio of time of ascent to time of descent is √2/√3, then retardation produced due to resistive force acting on the particle is (a) 4 m/s^2 (b) 2 m/s^2 (c) 3 m/s^2 (d) 6 m/s^2
2 years ago
Khimraj
3007 Points
ta/td = $\sqrt{(g-a)/(g+a)}$(ta/td )2 = $(g-a)/(g+a)$2/3 = (10-a)/(10+a)20 + 2a = 30 – 3a5a = 10so a = 2 m/s2Hope it clears........................................................................
2 years ago
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Guess my Number (Posted on 2004-12-20)
The Mad Hatter poured Alice another cup of tea and set the teapot down.
"May I please have one of those biscuits?" asked Alice rather timedly, rather taken aback by the sight of her present company.
"A biscuit?? Oh, I don't know. Those are the March Hare's special un-birthday biscuits, don't you know. Can she have a biscuit, March ?" asked the Hatter.
"A biscuit?? One of mine? Hurmph. Ok, I'm thinking of an integer from one to three (inclusive). If you get it, you can have a biscuit. If not, I'm afraid you'll just have to leave" said the March Hare, leaning back.
"Oh goodness, I don't know" said Alice worriedly. "Couldn't I just get a hint?"
"Oh, all right..." the March Hare sighed. "You get to ask me one yes-or-no question, that's all"
Alice thought for a moment, then smiled, knowing exactly what to ask.
See The Solution Submitted by Sam Rating: 3.7647 (17 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Only answer. | Comment 18 of 32 |
This question can not be answered satisfactory in the terms of a logic question, there is no definitive answer that in my personal opinion can be viewed as a realistic solution. It can however be answered by looking at it in it's true form, which is a riddle.
The statment clearly says "You get to ask me one yes-or-no question, that's all", the fact that is easily forgotten is there are three people. It states no where that any of the Hatter can not ask 1 or as many questions as he would like. allowing Alice to ask her 1 yes or no question getting a definitive answer.
Posted by Vinnie on 2004-12-23 20:38:27
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### Powers of negative numbers
When we have an exponent on a negative number, we can get a positive result or a negative result.
We will know this based on the exponent – whether it is even or odd.
### Powers with exponent $$0$$
Any number with an exponent of $0$ will be equal to $1$. (Except for $0$)
No matter which number we raise to the power of $0$, we will always get a result of 1.
### Powers with negative integer exponents
In an exercise where we have a negative exponent, we turn the term into a fraction where:
the numerator will be $1$ and in the denominator, the base of the exponent with the positive exponent.
## Examples with solutions for Powers - special cases
### Exercise #1
$112^0=\text{?}$
### Step-by-Step Solution
We use the zero exponent rule.
$X^0=1$We obtain
$112^0=1$Therefore, the correct answer is option C.
1
### Exercise #2
$5^0=$
### Step-by-Step Solution
We use the power property:
$X^0=1$We apply it to the problem:
$5^0=1$Therefore, the correct answer is C.
$1$
### Exercise #3
$(\frac{7}{125})^0=\text{?}$
### Step-by-Step Solution
We use the zero exponent rule.
$X^0=1$We obtain:
$\big( \frac{7}{125}\big)^0=1$Therefore, the correct answer is option B.
1
### Exercise #4
$(\frac{7}{4})^?=1$
### Step-by-Step Solution
Due to the fact that raising any number (except zero) to the power of zero will yield the result 1:
$X^0=1$It is thus clear that:
$(\frac{7}{4})^0=1$Therefore, the correct answer is option C.
0
### Exercise #5
$(300\cdot\frac{5}{3}\cdot\frac{2}{7})^0=\text{?}$
### Step-by-Step Solution
Due to the fact that raising any number (except zero) to the power of zero will give the result 1:
$X^0=1$Let's examine the expression of the problem:
$(300\cdot\frac{5}{3}\cdot\frac{2}{7})^0$The expression inside of the parentheses is clearly not 0 (it can be calculated numerically and verified)
Therefore, the result of raising to the power of zero will give the result 1, that is:
$(300\cdot\frac{5}{3}\cdot\frac{2}{7})^0 =1$Therefore, the correct answer is option A.
1
### Exercise #6
$7^5\cdot7^{-6}=\text{?}$
### Step-by-Step Solution
We begin by using the rule for multiplying exponents. (the multiplication between terms with identical bases):
$a^m\cdot a^n=a^{m+n}$We then apply it to the problem:
$7^5\cdot7^{-6}=7^{5+(-6)}=7^{5-6}=7^{-1}$When in a first stage we begin by applying the aforementioned rule and then continue on to simplify the expression in the exponent,
Next, we use the negative exponent rule:
$a^{-n}=\frac{1}{a^n}$We apply it to the expression obtained in the previous step:
$7^{-1}=\frac{1}{7^1}=\frac{1}{7}$We then summarise the solution to the problem: $7^5\cdot7^{-6}=7^{-1}=\frac{1}{7}$Therefore, the correct answer is option B.
$\frac{1}{7}$
### Exercise #7
$(8\times9\times5\times3)^{-2}=$
### Step-by-Step Solution
We begin by applying the power rule to the products within the parentheses:
$(z\cdot t)^n=z^n\cdot t^n$That is, the power applied to a product within parentheses is applied to each of the terms when the parentheses are opened,
We apply the rule to the given problem:
$(8\cdot9\cdot5\cdot3)^{-2}=8^{-2}\cdot9^{-2}\cdot5^{-2}\cdot3^{-2}$Therefore, the correct answer is option c.
Note:
Whilst it could be understood that the above power rule applies only to two terms of the product within parentheses, in reality, it is also valid for the power over a multiplication of multiple terms within parentheses, as was seen in the above problem.
A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms within parentheses (as formulated above), then it is also valid for a power over several terms of the product within parentheses (for example - three terms, etc.).
$8^{-2}\times9^{-2}\times5^{-2}\times3^{-2}$
### Exercise #8
$12^4\cdot12^{-6}=\text{?}$
### Step-by-Step Solution
We begin by using the power rule of exponents; for the multiplication of terms with identical bases:
$a^m\cdot a^n=a^{m+n}$We apply it to the given problem:
$12^4\cdot12^{-6}=12^{4+(-6)}=12^{4-6}=12^{-2}$When in a first stage we apply the aforementioned rule and then simplify the subsequent expression in the exponent,
Next, we use the negative exponent rule:
$a^{-n}=\frac{1}{a^n}$We apply it to the expression that we obtained in the previous step:
$12^{-2}=\frac{1}{12^2}=\frac{1}{144}$Lastly we summarise the solution to the problem: $12^4\cdot12^{-6}=12^{-2} =\frac{1}{144}$Therefore, the correct answer is option A.
$\frac{1}{144}$
### Exercise #9
$(3a)^{-2}=\text{?}$
$a\ne0$
### Step-by-Step Solution
We begin by using the negative exponent rule:
$b^{-n}=\frac{1}{b^n}$We apply it to the given expression and obtain the following:
$(3a)^{-2}=\frac{1}{(3a)^2}$We then use the power rule for parentheses:
$(x\cdot y)^n=x^n\cdot y^n$We apply it to the denominator of the expression and obtain the following:
$\frac{1}{(3a)^2}=\frac{1}{3^2a^2}=\frac{1}{9a^2}$Let's summarize the solution to the problem:
$(3a)^{-2}=\frac{1}{(3a)^2} =\frac{1}{9a^2}$
Therefore, the correct answer is option A.
$\frac{1}{9a^2}$
### Exercise #10
$10^{-5}=?$
### Step-by-Step Solution
First, let's recall the negative exponent rule:
$b^{-n}=\frac{1}{b^n}$We'll apply it to the expression we received:
$10^{-5}=\frac{1}{10^5}=\frac{1}{100000}=0.00001$In the final steps, we performed the exponentiation in the numerator and then wrote the answer as a decimal.
Therefore, the correct answer is option A.
$0.00001$
### Exercise #11
$\frac{1}{12^3}=\text{?}$
### Step-by-Step Solution
To begin with, we must remind ourselves of the Negative Exponent rule:
$a^{-n}=\frac{1}{a^n}$We apply it to the given expression :
$\frac{1}{12^3}=12^{-3}$Therefore, the correct answer is option A.
$12^{-3}$
### Exercise #12
$7^{-4}=\text{?}$
### Step-by-Step Solution
We must first remind ourselves of the negative exponent rule:
$a^{-n}=\frac{1}{a^n}$When applied to given the expression we obtain the following:
$7^{-4}=\frac{1}{7^4}=\frac{1}{2401}$
Therefore, the correct answer is option C.
$\frac{1}{2401}$
### Exercise #13
$\frac{1}{(-2)^7}=?$
### Step-by-Step Solution
To begin with we deal with the expression in the denominator of the fraction. Making note of the power rule for exponents (raising an exponent to another exponent):
$(a^m)^n=a^{m\cdot n}$We obtain the following:
$(-2)^7=(-1\cdot2)^7=(-1)^7\cdot2^7=-1\cdot2^7=-2^7$
We then return to the initial problem and apply the above information:
$\frac{1}{(-2)^7}=\frac{1}{-2^7}=\frac{1}{-1}\cdot\frac{1}{2^7}=-\frac{1}{2^7}$
In the last step we remember that:
$\frac{1}{-1}=-1$
Next, we remember the Negative Exponent rule ( raising exponents to a negative power)
$a^{-n}=\frac{1}{a^n}$We apply it to the expression we obtained in the last step:
$-\frac{1}{2^7}=-2^{-7}$Let's summarize the steps of the solution:
$\frac{1}{(-2)^7}=-\frac{1}{2^7} = -2^{-7}$
Therefore, the correct answer is option C.
$(-2)^{-7}$
### Exercise #14
$\frac{1}{2^9}=\text{?}$
### Step-by-Step Solution
We use the power property for a negative exponent:
$a^{-n}=\frac{1}{a^n}$We apply it to the given expression:
$\frac{1}{2^9}=2^{-9}$
Therefore, the correct answer is option A.
$2^{-9}$
### Exercise #15
$x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=\text{?}$
### Step-by-Step Solution
First we will rearrange the expression and use the fact that multiplying a fraction means multiplying the numerator of the fraction, and the distributive property of multiplication:
$x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=x^3\cdot x^4\cdot 2\cdot \frac{1}{x^3}\cdot x^{-8}=2\cdot x^3\cdot x^4\cdot \frac{1}{x^3}\cdot x^{-8}$Next, we'll use the law of exponents for negative exponents:
$a^{-n}=\frac{1}{a^n}$We'll apply the law of exponents to the expression in the problem:
$2\cdot x^3\cdot x^4\cdot\frac{1}{x^3}\cdot x^{-8} =2\cdot x^3\cdot x^4\cdot x^{-3}\cdot x^{-8}$When we applied the above law of exponents for the fraction in the multiplication,
From now on, we will no longer use the multiplication sign and will switch to the conventional notation where juxtaposition of terms means multiplication between them,
Now we'll recall the law of exponents for multiplying terms with the same base:
$a^m\cdot a^n=a^{m+n}$And we'll apply this law of exponents to the expression we got in the last step:
$2x^3x^4x^{-3}x^{-8} =2x^{3+4+(-3)+(-8)} =2x^{3+4-3-8}=2x^{-4}$When in the first stage we applied the above law of exponents and in the following stages we simplified the expression in the exponent,
Let's summarize the solution steps so far, we got that:
$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}= 2 x^3x^4 x^{-3} x^{-8} =2x^{-4}$
Now let's note that there is no such answer in the given options, a further check of what we've done so far will also reveal that there is no calculation error,
Therefore, we can conclude that additional mathematical manipulation is required to determine which is the correct answer among the suggested answers,
Let's note that in answers A and B there are similar expressions to the one we got in the last stage, however - we can directly rule out the other two options since they are clearly different from the expression we got,
Furthermore, we'll note that in the expression we got, x is in a negative exponent and is in the numerator (Note at the end of the solution on this topic), whereas in answer B it is in a positive exponent and in the denominator (and both are in the numerator - Note at the end of the solution on this topic), so we'll rule out this answer,
If so - we are left with only one option - which is answer A', however we want to verify (and need to verify!) that this is indeed the correct answer:
Let's note that in the expression we got x is in a negative exponent and is in the numerator (Note at the end of the solution on this topic), whereas in answer B it is in a positive exponent and in the denominator , which reminds us of the law of exponents for negative exponents mentioned at the beginning of the solution,
In addition, let's note that in answer B x is in the second power but inside parentheses that are also in the second power, whereas in the expression we got in the last stage of solving the problem x is in the fourth power which might remind us of the law of exponents for power to a power,
We'll check this, starting with the law of exponents for negative exponents mentioned at the beginning of the solution, but in the opposite direction:
$\frac{1}{a^n} =a^{-n}$Next, we'll represent the term with the negative exponent that we got in the last stage of solving the problem, as a term in the denominator of the fraction with a positive exponent:
$2x^{-4}=2\cdot\frac{1}{x^4}$When we applied the above law of exponents,
Next, let's note that using the law of exponents for power to a power, but in the opposite direction:
$a^{m\cdot n}= (a^m)^n$We can conclude that:
$x^4=x^{2\cdot2}=(x^2)^2$Therefore, we'll return to the expression we got in the last stage and apply this understanding:
$2\cdot\frac{1}{x^4} =2\cdot\frac{1}{(x^2)^2}$ Let's summarize then the problem-solving stages so far, we got that:
$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4} =2\cdot\frac{1}{(x^2)^2}$Let's note that we still haven't got the exact expression suggested in answer A, but we are already very close,
To reach the exact expression claimed in answer A, we'll recall another important law of exponents, and a useful mathematical fact:
Let's recall the law of exponents for exponents applying to terms in parentheses, but in the opposite direction:
$\frac{a^n}{c^n}=\big(\frac{a}{c}\big)^n$And let's also recall the fact that raising the number 1 to any power will yield the result 1:
$1^{x}=1$And therefore we can write the expression we got in the last stage in the following way:
$2\cdot\frac{1}{(x^2)^2}=2\cdot\frac{1^2}{(x^2)^2}$And then since in the numerator and denominator of the fraction there are terms with the same exponent we can apply the above law of exponents, and represent the fraction whose numerator and denominator are terms with the same exponent as a fraction whose numerator and denominator are the bases of the terms and it is raised to the same exponent:
$2\cdot\frac{1^2}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$ Let's summarize then the solution stages so far, we got that:
$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4}=2\cdot\frac{1}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$And therefore the correct answer is indeed answer A.
Note:
When it's written "the number in the numerator" despite the fact that there is no fraction in the expression at all, it's because we can always refer to any number as a number in the numerator of a fraction if we remember that any number divided by 1 equals itself, that is, we can always write a number as a fraction by writing it like this:
$X=\frac{X}{1}$And therefore we can actually refer to $X$as a number in the numerator of a fraction.
$2(\frac{1}{x^2})^2$ | 3,778 | 13,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 99, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-38 | latest | en | 0.720652 |
https://cloud.google.com/spanner/docs/mathematical_functions?hl=nb-NO | 1,621,395,295,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00544.warc.gz | 201,043,359 | 49,428 | # Mathematical functions
All mathematical functions have the following behaviors:
• They return `NULL` if any of the input parameters is `NULL`.
• They return `NaN` if any of the arguments is `NaN`.
### ABS
``````ABS(X)
``````
Description
Computes absolute value. Returns an error if the argument is an integer and the output value cannot be represented as the same type; this happens only for the largest negative input value, which has no positive representation.
X ABS(X)
25 25
-25 25
`+inf` `+inf`
`-inf` `+inf`
Return Data Type
INPUTINT64NUMERICFLOAT64
OUTPUTINT64NUMERICFLOAT64
### SIGN
``````SIGN(X)
``````
Description
Returns `-1`, `0`, or `+1` for negative, zero and positive arguments respectively. For floating point arguments, this function does not distinguish between positive and negative zero.
X SIGN(X)
25 +1
0 0
-25 -1
NaN NaN
Return Data Type
INPUTINT64NUMERICFLOAT64
OUTPUTINT64NUMERICFLOAT64
### IS_INF
``````IS_INF(X)
``````
Description
Returns `TRUE` if the value is positive or negative infinity.
Returns `FALSE` for `NUMERIC` inputs since `NUMERIC` cannot be `INF`.
X IS_INF(X)
`+inf` `TRUE`
`-inf` `TRUE`
25 `FALSE`
### IS_NAN
``````IS_NAN(X)
``````
Description
Returns `TRUE` if the value is a `NaN` value.
Returns `FALSE` for `NUMERIC` inputs since `NUMERIC` cannot be `NaN`.
X IS_NAN(X)
`NaN` `TRUE`
25 `FALSE`
### IEEE_DIVIDE
``````IEEE_DIVIDE(X, Y)
``````
Description
Divides X by Y; this function never fails. Returns `FLOAT64`. Unlike the division operator (/), this function does not generate errors for division by zero or overflow.
X Y IEEE_DIVIDE(X, Y)
20.0 4.0 5.0
0.0 25.0 0.0
25.0 0.0 `+inf`
-25.0 0.0 `-inf`
0.0 0.0 `NaN`
0.0 `NaN` `NaN`
`NaN` 0.0 `NaN`
`+inf` `+inf` `NaN`
`-inf` `-inf` `NaN`
### SQRT
``````SQRT(X)
``````
Description
Computes the square root of X. Generates an error if X is less than 0.
X SQRT(X)
`25.0` `5.0`
`+inf` `+inf`
`X < 0` Error
Return Data Type
`NUMERIC` is not supported directly as an input. You must first explicitly cast `NUMERIC` to `FLOAT64`. The output will be `FLOAT64`.
### POW
``````POW(X, Y)
``````
Description
Returns the value of X raised to the power of Y. If the result underflows and is not representable, then the function returns a value of zero.
X Y POW(X, Y)
2.0 3.0 8.0
1.0 Any value including `NaN` 1.0
Any value including `NaN` 0 1.0
-1.0 `+inf` 1.0
-1.0 `-inf` 1.0
ABS(X) < 1 `-inf` `+inf`
ABS(X) > 1 `-inf` 0.0
ABS(X) < 1 `+inf` 0.0
ABS(X) > 1 `+inf` `+inf`
`-inf` Y < 0 0.0
`-inf` Y > 0 `-inf` if Y is an odd integer, `+inf` otherwise
`+inf` Y < 0 0
`+inf` Y > 0 `+inf`
Finite value < 0 Non-integer Error
0 Finite value < 0 Error
Return Data Type
The return data type is determined by the argument types with the following table.
INPUTINT64NUMERICFLOAT64
INT64FLOAT64NUMERICFLOAT64
NUMERICNUMERICNUMERICFLOAT64
FLOAT64FLOAT64FLOAT64FLOAT64
### POWER
``````POWER(X, Y)
``````
Description
Synonym of `POW(X, Y)`.
### EXP
``````EXP(X)
``````
Description
Computes e to the power of X, also called the natural exponential function. If the result underflows, this function returns a zero. Generates an error if the result overflows.
X EXP(X)
0.0 1.0
`+inf` `+inf`
`-inf` 0.0
Return Data Type
`NUMERIC` is not supported directly as an input. You must first explicitly cast `NUMERIC` input to `FLOAT64`. The output will be `FLOAT64`.
### LN
``````LN(X)
``````
Description
Computes the natural logarithm of X. Generates an error if X is less than or equal to zero.
X LN(X)
1.0 0.0
`+inf` `+inf`
`X < 0` Error
Return Data Type
`NUMERIC` is not supported directly as an input. You must first explicitly cast `NUMERIC` input to `FLOAT64`. The output will be `FLOAT64`.
### LOG
``````LOG(X [, Y])
``````
Description
If only X is present, `LOG` is a synonym of `LN`. If Y is also present, `LOG` computes the logarithm of X to base Y.
X Y LOG(X, Y)
100.0 10.0 2.0
`-inf` Any value `NaN`
Any value `+inf` `NaN`
`+inf` 0.0 < Y < 1.0 `-inf`
`+inf` Y > 1.0 `+inf`
X <= 0 Any value Error
Any value Y <= 0 Error
Any value 1.0 Error
Return Data Type
`NUMERIC` is not supported directly as an input. You must first explicitly cast `NUMERIC` input to `FLOAT64`. The output will be `FLOAT64`.
### LOG10
``````LOG10(X)
``````
Description
Similar to `LOG`, but computes logarithm to base 10.
X LOG10(X)
100.0 2.0
`-inf` `NaN`
`+inf` `NaN`
X <= 0 Error
Return Data Type
`NUMERIC` is not supported directly as an input. You must first explicitly cast `NUMERIC` input to `FLOAT64`. The output will be `FLOAT64`.
### GREATEST
``````GREATEST(X1,...,XN)
``````
Description
Returns the largest value among X1,...,XN according to the < comparison. If any parts of X1,...,XN are `NULL`, the return value is `NULL`.
X1,...,XN GREATEST(X1,...,XN)
3,5,1 5
Return Data Types
Data type of the input values.
### LEAST
``````LEAST(X1,...,XN)
``````
Description
Returns the smallest value among X1,...,XN according to the > comparison. If any parts of X1,...,XN are `NULL`, the return value is `NULL`.
X1,...,XN LEAST(X1,...,XN)
3,5,1 1
Return Data Types
Data type of the input values.
### DIV
``````DIV(X, Y)
``````
Description
Returns the result of integer division of X by Y. Division by zero returns an error. Division by -1 may overflow. If both inputs are `NUMERIC` and the result is overflow, then it returns a `numeric overflow` error.
X Y DIV(X, Y)
20 4 5
0 20 0
20 0 Error
Return Data Type
The return data type is determined by the argument types with the following table.
INPUTINT64NUMERIC
INT64INT64NUMERIC
NUMERICNUMERICNUMERIC
### SAFE_DIVIDE
``````SAFE_DIVIDE(X, Y)
``````
Description
Equivalent to the division operator (`X / Y`), but returns `NULL` if an error occurs, such as a division by zero error.
XYSAFE_DIVIDE(X, Y)
2045
020`0`
200`NULL`
Return Data Type
INPUTINT64NUMERICFLOAT64
INT64FLOAT64NUMERICFLOAT64
NUMERICNUMERICNUMERICFLOAT64
FLOAT64FLOAT64FLOAT64FLOAT64
### SAFE_MULTIPLY
``````SAFE_MULTIPLY(X, Y)
``````
Description
Equivalent to the multiplication operator (`*`), but returns `NULL` if overflow occurs.
XYSAFE_MULTIPLY(X, Y)
20480
Return Data Type
INPUTINT64NUMERICFLOAT64
INT64INT64NUMERICFLOAT64
NUMERICNUMERICNUMERICFLOAT64
FLOAT64FLOAT64FLOAT64FLOAT64
### SAFE_NEGATE
``````SAFE_NEGATE(X)
``````
Description
Equivalent to the unary minus operator (`-`), but returns `NULL` if overflow occurs.
XSAFE_NEGATE(X)
+1-1
-1+1
00
Return Data Type
INPUTINT64NUMERICFLOAT64
OUTPUTINT64NUMERICFLOAT64
``````SAFE_ADD(X, Y)
``````
Description
Equivalent to the addition operator (`+`), but returns `NULL` if overflow occurs.
549
Return Data Type
INPUTINT64NUMERICFLOAT64
INT64INT64NUMERICFLOAT64
NUMERICNUMERICNUMERICFLOAT64
FLOAT64FLOAT64FLOAT64FLOAT64
### SAFE_SUBTRACT
``````SAFE_SUBTRACT(X, Y)
``````
Description
Returns the result of Y subtracted from X. Equivalent to the subtraction operator (`-`), but returns `NULL` if overflow occurs.
XYSAFE_SUBTRACT(X, Y)
541
Return Data Type
INPUTINT64NUMERICFLOAT64
INT64INT64NUMERICFLOAT64
NUMERICNUMERICNUMERICFLOAT64
FLOAT64FLOAT64FLOAT64FLOAT64
### MOD
``````MOD(X, Y)
``````
Description
Modulo function: returns the remainder of the division of X by Y. Returned value has the same sign as X. An error is generated if Y is 0.
X Y MOD(X, Y)
25 12 1
25 0 Error
Return Data Type
The return data type is determined by the argument types with the following table.
INPUTINT64NUMERIC
INT64INT64NUMERIC
NUMERICNUMERICNUMERIC
### ROUND
``````ROUND(X [, N])
``````
Description
If only X is present, `ROUND` rounds X to the nearest integer. If N is present, `ROUND` rounds X to N decimal places after the decimal point. If N is negative, `ROUND` will round off digits to the left of the decimal point. Rounds halfway cases away from zero. Generates an error if overflow occurs.
X ROUND(X)
2.0 2.0
2.3 2.0
2.8 3.0
2.5 3.0
-2.3 -2.0
-2.8 -3.0
-2.5 -3.0
0 0
`+inf` `+inf`
`-inf` `-inf`
`NaN` `NaN`
Return Data Type
INPUTINT64NUMERICFLOAT64
OUTPUTFLOAT64NUMERICFLOAT64
### TRUNC
``````TRUNC(X [, N])
``````
Description
If only X is present, `TRUNC` rounds X to the nearest integer whose absolute value is not greater than the absolute value of X. If N is also present, `TRUNC` behaves like `ROUND(X, N)`, but always rounds towards zero and never overflows.
X TRUNC(X)
2.0 2.0
2.3 2.0
2.8 2.0
2.5 2.0
-2.3 -2.0
-2.8 -2.0
-2.5 -2.0
0 0
`+inf` `+inf`
`-inf` `-inf`
`NaN` `NaN`
Return Data Type
INPUTINT64NUMERICFLOAT64
OUTPUTFLOAT64NUMERICFLOAT64
### CEIL
``````CEIL(X)
``````
Description
Returns the smallest integral value that is not less than X.
X CEIL(X)
2.0 2.0
2.3 3.0
2.8 3.0
2.5 3.0
-2.3 -2.0
-2.8 -2.0
-2.5 -2.0
0 0
`+inf` `+inf`
`-inf` `-inf`
`NaN` `NaN`
Return Data Type
INPUTINT64NUMERICFLOAT64
OUTPUTFLOAT64NUMERICFLOAT64
### CEILING
``````CEILING(X)
``````
Description
Synonym of CEIL(X)
### FLOOR
``````FLOOR(X)
``````
Description
Returns the largest integral value that is not greater than X.
X FLOOR(X)
2.0 2.0
2.3 2.0
2.8 2.0
2.5 2.0
-2.3 -3.0
-2.8 -3.0
-2.5 -3.0
0 0
`+inf` `+inf`
`-inf` `-inf`
`NaN` `NaN`
Return Data Type
INPUTINT64NUMERICFLOAT64
OUTPUTFLOAT64NUMERICFLOAT64
### COS
``````COS(X)
``````
Description
Computes the cosine of X where X is specified in radians. Never fails.
X COS(X)
`+inf` `NaN`
`-inf` `NaN`
`NaN` `NaN`
### COSH
``````COSH(X)
``````
Description
Computes the hyperbolic cosine of X where X is specified in radians. Generates an error if overflow occurs.
If X is `NUMERIC` then, the output is `FLOAT64`.
X COSH(X)
`+inf` `+inf`
`-inf` `+inf`
`NaN` `NaN`
### ACOS
``````ACOS(X)
``````
Description
Computes the principal value of the inverse cosine of X. The return value is in the range [0,π]. Generates an error if X is a value outside of the range [-1, 1].
If X is `NUMERIC` then, the output is `FLOAT64`.
X ACOS(X)
`+inf` `NaN`
`-inf` `NaN`
`NaN` `NaN`
X < -1 Error
X > 1 Error
### ACOSH
``````ACOSH(X)
``````
Description
Computes the inverse hyperbolic cosine of X. Generates an error if X is a value less than 1.
If X is `NUMERIC` then, the output is `FLOAT64`.
X ACOSH(X)
`+inf` `+inf`
`-inf` `NaN`
`NaN` `NaN`
X < 1 Error
### SIN
``````SIN(X)
``````
Description
Computes the sine of X where X is specified in radians. Never fails.
X SIN(X)
`+inf` `NaN`
`-inf` `NaN`
`NaN` `NaN`
### SINH
``````SINH(X)
``````
Description
Computes the hyperbolic sine of X where X is specified in radians. Generates an error if overflow occurs.
If X is `NUMERIC` then, the output is `FLOAT64`.
X SINH(X)
`+inf` `+inf`
`-inf` `-inf`
`NaN` `NaN`
### ASIN
``````ASIN(X)
``````
Description
Computes the principal value of the inverse sine of X. The return value is in the range [-π/2,π/2]. Generates an error if X is outside of the range [-1, 1].
If X is `NUMERIC` then, the output is `FLOAT64`.
X ASIN(X)
`+inf` `NaN`
`-inf` `NaN`
`NaN` `NaN`
X < -1 Error
X > 1 Error
### ASINH
``````ASINH(X)
``````
Description
Computes the inverse hyperbolic sine of X. Does not fail.
If X is `NUMERIC` then, the output is `FLOAT64`.
X ASINH(X)
`+inf` `+inf`
`-inf` `-inf`
`NaN` `NaN`
### TAN
``````TAN(X)
``````
Description
Computes the tangent of X where X is specified in radians. Generates an error if overflow occurs.
X TAN(X)
`+inf` `NaN`
`-inf` `NaN`
`NaN` `NaN`
### TANH
``````TANH(X)
``````
Description
Computes the hyperbolic tangent of X where X is specified in radians. Does not fail.
If X is `NUMERIC` then, the output is `FLOAT64`.
X TANH(X)
`+inf` 1.0
`-inf` -1.0
`NaN` `NaN`
### ATAN
``````ATAN(X)
``````
Description
Computes the principal value of the inverse tangent of X. The return value is in the range [-π/2,π/2]. Does not fail.
If X is `NUMERIC` then, the output is `FLOAT64`.
X ATAN(X)
`+inf` π/2
`-inf` -π/2
`NaN` `NaN`
### ATANH
``````ATANH(X)
``````
Description
Computes the inverse hyperbolic tangent of X. Generates an error if X is outside of the range [-1, 1].
If X is `NUMERIC` then, the output is `FLOAT64`.
X ATANH(X)
`+inf` `NaN`
`-inf` `NaN`
`NaN` `NaN`
X < -1 Error
X > 1 Error
### ATAN2
``````ATAN2(X, Y)
``````
Description
Calculates the principal value of the inverse tangent of X/Y using the signs of the two arguments to determine the quadrant. The return value is in the range [-π,π].
If Y is `NUMERIC` then, the output is `FLOAT64`.
X Y ATAN2(X, Y)
`NaN` Any value `NaN`
Any value `NaN` `NaN`
0.0 0.0 0.0
Positive Finite value `-inf` π
Negative Finite value `-inf`
Finite value `+inf` 0.0
`+inf` Finite value π/2
`-inf` Finite value -π/2
`+inf` `-inf` ¾π
`-inf` `-inf` -¾π
`+inf` `+inf` π/4
`-inf` `+inf` -π/4
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[{ "type": "thumb-up", "id": "easyToUnderstand", "label":"Easy to understand" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"Solved my problem" },{ "type": "thumb-up", "id": "otherUp", "label":"Other" }] | 4,436 | 13,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-21 | latest | en | 0.661896 |
https://solvedlib.com/question-4-4-points-if-a-120-360-and-b-400-i-100,350163 | 1,679,533,314,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00726.warc.gz | 594,080,160 | 16,191 | # Question 4 (4 points) If A-12.0 /-36.0 /and B = 4.00 i+ 10.0), what is the...
###### Question:
Question 4 (4 points) If A-12.0 /-36.0 /and B = 4.00 i+ 10.0), what is the magnitude and direction of C-A+ 2B? Show all calculations in the answer box. ▼11 Format の の EQ い Question 5 (4 points)
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5. [-12 Points) DETAILS ZILLDIFFEQMODAP11 5.2.015. Find the eigenvalues in and eigenfunctions yn(x) for the given boundary-value problem. (Give your answers in terms of n, making sure that each value of n corresponds to a unique eigenval y" + 2y' + (1 + 1)y = 0, y(0) = 0, y(7) = 0 n = 1, 2, ...
##### The relevant range of activity refers to the O geographical areas where the company plans to...
The relevant range of activity refers to the O geographical areas where the company plans to operate. O activity level where all costs are curvilinear. O levels of activity over which the company expects to operate. level of activity where all costs are constant (remain the same)....
##### Can average rate of change be negative?
Can average rate of change be negative?...
##### Construct a polynomial $P(x)$ with the specified characteristics. Determine whether or not the answer to the problem is unique. Explain and/or illustrate your answer.A fourth degree polynomial with zeros of multiplicity two at $x=2$ and $x=-3$, and a $y$ -intercept of $-2$.
Construct a polynomial $P(x)$ with the specified characteristics. Determine whether or not the answer to the problem is unique. Explain and/or illustrate your answer. A fourth degree polynomial with zeros of multiplicity two at $x=2$ and $x=-3$, and a $y$ -intercept of $-2$....
##### Why would any firm, whether global or otherwise, seek to collaborate with another firm, be it...
Why would any firm, whether global or otherwise, seek to collaborate with another firm, be it local or foreign? Explain in depth....
##### If; in a one-tail hypothesis test where Ho only rejecied the upper tail, Ihe p-value0558 and Zstat59,whal i5 Ihe stalistical decision if the null hypothesis tesled at the 09 level = of significance?What is the slalislical decision?Since the p-value (Type an inleger or & decimal Do not round )
If; in a one-tail hypothesis test where Ho only rejecied the upper tail, Ihe p-value 0558 and Zstat 59,whal i5 Ihe stalistical decision if the null hypothesis tesled at the 09 level = of significance? What is the slalislical decision? Since the p-value (Type an inleger or & decimal Do not round ...
##### Please help me out ASAP! pictures of question attached below. Needs an: -Income statement -COGS Schedule...
Please help me out ASAP! pictures of question attached below. Needs an: -Income statement -COGS Schedule -COGM Schedule thank you so much!! Superior Company provided the following data for the year ended December 31 (all raw materials are used in production as direct materials): Book ? Selling ...
Problem We wish to design a feedback controller for a missile. using hydraulic-powered ailerons that will keep the roll attitude close to zero. while staying within the physical limits of aileron deflection $and aileron deflection rate & The discrete-time motion dynamics is given by: S(k +1) = ... 1 answer ##### Problem 3. A profit-maximizing firm produces apples, and its production function is (L)- where L is... Problem 3. A profit-maximizing firm produces apples, and its production function is (L)- where L is the level of labor. Suppose the wage isw-1. The price of one apple is denoted by pa 1. Determine the marginal product of labor. Does the production function exhibit dimin ishing marginal product? Expl... 5 answers ##### Draw the mechanism of the following reaction, using the curved-arrow notation to indicate the reorganization of electron density: Be certain to show all intermediates, unshared electrons, and reversibility Or irreversibility. Draw at least one termination step.BrBrHzCH;C _H-Br:hvCH;(racemic)CH; Draw the mechanism of the following reaction, using the curved-arrow notation to indicate the reorganization of electron density: Be certain to show all intermediates, unshared electrons, and reversibility Or irreversibility. Draw at least one termination step. Br Br HzC H;C _ H-Br: hv CH; (racemic... 1 answer ##### Please write code in Python! You are asked to implement the following checksum formula to validate... Please write code in Python! You are asked to implement the following checksum formula to validate an identifi- cation number given to you. The formula works as follows. Using the original number, double the value of every other digit. Then add the values of the individual digits together (if a doub... 1 answer ##### I need 4 paragraphs which explains in detail please ill vote up thansk Write a detailed... i need 4 paragraphs which explains in detail please ill vote up thansk Write a detailed 4 paragraph discussion/summary on concepts discussed in Chapters 9 and 10 (Ullman). These concepts include how to connect to MySQL, what is a query, how is one executed, how to retrieve query results, updating... 1 answer ##### How to calculate the moles of diprotic acid H2X to reach the first and second equivalence... How to calculate the moles of diprotic acid H2X to reach the first and second equivalence points? Then used the moles of the diprotic acid to calculate the moles in the origional 100.0 Ml of solution. remmeber that you titrated 25.0 mL of the solution. information given: 4.9 mL of NaOH required to r... 1 answer ##### How do you plan for success in nursing school? How do you plan for success in nursing school?... 5 answers ##### How many paths from the upper left corner to the lower right corner of the grid to the right if you may only move down or right? Start with this: paths that start with go right" are mutually exclusive with those that start go down' Name the answer P(4,6). The mutual exclusivity tells usP(4,6) P(3,6) + P(4,5) How many paths from the upper left corner to the lower right corner of the grid to the right if you may only move down or right? Start with this: paths that start with go right" are mutually exclusive with those that start go down' Name the answer P(4,6). The mutual exclusivity tells us P(... 1 answer ##### Horizon Value of Free Cash Flows Current and projected free cash flows for Radell Global Operations... Horizon Value of Free Cash Flows Current and projected free cash flows for Radell Global Operations are shown below. Actual 2016 2017 Projected 2018 ... 5 answers ##### (4 points) radar gun was used t0 record the speed of runner at the times given in the table: Find the total distance covered by the runner in those five seconds. Fee= frce to use left or right endpoints midpoints_ or" strategic point choices as explored in class.Time (secouds Velocity (Ws7.,34 9,73 10.51 10,.76 10.81 (4 points) radar gun was used t0 record the speed of runner at the times given in the table: Find the total distance covered by the runner in those five seconds. Fee= frce to use left or right endpoints midpoints_ or" strategic point choices as explored in class. Time (secouds Velocity (Ws 7.,3... 5 answers ##### 2. For the following frequency distribution; Class Frequency 2 to 4 4to 6 6 to 8 8 to 10 10 to 12 12 to 14 a) Plot a Histogram, frequency polygon, and "less Than" cumulative frequency polygon_ 15 Points b) Compute the mean and sample standard deviation. 20 Points c) From the graphs in part a) estimate the median and mode. 10 Points d) Is this distribution symmetrical or Skewed? Why? What percent of the data values one expects to fall within 2 standard deviations from the mean? 10 Point 2. For the following frequency distribution; Class Frequency 2 to 4 4to 6 6 to 8 8 to 10 10 to 12 12 to 14 a) Plot a Histogram, frequency polygon, and "less Than" cumulative frequency polygon_ 15 Points b) Compute the mean and sample standard deviation. 20 Points c) From the graphs in part... 5 answers ##### MATHGYMWORKSHEETS pdf67 Whit n Ac)th mrrel <thr namrrator? 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##### In Exercises 114–117, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. If $f(7)=5$ and $g(4)=7,$ then $(f \circ g)(4)=35$
In Exercises 114–117, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. If $f(7)=5$ and $g(4)=7,$ then $(f \circ g)(4)=35$...
##### Jeabea "Atallaula LuiniLeHto - KreclantFraductRreciaol BRztctest
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##### Explain how to simplify $\sqrt[n]{a^{n}}$ if $n$ is even and if $n$ is odd. Give examples with your explanations.
Explain how to simplify $\sqrt[n]{a^{n}}$ if $n$ is even and if $n$ is odd. Give examples with your explanations....
##### Determine whether the statement is true or false. Explain your answer. Tabular integration by parts is useful for integrals of the form $\int p(x) f(x) d x,$ where $p(x)$ is a polynomial and $f(x)$ can be repeatedly integrated.
Determine whether the statement is true or false. Explain your answer. Tabular integration by parts is useful for integrals of the form $\int p(x) f(x) d x,$ where $p(x)$ is a polynomial and $f(x)$ can be repeatedly integrated.... | 3,123 | 11,254 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-14 | latest | en | 0.729969 |
https://www.jiskha.com/display.cgi?id=1409786495 | 1,503,305,946,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107744.5/warc/CC-MAIN-20170821080132-20170821100132-00382.warc.gz | 912,724,895 | 3,833 | posted by .
a square photo display board is made up of 60 rows of 60 photos each. the area of each square is 4 in. how long is each side of the display board?
90
120
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More Similar Questions | 524 | 2,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-34 | latest | en | 0.891797 |
https://www.goshen.edu/physix/310/gco/13.7.php | 1,429,333,477,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246633799.48/warc/CC-MAIN-20150417045713-00083-ip-10-235-10-82.ec2.internal.warc.gz | 992,204,594 | 2,479 | # Connection between classical and statistical thermodynamics
Work?....Heat?....
How should we understand these classical terms in statistical mechanics?
$$U=\sum_jN_jE_j.$$
So, $$dU = \sum_jE_jdN_j + \sum_jN_jdE_j.$$
Because $E_j = E_j(V)$, $\Rightarrow dE_j = (dE_j/dV)dV$, so the second term becomes... $$\sum_jN_jdE_j = \sum_j $N_j\frac{dE_j}{dV}$dV= - \sum_j Y\,dV.$$ That is, $Y=-\sum_j $N_j\frac{dE_j}{dV}$$. [The negative sign is a reminder that with energy levels of the particle in a box, $E_j \propto n_j^2/V^{2/3}$, as $V$ increases, the energy of level $j$ drops.]
So, in the statistical picture, we have: $$dU = \sum_j E_j\,dN_j - Y\,dV.$$
Classically: $$dU = T\,dS - P\,dV.$$
Holding $V$ constant, we have in the classical case, followed by the statistical case: $$$dU$_V = T\,dS = \delta Q_r = \sum_j E_j\,dN_j .$$
Apparently heating is associated with a change of the number of particles in energy levels, while keeping the energies of each level constant.
The remainder, $$dU - \delta Q_r = -\delta W_r = -Y\,dV = \sum_j N_j \,dE_j.$$
Particle in a box: $E_j \propto n_j^2/V^{2/3}$.
So, squeezing the box smaller increases the spacing between levels, and moves them all up.
Apparently work done *on* the system is associated with a shift in the energy levels (while keeping the number of particles in each state constant.) | 439 | 1,353 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2015-18 | latest | en | 0.871015 |
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Prove that $$\cos1^{\circ}$$ is irrational.
Is there any proof for it?
Note by Swapnil Das
2 years, 9 months ago
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Here is an elementary solution using strong induction and contradiction.
Assume that $$\cos 1^\circ$$ is rational. Then $$\cos2^\circ=2\cos^2 2^\circ-1$$ is rational. Note that $$\cos(n+1)^\circ+\cos(n-1)^\circ=2\cos 1^\circ\cos n^\circ$$.
Hence by strong induction $$\cos n^\circ$$ is rational for all integers $$n\ge 1$$. But this is clearly a contradiction as $$\cos 30^\circ$$ is trivially irrational. Thus we conclude that $$\cos 1^\circ$$ is irrational.
- 2 years, 8 months ago
Hint: Use Chebyshev Polynomials.
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A065482 a(n) = round( 2^n/n ). 2
2, 2, 3, 4, 6, 11, 18, 32, 57, 102, 186, 341, 630, 1170, 2185, 4096, 7710, 14564, 27594, 52429, 99864, 190650, 364722, 699051, 1342177, 2581110, 4971027, 9586981, 18512790, 35791394, 69273666, 134217728, 260301048, 505290270 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Harry J. Smith, Table of n, a(n) for n=1,...,300 FORMULA a(n) = A082894[n]/n. a(n) = floor((2^n + floor(n/2))/n ). MATHEMATICA Table[Floor[(Floor[n/2]+2^n)/n], {n, 1, 100}] PROG (PARI) { for (n=1, 300, write("b065482.txt", n, " ", round(2^n/n)) ) } \\ Harry J. Smith, Oct 20 2009 (MAGMA) [Round(2^n/n): n in [1..30]]; // G. C. Greubel, Jan 18 2018 CROSSREFS Cf. A000799, A053638, A082893-A082900. Sequence in context: A013588 A108150 A066015 * A323481 A319638 A179637 Adjacent sequences: A065479 A065480 A065481 * A065483 A065484 A065485 KEYWORD nonn AUTHOR N. J. A. Sloane, Dec 03 2001 STATUS approved
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Last modified September 16 18:24 EDT 2019. Contains 327116 sequences. (Running on oeis4.) | 533 | 1,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-39 | latest | en | 0.535837 |
https://couponsanddiscouts.com/normal-discount-rate-for-npv/ | 1,618,210,070,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00414.warc.gz | 291,313,438 | 9,145 | # Normal Discount Rate For Npv
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### What You Should Know About the Discount Rate
Sep 02, 2014 · As shown in the analysis above, the net present value for the given cash flows at a discount rate of 10% is equal to \$0. This means that with an initial investment of exactly \$1,000,000, this series of cash flows will yield exactly 10%.
https://propertymetrics.com/blog/npv-discount-rate/
### Discount Rate Formula: Calculating Discount Rate [WACC/APV]
Aug 16, 2019 · This second discount rate formula is fairly simple and uses the cost of equity as the discount rate: APV = NPV + PV of the impact of financing. Where: NPV = Net Present Value; PV = Present Value; Discount rate is key to managing the relationship between an investor and a company, as well as the relationship between a company and its future self.
https://www.profitwell.com/recur/all/discount-rate-formula
### NPV and Discount Rate: Property Investment Fundamentals
Aug 21, 2020 · Relationship between NPV and Discount Rate. Given the above formula, the discount rate affects NPV directly and in a negative way since it enters always in the denominator of each term in that formula, with the exception of the first term (CF 0). Thus, the higher the discount rate used in the NPV formula, the lower the resulting NPV value ...
https://property-investment.net/2020/08/21/npv-and-discount-rate-property-investment-fundamentals/
### Discount Rate in NPV | WACC vs Risk-Adjusted Rate
Apr 19, 2019 · Discount rate is the rate of interest used to determine the present value of the future cash flows of a project. For projects with average risk, it equals the weighted average cost of capital but for project with different risk exposure it should be estimated keeping in view the project risk.
https://xplaind.com/595242/discount-rate
### NPV Formula - Learn How Net Present Value Really Works
Dec 21, 2018 · The NPV formula is a way of calculating the Net Present Value (NPV) of a series of cash flows based on a specified discount rate. The NPV formula can be very useful for financial analysis and financial modeling when determining the value of an investment (a company, a project, a …
https://corporatefinanceinstitute.com/resources/knowledge/valuation/npv-formula/
### Go with the cash flow: Calculate NPV and IRR in Excel - Excel
NPV calculates that present value for each of the series of cash flows and adds them together to get the net present value. The formula for NPV is: Where n is the number of cash flows, and i is the interest or discount rate. IRR. IRR is based on NPV. You can think of it as a special case of NPV, where the rate of return that is calculated is ...
https://support.microsoft.com/en-us/office/go-with-the-cash-flow-calculate-npv-and-irr-in-excel-9e3d78bb-f1de-4f8e-a20e-b8955851690c
### A Refresher on Net Present Value
Nov 19, 2014 · If shareholders expect a 12% return, that is the discount rate the company will use to calculate NPV. If the firm pays 4% interest on its debt, then it may use that figure as the discount rate....
https://hbr.org/2014/11/a-refresher-on-net-present-value
### Discount Rates - NPV and Risk Modelling for Projects
However, government organisations tend to set higher discount rates than this. In the UK, the Treasury's rationale for its recommended standard discount rate is set out in the Green Book. The current discount rate is 3.5% in real terms, calculated using the Social Time Preference Rate (STPR) approach as summarised in the box below.
http://projectnpv.com/6.html
### What Is the Formula for Calculating Net Present Value (NPV)?
Examples Using NPV . Many projects generate revenue at varying rates over time. In this case, the formula for NPV can be broken out for each cash flow individually.
### Can one calculate an NPV without a discount rate? - Quora
The underlying principal of NPV is time value of money. In laymen's terms today's 1 dollar worth more than a dollar tomorrow. This is based on the possible earning the money can make during the time period. Discounting rate represent the annualise...
https://www.quora.com/Can-one-calculate-an-NPV-without-a-discount-rate
### Adjustment for Inflation in NPV Calculation
Mar 30, 2019 · Net present value = \$44.52 – \$25 million = \$19.52 million. Example 2: Inflation Adjustment using Real Cash-Flows and Real Discount Rate. Under the real method, we discount real cash flows using real discount rate. The relationship between nominal discount rate, real discount rate and inflation can be rearranged as follows:
https://xplaind.com/264707/npv-and-inflation
### The difference between net present value and discounted
To mitigate possible complexities in determining the net present value, account for the discount rate of the NPV formula. Bear in mind that different businesses and companies also have different methods in arriving at a discount rate, probably because each business accounts for their own specific business situation and investment realities.
### How do you calculate costs of capital when budgeting new
Jun 25, 2019 · A new project only makes economic sense if its discounted net present value (NPV) exceeds the expected costs of financing. Before budgeting for a new project, a …
### Net Present Value | Boundless Finance
The NPV Profile is a graph with the discount rate on the x-axis and the NPV of the investment on the y-axis. Higher discount rates mean cash flows that occur sooner are more influential to NPV. Since the earlier payments tend to be the outflows, the NPV profile generally shows an inverse relationship between the discount rate and NPV.
https://courses.lumenlearning.com/boundless-finance/chapter/net-present-value/
### Net Present Value (NPV) Calculator (for Fixed and Variable)
Choose the type of discount rate and fill in the initial investment, discount rate(s), the net cash flows for each period and a residual value, if applicable. The calculator will automatically determine the net present value. Calculator for Net Present Value (NPV) using a Constant Discount Rate
https://calculate.onl/npv-net-present-value-calculator/
### 80% OFF Reasonable Discount Rate For Npv Verified
[50% OFF] Reasonable Discount Rate For Npv - February 2021. 50% off (4 days ago) 4.3 Fair Value Using Net Present Value (Finance Meets . 5% 5 days ago Verified The analysts, the CFO, decides that reasonable discount rate is 5% per annum. So the net present value of these cash flows of 50, 50, 100 are now easily determined as \$50 million discounted at 5% for one year, plus \$50 million ...
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### Net Present Value (NPV) - Rate of Return Expert
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### Net Present Value (NPV) - Definition, Examples, How to do
Jan 22, 2018 · It means a rational investor would be willing to pay up to \$61,466 today to receive \$10,000 every year over 10 years. By paying this price, the investor would receive an internal rate of return Internal Rate of Return (IRR) The Internal Rate of Return (IRR) is the discount rate that makes the net present value (NPV) of a project zero. In other ...
https://corporatefinanceinstitute.com/resources/knowledge/valuation/net-present-value-npv/ | 1,858 | 7,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-17 | longest | en | 0.885815 |
https://kr.mathworks.com/matlabcentral/answers/1859873-i-need-cost-function-for-tuning-pid-controller-help-me | 1,674,881,950,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00367.warc.gz | 355,350,709 | 32,670 | # I need cost function for tuning PID controller help me
조회 수: 23(최근 30일)
Yazeed Ali 2022년 11월 22일
편집: Sam Chak 2022년 12월 27일
I have to apply some algorithms for tuning PID controller parameters (Kp,Ki,Kd) so for any algorithm i have to get separate cost function file or called objective function so if anyone has p;ease send me the file thanks
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### 답변(1개)
Sam Chak 2022년 11월 23일
편집: Sam Chak 2022년 12월 25일
Not sure what your system is, but I think you can try the one of the simplest cost function probably looks like this:
Tuning the gains is probably simpler by manual calculations (follow formulas in textbooks) or with the pidtune() command.
function J = costfun(Kp, Ki, Kd)
[t, x] = ode45(@(t, x) myode(t, y), tspan, ic, opts);
error = x - x_ref;
J = trapz(t, error.^2); % ISE
% J = trapz(t, t.*abs(error)); % ITAE
end
function dxdt = myode(t, x)
% Insert your system and the PID controller
dxdt = A.*x + B*u;
end
Edit: An example is included.
Example: Consider the system described by
.
The cost function J is given by
where
and .
A full state-feedback controller is used:
The optimal feedback gain matrix is found using fmincon().
costfun = @objfun;
nvars = 2;
A = -eye(nvars);
b = zeros(nvars, 1);
Aeq = [];
beq = [];
lb = [0 0]; % lower bound
ub = [2 2]; % upper bound
k0 = [0.5, 1.5]; % initial guess
[k, fval, exitflag, output] = fmincon(costfun, k0, A, b, Aeq, beq, lb, ub)
Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.
k = 1×2
1.0002 1.0006
fval = 2.0000
exitflag = 1
output = struct with fields:
iterations: 14 funcCount: 45 constrviolation: 0 stepsize: 8.2993e-05 algorithm: 'interior-point' firstorderopt: 8.0557e-08 cgiterations: 0 message: 'Local minimum found that satisfies the constraints.↵↵Optimization completed because the objective function is non-decreasing in ↵feasible directions, to within the value of the optimality tolerance,↵and constraints are satisfied to within the value of the constraint tolerance.↵↵<stopping criteria details>↵↵Optimization completed: The relative first-order optimality measure, 8.055695e-08,↵is less than options.OptimalityTolerance = 1.000000e-06, and the relative maximum constraint↵violation, 0.000000e+00, is less than options.ConstraintTolerance = 1.000000e-06.' bestfeasible: [1×1 struct]
function J = objfun(k)
tspan = linspace(0, 30, 3001);
ic = [1 0];
[t, x] = ode45(@(t, x) myode(t, x, k), tspan, ic);
Q = eye(2);
R = 1;
ifcn = Q(1,1)*x(:,1).^2 + Q(2,2)*x(:,2).^2 + R*(- k(1)*x(:,1) - k(2)*x(:,2)).^2;
J = trapz(t, ifcn); % ISE
end
function dxdt = myode(t, x, k)
dxdt = zeros(2, 1);
% my controller
u = - k(1)*x(1) - k(2)*x(2);
% my system
dxdt(1) = x(2);
dxdt(2) = - x(2) + u;
end
##### 댓글 수: 9표시숨기기 이전 댓글 수: 8
Sam Chak 2022년 12월 27일
편집: Sam Chak 2022년 12월 27일
I'm familiar with Grey Wolf. Why not consider posting the GWO code and the error message in a new Question? GWO was published by Seyedali Mirjalili in 2014. Optimization experts in MATLAB would be able to advise you.
I'm not sure if the proposed Cost function is helpful or not. Most likely it is because you cannot test with the GWO. However, it works with other optimizers such GA, PSO, ABC, etc., then we can safely assure that the proposed "Cost Function template" maybe effective.
I only verified it using fmincon. It does not guarantee to work with other optimizers. Let me know it doesn't work because I may consider removing it as I don't want other users to be misguided by the 'flawed' Cost Function. Thanks!
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Translated by | 1,241 | 3,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-06 | latest | en | 0.76486 |
https://whatdoesmean.net/what-is-chance/ | 1,722,837,407,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00619.warc.gz | 497,087,979 | 44,427 | # What is chance?
## What Does chance Mean
The accident refers to the combination of circumstances that it is impossible to anticipate and avoid . What happens by chance is unforeseen and therefore cannot be circumvented.
For example: "I tripped and, by chance, found a wad of bills on the ground" (in this case, the person came across the money by chance since he was not looking for it), "The collision caused several bricks to fall off from the wall, which accidentally fell on my car ” (the phrase is linked to the inappropriateness of having parked the car at the exact point where the bricks would fall, an event that the protagonist could not foresee).
Many people argue that, strictly speaking, chance does not exist . In the first example mentioned, it could be said that the subject found the bills since he tripped in the place where they were and before another person could notice the presence of the money on the ground. There was no will or intention on the part of the individual, but there was a temporal and spatial coincidence that rationally explains the finding.
The second example can also be explained from an apparently random sequence of events : since no one could foresee the accident, the driver parked his car without thinking that the wall might collapse; later, a collision in the same area caused a landslide that damaged the vehicle. Taking this event as a fluke emphasizes the person's supposed misfortune .
For mathematics, chance is linked to randomness (a process whose result is not predictable since chance intervenes). This means that the outcome of a random event cannot be known before it occurs. Instead, it is possible to speak of probabilities from statistics .
The limit of chance
Although it is possible to divide the human population into infinite groups, taking into account the most diverse parameters, the peculiar evolution of life in society that we have gone through as a species in recent decades reveals the existence of two well-defined groups: those who are take charge of their actions, and those who attribute their successes and failures to chance.
Believing that everything happens independently of our participation in the world constitutes a dangerous ideology, which turns life into a game of chance, in which it is not necessary to get involved with all our energies and with our senses wide awake, but it is enough to wait for it. lucky departure while we endure the terrible ones.
The exposure to which many are subjected today, making use of some of the tools that the Internet offers them, gives rise to innumerable examples of individuals who seek fame through various paths; New aspiring actors, dancers and singers are introduced every day, who publish videos of varying quality, length and theme, but with the same goal : to be discovered and become international stars.
Of this unstoppable cast of talents looking for an opportunity, only a few show an acceptable level of preparation and responsibility, while the majority make it clear that they do not understand the need to work towards their goals . When technical issues come into play, relying on chance to get the desired result is as absurd as riding a taxi every day to find a briefcase with money , just because it happened to someone.
For the good of the planet that so generously houses us and, consequently, for our own good, it is advisable to give a very small space to coincidences, always prioritizing the possibility of a precise cause for each situation we go through. Much of the decisions we make, if not all, derive from a fact that, sooner or later, will be the cause of a new one.
Go up | 713 | 3,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.968628 |
http://www.brightstorm.com/qna/question/4802/ | 1,371,683,685,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709805610/warc/CC-MAIN-20130516131005-00072-ip-10-60-113-184.ec2.internal.warc.gz | 357,006,283 | 10,837 | Quick Homework Help
525+0.21x=585+0.13x How to solve? ⚑ Flag
by Dylan138 at April 20, 2010
525+0.21x=585+0.13xsubtract 525 from both sides0.21x = 60 + 0.13xsubtract 0.13x from both sides0.08x = 60divide both sides by 0.08x = 750
McMayer April 20, 2010
0.21x-0.13x=585-5250.7x=60x=85.7
gibby. April 21, 2010
sorry i messed up it would be.08x=60x=750
gibby. April 21, 2010
I don't know
Sunny May 09, 2010
Subtract 525 from 585 and then solve for x
minime May 29, 2010 | 214 | 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} | 3.796875 | 4 | CC-MAIN-2013-20 | latest | en | 0.873539 |
https://www.tradingview.com/script/Q0eQz7ll-Fisher-Transform-Indicator-by-Ehlers-Strategy/ | 1,527,336,567,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.71/warc/CC-MAIN-20180526112331-20180526132331-00121.warc.gz | 850,319,711 | 70,531 | # Fisher Transform Indicator by Ehlers - Strategy
19597 15
Market prices do not have a Gaussian probability density function
as many traders think. Their probability curve is not bell-shaped.
But trader can create a nearly Gaussian PDF for prices by normalizing
them or creating a normalized indicator such as the relative strength
index and applying the Fisher transform . Such a transformed output
creates the peak swings as relatively rare events.
Fisher transform formula is: y = 0.5 * ln ((1+x)/(1-x))
The sharp turning points of these peak swings clearly and unambiguously
identify price reversals in a timely manner.
////////////////////////////////////////////////////////////
// Copyright by HPotter v1.0 01/07/2014
// Market prices do not have a Gaussian probability density function
// as many traders think. Their probability curve is not bell-shaped.
// But trader can create a nearly Gaussian PDF for prices by normalizing
// them or creating a normalized indicator such as the relative strength
// index and applying the Fisher transform. Such a transformed output
// creates the peak swings as relatively rare events.
// Fisher transform formula is: y = 0.5 * ln ((1+x)/(1-x))
// The sharp turning points of these peak swings clearly and unambiguously
// identify price reversals in a timely manner.
////////////////////////////////////////////////////////////
study(title="Fisher Transform Indicator by Ehlers Strategy", shorttitle="Fisher Transform Indicator by Ehlers")
Length = input(10, minval=1)
xHL2 = hl2
xMaxH = highest(xHL2, Length)
xMinL = lowest(xHL2,Length)
nValue1 = 0.33 * 2 * ((xHL2 - xMinL) / (xMaxH - xMinL) - 0.5) + 0.67 * nz(nValue1[1])
nValue2 = iff(nValue1 > .99, .999,
iff(nValue1 < -.99, -.999, nValue1))
nFish = 0.5 * log((1 + nValue2) / (1 - nValue2)) + 0.5 * nz(nFish[1])
pos = iff(nFish > nz(nFish[1]), 1,
iff(nFish < nz(nFish[1]), -1, nz(pos[1], 0)))
barcolor(pos == -1 ? red: pos == 1 ? green : blue )
plot(nFish, color=green, title="Fisher")
plot(nz(nFish[1]), color=red, title="Trigger")
DONATE/TIP
BTC: 1Lo1GoTNsPbrjGAgPgF4MYZPhBFdoVHaTm
Mr. Potter, what if the Fisher line is just crossing above the Trigger line and both are trending upward, however the candle is red instead of green that day. Is it still a buy signal? Thanks!
I created an account to say thank you. I love this study and I think it's applicable in even intraday data. Keep on wizarding harry!
What happens when xMaxH == xMinL? Divide by 0. If your period is low like 5 this happens with real data. Is the result 0 or undefined? When evaluation nz(nValue1) do you just ignore this term?
RBernstein7
In current version - yes.
nice induk! sps!)
is this AFL code?
PrasanthMathesh
No )
Hi,
whether the above code works for Fncharts?. I tried the above code and it did not work in Fncharts. Do we need to tweak the code in order to work in Fncharts. Is it possible to provide the code which will work for Fncharts.
thulasee77
I don`t know know what is a Fncharts.
HPotter
Thanks for the quick response. I mean Java enabled charts. I tried to replicate your code as per the below format as just inserting the code which you have mentioned did not fetching the details.
-------------------------------------------------------------------------
CREATING YOUR OWN INDICATORS IN FnCharts
-------------------------------------------------------------------------
This document contains the following information:
- sample indicator definitions
- the short description of indicator definition rules
- the list of predefined functions that can be used in indicator
definitions
- performance related tips
- error messages related information
-------------------------------------------------------------------------
SAMPLE INDICATOR DEFINITIONS
-------------------------------------------------------------------------
/* User-defined ROC (Rate of Change)
* - short version
* You can copy the following algorithm into the indicator definition
* window.
* Remember to enter the default value of the first parameter (i.e. 5)
* into the parameter field.
*/
var close = Close();
var roc = CreateArray(close.length);
var n = Param(1);
for(var i=n; i<close.length; i++)
roc = 100.0 * (close - close) / close;
/* End of user-defined ROC */
-------------------------------------------------------------------------
/* User-defined ROC (Rate of Change)
* - long version with comments
* You can copy the following algorithm into the indicator definition
* window.
* Remember to enter the default value of the first parameter (i.e. 5)
* into the parameter field.
*/
/* Obtain the array of closing price values using predefined Close()
* function and store these values in the 'close' variable.
*/
var close = Close();
/* Create the array to store calculated ROC values for every trading
* session. To perform this operation, use predefined CreateArray(length)
* function.
* Caution:
* The length of created 'roc' array must be the same as the length of
* afore-mentioned 'close' array.
* Predefined CreateArray() function will initially set values of each
* array element to zero.
*/
var roc = CreateArray(close.length);
/* Obtain the value of the first indicator parameter and store that value
* in the variable called 'n'
* Caution:
* The default value of the first parameter (i.e. 5) must be entered into
* the first parameter field.
*/
var n = Param(1);
/* For each trading session (numbered with 'i') calculate ROC value
* and store that value in corresponding element of the 'roc' array.
* Caution:
* Trading sessions are numbered from 0 up to length-1.
* session (most recent one) is numbered as length-1.
*/
for(var i=n; i<close.length; i++)
roc = 100.0 * (close - close) / close;
/* Take the array of ROC values, and add graph for that array using
* Caution:
* 'firstValidIndex' is the first index of the array element that has
* its value properly calculated.
*/
/* Add graph for horizontal axis (horizontal line at 0 coordinate) */
/* End of user-defined ROC */
-------------------------------------------------------------------------
/* User-defined MACD (Moving Average Convergence Divergence)
* - short version
* You can copy the following algorithm into the indicator definition
* window.
* Remember to fill-in all three parameter fields with default parameter
* values (i.e. 12, 26, 9)
*/
var close = Close();
var avg1 = ExpAvg(close,Param(1));
var avg2 = ExpAvg(close,Param(2));
var macd = CreateArray(avg1.length);
for(var i=0; i<avg1.length; i++)
macd = avg1 - avg2;
var signal = ExpAvg(macd,Param(3));
* window.
*/
var begin=Param(2)+Param(3);
for(var i=begin; i<macd.length; i++){
if((macd < signal) && (macd>signal))
else
if((macd > signal) && (macd<signal))
}
/* End of user-defined MACD */
-------------------------------------------------------------------------
/* User-defined MACD (Moving Average Convergence Divergence)
* - long version with comments
* You can copy the following algorithm into the indicator definition
* window.
* Remember to fill-in all three parameter fields with default parameter
* values (i.e. 12, 26, 9)
*/
/* Define three parameter variables and store actual parameter values
* in these variables.
*/
var param1 = Param(1);
var param2 = Param(2);
var param3 = Param(3);
/* Obtain the array of closing price values using predefined Close()
* function and store these values in the 'close' variable.
*/
var close = Close();
/* Calculate first exponential average of closing price values, using
* predefined ExpAvg(dataArray,period) function and store the result
* in avg1 variable.
*/
var avg1 = ExpAvg(close,param1);
/* Calculate second exponential average of closing price values, using
* predefined ExpAvg(dataArray,period) function and store the result
* in avg2 variable.
*/
var avg2 = ExpAvg(close,param2);
/* Create the array to store calculated MACD values for every trading
* session. To perform this operation, use predefined CreateArray(length)
* function.
*/
var macd = CreateArray(avg1.length);
/* For each trading session, calculate MACD value and store that value in
* 'macd' array.
*/
for(var i=0; i<avg1.length; i++)
macd = avg1 - avg2;
/* Take the array of MACD values, and add graph for that array using
* Note the value of the first valid index.
*/
/* Calculate values for the Signal line and strore them in 'signal' array.
* (the Signal line is an exponential average of MACD values.
*/
var signal = ExpAvg(macd,param3);
/* Take the array of Signal values, and add graph for that array using
* Note the value of the first valid index.
*/
/* Add the horizontal zero line to MACD chart.
*/
* Begin search from the array element, where both MACD and Signal values
* are properly calculated.
*/
var begin=param2+param3;
/* Find buy/sell signal occurrences and add them to the MACD chart.
* Buy signal occurs when the MACD line crosses the Signal line and moves
* upwards.
* Buy signal occurs when the MACD line crosses the Signal line and moves
* upwards.
*/
for(var i=begin; i<macd.length; i++){
if((macd < signal) && (macd>signal))
else
if((macd > signal) && (macd<signal))
}
/* End of user-defined MACD */
-------------------------------------------------------------------------
/* User-defined %R (Williams' Percent Range)
* - short version
* You can copy the following algorithm into the indicator definition
* window.
* Remember to fill-in all three parameter fields with default parameter
* values (i.e. 10, 20, 80)
*/
var n = Param(1);
var max = Max(High(),n);
var min = Min(Low(),n);
var close = Close();
var percentR = CreateArray(close.length);
for(var i=0; i<close.length; i++)
if(max-min != 0)
percentR = 100.0 * (close-min) / (max-min);
else
percentR = 100.0;
/* End of user-defined %R */
-------------------------------------------------------------------------
/* User-defined %R (Williams' Percent Range)
* - long version with comments
* You can copy the following algorithm into the indicator definition
* window.
* Remember to fill-in all three parameter fields with default parameter
* values (i.e. 10, 20, 80)
*/
/* Obtain the value of the first parameter (period) and store it in
* the variable called 'n'.
* Caution:
* The default value of the first parameter (i.e. 10) must be entered
* into the first parameter field.
*/
var n = Param(1);
/* Calculate maximum values of high price over the selected period
* (period) using predefined Max(dataArray,period) function.
* Use predefined High() function to obtain the array of high price
* values.
*/
var max = Max(High(),n);
/* Calculate minimum values of low price over the selected period
* using predefined Min(dataArray,period) function.
* Use predefined Low() function to obtain the array of low price
* values.
*/
var min = Min(Low(),n);
/* Obtain the array of closing price values using predefined Close()
* function and store these values in the 'close' variable.
*/
var close = Close();
/* Create the array to store calculated %R values for every trading
* session. To perform this operation, use predefined CreateArray(length)
* function.
* Note that each element of the created array will be initially filled
* with zero value.
*/
var percentR = CreateArray(close.length);
/* For each trading session, calculate %R value and store that value in
* 'percentR' array.
*/
for(var i=0; i<close.length; i++)
if(max-min != 0)
percentR = 100.0 * (close-min) / (max-min);
else
percentR = 100.0;
/* Take the array of %R values, and add graph for that array using
* Note the value of the first valid index.
*/
/* Take the values of the second and third parameter (Overbought and
* Oversold levels) and add corresponding horizontal lines to the %R
* chart.
*/
/* End of user-defined %R */
-------------------------------------------------------------------------
THE SHORT DESCRIPTION OF INDICATOR DEFINITION RULES
-------------------------------------------------------------------------
Indicator algorithms must conform to the following rules:
-they must constitute valid JavaScript code
-the following predefined JavaScript functions may be used to
simplify indicator construction:
Open(), High(), Low(), Close(), Volume(), OpenInt(), Param(number),
CreateArray(length), Min(dataArray,period), Max(dataArray,period),
-to add graphs to the indicator area, the following predefined functions
must be used:
The length of the 'dataArray' argument used in AddGraph function must be
equal to the length of the array obtained by using such functions as
Open(), High(), Low(), Close(), Volume(), OpenInt()
-for every indicator calculation, AddGraph function must be used 1, 2, or
3 times
function must be used
-you cannot comment JavaScript code using // string. Use /* ... */
-the following predefined functions cannot be used in indicator code:
__checkParams(), __GetValues(), __isIEBrowser()
They may only be used internally by the FnCharts program.
-------------------------------------------------------------------------
THE LIST OF PREDEFINED FUNCTIONS
-------------------------------------------------------------------------
Open()
Arguments: none
Returns the array of opening price values for the selected symbol.
High()
Arguments: none
Returns the array of high price values for the selected symbol.
Low()
Arguments: none
Returns the array of low price values for the selected symbol.
Close()
Arguments: none
Returns the array of closing price values for the selected symbol.
Volume()
Arguments: none
Returns the array of volume values for the selected symbol.
OpenInt()
Arguments: none
Returns the array of open interest values for the selected symbol.
Important information regarding above functions:
1. Arrays in JavaScript are indexed from 0 to length-1, so the first
element of an array A is A and the last element is A.
The first element of the array contains the oldest available data, and
the last element of the array contains the most recent data.
2. All afore-mentioned functions always return arrays of the same length.
For instance, the length of an array returned by Close() function is
the same, as the length of an array returned by Volume() function.
3. Arrays returned by Open(), High(), Low() and Close() will always
contain positive values (values greater than zero).
4. If there is no Open, High or Low data available for the selected
symbol, then arrays returned by Open(), High() and Low() functions
will contain Close values.
5. If there is no volume or open interest data available for the
selected symbol, arrays returned by Volume() or OpenInt() functions
will contain values 0.
Max(array,period)
Arguments: array - the array for which Max values will be calculated,
period - the period used in calculations
Calculates maximum value over the given period.
Returns the array wherein value of every element is calculated as:
value = max(array,array,...,array)
Min(array,period)
Arguments: array - the array for which Min values will be calculated,
period - the period used in calculations
Calculates minimum value over the given period.
Returns the array wherein value of every element is calculated as:
value = min(array,array,...,array)
SimpleAvg(array,period)
Arguments: array - the array for which average values will be
calculated,
period - the period used in calculations
Calculates simple average over the given period.
Returns the array wherein value of every element is calculated as:
value = avg(array,array,...,array)
ExpAvg(array,period)
Arguments: array - the array for which exponential average values
will be calculated,
period - the period used in calculations
Calculates exponential average over the given period.
Returns the array wherein value of every element is calculated as:
value = exp_avg(array,array,...,array)
StdDev(array,period)
Arguments: array - the array for which standard deviation values will
be calculated,
period - the period used in calculations
Calculates standard deviation over the given period.
Returns the array wherein value of every element is calculated as:
value = std_dev(array,array,...,array)
CreateArray(length)
Arguments: length - the length of the array to be created
Returns new array of the given length with every element initialized
to zero.
Param(n)
Arguments: n - parameter number
Returns the actual value of the given indicator parameter.
Argument 'n' must be in range between 1 and 3.
If the value of the given parameter 'n' is not defined then the
function returns the value of 0.
Arguments: array - the array of data,
index - index of the first array element containing valid
data
Adds a new graph to the indicator area. The graph is constructed
based on the data contained in the given array.
The length of the array must be exactly the same as the length of the
array returned by such functions as Close(), Volume(), etc.
Only the part of the graph that represents array elements with
indices greater than or equal to 'index' will be displayed.
The 'index' parameter is optional. If omitted, the function call
Arguments: value - the y coordinate of the horizontal line.
Adds a horizontal line to the indicator area.
Arguments: index - the index of the trading session on which the
signal occurred
Arguments: index - the index of the trading session on which the
signal occurred
Adds a new sell signal mark to the indicator area.
-------------------------------------------------------------------------
ERROR MESSAGES AND RELATED INFORMATION
-------------------------------------------------------------------------
If an indicator algorithm contains bugs, an error message and description
will be displayed in the indicator field and in Java Console.
Remember however, that error messages may contain the following
inaccuracies (due to the limitations of JavaScript and applet
collaboration):
- if you use Microsoft Internet Explorer then the information regarding
line numbers and error positions may be incorrect. This is because the
indicators' code is transformed to a single line before evaluation.
-------------------------------------------------------------------------
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A Less Frustrating Pencil ?
Author Message
Posted on: Wed Jun 26, 2013 8:25 pm
Posts: 5
Joined: Wed Mar 28, 2012 5:50 pm
A Less Frustrating Pencil ?
Could the pencil entries appear in the order they are entered and not rearrange themselves into ascending order?
Simple example: if you know the two cell answer to 10+ must be [3][7] or [4][6] , have pencil entries showing 34 in the first box and 76 (and not moving itself around to 67!) in the second. That way a quick glance would tell you that the first possibility is the first pencil number in each box, and the second...
Avril
Posted on: Wed Jun 26, 2013 9:28 pm
Posts: 2258
Joined: Thu May 12, 2011 11:58 pm
Re: A Less Frustrating Pencil ?
Yes, that's possible, but will require some (re)programming.
What do other people prefer?
Posted on: Thu Jun 27, 2013 3:07 am
Posts: 193
Joined: Tue May 24, 2011 4:55 pm
Re: A Less Frustrating Pencil ?
It may be neat in some specific cases (like avriljp's example) if you could determine the order of pencil entries yourself rather than having them automatically rearranged. But I think automatic rearrangement comes in handy for most purposes.
Depending on the puzzle, a puzzler will often want to arrange all pencil entries in ascending order in their respective cells because it helps identify candidate pairs and hidden muliplets in rows and columns (and, in the case of sudokus, 3x3 boxes) more easily. Automatic rearrangement into ascending order makes this happen even if you discover and enter the candidates in non-ascending order.
In larger puzzles where you use letter keys for entering numbers larger than 9, sometimes it is quicker and easier to hit keys without regard to order when you pencil in candidates. This is even true of smaller puzzles when you are working on a laptop keyboard without a numeric keypad.
To me the general benefits of automatic rearrangement outweigh the negative aspects it has in some situations. I think it should remain the default option. "Don't rearrange pencil entries in ascending order" could then be offered as a tickbox (as is currently the case with "Continuous error checking" and "Show coordinates") if enough puzzlers want it. Speaking only for myself, I doubt that I would use it much, if at all.
Posted on: Fri Jun 28, 2013 4:20 am
Posts: 226
Joined: Fri Jun 17, 2011 8:15 pm
Re: A Less Frustrating Pencil ?
I agree with Bram that the negatives seem to outweigh the positives for me. I'm imagining a scenario where one cell of a three cell addition cage has been narrowed down to two possibilities and that in turn implies the other two must equal either 15 or 16. [67] and [98] might represent the 15 option well, but the 7+9=16 option gets sort of lost.
In addition, I can imagine working through a puzzle and thinking I could exclude a number, realize I was mistaken and put it back only to find that [1346] has become [1463].
That said, if Patrick wants to add this as an option, why not?
Posted on: Fri Jun 28, 2013 3:31 pm
Posts: 116
Joined: Sat May 14, 2011 3:18 am
Re: A Less Frustrating Pencil ?
I often get frustrated with the automatic ordering. Like avrilp, I would like to have the option of determining the order. Many times I have wished there was a way of "stacking" the possibilities in vertical cages so that the leftmost column of guesses represents one scenario, and the columns in the middle or right represent another scenario.
For cages with a subtraction operatior, the guesses usually line up that way, but for addition or multiplication, not so much.
Regarding the ease of entering the digits in any order: If you provided the option to display the guesses in order of entry and I wanted the guesses in numerical order, I could always press the keys in that order.
So my vote is to provide the option to order guesses manually.
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https://allenact.org/api/allenact/embodiedai/mapping/mapping_utils/point_cloud_utils/ | 1,642,532,136,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00500.warc.gz | 136,991,371 | 13,836 | # `allenact.embodiedai.mapping.mapping_utils.point_cloud_utils`#
[view_source]
### `camera_space_xyz_to_world_xyz`#
``````camera_space_xyz_to_world_xyz(camera_space_xyzs: torch.Tensor, camera_world_xyz: torch.Tensor, rotation: float, horizon: float) -> torch.Tensor
``````
[view_source]
Transforms xyz coordinates in the camera's coordinate frame to world- space (global) xyz frame.
This code has been adapted from https://github.com/devendrachaplot/Neural-SLAM.
IMPORTANT: We use the conventions from the Unity game engine. In particular:
• A rotation of 0 corresponds to facing north.
• Positive rotations correspond to CLOCKWISE rotations. That is a rotation of 90 degrees corresponds to facing east. THIS IS THE OPPOSITE CONVENTION OF THE ONE GENERALLY USED IN MATHEMATICS.
• When facing NORTH (rotation==0) moving ahead by 1 meter results in the the z coordinate increasing by 1. Moving to the right by 1 meter corresponds to increasing the x coordinate by 1. Finally moving upwards by 1 meter corresponds to increasing the y coordinate by 1. Having x,z as the ground plane in this way is common in computer graphics but is different than the usual mathematical convention of having z be "up".
• The horizon corresponds to how far below the horizontal the camera is facing. I.e. a horizon of 30 corresponds to the camera being angled downwards at an angle of 30 degrees.
Parameters
• camera_space_xyzs : A 3xN matrix of xyz coordinates in the camera's reference frame.
• __Here `x, y, z = camera_space_xyzs[__:, i]` should equal the xyz coordinates for the ith point.
• camera_world_xyz : The camera's xyz position in the world reference frame.
• rotation : The world-space rotation (in degrees) of the camera.
• horizon : The horizon (in degrees) of the camera.
Returns
`3xN tensor with entry [`:, i] is the xyz world-space coordinate corresponding to the camera-space `coordinate camera_space_xyzs[`:, i]
### `depth_frame_to_camera_space_xyz`#
``````depth_frame_to_camera_space_xyz(depth_frame: torch.Tensor, mask: Optional[torch.Tensor], fov: float = 90) -> torch.Tensor
``````
[view_source]
Transforms a input depth map into a collection of xyz points (i.e. a point cloud) in the camera's coordinate frame.
Parameters
• depth_frame : A square depth map, i.e. an MxM matrix with entry `depth_frame[i, j]` equaling the distance from the camera to nearest surface at pixel (i,j).
• mask : An optional boolean mask of the same size (MxM) as the input depth. Only values where this mask are true will be included in the returned matrix of xyz coordinates. If `None` then no pixels will be masked out (so the returned matrix of xyz points will have dimension 3x(M*M)
• fov: The field of view of the camera.
Returns
`A 3xN matrix with entry [`:, i] equalling a the xyz coordinates (in the camera's coordinate frame) of a point in the point cloud corresponding to the input depth frame.
### `depth_frame_to_world_space_xyz`#
``````depth_frame_to_world_space_xyz(depth_frame: torch.Tensor, camera_world_xyz: torch.Tensor, rotation: float, horizon: float, fov: float)
``````
[view_source]
Transforms a input depth map into a collection of xyz points (i.e. a point cloud) in the world-space coordinate frame.
IMPORTANT: We use the conventions from the Unity game engine. In particular:
• A rotation of 0 corresponds to facing north.
• Positive rotations correspond to CLOCKWISE rotations. That is a rotation of 90 degrees corresponds to facing east. THIS IS THE OPPOSITE CONVENTION OF THE ONE GENERALLY USED IN MATHEMATICS.
• When facing NORTH (rotation==0) moving ahead by 1 meter results in the the z coordinate increasing by 1. Moving to the right by 1 meter corresponds to increasing the x coordinate by 1. Finally moving upwards by 1 meter corresponds to increasing the y coordinate by 1. Having x,z as the ground plane in this way is common in computer graphics but is different than the usual mathematical convention of having z be "up".
• The horizon corresponds to how far below the horizontal the camera is facing. I.e. a horizon of 30 corresponds to the camera being angled downwards at an angle of 30 degrees.
Parameters
• depth_frame : A square depth map, i.e. an MxM matrix with entry `depth_frame[i, j]` equaling the distance from the camera to nearest surface at pixel (i,j).
• mask : An optional boolean mask of the same size (MxM) as the input depth. Only values where this mask are true will be included in the returned matrix of xyz coordinates. If `None` then no pixels will be masked out (so the returned matrix of xyz points will have dimension 3x(M*M)
• camera_space_xyzs : A 3xN matrix of xyz coordinates in the camera's reference frame.
• __Here `x, y, z = camera_space_xyzs[__:, i]` should equal the xyz coordinates for the ith point.
• camera_world_xyz : The camera's xyz position in the world reference frame.
• rotation : The world-space rotation (in degrees) of the camera.
• horizon : The horizon (in degrees) of the camera.
• fov: The field of view of the camera.
Returns
`A 3xN matrix with entry [`:, i] equalling a the xyz coordinates (in the world coordinate frame) of a point in the point cloud corresponding to the input depth frame.
### `project_point_cloud_to_map`#
``````project_point_cloud_to_map(xyz_points: torch.Tensor, bin_axis: str, bins: Sequence[float], map_size: int, resolution_in_cm: int, flip_row_col: bool)
``````
[view_source]
Bins an input point cloud into a map tensor with the bins equaling the channels.
This code has been adapted from https://github.com/devendrachaplot/Neural-SLAM.
Parameters
• xyz_points : (x,y,z) pointcloud(s) as a torch.Tensor of shape (... x height x width x 3). All operations are vectorized across the `...` dimensions.
• bin_axis : Either "x", "y", or "z", the axis which should be binned by the values in `bins`. If you have generated your point clouds with any of the other functions in the `point_cloud_utils` module you almost certainly want this to be "y" as this is the default upwards dimension.
• bins: The values by which to bin along `bin_axis`, see the `bins` parameter of `np.digitize` for more info.
• map_size : The axes not specified by `bin_axis` will be be divided by `resolution_in_cm / 100` and then rounded to the nearest integer. They are then expected to have their values within the interval [0, ..., map_size - 1].
• resolution_in_cm: The resolution_in_cm, in cm, of the map output from this function. Every grid square of the map corresponds to a (`resolution_in_cm`x`resolution_in_cm`) square in space.
• flip_row_col: Should the rows/cols of the map be flipped? See the 'Returns' section below for more info.
Returns
A collection of maps of shape (... x map_size x map_size x (len(bins)+1)), note that bin_axis has been moved to the last index of this returned map, the other two axes stay in their original order unless `flip_row_col` has been called in which case they are reversed (useful as often rows should correspond to y or z instead of x). | 1,672 | 6,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-05 | latest | en | 0.745715 |
http://forums.wolfram.com/mathgroup/archive/2008/Nov/msg00022.html | 1,596,563,747,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00111.warc.gz | 49,370,135 | 7,956 | Re: newbie question
• To: mathgroup at smc.vnet.net
• Subject: [mg93238] Re: newbie question
• From: Peter Breitfeld <phbrf at t-online.de>
• Date: Sat, 1 Nov 2008 05:05:20 -0500 (EST)
• References: <geee9l\$ac9\$1@smc.vnet.net>
```polymedes schrieb:
> Hi I'm trying to apply the function ToDegrees (from Geodesy package)
> to an array of data. It does not seem to work on arrays. Can somebody
> give me an idea on how to do this conversion? Ideally, I'd like to
> import a file in the following format:
>
> 12,23,49 70,45,30
> 12,25,41 70,46,40
> 12,26,51 70,47,50
> 12,27,55 70,48,33
> .
> .
> .
>
>
> How can I convert the above numbers (from the given format, i.e.
> degrees-minutes-seconds) to decimal numbers?
>
I pasted your example data into a file data.txt. The I did the
following:
Out={"12", "23", "49", "70", "45", "30", "12", "25", "41", "70", "46", \
"40", "12", "26", "51", "70", "47", "50", "12", "27", "55", "70", \
"48", "33"}
You see, the Elements of the list raw are String. This may not be the
case for you, if your file was created in another way. Now I group
the data into tuples of three:
blocks=Partition[ToExpression /@ raw,3]
Out={{12, 23, 49}, {70, 45, 30}, {12, 25, 41}, {70, 46, 40},
{12, 26, 51}, {70, 47, 50}, {12, 27, 55}, {70, 48, 33}}
You must not use ToExpression/@, if your raw list elements aren't strings.
Then I feed the blocks to ToDegree:
ToDegree/@blocks //N
Out={12.3969, 70.7583, 12.4281, 70.7778, 12.4475, 70.7972, 12.4653, 70.8092}
Gruss Peter
--
==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de
```
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• Next by thread: Re: Frustration with opaque functions in Mathematica | 673 | 1,908 | {"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} | 2.6875 | 3 | CC-MAIN-2020-34 | latest | en | 0.809632 |
https://www.simplilearn.com/brief-introduction-to-swaps-and-swap-contracts-rrt3co39vd220-video | 1,542,277,604,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742666.36/warc/CC-MAIN-20181115095814-20181115121213-00003.warc.gz | 1,000,548,732 | 46,860 | # Brief Introduction to Concept of Swaps and Swap Contracts
Duration 03:00 946 Views
Swaps
Financial Risk Manager (FRM®) Part I of the FRM Exam covers the fundamental tools and techniques used in risk management and the theories that underlie their use.
Swaps
Welcome to the 30th session in your preparation for the FRM Part I exam. We are almost halfway through the Financial Markets module, and until now we have learned about futures, forwards and options. In this session, we will introduce a very interesting concept called swaps, wherein one can transfer the cash flow liabilities to another person and vice versa. For example, one can convert a fixed rate loan into a floating rate loan. This is done by entering into a swap agreement.
Agenda
The agenda of our discussion will be learning about swap contracts. We will begin by introducing the concept of swaps. We will then learn about plain vanilla swaps using both bonds and forward rate agreements. We will then learn about currency options. The calculations might seem a bit tricky, but as you practice, it will become quite easy. Questions from this section (especially about currency swaps) are always asked in the exam, so you are advised to fully understand the concepts as well as the numerical questions. We will end the session with a brief discussion on the comparative advantage of swaps and the risk involved in swaps.
Concept
Let us now understand what swaps are. Imagine that one person is holding a semi-annual bond that pays a 4percent coupon on a principal of Dollar100,000. This person receives a fixed payment of Dollar2000 every six months as coupon payments, as shown in the slide. Consider another person holding a bond that pays floating interest based on a six-month LIBOR. Suppose that the interest rates for the next 3 periods are 4percent, 4.5percent and 4.8percent; then this person will receive Dollar2000, Dollar2250 and Dollar2400 every six months. Now, they mutually agree and enter into a contract so that the person receiving a fixed payment will pay this fixed payment to the person receiving floating payment, and in return, the other person will pay the floating payment to the first person. The net payment is shown in the chart. Therefore, the contract enables both parties to interchange their cash flow liabilities. We also find that the net payment is positive for one party and negative for other. As such, the present value of the net positive payment to one party becomes the value of the initial contract known as the value of the swap. The agreement is termed as a swap.
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http://mathhelpforum.com/calculus/195241-volume-integral-cylinder-cone-print.html | 1,508,401,390,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823255.12/warc/CC-MAIN-20171019065335-20171019085335-00672.warc.gz | 213,495,582 | 5,359 | Volume Integral - Cylinder & Cone
• Jan 13th 2012, 01:25 PM
Juggernaut
Volume Integral - Cylinder & Cone
Hello all,
In mathematical terms, I've got a problem where I'd like to calculate the volume of a cylinder whose upper surface is bounded by a right circular cone. A brief discription of the geometries follow, but it's probably best to look at the attachments.
In the real world, a description of the problem follows: consider the cylinder an inground tank with radius 7.5 m and height 8.1 m. The tank is used to collect sand form the top edge [along the edge of the tank] such that the sand entering the tank creates the shape of half of a right circular cone of radius 14 m [and height 8.1 m]. Notice that the cone's base does not touch the far edge of the tank [in fact, it is 1 m away from the edge].
What is the volume of sand in the tank?
MY Attempt:
1. I decided to use a triple integral in cylindrical coordinates.
2. I defined my cylinder equation
3. I defined my cone equation
4. I identified the angles to integrate through
5. Compute the triple integral
But my answer is obviously wrong since the volume that I'm calculating is larger than the total volume of the tank.
Can anyone offer help? I've checked and re-checked. What am I not seeing? Is my method wrong? Or is my arithmetic wrong? Any help would be greatly appreciated!!
Also, I would have loved to type out the question in the thread -- but sadly, i dont know how to.
Juggernaut.
• Jan 14th 2012, 07:25 AM
Jester
Re: Volume Integral - Cylinder & Cone
First off, why are you using the entire base of the cylinder when there's no sand in the cone from 14 to 15.
Second, really a question, is the base of the cone entirely inside the circular base of the cylinder?
Next, why are you using
$z = h + \frac{r}{c}$ instead of $z = h - \frac{r}{c}$
Your formula says that when $r = 0$ then $z = h$ (correct) but as you increase $r$, $z$ increases instead of decreases.
• Jan 14th 2012, 08:17 AM
Juggernaut
Re: Volume Integral - Cylinder & Cone
Thanks a lot Danny! Responding to your questions...
1. "why are you using the entire base of the cylinder when there's no sand in the cone from 14 to 15. "
- Shouldn't I? Wont the surface component of the integral (i.e. the cone's surface) take this into account? If not, how would I qo about account for this?
2. "is the base of the cone entirely inside the circular base of the cylinder?"
- No, it cant be, the radius of the cone is far bigger than that of the cylinder. Am I missing something here?
3. "Your formula says that when r = 0 then z = h (correct) but as you increase r, z increases instead of decreases. "
- Great catch!! I should have seen that! Thanks!
However, based on comments 1 and 2, I'm guessing that I'm missing more to this question. Any further insights?
Thanks so much!
• Jan 14th 2012, 09:49 AM
Jester
Re: Volume Integral - Cylinder & Cone
Ya, here's some more. Since you are looking for volume then what I would do is
$\iint \limits_R z dA$
where $z = h - \dfrac{h}{a}\sqrt{x^2+y^2}$ (the cone with height $h$ and radius $a$) and R the region, in the xy plane, when the part of the cone inside the cylinder is projected down into the xy plane. If we view this downward what I see is two circles - one is the base of the cone and the other is the base of the cylinder. The region of integration in the area inside both circles. What I'm saying is that you're going to need two integrals to do this problem!
• Jan 14th 2012, 10:37 AM
Juggernaut
Re: Volume Integral - Cylinder & Cone
So, what you're saying is:
1. I need to do 2 integrals instead of 3?
3. Find the area of the base of the sand -- the area formed by the two circles you've described
4. Use this area to compute the integral you specified above
I'm tripping up over the dA in your integral above...How would I define it? Is it simply dxdy? I know it's the differential area element of the area in the two circles, but in my mind, I'd calculate this area and use it as a bound for my integral :S
Thanks so much for your help!!! :)
• Jan 14th 2012, 01:09 PM
Jester
Re: Volume Integral - Cylinder & Cone
What you want to calculate is
$\iint \limits_R \left( h - \frac{h}{a}r\right) r dr d\theta$
where $R$ is the region inside both circles
• Jan 15th 2012, 12:32 PM
Juggernaut
Re: Volume Integral - Cylinder & Cone
Quote:
Originally Posted by Danny
What you want to calculate is
$\iint \limits_R \left( h - \frac{h}{a}r\right) r dr d\theta$
where $R$ is the region inside both circles
I think I get it! My integral will look like this...correct? :
$\int_{0}^{\pi}\int_{15\sin\theta}^{14} \left( h - \frac{h}{a}r\right) r dr d\theta$
Where R is defined by the area inside $r = 14$ and $r = 15\sin\theta$, and ranging from ${0}$ to ${\pi}$ Correct?
• Jan 15th 2012, 02:15 PM
Jester
Re: Volume Integral - Cylinder & Cone
Not quite. First, there's some symmetry so integrate $\theta$ from $0$ to $\pi/2$ and multiply by two. Next, there's a point where these two circles intersect, call this $\theta^*.$ Then what you have is
$\int_0^{\theta ^*}{\int_0^{15 \sin \theta} \left(h - \frac{h}{a} r\right) r dr d \theta + \int_{\theta ^*}^{\pi/2}{\int_0^{14} \left(h - \frac{h}{a} r\right)r dr d \theta$
• Jan 15th 2012, 02:18 PM
Juggernaut
Re: Volume Integral - Cylinder & Cone
Amazing!! I see it!! I'll give it a shot, and I'll post back to let you know how it goes...Thank you so much for your help!! :D
• Jan 16th 2012, 10:41 AM
Juggernaut
Re: Volume Integral - Cylinder & Cone
So I've computed the integral, and I'm getting a volume of about 871 cubic meters. This seems like it's in the right ball park.
As a thought experiment though, lets assume that instead of a surface of a cone, it was a simple plane, defined by $z = h -\frac{h}{a}y$ that cut the cylinder...what scenario whould have more volume?
My thought was that the volume under the plane would be larger.
I did the calc explicitly, and found that my guess was wrong -- it was in fact he cone scenario.
Does this make sense? I can post my work if you'd like to see -- but what makes sense in your mind?
I'm not sure what you did but I agree that I think the volume under the plane is more. In fact it is easy to show that (of course for $h,a >0$)
$h - \frac{h}{a}y \ge h - \frac{h}{a} \sqrt{x^2+y^2}$ for all $(x,y).$ | 1,788 | 6,311 | {"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": 28, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-43 | longest | en | 0.908584 |
https://documen.tv/question/a-car-travels-30-miles-in-hour-what-is-the-average-speed-at-which-the-car-is-traveling-in-miles-22329275-94/ | 1,726,247,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00421.warc.gz | 183,867,538 | 16,013 | ## A car travels 30 miles in ½ hour. What is the average speed at which the car is traveling in miles per hour?
Question
A car travels 30 miles in ½ hour. What is the average speed at which the car is traveling in miles per hour?
in progress 0
3 years 2021-08-02T15:28:38+00:00 1 Answers 10 views 0 | 90 | 301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.93538 |
https://gogeometry.blogspot.com/2012/01/triangle-altitudes-parallel.html | 1,721,390,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00135.warc.gz | 226,867,700 | 14,141 | ## Friday, January 6, 2012
### Problem 715: Triangle, Altitudes, Parallel, Circumcircle, Angle, Measurement
Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 715 details.
#### 6 comments:
1. Angle B'A'C' = Pi - 2A = 62 deg
My guess is x = 62 deg.
Yet to prove.
Circle A'B'C'(9-point circle)passes through H
2. Pravin, you seem to be 100% correct.
/_BC'C=/_BB'C=90. Hence we conclude that B, B', C' & C lie on a circle whose diameter is BC. Now prove that H is the midpoint of BC and we're done for then HB=HC=HB' and hence /_HB'C=/_HCB'=/_C and /_C'B'A=/_A (for B,C,B' & C' are concyclic) and thus /_C'B'H=/_A=59
Now HC'=HB' if H is the centre of our new circle. Then /_HCB' also =59 and thus /_x=62.
All we now need to do is to prove that H bisects BC. Problem #714 must come into play in order to prove this.
Ajit
1. Why H is middle point BC ?
3. Proof of Problem 719 completed.
In the usual notation,
A’D = A’C’ = R sin 2B
A’E = R sin 2C
A’F = R sin 2B.sin 2C/sin(B-C) = A’D sin 2C /sin (B – C)
F, D, H, E being concyclic,
A’F. A’H = A’D. A’E
Follows
So A’H = A’E sin(B - C) / sin 2C = R sin(B - C)
Now BH = A’B + A’H = c cos B + R sin(B - C)
= 2R sin C cos B + R sin(B – C)
=R(2 sin C cos B + sin (B – C)]
= R sin (B + C)
= R sin A = a/2 = BC/2
Hence H is the midpoint of BC.
4. Pravin,
Please explain how this comes about:
A’F = R sin 2B.sin 2C/sin(B-C) = A’D sin 2C /sin (B – C)
Ajit
5. It is known that angles of triangle A'B'C' are 180-2A,180-2B,180-2C resp'ly.
Consider triangle A'C'F
<A'C'F=180-<A'C'B'=2C
<A'FC'=<CFB'=<AB'F-<ACB = B - C
By Sine Rule
A'C'/sin(B-C) = A'F/sin 2C
Circle A'B'C' is the nine-point circle and its radius is known to be R/2
chord A'C' = 2.(R/2).sin <A'B'C'= R sin (180-2B)= R sin 2B etc
Awaiting for a shorter proof of Problem 715 | 713 | 1,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-30 | latest | en | 0.850332 |
https://www.physicsforums.com/threads/integration-using-trig-subs-having-trouble-switching-back-to-the-original-variable.46155/ | 1,606,273,623,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00544.warc.gz | 819,212,646 | 14,876 | Integration using trig subs having trouble switching back to the original variable
Hi, I carried out the integration until the very end...I don't know how to convert the variable back to the original one.
$$\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^2-u^2}}$$
$$Let u = a\sin{\Theta}$$
$$du = a\cos{\Theta}d\Theta$$
The integral becomes...
$$\int_{R_{0}}^{R(\Theta)}\frac{a\cos{\Theta}d\Theta}{a\cos{\Theta}a\sin{\Theta}}$$
$$\frac{1}{a}\int_{R_{0}}^{R(\Theta)}\csc{\Theta}d{\Theta}$$
$$\csc{\Theta}d{\Theta} = \ln {|\csc{\Theta}-\cot{\Theta}|}+C$$
This is where I'm stuck. I don't know how to convert the thetas back into the "u"s. I haven't multiplied the answer by 1/a yet. I know that $$\Theta=\sin^{-1}{u/a}$$, but if I plug the $$\sin^{-1}{u/a}$$ into Theta, the expression becomes super messy and I really don't know what to do with it. | 291 | 850 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-50 | latest | en | 0.795242 |
https://codegolf.stackexchange.com/questions/154922/the-keyboard-shift-cipher | 1,702,121,261,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100909.82/warc/CC-MAIN-20231209103523-20231209133523-00504.warc.gz | 207,755,832 | 55,795 | # The Keyboard Shift Cipher
Given the following input:
• An integer n where n > 0.
• A string s where s is not empty and s~=[0-9A-Z]+ (alpha-numeric capitals only).
Using a standard, simplified QWERTY keyboard (as shown below):
1234567890
QWERTYUIOP
ASDFGHJKL
ZXCVBNM
Perform the following operation:
• Find the original row that each character is in on the keyboard.
• Replace the letter with the correct shifted equivalent for n based on its original position + n.
• E.G. s="AB" and n=2: A would become D and B would become M.
• If keyboard_row[position + n] > keyboard_row.length, wrap back to the start.
• E.G. s="0P" and n=2: 0 would become 2 and P would become W.
# Examples:
f("0PLM",1) = 1QAZ
f("ZXCVB",2) = CVBNM
f("HELLO",3) = LYDDW
f("0PLM",11) = 1QSV
f("0PLM",2130) = 0PHX
# Rules
This is slightly more difficult than it seems at first glance.
• Are we allowed to take the input as character-array instead of string? Currently assumed we, but forgot to ask.. Feb 5, 2018 at 14:51
• @KevinCruijssen shrug sure, it's not too outlandish. Unless it saves you a byte to break a tie I'm not complaining. Feb 5, 2018 at 22:43
# Jelly, 13 bytes
ØQØDṭ,ṙ€¥⁸F€y
Try it online!
### How it works
ØQØDṭ,ṙ€¥⁸F€y Main link. Left argument: n (integer). Right argument: s (string)
ØQ Qwerty; set the return value to
["QWERTYUIOP", "ASDFGHJKL", "ZXCVBNM"].
ØD Digits; yield "0123456789".
ṭ Tack, yielding ["QWERTYUIOP", "ASDFGHJKL", "ZXCVBNM", "0123456789"].
¥⁸ Call the two links to the left as a dyadic chain, with right
argument n.
ṙ€ Rotate each string in the array n units to the left.
, Yield the pair of the unmodified and the rotated string array.
F€ Flatten each, mapping, e.g., ["QWERTYUIOP", ..., "0123456789"] to
"QWERTYUIOPASDFGHJKLZXCVBNM0123456789".
y Translate s according to the mapping we've built.
• Jelly has keyboard-layout built-ins eh? Feb 5, 2018 at 14:13
• @MagicOctopusUrn No, only QWERTY right now :-P Feb 5, 2018 at 14:14
• 13 bytes? Which character set is that supposed to be? In UTF-8 it's 26 bytes! Feb 5, 2018 at 21:23
• @Cephalopod Jelly uses the Jelly code page. Feb 5, 2018 at 21:26
• I'm not going to complain about regex anymore Aug 22 at 14:52
# Python 2, 110 bytes
lambda s,n,y='1234567890'*99+'QWERTYUIOP'*99+'ASDFGHJKL'*99+'ZXCVBNM'*99:''.join(y[y.find(c)+n%630]for c in s)
Try it online!
This uses a big enough string (99 copies of each row) and the LCM between the rows lengths (630) to find the correct substitution avoiding individual correction between each row.
# Java 8, 159 158 bytes
n->s->{for(int i=s.length,j;i-->0;)for(String x:"1234567890;QWERTYUIOP;ASDFGHJKL;ZXCVBNM".split(";"))if((j=x.indexOf(s[i])+n)>=n)s[i]=x.charAt(j%x.length());}
-1 byte thanks to @OlivierGrégoire modifying the input-array instead of printing directly.
Explanation:
Try it online.
n->s->{ // Method with integer and character-array parameters, and no return-type
for(int i=s.length,j;i-->0;)
// Loop over the input character-array with index
for(String x:"1234567890;QWERTYUIOP;ASDFGHJKL;ZXCVBNM".split(";"))
// Inner loop over the qwerty-lines
if((j=x.indexOf(s[i])+n)>=n)
// If the current qwerty-line contains the character
// Set j to the index of this character on that line + input n
s[i]=x.charAt(j%x.length());}
// Replace the character at index i
// with the new character (at index j modulo length_of_qwerty_line)
• 158 bytes, at the cost of input-outputting the char[]. Feb 6, 2018 at 9:30
# Retina, 49 bytes
"$&"+T9odQW\ERTYUI\OPQASDFG\HJK\LAZXC\VBNMZ 0A Try it online! Takes input n and s on separate lines. Explanation: "$&"+
Repeat n times.
T9odQW\ERTYUI\OPQASDFG\HJK\LAZXC\VBNMZ
Shift all the characters one key to the right.
0A
Delete n.
# JavaScript (ES6), 101 99 bytes
Takes input in currying syntax (s)(n). Works with arrays of characters.
s=>n=>s.map(c=>(S='1QAZ2WSX3EDC4RFV5TGB6YHN7UJM8IK_9OL_0P')[(p=S.search(c)+n*4)%(-~'9986'[p%4]*4)])
### Test cases
let f =
s=>n=>s.map(c=>(S='1QAZ2WSX3EDC4RFV5TGB6YHN7UJM8IK_9OL_0P')[(p=S.search(c)+n*4)%(-~'9986'[p%4]*4)])
console.log(JSON.stringify(f([..."0PLM"])(1))) // 1QAZ
console.log(JSON.stringify(f([..."ZXCVB"])(2))) // CVBNM
console.log(JSON.stringify(f([..."HELLO"])(3))) // LYDDW
console.log(JSON.stringify(f([..."0PLM"])(11))) // 1QSV
console.log(JSON.stringify(f([..."0PLM"])(2130))) // 0PHX
### How?
We look for the position p of each character of the input within a string S where the keyboard rows are interleaved: the first 4 characters are '1QAZ' (first column of the keyboard), the next 4 characters are '2WSX' (second column of the keyboard) and so on. Unused positions are padded with underscores and the last ones are simply discarded.
col # | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
------+------+------+------+------+------+------+------+------+------+---
row # | 0123 | 0123 | 0123 | 0123 | 0123 | 0123 | 0123 | 0123 | 0123 | 01
------+------+------+------+------+------+------+------+------+------+---
char. | 1QAZ | 2WSX | 3EDC | 4RFV | 5TGB | 6YHN | 7UJM | 8IK_ | 9OL_ | 0P
This allows us to easily identify the row with p mod 4 and eliminates the need for explicit separators between the rows.
We advance by 4n positions, apply the correct modulo for this row (40, 40, 36 and 28 respectively) and pick the replacement character found at this new position in S.
# Jelly, 18 bytes
ØQØDṭẋ€‘}Ẏ©iЀ⁸+ị®
Try it online!
# C, 152 149 bytes
Thanks to @gastropner for saving three bytes!
j,l;f(S,n){for(char*s=S,*k;*s;++s)for(k="1234567890\0QWERTYUIOP\0ASDFGHJKL\0ZXCVBNM\0";l=strlen(k);k+=l+1)for(j=l;j--;)k[j]-*s||putchar(k[(j+n)%l]);}
Try it online!
Unrolled:
j,l;
f(S,n)
{
for (char*s=S, *k; *s; ++s)
for (k="1234567890\0QWERTYUIOP\0ASDFGHJKL\0ZXCVBNM\0"; l=strlen(k); k+=l+1)
for (j=l; j--;)
k[j]-*s || putchar(k[(j+n)%l]);
}
• Either I'm hallucinating, or the inner loop can be changed to for(j=l;j--;) but I don't know why without any other changed. Still, should get you to 149. Feb 7, 2018 at 21:12
• @gastropner Ah, yes, the search order doesn't matter, so it works. Thanks! Feb 7, 2018 at 22:11
# Red, 152 bytes
f: func[s n][foreach c s[foreach[t l]["1234567890"10"QWERTYUIOP"10"ASDFGHJKL"9"ZXCVBNM"7][if p: find t c[if(i:(index? p)+ n // l)= 0[i: l]prin t/(i)]]]]
Try it online!
Ungolfed:
f: func [s n][1
foreach c s [
foreach [t l] ["1234567890"10"QWERTYUIOP"10"ASDFGHJKL"9"ZXCVBNM"7][
p: find t c
if p [
i: (index? p) + n // l
if i = 0 [i: l]
prin t/(i) ]]]]
f(s,n)=[dropWhile(/=c)(cycle r)!!n|c<-s,r<-words"1234567890 QWERTYUIOP ASDFGHJKL ZXCVBNM",elem c r]
Try it online!
• You can use s#n= ... instead of f(s,n)= ... which is only an example notation used for the examples. Feb 6, 2018 at 8:16
$s=<>;for$i(1234567890,QWERTYUIOP,ASDFGHJKL,ZXCVBNM){eval"y/$i/".(substr$i,$s%length$i)."$i/"} Try it online! • Damn, I didn't see your answer. They're basically the same, feel free to use my optimisations and I'll remove my answer. Let me know, if not, I'll just remove this comment :) Feb 5, 2018 at 17:04 • @DomHastings They are different enough. Please keep both. I like to see variations in approach. I learn from all of them... Feb 5, 2018 at 22:22 # Japt, 20 bytes Running out the door to dinner so more golfing and an explanation to follow. ;£=D·i9òs)æøX)gV+UbX Try it # Perl, 595857 56 bytes Includes + for -p Give input on STDIN as 2 lines, first the string, then the repeat (echo 0PLM; echo 2130) | perl -pe '$a="OPQWERTYUILASDF-MZXCVBNM0-90";eval"y/HI$a/J$a/;"x<>'
• Wow, I can't believe you got 29 bytes off mine! I was pretty happy with it originally... Feb 9, 2018 at 8:14
# Perl 5, 85 bytes
84 bytes code + 1 for -p.
eval join"",map"y/$_/@{[/./&&$'.\$&]}/;",(1234567890,QWERTYUIOP,ASDFGHJKL,ZXCVBNM)x<>
Try it online!
# Clean, 144 119 bytes
import StdEnv
\n s=[l.[(i+n)rem(size l)]\\c<-s,l<-["1234567890","QWERTYUIOP","ASDFGHJKL","ZXCVBNM"],i<-[0..]&j<-:l|j==c]
Try it online!
Lambda function with the signature Int ![Char] -> [Char]
# Ruby, 101 bytes
->s,n{n.times{s.tr! '1234567890QWERTYUIOPASDFGHJKLZXCVBNM','2345678901WERTYUIOPQSDFGHJKLAXCVBNMZ'};s}
Try it online!
I'm honestly a little disappointed that I couldn't do better with 'cleverer' methods. The closest I got was along the lines of
a=%w{1234567890 QWERTYUIOP ASDFGHJKL ZXCVBNM}
b=a.map{|r|r[1..-1]<<r[0]}*''
a*=''
n.times{s.tr! a,b}
for a net gain of 7 characters. | 3,051 | 8,480 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.83064 |
https://help.scilab.org/docs/2023.0.0/pt_BR/unpack.html | 1,680,109,759,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00547.warc.gz | 333,638,638 | 3,391 | Scilab Website | Contribute with GitLab | Mailing list archives | ATOMS toolboxes
2023.0.0 - Português
# unpack
Uncompress a list of block-diagonal symmetric matrices.
### Syntax
`A=unpack(CA,blocksizes)`
### Arguments
CA
p-by-n real matrix of doubles, a compressed representation of `A`.
blocksizes
b-by-1 real matrix of doubles, the sizes of the blocks.
A
m-by-n real matrix of doubles, the entries of a list of block-diagonal symmetric matrices. n is the number of matrices stored into `A`.
### Description
This function is the reverse of the `pack` function. It uncompresses a list of symmetric block-diagonal matrices which have been compressed by `pack`.
### Examples
```// Define a compressed list of symmetric block-diagonal matrices.
// There are 3 matrices in the list, as indicated by the 3 columns in CA.
// Only the nonzero entries are stored in CA.
// Only the upper part of the symmetric diagonal matrices are stored in CA.
CA = [
2,1,2;
1,2,2;
2,1,2;
3,1,3;
1,3,4;
3,1,4
]
// Each of the 3 block-diagonal matrices has 2 blocks.
// Define the size of the two blocks:
// the first block has size 2,
// the second block has size 2.
blocksizes=[2,2];
// Unpack the matrices.
A=unpack(CA,blocksizes)``` | 334 | 1,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.795725 |
https://www.floridatoday.com/story/tech/science/satellites-planets/2014/03/13/happy-pi-day-one-and-all/6394433/ | 1,659,886,180,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00279.warc.gz | 695,338,566 | 54,104 | # Happy Pi Day, one and all
BrevardCounty
Today is Pi Day. Pi is defined as the ratio of a circle's circumference to its diameter, and its mathematical symbol is the Greek letter pi (π). The value of π to the nearest hundredth is 3.14 and today is 3/14 — hence Pi Day.
Pi is a special number which doesn't have an exact decimal representation. The calc program in windows confidently gives π as 3.141592 6535897932384626433832 795, but this is only an approximation. Instead the decimal expansion continues to infinity with a seemingly random sequence of digits.
In one year it will be 3/14/15, reproducing the first five digits of pi, 3.1415. The remaining digits continue, 92653, which could be construed as 9:26:53.
If we ignore the year, we can still have an exact pi time when all of the digits line up. Just use the three digits 159 as the time 1:56 and add all the remaining digits seconds with a fraction part 26.53 5897932384626433832795 ... and you'll have it all. When you consider relativity and quantum effects, it probably doesn't mean anything in a physical sense to specify time too precisely.
The orbits of satellites are computed using pi and some pretty sophisticated mathematics. We can make use of this mathematics without even knowing what is happening. Websites like heavens-above.com, and www.satflare.com/track. asp?q=19549#TOP show us what is happening on a world map.
Bright satellites tonight
The brightest satellite visible tonight is CZ-3 R/B, a Chang Zheng 3B rocket body launched in 2006 into a highly elliptical orbit: 206 km perigee, 15,536 km apogee with an inclination of 28.5 degrees to the equator. It is supposed to get to magnitude 0.9, which is just brighter than first magnitude. It will rise out of the west at 8:31 p.m. and swing low to the south, passing below Orion and Canis Major.
The inclination of 28.5 degrees means that the satellite doesn't appear north of latitude, 28.5 degrees, if viewed from the earth's center. Our own latitude, 28.366284 degrees is probably just within this band, so it might on rare occasions actually be north of us, but not tonight.
Thanks to Kepler's Second Law, we know that the satellite moves faster when it is closer to the center of the Earth (perigee), and slower when it is farther (apogee). In this case, the perigee is in the north, so we know the satellite spends most of its time in the far south at apogee.
Because the Earth is rotating in the same direction as the satellite, its ground track is pushed westward by the eastward motion of the earth's surface. When the satellite slows near apogee, this has a stronger effect and changes the shape of the ground track. The track in the northern hemisphere (closer to perigee) does not match the track in the southern hemisphere (closer to apogee).
This effect is even stronger with tonight's second brightest satellite. The IUS R/B(1) is a rocket body with a more extremely elliptical orbit. It has a perigee of 258 km with an apogee of 24,540 km at 27.0 degrees inclination. Although it doesn't reach the height of a geosynchronous satellite (42,164 km), it is still sufficiently far away that the ground track reverses direction and makes an apparent loop over the Earth.
Tonight's pass is closer to perigee, and hence it will be a bright 1.4 magnitude. It rises at 8:41 p.m. in the west and heads right toward Orion. It passes over Orion's chest at 8:49 p.m.
This week
at the Planetarium
FRIDAY
7 p.m., Mars Quest
8:15 p.m., Whales
9 p.m., Pink Floyd: Dark Side of the Moon
SATURDAY
7 p.m., Amazing Universe
8:15 p.m., The Living Sea
9 p.m., The Doors Laser Experience
WEDNESDAY
2 p.m., The Cowboy Astronomer
3:15 p.m., Whales
You may call the box office at (321) 433-7373 to hear the schedule details during off-hours or speak with the cashier when the planetarium is open to the public. To access the online calendar, visit www.calendarwiz. com/planetarium. More information about the Planetarium may be found at www.easternflorida. edu/planet.
The EFSC observatory's 24-inch diameter telescope is generally open for public viewing from 6:45 p.m. to about 10:15 p.m. on most Friday and Saturday evenings. Jupiter is still well placed for observation. Mars doesn't appear until late. | 1,060 | 4,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.910336 |
https://everything.explained.today/Equinumerosity/ | 1,686,116,853,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653608.76/warc/CC-MAIN-20230607042751-20230607072751-00679.warc.gz | 286,057,501 | 9,031 | # Equinumerosity Explained
In mathematics, two sets or classes A and B are equinumerous if there exists a one-to-one correspondence (or bijection) between them, that is, if there exists a function from A to B such that for every element y of B, there is exactly one element x of A with f(x) = y. Equinumerous sets are said to have the same cardinality (number of elements).[1] The study of cardinality is often called equinumerosity (equalness-of-number). The terms equipollence (equalness-of-strength) and equipotence (equalness-of-power) are sometimes used instead.
Equinumerosity has the characteristic properties of an equivalence relation. The statement that two sets A and B are equinumerous is usually denoted
AB
or
A\simB
, or
|A|=|B|.
The definition of equinumerosity using bijections can be applied to both finite and infinite sets, and allows one to state whether two sets have the same size even if they are infinite. Georg Cantor, the inventor of set theory, showed in 1874 that there is more than one kind of infinity, specifically that the collection of all natural numbers and the collection of all real numbers, while both infinite, are not equinumerous (see Cantor's first uncountability proof). In his controversial 1878 paper, Cantor explicitly defined the notion of "power" of sets and used it to prove that the set of all natural numbers and the set of all rational numbers are equinumerous (an example where a proper subset of an infinite set is equinumerous to the original set), and that the Cartesian product of even a countably infinite number of copies of the real numbers is equinumerous to a single copy of the real numbers.
Cantor's theorem from 1891 implies that no set is equinumerous to its own power set (the set of all its subsets). This allows the definition of greater and greater infinite sets starting from a single infinite set.
If the axiom of choice holds, then the cardinal number of a set may be regarded as the least ordinal number of that cardinality (see initial ordinal). Otherwise, it may be regarded (by Scott's trick) as the set of sets of minimal rank having that cardinality.
The statement that any two sets are either equinumerous or one has a smaller cardinality than the other is equivalent to the axiom of choice.
## Cardinality
Equinumerous sets have a one-to-one correspondence between them,[2] and are said to have the same cardinality. The cardinality of a set X is a measure of the "number of elements of the set". Equinumerosity has the characteristic properties of an equivalence relation (reflexivity, symmetry, and transitivity):[3]
Reflexivity: Given a set A, the identity function on A is a bijection from A to itself, showing that every set A is equinumerous to itself: .
• Symmetry: For every bijection between two sets A and B there exists an inverse function which is a bijection between B and A, implying that if a set A is equinumerous to a set B then B is also equinumerous to A: implies .
• Transitivity: Given three sets A, B and C with two bijections and, the composition of these bijections is a bijection from A to C, so if A and B are equinumerous and B and C are equinumerous then A and C are equinumerous: and together imply .
• An attempt to define the cardinality of a set as the equivalence class of all sets equinumerous to it is problematic in Zermelo–Fraenkel set theory, the standard form of axiomatic set theory, because the equivalence class of any non-empty set would be too large to be a set: it would be a proper class. Within the framework of Zermelo–Fraenkel set theory, relations are by definition restricted to sets (a binary relation on a set A is a subset of the Cartesian product), and there is no set of all sets in Zermelo–Fraenkel set theory. In Zermelo–Fraenkel set theory, instead of defining the cardinality of a set as the equivalence class of all sets equinumerous to it one tries to assign a representative set to each equivalence class (cardinal assignment). In some other systems of axiomatic set theory, for example in Von Neumann–Bernays–Gödel set theory and Morse–Kelley set theory, relations are extended to classes.
A set A is said to have cardinality smaller than or equal to the cardinality of a set B, if there exists a one-to-one function (an injection) from A into B. This is denoted |A| ≤ |B|. If A and B are not equinumerous, then the cardinality of A is said to be strictly smaller than the cardinality of B. This is denoted |A| < |B|. If the axiom of choice holds, then the law of trichotomy holds for cardinal numbers, so that any two sets are either equinumerous, or one has a strictly smaller cardinality than the other.[3] The law of trichotomy for cardinal numbers also implies the axiom of choice.[4]
The Schröder–Bernstein theorem states that any two sets A and B for which there exist two one-to-one functions and are equinumerous: if |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|.[3] [4] This theorem does not rely on the axiom of choice.
## Cantor's theorem
Cantor's theorem implies that no set is equinumerous to its power set (the set of all its subsets).[3] This holds even for infinite sets. Specifically, the power set of a countably infinite set is an uncountable set.
Assuming the existence of an infinite set N consisting of all natural numbers and assuming the existence of the power set of any given set allows the definition of a sequence N, P(N), P(P(N)), of infinite sets where each set is the power set of the set preceding it. By Cantor's theorem, the cardinality of each set in this sequence strictly exceeds the cardinality of the set preceding it, leading to greater and greater infinite sets.
Cantor's work was harshly criticized by some of his contemporaries, for example by Leopold Kronecker, who strongly adhered to a finitist[5] philosophy of mathematics and rejected the idea that numbers can form an actual, completed totality (an actual infinity). However, Cantor's ideas were defended by others, for example by Richard Dedekind, and ultimately were largely accepted, strongly supported by David Hilbert. See Controversy over Cantor's theory for more.
Within the framework of Zermelo–Fraenkel set theory, the axiom of power set guarantees the existence of the power set of any given set. Furthermore, the axiom of infinity guarantees the existence of at least one infinite set, namely a set containing the natural numbers. There are alternative set theories, e.g. "general set theory" (GST), Kripke–Platek set theory, and pocket set theory (PST), that deliberately omit the axiom of power set and the axiom of infinity and do not allow the definition of the infinite hierarchy of infinites proposed by Cantor.
The cardinalities corresponding to the sets N, P(N), P(P(N)), are the beth numbers
\beth0
,
\beth1
,
\beth2
, with the first beth number
\beth0
being equal to
\aleph0
(aleph naught), the cardinality of any countably infinite set, and the second beth number
\beth1
being equal to
akc
, the cardinality of the continuum.
## Dedekind-infinite sets
In some occasions, it is possible for a set S and its proper subset to be equinumerous. For example, the set of even natural numbers is equinumerous to the set of all natural numbers. A set that is equinumerous to a proper subsets of itself is called Dedekind-infinite.[3] [4]
The axiom of countable choice (ACω), a weak variant of the axiom of choice (AC), is needed to show that a set that is not Dedekind-infinite is actually finite. The axioms of Zermelo–Fraenkel set theory without the axiom of choice (ZF) are not strong enough to prove that every infinite set is Dedekind-infinite, but the axioms of Zermelo–Fraenkel set theory with the axiom of countable choice are strong enough.[6] Other definitions of finiteness and infiniteness of sets than that given by Dedekind do not require the axiom of choice for this, see .[3]
## Compatibility with set operations
Equinumerosity is compatible with the basic set operations in a way that allows the definition of cardinal arithmetic.[3] Specifically, equinumerosity is compatible with disjoint unions: Given four sets A, B, C and D with A and C on the one hand and B and D on the other hand pairwise disjoint and with and then This is used to justify the definition of cardinal addition.
Furthermore, equinumerosity is compatible with cartesian products:
• If and then
• A × B ~ B × A
• (A × B) × C ~ A × (B × C)
These properties are used to justify cardinal multiplication.
Given two sets X and Y, the set of all functions from Y to X is denoted by XY. Then the following statements hold:
• If A ~ B and C ~ D then
• ABC ~ AB × AC for disjoint B and C.
• (A × B)C ~ AC × BC
• (AB)C ~ AB×C
These properties are used to justify cardinal exponentiation.
Furthermore, the power set of a given set A (the set of all subsets of A) is equinumerous to the set 2A, the set of all functions from the set A to a set containing exactly two elements.
## Categorial definition
In category theory, the category of sets, denoted Set, is the category consisting of the collection of all sets as objects and the collection of all functions between sets as morphisms, with the composition of functions as the composition of the morphisms. In Set, an isomorphism between two sets is precisely a bijection, and two sets are equinumerous precisely if they are isomorphic as objects in Set. | 2,248 | 9,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-23 | latest | en | 0.91055 |
https://su2code.github.io/docs/Solver-Setup | 1,566,782,459,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330913.72/warc/CC-MAIN-20190826000512-20190826022512-00296.warc.gz | 658,616,951 | 6,823 | # Solver Setup
This guide is for version 7 only
This is a basic introduction on how to set up a simulation using SU2. We distinguish between single-zone computations and multi-zone computations. The following considers a single zone only. For an explanation on multi-zone problems, continue with Basics of Multi-Zone Computations.
# Defining the Problem
SU2 is capable of dealing with different kinds of physical problems. The kind of problem is defined by choosing a solver using the SOLVER option. A list of possible values and a description can be found in the following table:
Option Value Problem Type
EULER Euler’s equation Finite-Volume method
NAVIER_STOKES Navier-Stokes’ equation Finite-Volume method
RANS Reynolds-averaged Navier-Stokes’ Finite-Volume method
INC_EULER Incompressible Euler’s equation Finite-Volume method
INC_NAVIER_STOKES Incompressible Navier-Stokes’ Finite-Volume method
INC_RANS Incompressible Reynolds-averaged Navier-Stokes’ Finite-Volume method
HEAT_EQUATION_FVM Heat equation Finite-Volume method
ELASTICITY Equations of elasticity Finite-Element method
FEM_EULER Euler’s equation Discontinuous Galerkin FEM
FEM_NAVIER_STOKES Navier-Stokes’ equation Discontinuous Galerkin FEM
MULTIPHYSICS Multi-zone problem with different solvers in each zone -
Every solver has its specific options and we refer to the tutorial cases for more information. However, the basic controls detailed in the remainder of this page are the same for all problems.
## Restarting the simulation
A simulation can be restarted from a previous computation by setting RESTART_SOL=YES. If it is a time-dependent problem, additionally RESTART_ITER must be set to the time iteration index you want to restart from:
% ------------------------- Solver definition -------------------------------%
%
% Type of solver
SOLVER= EULER
%
% Restart solution (NO, YES)
RESTART_SOL= NO
%
% Iteration number to begin unsteady restarts (used if RESTART_SOL= YES)
RESTART_ITER= 0
%
# Controlling the simulation
A simulation is controlled by setting the number of iterations the solver should run (or by setting a convergence critera). The picture below depicts the two types of iterations we consider.
SU2 makes use of an outer time loop to march through the physical time, and of an inner loop which is usually a pseudo-time iteration or a (quasi-)Newton scheme. The actual method used depends again on the specific type of solver.
## Time-dependent Simulation
To enable a time-dependent simulation set the option TIME_DOMAIN to YES (default is NO). There are different methods available for certain solvers which can be set using the TIME_MARCHING option. For example for any of the FVM-type solvers a first or second-order dual-time stepping (DUAL_TIME_STEPPING-1ST_ORDER/DUAL_TIME_STEPPING-2ND_ORDER) method or a conventional time-stepping method (TIME_STEPPING) can be used.
% ------------------------- Time-dependent Simulation -------------------------------%
%
TIME_DOMAIN= YES
%
% Time Step for dual time stepping simulations (s)
TIME_STEP= 1.0
%
% Total Physical Time for dual time stepping simulations (s)
MAX_TIME= 50.0
%
% Number of internal iterations
INNER_ITER= 200
%
% Number of time steps
TIME_ITER= 200
%
The solver will stop either when it reaches the maximum time (MAX_TIME) or the maximum number of time steps (TIME_ITER), whichever event occurs first. Depending on the TIME_MARCHING option, the solver might use an inner iteration loop to converge each physical time step. The number of iterations within each time step is controlled using the INNER_ITER option.
A steady-state simulation is defined by using TIME_DOMAIN=NO, which is the default value if the option is not present. In this case the number of iterations is controlled by the option ITER.
Note: To make it easier to switch between steady-state, time-dependent and multizone simulations, the option INNER_ITER can also be used to specify the number of iterations. If both options are present, INNER_ITER has precedence.
## Setting convergence criteria
Despite setting the maximum number of iterations, it is possible to use a convergence criterion so that the solver will stop when it reaches a certain value of a residual or if variations of a coefficient are below a certain threshold. To enable a convergence criterion use the option CONV_FIELD to set an output field that should be monitored. The list of possible fields depends on the solver. Take a look at Custom Output to learn more about output fields. Depending on the type of field (residual or coefficient) there are two types of methods:
### Residual
If the field set with CONV_FIELD is a residual, the solver will stop if it is smaller than the value set with CONV_RESIDUAL_MINVAL option. Example:
% ------------------- Residual-based Convergence Criteria -------------------------%
%
CONV_FIELD= RMS_DENSITY
%
%
% Min value of the residual (log10 of the residual)
CONV_RESIDUAL_MINVAL= -8
%
### Coefficient
If the field set with CONV_FIELD is a coefficient, a Cauchy series approach is applied. A Cauchy element is defined as the difference of the coefficient between two consecutive iterations. The solver will stop if the sum over a certain number of elements (set with CONV_CAUCHY_ELEMS) is smaller than the value set with CONV_CAUCHY_EPS. The current value of the Cauchy coefficient can be written to screen or history by adding the CAUCHY field to the SCREEN_OUTPUT or HISTORY_OUTPUT option (see Custom Output). Example:
% ------------------ Coefficient-based Convergence Criteria -----------------------%
%
CONV_FIELD= DRAG
%
%
% Number of elements to apply the criteria
CONV_CAUCHY_ELEMS= 100
%
% Epsilon to control the series convergence
CONV_CAUCHY_EPS= 1E-10
%
For both methods the option CONV_STARTITER defines when the solver should start monitoring the criterion. | 1,286 | 5,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-35 | latest | en | 0.783114 |
http://www.kylesconverter.com/density/centigrams-per-cubic-millimeter-to-tonnes-per-cubic-meter | 1,534,889,706,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219109.94/warc/CC-MAIN-20180821210655-20180821230655-00145.warc.gz | 537,201,939 | 5,398 | # Convert Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter
### Kyle's Converter > Density > Centigrams Per Cubic Millimeter > Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter
Centigrams Per Cubic Millimeter (cg/mm3) Tonnes Per Cubic Meter (t/m3) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Tonnes Per Cubic Meter to Centigrams Per Cubic Millimeter
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Centigram per Cubic Millimeter:
Mass of 1 centigram per volume of a cubic millimeter. Equivalent density of 10 000 kilograms per cubic meter. 1 cg/mm3 ? 1000 kg/m3.
1 Tonne per Cubic Meter:
Mass of 1 tonne per volume of a cubic meter. Metric tonne of exactly 1 000 kilograms. Equivalent density of 1 000 kilograms per cubic meter. 1 t/m3 ? 1000 kg/m3.
Conversions Table
1 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 1070 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 700
2 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 2080 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 800
3 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 3090 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 900
4 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 40100 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 1000
5 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 50200 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 2000
6 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 60300 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 3000
7 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 70400 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 4000
8 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 80500 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 5000
9 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 90600 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 6000
10 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 100800 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 8000
20 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 200900 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 9000
30 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 3001,000 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 10000
40 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 40010,000 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 100000
50 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 500100,000 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 1000000
60 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 6001,000,000 Centigrams Per Cubic Millimeter to Tonnes Per Cubic Meter = 10000000 | 830 | 2,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.389488 |
https://oeis.org/A144786 | 1,632,743,229,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00492.warc.gz | 466,532,975 | 5,235 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A144786 If n is an oblong number A002378, then a(n)=a(j) where j is the number of oblong numbers in (0,n], otherwise a(n)=n. 10
1, 1, 3, 4, 5, 1, 7, 8, 9, 10, 11, 3, 13, 14, 15, 16, 17, 18, 19, 4, 21, 22, 23, 24, 25, 26, 27, 28, 29, 5, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 1, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 7, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 8, 73, 74, 75, 76 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS As a motivation, consider the greedy decomposition of fractions 1/n into Egyptian fractions, n=1: 2,3,7,43,1807,3263443,.. A000058 n=2: 3,7,43,1807,3263443,10650056950807,.. A000058 n=3: 4,13,157,24493,599882557,359859081592975693,.. A082732 n=4: 5,21,421,176821,31265489221,977530816197201697621,.. A144779 n=5: 6,31,931,865831,749662454731,561993796032558961827631,.. A144780 n=6: 7,43,1807,3263443,10650056950807,.. A000058 n=7: 8,57,3193,10192057,103878015699193,.. A144781 n=8: 9,73,5257,27630793,763460694178057,.. A144782 n=9: 10,91,8191,67084291,4500302031888391,.. A144783 n=10: 11,111,12211,149096311,22229709804712411,.. A144784 n=11: 12,133,17557,308230693,95006159799029557,.. A144785 n=12: 13,157,24493,599882557,.. A082732 k=13: 14,183,33307,1109322943,.. where the first few denominators of 1/n = 1/b(1)+1/b(2)+... have been tabulated. For some sets of n, the list b(i) of denominators is essentially the same: consider for example A000058, which represents primarily n=1, then in truncated form also n=2, and then n=6, n=42 etc. Or consider A082732 which represents n=3, then in truncated form n=12, n=156 etc. The OEIS sequence assigns the primary n to a(n). The interpretation of a(n) with ascending n is: n=1 is primary, a(1)=1. Decomposition of n=2 is equivalent to n=1, a(2)=1. Cases n=3 to 5 are primary ("original", "new"), and a(n)=n in these cases. n=6 is not new but essentially the same Egyptian series as seen for n=1, so a(6)=1. Cases n=7 to n=11 are "new" sequences, again a(n)=n in these cases, but then n=12 is represented by A082732 as already seen for n=3, so a(12)=3. Because the first denominator for the decomposition of 1/n is 1/(n+1), n+1 belongs to the sequence of denominators of the expansion of 1/a(n). The sequences b(.) have recurrences which are essentially 1+b(n-1)*(b(n-1)-1), looking up the oblong number at the position of the previous b(.). This is the reason why reverse look-up of the n via A000194 (number of oblong numbers up to n) as used in the definition is equivalent to the assignment described above. LINKS FORMULA a(n) = a(A000194(n+1)) if n in A002378. a(n) = n if n in A078358. EXAMPLE n=1 is not in A002378, so a(1)=1. n=2 = A000058(2), so a(2)=1 because there is 1 oblong number <=2 and >0. n=3 is not in A002378, so a(3)=3. n=6 = A000058(3), so a(6)=a(2) because there are 2 oblong numbers <=6 and >0. CROSSREFS Cf. A000058, A002378, A078358, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786. Sequence in context: A280490 A135347 A345708 * A247369 A009389 A091828 Adjacent sequences: A144783 A144784 A144785 * A144787 A144788 A144789 KEYWORD nonn,easy,frac AUTHOR Artur Jasinski, Sep 22 2008, Sep 26 2008 EXTENSIONS a(57)=57 inserted, a(61)=61 corrected and better definition provided by Omar E. Pol, Dec 29 2008 I did some further editing of this entry, but many of the lines are still obscure. - N. J. A. Sloane, Dec 29 2008 Comments that connect to Egyptian fractions rephrased by R. J. Mathar, Oct 01 2009 STATUS approved
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Last modified September 27 07:44 EDT 2021. Contains 347673 sequences. (Running on oeis4.) | 1,500 | 4,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-39 | latest | en | 0.73413 |
http://www.newton.dep.anl.gov/askasci/gen06/gen06327.htm | 1,418,938,659,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802767878.79/warc/CC-MAIN-20141217075247-00046-ip-10-231-17-201.ec2.internal.warc.gz | 692,502,215 | 4,756 | Individual Differences in Perception ```Name: neagh Status: student Grade: 6-8 Location: N/A Date: May 2007 ``` Question: Does everyone see things the way I do? i.e perception and senses? Replies: Obviously there are a lot of non-scientific answers to this question too, but I'm going to stick with some science. And I'm warning you now, it's some pretty weird science (but it's true!). There's a theory in physics called 'relativity'. Albert Einstein came up with relativity to explain a contradiction: how could it be that the laws of physics are the same for everyone, and yet the speed of light is constant? At first glance, this might not sound contradictory, but it is. Think of the speed of light: it's really fast, right? Well, what if you were already traveling at a speed just below the speed of light? Would the light just barely creep faster than you? The answer is that it *still* travels really really fast. How could this be? Einstein came up with some theories about how space and time "stretch" that explain how this works. The math is somewhat complicated, but trust that it works out pretty well. The bottom line is that depending on how fast you're traveling compared to someone else, the time and space you experience will be different. Time speeds up or slows down depending on your velocity relative to other objects. Experiments show that very precise synchronized clocks will show different times when one is left on Earth's surface, and the other is flown on an airplane for a long enough time. On the fast-flying plane, the clock slowed down, and ends up behind the clock on the ground. In everyday living, we all are traveling about the same speed relative to each other, so the differences are too small for us to notice. But scientists can measure them, and they can think of 'extreme' examples where the differences would be big. This means that everyone experiences slightly different time and space than everyone else. Two observers traveling very fast in opposite directions might observe, for instance a star exploding, at different times. Events that occur in succession might appear to occur right after the other, or a long time after the other. The *order* of events cannot change, but the time when they occur does. I don't know if this is the kind of answer you were expecting (maybe not?), but I wanted to share this phenomenon with you because it shows that science is sometimes very unpredictable. Sometimes things that you never expected -- and never could have predicted -- affect what you're doing. To answer your question in a more expected way, humans' senses all work generally the same way, but there are definitely differences between individuals. For instance, one person might be color-blind, and so they might see the world very differently than someone who sees all colors. Different people's senses of taste and smell can be very different (you could have a cold and not be able to taste/smell well, or simply have differently-sensitive taste buds). Any of our body parts can temporarily be not working -- or permanently -- or simply work better. The way the brain processes information from the senses also affects how you perceive them. For instance, if you lose your sight, your brain will devote more 'processing power' to your other senses. You might have more sensitive hearing or sense of touch, for instance. You can train your senses too -- think about a wine expert who can taste a wine and tell you precisely which vineyard anywhere in the world it came from. Burr NO! Every human will have some individual differences in the way they perceive the world, due to inherited differences in sensory abilities. Beyond that, other creatures, even those relatively closely related to us, will perceive the world in a vastly different way than we do. J. Elliott Click here to return to the General Topics Archives
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For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs | 860 | 4,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2014-52 | latest | en | 0.956159 |
https://www.physicsforums.com/threads/laplace-domain-to-time-domain.224290/ | 1,513,610,599,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948617816.91/warc/CC-MAIN-20171218141805-20171218163805-00540.warc.gz | 776,175,164 | 15,241 | # Laplace domain to time domain
1. Mar 25, 2008
### abhishek.akku
i have a question about laplace transforms. I had to find the maximum of a three exponential equation f(t). What I have done is took the first derivative of f(t) and equated it to zero. Now f'(t) is also a three exponential equation ( like f'(t)= e^at +e^bt +e^ct). So I have taken laplace transform for f'(t) = sf(s)-f(0) and obtained a 3rd digree equation in "s". I have solved the cubic equation for f(s) and found out three values of s. Now I need to find out "t" at which f(t) is max.
How can I convert "s= some constant" to "t" ? Can anybody help me out?? | 187 | 631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | longest | en | 0.963483 |
https://drawabox.com/community/submission/HV4IMHYC | 1,695,565,823,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506646.94/warc/CC-MAIN-20230924123403-20230924153403-00475.warc.gz | 235,912,173 | 15,846 | ## Lesson 2: Contour Lines, Texture and Construction
##### 12:30 AM, Tuesday November 15th 2022
I think I ran into quite the struggle in the dissecition exercise, especially with parts where the shadow or the dark parts of the pattern aren't quite as dark as pitch black if that makes sense. It's hard to achieve grey or somewhere in between white and black with the pen, so I was wondering whether this is standard practice, what's the "right" way to do shadows or colors with different shades, or whether this is just for practice?
In addition, I did find the form intersection exercise quite difficult for tricky intersections, like the side of a cone on a cube's corner, but I guess part of the exercise is to get myself acquainted to the difficulty.
Thank you so much for your help!
0 users agree
##### 7:22 PM, Wednesday November 16th 2022
I'll be the TA handling your Lesson 2 critique.
You're making progress towards understanding the concepts introduced in this lesson and hopefully this critique will help you in your future attempts.
• Starting off in the arrows section your lines are mostly looking smoothly and confidently drawn. You're doing a good job maintaining a consistent width as your arrows widen while moving closer to the viewer and with more mileage you'll become more consistent. It's good to see that you're trying to implement line weight, just remember that you want to keep your applications subtle and you'll become consistent with mileage. here are some things to look out for when applying it. I'd like you to experiment more with foreshortening in your future attempts, by utilizing it in both the arrows themselves as well as the negative space between their curves we can create a stronger illusion of an object moving through 3D space as demonstrated here.
• Moving into the organic forms with contours exercise you're doing a good job keeping your forms simple, plenty of people tend to over-complicate them. At least in this section you're redrawing your organic forms rather than just creating a single smooth mark, this causes them to be messier than they need to be so I'd ask that you not do this in the future. There are moments where your lines could be more confident as well, there's some wobbling in your contour curves. Remember that accuracy will improve with mileage so be sure to try and draw as smoothly and confidently as you can. Remember to draw through all of your ellipses including the small contour ones on the end of your organic forms as well. Speaking of contours I'd like you to try and shift the degree of your contours more. The degree of a contour line basically represents the orientation of that cross-section in space, relative to the viewer, and as we slide along the sausage form, the cross section is either going to open up (allowing us to see more of it) or turn away from the viewer (allowing us to see less), as shown here.
• In the texture exercises you're focusing largely on outlines, colour shifts and negative space rather than cast shadows created by forms along the texture itself (colour shifts are tied to patterns not texture, texture requires a change in the form's surface). This makes it difficult to create gradients with implied information which we could then use to create focal points in more complex pieces, by doing so we can prevent our viewers from being visually overwhelmed with too much detail. For more on the importance of focusing on cast shadows read here. I'd also like to quickly direct you to this image which shows that when we're working with thin line like textures if we outline and fill the shadow we will create a much more dynamic texture than simply drawing lines.
• It's quite common for people to feel like they don't fully grasp the form intersections exercise, if you feel like you may fall into this category try not to stress too much. This exercise is just meant to get students to start thinking about how their forms relate to one another in 3D space, and how to define those relationships on the page. We'll be going over them more in the upcoming lessons. When you're not redrawing your forms are looking pretty solid here, though you are skipping a step in your cylinders and cones, as mentioned here you should be drawing the minor axis.
I won't be moving you on to the next lesson just yet, each lesson builds upon each other and I'd like to make sure you understand a few of these concepts a bit more before potentially creating more problems down the road.
With that being said I'd like you to please re-read and complete:
• 1 page of the organic intersections exercise
If you need more of an explanation in regards to your texture question let me know when you hand in your revisions and I'll give you a more in-depth explanation. Ut's quite large so I don't want to bloat this critique more than necessary with something only directed towards a single section. Texture is also an introduction to the concept and not as important as organic intersections at the moment so I don't want to distract you from that.
Once you've completed the pages mentioned above reply to this critique with a link to them, I'll go over them and address anything that needs to be worked on and once you've shown you're ready for the next lesson I'll move you on.
I look forward to seeing your work.
Next Steps:
• 1 page of the organic intersections exercise
##### 6:27 AM, Thursday November 17th 2022
Hello,
Thank you so much for the in-depth critique! Below is my additional submission of one page of the organic intersection exercise.
https://imgur.com/a/BS7ofvl
Looking forward to hearing from you soon!
And regarding the texture question, I was asking more in terms of shadow coloring. In reference pictures, forms cast shadows, but depending on the light source's positioning, some shadows are darker than others. I find it quite difficult to capture grey, or lighter shadows, with the given tool.
Naturally I feel like cross-hatching would be the solution, but obviously since we don't want to use cross-hatching here, I was wondering whether I was missing anything.
##### 2:27 PM, Thursday November 17th 2022
These are definitely headed in the right direction, good work.
I recommend getting a brush pen if you're able to, they make filling large areas (like shadows) a lot easier and can be quite fun once you get the hang of them. It'll probably save you money on fine liners in the long run as well.
I'll be marking your submission complete and move you on to lesson 3, best of luck and give the explanation below a read.
Wall of text explaining Texture
Rather than being able to give you just one or two pointers about things to work on, texture is often a case of people trying to simplify the steps too much and it's easier for me to just explain the entire process.
First things first, open up this leaf texture picture, I find leaves are a good example and a texture that people are often drawn to and do incorrectly.
The first thing you may notice is that this image isn't in colour and instead in black and white, this is helpful because people often get distracted by shifts in colour and will try to darken an area in their drawing if the colour happens to be darker. We shouldn't rely on converting images to black and white but it is helpful and something you may want to consider when practicing.
Now if I handed a student this image and told them to use it for their texture exercises there are two typical outcomes I would expect.
The first is that they would draw all of the veins (or many of them if they aren't extremely patient), and this would be an example of focusing on outlines.
The second result that I would expect is that instead of drawing the veins themselves they would either fill the veins in completely with black or they would fill in everything but the veins completely with ink, and this would be them focusing on negative space.
This is where students get a bit confused at the start, they feel like they're looking at the image and drawing what they see but it comes out wrong. I should clarify that it's not necessarily incorrect, there is a time and place when obsessing over small details can be helpful but this is largely an exercise about learning how to imply information and thinking in 3D space. Remember that what we're learning here isn't observational drawing, it's constructional and while observation is definitely a part of the construction method there's an extra step that people tend to neglect in the beginning.
With that in mind let's go over the correct way to tackle these problems, bring up the leaf image (it's here if you closed it) and let's break down what we're working with. What people tend to neglect is that we're trying to think about the 3D space of the image we're observing and drawing. If we look at this leaf the veins are really just long organic forms, or cylinders if they're particularly rigid. The fleshy bit of the leaf could be thought of as either a plane or a thin box, and due to gravity's effect it will likely curve a bit (we don't need to think too much about it curving in this case because we're working so close up, but it's good to think of how the environment can have an effect).
To put it simply, a leaf is just organic forms that are intersecting with each other and a plane/box, just like the forms we practiced with in the form intersections exercise. With all of these forms in mind we can place a light source and depending on it's position and intensity we can create cast shadows much like we did in the organic intersections exercise.
An example of this can actually be seen in the picture itself. Let's just focus on the large main vein as well as the branching vein on the left. You'll notice that if you look along the bottom edge of the main vein there's a cast shadow, and on the right of the branching vein there's a shadow. From this information we can assume that the light affecting the leaf the most is somewhere to the upper left of the image and it's creating the cast shadows we want to draw.
This is an example of what drawing cast shadows might look like, it comes up early on in lesson 3 when Uncomfortable shows the process of drawing a leaf. You can see that he's not actually drawing the veins themselves, and instead just implying that they exist by creating cast shadows. Keep in mind that this leaf is fairly evenly lit and the point of the texture exercise is to work with light gradients but regardless it's a good example of what we're trying to achieve. There are times where capturing shadows doesn't immediately give the impression of what you're attempting to draw but it's just a single tool that you can use. You may not think that drawn leaf looks like a leaf, but if it was green, had a stem, and was attached to something that resembled a plant your brain would begin fill in the gaps until it goes "oh that's a leaf".
Implying information helps both the creator as well as the viewer, it saves the creator time from having to obsessively capture every tiny detail and it prevents the image from becoming too visually noisy and overwhelming for the viewer.
Craig Mullins is a painter whose work I appreciate a lot and I feel does an amazing job of implying information through shadow, colour and brush strokes. I highly recommend looking up some of his work if you'd like some examples of just how powerful implied information can be.
Hope this helped.
Next Steps:
Move on to lesson 3.
This critique marks this lesson as complete.
##### 3:04 PM, Thursday November 17th 2022
Thank you so much for the detailed wall of text! From what I understand, texture is just a construction of various organic forms, and the idea of drawing cast shadow is to imply the existence of these forms.
After which comes the decision on which shadows should you include, or which shadows will be enough for the implication, and which shadows will make the image muddier more than it helps with the implication.
I'll definitely come back to this message in the future whenever I want to understand texture more thoroughly. Again, thank you so much for your time!
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where k is the specific wear rate for abrasion of the rubber by the iron ore, V is the speed of the belt, μ is the coefficient of friction between the bed of ore and the belt surface, w is the width of the belt, and L is the overall (end-to-end) length of the conveyor. Make your assumptions clear. (b) In one particular conveyor…
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thermal power plant and other projects. A conveyor belt is the carrying medium of a belt conveyor system. A belt conveyor system is one of many types of conveyor systems. A belt conveyor system consists of two or more pulleys (sometimes referred to as drums), with an endless loop of carrying medium - the conveyor belt
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Conveyor capacity is determined by the belt speed, width and the angle of the belt - and can be expressed as. Q = ρ A v (1) where . Q = conveyor capacity (kg/s, lb/s) ρ = density of transported material (kg/m 3, lb/ft 3) A = cross-sectional area of the bulk solid on the belt …
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Conveyor Capacity Belt conveyor is required to convey certain quantity of material per hour. As shown in the figure given below, the material is accommodated on the belt forming certain cross-section of the material [S = S 1 (Upper Section) + S 2 (Lower Section)]. This cross-section multiplied
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• ### Grinding Equipment
The PC Hammer Mill, MTW, MTM Medium Speed Trapezium Mill, Ball Mill and the LM, LUM Vertical Mill cover all requirements of crude, fine and ultrafine powder production in the industrial milling field. Production of free combination from 0 to 2500 meshes can be realized. No matter which industry you are in, chemistry, energy, construction material or metallurgy field, SBM will always meet all your demands. | 2,261 | 9,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-39 | latest | en | 0.844178 |
https://itmyt.com/definition/check-digit | 1,708,945,451,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00719.warc.gz | 332,505,298 | 8,834 | # Check Digit
## Definition & Meaning
Last updated 5 month ago
## What is Check Digit?
### itMyt Explains Check Digit:
A test digit is a digit delivered to a String of numbers for errors detection Functions. Normally, the test digit is Computed from the other digits within the string. A take a look at digit enables Digital structures hit upon adjustments while facts is transferred from Transmitter to Receiver.
The test digit is the Decimal equal of the Binary Checksum utilized in binary structures.
## What Does Check Digit Mean?
A check digit set of rules calculates a test digit based on an original individual string, which include an account wide Variety. The receiver recalculates the test digit to affirm inFormation Access accuracy. If the recalculated man or woman string carries an appropriate take a look at digit, the statistics is errors-Free and may be used. However, a person string that doesn't include the appropriate take a look at digit indicates a Switch mistakes, which indicators that facts need to be re-entered and/or reverified.
When a take a look at digit gadget is used, blunders detection and information Implementation complexities and compromises are ineviTable. Simple take a look at digit structures easily understood by way of human beings can't Discover errors with whole accuracy, in contrast to complicated structures that use Greater complicated mistakes detection Algorithms.
A encouraged test digit function is left-padding with zeros, which never alters the authentic take a look at digit and lets in for the Software of digits with varied and dynamic length.
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1499- V12 | 567 | 3,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.884557 |
https://www.nasa2000.com.mx/geometric-mean-definition-examples-formula-uses/ | 1,701,955,979,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100674.56/warc/CC-MAIN-20231207121942-20231207151942-00801.warc.gz | 1,004,905,277 | 32,334 | If you’re familiar with logarithms, this can be a very intuitive way to look at it. For example, let’s say you wanted to calculate the geometric mean of 2 and 32. Thus, the geometric mean is also defined as the nth root of the product of n numbers.
Include your email address to get a message when this question is answered. Then convert 3% to a decimal and subtract it from 1 to get 0.97. Social login does not work in incognito and private browsers. To calculate result you have to disable your ad blocker first. This calculator will find the geometric mean of a set of numbers.
## Why Is Geometric Mean Better Than Arithmetic?
According to NYU corporate finance and valuation professor Aswath Damodoran, the geometric mean is appropriate for estimating expected returns over long term horizons. Computers use mind-boggling amounts of data which often has to be summarized using statistics. One study compared the precision of several statistics (arithmetic means, geometric means, and percentage in the top x%) for a mind-boggling 97 trillion pieces of citation data. The study found that the geometric mean was the most precise . Multiply the values you want to find the geometric mean for. You can either use a calculator or do the math by hand when you find the product.
Each percentage change value is also converted into a growth factor that is in decimals. The growth factor includes the original value (100%), so to convert percentage increase into a growth factor, add 100 to each percentage increase and divide by 100. Find the nth root of the product where n is the number of values.
The geometric mean is best for reporting average inflation, percentage change, and growth rates. Because these types of data are expressed as fractions, the geometric mean is more accurate for them than the arithmetic mean. The geometric mean is a type of average , usually used for growth rates, like population growth or interest rates. While the arithmetic mean adds items, the geometric mean multiplies items. Also, you can only get the geometric mean for positive numbers.
For example, say you study fruit fly population growth rates. You’re interested in understanding how environmental factors change these rates. Take the antilog of the quotient to determine the geometric mean. The antilog function is the inverse of the LOG function on your calculator and it converts the value back to base-10.
## How to Find the Geometric Mean | Calculator & Formula
The ratio of the corresponding observations of the G.M in two series is equal to the ratio of their geometric means. In this lesson, let us discuss the definition, formula, properties, and applications of geometric mean and also the relation between AM, GM, and HM with solved examples in the end. To make calculations easier meracalculator has developed 100+ calculators in math, physics, chemistry and health category. In surveys and studies too, the geometric mean becomes relevant.
• The Geometric Mean is the average value or mean which signifies the central tendency of the set of numbers by finding the product of their values.
• Geometric Mean is also used in biological studies like cell division and bacterial growth rate etc.
• If in an arithmetic mean we combine the numbers using the summation operation and then divide by their number, in a geometric mean we calculate the product of the numbers and then take its n-th root.
• Our geometric mean calculator handles this automatically, so there is no need to do the above transformations manually.
• Convert percentages to their decimal multiplier equivalents.
The geometric mean won’t be meaningful if zeros are present in the data. You may be tempted to adjust them in some way so that the calculation can be done. There are same cases when adjustments are justified and the first one is similar to the negative numbers case above. If the data is percentage increases, you can transform them into normal percentage values in the way described for negative numbers.
## Example: Geometric mean of percentages
Use your calculator to solve the equation and write down your answer. For example, the geometric mean is the only correct mean when averaging normalized results, which are any results that are presented as ratios to a reference value or values. One way to think of the geometric mean is that it’s the average of logarithmic values converted back to base 10.
The geometric mean has been used in film and video to choose aspect ratios . It’s used to find a compromise between two aspect ratios, distorting or cropping both ratios equally. Convert the number back to a percent by moving the decimal point 2 places to the right and subtracting 1 from it to find a total of a 3% increase in value. Geometric Mean is also used in biological studies like cell division and bacterial growth rate etc. Geometric Mean is used in the case when finding an average for set of numbers presented as percentages. Calculate the geometric mean from a set of positive or negative numerical values.
The arithmetic mean will not make sense in this case either. Levels of Measurement | Nominal, Ordinal, Interval and Ratio Levels of measurement tell you how precisely variables are recorded. The level of measurement determines how you can analyze your data. The average voter turnout of the past five US elections was 54.64%.
You can also use the logarithmic functions on your calculator to solve the geometric mean if you want. The Geometric Mean is the average value or mean which signifies the central tendency of the set of numbers by finding the product of their values. In mathematics and statistics, measures of central tendencies geometric mean of 2 and 32 is describe the summary of whole data set values. The most important measures of central tendencies are mean, median, mode, and range. Among these, the mean of the data set provides the overall idea of the data. The different types of mean are Arithmetic Mean , Geometric Mean , and Harmonic Mean .
## Logarithmic Values and the Geometric Mean
Count how many values are in the set you’re calculating the geometric mean for the value n. Use the n value to determine which root you need to take of the product. For example, take the square root if you have 2 values, cube root if you have 3 values, and so on. | 1,259 | 6,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-50 | latest | en | 0.911522 |
https://solvedlib.com/n/the-calvin-benson-cycle-starts-with-a-the-absorption-of,10257463 | 1,708,535,980,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00724.warc.gz | 551,168,576 | 17,882 | # The Calvin-Benson cycle starts with _________.a. the absorption of photon energyb. carbon fixationc. the release of electrons from photosystem IId.
###### Question:
The Calvin-Benson cycle starts with _________. a. the absorption of photon energy b. carbon fixation c. the release of electrons from photosystem II d. NADP+ formation
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lim y(t) Let y(t) be a solution Of @ Ju(1 ") such that y(0) Delermine lin 2 wthout fnding ylt) explicitly...
##### 0.270 is WRONG, Please provide correct answer. Conduct the hypothesis test and provide the test statistic...
0.270 is WRONG, Please provide correct answer. Conduct the hypothesis test and provide the test statistic and the critical value, and state the conclusion. A person randomly selected 100 checks and recorded the cents portions of those checks. The table below lists those cents portions categorized a... | 4,151 | 13,686 | {"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": 3, "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": 1, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-10 | latest | en | 0.840772 |
https://www.prepbharat.com/EntranceExams/JEEMain/MPF/mechanicalpropertiesoffluids-8.html | 1,713,330,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817144.49/warc/CC-MAIN-20240417044411-20240417074411-00361.warc.gz | 872,474,733 | 5,025 | Mechanical Properties of Fluids Questions and Answers Part-8
1. A big drop of radius R is formed by 1000 small droplets of water, then the radius of small drop is
a) R/2
b) R/5
c) R/6
d) R/10
Explanation:
2. When $10^{6}$ small drops coalesce to make a new larger drop then the drop
a) Density increases
b) Density decreases
c) Temperature increases
d) Temperature decreases
Explanation: Because energy is liberated
3. Which of the following statements are true in case when two water drops coalesce and make a bigger drop
a) Energy is released
b) Energy is absorbed
c) The surface area of the bigger drop is smaller than the sum of the surface areas of both the drops
d) Both a and c
Explanation: Both a and c
4. 8000 identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is
a) 1 : 10
b) 1 : 15
c) 1 : 20
d) 1 : 25
Explanation:
5. The surface energy of liquid film on a ring of area $0.15m^{2}$ is ( Surface tension of liquid = $5Nm^{-1}$ )
a) 0.75 J
b) 1.5 J
c) 2.25 J
d) 3.0 J
Explanation:
6. 8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of
a) 1
b) 2
c) 4
d) 6
Explanation:
7. If work done in increasing the size of a soap film from $10 cm\times6 cm$ to $10 cm \times11 cm$ is $2\times 10^{-4}J$ then the surface tension is
a) $2\times 10^{-2}Nm^{-1}$
b) $2\times 10^{-4}Nm^{-1}$
c) $2\times 10^{-6}Nm^{-1}$
d) $2\times 10^{-8}Nm^{-1}$
Explanation:
8. A mercury drop of radius 1cm is sprayed into $10^{6}$ drops of equal size. The energy expended in joules is (surface tension of Mercury is $460\times 10^{-3}N\diagup m$ )
a) 0.057
b) 5.7
c) $5.7\times 10^{-4}$
d) $5.7\times 10^{-6}$
Explanation:
9. A film of water is formed between two straight parallel wires of length 10cm each separated by 0.5 cm. If their separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done (Surface tension of water =$7.2\times 10^{-2}N\diagup m$ )
a) $7.22\times 10^{-6}Joule$
b) $1.44\times 10^{-5}Joule$
c) $2.88\times 10^{-5}Joule$
d) $5.76\times 10^{-5}Joule$
10. A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is $0.465 J\diagup m^{2}$ )
a) $23.4\mu J$
b) $18.5\mu J$
c) $26.8\mu J$
d) $16.8\mu J$ | 822 | 2,395 | {"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-2024-18 | latest | en | 0.813488 |
http://www.newhappyholidays.com/faq/often-asked-how-many-years-is-a-million-days.html | 1,652,998,471,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00364.warc.gz | 96,912,039 | 11,235 | ## How many years are there in a millionaire?
\$1 Million the Easy Way Putting aside someone’s \$40,000 in take-home pay every year—and earning that 10% return as described above—will get you to millionaire status in about 15 years. Halve those savings and you’re still only looking at 20 years. It will take more work for sure, but it’s a lot faster than 51.
## How many days is a million?
My friend Pat Singleton put this into perspective for me in terms I could really grasp. 1 million seconds equal 11 and 1/2 days. 1 billion seconds equal 31 and 3/4 years. 1 trillion seconds equal 31,710 years.
## How many years old are you if you’ve lived 1 million seconds?
1,000,000 seconds is equivalent to 0.031709792 years.
## What is 1million years?
A million years is called a megaannum, which is often abbreviated ‘Ma. ‘ This term comes from the word parts ‘mega’ which means ‘huge’ and ‘annum’
## How can I get rich in 5 years?
5 steps to becoming a millionaire, from a millennial who did it in 5 years
1. Get paid what you’re worth.
2. Save a ton of money …
3. Develop multiple streams of income.
4. Invest in what you know.
You might be interested: Question: How Many Days In A Year On Saturn?
## How many Trillionaires are there in the world?
As of 2018, there are over 2,200 U.S. dollar billionaires worldwide, with a combined wealth of over US\$ 9.1 trillion, up from US\$7.67 trillion in 2017. According to a 2017 Oxfam report, the top eight richest billionaires own as much combined wealth as “half the human race”.
## How long is a 1 billion seconds?
Answer: One billion seconds is a bit over 31 and one-half years.
## How many millions is 1 billion?
A billion is a number with two distinct definitions: 1,000,000,000, i.e. one thousand million, or 109 (ten to the ninth power), as defined on the short scale. This is now the meaning in all English dialects. 1,000,000,000,000, i.e. one million million, or 1012 (ten to the twelfth power), as defined on the long scale.
## How many seconds do we live?
According to the world bank, in 2016 the average person in the world lived for 72.035 years, that’s roughly 2.27 billion seconds.
## How long is a million seconds or 1 billion seconds?
Magnitude of the difference A million seconds is 12 days. A billion seconds is 31 years. A trillion seconds is 31,688 years.
## How many seconds old is a 10 year old?
Explanation: We cannot give an absolute exact answer without knowing the date and year of birth, because of the effect of leap year. Someone who is 10 will have experienced 2 leap years for sure, and perhaps 3. 1×365×24×60×60= 31,536,000 seconds in one year.
## What will happen in 100 trillion years?
By 1014 (100 trillion) years from now, star formation will end. This period, known as the “Degenerate Era”, will last until the degenerate remnants finally decay. The universe will become extremely dark after the last stars burn out. Even so, there can still be occasional light in the universe.
You might be interested: Question: How Many Days In Quebec City?
## How long is a billion years?
A billion years or giga-annum (109 years) is a unit of time on the petasecond scale, more precisely equal to 3.16×1016 seconds (or simply 1,000,000,000 years ).
## What are 20 years called?
Word Origin for vicennial C18: from Late Latin vīcennium period of twenty years, from Latin vīciēs twenty times + -ennium, from annus year. | 883 | 3,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-21 | latest | en | 0.948228 |
https://school.gradeup.co/ex-3.1-q1-aftab-tells-his-daughter-seven-years-ago-i-was-i-1njgt5 | 1,569,028,998,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574159.19/warc/CC-MAIN-20190921001810-20190921023810-00059.warc.gz | 651,496,842 | 30,478 | # Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Given: Seven years ago, Aftab was seven times as old as his daughter and after 3 years Aftab will be 3 times as old as his daughter.
To Represent the situations algebraically and graphically we need to find the linear equations for these situations.
Let present age of Aftab = x
Let present age of his daughter = y
Then, seven years ago the age of Aftab and his daughter must have been seven less than their present ages,
Age of Aftab seven years ago = x – 7
Age of Daughter seven years ago = y – 7
According to the question,
Seven years ago, Aftab was seven times as old as his daughter, So
x - 7 = 7(y - 7)
⇒ x - 7 = 7y - 49
⇒ x = 7y - 42
Now for finding different points of this equation, we can either take different values of x and put them in the equation to obtain values of y or vice versa
Putting y = 5, 6 and 7 in equation (¡),
we get,
For y = 5,
x = 7 × 5 – 42
= 35 – 42
= -7
For x =6,
x = 7 × 6 – 42
= 42 – 42
= 0
For y = 7,
x = 7 × 7 – 42
= 49 – 42
= 7
Thus we got 3 points to plot on graph for this equation.
Three years from now,
Age of Aftab = x+3
Age of Daughter = y+3
According to the question,
⇒ x + 3 = 3(y + 3)
⇒ x + 3 = 3y + 9
⇒ x = 3y + 6
Now for finding different points of this equation, we can either take different values of x and put them in equation to obtain values of y or vice versa
Putting x = 0, 3 and 6
Thus we got 3 points to plot on graph for this equation.
Algebraic representation
x - 7y = -42 (i)
x - 3y = 6 (ii)
Graphical representation:
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We strive to provide quality solutions. Please rate us to serve you better. | 577 | 1,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2019-39 | latest | en | 0.96874 |
https://inquiryintoinquiry.com/2021/03/30/animated-logical-graphs-69/ | 1,675,932,460,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501555.34/warc/CC-MAIN-20230209081052-20230209111052-00374.warc.gz | 326,668,436 | 45,977 | ## Animated Logical Graphs • 69
Re: Richard J. LiptonThe Art Of Math
Re: Animated Logical Graphs • (57)(58)(59)(60)(61)(62)(63)(64)(65)(66)
“I know what you mean but I say it another way” — it’s a thing I find myself saying often enough, if only under my breath, to rate an acronym for it ☞ IKWYMBISIAW ☜ and not too coincidentally it’s a rubric of relevance to many situations in semiotics where sundry manners of speaking and thinking converge, more or less, on the same patch of pragmata.
We encountered just such a situation in our exploration of the duality between entitative and existential interpretations of logical graphs. The two interpretations afford distinct but equally adequate ways of reasoning about a shared objective domain. To cut our teeth on a simple but substantial example of an object domain, we picked the space of boolean functions or propositional forms on two variables. This brought us to the following Table, highlighting the sign relation $L \subseteq O \times S \times I$ involved in switching between existential and entitative interpretations of logical graphs.
$\text{Peirce Duality as Sign Relation}$
• Column 1 shows the object domain $O$ as the set of 16 boolean functions on 2 variables.
• Column 2 shows the sign domain $S$ as a representative set of logical graphs denoting the objects in $O$ according to the existential interpretation.
• Column 3 shows the interpretant domain $I$ as the same set of logical graphs denoting the objects in $O$ according to the entitative interpretation.
### Resources
cc: Cybernetics (1) (2) • Peirce (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14)
cc: Ontolog Forum (1) (2) • Structural Modeling (1) (2) • Systems Science (1) (2)
cc: FB | Logical GraphsLaws of Form
### 8 Responses to Animated Logical Graphs • 69
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 483 | 1,898 | {"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": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-06 | longest | en | 0.811818 |
https://www.roadlesstraveledstore.com/what-is-1st-and-2nd-law-of-thermodynamics/ | 1,721,681,356,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00701.warc.gz | 829,189,238 | 14,905 | # What is 1st and 2nd law of thermodynamics?
## What is 1st and 2nd law of thermodynamics?
Key Points. The first law, also known as Law of Conservation of Energy, states that energy cannot be created or destroyed in an isolated system. The second law of thermodynamics states that the entropy of any isolated system always increases.
## What is Second Law of Thermodynamics class 11?
The thermodynamics second law states that the entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible in nature. The Systems which are isolated and spontaneously evolve towards thermodynamic equilibrium, the state with maximum entropy.
How does the second law of thermodynamics apply to living organisms?
Explanation: The second law says that everything goes from order to disorder, that is an increase in entropy. Living things die when the disorder in the system of the living organisms increases to the point where the system can no longer function. This makes a living organism an open system.
### What is Second Law of Thermodynamics formula?
The Second Law of Thermodynamics relates the heat associated with a process to the entropy change for that process. Therefore as a redox reaction proceeds there is a heat change related to the extent of the reaction, dq/dξ = T(dS/dξ).
### What are the 1st 2nd and 3rd laws of thermodynamics?
First Law of Thermodynamics; 2nd Law of Thermodynamics The Second Law of Thermodynamics states that the state of entropy of the entire universe, as an isolated system, will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative. 3rd Law of Thermodynamics
What are real-world example of the 1st law of thermodynamics?
One example of how the First Law of Thermodynamics acts is the perpetual motion machine. No one has ever built a machine that can continue to move forever without any external energy source keeping it moving. Every machine requires some input to continue moving. This input can be wind, chemical reactions, magnets,…
## What are the three rules of thermodynamics?
The three laws of thermodynamics define physical quantities (temperature, energy, and entropy) that characterize thermodynamic systems at thermal equilibrium. The laws describe how these quantities behave under various circumstances, and preclude the possibility of certain phenomena (such as perpetual motion).
## What does first law of thermodynamics stand for?
The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.
14/04/2019 | 568 | 2,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.940431 |
http://mathhelpforum.com/algebra/193103-how-solve-problem-algebraically.html | 1,480,796,598,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541134.5/warc/CC-MAIN-20161202170901-00384-ip-10-31-129-80.ec2.internal.warc.gz | 186,050,037 | 10,479 | # Thread: How to solve this problem algebraically?
1. ## How to solve this problem algebraically?
Qs: One number is 6 more than another. Their product is 520. Find the numbers.
-- Let the numbers be: x, x+6
So, x(x+6) = 520
or, x^2+6x=520
then what..?
The numbers are 20 & 26.
And if I put them in the above formula, it works..
20(20+6) = 520
or, 20x26=520
2. ## Re: How to solve this problem algebraically?
$x^2+6x-520=0$
Are you aware of any of the methods available for solving a quadratic equation?
3. ## Re: How to solve this problem algebraically?
One number is 6 more than another. Their product is 520. Find the numbers.
Let any letter of choice be the number we are searching for. I will use n for number but you can use x, y, z, or whatever letter you like.
Let n = one of the numbers
Let n + 6 = the other number
The words "Their product" should tell you right away that multiplication is what the question is leading to.
n times (n + 6) = 250
n(n + 6) = 250
n^2 + 6n -250 = 0
In the quadratic formula, let a = 1, b = 6 and c = -250.
If after visiting the site you have no clue what's going on, write back. I think it is important for you to try this one on your own. Keep in mind that x is typically used in the formula. Just replace my letter n with the typical variable x and follow the steps given in the link above.
4. ## Re: How to solve this problem algebraically?
Thanks for Both of the above posts. I've solved it using the factoring method e.g.
factors of 520 that will match to x, x+6 are 20 & 26
Since, x^2 + 6x -520 = 0
(-20)(26) = -520
(-20)+(26)= 6
So, (x-20)(x+26)=0
x-20 = 0, or x=20
x+26 = 0, or x=-26
x=20 ... this matches the equation, so the #s are 20 & 26.
Will try the formula way also. Thanks again... | 542 | 1,756 | {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2016-50 | longest | en | 0.893804 |
http://www.maa.org/press/maa-reviews/lecture-notes-on-diophantine-analysis | 1,498,528,281,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320887.15/warc/CC-MAIN-20170627013832-20170627033832-00397.warc.gz | 571,375,661 | 21,690 | # Lecture Notes on Diophantine Analysis
###### Umberto Zannier
Publisher:
Edizioni della Normale
Publication Date:
2009
Number of Pages:
237
Format:
Paperback
Series:
Lecture Notes 8
Price:
34.95
ISBN:
978-88-7642-341-3
Category:
Monograph
[Reviewed by
Felipe Zaldivar
, on
09/17/2009
]
Solving polynomial equations in integer or rational numbers is a classical arithmetic problem that can be traced back to the Babylonians but that was studied systematically by Diophantus of Alexandria (ca. 250 AD). Determining whether a given set of polynomial equations has solutions in integer or rational numbers, or finding all such solutions, or counting the number of solutions, are called Diophantine problems. Since a set of polynomial equations defines an affine or projective variety (e.g., over the field of complex numbers) the integer or rational solutions are points on this variety and the language of algebraic geometry gives an appropriate setting to formulate and study these problems, hence the name Diophantine geometry, which accurately describes this feature.
Zannier’s book is a self-contained introduction to some topics in Diophantine geometry. It starts with a preliminary chapter giving examples of classical Diophantine problems, especially “Pell’s equation,” treated using Dirichlet’s theorem on the approximation of irrational algebraic numbers by rational numbers, and sets the stage for the next chapters, where generalizations of this theorem by Liouville, Thue and Roth are proved or quoted and their application to more subtle Diophantine problems such as the S-unit equation or the Mahler-Thue equation are considered. This requires reviewing facts about heights of points in projective space over a field with a product formula. An important consequence is a special case of Siegel’s theorem on the finiteness of the number of integer points on a hyperelliptic curve.
Chapter 4 has new material not treated in any other book on Diophantine geometry, with the exception of the recent book by Bombieri and Gubler that considers some of these topics. Starting by studying the heights of points in powers of the algebraic multiplicative group Gm over an algebraic closure of the rational field Q, the author studies torsion points of small height on algebraic subvarieties X of this power of Gm, giving two elementary proofs of Zhang’s theorem that there is a lower bound for the heights of points on X away from a finite union of translates by torsion points of certain algebraic subgroups of a power of Gm. Zhang’s theorem is then applied to obtain a uniform estimation on the number of solutions of small height of the S-unit equation.
An appendix by F. Amoroso summarizes recent work on lowest bounds for the absolute height of algebraic numbers by Dobrowolski and Amoroso-Dvornicich for algebraic numbers that are not roots of unity in abelian extensions.
Zannier’s book is a very nice introduction to Diophantine geometry, including plenty of exercises dispersed along the text, most with hints for their solution with the aim of keeping the text self-contained. Each chapter ends with a section on notes pointing to further developments and historical comments. The book could be used as a text for a graduate class or for interested students learning this material on their own.
Felipe Zaldivar is Professor of Mathematics at the Universidad Autonoma Metropolitana-I, in Mexico City. His e-mail address is fzc@oso.izt.uam.mx.
1. Classical Diophantine Equations: linear and quadratic equations, Pell Equation, Diophantine Approximation, congruences. Supplements on Pell equations and irrationality of exp(n) and pi. Notes.- 2. Thue's theorems on Diophantine Equations and rational approximations: Description of strategy and detailed proofs. Later refinements. Supplements on integral points on curves and Runge’s theorem. Notes.- 3. Heights and Diophantine equations over number fields: Product formulas, Weil and Mahler heights, Diophantine approximation in number fields, the S-unit equation and its applications. Supplements on the abc-theorem in function fields and on multiplicative dependence of algebraic functions and their values. Notes.- 4. Heights on subvarieties of G_m^n: Torsion points on plane curves and algebraic points of small height on subvarieties of G_m^n. Structure of algebraic subgroups. Theorems of Zhang and Bilu and applications to the S-unit equation. Supplements on discrete and closed subgroups of R^n and on the Skolem-Mahler-Lech theorem. Notes.- 5. The S-unit equation. A sharp quantitative S-unit theorem; explicit Pade’ approximations and the counting of large solutions; counting of small solutions. Applications of the quantitative S-unit theorem. Notes.- Appendix by F. Amoroso: Bounds for the height: Generalized Lehmer problem, Dobrowolski lower bounds. Heights of varieties and extensions of lower bounds to higher dimensions; sharp quantitative Zhang’s theorem. | 1,074 | 4,940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.882777 |
http://math.stackexchange.com/questions/194703/finding-the-nearest-integers-to-real-numbers-defined-implicitly | 1,469,794,321,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00318-ip-10-185-27-174.ec2.internal.warc.gz | 160,730,621 | 19,548 | # Finding the nearest integers to real numbers defined implicitly
I was trying to bound the maximum cost of top-down merge sort: $$f(0) = f(1) = 0,\quad f(n) = n\lceil{\lg n}\rceil - 2^{\lceil\lg n\rceil} + 1,$$ where $\lg n$ is the binary logarithm of $n$ and $\lceil{x}\rceil$ is the smallest integer greater than or equal to $x$. Let $\{x\}$ be the fractional part of $x$. I made two cases:
• If $n=2^p$, then $f(n) = n\lg n - n + 1$;
• else $\lceil\lg n\rceil = \lg n -\{\lg n\}+1$ and $f(n) = n\lg n + \theta(1-\{\lg n\}) \cdot n + 1$, with $\theta(x) := x - 2^x$. The derivative of $\theta$ shows that $\max_{0<x\leqslant 1}\theta(x) = -(1+\ln\ln{2})/\!\ln{2} \simeq -0.9139$ and $\min_{0<x\leqslant 1}\theta(x) = \theta(1) = -1$.
So we have $n\lg n - n + 1 \leqslant f(n) < n\lg n - 0.91 n + 1.$ Considering $\theta(1-\{\lg n\})$ again, the lower bound is tight if $\{\lg n\} = 0$ or, equivalently, $n=2^p$.
What I want is to characterise the $n$ for which $f(n)$ is the closest to the upper bound: $\theta'(x) = 0 \Leftrightarrow x = -\lg\ln 2$, so $\theta'(1-\{\lg n\}) = 0 \Leftrightarrow \{\lg n\} = 1 + \lg\ln 2$.
How can I characterise the natural numbers $n$ such that $\{\lg n\}$ is as close as possible to $1 + \lg\ln 2$? For instance, could it be $n=2^p-1$? $2^p+1$?
EDIT In the comments, the real solution is $2^k\ln 2$, which raises the question of the integer solution. Assuming that the nearest integer is the solution, the first values for $k=1$ to $12$ are $2$, $3$, $6$, $11$, $22$, $44$, $89$, $177$, $355$, $710$, $1420$, $2839$. Is the assumption right? What is the general term?
-
lg? What's lg?? – Gerry Myerson Sep 21 '12 at 13:01
It is the binary logarithm. I am a computer scientist and the notation $\lg n$ is often preferred to $\log_2 x$, due to its frequency. – Christian Sep 21 '12 at 16:08
Ross Millikan's comment suggests taking $n = 2^p \ln 2$ rounded to the nearest integer. Certainly within any given dyadic interval $[2^k,2^{k+1})$ this will be the best (if we round a bit more judiciously). Are you asking for the rare cases where this approximation is much better than average? – Erick Wong Sep 21 '12 at 16:30
I would like an answer in the form $n=2^p+k$, where $p$ is a positive integer and $k$ is an integer. – Christian Sep 21 '12 at 16:54
I don't think your $x$ is well defined. $1+ \lg \ln 2 \approx = 0.47123$ but there are many $x$'s to choose from. Note that $2^{(1+ \lg \ln 2)}=2 \ln 2$, so $x=2 \ln 2$ is a solution with the nearest integer being $1$, but so is $x=2^{10} \ln 2\approx 709.78271289$ with the nearest integer being $710$
@Christian: you have said $x=2^n \ln 2$ for some natural $n$. For a given one, you can check whether $\{(n-1) \ln 2\}$ is greater or less than $0.5$. If you look through the $n$'s, you should be able to get as close to an integer as you like. I don't really understand what you are asking. – Ross Millikan Sep 12 '12 at 15:43
I think Ross has answered your question, but I'll try a different way to present it. If the fractional part of $\log_2n$ is close to $1+\log_2\log 2$, that says $\log_2n$ is close to $m+\log_2\log 2$ for some integer $m$, which says $n$ is close to $2^m\log2$ for some integer $m$. I doubt there's any way to characterize those integers $m$ for which $2^m\log2$ is close to an integer. | 1,173 | 3,311 | {"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} | 4.03125 | 4 | CC-MAIN-2016-30 | latest | en | 0.807829 |
https://admin.clutchprep.com/chemistry/practice-problems/73058/three-solutions-have-the-following-ph-a-ph-4-3-b-ph-6-8-c-ph-11-5-which-solution | 1,585,676,328,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00294.warc.gz | 340,106,529 | 21,857 | Chemistry Practice Problems pH and pOH Practice Problems Solution: 1. Three solutions have the following pH: pH ...
🤓 Based on our data, we think this question is relevant for Professor Gold's class at UTSA.
# Solution: 1. Three solutions have the following pH: pH = 4.3 pH = 6.8 pH = 11.5 a) Which solution contains the highest H3O+ ion concentration? b) Which solution is the most acidic? c) Which solution is the most basic?2. What is the difference between a strong acid and a weak acid?3. Calculate [H3O+] and [OH-] for the following three solutions a) pH = 9.5 b) pH = 3.6 c) pH = 7.0
###### Problem
1. Three solutions have the following pH:
pH = 4.3
pH = 6.8
pH = 11.5
a) Which solution contains the highest H3O+ ion concentration?
b) Which solution is the most acidic?
c) Which solution is the most basic?
2. What is the difference between a strong acid and a weak acid?
3. Calculate [H3O+] and [OH-] for the following three solutions
a) pH = 9.5
b) pH = 3.6
c) pH = 7.0
View Complete Written Solution
pH and pOH
pH and pOH
#### Q. At 25 C, how many dissociated OH ions are there in 1219 mL of an aqueous solution whose pH is 1.56?
Solved • Mon May 21 2018 07:28:53 GMT-0400 (EDT)
pH and pOH
#### Q. Calculate the poH of a 0.0321 M HCIO4 solution at the following temperatures
Solved • Fri May 04 2018 18:34:41 GMT-0400 (EDT)
pH and pOH
#### Q. What are the concentrations of hydrogen ion and hydroxide ion in household ammonia, an aqueous solution of NH3 that has a pH of 12.3?
Solved • Tue May 01 2018 17:23:50 GMT-0400 (EDT)
pH and pOH
#### Q. 0.90 g of sodium hydroxide (NaOH) pellets are dissolved in water to this solution? Express the pH numerically to two decimal places. pH = 11.75 What i...
Solved • Sun Apr 29 2018 13:08:04 GMT-0400 (EDT) | 584 | 1,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-16 | latest | en | 0.880105 |
https://transang.me/you-should-never-use-countif-function-google-spreadsheet/ | 1,714,031,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00263.warc.gz | 503,525,127 | 14,300 | Recently, I was asked to build a shopping order system for my sister.
Due to the complexity and ambitiousness of the requirement, I initially chose the google spreadsheet to build a simple system before starting with a web-based system.
Then I got to a case that I need to make a cell validation, which ensures there is no ID duplicated in a particular column.
## Failed solution
This stack exchange question appeared first when I googled.
Based on the accepted answer, I got the first solution by using the countif function. It works with the following expressions.
The solution works perfectly until my sister told me that the validation continuously fails even when no duplicated ID entered.
After debugging, I found out that the countif function does not work as expected in some cases, such as in the following example.
| A | B | C
------------------------------------------------
1 | 562572814105416318 | 2 | =COUNTIF(A1,A1:A2)
2 | 562572814105416000 | 1 | =COUNTIF(A2,A1:A2)
------------------------------------------------
Wow. Surprised!!. The countif function returns results that are totally different from what I expected.
All I can do is guessing.
I think that this weird behavior is the fuzzy matching functionality of the countif function due to its ability to check for a pattern with asterisks, number value comparisons (>=, =, <=, ...).
## Lesson learned
I did more debugging by removing digits one by one. Then I figured out that when the compared numbers have more than 15 digits, the countif function rounds all of them off.
| A | B | C
---------------------------------------------------
1 | 1.000.000.000.000.000 | 2 | =COUNTIF(A1,A1:A2)
2 | 1.000.000.000.000.001 | 1 | =COUNTIF(A2,A1:A2)
---------------------------------------------------
To ensure that all other functions do not compare the value in the same manner, I also checked other places where I use the vlookup.
Fortunately, the vlookup does not use fuzzy matching. Therefore, values are compared exactly.
| A | B | C | D
-------------------------------------------------------------------
1 | 1.000.000.000.000.000 | foo | foo | =VLOOKUP(A1,A1:B2,2,FALSE)
2 | 1.000.000.000.000.001 | bar | bar | =VLOOKUP(A2,A1:B2,2,FALSE)
-------------------------------------------------------------------
Of course, countif is very useful in many cases when we want to obtain various statistics from the data.
The lesson learned here is that you should never use countif when the criterion parameter (the second parameter) is dynamic. It is safe to use this function only when you know what the value of the criterion exactly (or statically) is.
## Solution
Eventually, I chose MATCH function as the final solution because of its true equality compare function.
This google spread sheet includes all the test formulas. You can open to check by yourself or make a copy and play with more formulas. | 662 | 2,987 | {"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} | 2.953125 | 3 | CC-MAIN-2024-18 | longest | en | 0.884989 |
https://socratic.org/questions/5947d8527c01490a7bca418b | 1,721,755,738,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00466.warc.gz | 467,942,625 | 6,827 | # Question a418b
Jun 19, 2017
(d) $4$ $\text{atoms C}$
#### Explanation:
Let's start by writing the equation for this reaction:
$\text{X" + "O"_2(g) rarr "CO"_2(g) + "H"_2"O} \left(g\right)$
Using the fact that $0.112$ ${\text{dm}}^{3}$ of compound $\text{X}$ was used up, we can use the ideal-gas equation to solve for the number of moles of $\text{X}$:
$P V = n R T$
• $P = 1$ $\text{bar} = 0.987$ $\text{atm}$ (standard pressure)
• $V = 0.112$ ${\text{dm}}^{3} = 0.112$ $\text{L}$ (given)
• $R = 0.082057 \left(\text{L"·"atm")/("mol"·"K}\right)$ (gas constant)
• $T = 273.15$ $\text{K}$ (standard temperature)
Plugging in these values and solving for $n$, we have
n = (PV)/(RT) =((0.987cancel("atm"))(0.112cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(273.15cancel("K"))) = color(red)(0.00493 color(red)("mol X"
What we can do now is find the number of moles of ${\text{CO}}_{2}$ present, using the given mass and its molar mass:
0.88cancel("g CO"_2)((1color(white)(l)"mol CO"_2)/(44.01cancel("g CO"_2))) = color(green)(0.020 color(green)("mol CO"_2
The number of carbon atoms in $\text{X}$ is equal to the ratio of moles of ${\text{CO}}_{2}$ to moles of $\text{X}$ (the coefficient in front of ${\text{CO}}_{2}$ is equal to the number of $\text{C}$ atoms in $\text{X}$). Therefore, the number of $\text{C}$ atoms in compound $\text{X}$ is
"atoms C" = ("mol CO"_2)/("mol X") = (color(green)(0.020"mol CO"_2))/(color(red)(0.00493"mol X")) = 4.1 ~~ color(blue)(4#
Thus, there must be $4$ atoms of $\text{C}$ in compound $\text{X}$, so option (d) is correct. | 594 | 1,603 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 35, "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} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.70417 |
http://www.questions4steveb.co.uk/html/PIC_projects/notes/0028_RS422-RS485-drive-with-one-PIC-pin.txt | 1,555,923,132,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-18/segments/1555578548241.22/warc/CC-MAIN-20190422075601-20190422101601-00408.warc.gz | 290,533,940 | 2,398 | ### RS422 / RS485 (differential) control with one PIC TTL output
RS422 or single master full duplex RS485 (where the PIC is the only 'master') Both RS422 and RS485 use a twisted pair cable to transmit signals over distances of up to 1km. Both use differential (A+, B-) signaling, but, unlike RS422, RS485 supports multiple 'masters' on the same cable. Full duplex uses 2 pairs, one for Tx the other for Rx, however bi-directional 'half-duplex' is supported over a single pair (so is more common, especially for long cable runs, which can be up to 1km !). A ground (shield) wire (C) is usually present. Driver voltages are typically +/-2.5v to +/-5v (max +/-6v) = the minimum 'specified' receiver difference is +/- 1.5v however most work down to +/-200mV (the max. voltage, referenced to chip ground, on any pin is -7 to +12). 'Idle' state is usually A more positive than B (which is taken as a '1') and (since the drivers are 'tri-state', idle is 'guaranteed' with the use of 'bias' resistors). The receiver is simply looking for the 'difference' between the A+ and B- wires. A difference of +200mV is taken as a '1' and -200mV as a '0'. The actual voltage levels on the two wires are thus irrelevant, it being the 'difference' that counts. This means you can, in fact, 'get away with' using TTL 0,+5 and +5,0 as the two states (so long as you 'bias' the other wire to +2.5v).
#### Termination
The 'ideal' RS422/485 system consists of a single linear cable (no branches) with 120 ohm resistors connected across the 2 wires at each end of the cable. Sometimes additional termination resistors are added between each wire and ground/V+ or ground/V-, however this is typically done only for 'long line' usage. RS485 can handle speeds of over 10 Mbits per second and line lengths of over 1 km. If you are operating anywhere near these values you must arrange your wiring close to the ideal and fit termination resistors HOWEVER at the more typical 4800/8600 baud and short distances, all the termaination resistors are often omitted.
#### The PIC drive
A PIC i/o pin will switch between 0v6 and Vdd-0v7 (for Vdd 5v, that's 4v3). If one of the RS422/485 wires is held half way between these two values (i.e. at 2v45) and the other driven direct by the PIC pin, we get a difference of 2v45-0.6 = 1v85 (well above the 200mV minimum) and 2v45-4v3 = -1v85, again well below the -200mV minimum). In practice the 'mid point' will be 2.5v (giving us +1.9 and -1.8) Note = the bus 'idle' state is PIC pin 'Hi' (so setting to 'input' mode (tri-state) using the weak internal pull-up (approx 22k) will work fine) Multi-master and RS484 half-duplex considerations The problem in a half-duplex or multi-master configuration is that the PIC is not the only device driving the lines. Any other master (or a half-duplex slave) will likely use the standard +/-6v, and whilst 6v is not a big issue, -6v is. In that situation, we have to add a Schottky diode (to prevent more than -0.3 appearing on the PIC i/o pin) plus a current limiting resistor (560R) to avoid blowing the remote transmitter (as it's -6v is 'shorted' to Gnd by the diode). This, by the way, means the same PIC i/o pin used as a input can receive the slave transmissions (as the diode 0v3 limit = '0' and 6v = '1')
#### Usage
RS485 is used for the 'Profibus' and 'Modbus' interface standards, and extensively used for CCTV camera control (for example, Pelco-P) | 938 | 3,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-18 | longest | en | 0.92518 |
https://numberworld.info/1342100 | 1,600,888,271,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212039.16/warc/CC-MAIN-20200923175652-20200923205652-00453.warc.gz | 509,022,532 | 3,920 | # Number 1342100
### Properties of number 1342100
Cross Sum:
Factorization:
2 * 2 * 5 * 5 * 13421
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
147a94
Base 32:
18ukk
sin(1342100)
-0.81224128102734
cos(1342100)
0.58332161060179
tan(1342100)
-1.3924416072797
ln(1342100)
14.109746109386
lg(1342100)
6.1277848763622
sqrt(1342100)
1158.4903970254
Square(1342100)
### Number Look Up
Look Up
1342100 which is pronounced (one million three hundred forty-two thousand one hundred) is a very special number. The cross sum of 1342100 is 11. If you factorisate 1342100 you will get these result 2 * 2 * 5 * 5 * 13421. 1342100 has 18 divisors ( 1, 2, 4, 5, 10, 20, 25, 50, 100, 13421, 26842, 53684, 67105, 134210, 268420, 335525, 671050, 1342100 ) whith a sum of 2912574. 1342100 is not a prime number. The number 1342100 is not a fibonacci number. The number 1342100 is not a Bell Number. 1342100 is not a Catalan Number. The convertion of 1342100 to base 2 (Binary) is 101000111101010010100. The convertion of 1342100 to base 3 (Ternary) is 2112012000102. The convertion of 1342100 to base 4 (Quaternary) is 11013222110. The convertion of 1342100 to base 5 (Quintal) is 320421400. The convertion of 1342100 to base 8 (Octal) is 5075224. The convertion of 1342100 to base 16 (Hexadecimal) is 147a94. The convertion of 1342100 to base 32 is 18ukk. The sine of the figure 1342100 is -0.81224128102734. The cosine of the figure 1342100 is 0.58332161060179. The tangent of the figure 1342100 is -1.3924416072797. The square root of 1342100 is 1158.4903970254.
If you square 1342100 you will get the following result 1801232410000. The natural logarithm of 1342100 is 14.109746109386 and the decimal logarithm is 6.1277848763622. that 1342100 is very unique figure! | 712 | 1,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.743729 |
https://cheapieshoe.com/does-shoe-size-equal-inches | 1,657,170,854,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00089.warc.gz | 200,654,814 | 12,956 | # Does shoe size equal inches?
4
Date created: Wed, Jul 28, 2021 5:54 AM
Date updated: Tue, Jul 5, 2022 2:25 PM
Content
Video answer: Shoe size comparison chart for men and woman
## Top best answers to the question «Does shoe size equal inches»
• Men's shoe size charts start from a US number of 6 (a European number of 39 and a UK number of 5.5). This size corresponds to an inch measurement of 9.25 inches (23.5 centimeters). If the measurement is provided solely in centimeters, multiply the length of your foot by 2.54 and apply to the chart. The biggest number on the men's chart is ...
• How does shoe size correspond to inches? The barleycorn is an old English unit that equates to 1⁄3 inch (8.47 mm). This is the basis for current UK and North American shoe sizes, with the largest shoe size taken as twelve inches (a size 12) i.e. 30.5 cm, and then counting backwards in barleycorn units, so a size 11 is 11.67 inches or 29.6 cm.
• The Romans decided three barley corns were one unit of measurement, and therefore one shoe size and this was formalized in England in 1324 by King Edward as one inch.
• US women's shoe size = (foot length in cm + 2 × 0.847) / 0.847 − 23 Explanation. One US size is equal to one-third of an inch (0.847 cm). The first 11 US sizes are not numbered. The following 13 sizes are assigned as children's sizes. The American shoe sizes for women start with Size 0 following the children's sizes.
FAQ
Those who are looking for an answer to the question «Does shoe size equal inches?» often ask the following questions:
### 👠 9.375 inches is what size shoe?
#### Women
US SizesEuro SizesInches
6.5379.0625"
737-389.25"
7.5389.375"
838-399.5"
### 👠 Does 13c in kids shoe equal size 2?
#### What's the difference between size 13 kids and adult shoes?
• There is not much difference between size 13 kids shoes and 1 adult. In terms of sizing there is only approx. 1/2 inch difference between 13C and 1Y. In the US, the kids shoe size system switches to the youth respectively men’s shoe size system after size 13 kids shoes. Then it restarts counting at 1 (Y).
### 👠 Does the size of shoe equal invhes?
The barleycorn is an old English unit that equates to 1⁄3 inch (8.47 mm). This is the basis for current UK and North American shoe sizes, with the largest shoe size taken as twelve inches (a size 12) i.e. 30.5 cm, and then counting backwards in barleycorn units, so a size 11 is 11.67 inches or 29.6 cm.
### 👠 Does your shoe size go by inches?
The barleycorn is an old English unit that equates to 1⁄3 inch (8.47 mm). This is the basis for current UK and North American shoe sizes, with the largest shoe size taken as twelve inches (a size 12) i.e. 30.5 cm, and then counting backwards in barleycorn units, so a size 11 is 11.67 inches or 29.6 cm.
### 👠 How does shoe size correspond to inches?
The barleycorn is an old English unit that equates to 1⁄3 inch (8.47 mm). This is the basis for current UK and North American shoe sizes, with the largest shoe size taken as twelve inches (a size 12) i.e. 30.5 cm, and then counting backwards in barleycorn units, so a size 11 is 11.67 inches or 29.6 cm.
### 👠 How long inches shoe size?
#### Men's Shoes Length and Width Charts
Length (inches)Length (centimeters)US Size
9 1/4″23.5 cm6
9 1/2″24.1 cm6.5
9 5/8″24.4 cm7
9 3/4″24.8 cm7.5
### 👠 How many inches shoe size?
• Size 6 = 9.31 inches Size 6.5 = 9.5 inches Size 7 = 9.69 inches Size 7.5 = 9.81 inches Size 8 = 10 inches Size 8.5 = 10.19 inches Size 9 = 10.31 inches Size 9.5 = 10.5 inches Size 10 = 10.69 inches Size 10.5 = 10.81 inches Size 11 = 11 inches Size 11.5 = 11.19 inches Size 12 = 11.31 inches Size 12.5 = 11.5 inches Size 13 = 11.69 inches Size 13.5 = 11.81 inches Size 14 = 12 inches Size 14.5 = 12.19 inches Size 15 = 12.31 inches
### 👠 Is shoe size in inches?
• It measures between 3 and 5 inches across the bottom of the ball of the foot, depending on your shoe size. For instance, in men's footwear, a size 10D is 4 inches across, while a size 14D is 4.5 inches across.
### 👠 Uk size 18 shoe equal what size usa?
• As you can see in the chart below, the US size is 2 sizes smaller than the UK size. For example, if you need a US women’s size 12, you would need to order a UK women’s size 16. For a US women’s size 18, this would be a UK women’s size 22. Simple, right?
Video answer: What does big kid shoe size mean?
We've handpicked 6 related questions for you, similar to «Does shoe size equal inches?» so you can surely find the answer!
What does child shoe measuring 10 inches equal?
#### How to measure your child's shoe size?
• Measure the foot of your kid from the heel to the longest toe using a tape measure. This is your child's shoe size. Differences in length can happen that is why you should always measure both feet.
What does men's shoe size 11 equal to europe?
#### Men's Shoe Size Conversion
UnitMen's Shoe Size Conversion Sizes
UK5.511
Europe3944-45
Inches9.311.1
Centimeters23.528.3
What does shoe size 38 equal?
If you've ever wondered "What size shoe is a 38?" (It is a men's size 7.5 in the US) or "What is a size 40 in shoes?" (that'd be a US men's size 9.5), read on to learn how to measure shoe sizes in different countries.
What does shoe size 44 equal to?
#### How do you calculate shoe size?
• To find your shoe size, start by placing a piece of paper on the floor and tracing the outline of your bare foot onto it. Then, use a ruler to draw a rectangle around the outline that touches the furthest point on each side of your foot. Finally, measure the length and width of the rectangle to get your shoe size.
What does shoe size us 7 equal to eu size?
#### Convert your US shoe size to EU or EU to US shoe size with this shoe size conversion chart. ... Euro Shoe Sizes: EU to US Shoe Size & US to EU Shoe Size.
US Men'sUS Women'sEU Shoe Size
5.57Size 37.5 EU
67.5Size 38 EU
6.58Size 39 EU
78.5Size 39.5 EU
### Video answer: Shoe size conversion table
What does shoe size us7 equal to eu size?
#### What are European shoe sizes compared to US?
• European shoe sizes are numbered in 30’s and 40’s. Your shoe size thirty seven in Europe would be the equivalent to shoe size five in the Unites States. That means U.S shoes are 32 numbers lesser than equivalent European shoes. | 1,790 | 6,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.914303 |
https://www.calculussolution.com/blog/arc-length | 1,656,719,491,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103947269.55/warc/CC-MAIN-20220701220150-20220702010150-00700.warc.gz | 716,573,917 | 11,758 | # Arc Length
## Primary tabs
One of the easiest and most important formulas in Mathematics is the formula for arc length. This simple formula has a lot of applications that we want to talk about in the future, but for now we want to explain what it is and how to calculate it.
One of the first things we learned in school about Mathematics was that the ratio of the circumference of a circle (the length around a circle) to its diameter (the length across the circle) is the same no matter the size of the circle. This ratio is always the number $\pi$ which is equal to $3.141593$. In other words,
$\frac{C}{d}\,=\,\pi\,=\,3.141593.$
But how do we find the length of only part of the circumference? This is called arc length and has the formula
$s=r\theta$
where $s$ is the arc length, $r$ is the radius and $\theta$ is the angle measured in radians. In the picture below the arc length is in red.
We can immediately recover our circumference formula by putting in $\theta=2\pi$ to get $s=r\cdot 2\cdot\pi$. Since the diameter is $d=2r$ we get $s=d\cdot\pi$. Since we know that $C=d\cdot\pi$ we see that the $s$ is the circumference when $\theta=2\pi$.
To use this formula one must be aware of the units of the angle. We recently talked about units and dimensional analysis. For $s=r\theta$ to describe the real world its units on the left side of the equal sign must be equal to the units on the right side of the equal sign. This means that $\theta$ must not have a dimension - it can't be a length or any other unit. The number $\pi$ is a dimensionless quantity that is the angle half-way around a circle. We call the number $\pi$ (3.141593) the number of radians half-way around the circle. The number $2\pi$ is the number of radians around the circle. The formula $s=r\theta$ requires that we use radians and not degrees.
Most of the time we think in degrees. We know that a circle has $360$ degrees and we could be tempted to just stick the number of degrees $\theta$ is into $s=r\theta$. This is incorrect and we must first convert the number of degrees to radians. To do this let's assume that $\phi$ is our angle measured in degrees and $\theta$ is the same angle measured in radians. To convert to radians we use
\begin{eqnarray}
\theta&=&\left(\frac{2\pi}{360}\right)\,\phi\\
&=&\left(\frac{2\cdot 3.141593}{360\mbox{ degrees}}\right)\,\phi\\
&=&1.7453\times 10^{-2}\,\phi/\mbox{degrees}
\end{eqnarray}
Let's do a quick example. Let's say we want to find the arc length of part of a circle with a radius of $2$ feet and an angle of $30$ degrees. Since the angle is in degrees we must first convert it to radians. Following our formula above we have
$\theta=1.7453\times 10^{-2}\,\cdot 30\mbox{ degrees}/\mbox{degrees}=0.52359$
(Notice how the degrees cancel to give a dimensionless quantity for the angle.) Now we can find that our arc length is
\begin{eqnarray}
s&=&r\theta\\
&=&2\mbox{ feet}\cdot 0.52359\\
&=&1.04718\mbox{ feet}.
\end{eqnarray}
That's it! This formula is very easy with the only thing we need to be aware of is the units of the angle. Making sure the angle is converted to radians is the only difficult part in finding an arc length. | 856 | 3,173 | {"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": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.929795 |
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